source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
Op amp with transistor in feedback I am currently looking for a high side current sense IC. I have noticed that most of them that are without galvanic separation, are using an op amp with bjt transistor like in bellow schematics: The datasheet of TSC101 describes more or less how to calculate the output current, but I would like to understand what exacly is happening in here and why is it this exact circuit always used for high side current sensing. I just mean the 1st opamp with bjt transistor config. I get the voltage follower part. If I were to build this from semidiscrete elements, can the op amp be powered from voltage ie. 5V and sense current on a lot higher potential with this circuitry?I would apreciate all help and explanation. <Q> Operation of this chip is clever but not too difficult to understand. <S> Current flowing to the load will cause a voltage drop between (1) and (2) and the drop will be proportional to the current. <S> Since the op-amp input impedance is very high, the voltage at (3) will be same as (2). <S> (More on this later.) <S> Initially the transistor will be off and voltage at (4) will be equal to that at (1). <S> The effect will be that (4) is higher than (3) <S> so the output of the op-amp will start to increase. <S> As the output starts to increase the transistor starts to turn on. <S> This allows current to flow through Rg3. <S> As the current increases the voltage at (4) starts to drop due to Rg1. <S> When enough current is flowing through Rg1, Q and Rg3 to pull (4) down to the same voltage as (3) <S> the circuit will stabilise. <S> Therefore the voltage across Rg3 will be proportional to the voltage across Rsense which is proportional to the current. <S> The Rg3 voltage is buffered out. <S> Rg2 is included to balance out any bias currents. <S> Both inputs will be affected the same amount so the errors cancel out. <S> They may also provide a little over-voltage protection and limit the current into any over-voltage protection diodes built in. <S> (Check the datasheet.) <S> If I were to build this from semidiscrete elements, can the op amp be powered from voltage ie. <S> 5V and sense current on a lot higher potential with this circuitry? <S> Nope. <S> With very few exceptions op-amp inputs must be within or very close to the supply rails. <S> They do seem to be using a trick on this device though as pins 3 and 4 can go to 30 V with a 24 V supply on the chip (judging by the schematic.) <A> Question One: Why is the transistor there? <S> Answer: <S> The transistor is there to guarantee that the Op-Amp can "settle out" so that the transistor pulls enough current through Rg1 and into Rg3 to get equal voltages on the V- and V+ inputs of the op-amp. <S> If it weren't there a lot of extra trickery would be needed to get the voltage difference across Rsense to cause a reliable, supply-independent current to flow down. <S> The thing where it could have been easier without the transistor is that Rg3 is also inside the package and again causes a voltage and possibly a 2-op-amp difference amplifier could maybe have achieved a same result. <S> However, many of these designs give the current the transistor pulls as an output pin allowing you to substitute your own resistor. <S> Apart from that, the transistor offers some options to accuracy and stability, depending on some details not supplied by your yellow block. <S> Question Two: <S> Can I do this, power the op-amp with 5V and then have the Rsense work at a higher voltage (like 12V, maybe?)? <S> Answer: <S> No. <S> Not without any other parts around it, which would make it infinitely harder. <S> An Op-Amp can only offer linear gain when its inputs are between the supply rails, and many can only do so when their inputs are even more to the middle of the supply range than the actual rails. <S> You would need a "More-Than-Rail-to-Rail-Op-Amp" and I have not heard of such a thing. <A> re Question 2: <S> In fact, the 1st (input) <S> "opamp" in the cct should be understood as a "high gain element" with balanced (i.e. symmetrical) behavior at both input pins. <S> And reasonably low bias current, <S> but I wouldn't expect this bias current to be very low. <S> If I had to take a guess at what is inside the design of such a gain element, it would consist of the two inputs feeding matching resistors (Rin, Rip) that feed matching NPN current mirrors. <S> Thus, (nearly) equal input bias currents (Ibn, Ibp) will flow into the gain element and the voltage drop <S> Ibn Rin (~=Ibp Rip) will keep the active elements of the input stage (current mirrors) below Vcc. <S> The difference in the outputs of the current mirrors is where the complicated part starts. <S> Off course, the analog IC designers live for such challenges :) <S> I hope this helps you form ideas for your 5V powered hi-voltage current sensor. <S> BTW: current mirrors are somewhat tricky to get well matched in the discrete world. <S> You can find (nearly) matching transistor arrays. <S> Good luck.
| As answered before, no, you couldn't wire up this cct with common opamps.
|
What does RU (UR) label on DC power relays mean? I'm trying to figure out what does RU or maybe UR label mean on the body of many many relays. It's labeled almost on all (I've seen so far) DC relays which make me very curious :). Any suggestions? <Q> It is a UL label. <S> From UL wikipedia article : The "Recognized Component Mark" is a type of quality mark issued by Underwriters Laboratories. <S> It is placed on components which are intended to be part of a UL listed product, but which cannot bear the full UL logo themselves.[6] <S> The general public does not ordinarily come across it, as it is borne on components which make up finished products. <A> RU is an Underwriter's Laboratories mark for certified products that are intended to be used inside of other devices. <S> Reference: <S> The UL site. <A> As a basis for approval, code authorities look for UL Listing Marks to verify that products have been investigated for installation in the field in accordance with model codes. <S> During inspections of electrical installations, code authorities may see a variety of certification marks, including the UL Listing Mark and UL Recognized Component Mark. <S> To properly approve the installation, it is important for code <S> authorities to understand the meaning of each of these marks as well as the important differences between them. <S> The UL Listing Mark is the most common certification mark seen and accepted by code authorities. <S> http://ul.com/wp-content/uploads/2014/04/ul_RecognizedComponentMarks.pdf <A> UL listings are for products . <S> ЯU is for components <S> and it means the component is pre-approved. <S> Say you design a DVR or smart-switch. <S> The DVR needs a UL listing to be sold at retail. <S> The smart switch needs a UL listing to be installed in mains wiring (NEC 110.2). <S> You send the unit into UL, and they go through the unit with a fine-tooth comb. <S> For instance there's a relay in there, they'll pop it off the PCB and do their tests - hi-pot, current, throw cycles, physical teardown, toxic-smoke test (if burned), etc. <S> This costs you money. <S> However, when they see the ЯU mark on the relay, they don't test it. <S> It's already tested and approved . <S> When your PCB is covered with ЯU components, listing is cheaper, faster, and there's less uncertainty. <S> Note that you can get a UL product listing from ETL, CSA, BSI, TUV, or any other Nationally Recognized Testing Lab. <S> But they'll just follow the same rules as UL. <S> A few Chinese junkmakers in the smart-switch space are trying to "go legit", and I notice they prefer ETL.
| It is the Recognized Component Mark UL
|
How do bitshifts work on the electrical layer? I'm a software developer and I understand how arithmetic and logical bitshifts work in principle. But how do they work on the electrical layer? I might have a completely wrong imagination; Let's say we have the binary value 0000 1011 and left shift it by 2. The outcome is 0010 1100, nothing special. But how do the bits "jump" to their second neighbors? Edit: Fixed a slip, where I wrote "right shift" while doing a left shift. Edit: Added ALU and DIGITAL-LOGIC tags <Q> In computers, arithmetic operations are performed by a specific integrated circuit that is placed inside the microprocessor, and is called "Arithmetic Logic Unit". <S> You can have a look at the Wikipedia article on ALU ( https://en.wikipedia.org/wiki/Arithmetic_logic_unit ). <S> As an example it mentions the 74181 ALU, which is a very simple Arithmetic logic unit, able to perform several types of operations, including the left shift, over 4 bits. <S> I've attached the logic schema of the 74181 ALU (see below). <S> Inputs are A and B, and result is F. <S> Each number is 4 bits long, so there is A3A2A1A0, B3B2B1B0, and F3F2F1F0. <S> The desired operation is selected with S3S2S1S0: <S> F = <S> A (op) <S> B 'op' is the desired operation. <S> If S = <S> S3S2S1S0 = 1100, then 'op' is left shifting. <S> This is how left shifting works in the particular case of the 74181: <S> Look at the upper left group of five AND gates and 2 NOR gates. <S> First AND gate is not exactly AND, because it has only one entry: <S> /A0. <S> Therefore its output is /A0. <S> Second and third AND gates have 0 output because S1S0 = 00 Fourth AND gate has output /A0*B0, because S2 = 1 Fifth AND gate has output /A0*/B0, because S3 = 1 <S> First <S> NOR gate has output A0. <S> Second NOR gate has output A0, because /A0*B0 <S> + /A0*/B0 <S> = /A0. <S> In conclusion, when you have S3S2S1S0 = 1100, the ALU makes F = A + A, which is the same as left shifting. <A> One approach is a parellel in parallel out shift register . <S> The data is loaded into the shift register, then shifted left or right by the desired number of places. <S> The down-side of this is that it can only shift by one bit per clock cycle. <S> An alternative is a barrel shifter which can shift by an arbitrary number of places in one go. <S> It's worth realising that shifting need be nothing more than how you wire the input lines to the output lines on a data bus. <S> Wired straight through is no shift. <S> Wire each input line to the output line that is one place to the left is a left shift by 1, and so on. <A> Look up something called a shift register . <S> One way to think about these is a chain of flip-flops. <S> Let's say these flip-flops take a snapshot of whatever value is on their input on the low to high transition of the clock, then transfer that snapshot to their output on the next high to low transition of the clock. <S> You connect the output of one flip-flop to the input of the next, etc, then connect them all to the same clock signal. <A> Ones and zeros are represented inside the CPU integrated circuit as voltage levels. <S> For example 0 volts is logic zero, and 5 volts is logic one. <S> Logic values are typically held in circuits called "flip-flop"s. <S> A typical CPU chip has millions of flip-flops. <S> And a typical CPU chip also has millions of electrically-operated switches. <S> So, in order to execute a shift function, the CPU turns on the appropriate switches to send the logic levels over to the NEXT flip-flop (or "bit position") in the register. <S> That is what the CPU chip is designed to do. <A> When you rotate the tires on your car how does that happen? <S> Well you take at least two off and swap them <S> yes? <S> Bit storage be it individual flip-flops or gobzillions of them packed in a ram. <S> Each bit of storage is readable and writeable. <S> otherwise what is the point? <S> A bit shift is nothing more than an operation that takes the tires off, reads the bits out of a register or memory location, and then writes them back in a different location, just like moving one tire from the back to the front, you read it out, then write it back in a different location. <S> There is no magic in logic, it is all very very simple in concept. <S> zero, one, and, or, not is all that is required to understand. <S> It is just at times a massive amount of and, ors and nots in parallel and/or series. <S> Some processors shift one bit at a time and in pseudo hdl it is not really different than in software alu_out[16] = <S> (operand_a[14:0],0) the 16 bit output is made of the lower 15 bits of the input re-positioned to the upper 15 bits of the output with a zero tacked on in the 0 position. <S> And that line of code may be in a switch like statement or tree of if then else statements that basically says if alu_operation = left_shift then do this. <S> And all of that is compiled into logic gates, the ifs and the assignments turn in to and, or, not gates or gates derived from and, or, not. <S> on a massive rats nest. <S> Some processors have a variable sized shifter that in one operation can rotate or shift however much you want (up to the whole word size obviously, otherwise what is the point) <S> granted that takes a lot more gates, <S> just that one operation each bit of the output for a 16 bit register has a 15, 16, or 17 to one mux depending, just for that one operation.
| Basically all math and logic functions are implemented by simply connecting bit registers in particular ways depending on what you want to do.
|
IRF3205 MOSFET gate threshold voltage higher than expected I have been trying to drive a 12 V LED array with an IRF3205 MOSFET.I wanted to do this: simulate this circuit – Schematic created using CircuitLab However, when I tested it, the LED was only being given 6 V and the MOSFET was dropping 10 V across the drain and source. Going through the things that could be wrong, I found that giving the gate 5 V causes the problem to be solved, and the LED gets the full voltage of the 12 V power supply. The LEDs draw about 100 mA when they are being driven correctly in this way. I looked up the datasheet , and saw that the gate threshold voltage is 2-4 V, meaning that this should work fine with the 3.3 V Raspberry Pi GPIO . What is going on, and is there a way I could drive this chip with a 3.3 V Raspberry Pi GPIO pin? <Q> A fundamental misconception many have with MOSFETs is that the gate threshold is where the "magic" happens. <S> The gate threshold is truly a threshold - it is the absolute threshold of conduction: where the device goes from totally of to jusssssssssssst a little bit on. <S> A meaningful diagram can be found a few pages deeper: <S> This shows various relationships between \$V_{GS}\$, \$I_D\$ <S> and \$V_{DS}\$. <S> With a 12V \$V_{DS}\$ voltage, you can see that even with 4.5V on the gate, the MOSFET will conduct only a small fraction of the 110A that the device is rated for. <S> Yes, these curves are characterized for pulse behaviour <S> but you get the idea - you need much more than the gate threshold voltage to really get the MOSFET "on" and working well. <S> If you want to drive a MOSFET from the GPIO line directly, you will need to find one which can sink sufficient current at that 3.3V drive level. <S> Consider the Fairchild FQP30N06L <S> device: <S> You can see that even with 3V, the device will conduct a lot of current. <A> If you want to get rid of MOSFET threshold and turn-on voltage problems, completely, I suggest you try this: R1 will dissipate about a watt while the LED's ON, so prudence would dictate using about a 2 watt resistor. <S> I like these. <S> and here's the LTspice circuit list <S> just in case you want to play with the circuit: <S> Version 4SHEET 1 <S> 1140 708WIRE 128 0 -16 <S> 0WIRE 304 0 208 0WIRE <S> 304 32 <S> 304 0WIRE 304 112 304 96WIRE <S> 128 160 80 160WIRE 240 160 208 160WIRE -16 <S> 224 <S> -16 0WIRE <S> 80 224 80 160WIRE <S> -16 <S> 336 -16 <S> 304WIRE <S> 80 336 80 304WIRE 80 336 <S> -16 <S> 336WIRE <S> 304 <S> 336 304 208WIRE <S> 304 <S> 336 80 336WIRE -16 384 <S> -16 336FLAG <S> -16 <S> 384 0SYMBOL res 224 <S> -16 <S> R90WINDOW 0 0 56 VBottom 2WINDOW <S> 3 32 56 VTop 2SYMATTR <S> InstName R1SYMATTR Value 91SYMBOL LED 288 <S> 32 R0WINDOW 3 28 68 <S> Left 2SYMATTR <S> InstName D1SYMATTR Value QTLP690CSYMATTR Description DiodeSYMATTR Type diodeSYMBOL voltage 80 <S> 208 R0WINDOW 3 24 96 Invisible 2WINDOW 123 0 0 <S> Left <S> 2WINDOW 39 0 0 <S> Left 2SYMATTR <S> InstName GPIOSYMATTR Value PULSE(0 <S> 3.3 0 100n 100n <S> 10m <S> 20m)SYMBOL voltage <S> -16 <S> 208 R0WINDOW 123 0 0 <S> Left <S> 2WINDOW 39 0 0 <S> Left 2SYMATTR <S> InstName V2SYMATTR Value 12SYMBOL <S> npn 240 <S> 112 R0SYMATTR InstName Q1SYMATTR Value 2N2222SYMBOL res 224 144 R90WINDOW 0 0 56 <S> VBottom 2WINDOW 3 <S> 32 56 VTop 2SYMATTR <S> InstName R2SYMATTR Value 240TEXT <S> -8 360 <S> Left 2 !.tran <S> .5 <A> I looked up the datasheet, and saw that the gate threshold voltage is 2-4v, meaning that this should work fine with the 3.3v Raspberry Pi GPIO. <S> That means 3.3 V could be below the gate threshold voltage. <S> Even if it were above, that's still not meaningful. <S> You have to actually read the datasheet instead of skimming the highlights and then making (wrong) assumptions. <S> On page 2, V GSth is clearly defined to be when the FET conducts 250 µA. <S> It's hard to even imagine how you think that is relevant when you want 100 mA. Just one line above that, the datasheet specifies R DS(on) at 10 V, which is what this FET is intended to be operated at. <S> Nothing says anything about what happens with only 3.3 V on the gate, other than sometimes maybe it will conduct 250 µA. Again, this is all clearly spelled out. <S> Sometimes you get ambiguous or poorly written datasheets, but this is not one of them. <A> You should not need to burn 2W in the resistor. <S> For driving 100mA into the diode you need to burn in a resistor : <S> P = VI ; <S> V = IR; P = I^2*R = .01*R <S> W <S> Assuming the LED needs about 3-4V to be on, that leaves 8V leftover to burn across this resistor. <S> V = IR; V = 8V, I = .1A; R = 80 <S> Ohms <S> Going back to the power burned in the resistor, I^2*R = <S> .1 <S> ^ <S> 2 <S> * 80 <S> = .01 <S> *80 <S> = .8WYou would also be burning 4V*.1A = <S> .4W <S> in the LED. <S> This seems a little wasteful to me, and a better way might be to reduce the 12V to something more like 5V. <S> You need a more responsive FET (lower Vt). <S> You could also just convert your output (0-4V) to (0-12V) using back to back BJT inverters, and use the current FET you have, along with 80Ohm resistor. <A> I have now completed the project using a standard 2N3904 NPN Transistor, which can handle the whole 100mA at 12v. <S> Thanks for everyone's advice.
| This FET is simply inappropriate for your application.
|
Audio DSP for hobbyists I'm several years away from finishing an EE degree and have yet to get any exposure to DSP. The few texts I've cracked open pre-suppose a level of sophistication i'm not at yet. Is there some stepping stone to start dabbling or is DSP all or nothing? <Q> No, you can start DSP with C, Matlab, or Octave to name just a few. <S> DSP is mainly about filtering, but also about signal processing and algorithms to manipulate sampled signals in the time and frequency domain. <S> There are also algorithms to detect\generate tones and chords which might be of interest for audio hobbyists. <S> Real time DSP involves making these processes run fast in an FPGA or DSP processor, which you would need hardware for. <S> I had some labs once where we built adaptive IIR filters that would filter out a tone real time from audio with only a few msecs of delay. <S> We implemented and tested the filters in C and Matlab before implementing them in hardware. <A> For $60, you can buy a " DM330011-MPLAB Starter Kit for dsPIC DSC " which combines a dsPIC <S> (essentially a PIC24 with added DSP instructions) with a codec.. <S> It is described as: " <S> This Starter Kit introduces users to the dsPIC <S> Digital Signal Controller device using its speech and audio processing capabilities. <S> The kit is USB-powered, has on-board debug circuitry and 24-bit codec for high-quality audio applications." <A> ADI ADAU1761 has dual A-D and D-A. with very little DSP knowledge <S> I used it to design and build a mic mixer, controlled by software (i write software). <S> The eval kit is $200. <S> Sigma studio software is easy to learn. <S> tie blocks together like a schematic. <S> EVAL-ADAU1761. <S> I think Audio is a good start. <S> You can generate sine, triangle, square waves manipulate and see on scope or hear on earphones. <S> I am still learning the science under the routines. <A> Uuuups - did not notice the question date :( <S> You can start with any Cortex M4 or M7 micro controllers with the DSP instruction set. <S> For example stm32F4xx stmF7xx or stm32h7. <S> ARM has quite nice CMSIS DSP libraries as well <S> https://github.com/ARM-software/CMSIS_5/tree/develop/CMSIS/DSP <S> As a starting point I would recommend any STM DISCOVERY board as they have microphones & audio amplifiers on board. <S> $50 (but with the 4" TFT on board and a lots of internal and external memory) <S> https://www.st.com/en/evaluation-tools/32f746gdiscovery.html or cheaper $25 https://www.st.com/en/evaluation-tools/stm32f4discovery.html
| A beginner could learn how to implement filters and also spectral estimation. All you need is some sampled data to play with and a way to manipulate that data, the rest is learning about the math (google is your friend).
|
Power Supply Issue, Voltage Drop Under Load I have the Ardupilot Mega flight control board (APM), which outputs 4.8V. I want to power CC3200 MCU with it, which needs at least 3V for stable operation. I power APM using a power module connected to an adequate power source and can output up to 2Amps for APM and all the peripheral devices connected to it. That's way more than enough. When CC3200 is operative under full stress, it can draw up to 200mA. I'm doing a level shifting on the 4.8V line, dropping it to 3.3V using two resistors 220Ohm and 100Ohm connecting them appropriately to form a voltage divider. 3.3V is the perfect voltage input for the MCU, which under full load should drop the voltage by maybe 0.1-0.2 Volts which is understandable. However, when i make this voltage divider using the resistors i measure the voltage on the CC3200 and it's 0.3V, instead of ~3.1V! I've tried 2.2kOhm + 1kOhm with same results. Tried an other module (WiFi bluetooth module) the voltage there goes to 2V, kind of better, but still this is 1.3 voltage drop! I also tried powering led's and other stuff from the APM, but in all cases the voltage drops heavily when a load is connected to it when using a voltage divider for level translator. Nothing's wrong with CC3200, normally i power it using 2 AA batteries. But i would much rather power it from the APM. But, if i power say the WiFi Bluetooth module without level shifting (it can handle up to 6V) the voltage remains almost perfectly stable to ~4.8Volts! This issue has been troubling me. Is there not enough current to power the external device when using resistors as voltage divider (unfortunately i don't have another level translator device to test this theory)? Not much else comes to mind, but i don't know how to make sure. I need the help of more experienced and knowledgeable people. What could be the reason? What is wrong? What am i missing? Thanks in advance! <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Voltage divider open-circuit. <S> Figure 2. <S> Micro attached in parallel. <S> Figure 1 shows the voltage divider when tested unloaded. <S> All appears well and 3.3 V is measured on the divider terminals. <S> Figure 2 shows the voltage divider loaded by the micro. <S> The micro is represented by a 33 Ω resistor which would pass 100 mA if connected to a stable 3.3 V PSU. <S> The combination of R4 and R5 in parallel is about 30 Ω (28.7 Ω). <S> Now our potential divider mid-point will give \$ 4.8 <S> \frac {30}{100 + 30} = 4.8 \cdot <S> 0.23 <S> = 1.1 V \$ <S> The moral of the story is that a potential divider will not give you a stable voltage under changing load conditions such as a micro whose current requirements will vary with time. <A> Take a look into buck converters. <S> In your application, you need to save as much energy as possible. <S> This resistor chain voltage divider will require a lot of energy. <S> You want to conserve as much as you can if you're flying a drone. <A> the resistor divider does not work because it has a too high source resistance = <S> about 100 ohms. <S> the concept of dynamic resistance is important here and try using a diode to drop the voltage by .7V. <S> You would require two diodes as the drop required is 4.5 to 3.3 or about 1.2V. the diodes have a dynamic resistance of much lesser than 1 ohm when used at currents of 25 mA or more. <S> If you are keen on knowing about dynamic resistance here is the data. <S> It is change in voltage / change in current. <S> Dynamic resistance of a diode is 26 ohms at 1ma <S> [Se grey and meyer]. <S> that is why a zener is much better than a voltage divider! <S> Zeners have dynamic resistance of about 10 to 15 ohms. <S> So if you have a 15V and want 5V, do not use a voltage divider but use a 10v zener in series that can drop the 10V. <S> the diode has a drop of 600mV at 1ma, 660mV at 10mA and 720mV at 100mA. <S> It has decreasing dynamic resistance of 26,2.6 and .26 at 1mA, 10mA and 100mA respectively. <S> thus it is way better than a voltage divider when you have to drop voltages. <S> use a 1N4007 {two in series} for your drop of 1.2 from 4.5 to 3.3
| You need to use a voltage regulator. Another option is to use a linear regulator, however, that will consume energy dropping the voltage turning it into heat, which isn't favorable.
|
What exactly does ARM sell to vendors? Assumptions: Computer architecture: Describes how the different modules of a processor interact with each other. A computer architecture is defined using vhdl files Computer Organization: Describes the physical layout of the processor modules on silicon. A computer organization is defined using a set of photo masks (and manufacturing process eg chemical, that goes at each step) Computer Organization, therefore, requires that the fab process be taken into account. ARM is not in the fabrication business, therefore it does not sell photo masks. My question(s): What exactly is ARM selling to a vendor (eg: freescale)? For a SoC (System On Chip), (eg: iMx6 ), which part is ARM and which is Freescale? Who did the integration? <Q> You're using those terms wrong. " <S> Computer organization" is a rarely-used term for the microarchitecture, and "computer architecture" is a superset of that. <S> Integrated circuit IP blocks come in two basic forms: <S> A soft macro is the RTL (VHDL or Verilog) that describes the functional implementation of the IP. <S> This is compiled into a gate-level netlist, which is then turned into a physical layout to produce the mask set for manufacturing. <S> Here's an example from Cadence -- an Ethernet MAC. <S> When you buy it, you get Verilog files, documentation, and a Verilog testbench for verification. <S> A hard macro is a physical layout of the IP suitable for a given process. <S> It's added to the larger chip layout as a single block, which saves some steps in the design process. <S> Here's another Cadence example -- an Ethernet PHY. <S> It's offered in 180nm and 130nm processes at TSMC, UMC, and SMIC, and is delivered to the customer in the form of GDSII layout files. <S> ARM sells both of these. <S> The MCUs I've worked on usually use soft macros of ARM Cortex CPUs. <S> We had some older product with ARM7 hard macros, but I don't know if they were hardened by ARM or us. <S> Today, ARM has hard macro versions of the Cortex-A series listed on their web site . <S> Most of their products are synthesizable (soft macros), though. <S> It looks like you can download the (soft) Cortex-M0 for free for non-commercial use on the ARM DesignStart site. <S> In an SoC, the ARM part is just the CPU. <S> (The designer can also buy peripheral IP from ARM, but it's not required.) <S> The SoCs I've worked on have a mix of third-party and internal IP. <A> What exactly is ARM selling to a vendor (eg: freescale)? <S> A license to sell chips that include ARM intellectual property. <S> For a SoC (System On Chip), (eg: iMx6), which part is ARM <S> The definition of the CPU core(s) and their interfaces. <S> ... <S> and which is Freescale? <S> Who did the integration? <S> Freescale, of course. <A> When I was involved in a chip design project which involved integrating an ARM, what we got was netlist Verilog . <S> That is, a complete list of the gates and connections that make up the chip, but not in a higher-level form suitable for easy modification. <S> We then did our own placement and routing. <S> This is usually necessary on each project as the shape of the chip area allocated to ARM will be different. <S> I believe multiple levels of license are available so you could buy a placed core in a standard shape already to save time. <S> (Placed netlists are exactly as they sound: a list of gates and their locations.) <A> ARM sells intellectual property (IP) to other companies. <S> In your example, Freescale purchases the rights to use ARM IP, namely the processor core. <S> What is the "core"? <S> The core is just about anything that isn't a peripheral, like a SPI driver or an ADC or DAC. <S> As part of the core, ARM included hardware systems to act as bridges between the actual processor stuff and the customer's peripheral. <S> As far as deliverables, ARM does not provide a complete HDL to the customer. <S> They provide high-level abstractions that the end user's HDL can link against (HDL wrappers), and they also likely provide the physical layout of the core. <S> ARM sells more than just processor cores, they also have a whole segment of complex IP that are not processors. <S> When ARM sells a processor core, they get money up front and/or they get royalties on all the devices sold with their designs in them. <S> It's a pretty good deal for them, considering there are about 10 billion ARM processors in the wild. <S> SoC makers like Freescale design their own peripherals for the ARM core.
| To sum up: ARM sells IP (designs basically) to customers to integrate into their end product The physical implementation, including the design of any peripheral logic included on the chip.
|
Using thermal overload protection in AC-DC PWM motor driver circut i'm having some difficulties with calculating the value of the thermal overload fuse in this circuit. The thermal fuse is used to protect the motor and the circuit behind it. The deal problem is that the thermal overload fuse is in the live mains circtut and not after the AC-DC PSU. And i cant seem to understand how to calculate the value. Could someone please help me understand what the formulas are and try to explain it to me? I have talked to other people and they say that its should be able to calculate the value. I already have the thermal fuse for this its this module( Moeller PKZM0-0,25) Supply voltage = 230VACMotor nominal current 3AMotor Voltage 12VDCAdjustable Range of overload fuse = 160mA - 250 mA <Q> That motor-protection is protecting the complete circuit - transformer, rectifier, PWM modulator and motor. <S> Your motor is rated at 12 V, 3 A. From the equation \$ P = <S> VI \$ we can calculate that it is a 36 W motor. <S> Discounting losses for a moment we can calculate the required current at 230 V AC to provide 36 W. Using the same formula and rearranging <S> we get \$ <S> I = \frac { <S> P}{V} = \frac {36}{230} = <S> 0.16~A <S> \$. <S> Allowing for some losses in the transformer, wiring and PWM <S> (in)efficiency I would probably set the trip current to 200 mA. <S> The protection device looks like the right choice. <A> You are using a 30W motor, right? <S> Suppose 90% efficiency of the driver, your power is 55W. <S> The current is then 33/230=0.14ARMS. <S> I would argue it's a little too tight. <S> Anyway the concept is that you are protecting the wires inside your walls from being burned by short circuit, so it's fine to take some margins. <A> The motor protection you have linked is a very professional way and common in machines. <S> The use of it is straigthforward when using induction motors with known motor plate - you have to adjust current on the protection device to match the nominal current on the motor. <S> In your case, there is a transformer, rectifier, PWM device and DC motor, which is slighly different than a motor, only. <S> You have to use some intuition to adjust it and to test. <S> Perhaps you could lock the motor apply full power to it and wait until it diconnects, it has to delay some 3-4sec.
| Take 100% margin for such fuse, otherwise it will trip from time to time.
|
How to build a circuit that generates a sine wave? How can I build a circuit to generate sine waves?What I am working on is building a 555-timer circuit that gives 50% duty cycle square wave, then using a low-pass or band-pass filter tuned to the frequency of the square wave to get a sine wave from the square wave (because square wave consists of infinite number of sine harmonics). I haven't tested the circuit in the lab but I've used LTSpice to simulate it and the results was acceptable for low frequency (~72 Hz) but bad for high frequency (~72 kHz).I used this guide from Texas Instruments to build the narrow-band band-pass filter: link .Here is the simulated circuit: and here is the result for frequency of 72 Hz: and for 72 kHz: The output of the 555-timer is a square wave with 50% duty cycle as wanted. How can I improve this design to get better results? Are there other ways to generate sine waves with controlled frequency? I don't want ready-made ICs that do this. I want to build a circuit. <Q> You can try to round off and then filter triangle waves with diode or active circuits. <S> At 72kHz, an LC oscillator would not be bad (wind your own high quality inductor or tapped inductor on a pot core if you like) and pair it with quality PPS or other film capacitors. <S> Or a Wien bridge oscillator. <S> If you want low distortion with an LC or Wien bridge you need to have a good AGC that controls the output level without adding too much distortion (so it will necessarily take many cycles to stabilize). <A> If you see what looks like a triangle wave from an op-amp circuit you are probablly slew-rate limited. <S> You need a better op-amp, the LM108 is an aincient part <S> (it's original manufactuerer doesn't even sell it anymore) and the datasheet doesn't even bother to specify important parameters like bandwidth and slew rate. <S> If you want a decent sinewave by this method you probablly also need more than one stage of filtering. <A> V3 looks like it's the wrong way around. <S> I'd use a low pass filter. <S> There are no harmonics below the fundamental frequency of the square wave! <S> I built one of these once for 1kHz sine out. <S> I used a 6 pole butterworth sallen-key low pass filter which ensured that there was very little 3rd harmonic passed. <S> Your op amp is saturating because when the fundamental is extracted from the square wave it has an amplitude of 4/pi times the amplitude of the square wave. <S> If you reduced the amplitude of the 555's output using a potential divider then the low slew rate of the op amp would cope. <S> Once the low amplitude sine wave is created it could then be amplified up if required. <S> The output of the 555 output amplitude reducing potential divider will have a DC offset and you may need to bias it down to 0V via a capacitor/resistor combination. <S> and then buffer with a unity gain buffer into the filter. <S> EDITThe vulnerability of a 555 when configured to produce a square wave in that way is that any loading on its output will alter the mark to space ratio away from 50%. <S> In my design I put a D-type flip flop in toggle mode (Qbar output connected back to D input) in between the 555's output and the amplitude reducing potential divider thereby ensuring an almost perfect 50% duty cycle. <S> Note that this type of toggle configured D-type flip flop divides the 555's output frequency by 2.
| DDS chip + filter is one way that's pretty common these days.
|
Synchronizing a diesel generator with a pure sine wave inverter As the title reads, I'm trying to understand the best (Practical) way to synchronize a diesel generator with a pure sine wave inverter, in order to implement an automatic synchronizer based on microcontrollers. I've searched alot for references, books and articles related to this topic and I came up (theoretically) with these conclusions : To Synchronize any two sources of single phase AC power , these conditions must be achieved: Phase angle between the two wave forms must match. Voltage on both terminals of the sources must match. Frequency between the two sources must match. Now, frequency synchronization seems to be achieved since both the inverter and the generator is designed to output 50 Hz pure sine wave,Phase angle condition can be achieved by studying the initial state of the inverter wave then start the inverter based on this method in order to make the phase difference between the two sources zero, for voltage It seems to be more complicated to match the voltage of both sources (since both sources are not designed by me /and seems to me that I can't control their voltage output/ neither the generator nor the pure sine wave inverter, I bought them both) so here lies the problem , how can this be done ? Also I want to know (please) what's the effect of slightly different values in each condition of synchronization, and is my approach to this goal totally wrong ? This is my first time doing something like this, but not the first time I deal with 220 volts AC, so any advice would be much appreciated. <Q> First, you have to control one of the sources. <S> I don't think there is anything to do about power network, but you must be able to control the diesel. <S> There should be some throttle something. <S> Actually, it could be a very nice control project. <S> You should measure the phase and by changing throttle keep the phase error zero. <S> Like a diesel PLL. <S> But a more practical way is to convert the diesel output to DC and then with an invertor just create any sine wave you want. <A> I don't think there is a simple way to add this sort of functionality to an existing inverter. <A> This is a good thing to work upon. <S> Actually, you need to have full control on one of the sources. <S> I have many ideas for which you can work on it. <S> But I want to mention something very important before proceeding. <S> First of all, the inverter must be giving purely sinusoidal output <S> otherwise there will be loss of power continuously during the parallel operations and heating may occur. <S> Now: I suggest you to go with controlling the Diesel generator. <S> As you have mentioned from your literature review, the parallel operation can be achieved by satisfying above conditions. <S> I have ideas that can be used for implementation of this project. <S> The frequency of the diesel generator can be controlled by the governor control which is a valve or choke like thing present on almost every generator. <S> You should dismount it and place a servo motor mechanism on it so that you should be able to electrically control the throttle of the governor ( basically you will be changing fuel intake in order to increase the speed of the generator). <S> Secondly, You should use an Auto Transformer whose tap can be changed by a similar servo motor. <S> This auto transformer will be used to balance voltages. <S> Then you can develop a light bulb test or dark bulb test like mechanism which is available across the internet. <S> Using all this, you can have a circuit breaker between the terminals of generator and inverter. <S> So your system will be continuously monitoring and changing parameters of voltage and frequency until they are balanced. <S> As soon as there is a match, the breaker is closed. <S> And You have successfully synchronized your generator with the inverter. <S> Note: <S> Always keep protections in your mind for example generator reverse power flow protection and over current protection etc. <S> This will reduce the chances of failure of the system and ensure safety. <A> The conventional way to do this (three phase sets) is to set the no load voltages to be the same (to within a volt or so), set the full load droop to be the same (at whatever 100% load is for each machine, this controls the reactive power sharing), and set the frequency to be as close as you can, then when the two machines drift into sync, close the switches... <S> The governor response then controls how much real power each machine contributes, and both machines run at the same frequency due to feedback from the torque produced by the rotor angle. <S> I would note that most inverters are actually current sources, you need one that has a voltage control loop as well to allow you to control the reactive power production and hence bus voltage. <S> Synchronisation of a small machine with the grid (which is effectively an infinite power sink) is actually much easier then synchronisation between a small number of machines of reasonably equal size, there is MUCH more inertia in the grid, so if tends to resist frequency changes. <S> I am not at all sure that synchronising a single phase machine is really worth the trouble, they tend to be small going on tiny, and the automatic phase locking once in sync will be a far more chancy proposition in the absence of a rotating three phase field.
| Buy an inverter which has the capability to synchronize to external ac sources built-in.
|
How do pick and place machines pick up components? I first thought they used a vacuum of air so they don't dislodge the components after they've placed them? <Q> Those are handheld ones, but the machine has one on a robot arm with XYZ control. <S> It's very simple, just a tiny suction cup with a vacuum that can be turned on and off. <S> The components are placed onto the solder paste which is slightly sticky, so small vibration won't dislodge them. <S> If the board will be double-sided, a small dot of glue is added before the component is placed. <S> This prevents the components falling off when the other side is being soldered. <S> Then the board goes through the oven, is subject to optical inspection to check that the parts have soldered correctly, and any parts that can't be handled by the pick and place machine are hand-fitted and hand or wave soldered. <S> This is usually only required for big connectors. <A> They use a vacuum pickup, typically with a spring-loaded tip so that the pickup has some compliance in the Z direction. <S> Since the force is limited by atmospheric pressure multiplied by the nozzle bore cross-sectional area, different sizes of nozzles are used for different parts. <S> Here are some typical nozzles used on one major brand of P&P machine: Not shown here, but they can also be rectangular (note the keying flats to prevent rotation). <A> Vacum pickup is indeed the normal way. <S> A machine will often have different nozzles for different sizes of component. <S> A camera is often used between picking and placing to check the position and orientation of the component on the nozzle. <S> Connectors will often have a small plastic or metal cap or a peice of kapton tape to provide a flat surface for the vacum to pick.
| Using a vacuum pickup .
|
Is it right to initialize a reg in verilog and apply condition with initial value of reg in Verilog? I have the little doubt related to initializing condition in Verilog. Like in given statement: module rf(out1,ack,en,a,f,c,d,e,clka); input [7:0] a,f,c,d,e; input clka, en; output reg [7:0] out1; output reg ack; reg[7:0] b[1:5]; reg [1:0] first=0; reg [2:0] k;initial begin for (k = 1; k <6; k = k + 1) begin b[k] = 0; endendalways @(negedge clka) begin if (en==1) begin if (first==0) begin first<=1; end if (first==1) begin first<=2; b[1]<=a; b[2]<=f; b[3]<=c; b[4]<=d; b[5]<=e; end endendendmodule I initialized reg first =0 ; Is it right ? As it is giving right result after simulation but is there any problem when we will synthesize it? I used the first condition because I wanted to execute statements written within (first == 1 ) execute after one clock pulse. Is it the right way? If not then what should I do if I want to execute few statements after one or two clock pulse ? Hope I explained my confusion clearly. P.S : module median_five_sh(out1,ack,reset,a,f,c,d,e,clka);input [7:0] a,f,c,d,e;input clka,reset;output reg [7:0] out1;output reg ack;reg en0,en1,en2,en3,en4,en5,en6,en7,en8,en9;reg[7:0] b[1:5],tmp;reg first;reg [3:0] i1,i2,n1,k;initial begin for (k = 1; k <6; k = k + 1) begin b[k] = 0; endendalways @( posedge reset) begin en0<=0;en1<=0;en2<=0;en3<=0;en4<=0;en5<=0;en6<=0;en7<=0;en8<=0;en9<=0;first<=0;i1<=0;i2<=0;n1<=0;k<=0;tmp=0;endalways @(negedge clka) beginif (reset==1) begin statement;en0<=1;en1<=1;..endendendmodule The above code is simulating and giving correct output but it is giving the error after synthesis. **Error: Signal en0 in unit .... is connected to following multiple drivers: * I wanted to execute statements written in always @( posedge reset) only once while initially. Basically its initialization of variables used in later statment. <Q> Initialising the registers at declaration is perfectly synthesisable. <S> It tells the compiler what the power-on value of the register should be. <S> Generally the initial value for the registers is always 0 anyway, and if you choose to have them set to 1, it will basically use bubble pushing optimisations to invert the register value and still use 0 as the initial value (but as far as your logic is concerned it would effectively be 1). <S> However, for anything other than a data bus (qualified by some valid signal), this is not recommended. <S> Why? <S> because of what happens if you have a reset signal somewhere else in your logic. <S> If half of your logic is reset at some point and you have a control signal that only has a power-on initial value and not a reset, then your two cores go out of sync - one <S> is in in a nice known reset state, the other is in whatever unknown state <S> it was when the reset occurred. <S> For qualified data signals, a don't care/unknown value doesn't matter as long as the valid-like signal is reset to a known state of invalid. <S> The better practice is to use a reset signal for all control and valid signals to have a reset value, either synchronous or asynchronous. <S> This eliminates the need for an initial power-on value requirement (you can still add it, but it's no longer required). <S> The power-on value will be determined by the synthesizer based on the requested reset value. <S> always @ <S> (<edge> clock or posedge reset) begin if (reset) <S> begin <S> //Reset value goes in here, this value also determines power-on value. <S> end else ... endend <A> It's generally poor practice to rely on that kind of initialization in synthesized logic. <S> For one thing, the initialization — if it supported at all — only applies immediately after power-up configuration of the FPGA. <S> This generally only works on SRAM-based FPGAs; other technologies don't support power-up initialization at all. <S> It is much better to have an explicit reset input to every module that puts everything into a known state, and this reset can be generated by any number of conditions, including power-up, a manual reset button, or various internal error condition detectors. <A> Initial blocks too, aren't synthesizable in ASIC flow. <S> The initialization values will become a part of the bitstream file(If <S> you are used to Xilinx terms), and will be loaded when 'Configuring' the FPGA. <S> As a matter of fact Xillinx recommends against adding an explicit reset, since the reset signal has to be routed to all your logic. <S> But if you plan to port it to an ASIC later then adding a reset(preferably synchronous) is a must. <S> I couldn't help but notice you are using two separate always blocks for a register. <S> That's why your synthesis tool is complaining of multiple drivers. <S> An always block synthesizes into a flop <S> /latch(if <S> you are describing sequential logic), with the logic inside the always block used for determining the D-input (& rst) for the flop. <S> Also tmp is assigned with blocking assignment. <S> It is generally said to be a bad idea to mix blocking and non blocking assignment in the same always block.
