source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
Problem with LT3750 charging I'm trying to build a flyback circuit to charge a 1uF cap to ~300, but am having a lot of issues. I'm just following the data sheet diagram for a 300V 3A charger (see pic below). I've never been able to get an output voltage of more than ~150V. And after moving things around, output voltage will usually be ~8V. The layout I followed is a combination of the data sheet recommendation and conversation with LT tech support. Components pretty much follow the diagram below. I've attached a picture of my very unprofessional circuit to see if maybe someone could point out a glaring problem I don't see. Other LT3750 circuits I've seen online are usually on actual circuit boards, but I'm not quite sure how to go about getting that done. Maybe someone could advise? It's quite frustrating to fail at making a proven design work. Any help would be greatly appreciated. <Q> first of all make sure the voltage rating is D1 and C4 is much higher than 300V. For a capacitor, the voltage rating would be at least double of the operation voltage for safety. <S> Second, try to build a PCB to test your idea. <S> I have experience to design a boost circuit using LT1930, from 3.7v to +-12v. <S> I never be able to get a good output with breadboard, either very noisy or not enough output. <S> After I deploy everything on a pcb board, the signal is so nice, clean and steady. <S> Please follow my suggestion: always using surface mount component(resistor,Capacitor), try to use solder paste for SMT. <S> If you don't know how to use it, please search youtube "hotplate solder paste", you will find it is so easy. <S> After I learned it, I very seldom use solder iron to solder chips. <S> Believe me. <S> It is amazing. <S> The performance on pcb is so good. <S> You will thank me. <S> In your situation, you can start from easy one, choosing slightly big components for soldering. <S> By the way for starting of smd, you can check this video: https://www.youtube.com/watch?v=yHRabJupfSo <S> It is not a very good video, but good to start. <S> I would find a chance to take the video if I populate a board next time. <A> Components pretty much follow the diagram below. <S> I hope they don't because that circuit diagram is wrong if you are using the DA2032 coilcraft transformer. <S> This circuit diagram (below) matches the coilcraft DA2032-AL transformer (the one you appear to have fitted) <S> : - In your diagram pins 6 and 7 are connected together but this will short one of the windings out: - Footnote to people who post schematics that supposedly represent their actual circuit but don't - this wastes my time and their's <A> your circuit has 2 times current sensed resistor [25mohm] vs 12mohm to reach 300v... <S> your circuit goes into constant current limit and cuts down voltage loop regulation by half and thus you will only see 150v max output....run your circuit in ltspice 4 and tune Rs and voltage feedback <S> Rvout/Rdcm resistor gain ratio...and you will be able to regulate any output voltage you want....
| If you exactly follow the example in the datasheet, the best shot is using pcb.
|
What to call the PCB that a daughterboard is plugged into? I'm looking for a concise term to describe something. Suppose I'm making a gizmo which has application-specific circuitry but also requires substantial computing power. Not wanting to reinvent the wheel, I decide to incorporate a pre-made board, call it PCB "B", which is a single-board computer like an Udoo x86 or Raspberry Pi, for said computing power. I design a larger PCB, call it PCB "A", which has "B" attach as a mezzanine board via pin headers, stand-offs, etc. PCB "A" has the application-specific circuits and maybe other stuff such as a power supply, connectors to the outside world, etc. What exactly do you call PCB "A"? Once upon a time, the physical relationship of "A" and "B" would have qualified "A" as a motherboard. Problem is, since the ubiquity of PCs, that term now carries specific connotations; in particular, it would tend to imply the CPU and computer chipset are on "A" when really they're on "B". Mainboard has the same problem, as it's generally understood as a synonym of motherboard. Backplane is not really applicable either, because "A" is not just a bus interconnection board. Various SBCs have their own terms like shield or cape to describe daughter cards, but while "A" may use some of the same pin headers, using these terms to describe "A" seems to misrepresent it as being diminutive to "B". Is there some generally understood term for PCB "A" that doesn't carry the wrong connotation? (For lack of anything better I might call it the "application board", but I prefer to use standard terminology where possible.) <Q> What exactly do you call PCB "A"? <S> A motherboard and the PCB that mounts on it is called a daughterboard. <S> I hear what you say about the implication of it being mistaken for a PC motherboard <S> but it still gets called a motherboard in my book. <S> PC motherboards have not cornered the exclusive use of the term. <A> In the case of computing you have a motherboard which as you say contain the main brains of the computer, and then daughterboards which tend to contain peripherals or other interfaces. <S> In your case however you have things the other way around. <S> You have the brains on a small module ("B"), and then the interfaces and peripherals on the larger board ("A"). <S> In this setup I would actually not go for daughterboard/motherboard, but rather "Compute Module" (or SoM/"System on Module"), and "Baseboard" or "Carrier Board" . <S> Your "B" board would be the module, and your "A" board would be the baseboard. <S> This is actually quite a common arrangement for things like FPGA/SoC modules where you have a baseboard that has various interfaces and connections, and then interchangeable modules which simply break out the FPGA pins to useful connectors and have the various power supply gubbins on them. <A> These are the terms I use or used. <S> Motherboard: mainboard, carrier board, primary board, or by function/application like processor board, FPGA board, memory board.
| Daughter board: secondary board, or module 1 or 2 of more than one, subcircuit, application board like transmitter/receiver board.
|
Maximum resistive load on 3 phases 400V, 16 A I have several resistive heating element and I want to draw the maximum power from a 400 V, 3 phase, 16 A socket. If I design the heating elements to connect between phase and null = 230 V RMS * 16 A, I get 3 * 3680 W = 11 kW max Power out of the socket. If I design the heating elements to connect between two phases I get 400 V RMS, but I can draw less than 16 A * 400 V. How much less? Is it 12 A per element between two phases? In which case I can draw 14,4 kW out of the socket. <Q> You are absolutely right that you can draw less current through each resistor when connecting phase to phase. <S> The actual current for each resistive element will be 9.23 A <S> not the 11 <S> A you calculated. <S> The current flowing in each line is limited by the wiring for each phase, which is rated at 16 A continuous. <S> You have two choices ... <S> Delta or Wye connected loads. <S> In the case of Delta connected circuit the line current (that flowing through the wiring) is <S> 1.732 * the individual resistor current when at least two resistive loads are connected to that line. <S> The converse being with a 16 A line current limit, you can only get 16/1.732 -- <S> > 9.23 <S> A flowing in each resistor. <S> This limits the total W to 3.693 kW per resistor and 11 kW total. <S> In the case of Wye connected load resistors <S> the line current is limited by the wiring to 16 A, the voltage across each resistor becomes 230 V -- <S> > <S> 3680 <S> W <S> so total power is 11 kW. <S> Net result is that you can only dissipate the same total watts in either Delta or Wye connected loads. <A> So this is what your run-off-the-mill <S> 3-phase heater looks like simulate this circuit – <S> Schematic created using CircuitLab <S> $$R_1= <S> R_2= <S> R_3= <S> R$$ Notice that at the center point, your neutral wire could be connected — but due to the 120° phase offset summing to 360°, that point is at a constant 0V, anyways, so this is, strictly speaking, optional¹. <S> Now, apply the Y-Δ transform : simulate this circuit For symmetry reasons, we can deduct that \$R_{12}=R_{23}=R_{31}=R_\Lambda\$. <S> $$R_\Lambda = <S> \frac{RR+RR+RR}{R}= \frac{3R^2}{R} = <S> 3R$$ <S> The important part here is that *we can't tell whether it's Y or Δ from the outside <S> **! <S> In other words, if your circuit breaker trips for \$R\$ in Y-configuration, it'll trip for \$3R\$ in Δ-configuration. <S> Since the power going into the two circuits can't be different <S> , it's impossible that you find one to give you more heating than the other. <S> ¹ <S> I was actually at a friend's place where one of the three phases in her multi-apartment building was dead – so she had no light in the kitchen. <S> Unless she turned on the heater – which led to the ungrounded center point of the Y-configured heater getting shifted, and thus, the dead P3 getting seeing enough voltage to turn on an incandescent light bulb dimly. <S> Scary stuff. <A> Well the fuse is 16A. <S> This is for each wire/phase in the supply. <S> So if a phase current exceeds 16A for a period of time it will trip. <S> What power can you draw from that? <S> It doesn't depend on your wiring, it can be calculated as a wye or delta but the result is the same: 11kW. <S> If you calculate it as a wye/star, you have 230VAC of voltage and 16A of current times x3. <S> P = <S> 230V <S> * <S> 16A <S> * 3 <S> = 11.0kW <S> If you calculate it as delta, you have 400VAC of voltage and line-to-line current of I_ll=16A * sqrt(3) = <S> 16A <S> * 1.732 <S> = 27.713A. P = <S> 400V <S> * 16A <S> * sqrt(3 <S> ) = 11.0kW. <S> These are for purely resistive loads. <S> If you have a phase-shift between current and voltage you will have a reactive part aswell. <S> The fuse are tripping on the sum of the resistive and reactive current, so having a reactive part (lagging or leading) will reduce the amount of real power that can be delivered. <S> The phase shift is cos(phi) and is often between 0.7-1.0 lagging, depending on the application.
| No matter how you connect the load resistances, there can only be a 16 A line current in each phase wiring run.
|
Can I wake ESP8266 with photoresistor like in this schematic? I want to wake ESP8266 when the light turns on, send information to the server, go to deep-sleep for a few seconds and then send the information again and keep doing that. If the server stops receiving the information, it would mean that the light is tuned off back again. This is my schematic: Will it work? I'm asking specifically about the part where I connect my photoresistor to the RESET pin. Additional question (optionally to answer): If I'd like to swap resistor with potentiometer, how should I connect it? <Q> I'd recommend not doing that. <S> I'd instead recommend using something that has a clear threshold and issues a single reset impulse instead of a constant high, so this doesn't happen. <S> In any case, RESET should be driven with a binary signal, and not something like " <S> well, I was 0.1 V below threshold, I'm 0.05 V above, so better reset". <S> Anyway, I don't know the ESP8266 very well, but are you sure you want to reset the device? <S> That's like switching your PC on and off with the power switch, just to wake it up. <S> Maybe the ESP8266 has a pin that can be used as interrupt instead and a sleep mode to wake up from, and your would prefer that. <S> Your idea seems to be that you hold the ESP8266 in reset mode while it's dark – I'm not even sure that is in any way power-efficient. <A> In theory: yes, with the right R6. <S> In practice, you will need a slightly different solution. <S> To reset the EPS8266, REST needs to be pulled below Vcc/4 (usually ~0.8V). <S> It has an internal weak pull-up and most pre-made modules (D1, NodeMCU, etc) add another external pull-up to avoid spurious resets. <S> Your photoresistor is probably over 1MΩ in the dark and in the 10..100kΩ range in light (measure!). <S> Between Vcc and REST, it acts like another, variable pull-up. <S> To pull the device into reset, you need to choose R6 low enough to overcome all these pull-ups, while at the same time keeping it high enough to avoid spurious resets. <S> In practice, this won't work. <S> What you could do instead is using a Schmitt trigger (e.g. 74AC14), voltage comparator (with open collector, e.g. LM393) or OpAmp to monitor the voltage divider and use their output to pull REST to GND when it's dark. <S> As an alternative, I would suggest that you use the CH_PD (ENABLE) <S> pin. <S> Instead of the permanent 10k pull-up (R1), use the same voltage divider and select R6 to be around 3-4 times the 'light' resistance of your phototransistor. <S> This should work without any additional parts. <S> The only drawback is that you can't use the RTC memory as technically your device is off, not sleeping. <S> PS: <S> To those not familiar with the ESP8266, the 'wake up' is actually a reset (usually triggered by the RTC pulling the reset line low) and the device has to be re-initialized after wake up. <A> As mentioned by Marcus, using an analog value for this isn't advised. <S> If you can I'd suggest changing to a photodiode, and then using it to trigger the one-shot circuit in this answer to a previous ESP8266 question <S> If you still want to use the photoresistor and a pot to have an adjustable threshold, you're probably going to want to use a comparator. <S> Something similar to this with pin 7 connected to the one-shot circuit. <A> You can keep the ESP in reset constantly when dark, but check that no additional current is drawn in that unusual state. <S> An option is to wake up with the light as you wrote, send data, set an automatic wake up to about a minute in the future, go to deep sleep. <S> If light stays on, you wake up one minute later and send data again. <S> If light goes off, you will not wake up and that's ok for you. <S> This is not the best. <S> Connect the phototransistor to both an attiny and to ESP, set the Attiny to wake up upon level CHANGE of the pin where you connect the trigger, wait for 1 second ( <S> in case the finger slips and light is turned on/off quickly by mistake), have the Attiny turn on the ESP using a P-MOSFET, have the ESP read the Schmitt output and send data. <S> You wake up only when needed, it costs only one attiny. <S> It fits on the back of the ESP-12. <S> You can also avoid the MOSFET and simply use the attiny to turn on the ESP from its power-off state. <S> http://ww1.microchip.com/downloads/en/DeviceDoc/doc8126.pdf 7.1, <S> sleep modes: you can wake it up from power down with pin changes. <S> 19.4 power consumption in power down: 0.2 uA. Years of use if you use a Li-Ion with MCP170x linear regulator to 3V. For info: in deep sleep some memory of the ESP is retained, about 60-100 bytes I think. <S> In power down mode, not.