| Yes, It is completely legal to initialise the registers to a value if your RTL targets FPGA, like the others have pointed out.
|
Coin cell battery retainer, problems and best practices I use Memory Protection Devices BK-913 CR2032 Through hole battery retainer. I use Adafruit's CR2032 PCB footprint, which is almost identical to the recommended footprint in the datasheet for the retainer. It is has a square 4x4 mm center pad defined as negative. Around the center pad is a 15.24 mm circle defined as "no solder mask". When the PCB is manufactured it looks like this: Is the square center pad and the surrounding circle electrically connected? If I understand the footprint correctly, they should not be? But the little breaks in the black square surrounding the center pad suggests that they are connected anyway? I have experienced some problems with this footprint, the connection with the battery sometimes fails. I have to readjust the battery, and some batteries only works if halfway inserted into the retainer. Not very reliable. It worked perfectly with the batteries I had, but when I got a new batch of the same batteries (same manufacturer, same supplier etc.), they are extremely difficult to make proper contact in the retainer. I guess the batteries from the new batch are a tiiiiny bit different physically..?! Are there any best practices for footprint design for coin cell battery retainers? For example, I have noticed that the pad under the battery retainer on the Texas Instruments SensorTag looks like this: What is the reason for this? Are the little solder dots there to help make better contact with the battery? How can I modify my footprint to be more reliable? <Q> Yes, the entire exposed area is electrically connected. <S> The intent is to place a small bump of solder inside the square to improve the connection to the battery. <S> The reason for the rectangular slots is to serve as thermal reliefs, similar to what is done for plated-through-holes. <S> Without the slots, you would have to heat up the entire circle of copper before any solder would melt. <S> And even then, the solder could flow over the entire area. <S> The narrow "thermals" made by the slots limit the transfer of heat. <S> Now you can get the square hot enough to melt solder without warming up the whole board. <A> When using coin cells, it is vital that the negative (bottom) side NOT be entirely covered with solder, otherwise a lack of connection can occur due to flux, corrosion, and other residue. <S> TI had a big problem with their CC2541 keyfobs that used a big slab of paste for the negative connection. <S> TI learned from this for the sensortag - they're using little dots of paste very strategically, small enough so that the top of the solder will be scraped off when installing a battery. <S> The preferred approach for the negative contact is, in order of preference:1. <S> Use a dedicated negative contact made for this purpose - Keystone <S> #2991 is one, but there are others. <S> We use this approach for things that have to last a long time. <S> This is what I use most of the time because it always works and allows a lot of routing flexibility beneath the battery.2. <S> Use ENIG surface finish for the bottom pad, ensuring that the bottom pad is the entire size of the battery <S> so you avoid this issue . <S> ENIG is a gold plating (about 2-3 microinches of gold, IIRC) on top of nickel. <S> The gold will help prevent corrosion.3. <S> Use another surface finish, but clean the PCB before shipping the battery and ensure that there's no chance of corrosion. <A> The center and outsides are connected. <S> It may just be like that to make it look cool, as it does not seem to serve a purpose. <S> If you want to make the connector more reliable, one thing you could do is remove the silkscreen around it. <S> That could keep the battery just barely above the contact if it is thicker than the copper. <S> An easier thing though would be, as TI does, to put a few solder dots on the contact. <S> An even easier way to fix it would to be putting solder over the whole middle section. <S> Solder tends to bulge in the center, so it would give a firm connection against the spring leaf I am sure is on top of the holder. <S> Just be sure not to put on a lot of solder, or the battery may not fit any more :) <A> It looks to me like you've got a ground plane that is interfering with the footprint. <S> I would add a keep out to the ground zone around the battery contact footprint. <S> Also, the CR2032 I've got here has a negative pad that measures around 18mm <S> so it would still be sitting up on your solder mask at the edges. <S> Try increasing the size of your solder mask keep out too.
| The solder should be placed so that it is relatively symmetrical to prevent the battery being inserted at a cant.
|
Can a low side N-channel MOSFET in a DC-DC converter limit inrush current? I want to add an NMOS just before the supply negative side and thereby controlling its slew rate so as to create a soft start. I've seen many examples but mostly with high side PMOS (high Rds on) or NMOS (additional circuitry to drive in high side). I doubt there must be some reason. Can we use a low side N-channel MOSFET in a DC-DC converter for limiting inrush current? Are there any drawbacks? <Q> This can definitely be done. <S> However, like any low side control, you should only do this if the return of your load is isolated from the ground of the source. <S> In many electrical and electronic systems, exposed conductors (like shields or chassis) are at ground potential. <S> Thus it's not safe to have two systems (your power supply and load, for example) with different "ground" potentials, because they could easily be shorted together. <S> This, combined with the convention of positive supplies (i.e. systems where ground is the lowest potential) probably explain why it is easier to find high-side inrush control solutions. <A> You can find both ways of achieving this. <S> You may also use a series resistor with input cap that you will bypass using a fet/thyristor/triac. <S> It was mentioned your "ground" <S> won't be exactly "ground" anymore <S> but it's close enough that it usually won't matter. <S> If you google for inrush current limiting you will find many examples. <S> Motorola AN1542 shows a simple method of high/low side limiting with mosfet using gate capacitance to do something useful. <S> Note that on AC/DC circuit the current limiting resistor is subject to a lot of power. <S> It may be only for an instant but a regular 1W resistor would die horribly. <A> This limits the current that the IC will allow the circuit to draw from the supply, ramping it up slowly. <S> This is the preferred way to do it, as a capacitor in mass production costs less than a penny, and mosfets can be in the dollar range.
| On Linear technology switching regulators / controllers, there is usually a soft start pin that you can hang a capacitor on. Using another mosfet just for soft-start is probably not economical, but would most likely be possible.
|
How to test SMD bluetooth modules without soldering them to a board I have made a costum PCB with components and a WT51822-S2 bluetooth module. After testing, there seems to be a low percentage of modules that have a fault in them. Since they are SMD modules, with connections on the sides, its hard to desolder them if faulty. They are soldered on easily with a normal solderin iron but desoldering, thats a whole other thing. Is there a way to test SMD modules without actually soldering them on the pcb? I have tried pressing them on (with force and duck tape) but some pins have contact, others dont. I have also made a brakeout board for testing, but it still requiers soldering. what to do? <Q> You can get special spring clips for castellated pads, too. <S> I don't know how many insertions/removals they'd be good for in a production test environment, however. <S> They're really meant as an engineering tool for evaluation and prototyping. <A> You need pogo pins like the ones used in this programmer board: <S> There are many manufacturers of such pins. <S> Another useful term to search for is "bed of nails. <S> " <S> That together with pogo pins should get you all you need to know. <S> Get some pins, build a board to hold them, arrange the pins so that you can press your board to test into the pogo pins, do your test. <S> You might consider a mechanical hold down to keep the test board in place if you have a bunch to do. <A> If you are going to be testing many of these modules, spending the extra (modest) amount of money on building the fixture will probably be less hassle for you overall. <A> Mill-Max sells spring-loaded pins that you can solder down to a PCB to break out the modules for testing. <S> Thn, you will just have to clamp down your bluetooth chip and the springs will keep the connections secure. <S> Many of them are also gold-plated, further improving conductivity. <S> here is a link to some that should work: <S> https://www.mill-max.com/products/socket/854-XX-XXX-10-001101 <S> These have the correct spacing for your chips (0.05") and are solderable through-hole for easy assembly. <S> Hope this helps! <A> Not sure if this would work (or how hard it would be to remove after testing) <S> but what about z tape? <S> https://www.adafruit.com/products/1656
| Sparkfun Electronics has a very good tutorial on how to build test beds.
|
Reliability problems with an array of 91 high-power LEDs I have designed and printed a 4-layer PCB that accommodates 91 infrared LEDs in a 7x13 rectangular layout. This will be used as a backlight for a machine vision project. I am having a problem where individual LEDs are burning out or perhaps becoming disconnected from the circuit in some way. I suspect the heat dissipation may be the cause of the problem. Image PCB Layout Each row of 7 LEDs (green LED text) is wired in series. The 12V supply (VCC powerplane) connects to the first LED. The next 6 are wired in series. Finally a current-limiting resistor (green R text) connects the last LED to the ground plane. Specifications: VCC plane: 12V, 2A supply LED: TSHG6200 . 100mA maximum ratedcurrent. Current limiting resistor: 20 ohms Solder: ThermoflowSn60/PB40 Total estimated power dissipation: 12V * 0.1A per row *13 rows = 15.6W. Size of array: 13 rows of 7 LEDs, approximately 7cmx 6cm Measurements With a 12V power supply, there is about 1.45V over each LED, and about 2.0V over the current limiting resistor, meaning a current of 100mA. Because this is right at the maximum allowable current, I put a big high-power potentiometer between the power supply and the VCC plane, and used this to regulate the input voltage to be slightly lower (11.5V or so). This gets the current safely below the maximum allowable amount. I am also using a Darlington pair to control the backlight with an Arduino. The backlight is on almost all the time, and occasionally is pulsed off for about 30ms. I don't think this is relevant to the problem but can provide more details if necessary. Problem After about 10-30 minutes of use, one or more of the rows of LEDs will go out. If I measure the voltage across each LED in the broken row, most LEDs are at about 0.8V and one has about 8.0V across it. No current is flowing. Sometimes resoldering the pins or tapping the LED fixes this. Sometimes it has to be replaced. In any case I only get another 10-30 minutes of use before another one goes out. Another observation is that the whole back side of the board is kind of sticky. You can see this in the picture above. I wonder if it is getting too hot and the solder is becoming compromised (perhaps exuding flux??). Question What should I try to improve the reliability? I've already tried running it at a lower voltage to get the current safely below the rated maximum. I wonder if I need to use a different kind of solder? Or some kind of heat sink? The LEDs get hot to the touch but not unbearably so. Edit, after trying suggestions Thanks everyone for the tips! I did something quite simple -- pointed a computer fan to blow air across the array -- and it worked fantastically! I guess this is really obvious to many of you but I was surprised at how enormous the difference was. Without fan: 25mA per row -> 39C 33mA per row -> 41C 40mA per row -> 48C 55mA per row -> 52C So we get into the "danger zone" of temperature well before reaching the maximum current per LED. With fan: 35mA per row -> 26C 60mA per row -> 30C 90mA per row -> 34C I ran it at 90mA per row and 34C for over an hour with no problems. Great! <Q> From the datasheet: Figure 1. <S> Absolute maximum forward current. <S> and further on: <S> Figure 2. <S> Derating LED current at increased temperature. <S> Current and temperature are your problems. <S> You are running at absolute max current with no wriggle-room and you are allowing the temperature to rise. <S> At 60° ambient the max current allowed falls off dramatically. <A> You have already hit on the answer: your LEDs are getting hot. <S> 15 watts may not sound like much, but it's building up and killing your LEDs. <S> I suggest you get a thermistor and attach it to the center of the board, then monitor the temperature as the system operates. <S> Even better, attach it to the body of one of the LEDs. <S> Because you're using this as a backlight, don't use narrow-beam LEDs. <S> Use relatively wide beam units, and space them apart so air can flow through. <S> If you can find a source of, let's say, 35 degree LEDs, install only every other one in a checkerboard pattern, soldering jumper as necessary. <S> You'll only get half the total brightness, but that's barely perceptible, and the improved airflow should be a big help. <S> You may also need to provide a fan with some ducting to keep the air flow through the array adequate. <S> And always include a temperature monitor. <S> While not directly applicable, this YouTube video shows the principles of cooling. <S> In your case, since you've got a forest of vertically standing LEDs, it is important not to let the LEDs touch each other, since this will block the flow of air. <A> A quick google for "LED heatsink" will find you a lot of suitable solutions for cooling down your burning-out LEDs. <S> I'm afraid that's more soldering, but <S> hey ho. <S> For more info, I've found that Wikipedia has a link to various LED cooling tech which may be of interest. <A> I have used arrays of high powered LEDs as an inexpensive way to UV cure conformal coating and an inexpensive light source for solar cell testing. <S> Initially, I attached the LEDs to a 1/4" thick piece of aluminum sheet to remove the heat, and the aluminum and LEDs got very hot. <S> One of the LEDs burned out and had to be replaced. <S> I initially used wet paper towels to cool the aluminum sheet, but this was a bandaid solution. <S> My solution for cooling was to place a copper heat pipe CPU cooler on the backside of the aluminum sheet. <S> This worked very well; the temperature of the aluminum sheet was maintained 30C and the LEDs stopped burning out. <S> These coolers are inexpensive, effective and easy to install.
| My initial recommendation though would be to turn the top of the PCB into a heatsink layer and use thermal adhesive to stick the LEDs to it.
|
How to lengthen a short pulse? i'm building a shot counter circuit for my airsoft gun controlled by an ATtiny26L MCU, where the shots are detected by an infrared diode + phototransistor couple. The PT outputs positive signal when it's triggered by the IR LED, which goes into a PNP transistor to invert the signal and then I have an electrolytic 1uF cap to lengthen the pulse, a 100k pull down resistor (initially there was a 1k), and a 1k resistor to regulate the output current to lenthen the cap's output: The circuit works flawlessly when I block the IR diody by hand or drop a BB through the detector. However if I shoot a BB through it doesn't detect it. The shot BBs trigger the detector for around 70 microseconds and if i'm correct there shouldn't be any trouble with a such big cap as 1uF lengthening the pulse, but maybe i'm doing something wrong? Or perhaps the phototransistor won't get triggered in such small time? Also my MCU runs at 1MHz. I'm glad to hear any tips. <Q> I would use mono-stable multi-vibrator for that. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The time it will stay in "triggered" position is: $$ t = ln(2) <S> R_3C_1 <S> $$You can find more on this topic here <A> Your positive-going signal from the opto-sensor will turn off Q allowing C to discharge through R. With 100k/1u you have a time constant of 0.1s <S> so the voltage will hardly budge in 70 us. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Quick discharge of capacitor. <S> The circuit of Figure 1 may behave better. <S> When the photo-transistor pulse goes high it will turn on Q1 which will discharge C1 instantly. <S> When PT turns off, Q1 will turn off and C1 will charge up via R2. <S> The time constant \$ \tau = <S> RC = <S> 10 <S> ^5 <S> \cdot 10^{-7} = 10^{-2} <S> = <S> 10~ms \$ should do the trick if your micro is polling at a reasonable rate. <A> You can pick Cx and Rx to get whatever pulse width you want, on whichever edge you like and pick either Q or /Q for regular or inverted output. <S> They in duals in various packages <S> (DIP if you like), for less than 50 cents <S> ea. <S> Edit: If you needed to stretch an extremely brief pulse, there are variants such as the 74VHC123A <S> which are much faster, but it's not necessary here.
| As others have said, you probably don't need to actually stretch the pulse, but if you want to, here is a very easy method- use a 74HC123 retriggerable multivibrator:
|
Possible error in schematic; unrealistic capacitor I am trying to get the parts for this electromagnetic coilgun , and I have gotten almost all of the parts. I cannot seem to find the 1,000 uF 1.5KV capacitors (C3 and C4) for the voltage ladder, and from my experience with electronics (and a little research) this is a ridiculous value. From the picture in the above link, it looks like a ceramic cap, but I don't know what it is. Is it a typo? If so, what do you think the right value is? Schematic: <Q> A bit of Googling finds this page , which has the following correction: Capacitors C3 and C4 Note that there was a typo for the values of capacitors C3 and C4. <S> The correct values for C3 and C4 are .01uf at 1.5KV DC or higher, ceramic disc. <A> M is possibly an abbreviation of micro-micro farad. <S> A micro-micro farad would be a pico farad. <S> See here for a discussion of micro-micro farads. <S> This answer is possibly the correct explanation of the marked values for C3 and C4, but is probably irrelevant, as RJR has found the correct values for the schematic in a later article that mentions that the schematic was marked incorrectly. <S> I'll leave this here as info on capacitor value markings, but please direct all upvotes to RJR's answer. <A> It is definitely a ridiculous value. <S> A 1000µF capacitor, rated for 1500v, would be very large, very heavy, and very dangerous. <S> A 1000MF of the same rating? <S> As big as a house, or more. <S> As the others have hinted to, the small size of the component dictates that it must be much, much smaller in capacitance. <S> 1000pF sounds reasonable. <S> It could be a translation, typesetting, proofreader, or even intentional error. <S> The Cockroft-Walton voltage multiplier stage will work just fine with 1000pF caps. <S> It would work better with a larger value, but then things get bulky and heavy. <A> There is a picture of the finished vero board in the link. <S> The 2 x 1,5kV capacitors can be clearly seen - 2 small blue devices. <S> Obviously not 1000uF.
| The correct value doesn't match any reasonable interpretation of the values marked in the original schematic.
|
FET Current limiter 30A and overload detection I plan to make a module for my motorbike to replace fuse by current limiters and overload detection system. I've heard nice things to do with J-FET. It should work up to 12V 30AMP. The detection should return the signal to a PIC Microcontroler. I have found the explaination below in litterature. It says that when it is overload the current falls to 8mA. Is there any ways to measure it and send a return signal to microcontroler? Thanks in advance! JFET current Limiter JFET current limiting circuit is shown in figure. Almost all the supply voltage therefore appears across the load. When the load current tries to increase to an excessive level (may be due to short-circuit or any other reason), the excessive load current forces the JFET into active region, where it limits the current to 8 mA. The JFET now acts as a current source and prevents excessive load current. A manufacturer can tie the gate to the source and package the JFET as a two terminal device. This is how constant-current diodes are made. Such diodes are also called current-regulator diodes. <Q> That won't work for multiple reasons. <S> 1) You can't get JFETs that handle much more than 10 mA 2) <S> If you current limit 30 <S> A overload on a 12 V system <S> , you'll dissipate 360 W of power and quickly fail your circuit 3) <S> Most systems like this limit current for a short time (< 1 ms), then turn off completely, and perhaps retry at intervals. <S> If you turn off too quickly, the voltage spike from the inductance of the will will also damage your circuit. <S> Take a look at this - http://www.nxp.com/files/training/doc/dwf/DWF13_AMF_AUT_T1023.pdf and look for eXtreme Switches to see if any of those would work for you. <A> I've Spice simulated & built all of the above, but the best option is a big honking power MOSFET w/ a big heatsink. <S> If you use the "analog mode", you will dissipate a lot of power. <S> The thing I haven't tried is a SMPS approach; "smart control" is unnecessary, as a proper regulator will inherently work or burn up. <S> OK, to turn this to an answer: JFET, NO, MOSFET, YES. <S> It will be large, I've done this. <S> Bipolar BJTs are current devices and don't turn off as well as the voltage device FET types. <S> JFETs are not made in the power ranges that would work. <S> But MOSFETs are available in "brick" packages. <S> My goal was 100A and I used MOS to get there. <S> The MOSFET bricks and a heatsink can dissipate the reject power. <S> I'm talking 1200V/100A devices used in megawatt <S> UPS's that I worked on. <S> To get away from the high power dissipation, a SMPS [Switch Mode Power Supply] method would be needed. <S> This makes use of a true switching OFF mode rather than imposing a dissipation. <S> However, conduction reject heat in the ON mode would still need to go somewhere, so big likely does not go away. <S> By "proper regulator" <S> I mean a circuit that regulates w/out an external "decision maker" [CPU]. <S> A lot of such burn up w/ thermal runaway when pushed beyond a SOA [Safe Operating Area] restriction, such as the Doubt 2-terminal BJT [Bipolar junction Transistor] circuit. <S> The Doubt circuit [Pat US3769572] looks promising but is best left to 20 mA or so applications. <S> I've tried them @ <S> > <S> 1000X. <S> Another problem is that these limiters use a series resistor as a sense element. <S> To get high currents, this sense R goes into the milliohm region. <S> You can, or will have to, make your own shunts by using a tapped heavy gage wire, i.e. the drain or emitter lead itself. <S> You will need a milliohm or microhmmeter for any high current shunt trimming. <S> TNX:) <A> I realize this is an old question. <S> But the correct answer to this question is "you don't want to do this. <S> " Fuses are specifically there to prevent wires from overheating and smoking the insulation or even starting a fire. <S> They are simple and reliable. <S> For example mosfet's usually fail with drain and source shorted together. <S> So this is just a terrible bad and unsafe and misguided idea. <S> It may be OK to put active current limiters between the existing fuse and load. <S> But not to REPLACE the fuse as this question says.
| You cannot replace a fuse with an active circuit incorporating mosfet's and whatnot, because the active circuit failure modes may be unsafe. I've tried this w/ the classic 2-terminal [Doubt patent] and it is only a "runaway resistor"; there is a 3 transistor version that works.
|
How to avoid damaging relay used for controlling motors? I've got a circuit with two DPDT relays controlling a motor - the diagram below is borrowed from a related but different question. RLY1 turns everything on and off, RLY2 switches the direction of the motor. I've found that after a brief amount of use the relays are damaged. Both relays have the same problem - on one set of contacts common/NO/NC are all connected, even when the relay coil is unpowered. (The other set of contacts is as expected - ie. common/NC are connected, NO is connected to neither of the other two - so only "half" of each relay is currently damaged). Questions : what is likely to have caused NO/NC terminals to be connected, and how should I have avoided that problem? Gory Details (possibly not needed): motor runs on 30V PSU that came with the motor is rated 30V/2A relays are rated for 2A/30VDC (model Meishuo MCB-S-205-C-M) relay is controlled from an ATMega MCU relays both have an IN4001 protection diode across the relay coil - not shown in diagram I did manage to "fix" one relay by tapping it on the table. Unknown if that will stay fixed or if permanent damage is done. The other relay doesn't respond to this treatment. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Snubbers on reversible motor. <S> Provided your relay contacts <S> are break-before-make you shouldn't be causing a direct short. <S> The problem is likely to be the inductive kick causing arcing of the contacts when they open. <S> This may look a little strange but is really just a rearrangement of the standard H-bridge protection as shown in Figure 2. <S> simulate this circuit Figure 2. <S> The same circuit in H-bridge configuration. <S> simulate this circuit Figure 3. <S> Redrawn to use relay configuration on OP's schematic. <A> Yes, you do have a problem. <S> Start here . <S> Then, when you close the switch or relay, if the motor is stopped, the motor will draw what is called stall current, which is much larger than running current. <S> This results in welding the contacts if they have not burned out when arcing. <S> From the linked article, The result is a large in-rush current at “turn-on” and arcing at “turn-off.” <S> When you are switching a motor load, typical industry practice is to derate to 20 percent of the resistive rating. <S> So, for a 2 amp motor, you should replace your relays with units rated at 10 amps minimum. <S> As to how the NO/NC contacts got shorted, frankly I haven't a clue, but if you take a hacksaw to your relays (carefully!) <S> you should see some strange sights around the contacts. <A> Although the PSU is rated 2A and the relays are rated 2A when the motor starts there is a very briff moment where the motor draws much more than 2A. Also, when the motor stops it might switch from being a motor to being a generator generating a current that can be higher than 2A. <S> The PSU might be capable of dealing with the higher 'reverse current' but the relays might not. <S> The solution, change your relays to other relays that can do, at leasy, 6A. <A> Been there, done that. <S> Our solution was appropriate for the era we were in (late 80's, early 90's). <S> We used N-channel Current-Sense MOSFETs (IRCZ44) for the bottom half of the bridge and SPST-NO relays (American Zettler AZ-2150A) for the top half of the bridge. <S> A small CPLD handled the logic. <S> Clamp diodes were installed across the top relay contacts (K to Batt +). <S> Startup was simple: close appropriate relay, delay, turn on proper FET. <S> Turn-off was a little more complex: turn off FET, wait, turn off relay, wait, turn on both FETs for braking if needed. <S> MOSFET current was continuously monitored and appropriate action taken when needed. <S> Mostly PWM-type current limit, with shutdown timer if the over-current persisted for too long a time. <S> This was used to drive a large linear actuator that changed the pitch of fan blades on large industrial earth-moving machinery. <S> The pitch of the fan was constantly changing to keep the engine temperature constant as the engine load varied. <S> We are talking about many fan-blade movements during every minute of run time. <S> The technique was extremely reliable. <A> While a redesign using FET transistors would make a more reliable solution you could also implement a simple solution by replacing RLY1 with a N-Channel MOSFET, and a 2K resistor to the gate. <S> The output that drives the relay coil can just drive the 2K resistor that feeds the gate. <S> The resistor slows the switching of the transistor enough to prevent generating RF noise and protects the MCU if the FET transistor fails with a short between the gate and the drain. <S> If isolation is needed between the control circuit and the motor circuit you can use an isolated FET Driver with a high input voltage 3-terminal regulator to supply the power to the FET driver. <S> FET drivers will specify their ability to drive an ampere or more but this type of circuit will only uses 20-30 mA <S> so it can easily be supplied by a 3-terminal regulator without any need for a heatsink. <S> A 47 OHM resistor between the driver and the gate will work in this case to keep the RF noise down. <S> The components could easily fit on a small PC board that would fit in place of the relay. <S> All that is needed is to make sure the direction relay is not activated or deactivated with power enabled and upgrade the relay to handle a minimum of 5 amps, 10 would be better to fix the other relay problem. <S> The FET should be a logic-level drive FET if driven directly by the MCU. <S> The FET should also be rated for at least 75 volts and handle more than 20 Amperes. <S> FETs can be obtained for a very reasonable cost that meet these requirements and will give you a very long life because the transistor will not be challenged even by surge currents caused by the motor. <S> Adding a capacitor across the motor as close as possible to the motor will help keep electrical noise down which is a good thing for all the components in the circuit. <S> I would go with a ceramic or foil capacitor 1 uF 150 Volts.
| Your big problem is that inductive loads, like motors, want desperately to keep their current at whatever level it currently is, and the result is that opening a contact will invariably draw an arc, which will damage the contact surfaces of the relay. One solution is to put diodes on the motor to shunt the current to the PSU.
|
Can I connect multiple SN74HC595 shift registers separately (without daisy chaining them) to the same Arduino? I'm sort of a newbie to this whole idea of shift registers and the Arduino shift register handling library. I've found many demonstrations online, where shift registers are daisy chained. But I need to connect multiple shift registers to my Arduino without daisy chaining them. My common sense says, Yes! you can, But I'm not quite sure and I don't have that IC on hand so that I can physically build the circuit and test. I'm really shy and uncomfortable to ask this question to you guys whether if it turns out a stupid question. So.. please mercy on my soul :) <Q> Yes. <S> Each shift register typically requires three pins (data, latch, clock), but if your application makes it possible to share clocks or both clock and latch, then your required pin count will be less. <S> Keep in mind that if two shift registers are connected separately (no shared pins), they cannot be updated simultaneously (footnote: you can use the SPI peripheral to drive shift registers, and this can actually run while you bit bang other pins, so you sort of can update shift registers simultaneously. <S> Sort of). <S> To expand on sharing latch/clock (assuming you are bit banging pins, and not using the SPI peripheral): <S> Shared latch : <S> the SRs will update (the internal registers will be latched to the outputs) simultaneously. <S> However, you will still have to clock the data to both SRs separately. <S> The sequence would be "unlatch, clock data 1, clock data 2, latch". <S> Shared latch/clock : you can update the SRs simultaneously and clock the data out in a single sequence. <S> In your clocking loop, you simply update both data pins. <S> So the sequence is "unlatch, clock both data streams, latch". <S> This is faster, but for many applications, this distinction won't matter. <A> Why do you need to connect without daisy chaining? <S> The fastest and easiest way to update all your shift registers, using the existing simple Arduino tools, is to daisy chain. <S> In software, what you do is allocate an array of bytes, as many bytes as you have 595s. <S> A single command will then shift the entire array out to all the 595s. <S> If one reason for not daisy chaining is that some 595s should not be updated, then simply load the corresponding byte with the old data, and that 595 will update to the same data, without any glitch. <S> If you have different processes for different 595s, then simply alias their own pointers into the output buffer, each process can then update only its own bytes. <S> Once the buffer transfer has been launched, it uses hardware, so is no extra load on the program. <S> It would need to be a very large pipeline of 595s so that the latency through them became significant. <S> Remember that if you drive all the latch pins in parallel, the outputs will all update at exactly the same time, regardless of their position down the pipeline. <A> While you can do this, if you need a lot of them and all of them are individual rather than chained, it might be easier to use an I2C GPIO expander, e.g. the PCA8574. <S> Each of these chips provide 8 pins that can be independently controlled. <S> They connect via the I2C bus and can be configured to one of 8 addresses, so you can use 8 of them with only 2 pins of your controller. <S> This can be expanded by also using PCA8574As (which have a different fixed part of their address) to 16 chips per bus. <S> Arduinos have a single hardware I2C port, but you can also control an I2C bus from ordinary IO pins, so can easily increase this to control more if necessary. <S> Of course, there is a downside to the pins being individually controllable, which is that if you want to change the output pins synchronously, you won't be able to. <S> But if that isn't a requirement, I think using GPIO expanders is a simpler and more expandable route to controlling large numbers of outputs. <A> The anwser is Yes. <S> The question is perfectly valid and it is applicable in certain scenarios where you need microsecond accuracy and out of gpios. <S> Daisy chaining two 595 chips and sending two bytes would take double the time of sending just one instead. <S> Say you have a full 8-bit information which should be frequently updated on one of them, and only control bits that needs less frequent updates on another. <S> You could afford to lose one gpio in such a scenario. <S> DS and SHCP could be shared. <A> But I'm not quite sure <S> and I don't have that IC on hand so that I can physically build the circuit and test. <S> you can do it, easier with HC164 than with HC595 but definitely doable: using the latch pin as a chip-select mechanism.
| If you don't daisy chain, you will have to repeat the SPI byte out function for different pins, which consumes excess time, as well as excess pins. You just need to assign separate STCP pins to each 595 in this case.
|
Single LED powered directly from AC mains Today I disassembled a power strip (power extension) and found a LED connected directly to AC with a normal resistor (not a power resistor). How is it working as I thought it would be burned in no time? The resistor value is 220k. <Q> Use Ohm's law : $$ {220\:\mathrm{V} \over 220\:\mathrm{k\Omega} } = 0.001\:\mathrm{A} = <S> 1\:\mathrm{mA} $$ <S> And power is the product of voltage and current, <S> so: $$ 0.001\:\mathrm{A} \cdot 220\:\mathrm{V} = <S> 0.22\:\mathrm{W} <S> $$ A 1/2 W resistor could connect across 220V just fine without burning up. <S> The LED is also a diode, and lets current flow only in one direction. <S> So half the time there's no current at all, so the power is actually half this. <S> And if you are in a 120V country, the power is even less. <S> That explains why it doesn't immediately burst into flames, at least. <S> However as others have mentioned, this plug strip has a number of other problems which might will result in other dangerous failures. <A> The resistor power dissipation is safe as Phil Frost stated. <S> However through my 54 year old eyes it looks like a SFR16 resistor that has a Voltage rating of 200V. <S> I would always put 2 resistors in series when using SFR16. <S> If the LED is just a standard one things will be dim. <S> A high efficiency or super bright one will provide a credible indication. <S> This stopped flicker and increased apparent brightness due to the doubling of average current. <S> The poor LED in your powerbox is getting negative mains pulses albeit current limited across it half the time. <S> Is the LED a special back to back LED or does it have an integrated reverse protection diode across it? <S> If not you have an unreliable product. <A> The problem here is when LED is in reverse voltage mode. <S> Reverse voltage can go beyond LED's specs and burn it.
| The actual resistor in the photo could be some generic or copy or ripoff so the voltage rating could be bad. What I always did was use a bridge rectifier made from 4 1N4148s.
|
Purpose of capacitor and diode in analogue input protection circuit of ADC? I am analyzing the following circuit used as an analogue input protection for an ADC ( image source: http://www.analog.com/library/analogDialogue/archives/43-04/process_control.html ): I don't understand the purpose of c5, c6, and D3. So my questions: 1- I'm guessing c5 can reduce spikes on the input. Is this correct? 2- I have no idea what D3 (1n4148) does. If a negative voltage is placed at the input of the circuit this diode would immediately fry (defeats the purpose of the circuit). 3- I also have no idea what c6 does? 4- Also the ref input is connected to 0.5 volts. Why? The op-amp is rail-to-rail, why is there any need to shift the output by 0.5 volts? P.S. Anti aliasing filter is placed on the ADC input and is not shown in this circuit. <Q> A partial answer 1 - Usually a low value capacitance is placed as close as possible to the connector for better EMI and RFI. <S> 2 - Me neither 3 - Also, don't know. <S> Although I guess it is a very slow filter, since the signals are industrial and relatively slow. <S> What confuses me is that it was placed upstream of the resistor divider. <S> 4 - The offset is done because the input ranges are symmetrical and the ADC input is single ended. <A> Others have addressed most of the points. <S> I just wish to address R5, S2 and D3. <S> The 250 Ω switched resistor is a common feature on industrial analog circuits such as PLC (programmable logic controller) inputs. <S> It converts a 4 - 20 mA (or 0 - 20 mA) signal to 1 - 5 V (or 0 to 5 V) for processing by the ADC. <S> Leaving switch S2 open makes the input a voltage input. <S> Closing S2 converts it to a current input. <S> (A resistor is, after all, a current to voltage converter.) <S> D3 will protect the circuit in the event of a reverse connection of a 4 - 20 mA signal. <S> There is little danger of the diode burning out as the current is externally limited. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Typical industrial 4 - 20 mA input. <S> Without R1 the circuit becomes a voltage input. <S> With R1 a 20 mA signal will produce a 5 V input to the ADC. <S> simulate this circuit Figure 2. <A> The use of so many elements seems to partially be for redundancy, though the various protection techniques will apply to slightly different voltages/frequencies. <S> It probably also helps shunt the high frequency energy in ESD. <S> The TVS will mostly help with ESD, but these devices can't dissipate much power, so larger diodes are also used. <S> D3 is a fairly big diode. <S> It can handle about 0.5 W continuous power. <S> I can only imagine it as being for protecting against negative input voltages. <S> Perhaps it is used to shunt the current away from the TVS, so that the circuit can handle a higher negative voltage continuously. <S> C6 looks like it serves as a low-pass filter when the input is a current signal. <S> It, along with the resistors, will set the low-pass cutoff frequency <S> An ADC will only digitize signals above some minimum voltage limit. <S> Shifting up the voltage would allow a slightly negative voltage to be properly digitized, or a zero input to be in the range of an ADC requiring a some minimum input voltage.
| The small-value C1 input capacitor helps filter out RF signals, and is likely high-breakdown voltage so it can handle ESD events. D3 is more likely to protect against reverse connection in this case.
|
Questions about CAN bus I am using the PIC18F2680 built-in CAN module along with CAN MCP2551 transceivers, and I am using the plib library for CAN2510 functions.My question are: Should we use twisted pair wires (like take CAN-Low and CAN-High wires and twist them)? The second is more related to the CAN protocol - If two transmitters start transmitting at the same time and are assigned the same low valued ID as mask then who will win the arbitration and if not resolved, how to win the arbitration supposedly if I could write a code for it? The third being: CAN works on CSMA/CD, so how is carrier sensing done? How can we use MCP2551 in the Proteus ISIS simulation tool? I am using the plib/CAN2510 library which is actually defined for PIC18Cxxx series. Can it be used for the PIC18F series or must I write my own functions for init, mask, filter, read, write, etc.? <Q> 1.Should <S> we use twisted pair wires (like take CANL CANH Wires and twist it). <S> Definitely, yes. <S> It also makes the twisted wires a more consistent transmission line. <S> If 2 Transmitters start transmitting at the same time and are assigned the same low valued ID as mask then who will win arbitration and if not resolved how to win arbitration supposedly if I could write a code for it? <S> CAN is a message-oriented protocol. <S> The case you describe is two transmitters simultaneously sending the same message, which does not make much sense as a CAN message provides up to 8 data bytes to distinguish "talkers" <S> so, in theory this situation should not happen. <S> Third being CAN works on CSMA/CD <S> so how is carrier sensing done? <S> As per Texas Instruments : <S> The CAN communication protocol is a carrier-sense, multiple-access protocol with collision detection and arbitration on message priority (CSMA/CD+AMP). <S> CSMA means that each node on a bus must wait for a prescribed period of inactivity before attempting to send a message. <S> CD+AMP means that collisions are resolved through a bit-wise arbitration, based on a preprogrammed priority of each message in the identifier field of a message. <S> The higher priority identifier always wins bus access. <S> That is, the last logic-high in the identifier keeps on transmitting because it is the highest priority. <S> Since every node on a bus takes part in writing every bit "as it is being written," an arbitrating node knows if it placed the logic-high bit on the bus. <S> The internal circuitry is designed so as to read the bus permanently, which makes the transceiver also read the data stream it sends on its own, see figure 4 in the document: <S> If there is a difference between what it reads and what it sends, it immediately stops transmitting and enters an "error" state. <S> The message is buffered for later retransmission. <S> I don't know about 4 nor 5 though. <A> CAN is a differential pair signal and could be twisted without issue. <S> The second and third questions go hand in hand. <S> CAN uses a very interesting system. <S> The bus has two states, dominant (0) and recessive (1). <S> The bus idles in the recessive state. <S> Every node on the bus can write to the bus and sense what is being written. <S> When a packet starts everyone writes a dominant state to the bus to indicate the start bit. <S> Then everyone starts writing their message address. <S> For example suppose the first bit of the address is dominant. <S> Every module that wrote a recessive state senses the dominant state and drops out. <S> This goes on until there is only a single module left. <S> The module that's left will finish its message. <S> That being said the CAN bus should never have two modules that can send a message with the same address. <S> Speaking hypothetically if such an event did occur and both modules got through arbitration. <S> Then the first instance where both a dominant and recessive state is written to the bus would be detected as a bus collision by the module who wrote the recessive state. <S> From here you have two options as the designer of your own bus. <S> Make the collision module shut up and let the other module finish transmission or make the collision module issue an error (hold the bus dominant for over 6 frames). <A> Answers to some of your question would be: <S> You need to twist the CAN-High and CAN-Low wires for reducing noise <S> I am not sure about the second question, but you could use the bit_error flag to restart the transmission. <S> And MCP2551 is not currently available in Proteus simulation tool.
| Twisting wires resolves part of parasitic influence of the environment by using common mode filtering .