| Photoresistors are slow devices, and you might be oscillating around the point where a reset is triggered, so this might not work that great.
|
Phase-locking pixel clock to HSYNC/VSYNC I am trying to capture pixel data going to a small B&W CRT display. The signals I have to work with are the TTL-level pixel data signal, HSYNC, and VSYNC. I know the pixel clock frequency (~16 MHz) but for my application, I have no access to the pixel clock signal. I want to sample the pixel signal at the appropriate time (during the middle of the bit period, not during the transition), so I figured I need to generate a new 16 MHz clock with some phase relationship to an edge of the HSYNC signal and use that to sample the pixel signal. I know how to use a PLL to multiply a clock signal and maintain a certain phase relationship between the input and output, but how do I maintain a similar relationship between a new 16 MHz clock and a signal that only occasionally has an edge (HSYNC)? Or is there a better way to solve this problem? <Q> One way to approach this would be to use your PLL (referenced to HSYNC) to generate a master clock at 3× or 4× the pixel clock, and then use a Johnson counter to generate new pixel clocks with 3 or 4 different phase values. <S> You can then select the phase that has the desired timing, either manually with a jumper, or electronically with a multiplexer. <S> There are ways to lock a PLL directly to an intermittent reference (i.e., the video signal itself), but since you already know the nominal dot rate, this shouldn't be necessary. <S> However, you could use the phase detector from such a system to help you automatically select the best phase of the Johnson counter for sampling. <A> Even with such old technology, it was common to derive all timing from one master clock. <S> If so, HSYNC is likely synchronous with pixel edges - easing your PLL design considerably. <S> A 'scope triggered on HSYNC while looking at TTL video would tell you if HSYNC is synchronous with pixel clock? <S> If it is, you also get some idea of jitter that your PLL will have to deal with. <S> And synchronous HSYNC eases your PLL design considerably. <S> Even in this easy scenario, you must contend with having no PLL input during vertical retrace. <S> In the case of non-synchronous HSYNC, VSYNC, your pixel clock reconstruction will be very difficult. <S> You might just have to over-sample pixels, and re-construct the output video from a frame buffer, introducing a one-frame delay. <A> This is (in principle) fairly easy. <S> You don't try to lock on your 16MHz clock. <S> Instead, you build a divider which has an output the same frequency as your sync. <S> For instance, if your horizontal period is 63.5 usec, this is 1,016 cycles of 16 MHz. <S> So you would feed the 16 MHz into a divide by 1016 chain, then sync the output of the chain to Hsync. <S> This is, "in principle" fairly straightforward, but the devil is in the details. <S> You must know EXACTLY how the clock is related to the sync, or the new 16 MHz will not be locked to the pixel positions. <S> You'll need to use a VCXO for an oscillator, or the large dividing factor will produce phase jitter on the VCO which may make the system unusable. <S> Finally, you'll need to experimentally determine the phase shift between the pixels and the 16 MHz, <S> which may or may not give you problems. <S> Note that you cannot do this with a fixed-frequency 16 MHz oscillator. <S> The oscillator frequency MUST be controllable, and should preferably be tunable over a very small range, like 100 ppm. <S> Hence the need for a VCXO.
| This sounds like a one-off project (most just trash B&W CRT technology).
|
Is there a problem with using delay functions when using internal oscillator in PIC ? I am doing a project that requires having the micro-controller to use the delay_ms() functions (compiler CCS C) to wait for a determined period of time. I am using PIC 16f628A and while the program ran as intended in Proteus but it ran about 85 times faster that that when I tried it on the board . I did choose to set the internal oscillator to the minimum possible frequency (48khz) to decrease power consumption and I specified that in the wizard and so the .h file have the line #use delay(internal=48kHz). What am I doing wrong ?. <Q> Summary: <S> You need to set OSCF (bit 3) to <S> 0 in the PCON <S> register in your code (i.e. during runtime) <S> when you want the PIC INTOSC (Internal Oscillator) to run at a nominal 48 kHz (actually anywhere between 31.4 kHz and 78.62 kHz) instead of the power-on INTOSC default frequency of 4 MHz. <S> Details: <S> I did choose to set the internal oscillator to the minimum possible frequency (48khz) to decrease power consumption and I specified that in the wizard and so the .h file have the line #use delay(internal=48kHz). <S> The problem is that none of the things you list actually set the INTOSC hardware to 48 kHz. <S> Based on what you said, it seems your software assumes the CPU will run at 48 kHz, but your hardware will still run at the default 4 MHz INTOSC frequency. <S> the program ran as intended in Proteus but it ran about 85 times faster <S> that that when I tried it on the board . <S> Yes, that's what I expect. <S> 85 faster x 48 kHz = 4 MHz (approx.) <S> This result suggests that your MCU was actually still running at the default INTOSC frequency of 4 MHz. <S> The important point is that you cannot configure that PIC to run at 48 kHz from power-on . <S> If you set the CONFIG BITS (a.k.a. Fuses) to one of the two variants of INTOSC setting, then the MCU will use the internal 4 MHz frequency at power-on. <S> Then, when you want to switch it to 48 kHz (perhaps at the start of your main() <S> but perhaps elsewhere in your code - it's up to you to choose) <S> you then set OSCF (bit 3) to 0 in the PCON register - that bit <S> is what switches the INTOSC frequency from 4 MHz to 48 kHz (after a short switchover transition). <S> See section 14.2.8 "SPECIAL FEATURE: DUAL-SPEED OSCILLATOR MODES" on page 101 of the PIC16F628A datasheet for more details. <S> Also note that the datasheet does not specify the accuracy of the 48 kHz clock ( <S> only the 4 MHz clock accuracy is specified there). <S> However the PIC16F628A errata shows that the 48 kHz clock can actually vary between 31.4 kHz to 78.62 kHz. <A> You may not correctly set the clock. <S> You can do the following test: do a while loop test:while(1){RB0=1;RB0=0;}Use oscilloscope to measure the speed. <S> Each instruct would take 4 clock cycle to execute if i remember correctly. <S> If the speed is right, which means clock has no problem. <S> otherwise fix the clock setting. <S> if the clock is right, please check the code for delay function. <S> I attached a code I used earlier for this chip: <S> I believe I got the source code from : http://www.alternatezone.com/electronics/dds.htm <S> delay.c <S> / <S> * * <S> Delay functions <S> * See delay.h for details <S> * * <S> Make sure this code is compiled with full optimization!!! <S> */#include <S> "delay.h"voidDelayMs(unsigned char <S> cnt){#if XTAL_FREQ <= <S> 2MHZ <S> do { DelayUs(996); } <S> while(--cnt);#endif#if <S> XTAL_FREQ <S> > 2MHZ unsigned char i; do { i <S> = 4; <S> do { DelayUs(250); } while(--i); } while(--cnt);#endif} <S> delay.h . <S> please note that you need to define the value of XTAL_FREQ. <S> in this example it is 4MHz <S> / <S> * * <S> Delay functions for HI-TECH C on the PIC <S> * * <S> Functions available: <S> * DelayUs(x) <S> Delay specified number of microseconds * DelayMs(x) <S> Delay specified number of milliseconds <S> * * Note that there are range limits: <S> x must not exceed 255 - for xtal <S> * frequencies <S> > 12MHz <S> the range for DelayUs is even smaller. <S> * <S> To use DelayUs <S> it is only necessary to include this file; <S> to use <S> * DelayMs you must include delay.c in your project. <S> * *// <S> * Set the crystal frequency in the CPP predefined symbols list in HPDPIC, or on the PICC commmand line, e.g. picc -DXTAL_FREQ=4MHZ or picc -DXTAL_FREQ=100KHZ <S> Note that this is the crystal frequency <S> , the CPU clock is divided by 4. <S> * <S> MAKE SURE this code is compiled with full optimization!!! <S> */#ifndef <S> XTAL_FREQ#define XTAL_FREQ <S> 4MHZ <S> / <S> * Crystal frequency in MHz <S> */#endif#define <S> MHZ *1000L / <S> * number of kHz in a MHz <S> */#define <S> KHZ <S> *1 <S> / <S> * number of kHz in <S> a kHz */#if <S> XTAL_FREQ <S> >= 12MHZ#define DelayUs(x) <S> { unsigned char _dcnt <S> ; \ _dcnt = (x)*((XTAL_FREQ)/(12MHZ)); \ while(--_dcnt ! <S> = 0) \ continue; } <S> #else#define DelayUs(x) <S> { unsigned char _dcnt <S> ; \ _dcnt = <S> (x)/((12MHZ)/(XTAL_FREQ))|1; \ while(--_dcnt ! <S> = 0) \ continue; }#endifextern void DelayMs(unsigned char); <A> Using delay_ms() is always critical if you want to have accurate timing. <S> It is difficult without seeing your code, but probably your processing takes too long. <S> A simple example: while(1) { delay_ms(100); do_some_calculation(); toggle_an_led();} You might think the LED will toggle every 100 ms, but if the calculation takes for example 10 ms, the interval is rather 110 ms. <S> Even toggling an LED itself has an influence on the timing.
| So, if you want to have accurate timing, rather use a hardware timer and not delay_ms().
|
What's the difference between a laser diode and ordinary LED? What are the discerning properties that distinguish them; determining their different characteristics? In other words, I'm trying to do a compare / contrast: What do they have in common? How are they different? <Q> Laser diodes use an optical resonant cavity which when injected with light above the "Lasing threshold" generates orders of magnitude more light than injected by the high Q resonance. <S> A integrated PD detects the output so that it must be regulated to avoid out of control heat rise. <S> A thick metal case is essential to dissipate the heat and higher power. <S> Natural effects of high Q emission of photons include coherence of the light with low jitter, whereas LEDs are low Q vibrations of band-gap vibrations resulting in incoherent phonon emissions of light. <S> Lasers are currently about 30% efficient whereas LEDs are currently up to 70% efficient. <A> Lazed light is different from normal light. <S> The photons emitted from a laser have temporal and spatial coherence, meaning that they are all traveling in the same direction with the same phase, which is why it shows up so focused. <S> LEDs, while monochromatic (emitting a single wavelength of light) do not have temporal or spatial coherence -- the photos are not phase aligned and are not all traveling in the same direction. <S> You can put some kind of plastic optics to route the light somewhat where you want it, but it is not lazed. <A> The biggest difference? <S> A population inversion. <S> Second difference? <S> Mirrors. <S> LED's emit light from recombination of holes and electrons across a P-N junction: <S> Source: Warwick Lasers emit light from exciting a population of particles to an excited state, which then causes emission of a photon: <S> Mirrors also help to maintain the excitation of a population inversion. <S> The gain of the lasing is increased with mirrors, photons are more likely to hit an excited atom causing the emission process to begin again. <S> Source: Olympus <S> So what about a laser diode? <S> Simply a diode with mirrors (and some other atomical arrangements to keep the emission on a narrower bandwidth) <S> Source: <S> Laser diode working applications <A> Normal colored led must be (in the same analoque) considered to be a huge bunch of much lower power radio transmitters with wide beam antennas and having all slightly different frequencies and fluctuating output powers - no possiblity to use in anything that needs exactly pure sinewave and only one and well <S> defined propagation direction is allowed.