|
How can you wire a 555 timer to gradually decrease sound volume? Say you are taking in an audio input from a headphone speaker and you would like the audio to gradually decrease until inaudible. Are there better, also very common chips that can do this more reliably? Let's say you wanted the volume to decrease over 1 minute? <Q> Specifically, a log digital potentiometer designed for audio (so that it switches on zero crossings), and just about any microcontroller. <S> Most digital pots are controlled via I2C (or occasionally SPI). <S> You can trigger the "fade" <S> however you like (button, etc), and you have very good control of fade length and linearity. <S> You can even do stereo with nearly gain-matched channels (there are digipots specifically for stereo audio). <S> This way, no analog weirdness, programmable fade characteristics, flexible triggering (you still haven't said how you want it to trigger), and basically two parts (plus a power source and a smidgen of passives). <A> A technique that can achieve what you want is passing your audio signal through a voltage-controlled switch. <S> The control voltage should toggle the switch on and off at a frequency f_c over the audible spectrum and the duty cycle should change progressively between 100% to 0%. <S> From a signal processing point of view, you are effectively multiplying your signal by a variable duty-cycle rectangular signal exhibiting the values 0 and 1. <S> The spectrum of this signal is a sum of terms of the form A_n·X(f-n·f_c) and the A_0 term is what you are interested in. <S> A low-pass filter should be used to filter out the higher terms, but this may be unnecessary if the elements after your circuit already have a low-pass response. <S> There are many ways to achieve this. <S> An analog switch could be controlled by any device able to produce a variable duty-cycle signal. <S> A 555 can certainly produce a PWM (Pulse-Width Modulation) signal, but you would also have to generate the linear (or logarithmic) <S> decaying function. <S> A simple microcontroller could do this job very easily. <A> Which is then tied to a CDS LDR (light dependent resistor) inline with the audio source, like a pot. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The preferred way to do it would be two LEDS/LDR combos, with two PWM sources that are inverse to each other. <S> See http://practicalusage.com/vc-1-digitalmidi-volume-control/ for a project that did this. <S> You need to couple the LED and LDR together. <S> Electrical tape always works. <S> This is done professionally though dedicated parts like photoresistive opto-isolators.
| Honestly, one of the least fussy solutions would be a small MCU and a digipot. A bit of a hacky way to do this would be a 555 or 556 Timer as a PWM source to a LED.
|
Run code once in lifetime of an embedded C program How can I make a code snippet run only once in the lifetime of a program? It can be turned off and turned on many times. The only option to run the code snippet again must be flashing the board again. The code is a Calibration Section which I don't want to run again. If I use EEPROM or Flash we will be setting a Flag to true or False. So When we first read that memory location what would be the random value in that memory area? What is the best method to implement this in embedded C? <Q> Your microcontroller might have some EEPROM, OTP memory, user fuse bits, where you can set a flag. <S> There is no "best method in embedded C", writing nonvolatile memory is different in every microcontroller. <S> edit: FLASH <S> Flash memory contents are erased while programming the device. <S> After programming, all bytes that weren't written contain 0xFF. <S> Consult the datasheet to find an area that can be safely programmed from within the running firmware. <S> EEPROM <S> Although it's not guaranteed in the datasheets, all EEPROMs I've seen so far contained 0xFF: <S> s when shipped from the factory (except ones preprogrammed with an unique MAC address, but that's explicitly documented). <S> Some can be write protected, permanently or reversibly. <S> OTP <S> One Time Programmable memory always contains well defined initial values, documented in the datasheet. <S> It's always a good idea to include a good checksum like CRC32 with the data written, to protect against data corruption caused by defective parts, transmission errors, cosmic rays, whatever. <A> You said: The only option to run that code must flashing the board again. <S> Others have said to use EEPROM to store a flag to indicate when the run_once() function has been run. <S> However, this has a draw back which is that if you reflash the microcontroller the ran_it_once flag in EEPROM has already been set and the run_once() function will not be executed. <S> If your microcontroller has embedded EEPROM then it may be possible to clear the ran_it_once flag when you reflash the microcontroler, if the programmer supports this. <S> A better way is to have version numbers in both EEPROM and the code. <S> When the code runs from power-up it should read the version number from EEPROM and compare it to the version number stored in the code. <S> If they don't match then the run_once() function is called, and the final act of the run_once() code is to write the firmware version number into EEPROM. <A> Choose a microcontroller that can write/erase its own program memory. <S> After executing the code in question, have the last part of said code replace the first instruction with a jump that bypasses it. <S> Optionally, you can also erase the rest (maybe replace with nop), so that there is absolutely zero chance of it ever executing again. <S> This message will self-destruct in 5..4... <A> As you are using this code for calibration, my suggestion would be to create a blast process which runs the calibration code as a first stage and not even have it on the finished production version of the board. <S> This is similar to the answer by apalopohapa, except different in the sense that you would have two separate program loads: have a blast process that flashes the first program load which runs all the calibrations and spits out the data from that. <S> Then take that data and incorporate it into the data of the second program load. <S> One benefit of this approach is that you absolutely minimize the amount of storage space you need - you don't need to store your once only code, only the data that it generates. <S> By having a blast process which loads two separate programs, you also insulate yourself a bit from bugs in the initialization code which could linger around otherwise. <S> You also have some additional flexibility if you want to re-run your calibration code: <S> instead of having to write extra code to clear out whatever bit denotes that your code has been run (which could accidentally get cleared already) you simply re-run your blast process.
| Some programming devices/software are able to erase or program EEPROM contents too. Each time that you modify the source code of the firmware you must increment the version number embedded in it.
|
How to to Use a Nylon Shoulder Washer? I'm somewhat curious as to how to use a nylon shoulder washer with a bolt. There is an example of this here ( https://electronics.stackexchange.com/a/45579/95654 ), but I can't figure out how it would stay attached to the board. The diagram does not imply that the washer is threaded, so the hole is large enough to fit the washer (without any teeth to hold it) and ensure that the bolt does not close any circuits, but then how does it not fall out if it was inverted/sideways? Even if there was a nut at the bottom of the bolt, I still can't see how it would be stable. <Q> Using the schematic option produces ugly results, but I hope the following diagram will help. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It shows a bolt passing through a single layer of material, with a sholder washer on each side, and a nut, all in cross-section. <S> When the nut is threaded onto the bolt, the washers are held in place. <A> (source: Heatsink mounting kit ) <S> The shoulder washer is intended for bolting a TO-220 to a heatsink (or some other metal part) while maintaining electrical isolation with the heatsink. <S> There is usually a nut on the other side of the heatsink, or the screw hole is threaded. <S> If you are attaching the TO-220 to a part of your PCB that has no traces, then you don't necessarily need the shoulder washer. <S> The PCB material under the TO-220 would act as an isolator. <A> I'm on my phone, so I can't link to a website that might show a picture. <S> A Shoulder Washer is normally used in conjunction with either a flat washer (thin material) or another shoulder washer (thick material). <S> The hole through the material is very slightly larger than the diameter of the raised part of the shoulder washer. <S> This shoulder keeps the washer centered in the hole.
| The most common use that I have had for shoulder washers is when mounting power transistors to a heatsink or metal chassis.
|
How do I get a position from an encoder? I have a motor that drives a rail which I need to know the position of. I have an AM102 encoder attached to the motor. But am unsure how to get a position from it. I have powered it up and attached the two outputs to my Teensy 3.0. In my program I can see the value changing as I manually move the motor. (Either 1023 or 0 if I recall correctly). I am wondering how I can use this to determine position. I'm guessing that I can determine position by counting how many times the value changes from 1024 to 0 and then multiplying that number by a distance (not sure how to get that value). But when the system is first turned on, how will it know the initial position? Is there a way to get this from the encoder or some other device/method? n.b. I also have an Atmega328P which I've been told may how me determine position how would I use this? Question summary: How do I get the distance multiplier? How do I get initial position? <Q> How do I get the distance multiplier? <S> You measure or calculate how far your device will go for e.g. one full rotation, e.g. in cm. <S> Then you take that length and divide it by the number of pulses per revolution the encoder delivers and you got your distance multiplier in cm/pulse. <S> How do I get initial position? <S> The encoder itself can only tell you the rotation angle relative to the "index" position. <S> If the motor can do more than a single rotation in either direction there is no way the encoder alone could tell you the absolute position after power on. <S> A common solution is to have a limit switch , either mechanical, inductive, or optical (IR), at one end of the rail. <S> This is then <S> the "0" position and any movement/rotation of the motor is measured via the encoder relative to that fixed position. <A> What you have is a quadrature encoder. <S> There are two outputs which are 90° out of phase. <S> (90° = quarter of a turn, hence quadrature.) <S> An absolute encoder can give position information on power-up. <S> An incremental can't. <S> Normal sequence is to run the motor and load back to a home switch and zero the encoder counter. <S> Thereafter it's a matter of keeping track of the count using an up-down counter. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> 2-bit rotary encoder waveforms. <S> The program logic is very simple. <S> Track the current state of 'A'. <S> If the state changes to 'high' then: Look at input 'B'. <S> If 'B' is low then count up. <S> If 'B' is high then count down. <S> You'll probably need to debounce the inputs to prevent spurious triggering. <S> Figure 2. <S> A simplifed encoder layout. <S> A and B photo-sensors looking at or through the encoder disc will produce the waveforms of Figure 1. <S> This encoder has only 5 pulses per revolution. <S> Your's has 1024. <S> How do I get the distance multiplier? <S> If one revolution of the encoder is n <S> mm of motion then one encoder count = <S> \$\frac <S> {n}{1024}~mm\$ motion. <S> How do I get initial position? <S> Use a homing routine and home sensor. <S> Interrupt driven To make the encoder tracking interrupt driven <S> you just need to have one phase ("A" in the example above) trigger the input. <S> The interrupt handler then looks at B to determine whether to count up or down. <A> But when the system is first turned on, how will it know the initial position? <S> Is there a way to get this from the encoder or some other device/method? <S> It can measure relative changes but has no idea where it's absolute starting point <S> is and doesn't retain any information through a power cycle. <S> See this wiki page for further information <A> you have to use another sensor to determine where you are, this will only tell you relative position from that calibration point. <S> as mentioned, a typical solution is a limit switch, move to that position then use this sensor from there. <S> with a relative sensor like this you always have to have some other solution for setting a known or initial position. <S> It is possible but risky to do this one time, and before powering off, save the position in non-volatile storage. <S> risky in a number of ways, assuming this sensor never slips or misses, assuming your logic/software never slips or misses, assuming you have enough power storage to have the time and energy to store before powering down, assuming that powered on or off nothing slips and changes your position, and on and on and on. <S> Why say this, because it is undesireable in some applications to sweep both ends to find the limit switchs. <S> I know some vehicles have actuators that control the heating and cooling vents, when they power up the first time they sweep the hard stops and calibrate, then they rely on battery power to remember that information, forever or until your car battery dies or is removed/replaced/disconnected for any reason. <S> you can use hard stops instead of switches, run the thing (gently) into the hard stop, this sensor will tell you that you are no longer accumulating counts. <S> then move it to the other hard stop. <S> that doesnt require a pair of switches.
| Upon power up, the controller drives the motor toward the end of the rail with the limit switch until the switch detects arrival at the end. The device you have is an incremental encoder and not an absolute position encoder.
|
How to measure the total current driven in a building by observing the electricity meter? Electricity meters show power. How can I measure roughly how many total current in amperes is driven at an instant in a building including a 3 phase AC motor? <Q> Electricity meters show power <S> No, they show power integrated over a time period (energy) <S> i.e. kWhours and not kW actually being consumed in a particular moment. <S> On this basis, you cannot know what the load current is in a particular "instant". <S> You could set a stop watch and record the change in kWhours. <S> Then convert this back to kW <S> (knowing the time period elapsed on the stop watch). <S> Then estimate average RMS current over that time period <S> but you'll have to take a stab at power factor which by no means will be either unity or constant over various load conditions that the 3ph motor might encounter over that time period. <A> Typical Power factors: Resistive load 1 Fluorescent lamp 0.95 Incandescent lamp <S> 1 Induction motor full load 0.85 Induction motor no load 0.35 <S> Resistive oven <S> 1 Synchronous motor <S> 0.9 <S> 120V = <S> 208V 3ph. <S> Current = 95000 <S> X1000 <S> / sqrt(3) <S> X V X <S> PF current= 95000000 / 1.732 <S> X 208V <S> X 1 current = 95000000/360.26 <S> current = <S> 263,698.44 <S> Amps or 263.7 kA <S> Resistive load worst case scenario <S> Electric motor PF <S> .85/.35 <S> = <S> 224.15 kA, 92.29 kA respectively...... <A>
| You can buy a meter that clamps onto the wire and will tell you precisely how much current is being drawn.
|
When will parallel LED voltage drops become an issue The following circuit will cause problems because of voltage drops: simulate this circuit – Schematic created using CircuitLab It should be changed to look like this: simulate this circuit However if we used both parallel and series resistors in this circuit, there is a potential for voltage drop issues between the LEDs. As we raise the value of the series resistor and lower the values of the parallel resistors, what determines the point at which voltage drop issues start to happen (and at what values?) In this scenario, all of the LEDs are the same brand, with a 2 volt drop +-2% and are run at 20 mA. simulate this circuit Edit---- Practical application: I designed my original circuit with 18 LEDs in parallel, each with a 220 ohm resister. Now, I want to change the voltage to 9 volts (or 12 volts). Will I need to switch out all of the original arrays or can I add one resistor in series? ...Then I became curious about the limits involved in doing this. <Q> It's not an easy calculation. <S> You can get some idea of the worst-case situation by looking at max/min forward voltage drops, and the slope of the current/voltage curve, and the LED temperature coefficient from the LED data sheet. <S> Without detailed statistics on LED voltage-current-temperature characteristics it will be difficult to guess what will typically happen. <S> Cheap consumer products often parallel LEDs directly with no resistors, but then nobody much cares if one LED is 30% brighter than the next and <S> the whole thing only lasts a few thousand hours or less. <S> Edit: <S> Here is a simulation you can play with. <S> I've altered the saturation current of the model for D2 to make the Vf high by 4% 40mA current splits with D1 getting 23.1mA and D2 getting 16.9mA. <S> That's if they are all held at exactly the same temperature. <S> If instead I assume that they are thermally independent and have a 50°C rise nominally, then the difference between the two would be 16°C and if the tempco is -1.7mV <S> /K <S> then that would cause another 1.7% difference between the Vf's, leading to more temperature rise etc. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> At any given level of LED drive current, adding another milliamp will increase the voltage by some amount. <S> The ratio of marginal voltage to marginal current may be called the marginal resistance. <S> If two LEDs in parallel have slightly different voltage drops, the difference in current flow will be roughly equal to the distance in voltage drops divided by the marginal resistance. <S> If the marginal resistance is small (as it is with some LEDs), the difference in current will be quite large. <S> Adding e.g. a 10-ohm series resistor will increase the marginal resistance of the LED+resistor combination by ten ohms. <S> That may not sound like much, but it's huge compared to the marginal resistance of some LEDs, and might thus reduce current variation by an order of magnitude. <S> If one were using resistors as the only current-limiting devices, there wouldn't be much point using N+1 resistors for N LEDs (versus simply using N resistors). <S> The approach may be advantageous, however, if one is trying to use a transistor-based circuit to control current to many parallel LEDs. <S> If one has ten parallel LEDs driven from a supply that produces a regulated 1 amp, the LEDs will receive an average of 100mA, but some might receive 150mA and others 50mA. <S> Adding a ten-ohm resistor in series with each LED would require that the supply be capable of producing one more volt than would otherwise be necessary, but would nearly eliminate variations in drive current. <A> The issue with the first circuit is that the three leds have a different forward current at the same forward voltage. <S> The drop across the three leds will be the same. <S> One led can't take 3 volts while it's parallel led only takes 2. <S> Unequal voltages can't happen in parallel circuits. <S> And as each has a different current across them, one may dominate the amount of current available, and may go into thermal runaway conditions, killing it. <S> Which kills the next one for the same reason. <S> Etc. <S> The third does not have this issue, as each is limited to the current by their individual resistor. <S> The voltage drop across each led and resistor will be equal. <S> Current may vary. <S> simulate this circuit – <S> Schematic created using CircuitLab Notice that V-Total is the same as the Voltage source, 5 Volts. <S> It will always be 5V. Notice that V-1 (R1 + D1), equals V-2 and V-3, but A-1 has a different current than <S> A-2 and A-3. <S> Yet V-4 and the nodes V-1/2/3 equal V-Total. <S> And A-1 + A-2 + A-3 = A-4. <S> The very basics of series and parallel circuits. <S> Re: <S> Your edit: <S> Practical application: I designed my original circuit with 18 LEDs in parallel, each with a 220 ohm resister. <S> Now, I want to change the voltage to 9 volts (or 12 volts.) <S> Will I need to switch out all of the original arrays or can I add one resistor in series. <S> Yes, you can. <S> You need to resize R4 so that it takes up the same current at higher voltage. <S> The formula stays the same. <S> Basic Ohms law R = <S> V <S> / I . <S> Since you know I, and the new V, just calculate for R. Assuming the values from above, we can change R4 and the Voltage Source but keep the same current of 15.45 mA. R = <S> (9 V Source - 1.909 V Forward ) / <S> 0.01545 <S> A = <S> 459Ω <S> Of course you need to chose the next standard sized resistor. <S> But we'll ignore that. <S> simulate this circuit <S> If you are just adding a resistor to an existing circuit like 2, then V Forward would be the old V Source , i.e. 5V.
| Highly depends on how much current can be pulled, i.e. the resistor value.
|
Characteristic impedance of a D-subminiature connector I would like to know what the characteristic impedance of two adjacent pins of a D-subminiature connector is, or how I can go about calculating it. For example, see below 50-pin D-sub, if I were to use pins 1 and 2, or 1 and 18, or 2 and 18 - each adjacent to each other - then what would the characteristic/transfer impedance be? I would like to use two pins for a impedance controlled balanced diff pair and want to know if this connector will pose reflection issues. It can be calculated using transmission line EM theory but I'm not sure how to include the effect of a grounded connector shield. I can't find this in MIL-DTL-24308 or any of the manufacturer specs. <Q> D-sub connectors are not designed to be used at frequencies where transmission line effects in the connector are important, so their characteristic impedance is not well defined and it is not specified or guaranteed by the manufacturers. <S> In order to make a D-sub connector with a well-defined \$Z_0\$, it would have to have a uniform geometry along the signal path. <S> The pins would have to remain the same distance apart, with the same diameters, and the dielectric material around them would have to be uniform. <S> Possibly in a cable-to-cable connection, these conditions may be approximately true, but in a PCB-mounted D-sub, they're not likely to be maintained as the pins turn down to connect to the board. <S> I found one such calculator here along with the estimating formula $$Z_0 = \frac{120}{\sqrt{\epsilon_R}}\mathrm{acosh}\left(\frac{s}{d}\right)$$ <S> Where \$s\$ is the center-to-center spacing between the wires and \$d\$ is the wire diameter. <S> The effect of the shield is difficult to predict because it is likely not going to be uniform along the signal path, and will have a different relationship to the two signal conductors depending which pins you choose to use. <S> Similarly, the presence of the surrounding pins (say, pins 2, 19, 34, and 35, if your signal pins are 1 and 18) will also cause some change in \$Z_0\$. <A> If you Google on "conductor spacing impedance" you can find online impedance calculators like this one and a small amount of effort will tell you that for 26 ga wire, the impedance is about 590 ohms. <S> Since the actual connector barrels are slightly larger than 26 ga, the actual impedance will be somewhat less. <S> However, you haven't specified your frequency range, so there's no way to tell if this impedance discontinuity, being of very short duration, will be a problem. <S> And if you're really worried, you can get combination D-sub connectors which will handle up to 2 GHz <A> Thanks. <S> The application is for 100Base-TX Ethernet RX/TX which needs a Z0 of 100-ohms for the balanced diff pairs. <S> As long as the bulk cable has the right impedance it will be fine. <S> 100BASE-TX is not that sensitive that you have to start obsessing over connector impedance.
| The calculation would be based on the pin diameter, pin separation, and dielectric constant of the material around the pins. If you did find a D-sub connector with a uniform geometry, you could get an approximate value for \$Z_0\$ using the same formulas that estimate \$Z_0\$ for twisted-pair wire.
|
How do you programmatically detect pin count on an MSP430G2553? I am using a Launchpad to program an MSP430G2553 controller. I use it for debugging. This controller has 20 pins. However, the end goal here is to run my program on an MSP430G2553IPW28R controller. This controller has 28 pins instead of 20. I'm using pins P3.0 and P3.1 on the 28 Pin Controller for some LEDs. And my launchpad controller doesn't have those pins. I need to determine which controller the code is running on. I will be using my launchpad's JTAG pins to program the 28 pin controller. So, the question is, how can I programmatically determine which of the 2 controllers my code is running on, and put some kind of #define in place to use pins P3.0, P3.1 when running on the 28 pin controller and pins P2.1 and P2.2 when running on the launchpad controller. I don't want to have comment and uncomment lines of code each time I switch from one processor to the other... Please note, I'm using Energia as my development environment. <Q> The method below will tell you if a pin exists on the controller. <S> So, to determine if we're using a 20 or a 28 pin controller, select a Digital Pin that is > 20 and pass it as the parameter to the function below. <S> If it returns true, you have a 28 pin controller. <S> If it returns false, you have a 20 pin controller. <S> I tested this using my MSP430G2553 Launchpad and have the following results: showPinExistence(2); showPinExistence(3); showPinExistence(5); showPinExistence(6); showPinExistence(7); showPinExistence(8); showPinExistence(11); showPinExistence(15); showPinExistence(18); showPinExistence(21); showPinExistence(22); <S> showPinExistence(23);void <S> showPinExistence(int pin) <S> { Serial.print("Pin "); Serial.print(pin); Serial.print(" Exists: "); Serial.println(pinExists(pin)?"True":"False"); Serial.flush();}bool pinExists(int pin){ bool lowResult; bool highResult; pinMode(pin, OUTPUT); digitalWrite(pin, LOW); lowResult <S> = digitalRead(pin); <S> digitalWrite(pin, HIGH); highResult = digitalRead(pin); return lowResult ! <S> = <S> highResult;} Here's the output for LaunchPad with MSP430G2553 (20 Pin): Pin 2 Exists: <S> TruePin 3 Exists: <S> TruePin 5 Exists: <S> TruePin 6 Exists: <S> TruePin 7 Exists: <S> TruePin 8 Exists: <S> TruePin 11 Exists: <S> TruePin 15 Exists: <S> TruePin 18 Exists: <S> TruePin 21 Exists: <S> FalsePin 22 Exists: <S> FalsePin 23 Exists: <S> False <S> Here's the output for MSP430G2553 (28 Pin): <S> Pin 8 Exists: <S> TruePin 9 Exists: <S> TruePin 10 Exists: <S> TruePin 11 Exists: <S> TruePin 12 Exists: <S> TruePin 13 Exists: <S> TruePin 14 Exists: <S> TruePin 15 Exists: <S> TruePin 16 Exists: <S> TruePin 17 Exists: <S> TruePin 18 Exists: <S> TruePin 19 Exists: <S> TruePin 20 Exists: <S> TruePin 21 Exists: <S> TruePin 22 Exists: <S> TruePin 23 Exists: <S> True I had to start with pin 8 because it didn't like when I attempted to set 2-7 as output pins :-) <S> Anyway, the point is that you can attempt to set pin 23 as an output pin and write and read its value to determine if the chip you're using is a 20 pin, or if it has more than 20 pins. <S> I don't have a 32 pin chip, but I would imagine that you could test on a pin greater than 28 to make a determination if it is a 32 pin chip. <S> Now, after ALL that I've said. <S> Since writing this, I discovered that there is now 28pin support on GitHub. <S> I am now using it and referring to pins with their proper P3_0, P2_0 notation instead of directly by Pin number and this seems to be working just fine now. <A> Depending on project complexity and size of the MCU I use from 1 to 3 pins for "PCB identification" with external pull-down resistors. <S> Resistors are mounted depending on the board revision. <S> Application inside the MCU looks at these pins at startup and determines what to do. <S> It helps even more if you have several products with some differences allowing you to have a single binary image for all hardware revisions. <A> IMHO the most preferred option is to look for a chip ID in the documentation. <S> Most modern chips provide an internal register with the chip type and revision. <S> You could use that to drive your logic. <S> However these only limit to the family die and do not distinguish between chip packages. <S> You could still try to test and verify if there is any differences.
| Alternatively you could use physical board changes, like adding pull-up/downs on certain pins that can be software tested or tying pins low/high. For MSP430 family I have been able to find some core IDs in the BSL documentation (page 37).
|
Purpose of Resistor - comparator I am studying this time delay circuit and am curious to know the purpose of the 4.7kΩ resistor.As I see it, it has no purpose. I've built and tested this circuit with and without it and see no difference. Anyone have any insight on this? <Q> One of the things it might be doing is reverse biasing the diode when SW1 is open, pulling it below cap voltage. <S> This would reduce the likelyhood of stray fields forward biasing the diode and charging the (peak detector) cap, as long as 4.7k was strong enough to drain off the field charge and maintain reverse bias. <A> The resistor makes the circuit behave predictably even if there is some leakage or shunt reactance across the switch. <S> For example, through capacitive coupling through long adjacent wires to the switch, or leakage due to the switch or wires getting wet. <S> Only a few hundred pF will actually create a noticeable (with care) effect at mains frequency, so it's not crazy. <S> You didn't test it with long wires or with leakage, so the issue didn't show up. <A> <A> The only purpose I can think of is to lightly load the transformer secondary during the timing operation. <S> Without it the transformer is only loaded on positive half-cycles. <S> Adding a 4k7 resistor across 24 V will only draw 5 mA or so <S> but it may be enough to make the circuit less susceptible to mains noise.
| The resistor prevents stray electric fields from charging up the 47 uF capacitor. Most of the time, it's not necessary, as the 150k resistor also performs this function, so it's really just insurance.
|
Can a Transformer Have Multiple Secondary Coils? Can a transformer have multiple, isolated, secondary coils from a single primary coil? Can each coil have a different voltage from the other? If so, is there any large negative effect, and what is the behavior as the load increases? Related: http://engineering.electrical-equipment.org/electrical-distribution/multiple-winding-transformers.html <Q> In ancient times, for vacuum tube equipment, power transformers often had three secondaries: 5 volts for rectifier filaments, 6.3 or 12.6 volts for other tube filaments, and a high voltage coil (perhaps ~250 V) for plate supply. <S> The high voltage winding would usually have a center tap (connection to the center of the winding) to allow the use of a full-wave rectifier circuit. <S> Many transformers have dual primary coils that can be connected in series for 240 volt operation, or in parallel for 120 volt operation. <S> It is also possible to use transformers "backwards" - using the nominal secondary winding as a primary, and the nominal primary as a secondary. <A> Yes <S> Yes, within reason (voltage will be determined by turn count ratio, and the primary will have the same turn count for all of the secondaries, which limits what you can do if the turn counts are small) <S> I am not 100% sure about this, but I think it will act like placing multiple single-secondary transformers in parallel. <S> I have seen setups like this in devices before, namely good quality (HP/Agilent/Keysight) isolated bench top power supplies, that have a single transformer and separate secondary coils to power the GPIB interface board (ground-referenced) <S> and each channel of the power supply (isolated from ground and from each other). <S> Many of these units will also have more complex primaries that can be reconfigured with a switch for efficient operation across a specific set of input voltages (100, 110, 120, 220, etc.) <A> Multiple secondary windings are very common in isolated DC/DC supplies for IGBT gate drivers. <S> look up pictures of multiple output flyback and IGBT gate drive supply. <A> As from ur question "a transformer have multiple, isolated , secondary coils from a single primary coil" -Yes Isolation in transformer means simply separation of different windings, From primary to <S> Secondary, even multiple secondary's. <S> Secondary can have multiple voltages 3v,5v,9v,12v,15v,18v etc without any problem. <S> All coils (Secondary) can be used at once or one or as per requirements without any Negative Effect.
| Yes - transformers can have multiple secondary coils of various voltages.
|
Operation Amplifier: + and - with equal voltage I understand the basics of Operational Amplifiers, but I am curious, what would happen if the + and - input pins have the same voltage, does the output pin (x) give a high or a low voltage output? simulate this circuit – Schematic created using CircuitLab <Q> Theoretically, in a perfect op-amp the output voltage would be mid-rail. <S> However, due to input offset voltages the output will be somewhere between the most negative rail and the most positive rail. <S> OP-amps have DC gains of millions so a 100uV offset at the inputs produces an open loop output level of hundreds of volts (limited by the power rails). <S> As a real example, the ADA4528 has an input offset voltage that is 2.5 uV maximum at ambient temperatures. <S> It also has an open loop gain that is at least 130 dB (3.16 million). <S> Result is potentially an output voltage in the range +/- <S> 8V. <S> However it's only a 5V device so most of the time the output will be clamped to the rails. <S> For those times that it is not against the rails the internal noises inside the op-amp will be rapidly end-stopping the output against the + and - rails. <A> The input offset voltage, which can be positive or negative, is the voltage that is added to the + pin to make it the same as the - pin. <S> Below you can see Vio for an MCP6001. <S> With regards to the output voltage, your circuit will operate as a comparator. <S> Therefore the output voltage could be 0V <S> or +V depending on whether Vio is positive or negative. <A> in an ideal opamp, the output voltage will become zero if both the terminals have same voltage.since it is made out of differential amplifier, the difference between the invertng and non-inverting voltage becomes zero.
| In a practical circuit, it would depend on the input offset voltage, Vio.
|
Does mutual capacitance increase for two positively charged plates when brought close? My understanding is that any two charge carriers will exhibit what is known as mutual capacitance. That is, there is an amount of charge that can be stored between any two objects, which is called mutual capacitance (typically just called capacitance ). As answered here , it seems very clear to me that this capacitance will increase when two objects of opposite charge come closer. (You generate a stronger electric field which draws more holes to the positively charged object and more electrons to the negatively charged object). As I see it, mutual capacitance exists between any two positively (or negatively) charged objects. According to the Wikipedia stub for Mutual Capactiance , no distinction is made about the charges, and it plainly says: All objects in the universe, conducting or non-conducting, that hold charge with respect to another exhibit capacitance. An object's capacitance increases when another object is brought closer to it. The following diagram demonstrates what my question is about: Since the right side has a larger (magnitude) charge than the left side, I'd expect the net field to go from right to left. (If I threw a proton in the field, it would repel more from the right and end up on or close to the left). But as I bring these two objects closer together I would expect that the capacitance would decrease because the right side would attract electrons from the left side, forcing out the holes in the left to some other part of the system. Conversely, if the objects were moved further apart, I'd expect the field to diminish to the point where both objects have no effect on each other. This would cause the capacitance to increase to its maximum. So, what is true? Does the mutual capacitance increase or decrease for these two objects as they come closer? Is it even possible for these two objects to have a capacitance if they have like charges (both positive in this case)? I may need to open up another question for this, but I'm trying to see how mutual capacitance is used for proximity/touch sensing. As it seems right now, I can't tell how this mutual capacitance can be monitored if only one object is connected to the sensor system, while the target (a human hand for example) is not connected. <Q> It is only a function of the geometry of the capacitor and the dielectric constants of the materials involved. <S> In general, for a parallel plate capacitor, the capacitance is inversely proportional to the distance between the plates. <S> So therefore if the plates are brought closer, the capacitance will increase. <S> However, the value of this capacitance depends only on the size of the plates and the dielectric properties of the material between the plates. <S> The value of capacitance is not affected by whatever charges may be present. <A> Q = CV. <S> If the two objects were dragged away from each other, capacitance decreases (as per the definition of what physically capacitance is) but voltage increases <S> thus Q remains constant. <S> But as I bring these two objects closer together I would expect that the capacitance would decrease because the right side would attract electrons from the left side <S> How are these electrons to cross the barrier of the insulation? <A> First let me try to interpret what you are saying: if there are two objects that are very far apart, therefore the field between them are very weak for a given net charge. <S> If an extra charge is brought to one of the object, since the field is so weak, the capacitance must change by a lot. <S> But it does not work this way. <S> Forget about capacitance for a moment. <S> When an extra charge is brought to one of the two objects, the field has changed. <S> And the potential between the two objects has changed. <S> If I measure the potentials or integrate using Coulomb's law to get the potentials and then do a little calculations:$$ \frac{initial.net.difference.of.charge}{initial.potential} = C_1 $$With final numbers:$$ \frac{slightly.larger.net.difference.of.charge}{slightly.larger.potential} <S> = C_2 <S> $$ <S> It turns out <S> \$ C_1 = C_2 \$ <S> and it is conveniently called Capacitance. <S> Quote from your comment: Say you take two metal spheres charged to +10Q each. <S> Originally they are far apart. <S> When moved closer it becomes increasingly difficult to add any more positive charges to either sphere because the other sphere helps in repelling Difficulty in adding an external charge to either sphere has nothing to do with the capacitance between the two metal spheres. <S> You need to take \$ C = <S> \frac{Q}{V} \$ similar to \$ R = <S> \frac{V}{I <S> } <S> \$. <S> If I don't apply a voltage to a resistor, does it mean the resistor has no resistance? <S> Does the potential difference increases when the distance between two objects decreases? <S> This is easiest to see with two approximately infinite parallel plates. <S> The electrical field is uniformly parallel and constant regardless of separation. <S> When integrated over distance, the potential difference V = <S> E <S> x d. <S> The potential difference actually increases with increasing distance.
| Mutual capacitance is independent of the actual charges that may exist.
|
how to get a breakout made for module that doesn't have one? I'm just starting out in this field and I got ideas for projects I want to build but I'm running into problems, not so much with how to find the part I want or how to get it to work. but how to physically connect it. I see modules like this and while I can find information on what pins I need to hook up what to, so the thing does what I want. I'm no where even close to being good enough as soldering to attach wires to modules like this. And most of the time I can't find an existing breakout board for the module I want to use. sometimes I can find one for other modules that claim to have similar functions but when I start reading though the details I find it really only shares the features I wasn't planning on using while the ones I did want aren't available. So all that is me explaining why I need to ask is there service or something that I could use where I can get the module I want on a board? I don't mind having to either buy the module separately and sending it to them and then paying them to assemble the board. or buying the module through them and paying extra for assembly. Or is there some trick to making them myself that I haven't heard about? <Q> To answer your question, yes there are services like that. <S> Places like SeeedStudio will allow you to design a PCB and get it printed, then you can choose from a list of 500 parts they have or you can send them your part and they will solder. <S> There are many other places like this, but this was the first I thought of. <S> Heres a link since their site it a little difficult to navigate: http://www.seeedstudio.com/service/index.php?r=pcb <S> You should note that this could end up being pretty expensive. <S> I'd say that the better bet (at least long term) is for you to learn to solder them. <S> Find old soldering boards or design your own and just practice a bunch until you feel confident doing it for production work. <S> Another option would be to find a nearby university and see if they have an EE club or robotics club and ask if anybody there can solder well. <S> You could probably pay them with pizza and/or beer and everybody would be happy. <A> For development, there is usually a development kit available from the module manufacturer. <A> You can use this free software to create a board with the pads you need, pretty simply by selecting the package type. <S> Then drag solder. <S> You can download the gerber file after building it and send it to a number of companies to have it manufactured, for probably $20 <S> 123d.circuits
| For (small batch) production, these modules can be soldered on other PCBs, all you need is SMD pads in the right places, slightly elongated if you want to hand solder.
|
How to distinguish PNP and NPN transistors? I need to know how to distinguish between an NPN transistor and a PNP transistor. If I pluck any transistor from a radio, how can I know its nature using a digital multimeter? <Q> This is pretty easy. <S> I used to do this routinely in high school when salvaging parts from unknown discarded boards. <S> As Spehro said in a comment, sometimes you can find a part number. <S> In that case, you can find the datasheet and get the parameters outright. <S> However, all too often there is no manufacturer's part number, or its just a short code, or its a in-house number. <S> Especially with the smaller packages, you're going to have to experiment. <S> First, make sure the ohmmeter is not set to some extra low voltage mode intended to not forward bias diodes. <S> Some meters have such a feature. <S> In this case, you definitely want to forward bias the junctions. <S> A bipolar transistor has only three leads, so only 6 possible two-wire measurements when taking polarity into account. <S> From the view of probing with a two-wire ohmmeter, a bipolar transistor looks like two diodes back to back. <S> There is one B-E and one B-C. <S> In a NPN, it takes positive voltage on the base relative to E or C to make the diodes conduct, and the other way around with a PNP. <S> The "N" and "P" in the names tells you the voltages required to make the diodes conduct. <S> Figuring out whether you have a NPN or PNP and which lead is <S> the base is therefore easy. <S> The next problem is to figure out which are the C and E leads. <S> On most packages, C is in the middle. <S> On a power package, C is usually connected to the case or tab or whatever. <S> Another way to test for C versus E is to measure the gain. <S> A transistor will still work with C and E flipped, but the gain will be higher when connected as intended. <S> I usually did this by connecting the meter across C-E. <S> With the base floating, there should be no current so the meter should read infinite resistance. <S> Now use your fingers to bridge C and B. <S> You should see a lower resistance than if you used your fingers to bridge C and E. <S> That apparent lower resistance is due to the transistor amplifying the base current. <S> Now run the same test with C-E flipped. <S> Most of the time, one orientation has a obvious higher gain. <S> If not, then you can run this test with a real resistor instead of your fingers. <A> Honestly it is better to type up the first line of numbers/letters you see on the front face of the transistor, add the word "datasheet" and search it up on google (and try to go for the PDF results). <S> You will able to determine if it is PNP, NPN, some other transistor like a mosfet, or a miscellaneous chip. <S> If you don't have Internet and want to test if it is NPN or PNP, set your multimeter mode to diode mode (should have the image of an arrow with a line at the end). <S> Using the two probes, probe the three legs all 6 possible ways (because two probes cannot probe the same pin) and put results as a table on paper with one probe on the same pin on the same row. <S> If your result comes out like this (not in order): <S> Not a number above zero, Not a number above zero <S> Not a number above zero, Not a number above zero <S> A number above zero, <S> A number above zero <S> Then you got a NPN transistor. <S> Else if your result comes out like this (not in order again): <S> Not a number above zero, A number above zero <S> Not a number above zero, A number above zero <S> Not a number above zero, Not a number above zero <S> Then you got a PNP transistor. <S> If your results do not match these, then you probably don't have either a PNP or NPN transistor as there exist many other types of transistors, and some basic computer chips can be made to look like a 3 pin transistor (voltage regulators, Hall effect switches, AM radio IC, temperature sensor). <S> If you don't want to use the Internet to determine the properties of a transistor, you could buy a Peak Atlas DCA55 semiconductor analyser , but I doubt that's necessary at this stage. <S> (But it's good for determining if the transistor is fake if you compare it with the datasheet online) <A> Note that because of the marvel of modern, inexpensive microcontrollers, and clever open-source software, there are small, inexpensive pieces of "test gear" which will identify and measure a wide variety of transistors, NPN, PNP, BJT, FET, etc. <S> and thyristors SCR, Triac, etc, resistors, capacitors, inductors, etc. <S> There are perhaps two dozen slightly different implementations of this design available on Ebay at any given moment with prices around US$15-25 <S> The more sophisticated ones have graphic display screens that draw the symbol of the device and identify the pins (Emitter, Base, Collector, or Source. <S> Gate, Drain, etc.) <S> Here is JUST ONE example of these gadgets. <S> There are many variations avaiable with nearly identical features and similar prices... http://www.ebay.com/itm/2015-12864-LCD-Transistor-Tester-meter-Diode-Triode-Capacitance-LCR-ESR-Meter-/271839788857 <A> this link <S> http://www.interfacebus.com/semiconductor-transistor-packages.html than you can build your circuit and find the type of your transistor wither is NPN or PNP <S> 1st: <S> NPN transistor 2nd: <S> PNP transistor and this one <S> i hope this can help you
| first thing find any number on the transistor this number is the key of transistors with the help of datasheet or the internet, ie you can find the type of transistor and the characteristics of it secend if you don't find any number on the transistor you can make this circuit to find the kind of the transistor, but befor that you shoult know the Transistor Package Styles chek You can also know all their characteristics.
|
Redirection of electrical current Sorry if this is about the newbiest question on this site, If I were to draw a graphite trace on a piece of paper (which from what I know would have some amount of resistance), and connect a multimeter and 9V battery to it as shown below: Assuming that an electrical charge flows from - to + and takes the path of least resistance, how would I redirect the charge to go from contact A to contact B, instead of contact A to contact C (effectively shutting off the multimeter), without shorting out the 9V and putting it at risk of catching fire? Thanks in advance! <Q> It would be more correct to say: "How can I change the path of the current from flowing from point A to point C to flowing from point A to point B. <S> " You would need to move the purple wire connected to the meter from point C to point B. For current to flow, you need a closed completed path. <S> You can trace the path with your finger by starting at the battery's positive terminal, moving toward the meter, going through the meter, moving toward the graphite trace, moving through the graphite trace then moving back to the battery. <A> how would I redirect the charge to go from contact A to contact B, instead of contact A to contact C (effectively shutting off the multimeter), Simply connect a wire from B back to the <S> + terminal of the battery. <S> without shorting out the 9V and putting it at risk of catching fire? <S> Make sure the resistance of the track from A to B is high enough that the battery can supply the current it demands, before you make the connection. <S> A graphite line on a piece of paper isn't a very good resistor, so if the line from A to B is long enough, you won't have a problem. <S> Assuming that an electrical charge flows from - to <S> + and takes the path of least resistance, This is two wrong assumptions in one sentence. <S> First, because electric current flows from higher potentials to lower potentials. <S> When a negative charge flows from left to right, we call that "current flowing from right to left". <S> It was an arbitrary choice made by the early researches in electrical physics before they knew that the charge carriers in metals are negatively charged, and we're not going to change it now. <S> Second, because current flows in all paths of a parallel circuit. <S> It flows in each path in proportion to the conductance of that path, so if there is large difference in conductance, one path will take the large majority of the current. <S> This is a common situation, and it's what we mean when we say "current takes the path of least resistance". <S> But still, current flows at least in small amounts in all paths of a parallel circuit. <A> Roughly, you could represent what you have drawn as this circuit: simulate this circuit – Schematic created using CircuitLab R is an arbitrary number that will depend on how wide the traces are and other factors. <S> Let's assume R = <S> 100K for the sake of argument. <S> I have assumed the meter is in voltage mode and it has an input resistance of 10M ohms. <S> Then you have 1.2M in series with 10M across the battery, and the meter would read about 7.4V. <S> If you short B to the + terminal on the battery, the meter reading will go down to about 5.5V (look up Thevenin equivalent and voltage divider if you want to know how to analyze it). <S> Some current flows through terminal B and some through terminal C. <S> If you short the meter out entirely (B open), the reading will go to zero, with 7.5uA flowing through the short. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Your circuit in schematic format. <S> How would I redirect the charge to go from contact A to contact B, instead of contact A to contact C (effectively shutting off the multimeter)? <S> You really mean how can I direct current (rather than charge). <S> This is traditionally done with a switch but can be done with transistors and other "controllable" conductors. <S> If your purpose is to demonstrate resistors and switches, you could make a SPDT (single-pole, double throw) switch using thumbtacks as contacts and a paperclip as the switching element. <S> Figure 2. <S> SPDT drawing-pin switch example. <S> Source . <S> Don't expect great results from the contact between the pins and the pencil lines.
| Another dodgy possibility is to draw the switch element on paper, cut it out, turn it face down and use it as the switching element. You can only direct current by connecting and disconnecting parts of the circuit.
|
Chokes vs. small capacitors on signal cables I know that many signal cables (usb camera to computer cables, etc.) have bulky ferrite chokes on them to prevent noise. Why do they all have ferrites instead of ceramic capacitors? Small ceramics also get rid of noise effectively, and would be much smaller (probably cheaper too?) than the ferrites. <Q> A common mode ferrite sleeve that is lossy at radio frequencies and doesn't attenuate the signal is rather more useful on a data cable than shunt capacitors that attenuate the signal and reflect rather than absorb RF. <A> Most EMC problems that products have are to do with Radiated EMC. <S> When you have normal tests done like say FCC A and B getting a radiated fail is not uncommon. <S> Radiated EMC is generally due to common mode issues. <S> The standard cable length of most products makes the cabling a good antenna system. <S> The ferrite common mode sleeve[s] that you often see on cabling deals with radiated EMC. <A> Ferrites on a cable is essentially a sign of a failure to do proper EMI design in the actual PCB. <S> Ferrites do not work very well on cables because the cable already typically has high impedance and you're essentially doing energy division.. <S> Capacitors are perfectly good solution for EMI, when they're on PCB itself. <S> Typically you would use three terminal flow-through caps to eliminate differential noise. <S> Capacitors may also eliminate common mode noise when connected from both signals to reference plane (usually GND) but this tends to kill your differential signal as well. <S> Or single-ended signal. <S> In short, capacitors are bad for your signal. <S> Trumping all that is a common mode choke. <S> These are available on bifilar configurations that work even on high-speed data lines such as gigabit ethernet. <S> Even better is having proper shielding and grounding strategy which will short the common mode noise back to the source instead of passing it to the cable. <S> Incidentally ferrites can work very well if you put them <S> INSIDE the case and use a cap to form a RC (well, LC) circuit. <S> Clip-on Ferrites have their place but they should be a last resort/stopgap solution or used for difficult cases such as high-speed V-by-one or LVDS signal cables. <S> Here's a primer from Murata on the subject: <S> http://www.murata.com/~/media/webrenewal/products/emc/emifil/knowhow/26to30.ashx edit Corrected some nonsense about capacitors across differential pairs.
| The fact that you see them so often should tell you something.. Without going to gruesome details of EMI design issues, it can be said that a ferrite on a cable is a) ineffective and b) expensive band-aid to a design which didn't pass the CISPR test requirements for consumer electronics.
|
voltage of input(+5V) is greater than the supply(+3.3V) provided for 74HC245 buffer In my design I am using 74HC245 buffers between a input signal of +5v and controller pins of +3.3V. What is the supply value I have to use for buffer (74HC245). I want to use +3.3V as the supply, but in data sheet it is mentioned that maximum VIH is Vcc. Please suggest me what happens if I provide +3.3V as supply voltage and feed input pins +5V?.I am providing the link for buffer data sheet below. http://www.nxp.com/documents/data_sheet/74HC_HCT245.pdf Thanks in advance <Q> If you want to translate 3.3V signals to 5V then you must power the 74HC245 with 3.3V, because at 5V it might not reliably detect a 3.3V 'high' logic level. <S> However it will also only output 3.3V, which may not be enough to reliably drive 5V logic. <S> Any 5V inputs will inject current through the protection diodes and out the Vcc pin into the 3.3V supply. <S> Depending on how 'stiff' the supply is, this may either exceed the protection diode's current rating, or raise the 3.3V supply to <S> ~4.4V (which would be bad news for anything on the 3.3V side that can't handle the higher voltage). <S> You could wire resistors in series with <S> the I/O's on the 5V side to limit diode current, <S> but they will slow down the signal rise and fall times - and make the 5V output level problem worse. <S> If you power the 74HC245 with 5V then you have similar problems, but on the 3.3V side. <S> So the answer is:- use a proper level translating buffer such as the 74LVCC4245A , which is powered by 3.3V on one side and 5V on the other. <A> In this instance you should be using logic family <S> that's 5V tolerant, traditionally LVC. <S> Also some "advanced" (AHC, VHC etc) logic families are 5V tolerant. <S> If you exceed VCC in HC logic, the protection diode will clamp the voltage to VCC so absolute input is usually defined as VCC + <S> 0.3V. <S> You may get away with using series resistor to limit the current but this is not recommended. <S> With the 5V tolerant chips there may be no input protection so you might cook the chip whatever series resistor you have if you exceed the input voltage limit. <A> Use resistors to limit the input current to less than the clamping current.
| If the input current is higher than the clamping current then you could destroy the chip.
|
How safe is a current transformer I acknowledge that the subject line indicates a "how long is a piece of string" style question, but I hope to make it more specific below. I have a "current transformer" (CT) purchased as part of an Efergy energy monitor unit. It appears to be a split core ferrite that closes around the primary (live) with a large number of turns on the secondary. The secondary has a 2.5mm audio jack attached for sampling. I assume there is a resistor inside the plastic case of the mini-CT to enable a voltage drop across it (facilitating sampling). No wires have been cut to attach the mini-CT, it is simply coupled to one live wire. The live wire is at 240VAC RMS and carries a current of several tens of amps to a household. Is it safe to touch all the connectors of the audio jack, with my feet touching the ground and the mini-CT attached around an AC live wire that's currently drawing a current? Would the advice of the previous sentence change if I reduced the ratio of secondary turns from 1:2000 (I don't know if this is the true number, I'm just guessing) to 1:10 or 1:1? <Q> There are two ways to get a shock from such a setup: Connecting to the live circuit. <S> This is unlikely as there would have to be insulation breakdown of the live conductor, the CT and the CT wire insulation. <S> A shock from high voltage generated by the CT during measurement. <S> Theoretically this is possible on an open-circuit CT when significant current is flowing in the primary. <S> In practice it doesn't seem to be much of a problem in small CTs. <S> To get a shock you would need to touch both terminals simultaneously. <S> The shock current would flow from terminal to terminal through your body and not to ground (through your feet mentioned in the question). <S> Is it safe to touch all the connectors of the audio jack, with my feet touching the ground and the mini-CT attached around an AC live wire that's currently drawing a current? <S> Bare feet? <S> No, not safe for any work near mains. <S> You should be wearing decent work shoes. <A> It is not a matter of a voltage between the current transformer and ground. <S> That is very unlikely unless there is some major breakdown of insulation of both the cable <S> your are monitoring AND the current transformer. <S> The RISK is from the voltage generated by an unloaded current transformer sensing some significant current. <S> And "tens of amps" very much qualifies as "significant current". <S> Unless proved otherwise, it is ALWAYS safe to assume that the bare output of a current transformer is DANGEROUS. <A> You need to consider: <S> Assuming that mains current (from say a wall outlet) is being sensed, the insulation / casing of the CT will have to be rated to block well over the line voltage. <S> (230V or 110V RMS). <S> If this is the case, the secondary of the CT should be quite safe to touch since it will be well isolated from the line voltage. <S> Assuming your turns ratio is large enough on the secondary side, the current in the secondary will be small and so will the voltage across the burden resistor. <S> In most "readymade" CTs (see below) <S> you cannot change the number of secondary turns, so your output volage range is fixed for a given input current range, and is typically <5V. <S> If you are making your own transformer, then you need to design the ratio suitably. <A> A CT can become unsafe if the burden resistor on the secondary becomes open circuit. <S> The output volts will try to go sky high but transformer core saturation will limit that. <S> The extra core heating could cause dangerous temperatures despite the load current being below the rating of the CT. <S> In fact you could overheat the CT without blowing the fuse. <S> Precations should be taken to avoid this potential safety hazard.
| It is NOT safe to assume that the current transformer has an INTERNAL burden resistor.
|
Using Li-Po or LiFePo4 for diy motorcycle battery Basically just after opinions on using LiPo or LiFePo4 for building a diy motorcycle battery. My current lead acid battery just died and i'm sick of replacing them after 1-2 years. Background info - Bike is a 2007 BMW F800S with a 400W alternator and 14.4V peak charge voltage. From what I can gather online, the system puts out closer to 13V at idle and has enough current to run everything off the alternator alone at idle. My problem - LiFePo4 has low current ratings, and LiPo has impractical cell voltages. Doing some research the lithium motorcycle batteries you can buy online all use LiFePo4 as a 4 cell gives the right voltage range, but they seem to most commonly have a 4 series/2 parallel setup (A123 cells) which gives about 140A peak - this is below the current lead acid battery which does 200A. LiPo on the other hand will easily do >200A from a small pack (easy to get packs with C ratings higher than 100) however 3 cell is impractical as the fully charged voltage is 12.6V (lower than alternator output) while a 4 cell pack would mean the resting voltage per cell is around 3.5V, which is close to the discharge voltage on these packs. Possible solutions; Use a LiFePo4 pack in 3 parallel config (along with a passive balancer) to give the current. The pack would remain connected at all times as the chemistry is relatively safe. this would mean 12 cells however which is getting expensive compared to lead. Use a 3 cell LiPo with diode isolator. Essentially use a diode to connect the battery only when the alternator is not running, and along with this use a simple regulated trickle charge balance circuit to keep the pack topped up. Use 4 cell LiPo and keep the cells at low charge all of the time by using a LiFePo4 passive balancer to keep them balanced at the lower cell voltage. At this point im leaning towards trying with A123 in 2 parallel and see if it has the cranking current to start the engine easily - this is the tried and tested solution - but would be very keen to give it a go with LiPo. Thoughts? EDIT The bike runs well and the alternator is fine, it just does not get ridden much at all so the lead battery discharges itself. I have a battery charger and can easily sit the battery on charge when it is not being used, but the battery sits behind a large fairing which is a pain to remove, so would rather not have to do this ever time I want to use it As far as I am aware, the existing alternator should easily provide the current needed to keep a lithium pack at full charge, as long as a balance circuit is used (passive current shunt). I have started the bike many times off a 3 cell LiPo and to start the bike uses very little of the pack (2200mAh) so I doubt the pack would ever see a high depth of discharge <Q> They are safer, cheaper, more rugged and drop in replacement. <S> Another real good reason is that the Lithium style batteries require very specific and stringent charging circuitry. <S> You would not be able to simply connect the Li style batteries up to the existing charging system that was designed for a lead/acid style battery. <S> If that was done you could risk battery explosion or fire. <S> If it seems like the lead acid batteries that you are having to replace are not lasting as long as they should <S> then I think you would do well to investigate this and make the necessary corrections. <S> Does the existing charging system overcharge the battery or apply too high of voltage? <S> Is the voltage regulator on the system set too low such that the battery never really fully charges and dies a slow death because of this? <S> Is your battery usage duty cycle to charging time not optimum? <S> (i.e. many starts with very short ride times). <A> An alternative you might consider is using a small bank of supercapacitors. <S> They're super simple to use, can last longer than electrochemical cells, and have much better behavior over temperature. <S> There's examples on youtube such as this video which prove it works on cars. <S> https://www.youtube.com/watch?v=z3x_kYq3mHM <S> The leakage is low enough to leave it sitting around for a while too. <S> Not quite as low as I would like, but extra parallel banks help solve that issue. <S> The associated costs used to be fairly obscene but they're much more reasonably priced now. <S> I would also recommend packaging the capacitors better than that example shows. <S> If you wanted extra capacity you could even parallel it with a battery that you keep on standby for extra power if you ever need to let it sit with the lights on. <A> 4 cell lipo pack will be 16.8V <S> full charged but that full charged value drops off rapidly in practice. <S> You'll find most batteries have a discharge curce similar to this where that initial voltage drops off pretty quick leaving you with a pack at nearer 14.4V. <S> I don't have any experience with bikes but car electronics are designed and required to handle much higher voltages in the case where a battery terminal comes loose with an alternator running producing huge voltages so I would have through 14.4V wouldn't be a problem. <S> This is only for a short period so you could always go with a system that only connects at crank if you're worried about the excess voltage. <A> If you keep the cells at low voltage you keep them mostly discharged. <S> When charging the cells, you could use a Buck–boost converter that would keep the voltage at the right level no matter what the alternator is producing. <S> You also have to make sure that the convertor has a current limiting option, like this one . <S> Advantages and disadvantages: <S> LiFePo4 cells are safer than LiPo cells. <S> The life of LiFePo4 cells is much better. <S> You can realistically expect twice the number of cycles or years if you don't do so many cycles. <S> LiPo cells are much lighter and are more compact for the same capacity. <S> LiPo cells have higher C ratings so for just starting the bike this could be a big plus. <S> In both cases it's recommended you have a BMS with balance capabilities and a current limiting charger. <S> When the batteries get full, the BMS disconnects the pack from the alternator automatically. <S> I would expect the pack to last for ~4 years <S> and then I would reconsider it. <A> If you are worried about the battery discharging, as stated, a decent motorcycle battery charger will work well. <S> There are several dealers of them specific to motorcycles (i.e Battery Tender as an example company, can't highly recommend them, however) that you connect once to your battery/electrical system in some way, and have a capped connector placed withing easy reach. <S> Using this, you can easily hide the connector (using zip ties, or my favorite, Velcro) out of sight and safely out of the way. <S> When you get back to the garage, you only have to reach, uncap, and connect. <S> This will negate the need to remove the fairing more than once (after the initial install, if you connect the wiring directly to the battery). <S> I used this setup (a battery charger with a 10' cord, and an SAE 2-prong connector with cap) on my bike for about 4 years. <S> Battery finally puked after the charger itself got disconnected from the wall. <S> As Michael Karas mentioned, you should definitely check his 3-point bullet list, along with perhaps <S> "Are there any other loads causing the battery to be used?" <S> i.e. very minor phantom loads that will eventually drain a battery no matter what over time. <S> Some systems keep things turned on no matter what (i.e. anti-theft alarms and other integrated systems).
| Lots of other applications use supercaps lieu of batteries for their bulk energy storage where long life and wide operating temperatures are a consideration. I would probably do it with a 4S LiPo pack because it would be cheaper and more compact. I would stick with the standard lead acid style of battery.
|
What constitutes a "sufficient" ground? Electricity novice here. Thanks for reading! I have a 3000 watt DC to AC inverter attached to a 12V battery bank. The inverter has "DC ground lug" on the exterior of the inverter's case which the manufacturer says should be "tied into the DC ground of the system". My "system", however, is a small box on wheels that I take around with me. There is no vehicle chassis or house ground that I can tie into. As I understand it, the purpose of the ground connection is to bleed off current in the event that the inverter's case should ever become energized, which in turn would trip a breaker. My questions are as follows: How much metal does the ground wire need to be connected to in order for the ground connection to work as intended? Is it enough for me to attach a strip of metal inside of my "mobile power box" and attach the ground wire to that? A car's chassis is massive and I see that it might conduct a lot of energy, but I'm not sure how much mass is enough. Can the ground in a system like this be tied into the negative side of the battery bank? The battery in this case is not tied into a vehicle chassis, so I'm not sure that would even make sense. Thanks in advance! <Q> @rdtsc provided a link to the manual which covers what to do with the grounding. <S> You are going to connect DC negative to chassis ground with heavy gauge wire (see manual), and you are going to have to decide what "chassis" means in your case. <S> Maybe if this box of yours is metal, you can just bond the chassis ground to the metal box. <S> But this is not really the important part. <S> More importantly, make sure you use a GFI somewhere between the inverter and your load. <S> Inside the inverter, since it is a UL 458 inverter, there is a relay which connects green wire to neutral only when the inverter is supplying power. <S> When it is in standby or pass-through mode (when you supply AC to the inverter so it can recharge its batteries) <S> the connection from neutral to GND is open. <S> In either case, there is exactly one place where GND is bonded to neutral. <S> Your 12 battery is not a shock hazard. <S> It stores an impressive amount of energy, and can be a fire hazard if there is a short, so make sure you have proper fuses. <S> But if you screw up the AC grounding, you can introduce a shock hazard. <S> Luckily it is not that complicated. <S> Just read the manual again. <A> In some jurisdictions a system that size would require being pushed to location and driving an earth peg (1.6m into the dirt) for the ground and connecting it to the lug you mentioned before it could legally be operated. <S> Same rules apply to generators. <S> Noting they you say: "tied into the DC ground of the system" This may refer to the solar panel earth as if the inverter was designed for a solar battery system. <S> Again in some jurisdictions, the solar array is required to be earthed with a not less than 4mm 2 earthing conductor which is taken into the main earthing system of the premises. <S> These arrangements prevent the inverter neutral floating above ground voltage and ensure that there is sufficient current return to operate RCD type protective devices in the event of a fault or electrocution. <S> The whole purpose of doing it properly is to avoid risk of electrocution. <S> Note <S> : In the manual linked in another comment it shows both the chassis ground and the battery negative connected to earth on pages 23 & 24. <S> If you are unsure of electrical work use a licenced and qualified electrician. <A> A ground is for a return current. <S> A return current should be able to flow through the ground without it affecting the design. <S> This can happen if: 1) <S> There is too much resistance creating a voltage in the current return path which can create a common mode problem. <S> For example, if you have 12A flowing through a 0.1 ohm ground cable, this will create a 1.2V voltage at the point of the current input (at the device) $/ V = <S> I*R $/. <S> 2) <S> The other problem is inductance, if you have a fast signal (or ground EMI) you need to have a sufficient reduction of inductance to ground out the high frequency signal. <A> All that these guys said is great, to dumb it down, especially for me, attach your ground to a screw in the bottom of your mobile box, attach the wire to it. <S> If this mobile battery box is used inside the house, it creates more problems. <S> I am doing the same thing, but I will be connecting the ground to the baseboard wall as my house is grounded already. <S> Your mobility is good as long as its outside or you don't mind burnt floors in case a discharge happens, rare, but it does. <S> If you don't provide an adequate ground, anything 3-prong might have difficulty working if there is a sensor reading that third prong. <S> Xbox One, Electric cars and the smart equipment are the example. <S> On the other hand "dumb" equipment like weed eaters, and toaster ovens don't care.
| That screw should go all the way through the box and be able to touch the ground, therefore creating your ground situation. If you do not have a suitable earthing system available that you can connect to legally then you need to drive an independent earth stake and connect both to that. So short story is measure the current through or voltage across the cable, and keep it lower than what your design needs.
|
how do I dispose of fluid from a lead acid battery? I am reconditioning a 12v lead acid battery, and a process I am trying requires me to remove aprox 2.5 to 3 oz of battery fluid from each cell. Before I do this, I'd like to know how to dispose of the removed fluid? Can I just neutralize it with baking soda? <Q> What you're removing is ~28% sulfuric acid. <S> Wear rubber gloves and goggles and add acid to water in small amounts while monitoring the temperature. <S> Have separate bucket of water nearby for yourself in case you get acid on your skin. <A> Are you not planning on putting the acid back in the cells after you recondition them? <S> Or are you planning on replacing it with new electrolyte? <S> Where are you getting the new replacement electrolyte? <S> Does the source not have a recycling system? <S> Can you take the electrolyte fluid to a place that sells (and recycles) auto vehicle batteries? <S> They are prepared to handle the material properly as they deal with it every day. <A> One way that is relatively safe (because of the small amount of acid you need to get rid of), is to dilute it very very much. <S> I've done it it the past and the possible dangers are low. <S> REMEMBER TO USE RUBBER GLOVES AND EYE PROTECTION. <S> Just add the acid gently to a very large amount of water (eg 5-10 liters of water). <S> Then you can possibly dump it even in the sewer. <S> Note that I'm suggesting that only because you have a small amount of acid which is clearly not worth it to pay someone to dispose it "properly" for you.
| Dilute your 18 oz with a quart of water, you can then neutralize it with baking soda with little difficulty.
|
Can I efficiently monitor sound card line in with a vu meter? I'm trying to monitor some audio input to my desktop pc with a vu meter. I've connected to the line in, and it's just a bog standard on-the-motherboard sound card. I'm having a problem understanding the various signal levels. I haven't actually bought the meter yet as I need two antique ones at some cost, so at this stage I'm looking at the feasibility and risks involved. From Wiki, consumer grade kit is designed for a nominal -10dB line level. Recording via Audacity, it appears that my sound card line input actually clips at -6dB so there is very little headroom. The typical analogue (it must be analogue) vu meter is calibrated -20dB to +3dB. The -10 mark is only approximately 1/7th of the way up the scale. To avoid the risk of clipping, I'd need to run at a level less than – 10dB. That doesn't seem a very efficient use of the meter real estate. I'm assuming that vu metering is for absolute readings and not arbitrary scales. I only have rudimentary knowledge of this area, but have I miss understood something? <Q> Yes, Paul, you are missing something basic: <S> The fact that dB is not an absolute measurement, but a relative one. <S> Could your VU meter is reading dBm instead of dB? <S> dBm is a measurement relative to 1 milliwatt, with 0 dB being a milliwatt. <S> If not, then the VU meter is relative to something else, specific to the application it was designed for. <S> As an example, the VU meters on a power-amplifier are usually relative to the maximum continuous power, with 0 dB being equal to 100 watts for a 100 watt amp. <S> So most of the scale is negative dB, with the positive dB being in the clipping region (an aside: it might be an issue different than clipping that sets the maximum continuous power for the amp). <S> With +3 dB as your VU meter's maximum, it seems like it is a relative reading, and not dBm. <S> In audio, most dB measurements are relative to full-scale, so you are almost always dealing with negative values. <S> If you can determine what the VU meter is measuring relative to, and also determine what Audacity's dB scale is relative to (possibly 1 volt p-p), then you can determine a constant value to add/subtract from the VU meter's reading to correspond to Audacity's dB reading. <A> By definition , your analog-to-digital input (or anybody's input) clips at 0dBFS <S> (the decibel ratio relative to Full Scale). <S> Likely the most accurate way to meter an input like this is with a software meter. <S> That means the meter is showing the ACTUAL level out of the Analog-to-Digital converter. <S> And THAT is what matters most. <S> Any other kind of metering scheme will have to be calibrated against the internal A-D converter, anyway. <S> Might as well skip the extra step and ambiguity. <A> Although VU meters measure dB, they measure dB relative to some standard level, normally 1mw into 600 ohms, or 0.775Vrms, so in practice they are an absolute measurement. <S> You still need to calibrate this level against your soundcard though. <S> However, VU meters are inappropriate for digital audio measurements - they are relatively slow response, effectively indicating average energy rather than peak levels. <S> With digital signals, it's important to measure peak levels, therefore PPM (Peak Program Meters) are more appropriate . <S> These were adopted in broadcasting because the AM modulation process clips (on negative peaks) as hard as a digital system at full scale, therefore detecting short term peaks was important. <S> In the UK, broadcast engineers were trained to "peak to 6" on the scale of 1 to 7. <S> The divisions on the PPM (IEC 60268-10 type II) are 4dB apart, with 0dB/1mw/600 ohms corresponding to "4" on the scale, <S> so "6" represents +8 <S> dBm. <S> This is still quasi-peak, with a response time of about 1ms, so it under-reports fast peaks. <S> So, when digital recordings became common, full scale was set to +18dB, providing 10dB of headroom above "6" on the PPM. <A> Can you post a link to the unit you´re looking at? <S> In any case, analogue VU meters are essentially current driven, so you can change the 0dB level for your setup by changing the shunt resistor on the meter.
| You need to know what your VU meter is relative to.
|
How do I drive 14.3Mhz clock input from 10MHz? I intend to use an IC which requires 14.3MHz clock input, but want to drive it from a stable 10MHz source - derived from GPS. How do I turn the 10MHz clock into the 14.3MHz that the IC requires? <Q> What you need is a PLL , a phase-locked loop . <S> It works by comparing one oscillator that you can control, with a reference oscillator. <S> There are numerous circuits that can do all of this in one package, sometimes even including a reference oscillator. <S> It is very common having to synthesize frequencies from a stable oscillator, so these are not unusual. <A> It is possible to change the order of multiplications and divisons to avoid frequencies above \$100~\text{MHz}\$. <S> If you want a pretty square wave, the last step should be a divison by \$2\$. \begin{align*}\frac{10~\text{MHz}}{2} & <S> = 5~\text{MHz} \\5~\text{MHz} <S> \cdot 9 & <S> = 45~\text{MHz} \\\frac{45~\text{MHz}}{11} <S> &= 4.090909~\text{MHz <S> } \\4.090909~\text{MHz} <S> \cdot 7 & <S> = 28.636363~\text{MHz} \\\frac{28.636363~\text{MHz}}{2} <S> & <S> = 14.3181818~\text{MHz}\end{align*} <A> If you want 14.31818181818 MHz from a source of 10 MHz, it is difficult. <S> The 14.31818 MHz is the american TV color burst frequency, the precise value is 315/22 MHz. <S> You may divide 10 MHz by 2, multiply by 9 and by 7 to get 315 MHz. <S> Then you divide by 22 to get the frequency you want. <S> May be more than one PLL is necessary to do that. <S> Another way is to divide the 10 MHz by 4 and multiply by 9 and 7 and finally divide by 11. <S> Of course it is theoretically possible to multiply by 63 and then divide by 44. <S> But this requires a very fast PLL oscillator for 630 MHz and also a fast frequency divider. <S> I suggest to divide by 22 first, then multiply with 63 and finally divide by 2. <S> But for a low phase jitter, separate multiplications by 9 and 7 may be better. <A> What sort of chip are you using that has that requirement, and what would be the allowable jitter? <S> If you could live with a large amount of jitter, one approach would be to use a device that turns both rising and falling edges into pulses (effectively doubling 10MHz to 20Mhz) and then discards 25 pulses out of every 88, or you could use a 25MHz or faster clock to drive a CPLD or FPGA which behaves similarly but uses the 10MHz reference to adjust how many pulses it needs to skip. <S> Both approaches would have considerable jitter, butdepending upon what is being done with the 14.3818Mhz clock that might be acceptable. <S> If using it for NTSC chroma generation, the effects of jitter might be minimized if the frequency were chosen so that alternate frames would have roughly alternating jitter.
| The trick is that it is easy to divide the frequency of an oscillator using a digital counter, so what you do here is to divide the 14.3 MHz oscillator by 143, the 10.0 MHz reference by 100, and then use the output from this comparison to make sure that the 14.3 source is running at an exact relation to the stable 10 MHz reference.
|
DPST switch control 5vDC and 220vAC I would like to make usb-charging sockets from two sides of the bed, and I don't like the idea of power supply working 24/7 so I decided the following scheme to have power ON while me or my wife are charging: simulate this circuit – Schematic created using CircuitLab But the power on my side socket will be ON, when my switch is OFF, if my wife's switch is ON (being continuously charged is not good for battery). So I have the idea to use DPST switch to also switch usb socket off: simulate this circuit Is it OK to use DPST switch between 220vAC and 5vDC circuits?Does it depends on it's characteristics, how to select right on?Would be there some kind of cross talk between circuits? <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Two-way switching circuit. <S> Mixing mains and low voltage on the one switch isn't recommended. <S> You can achieve what you want with a two-way switching circuit as shown in Figure 1. <S> The PSU (represented by a transformer in the schematic) will be on if both switches are up or both are down. <S> These type of switches are called "3-way" in North America. <A> Assuming this is a regular DPST switch, and both poles are isolated (which they are by default), then yes this should work. <S> Using a multimeter to test for connectivity between the opposite poles and etc. <S> when the switch is in the on position is a good precaution or else you may fry your usb connectors. <S> Make sure you buy a switch capable of handling your ac power current, or else it may weld closed. <S> Any switch should work for 5v DC. <S> An alternate solution is to have a switch on both 5v usb power outlets and one on the main AC power to the converter (assuming I interpreted your question correctly and you wanted the converter to be off when not used) <S> This would be better isolated. <S> Better yet, use a DPDT switch And have two converters (one for each socket)! <S> Point is there are many solutions. <A> Using a DPST switch like that, you need to be very careful about creepage distances between the mains and 5V USB terminals of the switch. <A> As mentioned by others though, make doubly certain that the terminals are well insulated and the switch is rated for the voltage and current. <S> Switched can be bought with plastic isolators between the pins molded in to the switch body. <S> I might actually steal that circuit now... <A> Many major mobile phone manufacturers are working to develop a rating system for charger energy efficiency in no-load mode. <S> For example, Sony's new standard chargers (EP-300, EP-310 and EP-800,etc) have a no-load power consumption ≤30mW.A <S> 2012 study at the Lawrence Berkeley National Laboratory found that an idle charger drew 0.26 watts on average; this figure goes up to 3.68 watts when a phone is attached and charging, and drops down to 2.24 watts when the phone is attached and fully charged. <S> Overall, the cost is a handful of pounds over 12 months. <S> Here is the typical Standby Power Characteristics of the Texas Instruments 5.3V 2A Mobile Phone Charger Design (PMP4432) : From this I conclude that it will be effecient to have two smartphone chargers connected to two separate wall points each controlled by their own switch. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Your second circuit is actually a nice little elegant thing, it allows the DC power supply to be turned off when not in use and allows each usb socket to be switched individually. If there's any possibility of mains voltage leaking across the insulation to the extra low voltage side on a damp day, then the arrangement could become very dangerous. Smartphones chargers are constantly being upgraded to be more efficient.
|
Measure temperature on a moving engine valve: transfer voltage and reading wireless? I am an automotive engineer with very very little or almost zero electrical/electronics experience. I am working on logging real time temperature values on an engine exhaust valve and mounting a thermocouple and routing the cable through the valvetrain and valve cover. Instead of that, someone asked if the thermocouple voltage values can be transferred wirelessly (a sender unit mounted on valve tip, i.e. top most on the pink thingy and a receiver on the spring seat) I would like to know if this is even possible and how difficult will this project be for an automotive engineer? <Q> It is an optical device that receives the heat/light emitted by the object whose temperature needs to be measured and then uses part of that "spectra" to calculate the temperature. <S> It relies on understanding the "black body" radiation from the object and its emissivity: - I've used it for temperatures as low as 600 degC on aero-engine turbine blades <S> but, in theory, can be used much lower. <S> You can pipe the radiation off in a suitable optical fibre and it's a non-contacting method. <S> It does rely on filtering a small fraction of the bandwidth of the emitted radiation and there are optical filters that can do this for you. <S> I've used an optical filter and the photo-diode's natural spectral response to select the 1.7 \$\mu\$m part of the emission. <S> If you can find a device and filter that is good for band-pass filtering <S> about 3 \$\mu\$m <S> then you should be in business. <S> The only area <S> I'm unsure of is the sub 600 degC region <S> but if you are looking to measure temperatures at or above this point it can certainly be done. <A> This is tricky, so the first thing you need to do is get someone that knows what they are doing with electronics. <S> The high temperatures and vibration make this difficult, regardless of how you eventually get the data from a sensor to the outside world. <S> I'd want to minimize whatever has to go onto the moving valve. <S> That probably means just a high temperature thermistor with flexible wires that will work long enough to give you a few useful readings. <S> Then put all the rest of the electronics where it's cooler, not moving, and not covered in dirty oil. <S> Some sort of wireless transmitter <S> and then a wireless receiver, even if only a transformer seems like a lot more trouble than its worth at first glance. <S> Again, I'd want to minimize the electronics in this harsh environment. <S> Keep it to just a thermistor and wires. <S> The reason I'm saying thermistor (varying-resistance temperature sensor) and not thermocouple is because with a thermocouple you are stuck with certain materials for the wires. <S> You want to choose the best material to withstand lots of flexing, which will be multi-strand copper. <S> You can't have both wires to a thermocouple <S> be the same material. <S> High temperature "platinum RTD" type of thermistors should work here. <S> I'm expecting there is oil splashing around, else I'd suggest IR temperature sensing. <A> I would suggest looking at making a blind hole (say via EDM plunge) in the valve and put a polyimide-insulated (check the expected temperatures) sensor deep within the valve itself. <S> A type K or N thermocouple (those are nickel alloy types, as opposed to, say, J or T) might have the required characteristics, but you'd have to check the fatigue life against the expected run time of the test. <S> Such a thermocouple can fit into a very small diameter hole and will give an accurate and reliable reading of the temperature at the junction as long as the connections last (and a break is easily detected within milliseconds). <S> I don't think electronics attached to a valve stem is going to be very practical, at least not at first look, however there are some RFID sensor chips that might be promising if the environment was not excessively harsh- if it's running much over 120°C probably not practical. <A> Yes, this absolutely possible. <S> Below I am going to summarize the proposed solutions with their associated problems and propose one idea of my own. <S> Solution 1: <S> pyrometer has been proposed by Andy aka and mentioned by Olin Lathrop. <S> The associated problem is sticky dirt blocking optical tract. <S> Solution 2: <S> copper wires have been proposed by Olin Lathrop. <S> The associated problem is much wear and tear on the wires. <S> Solution 3: transformer/coil has been mentioned by Olin Lathrop. <S> This can be used with a thermistor, or you can see if your valve has a useful slope of its magnetic characteristics in the region you intend to measure. <S> I would prefer this approach, especially if it works with valve as the core and no other stuff. <S> Maybe Olin will explain associated problems. <S> Solution 4: something like an RFID sensor chip. <S> The associated problem is the need of custom high-temperature sensor chip made of gallium nitride or some other temperature-resistant semiconductor material. <S> A major engine manufacturer can very well order such a chip, but you will certainly need professional consulting. <S> Also, I'll mention here that there is a lot of alternative options to power the RFID chip in the walve, including possibly piesoelectric or thermoelectric generation, so the power does not need to be transferred wirelessly, only the signal. <S> Note that first two solutions attempt to solve the original problem, not building of an LC circuit. <A> Your question is about the implementation of a particular solution to the problem of determining the temperature of a moving engine valve. <S> And you have received various solutions. <S> However, a simpler solution would be to install one (or more) thermocouple(s) on the valve guide (not being identified). <S> Although the temperature of the valve is not being measured directly, a correlation between the temperature of the valve and its guide can be determined, and readings "adjusted" to reflect the temperature of the valve. <S> Whether the thermocoule wires are to be routed, or its voltage values wireless transmitted outside the valve cover, either method can be implemented.
| A good way might be to use a pyrometer. I think this could be done with a ribbon or thin wire thermocouple attached to the top of the valve.
|
Does the twisting in a twisted pair reduce attenuation? i'm doing networking and trying to understand if the twisting in twisted pair reduce attenuation. I know that the cable can't be longer than 100 meters and the book says this From Microsoft Windows Networking Essentials Twisted-pair cable is the most commonly used cable type in networkstoday. It comes in multiple categories with different speedcapabilities. A twisted-pair cable used in a network includes fourpairs of copper wire. Each wire in the pair is twisted around eachother, and the four pairs within a cable are then twisted around theother pairs. The four twisted pairs are then wrapped in a polyethyleneor polyvinyl jacket. The number of twists per meter in these cables isdifferent for different categories of cables. Twists in the cable helpminimize both cross talk and EMI. Additionally, the number of twistsper meter determines the speed and frequency capabilities of thecable. Higher speeds and frequencies allow the cable to carry largeramounts of data. However, all the twisted-pair categories have amaximum distance of 100 meters. In other words, the cable can’t belonger than 100 meters between any two components. It is possible toextend this distance by using a repeater. The repeater amplifies thesignal, allowing you to run the cable another 100 meters. So this means that if i have higher frequency i have less attenuation? I mean i thought attenuation is the distance the data goes not how much data there is, so i think the answer to my question is no, but am i right or wrong? Thank you <Q> No. <S> The main effects on the characteristic impedance and the propagation velocity are from the diameter of the individual wires, the separation between the wires, and the dielectric constant of the material between the wires. <S> The main effects of the attenuation are from the lossiness of the dielectric, the conductivity of the wires, and the diameter of the wires. <S> The main limiter of the bandwidth is from the attenuation, which increases as the frequency of the signal increases (mainly due to the skin effect ) <A> Perhaps the manual meaning is: with more twists per meter the cable can carry higher frequency and higher frequency means larger amounts of data. <S> But I would say the precision of twists is important, the two wires have to be as closer as possible to reduce the free space window to minimum, this ensures low inductance -> higher frequency possible. <S> Larger number of twists just reduces noise susceptibility. <A> No. <S> The reason for twisting is to better cancel interference and noise from nearby sources (including the three other pair in the network cable itself). <S> The more twists per meter, the slightly higher the capacitance between the wires in the pair. <S> And higher capacitance is BAD for trying to send high-speed data. <S> That book was written by (or perhaps edited by) someone who didn't understand the fundamental electronic principles.
| If anything, twisting increases capacitance which increases loss (attenuation) at higher frequencies.
|
Does the IR2110 need a gate resistor? The datasheet of the IR2110 shows this as the typical connection: And in some cases even a gate-source resistor is used, as in this and this post But the app note shows this as the connection for a buck converter, no resistance whatsoever: <Q> Short answer, yes <S> you want a gate resistor. <S> Longer answer, it reduces EMI by slowing down the MOSFET which in turn reduces ringing and high frequency components of the switching waveforms. <S> On negative side this causes more power losses as the on and off times are increased. <S> You can see in one of the schematics that there's a diode across the resistor, typically switching the mosfet off can be done quicker without ill effects depending on topology. <S> None of these circuits actually seem to be SMPS circuits so unless there is fast switching involved, it may not matter. <S> Gate source resistance is probably just to keep the mosfet OFF if circuit is not functioning <S> so there's no 300V in output. <S> In SMPS applications you may run into dV/dT induced spontaneous turn-on as mosfet parasitic capacitances conspire to raise gate voltage above threshold voltage. <S> However in this instance a small cap is added from gate to source, not a resistor. <A> sometimes nm's (tens to hundreds of atoms) thick. <S> This creates a capacitor with some resistance. <S> The problem is this layer can be blown away if a low enough impedance source is used, the ramp up time can damage the layer. <S> ESD can also easily blow right through this layer (it doesn't take much joule heating in the material to create a hole because there isn't much material to start with and kV's of ESD create a lot of heat) <S> The is really dependent on the mosfet, and on how much you care about protecting the input. <S> There are many mosfets that have some kind of input protection. <S> If they don't have any, you'll need it. <S> The IR2110 datasheet's recommendation in the example schematic has limiting resistors, I would put them in. <S> With other drivers, I would check for some kind of limiting if they don't recommend them. <S> DC to DC converters need to be very fast, and I don't know what the circuitry is <S> but I'm willing to bet they limit the inrush current to cgs by design. <A> The gate resistance influences the Mosfet turn on and turn off time. <S> So the value or its absence depends on the Mosfet gate capacitance and the desired on/off time. <A> No, don't use it! <S> It's lamer's way to reduce EMI. <S> The real way is to ensure short trace, close return path, etc. <S> With resistor your rise and fall time will be longer and it will heat the MOSFET. <S> Why would you do it? <S> Even 10R may seriously harm your efficiency. <S> Current limiting must be integrated in the gate driver.