| Just to add something to already good other answers: Think Laser diode as a pure, high power non-modulated sinewave radio transmitter with nearly ideal one direction antenna, only its wavelength is small compared to radios.
|
Can I draw brief large bursts of current from a shift register, above output pin current rating (i.e. cap or IR sensor on the register output pin)? I am tinkering around with shift registers and I want to power an IR sensor from one of the output pins on the shift register. The register can push or pull 8mA/pin(recommended) or 20mA/pin(max) with a max of 50mA through the entire register. The IR sensor draws 5mA on average and 70mA during short bursts. I am going to add a 10uF cap across the power and ground of the IR sensor. Will this work, or will it destroy my register in the long run? Would having capacitors on the shift register output pins destroy the shift register, since they would require brief bursts of current to charge up every time? <Q> What are absolute maximum ratings there for? <S> No, what are they really there for? <S> They are to protect the manufacturer's reputation. <S> Engineers talk to each other. <S> I used this ACME part and it failed. <S> I used this NXP/TI/Analog/Maxim part, and it kept working! <S> The reason it kept working is because the data sheet accurately reflected what the part would do, and warned me against trying to make it do things that, in the long term, would cause sufficient numbers of devices to fail, that the manufacturer would get a bad name for selling things that didn't work. <S> So, can you draw 50mA from a pin that's rated 20mA absolute max? <S> Certainly. <S> For how long? <S> Maybe a few uS, maybe a month, you do the tests. <S> The manufacturer has already done the tests, thousands of hours of them, and decided that something about the part, the bond wire fusing, or electromigration in the diffusions perhaps, means that they expect that running at that current will change something vital about the device, like how well it works, or whether it works. <S> Use the limited current from your S/R pin to switch a driver that is specified to handle the IR diode current. <A> When the SR output switches from low to high, it will be driving a nearly/completely-empty capacitor that will instantaneously act like a short circuit and then start to charge. <S> It has a similar problem with discharging the capacitance when the output switches from high to low. <S> This will be putting stress on your SR's output stage beyond that in the manufacturer's data sheets and that may well diminish its reliability. <S> It's a cheap and simple step to drive a switching transistor stage with the output. <S> (You could use a MOSFET but also look at a ProFET, they might be available at whatever your voltage is.) <S> Then you have completely remove that possibility and any consequences from your design. <A> Putting capacitors on the output may not destroy your shift register, but it's certainly not a good design. <S> Why can't you buffer your register outputs with a FET device for a single channel that will sink 70+ mA. <S> Perhaps something like this: http://www.diodes.com/_files/datasheets/ds30734.pdf <S> Or if you need multiple outputs you could even consider the fabulous DMOS FET equivalent of the ULN2003: http://www.ti.com/lit/ds/symlink/tpic2701.pdf <S> You could even consider a 'Power' shift register like the http://www.ti.com/lit/ds/symlink/tpic6c595.pdf which includes everything in the one chip.
| As you are anticipating pulsing the diode, putting a capacitor across the diode is the worst thing you could do, as it increases the inrush current when you pulse the diode on.
|
Strange output of Temperature probes I've got 8 ds18b20 (knock-off I assume) temperature probes monitoring the temperature in a freezer. I'm outputting the value of those probes each minute to an RRDTool graph using a Raspberry Pi. However, occasionally I get some very strange readings and was wondering if anyone here would be able to help me figure out why. Here's a graph for one of the eight probes (they're all pretty much the same): AS you can see there are 8, somewhat evenly spaced spikes in the graph. The interesting thing is that I have 2 other probes also connected to the same 1-wire network that are just outside the freezer. They do not have any spikes at all: The probes are hanging in free-air inside the freezer, not touching any surface. Initially I thought that maybe there was some sort of electrical interference, maybe caused by the freezer's compressor. And this may still be the case, I'm not sure, but the fact that the 2 ambient temperature probes don't show any affect indicates to me that it's probably not interference. The external probes are pretty close to the freezer as well. Also there's the fact that immediately after the spike, the temperature doesn't return to the value that it had before the spike, but seems to be cooling down and slowly leveling out as it reaches the freezer's set temperature. I'd like to rule out electrical interference before continuing my investigation. If anyone knows whether this is possible, I'd really like to hear from you. Some more information that may help: It's a chest-freezer, It's not in optimal state as there's a small "leak"(?) there's water around it's base, I use old cardboard to absorb the water and it usually evaporates from the cardboard quite quickly. EDIT: So it looks like it's a daily thing. Here's some more graphs for the same sensor: I'm guessing that this is just a 'feature' of the freezer, and even though it looks highly inefficient, I'll just have to live with it. <Q> The drop in temperature during Sun AM establishes that the compressor works without producing pickup in your sensor, particularly if (as you indicate) the wires were not moved during the test, although the little glitch near the beginning of the trace may well be such pickup. <S> This pretty clearly indicates that the temperature swings you're seeing are real. <S> The regularity of the swings, especially in amplitude, and to some degree in timing, also argue that the swings are real, and the freezer controller is doing something odd. <S> The suggestion that this is a defrost cycle seems reasonable, except that I'd expect that more often, like once a day or something similar. <A> AS you can see there are 8, somewhat evenly spaced spikes in the graph. <S> could be someone opening the door, or defrosting, etc. <S> put a few sensors there and if they all behave the same, you can rule out a bad sensor. <A> The temperature is being controlled to better than 4°C. <S> You haven't given any specs on what the temperature should be, so this is obviously within spec. <S> Perhaps someone opened the freezer to cause the 8 events. <S> You haven't said anything about how controlled the conditions are for this test. <S> In short, you've given no reason that indicates this isn't part of normal operation, or that the result is out of spec. <S> There is no problem here.
| The 8 events could be some auto-defrost cycle.
|
Does reversing polarity in AC damage the equipment? (Sonoff Wifi Switch) Firstly, I am a beginner. Excuse me if this is a dumb question. I have ordered a Wi-Fi switch named Sonoff. I have a doubt whether the device will work if the polarity is reversed or will it get damaged? If I'm correct, AC doesnt have polarity right? Then why do the manufacturers specify Live and neutral on the components? So can I interchange Live and Neutral on this component? Thanks in advance!! <Q> I have ordered a Wi-Fi switch named Sonoff. <S> I have a doubt whether the device will work if the polarity is reversed or will it get damaged? <S> It will almost certainly not be damaged <S> IT WILL LEAVE YOUR APPLIANCE LIVE WHEN TURNED OFF. <S> If I'm correct, AC doesnt have polarity right? <S> Most AC supplies have one side referenced to ground and the other side at a significant voltage relative to ground. <S> If single pole switching or overcurrent protection are in use it is safer <S> if they are placed in the "live" side, that is the side that is away from ground. <S> This way when turned off the appliance will not be live. <S> If edison screw lampholders are used it is safer if the live goes to the center contact rather than the screw thread as the screw thread is much easier to accidentally touch than the center contact. <S> Then why do the manufacturers specify Live and neutral on the components? <S> Several reasons. <S> On some equipment, especially older equipment the neutral may be less well insulated than the live. <S> This is less common nowadays though as most equipment is designed for worldwide applicability and many countries use unpolarised sockets. <S> It allows single pole switching and protective devices to be placed in the live side. <S> In the case of a product like this it is desirable to keep the live-neutral distinction in downstream wiring even if the product itself doesn't care. <S> P.S. <S> I'm not seeing any evidence that the sonoff has ANY safety approvals. <A> L means live not load. <S> There are two power wires coming from your household (or office) fuse or distribution board. <S> Neutral happens to be the one that is cross coupled to your building earth and that is done for safety reasons. <S> Apart from that, the two power wires are interchangeable regards functionality. <S> Should you swap live and neutral? <S> No, do what the label says unless there are explicit rules in the data sheet or manual. <A> Several devices care about that, e.g. A lamp with a screw-in bulb, the outer screw shell should be neutral as you are more likely to touch it. <S> Other things don't care. <S> It particularly matters on switches. <S> Switches should be switching the hot wire, and leaving neutral through continuous. <S> (Since the alternative is sending hot continuous and switching the neutral - bad). <S> Even if this device is listed (tested) for polarity reversal, it passes power through itself downstream. <S> The downstream device may not be listed, e.g. A lamp. <S> In most US cities/states, the law is that you must use a part as instructed, since the listing (testing) is based on that.
| AC power is polarized in the sense that one leg is "neutral" and its voltage is near that of safety ground. It will almost certainly work
|
Simple wire loop for detecting AC Mains activity? Part of a project involves knowing if an AC motor is running or not, with that bit of information read by an Arduino. I understand there are several different paths I could take to detect this. The motor is controlled by relays with a 24V coil, so I could use a voltage divider of resistors across the coil of the relay. Or I could put a small-value current-sense resistor in series with the relay coil. However, this question is about another idea: The motor is a 120VAC (US mains) device. Could I wrap a couple of turns of small-gauge wire around one of the wires delivering mains power to the motor, rectify and smooth it and detect the resulting voltage using an opto-isolator? The schematic below shows an inductor, but that is really the couple of turns around the wire. simulate this circuit – Schematic created using CircuitLab Is this idea even viable for detecting (not necessarily measuring) AC power/current? <Q> Could I wrap a couple of turns of small-gauge wire around one of the wires delivering mains power to the motor, rectify and smooth it and detect the resulting voltage using an opto-isolator? <S> You are mistaken on how you think a magnetic field is produced by a wire and how to detect it with a coil. <S> Basically, a wire produces a magnetic field like so: - <S> Field lines need to "cut" through the plane of each wound turn as per below: - If you rotated the above coil through 90 degrees (moving the coil from the X=0 plane to the Z=0 plane), field lines no longer cut the coil and no voltage is induced. <S> Ditto if you rotated the coil from the X=0 plane to the Y=0 plane. <S> So, with wires wrapped round the conductor you get no cutting-action. <S> Unfortunately, the symbol for a current transformer does show it (incorrectly I might add) as wires wrapped around a central conductor but, in reality, the wires are wound in such a way as to run parallel with the current carrying conductor like so: - <A> You would need a current transformer, rather than simply making a few turns around the wire. <S> A current transformer consists of a ferrite ring with the secondary wound on it. <S> The primary wire, carrying the current to be sensed, is passed through the ring. <S> The seconodary winding must be terminated with a "burden resistor", and you measure the voltage across that resistor. <S> If you do a web search for "current transformer" you should find lots of information and illustrations. <S> Sensing the coil voltage on the controlling relay just tells you that the motor should be running. <S> The motor may not be running if the AC has failed, or the relay has failed, although power is available to the coil. <A> If you don't need an exact measure of the magnetic field from the current use a hall effect sensor . <S> DC or AC fields can be measure and the sensor is smaller. <S> The current in the wire can even be measured. <S> Here is an example
| Now if you wrap wires around the current carrying conductor your wire loops "cut" no net magnetic field lines and no voltage will be induced.