| Gate resistors should be used for input protection, the mosfet gate is a really thin layer of insulation
|
How is galvanic isolation established in an SMPS? I can understand how an isolation transformer can establish isolation; but since SOME SMPS supplies dont have transformers and they have rectifiers instead; I wonder how the AC and DC ends are isolated. edit: Here an SMPS with no galvanic isolation: http://uk.rs-online.com/web/p/embedded-switch-mode-power-supplies-smps/7516739/ Here is the datasheet(says there is no galvanic isolation): http://docs-europe.electrocomponents.com/webdocs/1072/0900766b810728dc.pdf How is that possible? I'm asking because people here all agreed SMPS has small transformers which isolates the input and output power. <Q> Sure they have transformers: - <S> The transformer looks different to a regular AC type <S> but it's still the largest single component on the PCB but a whole lot smaller than it would be for 50/60 Hz operation. <S> The transformer is the big yellow taped thing in front of the heat sink at the back/left. <S> SMPSs also use opto-isolators for feeding back a measure of the output voltage so that it can be regulated. <S> This is another significant isolation feature thus, output and input remain galvanically isolated to several kV and are therefore "safe". <A> I suppose this is a common misconception, among people who have studied DC-DC converters only. <S> The truth is that transformers are often used in AC-DC SMPS, specifically when isolation is required. <S> The difference is that the transformer takes the place of the inductor of a regular DC-DC converter, and energy is transferred magnetically to one or many secondary coils. <S> Feedback is often achieved with an opto-isolator, but there are other methods of feedback as well. <S> Typically a first stage will be a non-isolated power factor correction stage, which passes the 400V DC to the regulator with transformer isolation. <A> SMPS do have transformers. <S> They are not quite the same as the 50Hz transformers you find in linear power supplies, because they work at a much higher frequency, (like ~100kHz), but they have one. <S> Look inside a PC power supply: <S> The transformer is the block with the yellow tape and the writing on it, between the heat sinks. <S> Any AC-input supply also have rectifiers (usually as a bridge). <S> They are placed before the transformer for SMPS, or after the transformer for a linear supply, but they don't replace the transformer. <A> Generally you want isolation for safety, but for some applications isolation isn't required. <S> Some of the confusion is because the product page linked to in the question is all messed up. <S> Although the page says the product is is a SMPS, the title is wrong. <S> The page shows an inrush current limiter that takes AC input and outputs AC, limited to 10A. <S> It doesn't provide galvanic isolation since it's kind of a surge protector for a SMPS, not a complete SMPS. <S> The page describes the product as a "redundancy module", but that is a totally different product that can be added to an SMPS. <A> I can understand how an isolation transformer can establish isolation; but since SOME SMPS supplies dont have transformers and they have rectifiers instead; I wonder how the AC and DC ends are isolated. <S> Simple: they aren't .
| To answer your original question, yes, you can have a non-isolated AC-DC SMPS that uses an inductor instead of a transformer.
|
Why are my reed relays releasing 10x faster than spec? I have Magnecraft W171DIP-25 reed relays ( datasheet ). They are DPST relays with a built-in coil suppressor (according to datasheet). Their listed response time is 1 ms, and, following the logic of this question ( Do reed relays have shorter opening than closing time? ), I would expect the release time to be at least 1 ms. I measured the release time with an oscilloscope: 0.1ms. I'm not complaining, but why is this? I did it in two distinct (identical) setups: same result. I don't have a 5V supply available on the board, so I just ran them off of a 8V supply and stuck a 100 Ohm resistor in from of the relay (the coil resistance is 200 ohm like on the Mouser site , NOT 500 ohm as in the datasheet). I used a NPN transistor to switch the relay. <Q> The specified switching time is a maximum value, not a typical value. <S> Your relay may indeed perform better than the specification, but don't count on this for your design. <S> Some other relays of the same type may not perform quite as well, especially under extreme conditions (e.g, when unusually hot or cold, when switching larger voltages or currents, when switching rapidly, after ten years of heavy use, etc ). <A> As others have mentioned, the operating time specification is a maximum, and no indication of typical is given. <S> Often users will put a diode directly across the relay coil to prevent the voltage from rising much above the supply voltage. <S> This is the easiest on the switching transistor and the slowest. <S> Note also that putting a resistor in series with the coil as you did will actually speed up the close operation as it begins to approximate a constant current source, so it reduces the effect of coil inductance. <S> The time for the current to rise to 63% of final \$\tau\$ = <S> L/R <S> so the higher R is, the shorter the time constant will be (for a fixed final current). <S> It will also reduce the release time if the diode is across the resistor + coil rather than just across the coil. <S> Of course, the magnetic field is only part of the operation- even if it appeared and disappeared instantly, the contacts would take some time to move. <S> If you didn't put a diode in there, you should, to protect the transistor. <A> 1 msec is the guaranteed operating time. <S> There is no minimum listed, so there's no reason to worry about it. <A> What you measured was the time necessary to open a tiny gap between the relay contacts. <S> But a longer time is needed to get at least 90 % of the nominal contact distance and the full specified insulation voltage.
| You tested only one part and not a large sample from different batches of manufacturing. The relay will release faster if you allow the magnetic field to collapse more quickly- to do that requires a lot of voltage to appear across the coil.
|
Is there a potentiometer model for LTspice? I started designing a model for a three-terminal potentiometer in LTspice, since none are included and it's such a common component. Drawing the .asy symbol and the wiper terminal, it dawned on me that this was going to be more complicated than it appeared. How would the various tapers be modeled? How would this taper be "controled" during simulation? It looks like a subcircuit and library at least, is in order. Before I reinvent the wheel, has anyone done this already ? Thank you. <Q> ...has anyone done this already? <S> Yes, someone has already done this. <S> (I believe his name is Helmut Sennewald). <S> The Yahoo LTSpice group has a set of potentiometers that work very well. <S> You will have to register a Yahoo account and join the group to download them (by the way, I highly recommend doing this if you want to pursue LTSpice, the Yahoo group has one of the larger collection of third-party LTSpice models). <S> The relevant files are potentiometer_standard.lib and potentiometer_standard.asy , as well as some other supporting files. <S> The models provide linear, log, and other models, as well as a potentiometer symbol. <S> The following is an excerpt from the readme file. <S> pot_lin : ideal linear resistance dependencypot_pow : ideal power function resistance dependencypot_plog : ideal positive logarithm function resistance dependencypot_nlog : ideal negative logarithm function resistance dependencypotr_tab: arbitrary(table) <S> based resistance <S> dependencypot_piher_plog : <S> pseudo logarithm function resistance dependency, Piherpot_radiohm_plog : measured pseudo logarithm fucntion resistance dependency, Radiohm <S> How would this taper be "controlled" during simulation? <S> These pots have a wiper property which can be easily parameterized as a regular LTSpice parameter. <S> For example, you might say wiper={GAIN} , and then add a directive such as .step param GAIN 0 1.0 0.25 . <A> Tried to follow the suggestions above but took me a awfully long time to create a potentiometer that looks like a potentiometer and that can be instantiated from the main schematic. <S> So, for the benefit of anyone that may be as dumb as me... Just copy these 3 files to a directory in the LTspice search path (erase any initial spaces in every line). <S> Hope the names are self-explanatory. <S> potentiometer_test.asc Version 4 SHEET 1 880 680 WIRE <S> 272 48 0 48 WIRE 528 48 272 48 <S> WIRE 272 80 272 48 WIRE 528 80 <S> 528 48 WIRE 0 96 0 48 WIRE 0 <S> 192 0 <S> 176 WIRE 272 208 272 176 WIRE 528 208 <S> 528 176 <S> FLAG 272 <S> 208 0 FLAG 0 <S> 192 0 <S> FLAG 320 <S> 128 out1 FLAG <S> 528 208 0 <S> FLAG 576 128 <S> out2 SYMBOL voltage 0 <S> 80 R0 SYMATTR InstName V1 SYMATTR Value 10 SYMBOL potentiometer 272 176 M0 <S> SYMATTR InstName U1 SYMATTR <S> SpiceLine2 <S> wiper=0.2 <S> SYMBOL potentiometer 528 176 M0 <S> SYMATTR InstName U2 SYMATTR SpiceLine R=1 <S> SYMATTR <S> SpiceLine2 <S> wiper=0.8 <S> TEXT 140 228 <S> Left 2 ! <S> .op <S> potentiometer.asy Version 4 SymbolType BLOCK <S> LINE <S> Normal 16 -31 <S> -15 <S> -16 <S> LINE <S> Normal <S> -16 <S> -48 <S> 16 -31 <S> LINE <S> Normal <S> 16 -64 <S> -16 <S> -48 <S> LINE Normal 1 -9 <S> -15 <S> -16 <S> LINE Normal 1 0 <S> 1 -9 LINE <S> Normal 1 -94 <S> 1 -87 <S> LINE <S> Normal -24 <S> -56 <S> -16 <S> -48 <S> LINE <S> Normal <S> -24 <S> -40 <S> -15 <S> -48 <S> LINE <S> Normal <S> -47 <S> -48 <S> -15 <S> -48 <S> LINE Normal <S> -16 -80 <S> 16 -64 <S> LINE <S> Normal 1 <S> -87 <S> -16 <S> -80 <S> WINDOW 0 <S> 30 -90 <S> Left <S> 2 WINDOW 39 <S> 30 -50 <S> Left 2 WINDOW <S> 40 31 -23 <S> Left 2 <S> SYMATTR Prefix X SYMATTR ModelFile potentiometer.lib <S> SYMATTR SpiceLine R=1k <S> SYMATTR <S> SpiceLine2 wiper=0.5 SYMATTR Value2 potentiometer <S> PIN 0 <S> -96 NONE 8 PINATTR PinName 1 PINATTR SpiceOrder 1 <S> PIN 0 0 <S> NONE 8 PINATTR PinName 2 PINATTR SpiceOrder 2 <S> PIN <S> -48 <S> -48 <S> NONE 8 PINATTR PinName 3 <S> PINATTR SpiceOrder 3 <S> potentiometer.lib <S> * <S> This is the potentiometer <S> * <S> _____ <S> * 1--|_____|--2 <S> * | <S> * 3 <S> * .SUBCKT <S> potentiometer 1 <S> 2 3 .param <S> w=limit(wiper,1m,.999) <S> R0 1 3 {R*(1-w)} <S> R1 3 2 {R*(w) <S> } .ENDS <A> Google LTSpice potentiometer, there are lots of examples with varying degrees of complexity. <S> Most use a sub-circuit along these lines: <S> * <S> This is the potentiometer* _____ <S> * <S> 1--|_____|--2 <S> * <S> |* <S> 3*.SUBCKT potentiometer 1 <S> 2 3.param w= <S> limit(wiper,1m,.999)R0 1 3 {Rtot*(1-w)}R1 3 2 <S> {Rtot*(w)}.ENDS <A> To vary a parameter (such as a component value), you can use the .step command to do a parameter sweep. <S> If all you want is a two-terminal variable resistance, you can use a normal resistor for this. <S> If you need three terminals, Steve's answer seems like a good one. <A>
| Under "Special Functions" there is a voltage controlled varistor that you could use instead.
|
Voltage divider output with analog voltmeter I'm using a voltage divider to bring down 24V to 5V and attaching the output to an analog voltmeter gauge. So my question is should I be counting the resistance of the analog gauge into the value's of the resistors of the voltage divider and vary as needed to get my 5V? (see attached diagram) <Q> Yes, you certainly should. <S> The resistor in parallel with the coil increases the loading on the circuit under test unnecessarily. <S> The optimum value for R2, therefore, is infinity! <S> Assuming that this is a fixed range meter rather than a multimeter the normal solution would be: Measure the coil resistance and find its full-scale deflection voltage. <S> Since \$ \frac {V_{TOTAL}}{V_{METER}} = \frac {R_{TOTAL}}{R_{METER}} \$ you can work out the series resistor. <S> The result will give you the minimum possible loading of the circuit under test by your meter. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Basic DC multi-voltmeter. <S> Usually a rotary selector switch would connect the red lead to the appropriate point in the resistor chain. <S> If you look at the specifications of any analog multimeters you will generally find that they had a resistance of 20 kΩ/V on the DC voltage range. <S> Changing range just switched in the appropriate series resistance. <S> Figure 2. <S> Simpson analog meter clearly showing 20 kΩ/V on front panel. <S> Image source: Simpson160.com . <S> In contrast, most digital multimeters have a constant input impedence of 1 M on all DC voltage ranges. <A> The voltmeter will provide a parallel resistance to R2, changing the value of the voltage divider, hence the change in Vout. <S> The value of the voltmeter will vary based on it's internal resistance, range, the resistance of the leads used, etc. <S> A digital voltage meter will have a significantly higher resistance, closer to 10 Megaohms. <A> In theory, you should factor in the resistance of your voltmeter, but in practice the latter's resistance will be so much higher than your R1 and R2 values that it effectively has no effect and does not need to be taken into account. <S> What you should consider though is the 'load' you will connect at the 5V point that you are trying to achieve. <S> I am sure that you are not just creating a divider to display a voltage on a meter...
| Calculate the series resistance required for full-scale voltage in your application.
|
Mystery issue driving load through NFET with microcontroller U1 is a microcontroller driving M1 to drive a load. I am running into some bizarre behavior and I'll try to include all the relevant information. Hope it's not too much. M1 is a logic-level FET. U1 is running off a 3.3V supply derived from the 3.7V battery in the system. There are no other power sources connected, like benchtop power supplies. All share a common ground. Although I have drawn the load as a resistor, that was more of I didn't see another option in CircuitLab for generic black-box load. I literally am driving a black box load as well, one of those high-voltage generators you can buy on eBay like this one . My understanding is that the first stage input inside their cavernous epoxied insides is some sort of blocking oscillator, and so it's entirely possible this load is not purely resistive. Just keep it in mind, but so far I don't think it's inductive as I haven't had any reliability issues with like back-EMF destroying the FET as if there was a solenoid inside the box. The load draws high current, about 2A. Here's what's happening: I am doing this on a breadboard. If I hook VGATE directly to the 3.3V supply to U1 (what the GPIO is pulling to), it works fine. When I toggle a pushbutton that drives an ISR to pull the gate high, it does so very briefly and then the IC resets (I have the code blink a different LED at startup so I know it's reset). This made me suspect voltage droop on the supply lines due to the sudden surge from the battery. To fix the suspected supply droop, next I added a ton of decoupling capacitors on both the input side of the LDO linear regulator and the output side. This did seem to help a bit but still would only drive the load for less than a second then reset. I added an LED to the output and surprisingly this seemed to help. Perhaps it loads down the GPIO a little bit more, reducing voltage slew rate on VGS and thus reducing supply droop? Now here's where things get weird: If I connect a Logic Analyzer probe to VGATE (I wanted to see what the waves looked like), it completely works! Damn you, Werner Heisenberg. I have done a few experiments and it doesn't matter if both the signal probe and the ground are connected or just one (either, doesn't seem to matter). Yes, a single wire with no return path for current is somehow affecting this. Could this be some weird antenna thing going on, or noise coupling onto the wire which might be doing something? I will say that since I am breadboarding, my wires are getting a bit long, nothing longer than 6" though. It also doesn't matter if the logic analyzer is plugged into my computer or not (USB logic analyzer, Saleae Logic Pro, if it matters). I was thinking at first maybe there was some different ground potentials going on, though I'd expect the ground in the analyzer to be isolated from my PC USB port. Either way, with a single wire connected I don't see how current can be flowing in or out of that wire anyway and thus doing anything. Anyway, I'm banging my head against the wall trying to figure this out and figured sometimes it's not good to work in a vacuum. Thanks to anyone who can help me out. EDIT: Testing the idea again that CGD coupling might be turning off the FET, I tried to make the cap to ground there larger with a cap on VGATE to ground. Though this didn't work, I notice that if I just touch the top of the cap (it is electrolytic), the circuit works again no problem. So, touching my body to the top of the cap fixes the issue. Putting a wire from a logic analyzer on that net (either GND or VGATE) fixes the issue. And putting an LED load on the GPIO pin also helps (but doesn't completely fix it). What theory could possibly explain all three and why they fix it? EDIT2: Getting somewhere... taking Spehro's suggestion that the di/dt from switching such a large load could inductively coupling, I put a large RC filter on VGATE so that would limit the di/dt switching. What I find is that, all else unchanged, the LED on the GPIO slowly dims brighter until the FET is fully on, and when the HVGEN sparks, the micro resets. If you record video of the circuit on a phone, the picture is noisy due to the large EMI. I basically think the EMI from the sparks is resetting the microcontroller independent of di/dt spikes or other events like this. For some reason when I touch the capacitor on VGATE, the problem goes away. Is it possible I'm a large body that "absorbs" the EMI somehow and thus the micro does not reset? If so, how can I fix this permanently, short of touching the circuit at all times? ;) Is there some way without enclosing the entire PCB in a Faraday cage? <Q> Note that in the IRF530 spec sheet (an older IR brand) the gate threshold voltage (gate turn on) is shown as 2V min to 4V max. <S> So your 3.3V I/ <S> O line might not be so well suited. <S> The mosfet may be listed as a logic level part, but this may only be related to 5V logic <S> I/O's. <S> So with a marginal gate drive any additional noise on the gate (inductively or capacitively coupled) <S> could make a big difference.... <S> See: irf.com/product-info/datasheets/data/irf530.pdf . <S> If the microcontroller you're using is capable of running at a 5V level try that to confirm the gate drive issue. <S> You can also place bypass caps (in this case higher values, maybe 100uf or more) <S> right at the "Load" <S> +V point to ground. <S> This would help to provide initial high current to the load and further prevent negative spikes from getting to the regulator. <S> (If that is the cause of the Reset issue) <A> While all of the answers here were helpful in debugging, none of them turned out to be the complete story. <S> In short: I needed to shield the microcontroller from EMI with a Faraday cage. <S> As noted in my second edit, putting excessive gate load on the MOSFET to limit the di/dt showed that it wasn't the spike in current (since I limited it by limiting Vgs slewrate), but rather when the device sparked, the EMI from the device caused it to reset. <S> Spehro's suggestion to not do this on a breadboard was apt, and proper layout on a professional PCB (including lots more decaps and proper star grounding techniques) <S> did dramatically help the issue. <S> These techniques alone would allow the device to run without resetting, as long as the HVGEN was not directly on top of the microcontroller. <S> However, when placing the unit directly on top of the PCB (the HVGEN unit is fully insulated), there was still enough EMI that the proper layout could not overcome and it would reset again. <S> The only thing that fixed it for good even with it sitting straight on top was an EMI shield like this one . <S> Thankfully I put a footprint on my PCB for this, just in case, hoping I wouldn't have to populate it, but doing so was the trick. <S> A full Faraday cage around the microcontroller fixed the issue. <A> You know you shouldn't be doing this kind of thing on a breadboard. <S> Try adding a 10K or so resistor in series with the gate. <S> You may be getting spikes coupled from the drain to the gate which are upsetting the micro, or spikes below ground due to inductance from your breadboard setup. <S> You can diagnose the latter if you put a pushbutton switch from the gate to the +3.7 (disconnect the port pin) with the micro running-just that one connection broken- and see if pushing the switch a few times resets the micro (the 10K will discharge the gate charge). <S> Also make the connections between the source, the battery (-) and the ground wire running to the breadboard outside the breadboard, or at least use an isolated set of connections to tie them together at one (and only one) point, which is then jumpered to each of the three.
| Or there is a bypass capacitor in the black box which if you switch too rapidly will cause the micro to reset.
|
How is it physically possible that there is no voltage drop across two pins of this DC jack? I have a DC jack with three terminals. Terminal 1 is tip (+), terminal 3 is sleeve (-), and terminal 2 is a switch, my understanding of which is such that if there is no adapter plugged into the jack, terminals 2 and 3 are connected. When I plug a DC adapter into the jack, I get the voltage drop I would expect across terminals 1 and 3, about 9V. However, there is zero voltage between terminal 2 and both 3 and 1. How is it possible that there is no voltage drop between either? Shouldn't there have to be a 9V drop between terminal 2 and 3 if there's 0V between 1 and 2? <Q> In this diagram used for illustration purposes Pin 3, the Switched Pin, is physically disconnected when the plug is inserted. <S> Pin 2 is physically moved by the barrel. <S> Its normally made of a spring metal that can bend and return to its normal position when the plug is removed. <S> The voltage seen is 9v, with +9V on the center pin/Pin 1, and 0V/Ground reference on pin 2 (for a center positive adapter). <A> Your reasoning about voltages is correct. <S> What's happening when you insert a connector is that the normally-closed contact opens, and is now floating. <S> That means it should measure open , not 0 volt . <S> I suspect that your multimeter does not show the distinction clearly enough. <A> Because trying to measure the voltage changes it! <S> Your meter has a finite input resistance (usually 10 megohm). <S> Pin 2 is not connected to any of the other pins, so there is no reason to expect it to have any particular voltage relative to them. <S> When you connect your meter between 1 and 2 it pulls the voltage between 1 and 2 down to zero.
| No voltage is measured with the switched Pin because it is no longer connected with anything. Similarly when you connect it between one and three it pulls the voltage between 2 and 3 down to zero. The 3rd picture is a Phone connector with, but the concept is the same in a DC barrel jack.
|
Low power device in sealed unit -- design help Main question: can I build an arduino-based sensor that'll use such low power that something like inductive charging will suffice to power it? Details: I've got a "holding tank" on my boat ... a 3-foot tall polyethylene tank that's typically part-full of human waste mixed with salt water, which is a pretty horrible environment for electronics. I want to be able to know how full it is. External sensing hasn't worked well, so I'm considering putting some assembly into a piece of PVC plumbing pipe, with sealed ends, and sticking this in the tank. My plan is to use capacitive sensing (as in this design ) with temperature compensation using a low-budget temp/humidity sensor I've got lying around. I plan to get information from this sensor to a display via bluetooth or some similar coupling. I typically need to know the tank-fullness about once or twice a day ... this doesn't require continuous monitoring. My plan is therefore to use something like sleep mode most of the time, waking up now and then (or perhaps waking up in reponse to some sort of trigger -- I'm not clear on how to do this), and powering the thing from something like a lithium battery. But I need to recharge the battery, even if it's a very low power system overall. I was thinking that I could use inductive coupling (from a coil of wire outside the poly tank to a coil of wire inside the PVC pipe), and then take the resulting current and rectify and filter it a bit and use that to recharge the battery. Any thoughts on whether this is even slightly plausible? The PVC is probably 1/4" thick, and the poly tank is about 3/16"...so the inductive coupling would have to be over a distance that's on the scale of 1 inch, more or less, and Whether there's an easy way to waken the system from low-power mode in response to some external signal, and Whether using bluetooth, when the processor is going to be in sleep mode 99% of the time, is plausible, or am I going to be spending all my power in reestablishing the bluetooth connection? <Q> A few ideas come to mind: <S> Your tank is polyethylene. <S> You should be able to sense liquid level through the wall of the tank using your capacitive sensing. <S> If that doesn't work it won't work in a PVC pipe. <S> Advantages: <S> no wireless, no battery, no retrieving the unit out of the tank ... <S> I'd consider an ultrasonic level sensor. <S> These are used in all sorts of industrial applications, quite reliable and can go in the top of the tank where it shouldn't get dirty. <S> If it does get dirty you may have problems. <S> Advantages: simple, analog output for level display, easily retrieved for adjustment, etc. <S> I've never considered the matter, but is there any risk of methane buildup making the airspace an explosive atmosphere? <S> If so you need to take appropriate measures with any electrical devices inside the tank. <S> From memory ATEX equipment is limited to 10 V or so at 8 mA. (Someone will jump in with the correct numbers.) <S> This is low enough to prevent a spark in any circumstances. <A> I wanted to put this in a comment, but apparently I don't have the cred to do so yet. <S> Seems backwards I can answer but <S> not just comment... <S> I know this takes all the engineering fun out of it, but have you considered using a clear polyethylene tank like this one <S> so you can visually see the level? <S> I'm not sure if that would work in your setup or if your tank has more to it than described. <A> I just wrote in another answer : <S> If you use a differential pressure sensor you can build a solution that will be very robust because it has no moving parts, is not impacted by temperature/humidity, water cleanness is irrelevant (both with regards to salts and transparency), and no electrical sensors are in contact with the fluid (so if you want to meassure level in an acid tank, no problem). <S> and I think that solution is just as applicable in your case.
| You do not need to put the sensor or control unit into the tank, you can get away with just inserting an empty, sealed pipe of some kind.
|
Merge polygon pours in Altium Designer I need to know if there is a way to merge two or more polygon pours of the same type, in order to have only one. So two objects that become one. Is this possible?Because anytime I need to redefine a polygon I do not want to start again if I only need to add and improve a little part of it. <Q> At the same time, it's possible to achieve a similar effect while keeping the polygons distinct (without actually merging them). <S> Make the polygons pour over each-other. <S> No isolating gap between polygons that are the same net. <S> Right-click on the polygon pour. <S> Select Properties... <S> at the bottom of the context menu. <S> Group the polygons. <S> Now they can be dragged around together. <S> To adjust the shapes of individual polygons, you would have to temporarily break-up the union. <S> You can't adjust the polygons while they are inside of a union. <S> Select the polygons which you wish to group. <S> Right-click. <S> Select Unions <S> → Create union from selected objects . <A> As of Altium 17, this function is available: Select both polygons <S> right click <S> "Polygon Actions" "Combine selected polygons" <A> No, you cannot merge them, but of course you can change them. <S> Click SHIFT/CTRL and drag a vertex. <S> It will take some time to get the hang of it, but once you understand how it works, it's quite comfortable. <S> You can even switch corner mode by pressing Shift-Space during the modify operation.
| I'm not aware of a function that lets you merge polygons.
|
Directly driving 5V source MOSFET H-Bridge with 3.3V logic I am trying to design a H-bridge circuit for my project to drive a thermoelectric cooler (TEC). Basically, I am trying to control the TEC with a voltage source driven by PWM. Now the problem is, I have a 5V DC source but my microcontroller ( ESP8266 ) uses 3.3V logic. Since the space for components is limited, I would like to drive the MOSFET directly from the microcontroller. I have built a circuit based on some things I have found on the internet but am not exactly sure if it is correct. The TEC I am using is expected to draw up to 1A at 1.7V (maximum); the MOSFETS I am using are ROHM RZR040P01TL (p-type) and RUR040N02TL (n-type) which are rated at 4A, have Vgs(th)~=1.3V and built-in GS protection diodes. Since I have never designed an H-bridge before, my first questionwould be about the design. Does it look like something that wouldwork or I am missing something here? When I increased the PWM signal abovezero, there was a high pitch noise coming from the MOSFETs. Is thissomething to be expected or does that signal an issue in the circuit? I have tried sending low duty cycle PWM from the microcontroller (1-10%) which seemed to be working fine - while the MOSFETs were still buzzing, the TEC was working fine and there was around 0.4V voltage accross the terminals (which you would expect). However, when I increased the duty cycle to 20%, one of the N-type MOSFETs blew out. I am guessing that this either happened because the MOSFET overheated or because I have done something horribly wrong in my circuit design. I would appreciate some advice on building MOSFET H-Bridge circuits in general as well as some specific recommendations for this particular design. Original design Update #1: I have redesigned the circuit to have each MOSFET switch separately as @WhatRoughBeast suggested. I also have added two N-type MOSFETs to control the P-type gates which should solve the voltage difference problem. This appears to have solved the noise issue - now buzzing only appears if I reduce the switching frequency to 5kHz or lower. In the current configuration I am trying to control the direction and voltage by: having B1 and B2 set to LOW, A2 to HIGH and PWM-controlling A1 having A1 and A2 set to LOW, B1 to HIGH and PWM-controlling B2 While the MOSFETs appear to be working now (as in not overheating and blowing up), it seems I have another problem - with 5V supply, whatever PWM duty cycle I use, the TEG always receives the full 4.5V (using 10-20kHz). <Q> You must drive this kind of bridge with at least 5V (or close to- 4.7V will guarantee less than 1mA of conduction according to the datasheet ). <S> As you have it, when an input to the bridge is at 3.3V, the N-channel MOSFET is 'on', but the P-channel is also pretty much on with 1.7V nominal drive, and it will typically conduct several amperes of current, which will fry it or the N-channel or both, depending. <S> You can use a voltage translator or a MOSFET driver. <S> The latter will be capable of much more drive current and will result in less heating, but will cost more. <A> Another way to look at this is to see what happens to the transistors as the gate voltage goes from zero to 5 volts. <S> Spehro has pointed out the appropriate rating on the data sheet. <S> If the voltage is less than 0.3 volts, the p-type will be on (gate-source voltage 4.7 or more) and the n-type will be off. <S> For voltages greater than 4.7 the p-type will be off and the n-type will be on. <S> For any voltage greater than 0.3 and less than 4.7, both transistors will be on, and one or the other is going to get very hot. <S> This assumes, of course, that you use the minimum Vgsth. <S> Since this is specced for 1 mA, this is very conservative, but it's pretty clear that using 1 volt will get you in trouble. <S> The condition of both FETs being on simultaneously is call "shoot-through" for obvious reasons. <S> You have two possible routes to go. <S> This will transition through your danger zone very quickly, and the shoot-through will only last a few 10s of nanoseconds. <S> The other possibility is to stagger the timing of the two gate drives so that only one is driven on at any time. <S> This is often built in to integrated bridges and bridge drivers. <S> It will take a good deal more thought and effort than doing it the way you're doing it now. <A> There are alternate ways of doing this. <S> Right now you have two sync buck circuits. <S> You could replace this with a single buck controller which does not need to be synchronous @ 1A output. <S> On top of that you'd need a set of mosfets which can reverse the TEC polarity. <S> In essence one pair of N-MOSFETs which will connect + to 1.7V and - to GND. <S> And a second pair which will connect + to GND and - to +1.7V. <S> This is probably easier to pull off than two separate syncbuck circuits as you only need one PWM signal and two on/off signals with open-collector NPN transistor to let the topside Mosfets have full +5V gate voltage. <S> 1.7 + 1.3 <S> = 3V <S> but with 5V it's firmly in the saturation region with minimal RDS on.
| The first is to get a gate driver.
|
Implement ac-dc converter for low frequency and very low voltage without external powering? Iam using a basic cascade voltage multiplier to boost and rectify the 0.3v ac signal. The circuit rectifies but the boost is very less. So I thought of using LTC3105 after this stage to get a regulated dc output voltage. Problem in simulation is it is taking really long time to simulate moreover LTC3105 is not boosting up below 0.5 v but according to datasheet LTC3105 starts up at 0.25 v. Can you please help me out with this? Implement ac-dc converter for low frequency and very low voltage without external powering? - ResearchGate. Available from: https://www.researchgate.net/post/Implement_ac-dc_converter_for_low_frequency_and_very_low_voltage_without_external_powering [accessed Apr 21, 2016]. <Q> This circuit as it is will NOT work because the forward voltage drop for the 1N5817 is about 0.55 volts at its lowest. <S> Unless you use external power, the suggestion of a step-transformer by @Bruce Abbott is a good idea. <S> You have almost no options at this low of a voltage. <S> Use a tiny audio transformer and see if the speaker output (measures a couple of ohms dc) can be your signal input, the measure the secondary windings (many ohms dc) and see what kind of boost you will get, and is it enough. <S> Increase C6-C9 <S> to double or triple their present values. <S> You are working on the fringes here. <S> The 140NEX transformer should work fine. <S> It could increase the input signal a lot, but your voltage multiplier is going to load it down a lot. <S> The big issue is if this gives you enough to run the circuit. <S> Now it comes down to how strong your raw 0.3vac <S> is (how much current can it supply). <S> If it can maintain 100 mVAC under load I think this circuit may work. <S> It depends on the LTC3105 start-up current. <A> The LTC3105 is a complex IC which can take a while to simulate, particularly at startup. <S> The basic problem with your circuit is that the rectifier is very inefficient at such low voltage. <S> Putting a booster after it doesn't help because the rectifier can't supply enough current to run it properly. <S> There are two reasons why the rectifier doesn't work well. <S> Coupling capacitor values are too low, causing excessive AC voltage drop. <S> You should increase the values of C8 and C9 to 1000uF or higher. <S> Connecting C8 directly to V2 instead of in series with C9 will also help. <S> Your AC input is 0.3V peak . <S> Even at low current the 1N5817 has a voltage drop of about 0.2V, so most of the input voltage is wasted. <S> Schottky diodes have lower voltage drop than most other types, so to get much lower voltage drop you might have to use active rectification <S> (transistors switched on and off in sync with the AC waveform). <S> Instead of trying to rectify the low voltage directly you could step it up with a transformer. <S> This may be quite bulky at 20Hz, but a lot simpler than an active rectifier. <A> Just to answer the simulation question, adding a large resistance to any nodes that have only caps on them can help speed up the simulation. <S> In addition there are some simulation options that allow you to avoid certain numerical solution techniques. <S> I havent used these options much <S> so I cant add anymore to that. <S> Also, the more stable a circuit is the more easily it can be simulated, so <S> just the fact that it is taking a long time to simulate can mean there are some problems. <S> Then again simulating an isolated flyback converter usually takes longer when there is no resistive path to ground and that doesnt have much to do with the stability of the circuit. <S> Well...anyways. <S> Normally though this idea of a stable circuit or a properly functioning circuit does lead to faster simulations. <A> I have used 140NEX transformer. <S> I tried doing the simulation with the following values. <S> DC primary resistance 5.30 ohms and secondary resistance <S> 116.5 ohms and inductance of (326mH and 9.36H).Parallel resistances (primary-2.83k and secondary-79.0k). <S> I am getting a output value of about 5 v. <S> Can you please let me know if the specifications I used in the simulation are suitable or not?
| In your circuit the high ripple and voltage drop under load stops the LTC3105 from starting up properly, so simulation is very slow.
|
What is so special in the power MOSFETS IRFD120 and IRFD110? I was trying out a hobby project and I was using the IRFD120 and IRFD110 power MOSFETs to control a motor. The transistors worked pretty fine and we simply ran out of them. When I went to buy some more I found out that they are outrageously expensive for a transistor. However, I found the FQP30N06L and FQP27P06 power MOSFETs for normal cheap transistor price. This raised my "sanity" flag. I thought that I knew which parameters matter, yet it seems that there is another parameter so important that it affected the price this much price. I ran a small specs comparison and I got this (comparing N-channel only for simplicity): FQP30N06L: Current Rating = 60 A Voltage Rating = 30 V IRFD110: Current Rating = 1 A Voltage Rating = 100 V This clearly shows that the FQP30N06L can withstand more power, yet it is ~5 times cheaper than the IRFD110. Is it only the 100 V rating that is making the IRFD110 so special? I am asking this question because I need to buy large quantities and I haven't used the FQP30N06L yet. Would it be a good idea to replace the stock of IRFD110 and IRFD120 with FQP30N06L and FQP27P06 for hobby project related applications similar to this: youtube.com/watch?v=Y-afnY32RrY? <Q> But I think in this case, the reason is the strange package the IRFD110 uses. <S> This is non standard and certainly uses specific production lines. <S> And usually, the reason is just that: they use older technology that is more expensive to produce than newer technology. <S> The older products then become more expensive, although they are not better speced (usually, they are worse, as you have noticed for the current handling ). <S> Now, if you look for a replacement MOSFET, look at the datasheet <S> The voltage handling (both in VGS and VDS) should be the same or above (in your case, it's below, so double check that you can do this in your circuit) <S> The current handling should be the same or above. <S> The VGs gate threshold should be similar (in your case, it slightly below, but that is less of a problem than if it's above) <S> The RDSon should be below <S> The gate charge should be in the same range (if it's a lot more, you may have to use bigger drivers). <S> Make sure the dissipation is appropriate. <S> If you check all this, I think you're clear. <A> That package is not widely available from different makers (though I see plenty in stock- eg. <S> 20.5 cents Qty 50 from Future). <S> If you want the lowest price, now and in the future, newer parts in SMT will likely be the best value, with the bulky TO-220 package being the main through-hole alternative. <S> The original part has quite a high Rds(on), a not-so-low gate charge and an oddball package. <S> Not great performance. <S> Generally, all other things being equal, the die size (and thus manufacturing cost) of a MOSFET is related to maximum Rds(on) and Vds rating. <S> But the acreage of silicon does not necessarily track selling price, especially for parts on extremes of the life cycle and in tiny quantities. <S> You may find the price dips as a part is going obsolete, as the stock is dumped onto the market, followed inevitably by unavailability at any kind of reasonable price (brokers and middlemen as well as users doing lifetime buys soak up the stock, with the former looking to make a tidy profit by waiting until the users run out and are forced to choose between huge engineering and testing costs vs. paying a fortune for obsolete parts). <S> MOSFET technology in particular has been evolving steadily, and newer better and cheaper parts are regularly introduced. <S> Try doing a parametric search at a distributor- <S> of course you have to understand all about the circuit to know which parameters are important- <S> Vds, Rds(on) at what gate voltage, gate charge, power dissipation capability in a realistic situation (not the stupidly high number on the datasheet), and a number of other factors. <S> Obviously if you can relax the Vds requirement to 50V or 60V you can get a better performing MOSFET and/or a better price. <S> If the power dissipation turns out to be very low, you might even be able to use an SOT-23 MOSFET which could be very small and cheap. <S> But it`s always possible, in any given circuit, there is some parameter that does not allow safe substitution and serious problems will arise, as they tend to do if they have not been fully examined quantitatively. <A> Would it be a good idea to replace the stock of IRFD110 and IRFD120 with FQP30N06L and FQP27P06 for hobby project related applications? <S> Nobody but a fool is going to say yes to that question. <S> It's all down to the application, the load, the power rails, the switching frequency and the driving circuit, none of which is specified in your question. <A> It sounds like you're being ripped off. <S> In Digi-key the IRFD110 is considerably cheaper than FQP30N06L. 21p versus 35p each for 1000 pieces. <S> That being said, the different manufacturers do use different pricing strategies. <S> The right cost of a component is as much as you can possibly ask to maximize the profit. <S> c.f. Apple.
| The 100V could be a reason: high voltage MOSFET tend to be quite expensive.
|
Generating a (ideally several) PWM signal from GPIO, alternative to software-based pulse Scenario : I am trying to control precisely (without glitches) several servo-motors from a computer through GPIO. Final position, speed and ideally acceleration need to be controlled. Currently I generate the PWM signal by software, but software seem not precise enough (due mostly to kernel context switches, the PWM signal is some times several micro-seconds late, producing some "shaking").I know several way to improve, for instance implementing the signal from a driver (kernel) context, but the problem will inherently stay. Question: How to generate a PWM signal, based on software indications. Which solution is recommended? Building the logic myself: That seem an overkill: 2 clocks, plenty of logical gates.. Even if probably achievable, the final result would take too much space. Using an intermediate pic, dedicated to provide precise PWM, the protocol between the computer and the pic to be defined. Could be a solution, I am not sure however of the result or if there are better ways. Looking over internet shops by PWM, servo controller, ... I did not found any chip which integrate this feature. Probably I miss an important concept. other? <Q> Best practical solution is to buy of-the-shelf servo and control it from your software by communication lile UART or ethernet. <S> It's because the people who make servo do it better than you and cheaper then you. <S> And they actually have it done, while you are just starting. <S> Still, let me explain what is inside a typical servo. <S> There is a microcontroller with built-in pwm peripheral, that is set on startup to certain frequency and the duty cycle is updated each cycle according to control loops outputs. <S> Normally one MCU can control two motors, sometimes three or even more. <S> The servo firmware normally has current, velocity and position loops, and depending on vendor advanced control algorithms (error correction, ecam, input shaping, you name it). <S> More advanced architecture has one master controller with cool cpu inside and separate power stages connected by communication to the master. <S> A decent dual axis servo costs around $500. <S> I am sure, time you are going to waste building it by yourself costs more. <S> If you need hints on such servo, i suggest for Galil (simple, but almost high end), Adtek QS9 (chinese high end, very simple and practical), then you have beckhof, yaskawa, <S> elmo- good, but more expensive and complicated. <A> Once you run out of hardware, add another CPU. <S> You need slow enough PWM and enough bits to handle the 1-2ms pulses with sufficient resolution- <S> not all micros will have this. <S> Its also possible to do firmware PWM with a small-ish CPU-- <S> let's say you want <1usec jitter and resolution, you could update several servos one at a time within a (say) 20ms frame. <S> You would have to worry about not overflowing whatever buffering is available in the micro (say by interleaving communication). <S> And finally, you could do a hybrid by steering hardware timer pulses from a single timer through a demultiplexer to individual servos. <S> This might be the easiest of all because the micro would have lots of time to deal with communications (assuming it is interrupt driven). <A> You are getting jitter out of a software implementation because generally computer kernels are not realtime. <S> As such, anything may get prioritized before your PWM switch. <S> You may reduce the problem by elevating you controller's priority, but the problem still won't go away, since drivers and other kernel modules may still have priority. <S> To really fix the problem in the computer side would require using a realtime kernel. <S> Linux has the RTAI project. <S> MATLAB has its own simple OS to run systems on. <S> But generally my opinion is that it's usually simpler to just have a MCU (microcontroller) <S> do the hard work in bare metal. <S> A simple Arduino will be cheap/easy/simple enough. <S> If you want more PWM channels you may try an ARM board (there are some ARM-based Arduinos now too). <S> If you really need the controller to be on the PC side, then either you have to go realtime <S> or you must design you controller to allow jitter when sending information to the MCU (if using USB you may get delay jitter), <S> but then the PWM waveform will still be unaffected.