|
Driving a fan using a mosfet, do I need a diode? I am using this BS170 MOSFET to drive a fan using a pwm signal generated by a 3.3V MCU according to the following schematics: (I know about the wrong symbol it's just for the packaging on the PCB) So my gate gets connected to ground using the pulldown resistor and to the PWM signal. I have two questions: Should i still use the D1 diode for back EMF protection or i can relay on the internal one in my transistor. Is my transistor is suitable for a fan using 0.32A. It should handle 0.5A so it should work right? <Q> 1) Depends on the fan, if it has inductance (A DC brushed motor) <S> then you may want some kind of over voltage protection like D1. <S> 2) <S> It says 500mA right in the datasheet, your fan can't be more than this. <S> A 320mA fan will be fine. <S> The mosfet also has a 5Ω RdsOn which means its like a 5Ω resistor when its on. <S> At 320mV this will cause the voltage of the fan to be 1.6V, a fan with a controller may or may not be able to tolerate this. <A> For your first question, yes keep the diode - if it isn't needed because of something special about the fan, it will do no harm. <S> If it is needed then leaving it out will damage the reliability of your circuit. <S> The diode in your MOSFET is not useful to do the same thing. <S> For the second question, personally, I wouldn't use this MOSFET for this job - it's a 'small signal' MOSFET, and you would almost certainly be much better with a power MOSFET of some kind: <S> The Rdson is too high - at 0.32A <S> , you'd be dissipating 0.32 <S> *0.32 <S> *5 = 512mW which is more than the device is rated to dissipate. <S> And that's before you start to think about switching losses caused by your PWM switching. <S> There are no characteristic graphs for Vgs as low as 3.3V - although you're above the threshold at 3.3V, you're clearly not operating the device in the way the designers intended. <S> Additionally you should almost always add a resistor in series with the gate of a MOSFET, to control the turn on speed/gate current. <S> Here's a post about choosing a FET: <S> Selecting a MOSFET for driving load from logic <A> For the first question, I believe depending if your using a threshold voltage to manage the power to the fans, a diode will be useful. <S> For the second question, If I understand the question correctly, you want to manage the current to the fan to prevent overloading to the first Fan or the second actually. <S> Seemingly, if the circuit is trying to maintain two fans that require different amounts of power, I suggest testing that Fan 1 can operate until a certain threshold is met and when the Fan 2 is needed to start. <S> Overall, according to the specifications, the maximum threshold voltage is 3.0V, so I would work around the voltage to determine the amount of current to each fan. <S> Consider adding/removing resistors based on the 3.0v threshold.
| If its a brushless DC motor (like a PC fan) then it already has a circuit built in for overvoltage protection or it doesn't need it.
|
Reversing a power supply with an spdt switch? I'm currently trying to turn a 12 rpm motor from clockwise to counter clockwise and vice versa, I just wanted to know if it was possible to do with 1 spdt toggle switch, Ideas? <Q> so you are 'wasting' one contact on the switch and replacing it with another contact or transistor(s)). <S> If you have two supplies (eg. <S> a bench supply with <S> +/- <S> outputs) - not likely: simulate this circuit – <S> Schematic created using CircuitLab <S> More likely situation (only one supply available): simulate this circuit Since SPDT relays are much cheaper and more available than DPDT relays <S> (eg. <S> JS1a type) <S> this would be a better solution if your switch can handle the motor current (and a bit more for the relay). <S> If you want to add suppression to the motor to minimize contact arcing, connect a bridge rectifier with the 'AC input' terminals across the motor, and the + terminal to the positive supply and the - terminal to ground, as shown above. <A> In the general case, it takes a DPDT switch to effectively flip two leads: <S> If all you have access to are the two wires from the power supply and the two wires to the motor, then no, it can't be done with a SPDT switch. <A> Using ONLY spdt switch, it's not possible. <S> But, you can using four npn-transistors to build a simple H-bridge circuit to change the polarity of the motor. <S> Here's a simple circuit to do that. <S> Here, S1 is the desired SPDT switch. <S> To make your life easier, you can also use motor drivers like l293d that has two H-bridges built-in. <A> You can, granted you have a Bipolar supply. <S> If this contraption is battery powered then you can form a Bipolar supply by tapping off the centre connection of a series arrangement of two batteries.
| You can use a dual supply, use the switch to control the motor and a half bridge, or use the switch to control a SPDT relay (or a full H bridge/DPDT relay but in the latter two cases you only need a SPST switch
|
How is voltage exactly created in a citcuit? I just began studying electricity and concept of voltage and this particular concept of voltage is so confusing. I have searched answer for this question everywhere and every answers stops short at saying that voltage is created by chemicals in a battery and nothing beyond this. You don't need to go deep into explaining chemistry but I just need an idea of how electric potential is created in batter and how it "forces" electrons in circuit to move. Like how is charges spearate in first place. If battery is not one to supply electrons in first place, then how come potential difference is created in battery which would then force electrons in wire to flow. If anyone can shine light on this, I really appreciate it. <Q> You should start from a physics course. <S> Learn what charge is, what is electric field, what is electric potential. <S> Basically negatively charged particles are attracted to positively charged particles. <S> So if you have them separately, they will flow one to another and release some energy in the way. <S> Circuits are just engineering of physics. <S> Chemical reactions in battery separates atoms to ions (positive charge) and electrons (negative charge), while wires allow them to flow. <A> Physics won't give you a 'why' answer, but it can give a 'how', or a 'sort of like this', if you're happy to think in terms of lower level things that can't themselves be explained. <S> Taking your specific case of a chemical battery. <S> A battery is two different conductive materials (electrodes), separated by an ionic fluid (electrolyte). <S> For instance a zinc plated nail and a copper coin pushed into a lemon makes a battery . <S> This battery operates by exactly the same principles as any other. <S> The electrodes will tend to dissolve in the electrolyte to different extents. <S> Solution involves the movement of some of the electrons on atoms to make ions. <S> Some atoms hold tight to their outer electrons, some less so. <S> This difference makes for a difference in the energy that is released or absorbed when atoms go in or out of solution, and also makes for a difference in the number of excess electrons (charge) that appears on each electrode as it reaches equilibrium with the electrolyte. <S> You can think of the different charges on the electrodes as causing a different 'pressure' of excess electrons. <S> Some languages in fact use the word 'pressure' for voltage. <S> If the electrodes are connected by a wire, this voltage difference causes electrons to flow along the wire. <S> You can see that this energy is intimately related to the difference in the energy of solution of the different electrode materials. <S> A lemon battery works, but does not work well. <S> Most of battery technology is about finding electrodes with a large energy difference, that are cheap to make and practical to build into a battery, hence lead, nickel, lithium, and electrloytes that work well with them. <A> The voltages in a circuit are (usually) there because the circuit needs to be powered for it to work. <S> Maybe your question is more like: many circuits generate their own internal voltages. <S> Especially many power supplies generate for example 5 V. <S> How do they "know" what 5 V is ? <S> Although as explained in the other answers, certain chemical reactions like used in batteries result in a certain voltage. <S> But as batteries wear out and may contain nasty chemicals, this is not a very practical way to generate a specific voltage. <S> Also that voltage changes a lot over temperature and lifetime and charge level of the battery. <S> So instead most circuits use a special circuit as a reference, that circuit is called a bandgap reference circuit . <S> It uses the physical properties of diodes and/or transistors to generate a (fairly) constant voltage. <S> Such a circuit is used in almost every voltage regulator circuit and other circuits where a reference voltage is needed. <S> Using voltage divider circuits (consisting of just a couple of resistors) and amplifiers that bandgap reference voltage can be reduced or amplified to practically any value needed.
| More strictly, voltage is a measure of how much energy is involved when charge moves.
|
Charging a supercapacitor with a zener diode I have a dc source which gives 5.5V and I want to know how to give only 2.7V to the supercapacitor for charging.I'm thinking about this circuit : What are the limitations of the circuit? How can I improve the circuit ? Will it work ? Thanks for your help. <Q> The 1N5223 zener is nominally 2.7 volts and can be as low as 2.565 volts or as high as 2.835 volts so if you can live with this variation then that's fine. <S> The BAT54 being schottky will look like a small volt drop as the capacitor reaches full charge so the only problem here is what the residual current taken by the capacitor is - the higher the residual, the more the BAT54 drops voltage. <S> It could be about 0.1 to 0.2 volts at 100 uA. <S> Is this good enough? <S> Only the OP can decide. <S> If you want something more accurate I'd use a linear voltage regulator to produce 2.7 volts and feed that to the supercap. <S> If you had a couple of more volts available at the input there is this Linear technology solution: <S> - LT describe it as: - The LT3663 is a 1.2A Step-Down Switching Regulator with Output Current Limit that is ideal for use as a supercapacitor charger since it provides adjustable output voltage and adjustable charging current limit. <S> I'm sure there will be switching or linear solutions out there that will do the job rather than rely on the tolerance of a zener diode (+/- <S> 5%) and the unknown leakage current of the supercap. <A> It should work (i didn't look on zener datasheet, but you probably took 2.6V or something). <A> I believe not, here is why: I don't know how precise your 2.7V has to be, <S> so you're going to have to make a decision here. <S> :) <S> The zener voltage is not precise. <S> I had a look at Fairchild's datasheet , It says that Vz = 2.565V Min / 2.8V Max @Iz, add to this the temperature drift <S> you're gonna be off. <S> Moreover, even if you have a precise 2.7V zener diode, you have to take the voltage drop across your D1 into consideration. <A> I tested my system IRL <S> and it works, I have the 2.7V I want. <S> Using the LT3663 is a great solution but as I answered to some people, I have to keep the price of system low and therefore keep the number of component as little as possible. <S> Thanks everyone for your help.
| But i would suggest a bit more interesting circuit: a constant current charger (PNP biased with a resistor) and a comparator that would measure supercap voltage and switch the PNP off once 2.8 is reached.
|
typical value of contact resistance: 1) banana connector, 2) solder joint Two separate questions: 1) What is the typical value of contact resistance in electrical connectors ? I am interested mostly in a banana connector for delivering power supply from the source to the PCB. 2) What is the typical value of contact resistance of a "good" handmade solder joint? This is assuming the contact was soldered properly (in particular, enough solder was used). Appendix 1. My particular situation My particular situation: 1.8 V, 1 A DC are delivered from the power source to the PCB and I am going to use a thicker copper wire to decrease IR drop. I calculated 1 mm diam copper wire, 50 cm length has a resistance of about 10 mOhm (so, 10 mV IR drop at 1 A). So, if I buy a thicker copper wire, solder it to banana jacks which are to be inserted into the power source and the PCB, I'll have four parasitic contact resistances: two for the banana connectors and two for the two solder joints. So, I wonder, does it make sense to fight for 10 mOhm (if thicker wires are used) if contact resistances could be greater. This is a test PCB (to test a chip), no VRM on it. Appendix 2. Some contact resistance values for banana connector found in the Internet URL1 , page 2: 10 mOhm maximum. URL2 , page 5: 0.5 - 1 mOhm maximum. URL3 , page 4: 0.5 - 1 mOhm maximum. URL4 : 1 mOhm maximum. Appendex 3. Questions about conact resistance found at Stack Exchange Relay with contact resistance less than 10mohm How the relay contact resistance is measured? What kind of effects does a relay have on signals? What causes contact resistance? <Q> 10 mΩ for a full-size banana jack sounds really high to me. <S> However, the obvious things for you to do are: Measure it. <S> Put a known current thru the connection and measure the voltage across it. <S> This will tell you what you are actually getting, not the worst case it could be. <S> Look at a datasheet. <S> Find a reputable company that makes banana connectors and see what they rate the maximum contact resistance as. <S> These things should have been obvious, and basic research you should have done before asking. <S> As for the solder joint, there is no mechanical "contact" there, and the problems associated with it. <S> It's just a solid metal connection. <S> Most likely the weak link is the copper trace on the board after the solder joint. <S> For starters, analyze a solder joint as if only the copper were present, but directly connected. <S> In a proper solder joint, the copper parts are touching anyway. <S> Solder has higher resistivity than copper, but there is also a lot more of it in parallel around the joint. <S> The gap between the solid conductors is also very small. <A> Measuring is a good idea. <S> But please consider to use the kelvin (4-wire) measurement. <S> Otherwise you will influence the measurement with the resistance of the measurement leads + contact resistance and so on. <S> Explanation of the kelvin measurement can be seen in this link . <A> The range if resistance values you've found quoted for banana connectors instructive. <S> The resistance is very variable in detail, it depends on surface finish, contact pressure, and surface contamination. <S> This is because they do not get contaminated like base metals. <S> However, connecting banana plugs 'wipes' the contact surfaces, so even with base metals, a fresh connection will be good. <S> As you want to connect a relatively low voltage high current power supply, it's worth considering the resistance of the wire-to-connector connection, and the length of wire itself, both of which are in series with the connector. <S> It would be quite easy to end up with an order of magnitude more resistance end to end than was present in the mating connector surfaces themselves. <A> Typical solder joints are in the range of a few hundred micro-ohms (or at least they were every time I have made a measurement). <S> But of course it varies with the size of the joint, type of solder used, thickness of the solder, etc.
| You will get the most consistent, and lowest, contact resistances from connections where both the plug and socket are gold plated.
|
why need to connect ESD diode close to the connector? It is noted that TVS should be placed close to the connector.What is the significance of placing close the connector? Added the snapshot for reference <Q> To prevent induced or conducted (due to breakdown) spikes on other traces. <A> The App Note explains it already. <S> " <S> However, I ESD can still be "steered " towards the TVS by making L4 much larger than L1. <S> This is accomplished by placing the TVS as near to the ESD Source as the PCB design rules allow while placing the Protected IC far away from the TVS." <S> Re-read it carefully until understood. <A> Actually, it is not always the best solution to put it as close as possible to the source. <S> The best solution depends on the shielding of the product. <S> If the return path has a low impedance path to a large conducting surface, the statement is correct. <S> The esd will flow directly to the conductor with a relatively large capacitance to ground. <S> It will transfer the energy away from the circuit more easily. <S> It will limit di/dt, avoids breakdown of TVS, limits emmission, etc. <S> The negative part is that the peak voltage at the point of impact is higher (dv/dt aswell), but your circuit connected to the TVS is protected. <S> Stick to the other hint in the list!
| If the product has no direct path to loose the energy, it is good to limit the current by adding inductance and/or resistance between the source and the TVS (thin trace or ferrite).
|
Driving blue LED with 3.3V using an op-amp I'm trying to design a circuit that drives six LEDs - two red, two green, and two blue - off a 3.3V GPIO pin. Since the voltage drop on a blue LED is in that range, I'm attempting to step the voltage up with an LM318N and then drop it with a resistor. However, all the red and green ones are working when set up like this, but the blue ones are not. Is this a reasonable setup to accomplish what I'm going for, or is there something wrong with the way I'm approaching this? <Q> LM318 is only specified to drive to within 3v of either rail, it's not a rail to rail (R2R) output opamp. <S> As you've got it supplied from the +5v pin, its output current into a 3v load will be low at best, and might be zero. <S> For a drop in replacement, find an amplifier that's got a R2R output. <S> LM339 makes a nice low current FET driver as well, output pulls close to ground, and the inputs are very flexible. <A> I don't think you need to boost the voltage to drive the LED. <S> Just look for a blue LED with a low Vf. <S> Also, there is no reason to use an op-amp. <S> Use this circuit <S> : simulate this circuit – Schematic created using CircuitLab <S> You may need to tune the resistor values a bit. <S> This is a current source circuit. <S> When the GPIO is high, you will get a fairly constant current of roughly 10mA through the LED over a wide range of voltages. <S> This circuit performs much better than a simple current limiting resistor when you barely have enough voltage to drive the LED. <S> The basic idea is that R1 and R2 are a resistor divider from the GPIO voltage of 3.3V. <S> They produce around 1.1V at the base of Q1. <S> Q1 is not saturated. <S> It is acting as an emitter follower, producing 1.1 - 0.6V at the emitter. <S> This means that the emitter current is around 0.5V / R3. <S> The things you might want to do are lower R2 to get a smaller voltage at the base, and change R3 to change the operating current. <S> Right now it is around 10mA, very roughly. <S> For example, a single cell lithium ion battery voltage, or 5V from USB. <S> The LED will only turn on when the GPIO is high, no matter what voltage you use for the diode. <A> A single transistor configured as a common emitter is enough to allow you to drive the LED from a Pi GPIO and using the Pi +5v line. <S> Q1 can be any jellybean NPN part such as a 2N3904, 2N4401 etc. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> There are several flaws in the circuit. <S> Firstly you are loading the GPIO pin quite heavilly. <S> When unloaded the pin will be at damn near 3.3V <S> but when loaded down heavilly it may be substantially lower than that. <S> Secondly as others have pointed out your op-amp is nowhere near a rail to rail device. <S> In general using an op-amp for this seems rather silly. <S> I would suggest you use a transitor switch configuration and use the transitor to control all the LEDs, not just the blue ones.
| Also, since this circuit maintains a regulated current irrespective of voltage, you can potentially use an unregulated voltage for the LED. It might be simpler to use a switch instead, a simple transistor or FET, or a 5v rail gate with 3.3v capable input like an HCT74xx part.
|
How does a negative current start in the inductor in ac circuit? As shown in this image, how does a negative current appear at the beginning, when the voltage starts to increase ? I understand the current lead lag mathematics behind it. I just can't figure out, what is going on with the electronics inside the wire. How can an increasing voltage, cause a reverse current ? Update Thanks to Andy aka for pointing out the right solution with correct references. I realised that the answer to the anomaly lies in the equation of the inductor. Since for an inductor v = L * di/dt Therefore current = Integral (v/L * dt) Integral of Vsin(wt)/L = -Vcos(wt)/L And the TRICKY part is to take integrals between 2 points A and B So the current at a time t = -Vcos(wt)/L - (-Vcos(0))/L = -Vcos(wt) + V/L The cos(0) is the DC component that "lifts" the entire current wave above the 0-line so the current keeps flowing only in positive direction (assuming a 0 resistance path). This is mentioned here - http://licn.typepad.com/my_weblog/2012/03/to-zero-cross-or-not-to-zero-cross-john-dunn-consultant-ambertec-pe-pc.html Observation This is probably why the SMPS of my CPU cabinet makes a loud noise right when I turn it on. The sound persists for about a minute and then fades away. Because the voltage supply starts with a 0V at t=0 and the ac current flows only on the positive side of the graph above the 0 level, giving a higher dc value. Notes I tried reading a few more online tutorials on circuits to understand the very basics of RL and RC AC circuits. Every RL and RC ac circuit has 2 currents. One is the steady-state current I(ss) and another is the transient current I(tr). The phasor method calculates only the steady-state current of it. Which many believe to be the "real" behavior of the circuit, while it is not. This link shows the calculation of the transient current in a RL ac circuit - http://www.ee.nthu.edu.tw/~sdyang/Courses/Circuits/Ch09_Std.pdf I have created a partsim simulation of the RL ac circuit, which shows the whole thing nicely - https://www.partsim.com/simulator/#69215 <Q> When you apply a sinewave voltage to an inductor, current has no option other than to begin at zero - it cannot suddenly adopt a negative value as would be implied in the steady-state waveforms. <S> In fact it has a name - it's called inrush current and particularly in motors, inductors and transformers it can cause core saturation because, the current rises above the peak of the normal steady-state value: <S> - This is why some inductive loads are connected to the AC supply when the voltage waveform is at its maximum peak (negative or positive) because that is when the current would naturally flow through a zero-crossing point. <S> Without core saturation (or other losses) the current would remain having a positive average DC value. <S> Here's a picture that shows what I mean: - Picture taken from here . <S> When voltage is applied at a zero crossing <S> the current will remain having a dc value for some time and <S> this dc offset will ebb away with losses due to R1. <A> What is shown in the diagram is about one period in steady state . <S> It does not show the conditions when the sinusoidal soucre is switched on at t=0 ( transient state ). <S> I.e. you have to imagine that many more cycles have happened before. <S> So at t=0 the coil is already energized and current and voltage phases are stabilized. <S> "Start of the alternating current cycle" does not mean that alternating current is switched on at that moment. <S> It is just the start of the repeating cycle . <A> An inductor is a energy storage device <S> but it stores energy as magnetic energy. <S> As opposed to a capacitor which stores energy as electrical energy (electrons). <S> If you start with thinking about current through an inductor what you have to think about is the rate of change of current through the inductor resulting in certain amplitude of voltage across it that opposes the rate of change. <S> So it generates a negative voltage across it. <S> When I say negative <S> I mean the opposite polarity of a voltage that would be generated across a resistor say. <S> This is essentially Lenz's Law that is important to understand to also understand Faraday's Law. <S> At a basic physical level its all about momentum and the fact that if you have mass you cannot move through a medium without something else acting against you. <S> Natural systems do not like change. <S> In the case of inductors and coils this is really great for us as it allows us to generate power and convert power.$$emf = -L\frac{di}{dt} = <S> -\frac{d\Phi}{dt} <S> $$\$ \Phi \$ is the magnetic flux
| If you force a voltage across a capacitor its the rate of change of voltage that gives rise to a current that will oppose the rate of change.
|
DC motor speed control when having relay change directions im building a circuit atm that drive a dc motor (24V, 5Amp) in both direction.(Not using H-bridge because of the high amp motor, I cant find anything at the local store at least).For now the circuit look likes this: Which when the Arduino set pin 7 and 8 to HI-LOW, motor got 1 direction. And when LOW-HI is the opposite.(Relay switch the negative and positive connections) However with this method I cant control the speed of the motor. I would love to make the motor slowly increase it speed. I'm thinking to use a MOSFET and connect it to the PWM pin on the Arduino this method only works for 1 direction but not when reverse. I'm lost, can someone point me a direction or giving me some link to go to. Thanks you first in advance. <Q> The mosfet PWM drive and relays should not be a problem if your motor changes its direction when the polarity changes; there exist one direction only motors that have no permanent magnets but a magnetizing coil. <S> you have carefully thought what happens if you change the direction at full speed <S> you have added some spark supressors to make your relays to live longer. <S> A motor this big haves substantial stray inductance that causes sparks in opening relay switches. <S> See the addendum: you know that a motor can take many times more current when started compared to a steady run. <S> Addendum for spark supression: If one tooks off a DC supply wire from a running motor, a spark is seen. <S> This is common for heavily inductive parts such as relays, motors, electric magnets ets... <S> The spark is caused by induction. <S> In coils that are connected to a DC voltage it means that the coil current never starts or stops in zero time, it happens only gradually. <S> When a DC supply wire is taken off, the current still continues and degrades gradually until all magnetic energy is dissipated. <S> The current chooses that way which breaks most easily. <S> Often it's the air and that we see as a spark. <S> Suppress <S> the spark by giving an easier way for that inductive spike. <S> Take two more than 12 V zener diodes, say 15 V each. <S> Connect them series, but opposite directions. <S> Connect this dual zener supressor parallel with your motor. <S> It does not short the 12 V supply because 15 V is needed for any current. <S> But your relay gets no welding treatment when the contact opens because the inductive spike current is wasted in a zener diode. <S> Accepted supplementary note: <S> When the a running motor is suddenly supplied with reverse voltage, the supply current will spike. <S> An inductor at the power source will limit the momentary current spike and this will be less abusive to the power source. <A> You can't control speed directly with the relays, because they are mechanical and you can't switch them in high-speed. <S> You pretty much have two solutions: use an H-bridge or use the relays to switch direction and a single transistor to control speed. <S> There are H-bridges that control a 24V, 5A motor. <S> You usually can't find them on a local store. <S> If local store components are a requirement, you have the option of building your own H-bridge with discrete components. <S> The harder part will usually be the high-side, since your Arduino is at most 0-5V and your motor is 24V, and P-MOSFETs are harder to find in bigger amps, but you can usually work it out with optocouplers. <S> By the way, you usually want to isolate with optocouplers anyway. <S> Also remember to properly calculate the driving requirements, because the Arduino may not handle the gate current capacitances. <S> The bottom-most circuit in this thread looks like a nice starting point. <S> Since you already have the relay switches working, it may be easier to go for the second option. <S> For that, the simplest way is to use an N-channel MOSFET between the relays "ground" and the actual ground. <S> Don't forget to properly calculate the driving requirements too. <S> Since you didn't provide actual schematics, it's hard to give better advice, but the schematic from this page looks good (reproduced for completeness): <S> You can change the DPDT relay for what you already have, and the TIP120 for a more powerful MOSFET. <A> R1 keeps M1 turned off if the Arduino's PWM output is floating or disconnected. <S> Lm and Rm represent the internal inductance and resistance of the motor. <S> D1 is a fast switching Schottky diode which prevents high voltage spikes (caused by magnetic energy stored in the motor's inductance) that would otherwise destroy the FET. <S> It also improves efficiency by recirculating current through the motor while the magnetic field is collapsing. <S> To change direction you just have to reverse the motor leads like you did before. <S> The relay contacts must be wired between the speed controller and the motor, like this:- simulate this circuit <S> The motor goes forward when one relay is operated, and reverse when the other relay is operated. <S> When both relays are released the motor has a short across it, so it will stop quickly and resist being rotated mechanically. <S> Otherwise it will draw a very high current as it reverses, because while still spinning in the original direction it generates a voltage which adds to the applied voltage.