| One solution is to use small CPUs with hardware PWM.
|
How to efficiently power wireless sensor node from a 9V battery? Context I have recently set out to add some wireless features to my entry-level smoke detector. Problem In order to achieve that, I will be adding a custom pcb embedding an ATmega328P and a nRF24L01, both running at 3.3V. However, the detector is currently powered by a 9V battery, which is not ideal to work with. Additionally, I would like to avoid making any modifications to the actual hardware, and for ease of maintenance, I do not wish to add a second power source. Question I am trying to find with the most efficient 9V to 3.3V solution in the case of a wireless sensor node that will be spending most of its time sleeping. A similar question was asked at Shut down regulator during sleep , but the 9V battery restricts my choice of regulators. Thoughts and reasoning Because of their poor efficiency, linear regulators are of course ruled out. Which leaves us with switching regulators. The system will be asleep most of the time, consuming then only a few micro amps. This is apparently something traditional switching regulators are not good at. Having such a light load decreases their efficiency from 90% to 50% sometimes. As a solution, some of them now have a feature called "discontinuous mode" where the switching frequency varies proportionally to the load current. However, how efficient is this feature really when dealing with loads < 50 µA ? Charge pumps seem to be a solution to the poor efficiency of inductor-based dc/dc converters, but I couldn't find any accepting an input of 9V. I however like the idea of using capacitors instead of inductors to store the power. That's the road this guy has taken: https://hallard.me/ulpnode-low-power-secret/ . His solution looks like a good starting point. I am ok with the 3.3V line dropping regularly as low as 2V, as this value is still in the operating conditions of the components I am choosing. I could use the output from the low voltage detector IC as an interrupt source to periodically wake up the ATmega from deep sleep instead of the watchdog. And by disabling the watchdog, I am saving a little bit more power. Some switching converters do have a feature called true disconnect, in which the input gets physically disconnected from the output path. Is this really saving power by the way? Proposed solution simulate this circuit – Schematic created using CircuitLab U1 is a dc/dc converter with true disconnect and a 2 to 10V input range. R1 pulls up the !SHDN line to make sure U1 is enabled at power-up. C1, whose value is still to be determined, acts as a reserve capacitor, providing power when U1 is shutdown. U2 is an undervoltage supervisor IC with an open-drain output R4 pulls !ENABLE pin-high to deactivate U2 at startup. R2 and R3 form a voltage divider used to set a threshold of 2V for U2. R5 and R6 form a voltage divider used to bring the voltage of the !SHDN line down to approx 2V to meet the requirements for the ATmega input pins maximum voltage. PD3 changing will trigger an interrupt, used to wake-up the ATmega from time to time, when C1 discharges. PD4 ouput can be set to LOW to enable U2. At power-up, U1 will start slowly, taking a couple of ms to feed Vcc with 3.3V. U2 will stay disabled. The ATmega will then start and proceed to setup. When desiring to enter into sleep mode, the ATmega will power down the nRF24 tranceiver, enable it's interrupts, shutdown some of its features (like BOD, WDT, OSC, ...), pull-down PD4 to activate U2, and finally start sleeping. With Vcc close to 3.3V, U2 will quickly shut down U1, and the system will start living off C1 slowly discharge. Eventually, Vcc voltage will drop below 2V causing U2 to pull high !SHDN for approx 190 ms. U1 will then charge C1 and bring Vcc back to 3.3V. The interrupt on PD3 will cause the ATmega to wakeup and provide it with an opportunity to send a wireless message, or not. So, Is anything wrong with my reasoning? Are there any issues with the proposed circuit design? Are you aware of different (and perhaps simpler) ways of achieving this? Thanks a lot, any input on any of the above will be greatly appreciated. <Q> This excellent application note (refers to the MSP430, but is applicable in general to low power) explains this. <S> An important part of that application note states: <S> Designers must deviate from conventional thinking that efficiency is the most important figure of merit in a power system. <S> In a battery-powered system, battery current drain is the main concern. <S> Using a linear regulator with 1\$\mu\$A or less <S> \$I_Q\$ (sometimes listed as ground pin current) in this type of system will be both simpler and more efficient; the lower the duty cycle of the circuit, the greater the effective efficiency. <A> That's clever... <S> Now let's do some math: <S> The datasheet of the LTC1515 indicates, in shutdown mode and with Vin>5V, a 25µA max current. <S> That's something you can't avoid. <S> It is a max spec, and the typical value is not given <S> so let's say it's 5µA. <S> Then you have the TPS9837P which is always running, and it uses 6µA. With the efficiency of LTC1515 which seems to top out at 50% for 3.3V output and 6V input (and would be a lot lower for 9V input, I think), that makes ~3µA. <S> So, at least you have average 8µA consumed by this circuit, without the load itself. <S> This is quite good, actually. <S> Now, let's say we try to keep it <S> simple: type "nanopower regulator" in google, and you see the LTC3388. <S> It has less than 1µA at no load. <S> It is simpler to implement than the scheme you propose. <S> It has better efficiency when the load is active. <S> And the best part: It's not more expensive than the LTC1515. <S> I didn't do much research. <S> There may be more interesting choices than the LTC3388, depending on what you need exactly (LTC is damn expensive, there may be better compromises). <S> But if I were you, I'd try to keep it simple. <S> Edit Note <S> I had a deeper look at the LTC1515, because I had a doubt about its efficiency at high input voltages, and I actually realized in your case <S> , it's not even better than a linear regulator (see top of datasheet page 4). <S> It seemed indeed strange to me that it would be more efficient, because there are very few integrated charge pumps that can work as voltage halvers. <S> And those kind of pumps need at least two flying capacitors. <S> So you'd better take a cheaper linear regulator with very low quiescent current if you don't want the switching buck regulator, because LTC1515 provides no benefits in your case. <S> But it's not that inefficient actually. <S> Because your circuit spends most time sleeping, and given the sleep consumption of the Atmega328P, the consumption will be dominated by the quiescent current of the main regulator, not really by its efficiency. <S> So a linear regulator may be actually acceptable. <S> Do some math to check it, it depends on the duty cycle of the work/sleep states in your specific application. <A> Let's look at LTC1515, or switched-capacitor voltage converter in general. <S> This is easiest to explain if restricted to step-down operation which you are using it for. <S> How can you charge a capacitor to less than the input voltage (without inductor)? <S> You charge it through a resistive element and stop when the voltage reach the required level. <S> The full voltage differential is present across this resistive element. <S> Therefore a step down switched-capacitor voltage converter has the same efficiency limitation as a linear regulator. <S> Because of the extra complexity, with comparable technology, the switched-capacitor converter would lose out to a simpler linear regulator. <S> (One way to make a step-down switched capacitor converter more efficiency would be to charge a series of capacitors, then have a network of switches to shovel the connections such that the capacitors discharge to the output with closely matched voltage levels.) <S> I would suggest revisiting the linear regulator, find a really low quiescent current one and create a solution as the benchmark. <S> For example, along the way, you would need to define operating conditions such as: for every 10s, the circuit would go active for 1ms using 100mA. <S> If this is indeed the operating condition, this works out to be 10uA overall average. <S> Looking at just the resistors you have with your existing schematics, the current used by those resistors far exceeds that. <S> (I know you can refine those resistor values further, and a benchmark would be a useful tool for guidance for this and other design decisions.)
| Using a linear regulator can actually be more efficient in ultra low power battery applications.
|
Why aren't there electric cars with solar cells built on top of them? I think it would be nice if leaving the car outside on a sunny day would charge up the batteries. It would make possible long-range weekend trips to the countryside because during the 2-3 days of vacation the car would charge up. Or wouldn't? Please help me solve this mistery. I can't come up with numbers, I'm only guessing. I have the same question for laptops and cell phones. Wouldn't be nice if leaving them under the sun would charge them under a couple of hours? EDIT on 20-11-2016:There are custom-built solar-power-augmented EVs in the wild: The next generation Prius will have optional rooftop solar panel although it won't charge the main battery: http://www.autonews.com/article/20160616/OEM05/160619900/next-generation-toyota-prius-has-solar-roof-for-europe-japan <Q> First, let's have a look at current electric cars made for everyday's use. <S> I've choosen the Tesla Roaster, Model S and X and BMW i3: <S> Roaster S X i3Capacity [kWh] 53 85 <S> 75 <S> 18.8Range <S> [km] <S> 350 <S> 500 <S> 381 <S> 190Mileage <S> [km/kWh] <S> 6.6 <S> 5.9 <S> 5.1 <S> 10.1 <S> The BMW is trimmed to high efficiency, while the Teslas are larger and are a bit more racy. <S> And keep in mind, this is official data. <S> If this is as realistic as mileage data for petrol driven cars, then... <S> well... <S> Sunlight at the equator gives us about 1000W/m², but the efficiency of available solar cells is somewhere below 30%. <S> So you get just about 300W <S> /m². <S> Next, the sunlight has to hit the solar cell perpendicularly to get the max. <S> power. <S> But if you place the cells flat on the floor / car, they generate just a fraction of their max power most of the day... <S> In addition, weather reduces the power. <S> I've found a website which takes meteorological data into account to calculate the monthly energy output of solar cells. <S> The site is german (you can switch to english, but I didn't find this calculator on the english version). <S> It takes orientation (both 0 for flat on the floor), peak power in kWh (0.3) and position (click on map), and I got this min/max values for winter/summer. <S> Daily data calculated by division by 30: Monthly [kWh] Daily [kWh]California: 25 - 60 0.83 - 2.00New York: 15 - 43 0.50 <S> - 1.43Germany: 8 - 48 0.26 - 1.60 The most economical BMW i3 makes between 2.6km (1.6 miles) in german winter to 20.2km (12.6miles) in californian summer per daily charge from 1m². <S> I quess this car can not have more than 3m² of solar cells. <S> And keep in mind: This are average values. <S> What if one or more months are unusual dark? <S> And never park in a parking garage, in shadows of trees or buildings, ... <S> So, today, solar cells don't give enough energy per day for a car. <S> But they would cost a lot, while a full charge from the wall outlet is quite cheap. <A> In order to charge a car in any reasonable time frame, a very large solar array would be needed. <S> Much larger than the car. <S> On a typical car you would be able to get less than 1kW of solar panels. <S> It would take weeks to recharge the battery at that rate. <S> I do think it would be a good idea for utility vehicles on larger estates if they are not driven continuously. <S> You could just drive it around all the time and not worry about parking it at a charger. <S> Also, their batteries are a lot smaller than passenger cars, and they are not aerodynamic. <S> Often they have a rectangular roof canopy that would be perfect for solar panels. <A> Solar power at full sun is ~~ 1000 Watts/m^2. <S> Nest efficiency going is around 30%. <S> Most are rather lower. <S> 1000 <S> Watt x 30 % = 300 Watts/m^2 or 0.3 kWh per hour of full sun per m^2. <S> Full sun equivalent ours/day are lower than most people expect. <S> Typically 2 or 3 SSH per day in winter and around 6/day in many places. <S> Kabul is good at ~ 8 SSH/day midsummer, but then, you'd have to live there. <S> If you had say 3 m^2 of PV panels on a car - rather hard, and 6 SSH and 30% efficiency - all on rather enthusiastic side. <S> Energy/day ~= <S> 3 m^2 x 300 W/m^2 <S> x <S> 6 hours = 5.4 kWh/day <S> IF you could store this t 100% efficiency (you cant) and apply it to the wheels at 100% efficiency (you cant) then you'd get about 10 kW x 1/2 hour <S> OR 20 kW x 1/4 hour. <S> 20 kW may be OK for town driving. <S> So you may get 10-15 km on flat if all the lights were green. <S> Useful but not stunning. <S> In winter and cloudy days etc .... :-( <A> http://sunwindsolar.com/wp-content/uploads/2013/01/SOLARCAR.jpg <S> Here. <S> I saw that probably ten years ago. <S> The problem is in power. <S> Sun provides like 20kW of power to the whole car. <S> Maybe with today's technology you could gather 5kW out of it. <S> For driving it's not enough. <S> For charging- could be fine, but how many people would buy it? <S> Most of us want charging as quickly as on fuel station... <S> So for now it's nice, but has no market.
| Also, existing high-efficiency solar panels cannot be molded into sleek shapes that would allow a car to be aerodynamic. Depends on size of car etc.
|
What happen to electricity when you turn off the lights Sorry if this is the wrong place, but hopefully someone here can help me OK so the way I have been taught is that turning off the lights doesn't really actually save electricity due to the fact that all electricity is is electric fields pushing electrons. So when you shut off a light and stop the electron flow, the electric field is still there. So to my understanding the only real way to save electricity is to save it in a battery of some sort. Otherwise it just dissipates in the wires? Can someone pleas confirm or deny this. All that happens when I try and google it is I get a bunch of "10 awesome power saving tips" that don't really help me at all Again sorry if this is the wrong place. Hopefully this won't take up much of anyone's time. <Q> What turning the light off means is that you don't consume power in your home (so you don't get billed for it), so the power never gets generated at the power station, and fuel never gets burnt there to make it (which the power company then don't get billed for). <S> That's what 'saves' means in this context. <S> Most people would call what a battery does 'stores' rather than 'saves'. <A> Think of electricity as a handy on-demand way of moving work done in the generating plant into your home's lights and appliances. <S> That means they consume a little less fuel, and in the aggregate if a lot of other people go to bed too, some of the generators could probably be shut down for the night. <S> Of course reacting to demand changes is itself a challenge. <S> Some types of generators can be started or stopped more readily than others, so they make the adjustments with those first. <S> There are also some ways to store excess generating capacity at one hour for use at another - pumping water up hill (that can later generate power running back down), charging huge battery banks, spinning up flywheels - all expensive and having their inefficiencies, but in use or contemplated to some extent. <S> The other approach is to try to activate and de-activate various industrial power uses that don't have to be done at a particular time, to use the capacity of generators when they are under-loaded, and not use it at the peak of a summer day when all those air conditioners threaten to bring the grid down. <S> As for the power dissipated in the wires, at least for well-behaved loads that basically scales with the amount of work they are moving from one place to another, so when you turn off the light, most of the power lost along the way in moving that work to you stops being lost as well. <S> The only degree to which turning off the light is not effective comes from the fact that a practical generator's fuel consumption is only loosely correlated to its electrical load, and the overall operating expense comes not only from usage, but also from having it available to use. <S> And secondarily, to a very tiny extent, the miniscule power lost to imperfect insulators, induction, etc by having a grid of power lines, transformers, and compensating capacitors cycling at 50 or 60 Hz regardless if they are loaded or not. <A> In most cases,Modeling Electricity current with a water is a successful example. <S> Assume a bottle of water as a source,you when drinking it as a consumer,and your mouth as a conductor,when you turned it on(take off bottle's plug & drink) <S> you consume the source and when you stopped drinking IT is like YOU Then turned oFF the lights. <S> ________________________________ RM edited to above from " <S> ... is like hen turned on the lights. <S> " Is that what was intended? <A> Water analogy is imperfect but good enough. <S> Pressure = voltage, pipe flow resistance = resistance. <S> Current flow = current flow :-). <S> In a properly constructed water based energy transfer system power = pressure x current = <S> voltage x current energy = integral of power over time. <S> If you have a hydraulic motor on eg a loader it is powered when there is flow and not powered when the flow is turned off. <S> On both systems pressure/voltage without current/current does not result in energy transfer. <S> ___________________________ <S> No hens were harmed in the turning on of any lights.
| If you turn off the switch, you break the circuit, and your lights no longer request this work of the generators.
|
Why a neon lamp indicator on power strip switch flickers in the dark? Neon lamp indicators embedded in power strip switches indicate whether the switch is ON or OFF. When the neon lamp is ON in a dark environment (lights of the room are off) the lamp start to flickering and some times it goes off for a few seconds. When I switch the lights of the room on, it stops flickering and remains bright. Even when I unlock my phone in the dark near the lamp, it stops flickering! Is the mechanism of making brightness in neon lamps is related to the luminous intensity of the around environment? UPDATE: I captured my observations with my iPhone and uploaded it to the YouTube. Here is the link of the video: https://youtu.be/1jlUmEfGHZA All lights of my room was turned off. There was no connected plugs to sockets of the power strip. The only light was screen brightness of my MacBook Pro that was not connected to the power supply. I was decreasing and increasing the brightness of the screen during the video. The two last lights are the brightness of my phone. <Q> This is a common phenomenon: neon pilot lights have a limited lifetime, and after many years of use, they begin to flicker, then they finally go dark. <S> They no longer can operate at line voltage, but instead require a higher voltage for stable operation. <S> Also, neon pilot lights can act as photosensors. <S> Try this with a flickering neon bulb: <S> shine a red LED on it. <S> Then shine a blue LED on it. <S> If both LEDs are roughly the same brightness, then the blue LED should have a much greater effect. <S> You're seeing the same photoelectric effect that won Einstein the 1921 Nobel, leading to the "photon" concept. <S> All gas-discharge tubes suffer from a common problem: the gas atoms tend to become embedded in the metal surface of electrodes, so over many years the gas pressure falls as well as the gas mixture being altered. <A> To quote from the great wiki - https://en.wikipedia.org/wiki/Neon_lamp <S> "When the current through the lamp is lower than the current for the highest-current discharge path, the glow discharge may become unstable and not cover the entire surface of the electrodes.[6] <S> This may be a sign of aging of the indicator bulb, and is exploited in the decorative "flicker flame" neon lamps. <S> However, while too low a current causes flickering, too high a current increases the wear of the electrodes by stimulating sputtering, which coats the internal surface of the lamp with metal and causes it to darken. <S> The potential needed to strike the discharge is higher than what is needed to sustain the discharge. <S> When there is not enough current, the glow forms around only part of the electrode surface. <S> Convective currents make the glowing areas flow upwards, not unlike the discharge in a Jacob's ladder. <S> A photoionization effect can also be observed here , as the electrode area covered by the glow discharge can be increased by shining light at the lamp." <A> Responding to old post <S> but it comes up at top of google search: <S> Neon lights can't work in complete darkness. <S> This is the "neon lamp dark effect". <S> I know it sounds fake, but that's why neon lamp bulb makers sometimes add a little ionizing radiation inside the bulbs. <S> Dark Effect: <S> All ILT Neon lamps are subject to a condition called dark effect. <S> Dark affect is defined as a drastic increase in the amount of voltage required to make a lamp glow when the lamp is in a dark environment. <S> Neon lamps can also become erratic in total darkness. <S> To prevent dark affect an external source may be run near the neon lamps or in some cases, custom Neon lamps can be supplied with a radioactive gas, often Krypton 85. <S> ( Source: International Light Technologies, Inc. - ILT ) <S> Yes it is bizarre and a trouble-shooting nightmare for the unaware, lol. <S> But you can use a simple neon bulb as a photosensor. <S> But also they do flicker more and more as they age and die: but they not really dead, they just need more power than the device gives. <S> If you ramp up the juice (in a different device) they will work for a while longer.
| Because the lamp is photosensitive, it may require many additional volts to start if no light is present.
|
why do I get noise on the output of the LM386 when I connect speaker but not when I connect a resistor? I recently made a circuit to test out LM386 audio amplifier for the first time on solderless board. I relied on the datasheet circuit shown below: First I connected a 7.5ohm resistor at the output instead of a speaker. I get a clean sine wave at the output. The upper wave is the input while the lower wave is the output. However, when I replace it with this speaker, the output (and input) seems to become corrupted. Here is the scopeshot: (1) Why could this be happening? Besides this I am trying to figure out a few more things. (2) Why do we have the output connected to ground via a 0.05uF capacitor and a 10ohm resistor in most schematics in the datasheet? What purpose does this serve since this is clearly not a filter. (3) This is clearly a single supply amplifier topology. However, even if I provide it with a 100mV p-p signal, the negative cycle still shows at the output. How come??? (4) Why is output always capacatively coupled to a speaker? Is it because the DC signal may overheat the speaker? (5) The bypass pin is used to prevent signal degradation as per the data? It says this pin may grounded or a capacitor used. How do I know which one I should do? Finally, I made another interesting observation. Initially the output was completely corrupted, it was looking like this: Then I realized that oh there is noise on the supply rails. When I added a 0.1uF capacitor, the noise went away and the amplifier output became a sine wave (when driving a 7.5ohm resistor). The noise looks like this: I used AC coupling to be able to look at the noise. (6) Why does the supply noise above cause the amplifier to completely fail? Somehow without the smoothing capacitor, both input and output become corrupted. <Q> One aspect of the question: <S> Why do we have the output connected to ground via a 0.05uF capacitor and a 10ohm resistor in most schematics in the datasheet? <S> What purpose does this serve since this is clearly not a filter. <S> This is the "Zobel network". <S> It is for stability against oscillation; it provides a load for the amplifier at high frequencies (which the speaker doesn't, being an inductor). <S> This feature is very commonly seen in the designs of audio power amplifier output stages. <S> However, even if I provide it with a 100mV p-p signal, the negative cycle still shows at the output. <S> How come? <S> Though single supply, it is internally biased. <S> The thing to remember is that this is not an OP amp, even though the schematic symbol closely resembles one. <S> It helps to look at the representative schematic of the internals that is shown in some of the data sheets. <S> You will see certain features, like 50 \$\text{k}\Omega\$ biasing resistors on both inputs. <S> These generate a voltage when the bias current flows through them out of the transistor bases. <S> Note how the speaker load is coupled with a big capacitor; that's to remove the DC offset on the output. <S> Why is output always capacatively coupled to a speaker? <S> Is it because the DC signal may overheat the speaker? <S> That's basically it. <S> But not only overheat, but DC through a speaker generates a magnetic field in its coil, which reacts against the speaker magnet, producing a mechanical displacement of the speaker cone. <S> That could cause distortion, and lowered headroom. <A> I will answer your questions in the order that you asked them. <S> 1) <S> First you need to make sure you have a 470uF to 1,000uF capacitor from the power pin to ground pin. <S> That will get rid of the 'self-oscillation' effect. <S> 2) <S> The .05uF and 10 ohm resistor on the output act as a phase-shifter, to cancel out the internal phase shifting caused by the LM386's own design. <S> It is common to see this on many IC amplifiers. <S> 3) <S> The capacitor in series with the output not only blocks any DC voltage from going through the speaker, it effectively creates a new 'AC' based signal, with no DC component. <S> 4) <S> Because there is a DC voltage at the output equal to 1/2 your power supply, there has to be a DC "blocking' capacitor with a high enough value to allow some bass to come through. <S> 5) <S> You can let the bypass pin float, but a 10uF capacitor to ground (the pin gets the + lead) may help with noise. <S> Use it only if needed. <A> (1) Why could this be happening? <S> Instability caused by inductance of the wiring and a possible 'ground loop' (ground wire shared between input and output signals). <S> The circuit is oscillating because at a particular frequency the phase shift reaches 180°, turning negative feedback into positive feedback. <S> It only happens with the speaker because its inductance causes an extra phase shift. <S> The datasheet doesn't say anything about bypassing the power supply, but you should do it anyway. <S> A 100uF and 0.1uF capacitor in parallel should be enough (placed as close as possible to the IC). <S> Arrange your wiring so that power supply negative goes first to the LM386 GND pin and speaker, then to the input circuit. <S> This prevents the relatively high output current from generating a voltage in the sensitive input ground. <S> (2) Why do we have the output connected to ground via a 0.05uF capacitor and a 10ohm resistor in most schematics in the datasheet? <S> What purpose does this serve since this is clearly not a filter. <S> It bypasses the speaker at high frequencies, reducing the effect of its inductance. <S> (3) <S> This is clearly a single supply amplifier topology. <S> However, even if I provide it with a 100mV p-p signal, the negative cycle still shows at the output. <S> How come??? <S> The input has a PNP transistor connected in Emitter Follower configuration, so it can go below ground by about 0.6V (Base-Emitter voltage drop). <S> Since the amplifier has a gain of 20, this is more than enough to get full volume at the output. <S> (4) Why is output always capacatively coupled to a speaker? <S> Is it because the DC signal may overheat the speaker? <S> The output sits at half the supply voltage. <S> Connecting the speaker directly from there to ground would cause a high DC current to flow. <S> This could overheat the amplifier and speaker coil, as well as biasing the speaker cone in one direction which reduces its dynamic range. <S> (5) <S> The bypass pin is used to prevent signal degradation as per the data? <S> It says this pin may grounded or a capacitor used. <S> How do I know which one I should do? <S> The bypass pin connects to an internal power rail and must not be grounded. <S> The unused input can be connected to ground to prevent it from picking up noise.
| This is clearly a single supply amplifier topology. Putting a capacitor on the Bypass pin may also help.
|
How does a thermoelectric generator generates electricity? I want to know how a thermoelectric generator generates electricity. I know that it converts 'heat flow' to electricity.But some people say it consumes heat and converts it into electricity.The rest of the heat is dissipated through the other side. Which is true? <Q> It converts some (a small fraction) of the heat flow to electricity. <S> It absorbs X amount of heat from the hot side, and rejects X-delta amount to the cold side. <S> The output voltage is related to the temperature difference between the hot and cold sides. <S> The output current is related to delta, the heat flow difference. <S> Because present devices are very inefficient, without taking very careful measurements, measurements that are beyond most hobbyists to do accurately, it would be easy to conclude that it doesn't consume any heat at all. <S> X-delta is very similar in magnitude to X. A small difference between two big measurements <S> has been the bane of experimenters since antiquity! <S> It needs both. <A> It takes a part of the heat flow and generates electricity from it. <S> However, there must be a heat flow ; otherwise you'd have a perpetual motion machine of the second kind , and that is impossible. <A> This seems to really be a physics question about heat engines. <S> Anything that converts energy from a heat difference into some other form is a heat engine. <S> In this case, you are apparently asking about one of the ways of converting a heat difference into a electric potential directly. <S> What the output is (electric power, rotating shaft, etc) makes no difference. <S> When the input is a heat difference, you have a heat engine. <S> Heat engines are limited by the Carnot efficiency : Efficiency = <S> (T hot - T cold ) <S> /T hot <S> The temperatures must be in a linear absolute scale, like degrees Kelvin. <S> For example, to extract power from reservoirs of boiling water (373°K) and ice water (273°K), the best you can ever do is 27% efficiency. <S> This means that if you were to extract 10 W from this setup with a theoretical perfect heat engine, there would actually be 37 W of heat flowing from the hot reservoir to the cold reservoir. <S> This basic physics applies to heat engines with electric output too, like thermocouples and thermionic generators. <S> The main difference is that real heat engines won't achieve 100% of the possible Carnot efficiency. <S> In a thermistor, for example, there is significant heat flow along each wire that is not converted to a electric potential. <S> The heat gradient along the wire produces the electric potential, but also provides a heat leakage path.
| So people that say 'it converts heat flow to electricity', or 'it converts heat to electricity', are both correct, up to a hand-wavy approximation.
|
Are there any advantages to either DC barrel connector polarity? I was putting together a fairly basic LBB at the weekend to control a handful of DC devices which got me thinking. Why are both pin-positive and pin-negative DC barrel connectors used? Are there any concrete, electrical advantages to either design? Or is it just for manufacturer lock-in? Pin-positive seems to me to be the obvious one to use from a safety point of view, but a quick audit of the supplies in my spares box revealed a not insignificant number of pin-negative supplies. <Q> Image from: http://www.canadarobotix.com/image/cache/data/products/400/437-1-800x800.jpg <S> On the figure above, on the left, a connector with no plug inserted. <S> Pins 1 and 2 make contact. <S> On the right, with a plug inserted, pins 1 and 2 makes no contact, and the plug outer ring makes contact with pin number 1. <S> Some connectors have a third pin that can make contact with one of the others when no cable is inserted and is disconnected when you put a cable in. <S> This is often used to handle dual power (mains brick and battery, for example) or to signal to a processor that a cable is plugged in or no. <S> This third pin is usually connected to the outer ring. <S> If you need to disconnect the positive battery pole, you will have to make your polarity negative-tip. <S> This is one of the reasons I can think of having this kind of polarity plug. <A> "Pin-positive seems to me to be the obvious one to use from a safety point of view" I don't see that point regarding safety. <S> I think either is equally safe. <S> The danger is in mixing up the polarties especially for devices that cannot handle inverse polarity properly (this can be fixed with a diode and a fuse). <S> In my opinion the center negative solutions are just WRONG . <S> But if there is, please let me know. <S> The center pin should be the positive and if everyone stuck with that the world would be a better place. <A> Voltage is relative, not absolute. " <S> plus" and "minus" were assigned purely arbitrary. <S> No physical significance whatsoever. <S> Outer barrel is very easy to short circuit to the chassis, as it often partially sticks out. <S> So the only thing that matters is "what has been chosen as GND for this particular LBB?". <S> For circuits with positive GND a center-negative PSU is a no-brainer. <A> Most automotive DC-DC adapters are not isolated. <S> Otherwise, there is no great advantage one way or the other, however my observation is that the vast majority of adapters manufactured in the last decade or so are centre-positive, so there is often some advantage in going with the crowed. <S> On the other had, we were able to buy up a large supply of center-negative adapters a few years ago at a very attractive price and use them with a product that can accept either polarity. <A> The solution to this problem is AC, put a diode bridge between your connector and device so that it can use either flavor of DC wallwart.
| Center positive is no safer than negative. If your LBB might be used in a car, and by the sort of person who would buy a non-isolated adapter, then center-positive is safer, since the exposed metal could otherwise short to exposed metal (if you can actually find any in a modern car!) and blow the fuse in the adapter. Most automotive outlets are 12V negative ground. There is no reason for it.
|
Can I charge a 12v toy with USB? I have this small palm sized toy drone. It uses a 'charger' which houses 8 AA alkaline batteries. The charger plugs into the drone and the charger is activated by a toggle switch. I want to replace the charger with a USB cable that plugs into my PC monitor or even wall socket. Can I hack this existing charger into using a USB power source instead of 8 1.5v alkaline batteries? A little bit of 'f it, I'm gonna try it' resulted in this. <Q> Edit: Hold on, wait. <S> Normally 8 AA batteries equals 12 volts DC. <S> But looking at the parts involved, it's possible the batteries are in series-parallel. <S> I would reassemble it as-supplied, put the 8 batteries in, and measure what's coming off the end of the batteries. <S> If it's 12 volts (all batteries in series) nothing easier. <S> Dive into your shoebox of old wall-wart power supplies. <S> One of them is bound to be 12 volts DC, since that is the most common voltage. <S> Hack off the wires... <S> or if you want to retain it for its original use, visit Radio Shack for a socket compatible with the plug <S> (the 12V DC ones are fairly standard). <S> These can also be bought new if you never save them, or ask your friends or neighbors if they have any. <S> If it's 6 volts (2 parallel strings of 4 each), you can use a common wall-wart that is 6 volts instead of 12. <S> These are far less common, but you can buy new ones. <S> But try this: test the product again, only use NiCd or NiMH AA batteries. <S> These cells are 1.2 to 1.25 volts per cell, or in a stack, 4.8 to 5 volts. <S> Exactly the voltage of USB! <S> So if the charger and drone work normally and to spec on NiMH/NiCD batteries, they should adopt straightaway to USB charging. <S> I would use a 2A supply like an iPad charger, and not a computer port. <S> The general concept here is to supply power in a way which emulates and replaces the AA batteries, and do not tamper in any way with the factory as-designed circuitry which is responsible for safely charging a lithium battery. <A> Are you able to measure the output voltage of the charger or check how many battery cells are in the drone? <S> Otherwise a cheap boost regulator should do the job, nothing fancy as the voltage from a AA varies from over 1.6V down to less than 1V per cell as you drain it, so you've got some margin to play with. <A> USB nominally outputs 5V. <S> Even a near flat 12V battery could easily damage your USB port. <S> So only plug into chargers that you can afford to destroy.
| Definitely do not plug into a computer unless you want to destroy the USB port or worse still the computer. If the drone has a single 3.something volt Li-Ion battery, the charger can probably be persuaded to work on 5V (if the AA's are arranged to give 6V, you might not even need to mod the charger at all).
|
Should a regulator be added to a circuit where the power source voltage matches the ICs? I am trying to design a very simple circuit consisting of an accelerometer IC and a Bluetooth IC, all rated at 1.8 up to 4.something volts and thinking to use a single CR2032 cell. Seems like these usually come at 2.8 - 3V and about 250mAh current, depending on manufacturer and other factors. I have already planned to add required capacitors around each IC, but I am wondering if I should still use a voltage regulator with this circuit? So far the only good reason I could think of would be to prevent frying the ICs in situations where the user would manage to somehow cram up 2 batteries instead of one in the enclosure, aka making the circuit idiot-proof. <Q> If those chips are meant to run over that full range, then I would take advantage of that. <S> They were probably designed that way to address battery-powered applications. <S> If you want over-voltage protection, put a zener diode across the battery input. <S> That would also give you reverse-battery protection and won't draw additional power (except possibly for a little leakage). <S> Select one whose zener voltage is greater than one battery and is less than two batteries. <S> The current carrying capability should be greater than the current capability of the battery. <A> Most available linear regulators have a "drop out" voltage by which the output will be less than the input. <S> This can range from under 100 mV for good examples lightly loaded, to well over a volt for others. <S> So putting one in means you reduce the available voltage, which may not be something you want if it is already in range. <S> If all of your parts can tolerate the open-circuit voltage of a new CR2032, then you probably do not want a linear regulator. <S> There are, however, some tiny switching regulators designed to save power when running devices that can operate in the sub-2v range, from a coin cell by efficiently reducing the voltage to make them consume less energy. <S> Good low power design of course is an art that extends beyond the supply - you need to make sure never to leave a voltage across a pulling resistor, and to sleep your processor and its main clock the overwhelming majority of the time. <S> More importantly, if you ever think of using a different chemistry, such as lithium ion or lithium polymer, then the voltage of a fresh cell may be around 4.2v and so likely too much for some of your components. <S> In that case you might try using a very low dropout 2.8v or 3.0v linear regulator, or else looking at the switching solutions. <A> From your description, you do not need a voltage regulator on a single 3.0V lithium primary cell to prevent damage to your two devices. <S> Note that there are still some reasons you may consider regulating the cell down to a lower voltage: <S> The quiescent current of the circuit may be lower at a lower V DD (though you have to add the regulator current) <S> The circuit may change performance as V BATT changes (due to state of charge, or large pulse currents) <S> The circuit outputs an analog voltage proportional to V DD <S> You want a lower transmit power from your Bluetooth <S> That being said, you probably don't need to include a regulator for your application.
| Adding a voltage regulator would dissipate more power (although not much, if it's a good switching regulator), which isn't desirable in a battery-operated device.
|
Why is one return conductor for each forward conductor in a ribbon cable better than a shared return conductor? I don't understand why second way of implementing is better than the first one? I have my own basic logic regarding this topic, but i am not confident if what i think is correct! so i was looking forward to have some general view on this topic! <Q> Ask yourself which cable is likely to have the lowest loop area: <S> A large loop area has greater inductance and can emit more EM interference. <S> It can also receive more EM interference. <A> If you were transmitting something of the order of a 10s to 100s of kHz, number 1 may be perfectly adequate, but number 2 does have the advantage that each signal has a return path that is closer and therefore reduces the total current loop. <S> This is useful in minimising radiated emissions (and radiated susceptibility too), to say nothing of crosstalk . <S> I would not say they are perfectly balanced in number 2 (Signal 1 has only return 1 <S> , signal 2 has return 1 and return 2). <S> That said, the signals are shielded more effectively in this configuration. <S> In number 1, the single return would have to carry all the return currents and would need some careful checking to ensure the path is capable of the current loading. <S> In a high speed world, things would have balanced return paths for both impedance control ( <S> the distance to the return path is an important component of the track impedance) and to separate the return currents for various reasons. <A> The second alternative provides far better shielding. <S> Since the current in signal <S> N and return <S> N flows in opposite directions, their EM fields cancel each other out. <S> So in the first example, when signal 1 is active, signal 2 is more exposed to crosstalk as it is in the second example. <S> As the picture title says, a single return path creates unequal (and much larger) loops than dedicated return paths. <S> If the cable in example 1 is exposed to external EMI source, signal 1 will receive four times as much interference as signal 4 will. <S> In the second example, both signals will receive the same, minimal amount of interference. <A> Basically, when the path that current takes encloses an area, this creates inductance. <S> Inductance is like electrical inertia, it prevents the current from changing quickly (like how regular inertia slows down speed changes), this electrical inertia rounds off square pulses and attenuate fast signals, worse <S> still the current still want's to flow even when the load is removed, creating voltage spikes. <S> Loops also act as good antennas and will broadcast their contents all over the place. <S> These effects get worse the larger the loop area (larger inductance). <S> Having a return right next to each signal keeps the loop area really small and keeps the signal lines isolated from each other - each return is essentially a shield. <A> I'm surprised no one has mentioned differential signalling. <S> https://en.wikipedia.org/wiki/Differential_signaling If you want fast clear signals, you use this technique which requires the "return" line be used in an active manner. <S> In this case, you wouldn't want multiple drivers of the single return line. <S> From source:"The technique minimizes electronic crosstalk and electromagnetic interference, both noise emission and noise acceptance, and can achieve a constant or known characteristic impedance, allowing impedance matching techniques important in a high-speed signal transmission line or high quality balanced line and balanced circuit audio signal path." <A> In addition to the EMI issues, wires which run parallel to each other will have a certain amount of capacitive coupling. <S> Capacitively coupling signal wires to ground will increase the amount of energy lost to the resistance in the cable or signal source, but that can be dealt with much more easily than crosstalk caused by capacitive coupling of signal wires to each other. <S> In clocked parallel protocols which are slow enough that data wires will have time to stabilize before their clock arrives, it may not be necessary to worry too much about crosstalk between data wires, but modern protocols are more prone to send lots of data quickly over a few wires than to send a smaller amount of data more slowly over each of a large number of wires.
| If each forward conductor has its own return wire then this potentially minimizes each circuits loop area.