| Before switching directions you should stop the motor by slowly ramping the PWM down to zero and then releasing both relays. Inductive current limiter can be used too. First make a unidirectional PWM speed controller, like this:- simulate this circuit – Schematic created using CircuitLab M1 is a 'logic level' FET rated to turn fully on with 5V on the Gate and easily handle the applied voltage and current.
|
How to prevent a voltage drop when a motor turns on? I have two motors (looking like this one ) connected in parallel and attached to a 9 V battery; initially motor A is switched on and Motor B is switched off. When I turn on motor B, motor A's speed slows down - and when I turn motor B off, motor A speed is back to previous value. How do I prevent this behaviour, i.e. what should I do to keep motors running at same speed no matter if the other motor is on or off? <Q> Use a better power supply than a 9V battery- <S> one that does not change output voltage significantly regardless of the number of motors. <S> You could also add a regulator. <S> For example, a 12-volt SLA (Sealed lead-acid) battery with a suitable regulator for the currents involved would be almost perfect. <S> Even without the regulator it would be pretty good (provided your motors were happy enough with the higher voltage). <S> The problem you note will only get (much) worse as the battery is depleted and its internal resistance rises. <A> <A> In case you wan't to keep the circuit simple (no regulator) <S> you should look into the C-rating of the used battery. <S> There are some LiPo for drones which can deliver around a 1kW in power which is hopefully enough for your two motors. <S> The C-rate is a measure of the rate at which a battery is being discharged. <S> It is defined as the discharge current divided by the theoretical current draw under which the battery would deliver its nominal rated capacity in one hour.[32] <S> A 1C discharge rate would deliver the battery's rated capacity in 1 hour. <S> A 2C discharge rate means it will discharge twice as fast (30 minutes). <S> A 1C discharge rate on a 1.6 Ah battery means a discharge current of 1.6 A. A 2C rate would mean a discharge current of 3.2 A. Standards for rechargeable batteries generally rate the capacity over a 4-hour, 8 hour or longer discharge time. <S> Because of internal resistance loss and the chemical processes inside the cells, a battery rarely delivers nameplate rated capacity in only one hour. <S> Types intended for special purposes, such as in a computer uninterruptible power supply, may be rated by manufacturers for discharge periods much less than one hour. <S> Straight from Wikipedia <A> The simplest solution is to keep the motors isolated . <S> This means using a power supply/battery for each motor. <S> This way you can turn each motor on or off, without affecting the other motor. <S> As to keeping the motors running at the same speed, your control circuitry could become very complicated, depending on the "degree of accuracy" you want the speed to be the same.
| use a beefier battery or a power supply; use a motor that consumes little current; use a separate power source to power the motors....
|
Diode pin labels swapped in this circuit? I'm curious why "bl_anode" is shown connected to the "cathode" of the diodes and the two "bl_cat1/2" are shown connected to the anode sides? Shouldn't the labeling be the opposite? My understanding is the arrow in a diode diagram is the anode and the line the cathode <Q> Hence the net-name sounds fine to me and has zilch to do with D4020 and D4021. <S> Another hint is the "LCM" might stand for liquid crystal module i.e. a module containing a liquid crystal display AND a back light. <A> It looks like the labels you are referring to are with respect to the rest of the circuit. <S> Basically, this is a power supply, and its outputs are intended to be connected to cathodes and anodes of other things. <S> The fact that internally the cathodes of diodes are connected to the "anode" output is immaterial. <A> This is how a cellphone display backlight gets it DC supply from a step-up switchmode power supply. <S> Actually there are two parallel power supplies with different inductor sizes to choose from for adjustable brightness. <S> BL anode is the anode of the background light source (not drawn here). <S> The diodes are fast schottky diodes that belong to the step-up switch mode supply circuit. <S> In that switchmode topology the +output comes from the cathode of the diode. <S> Addendum <S> : The parallel supplies produce a current pulse in their turn to the output. <S> The current pulses carry different size charges. <S> Output capacitor keeps the output voltage quite stable (=no flashes and minimal radio interference). <S> By interlacing a proper ratio between big and small pulses the average current is as wanted. <S> No reason, why the supplies can't operate even at the same time. <S> But to be sure, a datasheet is needed. <S> I haven't found one. <S> Step-up principle is used to bring the wanted charge into proper voltage level for many leds in series. <S> Cellphone battery has too low voltage as-is.
| "BL" possibly means "back-light" and maybe it is a device that uses an anode and a cathode.
|
How to know or find if a potentiometer is UL certified We have selected a 100kohm potentiometer to dim LED's that are connected to a class 2 power supply. However, according to UL; all selected parts need to be UL recognized for the device to be UL approved. I can not find any information regarding whether certain potentiometers are UL recognized or not. The potentiometer that we have selected is P231-QC20BR100K but it can be any panel mounted 100K pot pretty much I have a UL lookup where I can find a manufacturer and see if they have UL approved parts but it seems backwards and I have yet to find one that does. Here is the UL SAM http://www.ul.com/global/documents/offerings/industries/lighting/Downloads/SAM.pdf I know if I select parts out of the SAM then I am good to go, but I am looking for a more basic solution. <Q> Not every component needs to be recognized. <S> The power supply as a whole will be listed / recognized. <S> It's up to the evaluation to the appropriate standard to determine if the component is critical. <S> Potentiometers are not usually recognized unless they have switches or are needed to seal out water and dust etc. <A> I contacted a mean well engineer for this issue and his response was <S> "The potentiometer does not require to have UL cert because the PWM-60-12 CofA does not require. " <S> Datasheet - http://www.meanwell.com/productPdf.aspx?i=256 <A> look at the part for the UL mark; call the oem; check out the datasheet; call UL? <S> submit it a UL certification yourself.... <A> You can go to UL.com and get to their on-line certifications directory <S> then type in the name of the manufacturer to see which (if any) of their components are <S> UL listed or recognized. <S> If they are not on there, they are not. <S> As said, if a manufacturer goes to the trouble and considerable expense to do so, they are going to make it obvious. <S> That said, if you are connecting to the load side of a class 2 power supply, you may not need to use UL listed or Recognized components. <S> Have you talked to UL about this?
| UL listing or recognition stipulates that the UL mark must be on the product or data sheet.
|
Connect emergency button to arduino I need to connect this kind of button to my Arduino for a project : http://www.ebay.com/itm/NC-N-C-Emergency-Stop-Switch-Push-Button-Mushroom-Push-Button-4Screw-Terminal-KG-/172356802290?hash=item28214416f2 This one is 600V, so I guess I cant connect it to the 5V of my arduino.I don't know if if fits....Do I need an external battery? How to wire it? The other option is to self-build a button with one of these connectors : http://www.automation24.fr/commande-et-signalisation/element-de-contact-etroit-eaton-180792-m22-fk10-i132-3060-0.htm What do you think? Thanks ;) <Q> They are just switches - you connect them to an Arduino as you would any other switch. <S> You will probably need a pull-up resistor, unless the Arduino has internal pull-ups that you can enable. <A> (The same thing goes for the 10A rating.) <S> You don't need to do anything special. <A> Switches rated to handle 600V will often not be suitable for TTL level circuits, at least not reliably. <S> They are open to air and will eventually build up a resistive film on the surfaces. <S> Opening and closing contacts generally burns that film off, but at 5VDC there is not enough energy to do so. <S> For TTL level circuits, you need to find a button that uses what are called "reed switches" as contacts, or gold flashed contacts.
| You can hook this switch up to your Arduino just like any other switch. The 600V rating on that switch is the maximum voltage that the switch can be used with.
|
Removing noise from accelerometer data I am new to signal analysis and I need to remove noise from an accelerometer recording. 2 accelerometers are mounted a machine and recorded vibrations at 500 Hz. The aim is use vibrations to differentiate working situations, we expect increased vibrations on certain situations. The figure below shows recording for one channel. Machine started to work at 250 s and stopped at around 3100 s, the recordings before and after shows noise from other sources. These noise sources are also exists during the machine's working time. Figure below shows fft for noise and signal+noise. What I need to do is remove the noise from recordings. What kind of filter should I use? <Q> From the looks of this data you could simply clamp any samples between 0.7 and 1.3 to 1.0 so that you get a flat line until the machine starts operating. <S> Alternatively (or additionally) only enable data collection when you see a short series of samples above a particular threshold. <S> Looking at your frequency plot, you show that the noise floor is wideband and spans all the way across your target range. <S> This means there's no frequency-based filter (e.g. low pass) that you can use in this scenario. <A> your approach will depend on how much you know about the noise. <S> if you know its frequency composition, for example, you can decompose it via FFT, reset the amplitude for the frequencies where you think the noise is present, and do a reverse fft. <S> if you have additional measurements of the same physical attributes, you can fuse those measurements. <S> if you don't know much (other than that the noise is of higher frequency), you can filter it. <S> each strategy has numerous ways to be implemented, however. <A> You do that by creating band-pass filters centered around the parts of the spectrum where your signal has the highest power level. <S> The result is that you throw out the noise across most of the spectrum. <S> You also throw out some of your signal, but only in the parts of the spectrum where it had much lower power levels. <S> From your FFT graph, your signal has the highest power levels at 150Hz and 200Hz. <S> There is also a little more power at 50Hz and 225Hz. <S> If you take your FFT data array and zero out all the samples from 10Hz to 40Hz, 70Hz to 120Hz, 230Hz and onward, and then take the inverse FFT you will get your original signal, with some minor distortion, and most of the noise removed. <S> If you are using MEMS accelerometers, the RMS noise is often proportional to the square root of the sample rate. <S> So sampling at a lower rate can often lower the noise power. <S> If the noise were occurring at some particular frequency you could just create a notch filter at that frequency. <S> Or even simpler, take the FFT of your results, set the values in the FFT data array at the noise frequency to 0, and then take the inverse FFT to get your original signal minus noise. <S> The problem is that your FFT graph shows the noise amplitude as pretty flat across the in the frequency domain. <S> So it would be classified as white noise. <S> Therefore you can't use a notch filter to remove the noise. <S> If you just want to detect what operating mode some machine is in based on the vibration (and you don't really care about the shape of the signal itself), then you can just put a band pass filter at the frequency you know the vibration to be occurring, and then trigger based on the amplitude going over a certain threshold.
| You can increase the signal to noise ratio by using a filter that has a higher gain in the portions of the spectrum where your signal has higher power, and lower gain where your signal has lower power.
|
How to prevent voltage drop when a Motor turns on? In my arduino project I have 2 motors connected to a 9V battery , Motor A is switched on and Motor B is switched off, when I turn on Motor B motor A's speed slows down and When I turn motor B off motor A speed is back to previous speed. How to prevent this behavior ? , How to keep motors running at same speed no matter other motors on or off , The motor i am using Please help. :) thanks :) <Q> You need more power. <S> The battery doesn't have enough power to run both motors at the same time, use 2 9v batteries in parallel. <A> You can look on battery as on "ideal power source" in serie with "internal resistance" (which grows when battery is more used). <S> Then you connect the motor to the serie. <S> The current goes thru the circuit and the voltage is divided between the internal resistance and the external resistence - the motor (the higher resistence, the bigger part of voltage is on it). <S> So the motor gets only part of the voltage, that the battery have with anything connected. <S> If you add another motor paralel to previous one, the external resistence drops (to half, if the motors are the same), so the voltage for external rezistence drops as well and voltage for internal resistence increases acordingly. <S> So even two or three batteries in parallel may show this effect, if your motors are to take big current each. <S> Or maybe the effect would be small enought for you to not care. <S> If you really need to keep the speed, then you need to compenzate for the voltage drop. <S> Easiest way is put some DC/DC convertor/stabiliser to way to each motor separately ( be it step-up or step-down does not matter in principle) but it would take the energy from battery faster as the convertor burns some of it to heat anyway (and will cost you more), but if the battery is powerfull enought (or the convertors have wide enought range), then your motors would run the same speed regardless of thier actual count. <A> You are reaching the current capacity of the battery you have chosen. <S> The generic 9 volt battery (about 400 to 1200mAh) is not designed to provide as much current as a C or D size battery (about 4000 to 11000mAh) <S> over the same amount of time. <S> This is a case were you need to specify what is good enough before making a decision. <S> If you are doing this empirically, simply try using enough of a larger capacity batter to develop the necessary voltage. <S> It is expected your results will improve.
| If the internal resistence is small (good battery with high capacity and high current possible), then the difference is small and the voltage on battery drops only a little, if the battery is weared (high internal resistence) and those motor are hungry (low resistence), then the drop of voltage is big.
|
Controlling voltage on source and drain of MOSFET I learned that MOSFET works by controlling the voltage on the Gate terminal. For NPN MOSFET as an example, when a voltage on the gate is high enough, it lets the current to flow through the MOSFET, and when a voltage is low, the MOSFET is off. But what happens when a voltage changes on the source or drain terminal? I am studying the voltage level shifter and could not understand the logic behind it. When voltage on the source is high or low, what happens and why? What about when voltage on the drain is high or low? Thanks! <Q> The controlling factor is not 'the voltage on the gate', but 'the voltage on the gate with respect to the source terminal , Vgs'. <S> As the source voltage changes, so will Vgs. <S> That type of FET is an N-channel FET, NPN is only meaningful for bipolar transistors. <A> The gate voltage here is with respect to source and not ground. <S> So if your gate voltage is 5v wrt ground and source voltage <S> also 5v wrt ground your n-channel MOSFET(nmos) <S> would still be in off state. <S> In such case you must either increase gate voltage to >5 Volts(above forward bias voltage) or reduce the source voltage below 5v to make it work. <S> Next, when proper bias voltage, I.e Vgs is applied, the Vds (drain to source voltage ) can be used to control the current through this MOSFET. <A> Voltage is allways between two points. " <S> Gate voltage" is nothing until someone tells what is the other point. <S> Voltmeter has 2 measuring probes for that reason. <S> N-channel (not NPN) mosfet gets its external control voltage between gate and source. <S> Mosfet acts as controllable current valve. <S> The controlled valve inside the mosfet is between drain and source. <S> The load is connected typically in series with the mosfet, connection at drain. <S> The free load terminal is connected to power supply + terminal. <S> Mosfet's source is connected to the minus terminal of power supply. <S> Mosfet controls how much current is allowed to flow through the load. <S> Load current grows as big as the power supply voltage and load recistance allows. <S> Load current drops gradually to zero if one reduces the control voltage gradually. <S> In many applications it's wanted to break the load current soon. <S> In that case the G-S voltage should be reduced to zero as fast as possible. <S> Voltage between drain and source is not a way to control the mosfet. <S> I's a cosequence "how much is left" when mosfet lets a certain current to flow thru D->S and the load takes its part due Ohm's law. <S> Together load voltage and Vds are the supply voltage.
| When one applies a control voltage, typically a few volts between gate and source (+ at the gate), the mosfet opens ad effectively the load is now connected to the power supply.
|
What is this amber like substance on capacitor and inductor? What is the mysterious substance on the inductor and (failed) capacitor? What is its purpose, and how do I replace the failed capacitor with it there? The substance is hard and translucent, but I can make a very small impression with my finger nail. This comes from a Belkin n-wireless router I am trying to fix, if that helps. Thanks in advance! <Q> If you want to remove it, use a heat gun on a very low setting, if it is indeed hot glue it should allow you to scrape it off. <S> If its epoxy you have to nip at it to try and scrape enough off to allow you to remove the bad capacitor. <A> That looks like some sort of glue, used to fix the components firmly to the board. <S> You should be able to remove the capacitor by heating the glue with a hot-air gun and when warm, try break the seal between the capacitior and coil. <S> Then de-solder the capacitor and remove. <A> The translucent material on these components is there to provide mechanical stability and serves no electronic purpose. <S> It is unlikely that the inductor has failed. <S> You should be careful that its windings remain insulated from one another - their copper is coated with an insulating enamel to keep each turn insulated from its neighboring turns. <S> Your photo shows unclearly how firmly the translucent material holds the electrolytic capacitor. <S> It is likely soldered to the board with two leads. <S> These should be de-soldered, and all solder sucked away (not an easy job). <S> You might be able to slice through its black plastic sleeve to the aluminum casing. <S> It is possible that the translucent material only has a hold of the capacitor by its sleeve. <S> Try to keep the aluminum casing intact - this capacitor contains volatile material that should not contaminate you, or the board's components.
| What you are referring to is simply a magnificent blob of hot glue or more colloquially known as "hot-snot" to prevent the capacitor from breaking off with excessive vibration.
|
Why should each row of cells in battery grid have equal number of cells for maximum power transfer? How a battery grid can be made using \$Z\$ cells of \$r \space \Omega\$ internal resistance which can deliver a maximum power to a load of \$R \space \Omega\$ resistance? In a battery grid with \$N\$ cells in series and \$M\$ rows in parallel \$E_{eq}=nE\$ and \$r_{eq}=\frac{Nr}{M}\$. Total cells \$NM=Z\$ and according to maximum power transfer theorem \$\frac{N(r)}{M}=R\$ My teacher said that for maximum power transfer, each row of cells should have equal number i.e. \$N\$ number of cells. Why is it so? Why can't each row have different number of cells? <Q> Under these condition if you removed (or added) a single cell in any given row, you change the matrix of internal resistances and move the maximum power point current flow. <A> <A> Obviously the cell numbers must match or there would be high circulation currents discharging the cells <S> The resistance matching maximizes power transfer but not efficiency and certainly causes a serious temperature rise but in theory is the maximal power transfer. <S> With 10mohm typical ESR and 3.7V, this implies a 37 Amp rate with Pd=13.7W= <S> I²R heating up the battery. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Further analysis would show voltage inbalance per cell if the internal resistance, Ri is not balanced which results in thermal inbalance under high currents and unbalanced aging effects, so ESR is often matched <<1% when new and mismatch >>1% causes rapid aging of the hottest cell with highest Ri. <S> The total cell voltage *N and Ri ought to be balanced for ideal charging, which becomes critical at high current rates. <A> My teacher said that for maximum power transfer, each row of cells should have equal number i.e. N number of cells. <S> Why is it so? <S> If you have a row of (say) ten perfect cells and each cell has a terminal voltage of (say) 1 volt, then the row voltage is 10 volts. <S> If you put that in parallel with a row with a different number of cells then you will create infinite current circulating around the cells. <S> However, it's still a bad situation if each cell has 0.1 ohm resistance because there is still a voltage mismatch when you connect rows up and you will get a large circulating current that may damage or massively reduce the energy storage in each row.
| IMO what your teacher actually meant to add was that all the cells are identical, so the internal resistance values are all the same and in parallel in any given row. If you don't have an equal number of cells you'll have an imbalance in the flow of electrons which results in some cells not being used to their full (electromotive) potential.
|
How to "shift" a logic LOW of 2,5 V to GND? I'm trying to control a 24 V LED strip with a device that has its own control mechanism and PWM output. (The output PWM frequency is 1 kHz.) The device itself requires 5 V, and since it does not draw that much power so I'm planning to use a standard voltage regulator there, not that interesting. My problem here is that the device's PWM output logic is rather weird. It's logic LOW is 2,5 V while a logic HIGH is 5 V. Here is my schematic: As for the N-channel mosfet, I'm planning to use the IRLB8721 ( datasheet ). But this one won't turn off on a 2,5 V logic LOW by my device. The question is; how can I "shift" this logic LOW down to GND? Almost all Google searches I've tried lead me to logic level shifters that most often let 3,3 V logic devices connect with 5 V logic devices and vice versa... I did find two interesting topics here on the Electrical Engineering Stack Exchange with almost the same problem, but I could not figure out how to translate the solutions posted there to my specific problem: Translating to “below ground” logic levels -5/0 volt to 0/5 volt logic levels? <Q> How about some simple bipolar inverters like below? simulate this circuit – Schematic created using CircuitLab <S> With 5V at input, Q1 is turned off <S> so R4 pulls the gate high, turning on M1 and your LED strip. <S> With 2.5V at the input, Q1 is turned on which turns on Q2, which pulls the M1 gate low which shuts off M1 and your LED strip goes off. <S> You'll need to refine the circuit, check the resistor values, find a better M1, make sure response is fast enough for your PWM frequency. <A> If you want a noninverting translator, one transistor will do it. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is a so-called 'grounded base' configuration, it gives the voltage amplification you require, but no current gain (the 'Device' will have tosource 0.5 mA in order to put 5V across the R3 load resistor).The R1 <S> , R2 values set the base at 3.1V quiescent (it should switch atabout 3.7V from 'Device'). <S> This circuit does not try to be a fast switch, but a simple one. <S> R3 andthe 'Device' output current determine how fast the gate capacitance ischarged and discharged. <A> That is probably not an output with a logic digital signal, or else it might be the first device in the world that outputs such a signal. <S> Could the output be a darlington output, which is capable to drive the leds ? <S> What happens to the PWM signal when you connect a resistor to 5V or to GND ? <S> Do the test for example with a 1k resistor. <S> When you know the impedance for a high and low, then you can decide how to translate that signal. <S> Perhaps a optocoupler can be used to translate the signal or perhaps a transistor with a few diodes as suggested. <A> At first check; Is it really +5V <S> =high +2,5V <S> =low. <S> Assuming yes <S> : Use a comparator that outputs 0V or +5V directly to the gate of the mosfet. <S> Comparator tests, if device X outputs more than 3,75 V. <S> If yes, then give +5V to mosfet's gate. <S> A proper comparator that operates from single +5V supply, can produce 0V and +5V output levels and can sense properly +5V is tricky. <S> Here it is as a discrete part implementation <S> Parts must calculated starting from the end R5 must pull the gate down fast enough Q3 must push enough collector current to lift the gate near +5V Q3's base current = <S> the collector current/hfe for reasonable turn off speed, R4 must take at 0,7 V at least as much as Q3 needs base current; that is R4 < or = <S> (0,7V/ Q3's base current) <S> Ri and R2 produce the treshold 3,75 V to Q2 R3 gets about +3,05 V. R3 = <S> 3,05 V/(Q3's base current + 0,7 V/R4) <S> the 3,75 V reference should not teeter. <S> Let (R1 parallel R2) to be five times R3 or less. <S> Add a capacitor parallel with R2, if needed. <S> solve R1 and R2 from the wanted voltage and max parallelled resistance. <S> At first take Q1 = <S> Q2 = <S> 2N1711, Q3 = <S> 2N2905. <S> 2N3704 <S> and 2N3904 should be ok, too. <S> Simulate before building a prototype <A> Well, you have a whole lot of possible answers. <S> I'm going to assume you still want to use your MOSFET device, the IRLB8721. <S> A \$V_{GS}\gt 6\:\textrm{V}\$ looks like it operates very solidly then. <S> The total gate charge is about \$12\:\textrm{nC}\$ <S> and I've set up a circuit that should deliver that charge in under \$20\:\mu\textrm{s}\$, which may be fast enough given the PWM rate. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The gate voltage should switch between \$0\:\textrm{V}\$ or \$7.5\:\textrm{V}\$ when the input PWM switches between \$2.5\:\textrm{V}\$ and \$5.0\:\textrm{V}\$. <S> The circuit includes hysteresis, will keep the drive voltage to the gate of the MOSFET safely low enough, is widely tolerant of BJT \$\beta\$ <S> (anything from 100 to 300 is fine), accepts reasonable variations in the saturation current of the BJTs, and provides about \$500\:\textrm{mV}\$ width for the hysteresis band, which should provide reasonable noise immunity as well. <S> It doesn't use boutique parts and should be relatively cheap. <S> Figure about an additional \$4\:\textrm{mA}\$ load on your \$24\:\textrm{V}\$ supply rail. <S> (It doesn't need or use anything from the \$5\:\textrm{V}\$ supply rail.)