|
Cheap, low-quality constant-current sink over wide voltage range I'm looking for a cheap (as in: low component count, low PCB area, low component cost) way to put >= 1mA of load on a voltage that may be anything from 1.2 to 17 volts (minimum required load on a LDO output). The easiest solution is a simple 1.2kOhm load resistor, but the power consumption becomes enormous at higher voltages (240mW at 17V, which leads to a huge package size). A JFET-based current-limiting diode seems ideal - but I can't find a vendor that actually sells one. So my idea is to use a BC817-16 NPN transistor, which has current gain between 100 and 250 , send 10-15µA flowing into the base from a nearby 3.3V rail via a 220k resistor, and just watch 100 to 250 times that (1mA - 4mA) flow from the collector to the emitter, no matter the collector voltage: simulate this circuit – Schematic created using CircuitLab The circuitlab simulation indicates a current of about 4mA for 17 volts, which gives a more than acceptable power dissipation. However, one of my (much more experienced) colleagues strongly advises against this technique, claiming that there will be runaway effects and the transistor will blow, so... will this circuit work reliably? Any other suggestions? <Q> No, don't try to use the constancy of HFE to run your transistor, it varies with temperature, voltage, device, too many things. <S> This is a standard technique for biassing transistors and creating current sources. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The two diodes maintain around 1.4v on the base, or 0.7v across R1. <S> With RE=130ohms, that's around 5mA. <S> It will vary slightly with VBE and the diode drops. <S> Change the value of R1 for other sink currents. <S> R2 needs to be low enough to conduct at least Iout/HFE with 1.9v across it, ideally 5x that, so 5mA Iout, 100HFE, IR2 should be at least 250uA, R2=6.2k. <S> Depending on the VCEsat of the transistor, this will sink current down to around 1v. <S> The transistor needs to be heatsunk to tolerate 16.3v * 5mA of power dissipation within it. <S> You can replace the two diodes with another resistor if you want, which gives more flexibility for Vre, at the cost of a little thermal and HFE stability. <S> If you are happy to sink current only over the range 3v to 17v, then you can connect the base directly to the 3.3v, with 2.7v across R1, resized accordingly. <A> Yes, you can do that, you can even buy transistors with more closely defined hFE. <S> For example, the 2SC1815Y is guaranteed to have an hFE within a 2:1 range (120~240). <S> It might vary another 2:1 from -25°C to 100°C, so that's a 4:1 range total, plus perhaps 50% for Vce variation. <S> In SMT, the BCX70x is even more tightly specified (< 2:1 at 2mA). <S> It's generally frowned upon by folks who have never done the math, but it will work if you're not that picky (5-6:1). <S> Expect a lot of clucking. <S> This crappy method of biasing transistors has been used in consumer electronics in the past many times. <S> The JFET regulator diode probably won't work because the 1.2V is too low to get it to regulate well. <S> Using a dual transistor (eg. MBT3904 ) with the below circuit gives you a lot better control. <S> The current will probably change 10-15% over temperature (simulate it if that matters to you). <S> I did the below for 2mA nominal. <S> The self-heating will cause the current to increase somewhat over time, depending on the voltage, but it still should be pretty good. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The J505 is still available I believe here; Link to RS . <S> It's a JFET device with gate connected to source like this: - And has a VI response like this: - According to the data sheet it typically draws 1 mA and works across a range of voltages from about 1.2 volts: - <S> All the way up to 100V
| It's best to use a dual so they will track thermally. As you have 3.3v available, use emitter degeneration like this.
|
How can I tell if a serial connection is sending bytes okay? I have a Teensy connected to a Pololu motor controller. If I send bytes over USB (using Pololu Serial Transmitter v1.3) I can get it to work (For instance, 0x85 0x00 0x20 makes the motor run at slow speed). However when I attempt the same thing in my sketch I cannot get it to work. This leads me to the conclusion that there may be a problem with my serial wires or with the motor control board itself. So I'd like to be able to test that the serial connection is working, can I accomplish this with a multimeter or something? My code, just in case there's something wrong with that is: #define rxPin 10#define txPin 9SoftwareSerial smcSerial = SoftwareSerial(rxPin, txPin); loop: smcSerial.write(0x85);smcSerial.write((byte) 0x00);smcSerial.write(0x20); <Q> If you are using the auto-baud rate setup for Pololu motor systems you first need to send "a baud rate indication byte", basically you send an 0xAA first, and after any power on or reset. <S> See descriptions of serial settings here: <S> https://www.pololu.com/docs/0J44/6.1 <S> It does not seem to say if the "indication byte" is acceptable with every transmission. <S> To physically verify transmission from your Teensy controller you could also put an LED and resistor on the transmit line (say with a 500 ohm resistor to gnd or V+). <S> The LED should flash briefly with each transmission packet, though at higher baud rates the flashes will be quite short, (so if that is a problem do some testing at lower rates for verification). <A> I cannot comment yet, so I'm posting this as an answer: <S> Did you forget smcSerial.begin(9600) in your setup() <S> function? <S> Also, make sure you pass the proper baud rate to <S> begin() , 9600 is just an example. <S> And no, a multimeter would be of no use troubleshooting this problem. <A> Traditionally DSO are quite pricey. <S> But recently there have been a few hand-held designs coming out of the hobbyist/maker market that are good enough to troubleshoot most digital communications. <S> Google "DSO nano" and "DIY DSO". <S> If even hobby grade DSO seem too much to spend on you can always use some software to transform your sound card into an oscilloscope. <S> In fact, this is what I've done. <S> Note that your sound card is crap at capturing exact voltage values because the input is capacitively coupled. <S> But it's good at capturing digital serial data - enough that you can actually decode the data yourself. <S> I've only ever used XOscope on Linux <S> but you can google "software oscilloscope mac" or "software oscilloscope windows" in you <S> need similar software on other platforms.
| Overall it may be best to just set the system up to a known baud rate ahead of time, or use the default (which seems to be 9600). My go-to tool for testing serial data and/or pulse modulation is to use a digital storage oscilloscope.
|
Antistatic mat , how it works? I have a large anti-static mat lying on the table of my workdesk, which looks like this. And it's connected to ground through the wall outlet from one point, meaning it's an open-loop: Let's say I connect myself to the mat using an anti static wrist strap , how is it supposed to help me discharge my static charge when there is no closed loop between me, the mat, and the ground? And lets say I am using a power supply which I am feeding from the mains, and the positive clip of the supply touches the mat accidentally. Wouldn't it be a dangerous scenario for me to be in contact with the mat? I am quite puzzled with the usage of the anti-static mat and nature of the electron movement when there is no closed path. Have a nice day people :) <Q> First, the mat (and the connection to mains earth!!) should have a resistance that is high enough to cause no (extra) harm to you when 'normal' Voltages (let's say up to mains level) are involved. <S> Think Mega-Ohms, and preferably more than one resistor in series to counter single-point-failure issues. <S> As for the open/closed loop part: think of your body as one plate of a capacitor, the other plate being the earth (in the form of all earthed metal in your surroundings). <S> Now when you touch your mat, there is a closed path. <S> The part that you think of as open (you and the earth) is a capacitor. <S> Another way to look at it is that for static electricity (as opposed to the more-or-less continuously current carrying electronics we are used to) a closed path is not a requirement. <S> An excess of charge tends to spread itself out and preferably find a place with a shortage of charge to combine with. <S> This is a temporary phenomenon, which lasts until the charges are evened out. <A> Let's say I connect myself to the mat using an anti static wrist strap, how is it supposed to help me discharge my static charge when there is no closed loop between me, the mat, and the ground? <S> As Wouter says, your body can behave like a capacitor, storing electric charge. <S> A good example of a body storing charge is in the winter, removing a sweater or shuffling your feet across a carpet, then touching a doorknob - ZAP! <S> That tiny blue spark is thousands of volts of static electricity jumping off your charged body. <S> So if you put on the wrist strap or touch the mat, those thousands of volts are harmlessly coupled to ground. <S> Current does flow - from your body, through the mat, and to ground - simply because your body has substantially more charge on it than ground does, which is an imbalance - and charge prefers to equalize. <S> And lets say I am using a power supply which I am feeding from the mains, and the positive clip of the supply touches the mat accidentally. <S> Wouldn't it be a dangerous scenario for me to be in contact with the mat? <S> If the mat were very conductive, yes. <S> But the mat isn't - it's only very slightly conductive - the equivalent to Megaohms of resistance. <S> And it has to be, to dissipate those thousands of volts safely and slowly. <S> The resistance is so large, that any type of "shorting out a benchtop power supply" would be imperceptible by us. <A> The lumped element model we all learn where current flows in loops and components have no net charge is just that, a model. <S> For many purposes it's a very good model but it doesn't cover every case. <S> In reality net charges <S> can and do build up on objects. <S> These charges are small but the voltage (potential differences) can be very large. <S> We call this "static electricity". <S> When two conductive objectsat least one of which is carrying an electric charge come sufficiently close to each other the insulation between them will break down and the voltages will equalise. <S> That breakdown can cause serious damage to electronics. <S> The point of an antistatic mat or wrist strap is to provide a path to earth that is neither a good conductor or a good insulator. <S> This allows the charge to bleed away slowly and the item to reach an equipotential with the earth without causing damage. <A> The wrist strap (connected to the mat) ensures that you and the electronics (which are on the map) are on the same potential level. <S> Without a difference in level, there's no discharge. <S> That's all cool, but it only allows the small local system (you + mat + electronics) to be on the same level. <S> That other guy next to you might be on some entirely different level. <S> If he touched your stuff, things could still fry. <S> If all mats are on ground level, they are all on the same level. <S> the wrist strap pulls your potential to the mat the ground wire pulls the potential of the mat (and thus you) to ground, where all mats are
| The antistatic mat is very slightly conductive.
|
Why put a voltage divider between a battery and the battery pin of an RTC? I bought a PCB module including a DS1307 real-time clock to evaluate if that RTC is right for the system I'm designing. Here's part of the schematic: The DS1307 works with a battery voltage between 2 V and 3.5 V. I use the module with a 3 V coin cell battery. Resistors R37 and R38 make a voltage divider that turn the 3 V of the battery in about 2.5 V, and feed such voltage to the VBAT pin of the DS1307 real-time clock. What's the use of such voltage divider? Why not connect the 3 V battery directly to the VBAT pin? EDIT : I added the 5 V flag to the voltage live on the top of the picture. <Q> That module design with the voltage divider is TERRIBLE. <S> It puts a permanent load on the battery when the 5V is off shortening the battery life significantly. <S> In addition it reduces the usable voltage range of the battery again shortening its life span. <S> And lastly the circuit forces a charge current into the battery when the 5V is on. <S> Almost all coin cells are NEVER supposed to be charged. <S> Here is how the circuit should be changed: <A> If the system is powered from a 5 volt supply, the divider is necessary to limit the voltage at VBAT. <S> The voltage after the diode D4 will be somewhere between 4 and 4.5 volts depending on the current and diode, which is above the 3.5 volt maximum. <A> This board design is indeed full of silliness -- in fact, the DS1307 RTC chip has a UL recognition to not reverse-charge lithium coin cells that is rendered void by this circuit! <S> For a standard, non-rechargeable coin cell, simply remove R38, R39, and D4, and replace R37 with a 0-ohm jumper. <S> As to why the circuit was added? <S> Still, it is a poor design -- a better solution would have been to use a fixed 2V shunt ref such as an LM4050-N-2.048 as if it were a low-current Zener <S> (Zeners and forward-biased diodes are both notoriously sloppy at low currents) with a Schottky in series with it to prevent the battery from trying to power up the rest of the board, as well as a 1K resistor for supercap applications to limit inrush currents. <S> simulate this circuit – <S> Schematic created using CircuitLab
| I suspect the board was originally designed to work with supercapacitors or rechargeable coin cells, and the DS1307 lacks on-chip support for charging these.
|
Using a battery charger instead of a battery? I have built a simple circuit with two diodes that light up intermittently. It requires 9V DC voltage to run. I don't have a DC source at home, and would like to avoid buying 9V batteries repeatedly to power my future projects, as portability is not necessary. I however have a battery charger (wall adapter with battery slots), which can also recharge 9V batteries. I thought that battery powered devices are often able to run plugged in without the battery, so this should be the case here, too. So I tried to connect the wires where the 9V battery to-be-recharged would normally fit, but it's not working as expected: D2 stays on indefinitely, while D1 flashes for a brief moment about once a second. Note: The charger has a led indicating whether the battery in the respective compartment is charging. It does NOT light up when the circuit is connected. Normally, they are supposed to flash intermittently, about one second each. (The circuit works correctly, tested with a 9V alkaline battery). I don't understand why this happens, and would greatly appreciate some help! <Q> After your edit to mention that it works correctly on a 9V battery, I expect the problem is that your 9v charger has some "smarts" and only charges when it "thinks" it sees a 9V battery attached. <S> The charger could be checking a few things to see if it should start the charge cycle. <S> Existing voltage - the charger checks to see if there is a partially charged battery attached and only starts charging if there's some voltage coming in from the battery connector. <S> A "dead" battery isn't really zero volts (it would be 6 or 7V for a 9V battery.) <S> Check for a load, and shut off if there's no load attached <S> - the charger thinks there's no battery to be charged and shuts off. <S> Stop the charge cycle if the load is too low <S> - it thinks the battery is charged. <S> Number 3 is the most likely, follow by number 2. <S> Thought of a number 4: <S> The charger is putting out pulsating DC and keeps resetting your circuit. <S> In this case, the capacitor that @JIm Dearden mentions would help. <A> As already stated, a modern 'smart' battery charger won´t make a good power supply. <S> A smoothing capacitor across the output may work, but maybe you should consider some other options, especially for the future: <S> You could: 1) <S> Buy a couple of PP9 rechargable batteries to use with your projects - and so make proper/full use of your charger. <S> so could be used for a variety of projects. <S> Or better: 3) <S> Build a small, variable, power supply. <S> There are plenty of simple circuits out there (eg based around the LM317). <A> Your circuit will work at 5V too. <S> It may flash a little faster and the LEDs may not be quite as bright but both those issues can be adjusted by tweaking the values of the 75K and 820 ohm resistors accordingly. <S> I mention this because of the plentiful supply of cheap 5V USB device chargers around these days. <S> One of these may be a good alternative to powering your circuit.
| 2) Buy a 'wall-wart' power supply - they are quite cheap and most have multiple voltage outputs If you are going to experiment with electronics, a variable power supply is pretty much essential.
|
I can't understand the logic with power = voltage * current I mean, if I have a 2000W device and plug it in a 220V outlet, it will produce about 9,1A. But if somehow the outlet is not delivering 220v, but less, like 180V, according to the equation, it will produce 11,1A. But how decreasing voltage increases current if voltage is the 'provider' of current? <Q> If the device in question actually draws 2000W fixed, then you are correct. <S> For example, if I have a 48V switching supply connected to a 1.152\$\Omega\$ resistor, the input current will increase as the input voltage decreases to maintain the same output power. <S> However most devices do not behave that way . <S> The current will decrease as the voltage decreases, and the power will decrease even faster than the current (with the square of the voltage for a resistor, somewhat less for most heaters because the resistance drops with decreasing temperature). <S> If I have a 220V heater rated at 2000W it will be about 24.2 ohms. <S> If it stays 24.2 ohms, on 180V it will draw 180/24.2 = <S> 7.44A and dissipate about 1339W <S> (P = I * V). <S> You can also calculate it from \$P = <S> (\frac{180VAC}{220VAC})^2 \cdot 2000W\$ <S> = <S> 1339W. <A> Decreasing the voltage doesn't increase the current. <S> It decreases the current and power (in a resistive model for simplicity). <S> I= <S> V/R. <S> The resistance is nominally constant, so the lower the voltage the lower the current. <S> Thus the product of voltage times current is also lower, in fact it decreases by a square law. <A> That is not the case. <S> That is, your device doesn't always put out 2000W or whatever the maximum output power is. <S> I am assuming here that your device supplies power, e.g. power supply, not a load in this example... <S> Your device can deliver up to 2000W at 220V <S> , that is its maximum capability. <S> That is the max power it can provide. <S> But even if you were supplying the device 220V, it doesn't mean it has to output 9.1A (or 2000 Watts in terms of power). <S> The load will determine how much current it needs to operate. <S> You can have a load that will draw 5 Amps, that just depends on the load. <S> If that's the case, then your device will be providing \$ P= <S> (5A)(220V)= <S> 1100W\$. <S> You would reach the device's max rating if you keep adding loads or a single load that takes up the rest of the power. <S> Let's say you add a load, on top of the previous one, that draws 6 Amps. <S> Now, that last load is using up 1320 Watts. <S> Since you already had a load at 1100 Watts, this last load will max out your device (1100W+1320W= 2420 Watts). <S> Your device will provide as much as it can which is 2000 Watts. <S> Although pushing it past its max rating will probably burn it out. <S> Hope it helps.
| If you lower the voltage, the current also goes down.
|
Double-throw switch with no common connection? I need a DT, ON-OFF-ON switch to control two independent circuits with no common connection. I need something that behaves like either A and B, or C and D in the diagram below. Does such a thing exist? This seems pretty straightforward, but I haven't been able to find one. In the diagram, A closes one circuit and B closes the other; the other possible configuration would be one in which C closes one and D closes the other. update: Thanks for the suggestions. Here's how the switch I bought works and how the pins are configured. I just need to put a jumper across 2 and 5, which are common, and connect one circuit to 1 and 4 and the other circuit to 3 and 6. <Q> simulate this circuit – Schematic created using CircuitLab <A> An ON-OFF-ON switch is a three position switch, with the middle position disconnecting everything. <S> If that's what you want, then you'll need a Double-Pole-Double-Throw switch with a center-off position. <S> When you go to wire it, if it's a conventional switch, the wiring should go like this: <S> In any case, you should test continuity with the toggle in its various positions in order to decide which circuit goes where. <A> You just described a common DPDT switch, just with center-off. <S> It's even simpler. <S> Here I've drawn in how to do it on your switch's diagram. <S> I also scrawled in an old fashioned knife <S> switch with brass blades and a purple handle (for Prince) <S> so you can visualize what's going on. <S> Purple is an insulator.
| There's an excellent selection at DigiKey A Double-Pole switch comprises two independent switches in the same housing, and most DPDT switches have the commons as the center terminals and the made contacts on the opposite side of the switch with reference to the tip of the toggle.
|
Calculating Base Resistor for Darlington Pair So I know lots of people have asked this question but the more answers/threads I read the more confused I get because people explain it so differently. My confusion is why do I have to even use the hFE Current Gain in any part of my calculation for the base resistor value? My thought process is since Vbe is 1.4V (0.7*2 transistors) and my arduino output is 5V then lets say "I choose" to have Ib be 2mA then by ohms law R=(5-1.4)/2mA=1.8K ohms. Does this thought process not work? The purpose of the resistor is to limit current to the transistor, well if I decide to limit it to 2mA then I have all my parameters to choose R without having to use hFE at all. <Q> hfe is specified for a certain voltage and current. <S> For the TIP120, this is a Vce of 3 volts. <S> Problem is, you don't want to run your transistor at 3 volts. <S> You want the transistor to be in saturation, with Vce as small as possible. <S> Looking at saturation numbers, you'll see a specified base current and collector current. <S> As a good rule of thumb, use the closest numbers to your application. <S> Then, you can compute your base resistor as you suggest. <S> When using single transistors, the general rule is to assume an hfe of 10 to drive the transistor into saturation. <S> This is not a hard and fast rule, but it will consistently give solid performance and low saturation voltages. <A> The TIP120 incorporates two base-emitter resistors (120 ohms and 8K), which minimize leakage, but reduce the effective hFE (and, more importantly here, give it a kind of threshold that is relatively high compared to an ordinary darlington). <S> Figure 2 in the datasheet shows typical Vce with Ic/Ib = 250, as does the guaranteed Vce(sat) numbers in the datasheet - (both guaranteed numbers are Ic/Ib = 250) <S> so I would suggest that as a useful guideline. <S> You will get almost no improvement in Vce(sat) numbers by going for more than 1/250 of the collector current, and you may get significantly worse if you go much less, so it's safest to go with 1/250 in the 1A-3A 'wheelhouse' of this part (here is the only place <S> the hFE curve <S> Fig 1 comes in handy), perhaps a bit more at higher or lower currents. <S> That's probably close enough unless you need to operate at low temperatures where hFE will drop. <S> So, at 1A out you need about 4mA base current. <S> The microcontroller output won't be quite 5V when supplying 4mA, so if you use (4V-1.5V)/0.004 = 625 ohms (pick the next lowest E24 value 620 ohms) and you should be fine. <S> Note that the 1.5V is the typical value from the graph in Fig 2 above, I didn't calculate it. <S> The worst case value is 2.5V at 3A current (vs. 1.7 <S> typically)- using that would be more conservative again, but it gets you down into the 330 ohm range and is probably unnecessary for an ordinary design. <A> Transistors are current controlled devices that is Ic <S> =Ib*Hfe so to give the right value to base resistor you should start from current required by pump so you will know Ic. <S> From Ic you could define Ic= 10*Ib for giving minimum Vce drop at saturation, but with darlingtons probably you could use Ic=100*Ib. <S> Anyway mosfet are better switches in these applications because voltage drop is much less. <A> Well, first off, the LM137 is a negative voltage regulator and it's a 3 terminal device, so <S> I don't know why that ground - or a positive power supply for that matter - is connected to it. <S> Assuming you made a trypo <S> and it's supposed to be an LM317 <S> , it can't source more than about 1.5A, so if we assume your motor draws about an amp once it's running, we can start to think about the TIP120 as a switch. <S> More often than not, when you design, you have to work backwards, and since we know the motor's the load and we have to - somehow - drive it, this is one of those cases. <S> I have some other business to take care of <S> so I can't finish this now, but I will tomorrow. <S> It's tomorrow, and it looks like Spehro pretty much covered what I was going to, so the only other thing I'd advise you to watch out for is whether the LM317 can supply the starting/running current for the pump motor. <S> If it can't supply the starting current, then a big enough BFC from the LM317's output to ground to store the charge to start the pump before the TIP120 kicks in will be required. <S> If it can't supply the running current, then it's a new ball game.
| You are, in one respect, correct in not trusting hfe, but not for the reason you think.
|
What is the use of an op-amp whose output and inverting input are connected to ground? I am in first year of Engineering school and I was given an assignment containing this circuit, which drives pressure sensors in a pitot tube : I am struggling to understand the whole circuit, and more precisely the first op-amp, which output (pin 1) and e- (inverting) input (pin 2) are connected to ground. What is its use? How can such an op-amp have an influence on the overall circuit, if its output is not used? <Q> The first OP-amp is actually creating the circuit ground. <S> The 7810 creates a stable 10 volt, which is then divided by the voltage divider R2 and R3, filtered by C3 to make a stable 5 volt level relative to the most negative level. <S> The OP-amp then buffers this, and the rest of the circuit uses its output as the reference ground. <S> Remember that ground in a circuit like this is just a convenience, a node that is used when referring to other voltages. <A> While I agree with @pipe and in fact upvoted his answer, an additional nuanced reply is that a ground is more than "just" a reference. <S> What I mean by this is a ground <S> is not just a voltage, but something that can source & sink current and stay at the same potential. <S> The ground created by that op-amp can both source and sink current and remain roughly half way between the rails of the +12 volt source. <S> If the design were to have just used another regulator like a 7805, that device would have only sourced current and thus would only have put out the right "mid voltage" value when current was flowing out of it's output. <S> Rather more restricting that the circuit shown. <A> (TL,DR: see paragraph 5) <S> The meter module requires the voltage on its GND pin to be between its V+ and V- supply pins. <S> The 7810 regulates the input to 10V between nodes labelled +5V and -5V. R2 and R3 <S> provide a mid-point voltage with Thevenin impedance of 2.5 kOhm <S> (= 10K // <S> 10K) in parallel with 100nF. <S> Thus any (DC) current drawn from or supplied to this node will push the voltage by 2.5V per milliamp. <S> The GND node will be carrying currents from: Meter VIN+, R9, R11, R15, R16, R13, A2 and C5. <S> These will probably sum to less than a milliamp, but the meter may draw varying current through each measurement cycle. <S> Amplifier 1 acts as a voltage follower for the R2 R3 chain. <S> It will act to hold its output, the node labelled GND, at the mid-point of nodes labelled +5V and -5V. <S> Looking at it from a different perspective, it acts to pull the mid-point of its supplies towards its output voltage. <S> It will have a closed-loop output impedance of a few ohms, so current drawn on the GND node will have little effect on the voltage between GND and the +5V and -5V lines. <S> Amplifiers II-IV are all configured as simple differential amplifiers. <S> II and III have a gain of 100V/V and Zin of 10K. IV has a gain of 20 and Zin of 50K. Having C5 connected directly to the output of amplifier IV <S> is an error. <S> OPAs are not specified to be stable with a large capacitative load. <S> It would probably be better to put it across meter VIN+ and GND, with 10K or so between the wiper of A2 and the VIN+. <S> The gain of the circuit will directly depend upon the output voltage of the 7810. <S> If the meter has an external reference input it would be best to connect this to a fraction of +5V, this giving a ratiometric reading. <S> The offset voltage of all four amplifers will contribute to the signal. <S> The amplifiers will need good DC and 1/f noise specifications. <A> A simple and direct answer to your main question, is that this is one way of providing the differential voltages that the op-amps need (+ & - 5v). <S> By floating the ground (to +5v) the single 10v source can supply + & - 5v! <S> You should now be able to understand that the output of op-amp I, is being used .
| It converts and displays the voltage between its IN+ pin and GND. It creates a virtual ground (or reference ground).
|
Interfacing open-collector optoisolator to 3.3V microcontroller I have a bit of a puzzle on my hands trying to interface a Fairchild MID400 optoisolator (popular "logic level" optoisolator for mains detection) to a 3.3V GPIO pin. The MID400 is a 5V part [*] and has an open-collector output which is pulled up internally to a 5V logic high level. This is the aspect that's causing me problems as otherwise I could just pull it up externally to 3.3V and all would be well. The microcontroller and optoisolator are metres apart and I would like to perform the level shift on the optoisolator end of the wire where only 5V is available. I can't see how it can be done using diodes or resistors but happy to try out anything suggested. Switching frequency is low, rise/fall time unimportant. Thanks for any suggestions. [*] Just in case anyone should happen across this question wondering if the MID400 might work at 3.3V albeit out of spec, I tried it and it doesn't at all. <Q> That open collector output V o is not pulled up internally according to the datasheet. <S> Zero mention of it, and every diagram shows a simple npn open collector. <S> Power the VCC at 5V, and pull up V o <S> to 3.3V <S> vía a external resistor or your microcontroller internal pull-up resistor. <S> Worst case, use a voltage divider to bring the signal at V o down to 3.3V. <A> That being said, how about this? <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If something in the signal source fries this will offer you some cheap protection. <S> If the input can source more thatn 1A install a fuse to protect the diode.
| Nowhere in the datasheet is it stated that the output is internally pulled up, so you can design your pull-up network however you'd like. If you are concerned about damage to your controller you can limit the input high with a 1N4728A and a 10K current limiting resistor.
|
Why isn't Voltage considered in mAh ratings? The capacity of a battery is measured in mAh, for example a 1000mAh battery can provide 1A for 1 hour. Why is voltage not considered in this equation? 1A at 5V draws twice as much power (P = V*I) as 1A at 2.5V, which would surely drain the battery twice as quickly? The only reason I could think is the mAh is calculated on the output voltage of the battery in which case how can we compare batteries using mAh ratings? <Q> There are two reasons why batteries are rated for capacity (A·h) instead of energy (W·h): <S> Cells using the same chemistry will have equal (or very close) voltage ratings, so voltage can be factored out when comparing their capacity. <S> In many chemistries, the voltage changes significantly during charge-discharge cycle, but also with temperature and load. <S> Factoring that variable voltage in the battery energy will make many calculations complex: <S> a cell discharged to 50% of its energy would require e.g. 80% of its nominal energy to recharge back to 100%, while a cell discharged at 0% may require 200%. <S> With capacity, charge and discharge figures are much closer, and the losses are nearly the same regardless of the charge level. <S> the same cell at 50% energy would gain some energy by simply being heated up. <S> Its remaining capacity, on the other hand, would not change significantly. <S> the same battery would deliver 95% of its energy to load X, but only 70% to load Y. This is also true for the capacity in some chemistries, but to a lesser extent. <S> As you can see, if you say "I have a batter of 1 W·h", you'll need to specify in which conditions this energy has to be consumed, while "a battery of 1 A·h" characterizes the battery itself, not the environment where it will be used. <A> The marketed mAh value is obtained as follows: <S> The manufacturer puts the battery in a constant-current sink circuit (e.g. 100mA), and checks the voltage of the battery, which then decreases slowly. <S> Once the voltage reaches a minimum threshold (let's say 1.2V for an alkaline battery), The manufacturer multiplies the current by the time it took to reach this value and says it's the capacity of the battery. <S> So the mAh value actually depends on at least those two parameters: current drawn and voltage threshold (it also depends on temperature, etc...). <S> But if the current drawn is smaller, they show better capacity, for example. <S> Now for the voltage: The manufacturers usually provide graphs that show how the voltage evolve during this test. <S> Here is one: <S> So, the mAh is given for the voltage of the battery all througout this test. <S> Now, if you want to have the total energy that the battery can give (expressed in Wh : Watt-hour), which is the value you seemed to be interested in, you integrate this graph (not easy), but it's not really necessary anyway, since the voltage stays within reasonable bounds. <S> Just multiply the mAh by the average battery voltage. <S> To sum up and go back to you question If you want to compare batteries: if they are specified with the same voltage, you can directly compare the given mAh values. <S> If they have different nominal voltages, multiply the mAh values by the voltage of the batteries, you'll have mWh values that you can compare to each other. <S> Now, whether the battery energy is drawn "quicker" at 5V than at 2.5V is not really relevant. <S> It depends solely on the load type. <S> Finally, why the manufacturers indicate the mAh rather than the mWh ? <S> I don't really know. <S> It's like that. <S> Maybe because the mAh value is easier to get from the above graph. <A> As I understand it, most lithium ion batteries are rated at 3.6 volts. <S> As others have pointed out, the actual usable power is less than the full capacity and how they are used also effects the power they will provide. <S> For instance, the slower you use the power, the more total power the battery will provide. <S> The quicker you draw it down, the less power it will provide. <S> So that mah rating is only nominal, the actual power provided will be different.
| Since the voltage is standard, you can compare the mah of two lithium ion batteries and determine how much relative power they will each provide.
|
wav file audio samples are signed, how does DAC deal with signed numbers when it does not have negative supply? I was looking into wav audio file specification, they may hold the pulse coded modulation data as signed our unsigned integers as well as floating point values. How does a DAC deal with signed numbers? Do they need a dual supply to output signed numbers? What would be an appropriate way to convert floating point numbers into integer when sending them to DAC? I assume e.g conversion onto 8 bit would be simply multiplying by 255 but I am not sure since floating point numbers are signed. Yes I realise that there are multiple types of ADCs. <Q> Lets say you have a DAC with a reference of 2.5V operating from a single supply (i.e. the full span output is 2.5V) <S> that accepts 2s complement encoded values at the digital input. <S> The zero point in such a device is at \$\frac {V_{span}} 2\$ or 1.25V in this particular application. <S> If a device takes 2s complement as an input, then the values below 0 are encoded as less than the mid-span voltage; specifically, \$V_{out} = <S> V_{midspan} <S> + Code_{in <S> } <S> * V_{bit}\$ where \$V_{bit}\$ is the resolution; \$ V_{bit} = \frac {V_{span}} {2^n}\$ where n is the number of bits for the converter. <S> \$V_{code}\$ is the signed value of the input. <S> Note that on a split supply (+/- <S> 2.5V for example) <S> midspan is 0V. <S> There is an excellent application note that explains the various types of data formats used by DACs and ADCs. <S> ADCs universally (all the ones I have seen) output fixed point values. <S> There are complex devices that have DSPs inside that can do the floating point to fixed point conversion on board. <A> Simply: <S> convert float to integer, by multiplying the float number. <S> (multiplier must be high enough to get the complete resolution of the float. <S> add the average value of the integer signal to the signal ( or the highest absolute value) <S> use just the highest bits of the integer to satisfy the DAC <S> The DAC doesn't need negative values. <S> The alternate component is still stored in the signal, but now with just positive numbers. <A> Audio is inherently AC. <S> That means the DC level doesn't matter and can be arbitrarily subtracted off and something else added. <S> In fact, frequencies below 20 Hz can be arbitrarily eliminated even in "HiFi" audio. <S> The WAV file is specifying the audio signal values as if the signal had zero DC offset. <S> That makes sense when you think about it, as it's the logical choice for communicating arbitrary audio signals. <S> However, that does not mean your D <S> /A converter needs to produce negative voltages. <S> If the D <S> /A converter has a output range of 0-5 V, for example, then you simply declare the midpoint of 2.5 V to be the WAV file zero level. <S> Note that this is little different from the arbitrary scaling you need to apply and that you don't seem to have a problem with. <S> You scale the min to max WAV file values for whatever representation that WAV file uses, to linearly map to the D/A output range. <S> There is no special case because one or both of these ranges include negative values. <S> Either way it's still just a simple linear scaling with offset. <S> In the case of a 0-5 V A/D, the midpoint is 2.5 V, which represents the audio zero level. <S> If the A/D is followed by a amplifier, then it will alter this range again. <S> Eventually a power amplifier driving speakers will eliminate the average DC level <S> so there is no net DC going to the speakers. <S> Electrically, DC is easily eliminated from a audio signal by a high pass filter , which can be as simple as a capacitor in series followed by a resistor to ground.
| Floating point numbers would normally be converted to integers within the range of the converter for a DAC.
|
Eagle/Flatcam/LaserCNC: How to have "open pad" for futur drilling I'm using Eagle, flatcam and a CNC laser machine.My problem is when I'm trying to export my board, all my pads are filled !It means, when my board will be engraved (chimical method), I will not find any holes in the middle of the pad to use for drilling (manual) Eagle board: Flatcam preview: Do you know where I can change this to have "open pads" ?It's an eagle parameter ? flatcam parameter ? Thanks for your advice, I'm a newbee <Q> Unfortunately there is no way to do this directly. <S> The Eagle CAM processor specifically fills in hole areas with copper so that if there is any misalignment in the drilling stage of manufacture you don't end up with gaps in the copper next to the hole which could cause problems in the through-hole plating step and during assembly. <S> I understand for home assembly it would be a useful feature to create self-aligned holes, but there are no options in the CAM processor to do it. <S> Even the DXF export ULP has the same behaviour (fills holes). <S> You can do it by printing the design to a PDF showing only the copper layer. <S> This does result in the holes showing up as unfilled. <S> However this is only useful for home chemical etching using <S> say the toner-transfer method. <S> I don't think there is a way to import the PDF into Freecad for milling. <A> Since I don't use Eagle or Flatcam, I'm going to make an oddball suggestion for a solution that will work regardless of the software you used to generate the gerber and excellon files. <S> Download <S> Gerbv and install it. <S> Or, for Linux users, install gEda since Gerbv is part of that package. <S> Open the gerber file for the bottom side of your PCB in Gerbv <S> Open the excellon (drill) file in Gerbv. <S> Check that the gerbers and excellon data line up (drill holes on the pads and not floating off in space somewhere.) <S> Set the color of the drill layer to white and make it the top layer. <S> Set the color of the gerber layer to be black. <S> All of your traces will appear to be gone. <S> Don't panic. <S> They are just black lines on a black background. <S> Export the resulting file as an SVG. <S> The black background doesn't get exported, so the resulting svg has a transparent background with black traces and white drill holes. <S> Use any SVG viewer/editor to display the final file. <S> This is pretty much universal, so long as Gerbv can understand your gerber and excellon files. <S> Gerbv couldn't read some files from I got from Maxim, so there are exceptions. <A> Check out LineGrinder instead. <S> Works with your Gerber files directly. <S> It's excellent software <S> and I've used it successfully to do what you're trying to achieve. <S> It can not only use your Excellon drill file to mark drill locations, but can even reference pin holes for flipping if you are trying to do double sided.
| I used Inkscape Print the SVG and use the printout as your mask (photoresist or laser toner transfer) for etching the PCB. You can get creative in eagle and redesign all your pads to be made of polygons shaped with a hole in the middle and a much smaller pad off to one side, however this would be far more trouble than its worth and would likely be a DRC nightmare.
|
What features give reason for such a wide range of prices for digital multi-meters? Ever since high school, I've been using the same multi-meter. It works well for low voltage purposes, but I'm scared to use it for anything else (high voltage, unless it's a little zap from an electric fence which I think is high voltage) because I purchased it in 1993 or 1994. It sometimes flashes on/off, and loses its display. I'm doing some shopping for a new one. One feature about my current one is that I always need two hands to touch the wires because it has leads at the end of it. So I'd like a way to keep the multi meter leads on the wire somehow. I also want one that stands up. There are SO MANY different kinds! I'm looking for some electrical engineering advice on why there is such a HUGE range in prices and what I should buy and get my money's worth. Spending up to a few hundred won't be an issue, but I'd like to know if there is a reason to spend that much. If you know about features that clearly give reason to buy a more expensive one, please let me know. It might also be good to explain when you'd pay more for a multi-meter for low voltage (or low current) situations, or vice versa, during high voltage (or high current) examples? Some time ago, I also remember reading something where some multi-meters prevent the end user from damaging the device. Say for example, if you have current running through a circuit, and you turn on the Ohm Meter to measure resistance. In the past, I was always told to turn the power off my circuit when measuring resistance in a circuit because it uses the voltage inside the multi-meter itself for the actual voltage when measuring, and it could be damaged if there was current in the circuit when measuring. I've never made the mistake of doing this, but I would think the newer ones (or more expensive ones) would prevent you from damaging the multi-meter if this ever happened. Especially if they need to pass federal electrical regulations, whatever those are. Seems like a trivial thing to prevent when engineering the product. Sorry for a loaded question, but I'm sure your answer will result found in plenty of Google search results if you have a good answer. <Q> What features give reason for such a wide range of prices for digital multi-meters? <S> THE most important feature is stability aka the voltage reference internal to the device. <S> A lot of really good multimeters use the LTZ1000. <S> Here's a picture: <S> - Notice how the device is mounted on spindly bits of PCB legs to avoid stress on the die causing random ppm shifts in the output voltage. <S> If you read up on it you'll also find that it has a resistive heat generator and temperature sensor internally so it can be oven controlled. <S> The LTZ1000 boasts temperature drifts of 0.05 ppm per degree centrigrade. <S> The best "off-the-shelf" device I've found is a maxim device at 2ppm/ <S> degC or the LTC6655 also at 2 ppm/degC. Long term stability is also top of the list when it comes the the voltage reference. <S> Next is very tight precision low drift resistors in fact matched resistor pairs like you can get from Vishay and Linear technology. <S> These are not cheap and turn a run-of-the-mill meter into something like a decent fluke or HP meter. <S> Forget features <S> - they are ten a penny in comparison. <A> There are several reasons to spend more money on a multimeter. <S> Safety. <S> If you are doing work involving "mains" electric supplies <S> then you should be aware of "overvoltage categories". <S> If you are only doing portable appliances then "CAT II" should be ok, if you are doing installation work you really want "CAT IV". <S> I would also avoid anything not bearing the brand of a reputable western company as it is all too easy to fraudulantly claim safety compliance. <S> Features, do you want a current clamp? <S> frequency measurement? <S> an interface to connect to a computer? <S> a crude but sometimes useful osciloscope? <S> Precision and accuracy, this is where multimeters can get REALLY spendy. <S> The more digits a meter has the more accurate and stable all it's internal components need to get and the more attention needs to be paid to noise and distortion. <S> Regarding leads/probes, most multimeters use standard 4mm shrouded sockets (some very cheap ones use a non-standard semi-shrouded 4mm, old ones used unshrouded 4mm or even 2mm). <S> The leads that come with a meter will usually be a fairly basic set with fixed, bulky heavily insulated probes and no clips or other accessories. <S> Don't be afraid to buy (or even make) alternative probes/ <S> leads that are more suited to your application. <A>
| Basically, quality, fully characterized and calibrated parts are expensive, there is the labor in calibrating each unit, more expensive units are usually built more rugged too, tougher plastics, better connectors, proper input protection (gets replaced by the dreaded transistor tester), the higher end ones tend to have extra features as well, like statistics and logging.
|
Switching power supply only works when warm Update With the multimeter and a steady hand, I've found the 3843B PWM controller Vcc pin is 12V when the supply is on, and 8.2V when it is plugged in but cold. Datasheet says it needs 8.4V Vcc to start-up so heating it is lowering its startup voltage or increasing the Vcc just enough. To test the theory I supplied external 9V power to the pin, to "jump-start" the power supply. It worked! So now, I think I need to find where this 8.2V is coming from so I can work out how to bump it just that little bit more, maybe by replacing a diode or something. How this "start-up current" is usually generated on power supplies? Maybe from the pictures you can point to some probable locations I should look? What could be the cause of the low start-up voltage? Thanks in advance. Okay, this is a weird one. First time on this Stack Exchange, work with computer science, and have only basic electrical knowledge, so beware of any possible misunderstandings below. Some context I have an "universal" laptop power supply, one you choose the desired output voltage with a switch, it goes from 12 to 24V. Since some time ago it started working only when warmed up. So to start using it, I had to warm it up with a hair-dryer and then couldn't leave it turned off for too long, or it would cool down. I decided to fix this for good because it was getting worse over time (needed more temperature to start-up), it looks fun, and I am broke. So I opened it up, armed with a multimeter and started probing. It takes the 110/220V input AC, rectifies it and then passes it through a MOSFET (gate, I suppose, connected to some signal generator, the frequency of which maybe determined by the voltage selector) and onto the main coil, which gives our output voltage which is rectified again, filtered and then out. So the problem is, whatever drives this MOSFET wasn't working when cold. I narrowed the problem to 3 ICs and did the logical: warmed them one-by-one with a lighter to find which one was problematic. TL;DR: The actual question So I narrowed the problem down to an IC called "3843B" which turns out to be a "high performance current-mode PWM controller". So, heating this IC (or something on its close vincinity) gets the power supply going. After it starts working, then everything is OK. Below is a photo of it. What could be causing this? Which of the suspect components in the photo could have a failure mode which makes it work only after it's hot? What can I do or which components do I replace? Close-up of suspect Board overview <Q> Could be a bad chip, or could be a bad solder joint. <S> Easy to find out if it's a bad joint- <S> just take a soldering iron and reheat all of the joints and leads in that area (including pin 6 on the IC- solder looks a little sparse there). <S> Use a fine tip on your iron, and add a little flux to aid in the reflow (a tiny bit of added solder might help too, but not too much). <S> Start with the IC, and test the PS after things cool off. <S> Then move on to other joints, one at a time. <S> Moderation! <A> Check the below-pictured resistor out (soldering and value)- just a hunch. <A> You may follow the good instructions of the answers and comments given, but it nothing solves your problem, most likely the chip has bad gold wire bonding inside it. <S> This happens when wire bonding process was poor during chip assembly. <S> In this case you got to replace the smd chip. <A> It's hard to guess without oscilloscope and the device. <S> There is an optocoupler, four-legged ic right next to the diode bridges, would start checking it <A> I don't have enough experience but sometimes a NTC resistor is used to limit inrush current and its value only gets down when it's "warm". <S> I wish this is helpful <A> probably the black capacitor next to the 2nd rectifier. <S> These tend to have a higher leak current than the specs when they dry. <S> The PWM controller would only produce low duty cycle or interval because the feedback mechanism detect a short circuit in the DC side.
| It could be stitch crack or poor intermetalic weld on chip bond pads. Use as low a temperature as you can and don't heat a joint to the point that other joints unsolder themselves. It may be this component at the top:
|
Why don't we use 7075 aluminum Ethernet and USB cables? Since the skin depth of copper at 1Ghz is about 2.3 micrometers, it doesn't seem like aluminum would provide a worse signal to noise ratio. 7075 aluminum is also much more resilient and stronger than pure copper, so it should be able to bend more without breaking. The main problems seem to be splicing wires (due to oxidation) and Power over Ethernet, but for other uses, aluminum wires seem like a cheaper and lighter (for aerospace) solution. So why do we still use copper for data? <Q> Aluminum oxide is stable, hard (as sapphire, because it IS sapphire, akaalumina, Al2O3), and nonconductive. <S> Oxide grows spontaneously on contactwith air, so an aluminum electrical connection is often unreliable. <S> Weldingworks, and some (fluoride-based) fluxes can allow soldering, butfor crimp connections, you need antioxidant pastes and/or odd mechanicalcontrivances. <S> Reliable aluminum electrical connections are messy or bulky. <S> Copper is compatible with a variety of insulation-displacement connectionschemes (basically, just a hard clip that dents the copper but cuts through plastic insulation), that stay reliable for years. <S> The clip parts canbe made of copper alloys, so there are no dissimilar metals issues. <S> Copper oxide is neither hard, nor insulating (it's a semiconductor), so copper wire just makes a better connection. <A> Rdtsc gave a very good link in a comment. <S> That list is a run down of all the problems you will encounter when using aluminum wire. <S> The short of it is that aluminum has poor mechanical properties for use as wire. <S> A short list of its worst properties: It isn't ductile enough (fatigues and breaks easily) <S> It is too maleable (squeezes out and "flows" under pressure leading to bad connections) <S> It expands and contracts more with temperature changes, which lead to bad connections. <S> It also has poor chemical properties - the oxidation that Whit3rd mentions. <S> Oh <S> , yeah. <S> It is also not easy to solder. <A> Looking at a conductivity table <S> 7075 aluminum alloys have worse conductivity than pure aluminum, nearly half for some of them. <S> Also, Wikipedia says the 7075 alloys are rather expensive (relative to other aluminum alloys), but I don't know how they compare to copper price-wise. <S> These issues together probably explain why despite the 7075 alloys being available for over 70 years now <S> they don't seem to have been used in any electrical applications . <S> As @Harper correctly mentions below, the 8000-series aluminum alloys are the ones usually used in electrical applications . <S> The current way to make cheap data cables is copper-clad aluminum (CCA) . <S> You can help yourself to the ASTM B566 standard and see what's the exact composition of the ideal CCA cable (which doesn't mean that your Won Hung Lo manufacturers will even stick to that.) <S> It's about 10-15% copper and the rest aluminum. <S> For a summary of CCA cable properties from a manufacturer thereof (adhering to the aforementioned standard) see this page , for instance. <S> There's also an ISO 13832 covering CCA, which seems to just incorporate the aforementioned B566 standard when it comes to CCA. <S> UL also references that B566 in their testing services. <S> Do note however that no CCA cable can be currently and legitimately be advertised as Cat 5e/6. <S> That's because those standards mandate copper ; more details over here . <S> Although I know of a ISPs with millions of users using CCA [for final customer-premises connection] and it seemingly pays off for them... <S> even at 1Gbps speeds. <S> But they can afford to test cables to their own standards in-house. <S> Also note that CCA cables also don't conform to some [US mainly] building code standards; the ISPs I made reference to are located somewhere in Eastern Europe <S> (although in the EU); they are also using mostly Huawei equipment. <S> In contrast, the ANSI/SCTE 100 (2010) <S> standard for 75-ohm coax however allows for B566 CCA cable core (the 10%-copper grade); and it's good for 5MHz-1GHz operation (with -20dB SRL). <S> I don't know (or care) <S> much about USB cable standards to tell you exactly what they allow or require... <A> Cheap 75 ohm coax as used for TV typically has loose aluminum wire braid shield with thin thin aluminum wrap. <S> I think the centre conductor is copper-plated steel. <S> Nasty cheap stuff, but usable when crimped. <A> For very fine 24-gauge ethernet wire, there's just not enough metal there to make much of a difference in total cost of cable fabrication.
| To put a simple summary to the existing answers: Aluminum isn't commonly used in wire because while it is a fair conductor, it makes a mechanically poor wire.
|
How to turn a pulsed signal into a continuous one? I have a pulse signal at 1 hz which goes high +5V and low 0V. How do i turn it into a continuous one? +5V. i have searched in different places and what i found says that there are different methods such as using an op amp as an integrator, then i also found that it can be used a diode, resistor and a capacitor, others suggest that a sample and hold circuit can also work. However none of these answers seem to be clear to me. i am still a novice in this matter. Would somebody help me with an answer that has a schematic i could test?. I have also found that a retriggerable monostable can work but i am not sure if this is the right approach, as i need the "continuous signal" to go low immediately after the input pulse is interrupted. If possible i would like to do this using logic gates such as the ones found within 4000 cmos series. Thanks for taking your time reading my doubt. <Q> You are failing to see the error in your belief. <S> If the pulse is high (as normal) and then goes low (as normal) and while low it is "interrupted", the signal will still be low and you won't have known the signal was interrupted until you realize that it has failed to go back high some time later. <S> It's like <S> when your phone rings - it does so in bursts - you go to pick it up when the bell falls silence <S> but you don't actually know that the person ringing you has hung up during that silence. <S> So forget about "immediately" and use a re-triggerable monostable. <S> Here's the waveforms: - Here's a retriggerable monostable circuit: - Picture stolen from here . <S> Basically, a re-occuring pulse fed to the input (the little circle to the left in front of the two resistors feeding a BJT) keeps the 555 from timing out. <A> If you have a 1 Hz square wave (high 500 ms, low 500 ms), then it takes at least 500 ms to tell that the pulses have stopped. <S> You can't tell "immediately" that the signal has stopped. <S> When the signal goes low, for example, you have no way of knowing the difference between the signal having gone away, or just entering the low phase. <S> You can only tell after 500 ms because then the real signal will go high, but if the signal went away then the voltage will stay low. <S> When this device is hit with a pulse, its output goes on. <S> With no other input, it stays on for some time, then turns off after the delay time has elapsed. <S> The "retriggerable" part means that if it is hit with another pulse while on, the delay until off time is reset and the output stays on longer. <S> In your case, you need to set the on-time at least as long as the maximum time between pulses. <S> If the device only triggers on a particular edge, then that time needs to be at least one second. <S> Of course you want to make it a little longer for some margin, like 1.2 seconds or 1.5 seconds. <A> " I would like to do this using logic gates such as the ones found within 4000 cmos series." <S> Using and EXOR gate (4070) and an RC delay you can separate the incoming 1 Hz clock into its edges to give very narrow pulses at twice the frequency (2Hz). <S> Then a JK flip flop (4027) can recombine the two pulses back to the original 1Hz clock. <S> Now you have access to the SET and RESET inputs of the JK <S> flip flop wich will allow you to set the output HIGH or LOW and monitor these control signals as the RS inputs override the incoming clock. <A> Can't a simple grounded capacitor of <S> 1 uF or 0.1 uF <S> between the input line and ground do the trick?It would cause a small delay thought. <S> Not exactely "immediately". <S> The higher the value of the capacitor, the longer the delay.
| In any case, to turn a pulse train into a continuous signal (within the fundamental constraint described above), you can use a retriggerable monostable multivibrator , more commonly referred to as a retriggerable one-shot .
|
Component with multiple VCC and GND pins. Should I connect all of them? I want to make a really small arduino using the Atmega 328p and I noticed that it has 3 GND pins and 2 VCC pins. I measured the pins and it seems that all GNDs are connected and all VCCs are connected to each other. Can I rely on this feature to make my arduino schematic?Is it a good idea?Can I rely on the same idea for other components with multiple VCCs and/or GNDs? Cheers <Q> Connect all of them and more importantly apply very localized decoupling capacitors on them all. <A> Generally, a device with multiple supply/ground pins intends for those pins to be externally connected to the appropriate supply/ground. <S> It is particularly important to connect the ground pins together using wide traces or polygons/planes to avoid ground loops. <S> The datasheet , section 1.1, for the 328p indicates three specific power connection types: VCC, GND, and AVCC. <S> While this device shows a direct connection between supply/ground pins, some devices require connections between those pins externally and may be damaged if some pins are left floating. <S> It is best to assume that pins of the same supply must be connected together unless specifically noted in the device datasheet. <S> Additionally, considering decoupling: multiple supply pins allows multiple bypass capacitors (usually 1000pF to 0.1uF, depending on the application) to be placed around the device. <S> For a low speed device with short connections between supply pins and an uninterrupted ground plane around the device, a single 0.1uF capacitor may suffice. <S> Higher speed devices may require one capacitor per supply pin or even multiple sizes of capacitor on each supply pin. <A> Connect them all. <S> The reason why multiple pins of the same potential are brought out of the die is for two reasons. <S> 1) this reduces the resistance. <S> When you do that you reduce the voltage bounce that occurs from current pulses that are formed upon switching. <S> So you get a cleaner voltage source on the internal power rail. <S> 2) it reduces the inductance of the wire bond leads from the bonding shelf to the die. <S> Reducing the inductance makes any decoupling capacitance more effective and therefore reduces any internal rail bounce resulting in again, cleaner internal power voltages.
| Thus, all of the VCC pins should be connected together, the GND pins should be connected together with care for a low impedance connection, and the AVCC pin should be connected (with optional filter described in the datasheet) to VCC. Multiple external connections allows more current to safely flow into/out of the device.
|
Two identical DC power supplies in parallel for redundancy? I'm trying to design a power backup system for my saltwater aquarium. In short, the system will be powered by a 12VDC power supply while the mains power is on. On power failure, the system will automatically switch to a battery source until the mains comes back on. At this point, it will switch back to the 12VDC power supply. I believe I have all that worked out, no issues. The only thing I'm stuck on is the 12VDC power supply that runs the system under normal conditions. If this power supply fails, the system will detect this as a power failure and switch to the battery bank. The batteries will eventually run out because they're meant for short term life support, not for indefinite running. At this point, it seems like the greatest weakness in the system is the 12VDC power supply that runs the system during normal operation. So, my question: would it be safe to wire two identical 12VDC power supplies in parallel for redundancy purposes only? Note, I am NOT trying to gain additional capacity from this configuration, only redundancy. One power supply could very easily handle the entire load. The power supplies in question will be 12V 40A, and the consumption of the system tops out around 300W. Thanks in advance for any help regarding this. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Parallel PSUs with power indicators. <S> V+ will be about 0.7 V below the power supply output voltage due to diode drop. <S> Both PSUs on all the time and, if similar type, will probably share the load 50/50. <S> simulate this circuit Figure 2. <S> Relay switched PSU. <S> Both PSUs on all the time but one on standby. <S> simulate this circuit Figure 3. <S> DPDT relay switches mains on second power supply (preventing it from wasting power when PSU1 is running). <S> A second contact switches the DC side. <S> Note my schematics show F1 as a common point of failure. <S> The same is probably true of your home supply. <S> You need to take that into account when designing the system. <A> You should re-think your topology if reliability is really so important. <S> Your most likely point of failure is the switchover relay, not the 12 V power supply. <S> I would set up the system to always run from the 12 V battery, just that the power supply charges the battery when line power is available. <S> It is always connected to the battery, but the battery is charged when the alternator is running. <S> Lead-acid batteries can be held at their "float charge" voltage indefinitely without damage. <S> The alternator and regulator system in a car is usually set to 13.6 V. Since you'll generally be charging for long periods of time, you can be a little more conservative, like 13.0 - 13.2 volts. <S> To achieve this, get a "12 V" power supply that can be tweaked a little. <S> Many can. <S> Put a Schottky diode between the power supply output and the 12 V lead-acid battery, then adjust the power supply for the desired float charge voltage at the battery. <S> The actual power supply voltage will be a little higher due to the diode. <S> You can even add redundant power supplies, each with their own diode if you like. <S> You have to make sure that your 12 V equipment can handle the full range of voltage from the battery almost depleted to the battery full and being float-charged. <S> 10-14 V would be a good target. <S> 9-14 V even better. <S> This system can even support multiple power supplies and multiple batteries. <S> Here is what it would look like: <S> This scheme allows for as many power supplies and batteries as you want. <S> There is one diode per power supply per battery for charging. <S> That's what D1-D6 are in this example. <S> Then there is one diode per battery to the final load. <S> That's what D7 and D8 are in this example. <A> No, it is generally not safe to parallel two power supplies (even of the same model) unless they explicitly support such a mode of operation. <S> Some power supply chips (such as the LTM4625 ) are designed to be placed in parallel on the same circuit board. <S> I can't quickly find a wall-powered supply that supports parallel operation. <S> Such a supply would also require the ability to operate with its output being powered by the other supply when the wall power fails. <S> You may need to adjust your specifications or simply have larger battery backup capacity to meet your requirements. <S> Having an alarm when battery power is in use is standard for computer/IT UPSs, which would alert you if the 12V supply or the wall power fail.
| If configured correctly bench power supplies can be connected in parallel for load-sharing. This is exactly how the 12 V system in your car works.
|
Driving LEDs: Brightness, efficiency and heat generation I'm planning on building a low-cost LED fixture and I'm a bit confused with how I should drive it. So, if we consider these 2 led cobs: 1W 350mA 3.2-3.4V 100-110LM 3W 700mA 3.2-3.4V 260-280LM If I drive them both with a 300mA constant current driver, I have the following questions: Would the 3W led generate less heat? Would the 3W led generate more light? (lumens) Would the 3W led be less efficient? (lumens/watt) For reference: leds , driver <Q> It's impossible to know the behaviour at a current that is not the specified operating point <S> if there's no datasheet containing an "intensity over I_f" graph – if this is of concern to you, buy LEDs that come with such a datasheet. <S> Yet another thing I'd like to point out that you don't say anything about your 300mA supply – it might look that an LED that is more efficient at 300mA would be the power-efficient choice, but if it happens to have a lower forward voltage at that current (is there at least an I/U graph?), you'll just "burn" that additional energy (voltage = energy per charge) in a linear regulator. <S> Now, it's really impossible to tell what kind of regulator you have picked there – <S> yes, this looks a bit like the mains voltage gets rectified on the input side, and then chopped up and downconverted by the transformer type thingie there, but I can't tell the least about how exact the current regulation is ( <S> but I do have a hunch that without any secondary side sensing, it's not going to be that exact) or how efficient the conversion. <S> So: if you're at all concerned about operational specifics of your LED system, this is not the supply of your choice. <S> Generally, people will sell their LEDs at the point at which they produce the maximal amount of light without thermal damage (or without exhaustion of recombination opportunities in the semiconductor), and that will be pretty close to the point of maximum efficiency. <S> So, if you want to drive an LED specified for 700mA at 3.3V with less than its maximum power, you typically use PWM to switch it off and on with a configurable duty cycle, driving it at an "efficient" 700mA when on. <A> As a rule of thumb, leds are usually more efficient when run at lower powers (but not by a whole lot). <S> The problem I think you're going to have is that the efficiency differences between manufacturers is going to be greater than between power levels. <S> Some cheap leds can only do 40lm <S> /w (even though they're advertised as 100lm/w) while some pricey cree leds can do 160lm/w. <S> But what you will probably find is that the 1w and the 3w leds may actually use the same led chip. <S> I used to work for a lighting company and we saw this all the time from cheap manufacturers. <A> Take this guy , advertised as 10W at its hard limit of 3000ma. <S> But all the specs in the sheet are based on driving it at 1050ma, which yields about 3.1W. <S> This yields 400 lumens or an impressive 129lm/w (I'm reading out of bin V2, <S> 85C which is conservative, other bins/temps go as high as 172 lm/w). <S> However running at the (advertised) hard limits of 3000ma, it only gives 91 lm/w. <S> Whoops. <S> So in that case you get a 41% efficiency gain by driving the device gently. <S> It's funny, the curve doesn't look that bent! <S> But it is multiplied by the amp-volt graph, which is bent too. <S> Try 480ma @ <S> 2.78V <S> (from the amp-volt graph)... 50% rated lumens so 200lm, @1.334w or 150 lm/w. <S> Crazy. <S> And that's the lowest bin, if we get into the high bins, low temps and underdrive, we could hit 200lm/w. <S> Having looked at quite a few LEDs, this is very typical. <S> LEDs make heat for 2 reasons: being overdriven to get max lumens from minimum $ in emitters; and phosphors are inherently inefficient. <S> All things being equal you are better off driving the larger device at 1/3 power.
| In my experience, LEDs are marketed at their maximum safe overdrive, but spec'd at their most efficient output, which is much lower.
|
Two bulbs of different wattage in one AC circuit If I connect two bulbs of different wattage in one circuit, can that lead to problems? Is that a bad idea?Does it make a difference if the circuit is serial or parallel? <Q> Parallel All the mains bulbs in your house are probably in parallel so that should satisfy you that that works. <S> Each lamp will draw the current it requires. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Your house wiring. <S> This system works well because the lamps are fed from a constant voltage supply. <S> Let's do some easy maths using a 100 V supply. <S> \$P = VI\$, where P is power, V is voltage <S> and I is current. <S> Using Ohm's law, \$ V = IR \$ we can modify the power formula to get \$ P = <S> I^2R = <S> \frac <S> {V^2}{R} \$. Rearranging the last one gives us \$ R = <S> \frac {V^2}{P} \$. <S> This is hot\$^1\$ filament resistances were calculated. <S> Series connected similar lamps <S> If you put equal lamps in series they will divide up the available voltage between them and the current will be \$ \frac {1}{n} \$, where n is the number of lamps.\$^2\$ simulate this circuit Figure 2. <S> In this circuit both lamps have the same resistance so will drop half the supply voltage across each. <S> Since the total resistance is doubled the current will also be halved. <S> Since P = VI then the each lamp will be running at 1/4 power. <S> Series connected dissimilar lamps simulate this circuit Figure 3. <S> More voltage is dropped across the higher resistance / lower wattage lamp. <S> It will glow the brightest. <S> If you put unequal lamps in series the one with the highest filament resistance will limit the current to the others. <S> It may also glow the brightest (if they're all the same voltage rating). <S> Power calculation for each lamp: <S> 100 <S> W: <S> \$P = \frac {V^2}{R} = \frac <S> {33^2}{100} = <S> 11~W\$. <S> 50 W: <S> \$P = \frac {V^2}{R} = \frac {67^2}{50} <S> = <S> 22~W\$. <S> Note the possibly surprising result that the lower powered lamp will be brighter and consume more power. <S> \$^1\$ Lamp filament resistance increases dramatically with temperature. <S> I have not factored this into the calculations. <S> \$^2\$ Again, the current calculation will be higher than calculated here because the filament resistances will be lower when cold. <A> Assuming that you are connecting two light bulbs of the same voltage rating, but different power, to the same AC source, you have two cases: <S> Series: <S> the bulbs are both dimmed and cannot operate at rated power because they are sharing the voltage of the supply (voltage divider); this is usually not done for household lighting <S> Parallel: both bulbs will operate at their rated power like normal; this is effectively what happens in a chandelier or when multiple lamps are powered by the same light <S> switch <A> Constant-voltage bulbs include pretty much any incandescent bulb or packaged bulb-like product or integrated fixture in common use on mains voltage, including whole Christmas tree light strings as a product, and constant-current bulbs packaged to be constant-voltage, such as LED strips. <S> Mixing is fine. <S> You can parallel any light intended for mains voltage. <S> You can parallel any 12V lighting including LED strips of any length. <S> If you are using constant-current bulbs, then you drive them in series , so you only need one constant-current power supply. <S> This typically comes up when building an array of raw LED emitters. <S> Most people keep the entire array matched, but you could mix them if current is the same. <S> Many lamps must be driven constant-current, or they will burn up. <S> Including fluorescent, neon , mercury vapor, metal halide, LPS, and HPS . <S> It may make sense to "series" them to save money on power supplies, but if one bulb blows, they all go out. <S> Some fluorescent ballasts do just that. <S> That's not a big problem for LED emitters, which will likely outlast the driver circuit. <S> Incandescent lights are a relatively linear load, they work well in parallel but can also go in series if they are matched. <S> For instance Christmas tree lights, electric railway lighting (5x120V=600V) and street lights in a few older cities, notably Detroit.
| If you are using constant-voltage bulbs, then you should parallel any mixed bulbs of the same voltage.
|
Light an LED when load is in circuit I want to have a LED light up when a load is in circuit, and go out when it is not. Initially I set the circuit up so that the LED is in series with the load but this didn't work. I think possibly because the voltage drop across the resistor was too high. I could adjust the input voltage to match but I'm not sure I entirely understand the impact of doing so, especially given that the load might be variable. It occurred to me that maybe I can use a transistor in positions X or Y in the diagram, but I have no idea what type of transistor to use or which pins I should connect where. Ideally I'd also want to have a diode in series with the load to protect the supply against any voltage coming back the other way, how can I work out how much to increase the supply voltage by to compensate for this? (And can I just use the LED here after all?) If it matters the load is a bunch of cells to be charged (4x 3.7v Li-on hooked up in parallel for charging, or series when in use) +VCC is currently 4V but I can adjust this. EDIT:I've just realised that actually I can just move the"switch" after all, despite previously stating that I couldn't Basically there are spare pins on the plug that I can use for this purpose. <Q> This? <S> The PNP transistor makes it so the LED can't run if there is no load. <S> Maybe this (Second circuit by @transistor <S> because... <S> Me stoopid) <A> Just connect the LED after the switch, not before it. <S> That way it is powered from the same switched net that the load runs from. <S> Of course you need a resistor in series with the LED to get the proper LED current at whatever power voltage you are using. <S> Here is what I'm talking about: <S> To size R1, you need to know what the power voltage is. <S> Let's say that it's 12 V for example, the LED is a typical green T1-¾ type that can take 20 mA current. <S> You decide to run the LED at 10 mA since this is only for indoor indicator use <S> and you don't want to be pushing any limits. <S> Such green LEDs drop about 2.1 V. <S> That leaves 9.9 V across R1. <S> From Ohm's law, R1 = <S> (9.9 V)/(10 mA <S> ) = 990 Ω, so 1 kΩ it is. <S> Just to be sure, you should check the power dissipation. <S> (9.9 V)(10 mA <S> ) = 100 mW. <S> So a 0805 resistor will get hot but should still be within spec unless you're doing something unusual. <S> Modern LEDs can be plenty bright with just 5 mA thru them. <A> Padding..........................
| If this is in a closed chassis that can get hot, then you may want a higher power resistor or use lower LED current.
|
How do transistors have three legs when there are two circuits going through? I keep seeing a not-transistor outlined like this: NOT transistor http://www.waitingforfriday.com/images/thumb/4/44/Slide12.PNG/400px-Slide12.PNG But every picture of an actual transistor looks like this: Transistor http://nefarius.at/wp-content/uploads/2012/04/P4140062.jpg I can't help but notice four lines exiting the diagram, but only three legs exiting the physical transistor. The same seems to be the case for all the transistors I've looked at. What's the missing leg? <Q> You mis-identified what a transistor is. <S> The transistor is one component in that circuit that you referenced. <S> The symbol for an NPN transistor is: Look closely and you will see one of these in there. <S> Each of the "squiggly" components is a resistor. <S> You posted a circuit consisting of 2 resistors and 1 transistor. <A> There is no missing leg. <S> You are slightly confused because the schematic doesn't show the bounds of the transistor itself. <S> The connection "Out" in your diagram doesn't come out of the transistor. <S> It is connected to the wire between the top resistor and the transistor. <S> Imagine the housing of the transistor around the schematic representation (like in the diagram below where the dotted line represents the housing.) <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The transistor has only three legs. <S> The output comes from the junction between R1 and the transistor. <S> After seeing your comments, you are still wondering about two circuits. <S> The "circuits" you have in mind are as follows: <S> Base of the transistor (from the left) to the emitter (bottom) <S> Collector of the transistor (top) to the emitter (bottom) <S> I've labeled the diagram and added arrows to show the two current flows. <S> When a voltage on the base causes current to flow through the base to the emitter (1.), the transistor allows current to flow from the collector to the emitter (2.). <S> The current flow through (1) can be very small. <S> Transistors have what is called current gain, whereby a small current through (1) causes a larger current through (2.) <S> If no current flows through (1) then no current flows through (2.) <S> So, for the two "circuits" you are thinking of, both must use the emitter. <S> This is called a common emitter circuit. <S> The wikipedia link includes a lot of info on the common emitter circuit, as well as links to the common collector and common base circuits that can also be built with BJT transistors. <A> In the schematic you've shown, input is connected to the ' base ' terminal, through a resistor. <S> The terminal called ' emitter ' is connected to the ground. <S> The ' collector ' terminal has 2 connections, the resistor and the output pin.
| A transistor is actually a 3-terminal device.
|
Working display connected by GPIO pins to microcontroller? Would it be possible to add a normal HDMI port or a display to any microcontroller with the GPIO pins? (I didn't buy anything yet, I'm just getting information before I buy...) EDIT: Thanks for all the information guys, a moderator should probably close this discussion now... <Q> HDMI (and HD video in general) requires a very large bandwidth, which implies a fast processor that is capable of generating or manipulating that much data. <S> This excludes microcontrollers. <S> If you want to drive HDMI, the smallest chip that could do so would be a small FPGA, an ARM Cortex A series, or similar. <S> Microcontrollers simply don't have the required clock speed, interconnect bandwidth, or memory. <S> Also, HDMI requires differential signal drivers, which are exceedingly rare if not nonexistent on microcontrollers. <S> Now, that said, it may be possible to couple a microcontroller to an HDMI transmitter, but you will still be hampered by the slow clock speed and very limited RAM. <S> You will be better off using either an ARM A8/9 or an FPGA (probably with external DDR). <S> Typically, when you need a microcontroller-attached display, you would use one of the following options: a small LCD display, typically 16x2 or 20x4 monochromatic line display. <S> These can be driven directly via GPIO pins, or through a serial/I2C "backpack" which reduces the number of needed IO lines. <S> an I2C/SPI OLED. <S> These are often 128x64 pixels, monochromatic. <S> a SPI TFT LCD. <S> Full-color, typ. <S> 1.8-~4in diagonal, sometimes with resistive touch support. <S> It is possible to directly drive VGA at limited resolutions and color depths from a MCU, but it requires the majority of available processing cycles and memory, so it is typically more often a novelty rather than a practical solution. <A> Any display controlled by an HDMI port is not going to be possible to get working on a "few GPIO pins". <S> You would really need a microcontroller or an SOC with a built in HDMI controller. <S> Keep in mind that such MCU or SOC would also have to have support for connection of SDRAM or DDR memory to support the image source for the HDMI controller. <S> For most other types of simpler microcontrollers you can use a few GPIOs to support small displays such as monochrome character LCDs. <S> The 16 character by two lines is a very common type. <S> In similar manner you can support small graphics displays that support display refresh memory right in the display controller chip onboard the display module. <S> A particularly popular type of display is the back lit 1024 x 256 pixel display. <S> The graphics displays that are easiest to work with support one bit per pixel. <S> Color displays are also possible but are harder to work with and may support a few more bits per pixel. <A> Simply, no. <S> HDMI is a quite specialized kind of digital output format which cannot be implemented with a few GPIO pins from ANY microprocessor or microcontroller. <S> There are three major reasons for this. <S> 1) <S> The electrical signals are current-mode, low-voltage differential signals which are quite different from GPIO. <S> 2) <S> The data rate is quite high (up to 18GBPS) which is quite beyond what GPIO interfaces are designed to handle. <S> 3) <S> The timing of the signals is critical. <S> Another thing that is difficult to impossible to achieve with ordinary GPIO ports. <S> To be sure, there ARE chips that are specially designed to generate (and receive) HDMI. <S> A prime example is the Raspberry Pi which is based on a Broadcom "System on a Chip" (SOC) which has integrated HDMI output. <A> unfortunately, no, even if the micro could handle the immense bandwidth (it can't), most gpio pins have a limit of 20-60MHz, far short of the 1000MHz or so needed by hdmi, although there are micros with dedicated hdmi outputs, but you're looking at a $20 part with 400 BGA pins
| And there are a growing number of mcirocontroller chips which have HDMI output built-in.
|
Magnifiers for SMD soldering I need some some Magnifier for SMD soldering so I bought these Magnifying glasses: http://www.ebay.com/itm/381555608061?_trksid=p2060353.m2749.l2649&ssPageName=STRK:MEBIDX:IT It is good for PCB inspection but it's not for SMD soldering: for 10x magnification glasses need to be only few centimeters away from PCB and for greater magnification, distance is of course smaller so it is not possible to use it for SMD soldering. My question is, is there any cheap solution? I'm not good at optics, but smaller magnification mean larger distance, right? What do you think about these glasses: http://www.aliexpress.com/item/Magnifying-Glasses-2015-Rushed-Top-Fashion-Glass-Surgical-Loupes-1x-1-5x-2x-2-5x-3/32381736388.html?spm=2114.30010308.3.52.SCX8Jl&ws_ab_test=searchweb201556_9,searchweb201602_1_10034_507_10032_10020_10017_10005_10006_10021_10022_10009_10008_10018_10019,searchweb201603_6&btsid=8d9303e9-107a-4a6d-b3b4-eabeec9ed4cc ? Here are specifications: Color: Black Material of Lens: Acrylic Lens Battery: 3 * LR1130 Button Batteries (included) Lens: 1.0X / 1.5X / 2.0X / 2.5X / 3.5X Field of View: 1.0X: 250~350mm 1.5X: 200~300mm 2.0X: 175~275mm 2.5X: 150~250mm 3.5X: 80~120mm Size of Lens(1pc): 85 * 8.5mm (L * T) Item Size: 220 * 145 * 55mm / 8.66 * 5.7 * 2.16in (L * W * H) Package Size: 22.0 * 17.7 * 4.7cm / 8.66 * 6.9 * 1.85in (L * W * H) Package Weight: 270g / 9.5oz What does "3.5X: 80~120mm" mean exactly? <Q> You can purchase an inexpensive USB microscope that can work very well. <S> The main drawback of the inexpensive unit that I use is a small lag when displaying the image. <S> But I quickly got used to the delay. <S> I have two identical units purchased from eBay and I assume that Asian suppliers such as Aliexpress or Banggood have them for even less cost. <S> Use it with an old netbook computer. <A> To get good magnification at a distance, you need optics with multiple lenses, essentially small telescopes attached to glasses. <S> These are often used by dentists and surgeons. <S> You can find these by searching for "loupe glasses". <S> You'll see many for $200 and up and a few in the $20-$60 range. <S> Look for ones that specify the appropriate operating distance. <A> Personally, I don't bother with magnifiers at all (even when I'm soldering 0402 parts), it depends on how good your eyesight is, SMD parts are small and the magnifier is just there to make them easier to see, <S> but in the end it's more of a personal preference, use as much magnification as you need to be able to comfortably see the parts.
| If possible bring some smd parts into a store that sells magnifiers (some office supply stores have them) and have a look at them under the magnifiers, if you can see the parts clearly and you like the magnifier, go for it There are many such microscopes available.
|
Working with current feedback opamps [CFA] Before entering in to problem: My application is a laser pulse detection. In this process, when a laser pulse falls on a photo diode, it generates current. Later stages include a I-V conversion, gain, difference amplifier and comparator. My current problem with I-V conversion can be seen here . My pulse characteristics: pulse width min : 10 ns max : 150 ns rise/fall time : 2 ns - 5 ns pulse to pulse width :22 µS How I selected the opamp for gain and difference amplifier stages Key parameters: BW - bandwidth and SR - Slew Rate From this literature I found my BW can be calculated by the formula: BW = 0.35/( rise time ) = 0.35/5 ns = 70 MHz Now the slew rate is to be calculated: SR = 2 * 3.14 * BW * Vp = 2 * 3.14 * 175 MHz * 4 V = 2198 V/µs So matching both these parameters is a CFA AD8003 from Analog. When I go for high slew rates and high BW almost all opamps avaliable are CFA's! What is worrying me? Without any notion when I used CFA for my difference amplifier stage, with unity gain using a 500 ohm resistor for R f and R b , the results seemed stable, but it is this statement from Texas Instrument technical literature , clearly saying: the first difference between voltage feedback and current feedback—theinput impedance of current feedback op-amp inputs is very different.Because the inverting input has low impedance, current feedbackamplifiers are not good for balanced systems such as differentialamplifiers CFA as difference amplifier So my results are like this for a difference amplifier followed by a comparator. The pulse here is 10 ns width and 1 ns rise and fall times and repeating at a 150 ns rate. Why is the pulse quite expanded? Which capacitor stage has done it I didn't understand. CFA as amplifier with gain The results for the above circuit are also not as desired (I just see 0 V output), why so? <Q> There is nothing is wrong with your circuit. <S> The wide pulse width is because you're feeding your diff_amp_out to a comparator. <S> It can only supply you with a pulse of full voltage if the input (in this case) is even slightly above zero volts. <S> I like how your scope shows the tiny delays from stage to stage. <S> You could adjust the 'trip' level of the comparator, but the output will always be +5 V. <S> If you just want to amplify diff_amp_out, add another AD8003 op-amp and set your gain to fit your needs. <S> At least you keep your true signal pulse width . <S> Admittedly, working with these nS pulses is tough. <S> All video and RF amplifiers are the CFA type. <S> The VFAs are best for audio and precision DC measurements. <A> You have to start from the photodiode + input transimpedance amplifier design. <S> To get that speed easily you likely should limit yourself to a small (1 mm 2 or so) low capacitance photodiode. <S> Then you should get into tricky high-speed transimpedance photodiode amplifier design (you can google a lot), e.g.: http://www.linear.com/solutions/5729 <S> http://users.ox.ac.uk/~atdgroup/technicalnotes/Getting%20the%20best%20out%20of%20photodiode%20detectors.pdf <S> It is all about noise/BW/DC limit optimization. <S> If your signal is strong enough, so you don't care about noise - it is easier. <S> In the simplest case of a very strong signal you can use just shunt resistor. <S> If you don't care about the DC component you can use an RF pulse transformer (balun) to convert impedance 1:4, get signal into 50 ohm coax (12.5 impedance visible to the photodiode) and use the Minicircuits ZFL-1000 amplifier or similar. <S> For best signal/noise at high speed one can use an avalanche photodiode, PMT (depending on the wavelength) or even SiPm (Sensl, etc.). <A> I have some doubts about your design. <S> First, it seems, the whole amplifier AD8003 has no meaning: you first attenuate input signal 2 times by R4 R6 <S> then you amplify it 2 times by U1 R3 R5. <S> Why not to send the input straight to high speed comparator? <S> The rise / fall times of your optical signal are so short that you can not improve signal quality by AD8003 <S> , you only make it worse, as we see from your pictures. <S> Second, R2 in parallel with R4 R6 has no meaning. <S> So - the basic question is: what for do you use AD8003 stage?
| CFA amps are designed for current feed back, which lowers the input impedance, but greatly increases its bandwidth and high frequency response compared to VFA.
|
How can I ensure low-resistance ground contacts to an aluminum enclosure? I am building an electronics project with an aluminum enclosure. I need to ground it—and very well—for human safety. (The system handles 300 A of current, and I want to be sure that if there is an internal failure in the unit causing a panel of the enclosure to become energized, a fuse will blow rather than the panel becoming lethal to touch.) Of course aluminum forms a 4 nm layer of aluminum oxide within 100 picoseconds of contact with air , and aluminum oxide is an electrical insulator with a resistivity of 1x10^14 Ω·cm . Putting those together means that there is a 31.5 megaohm resistance between a piece of aluminum and a 0.5" conductive disc (such as a metal washer) in perfect contact with it if the aluminum oxide layer is intact. I know that (for example) mechanically tightening things, particularly using a locking washer with teeth, can easily penetrate the aluminum oxide layer, but I am looking for industry specifications or best practices on choosing washers and torquing the connectors to ensure this happens. These ground connectors might also need be removed and replaced when servicing the equipment, and I want to provide specifications for reinstalling them to make sure this is done properly to ensure the aluminum oxide layer is pierced and conductivity is maintained. As an example, I'd like to be able to write in the repair manual—with calculations based on sources I can cite—something such as "replace the lock washer with a new one, part number XXXXX, then tighten the bolt to YYYY Newton meters of torque, which will ensure an electrical resistance to the enclosure of less than 2.5 mΩ. Verify this by taking a four-point resistance measurement between test points 17 and 29, which should be 5.0 mΩ or less." I'd also be interested in things such conductive greases, which I could put on self-tapping studs, that might work particularly well on aluminum to prevent the formation of the oxide layer, and could then be left in place. Can anyone offer some pointers to resources I could use to learn best practices in working with aluminum as a conductor, and which I could use to develop a set of guidelines for repair and rework of electrical conductors in contact with aluminum? <Q> I am going to elaborate a bit so this becomes an answer. <S> Typical avionic equipment has a typical bonding requirement of 2.5 to 5\$m\Omega\$ around a chassis. <S> There are different classes of bonding and earthing with their own set of requirements, depending on the threat and application. <S> There is an excellent survey from NASA on methods used over the years and the rationale behind them. <S> As already noted, MIL-HDBK-419A Volume 1 and Volume 2 contain a wealth of application assistance. <S> On the subject of oxidisation, it is common to use a chemical conversion coating to prevent aluminium from oxidising in both aircraft and on board ships; this has an advantage of reducing overall corrosion artefacts as a mated face could be the same metal as the new coating. <S> Note that galvanic corrosion (also known as dissimilar metal corrosion) can be a major issue in aircraft, and we seek to minimise (or eliminate) it as it adds costs to the system operator because galvanic corrosion will eventually require repairs. <S> It is not always apparent that a great deal of assistance can be available for this type of issue, unless you just happen to be in (or have been in) one of the industries that require it. <A> http://www.we-llc.com/docs/librariesprovider3/default-document-library/code-compliant-weeb-info-for-inspectors.pdf?sfvrsn=0 <S> I would be tempted to have the enclosure anodized, as I'm not sure that the Aluminum oxide layer is the equivalent of anodization. <S> I'm sure the makers of the WEEB washers could tell you. <A> You could also try capacitive discharge welding or short arc welding that is currently employed for steel electrical boxes and can be done also on aluminium. <S> This is a very quick and sure system. <S> Generally who builds boxes for electronics could do this, ask him.
| Remember also that alumina is very brittle and aluminium is very soft so when you tighten a bolt the local pressure breaks the layer and makes a good contact to base metal. There are specialized washers designed to produce gas-tight connections through anodization layers.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.