| It's also easy to tweak the gate drive for other MOSFETs which may require higher \$V_{GS}\$, especially given access to a \$24\:\textrm{V}\$ rail (as you currently have.)
|
Are Schottky diodes the best option for this application? I want to make bridge rectifiers for some toy field phones so that the wires that connect one phone to the other will work regardless of polarity. I was wondering if Schottky diodes would be the best option because of there low forward voltage drop? But I don’t know if the higher reverse leakage of a schottky would be detrimental to proper function. 6v dc (from AA batteries) would be the maximum voltage involved. If they are what would be a particular diode with the lowest forward voltage drop? Thanks. <Q> Schottky diodes are best used for fast switching application and where the voltage drop due to diode needs to be as less as possible. <S> Disadvantage is the high reverse leakage current (and the current is temperature dependent) and <S> lower reverse breakdown voltages. <S> If your project is battery operated only and need to save each bit and byte of energy for long term, then opting for normal diode <S> with least reverse leakage current will be better choice. <S> Also other factors such as price may factor. <S> Schottky diodes are expensive. <S> So, its a trade off among reverse voltage rating, speed, leakage current and price. <S> So, for your case I would really see further down on the requirement of minimum battery voltage needed for the circuit to operate and battery life time expected. <S> Only, then a good decision can be made. <S> Also, if space is not a constraint, opting for PMOSFET as blocking diodes (even NMOS will do) also help in getting out all the juice off battery to the circuit as the forward drop of FETs will be in a few 10s of mV typically. <S> Using a diode to ensure current flows in only one direction, without causing a voltage drop <A> I was wondering if Schottky diodes would be the best option because of there low forward voltage drop? <S> Probably. <S> A typical 6V dry cell battery has useful capacity down to ~4V <S> (1V per cell) and your telephones are probably designed to work down to this voltage only (if they could use a lower voltage then why have 4 cells?). <S> A silicon bridge rectifier drops ~1.4V (about the same as having one flat cell) which may significantly reduce the operating time. <S> A Schottky bridge drops <S> ~0.8V. <S> Here's an example 6V AA discharge graph. <S> I have copied the original curve to show what you would get with 0.8V and 1.4V drop through the rectifier. <S> In this example (80mA average current draw) the ~12 hours run time is reduced to ~7.5 hours with a Schottky bridge and 2 hours with a silicon bridge. <S> In practice it might not be quite this bad because the current draw will probably be less at lower voltage, but you get the idea. <S> I don’t know if the higher reverse leakage of a schottky would be detrimental to proper function. <S> Leakage current also increases as voltage approaches the diode's rating, so choose a rectifier that has much higher voltage rating than needed but not much higher current, eg. <S> 1A 30V rather than 3A 10V. <S> Finally, consider the current draw of your phones. <S> If it is much higher than the leakage current (likely) then it shouldn't be a problem. <S> A higher current diode may have lower voltage drop which is more important. <A> When you say best, what do you mean? <S> Cheapest, best performance, easiest? <S> Usually it's a balance of things. <S> Shottky is cool if only need simplicity. <S> Otherwise you need more complex circuit.
| High leakage is generally only a problem if the diode is rated for much higher current than needed or is being operated at high temperature.
|
High Capacity Phone Batteries - are they for real or a scam? I recently bought a new, allegedly high capacity battery for my Samsung Galaxy S5. It was being sold as a 3800mAh battery whereas original is a 2800mAh battery.On inspection I read the following from the cases of the 2 batteries; Original Samsung:- 2800mAh, 3.85V, 10.78Wh High capacity replacement:- 3800mAh, 3.85V, 10.78Wh If Watts = Amps x Volts then... Watt-hrs = Amp-hrs x Volts Therefore Amp-hrs = Watt-hrs / Volts or 10.78 / 3.85 = 2.8Ah (2800mAh) This equates to 2800mAh (2.8Ah) for both batteries so the replacement is wrongly labelled and sold as a 3800mAh Higher Capacity battery to Samsung Galaxy users. Or have I got something wrong here? <Q> You are correct. <S> The true measure of a battery's capacity is Wh. <S> As you noted, the information on the new battery is contradictory. <S> Either the Ah is wrong and it is higher capacity or <S> the Wh is wrong and it is the same capacity. <S> 25% increase in capacity should be pretty easy to tell over a few days. <A> Fantasy capacities are quite popular. <S> You have already found that the new battery does not last longer than the old one. <S> You'll find out half a year down the road that it will last a whole lot <S> less than the old one: the fantasy batteries tend to invest quite a lot less into the materials making a lithium ion accumulator actually long-lived (its chemistry is rather brutally corrosive and so needs a lot of mollycoddling at the cost of price and capacity). <S> So given today's market, you are likely better off looking for a battery of the original capacity or less. <S> That makes it more likely you'll get a battery that started out with significant life time expectations. <S> The only problem is that you might get 10-year old stock. <A> I think they play on flexibility of electro-chemistry of the batteries. <S> From my limited experience with Li-Ion batteries, the charge voltage level is not something of hard bounded. <S> Many actual battery specifications (for phone/tablets under development, usually not available to public, only to developers) list the charging voltage at 4.200 and even 4.300V. The charge can be done at slightly higher voltages resulting in substantially higher amount of stored energy, but at the expense of battery lifetime. <S> So it is very possible that the manufacturer can really demonstrate the claim by using a special overcharging, but they wouldn't mention the resulting reduction in lifetime. <S> Since your phone is not aware of the "new" charging requirements, it would continue to charge it to old voltage, and therefore the perceived capacity will remain the same.
| The easiest way to tell is to try the new battery.
|
A question about a debouncing circuit I was reading a tutorial on debouncing the outputs of a rotary-encoder which are read by a micro-controller.Below is the circuit given: The writer of the tutorial claims the following: "In the above circuit we use a 1uF capacitor with an SN7414 Schmitt-trigger input NAND gates that clean up the switch-contact noise." I have couple of questions regarding this claim and the circuit. 1-) I can understand the reason for the use of capacitors, they act as RC low pass filters and can filter bounces.But how does the SN7414 Schmitt-trigger help debouncing here? Is that really needed or does it improve debouncing? 2-) In the above circuit the resistor and the capacitor forms a low-pass filter with a cut-off freq. fc.What is the motivation/idea when it comes to sizing the capacitor here. Why 1uF?I mean let's say we have a rotary-encoder and it is recommended to be used with 10k resistors.Which parameter of the encoder from its data-sheet should be taken into account when sizing the capacitor.Is there a method or reasoning between choosing the capacitance and a parameter in the data-sheet? <Q> The schmitt trigger doesn't help debouncing in itself. <S> But it is required because after the RC-filter, the signal will have slow rise/fall times (this is obviously the purpose). <S> And signals with slow rise and fall times are not appropriate for feeding regular logic inputs. <S> If the gate inputs stay "inbetween states" for a long time, you'll experience higher power consumption and may see instability problems. <S> This is fixed by the schmitt triggers. <S> It also depends on the values of the pullup resistors. <S> Basically, take a RC time constant (here 10k <S> * 1µ = 10m) <S> and it should be higher than the max bouncing time. <A> 1) Why use a Schmitt-trigger input device as a buffer. <S> With that RC low pass circuit you are going to get a fairly slow rise time on the signal. <S> The signal will also have noise on it. <S> If you have an input stage without much hysteresis then the combination of a slow rise time and noise can cause the signal to cross the threshold a couple of times and so generate the exact type of noise you are trying to avoid by using a denouncing circuit. <S> By using a device with lots of hysteresis on the input you eliminate this risk. <S> Also some logic parts don't like slow ramp rates on their inputs, especially on clock inputs. <S> I've had discrete D-type latches which have failed to clock reliably due to the clock signal coming from an RC rising too slowly. <S> 2) Why that size capacitor? <S> You want a time constant that is fast in comparison to the rate at which the signals will be changing and too small for a person to notice but slower than the mechanical vibration of the contacts. <S> A few milliseconds seems like a reasonable time period to aim for. <S> The exact value isn't that important, 10uF may be a little slow and 10nF would probably be a little fast but just about anything in between would probably work fine. <A> A quick look at major application of Schmitt trigger: used to filter noise in digital lines: The one used in the OP question is inverting schmitt trigger basically to improve the transition response of the circuit: <S> Yellow: Input Green: Output <S> The capacitor size depends on the debounce time required by the encoder. <S> In the datasheet of the encoder below spec. <S> or similar will be mentioned: RC time constant to be sized accordingly. <A> we use a 1uF capacitor A circuit like that most likely will work <S> but it is wrong : I have seen cases where this type of circuit resets a MCU by pulling the input below ground. <S> Three tricks here: <S> You should have a resistor between the capacitor and the switch. <S> That defines the current during discharge. <S> If undefined, that discharge current plus stray inductance, can bee quite nasty. <S> You should use as small of a capacitance as possible. <S> To limit the amount of energy dissipated when the switch is pressed. <S> 0.1u would be my upper limit. <S> 11n or 1n preferred. <S> Use a switch with high contact resistance, if you want to eliminate thee resistor recommended in 1) above. <S> All of this is to reduce the discharge current <S> and it's slew.
| The capacitor can be sized easily, and depends on the "boucing time" indicated in the encoder datasheet.
|
Temperature raise inside a case I'm using a D2PACK triac model ACST830-8G datasheet According to datasheet (over 1cm2 cooper) it supports 2A, considering Tamb = 43° TABLE GRAPHICS I want to use this triac inside a case and without heat sink. Inside a small case, can I assume this triac will handle 2A at 43 °? I worried about this point because, as I understand, it will not be in "free air convection" as the second image says. <Q> Free air <S> convection assumes that there is enough air around the part (or in the room) that the ambient temperature of the air would not be affected by the heat from the part. <S> If the part is in an enclosure then you introduce a thermal barrier between the air in the room and the enclosure and heat the enclosure air up. <S> So interpreting the graph, the air in your enclosure will probably be more than 43C while running at 2A because of the added thermal resistance. <S> The temperature range of the part is 125C (which you probably wouldn't the part or other parts in you enclosure to reach but it will still operate). <S> One thing you will want to avoid is thermal runaway where the heat of the part increases the internal resistance and leads to more heat being generated by the part until you have a meltdown. <S> This will be determined by the thermal resistance of the enclosure. <A> Then no, you can't assume \$2\:\textrm{A}\$. <S> Imagine that you place this inside a thermally sealed environment (perfectly so.) <S> Then no heat energy would ever be lost and the temperature would rise forever. <S> Reality will be between these cases, obviously. <S> But if your circumstance is worse than "free air convection" then the temperature rise of the die will be higher and you must derate on the basis of what you know about the differences. <S> If you can't do a finite element analysis, plus convection model, based upon a detailed physical model of your circumstances, then the next best here would be for you to make a test and find out what happens in the exact circumstance that you make and use. <S> But if worse than "free air" you definitely will need to lower that figure. <A> You'll get some heat movement even thru plastic. <S> Copper foil, the standard 1 ounce/foot^2, has thermal resistance 70 degree C per watt, per square of foil.. <S> any size square.... <S> the heat flowing from one edge to the opposite edge of the square. <S> Not perpendicular to the foil, but with the layer of foil. <S> The plastic epoxy has approximately 100X that thermal resistance, but some heat does flow. <S> Suppose we have a plastic blob, 1,000x <S> the thickness (which is 1.4 mils) of standard copper foil....thus 1.4 inches. <S> The copper would have 70/1,000 or 0.07 degree C. <S> The plastic would be 100X thermal resistive, or 7 degree Centigrade per watt, if the plastic blob is a cube, 1.4 inches on a side.
| The a good way to determine what the internal temperature would be to experimentally verify it with the design in the enclosure or find the max power dissipation of the design and put a dummy load in the enclosure and measure the temperature.
|
Should I separate grounds with optocoupler and dc converter? I have a 24V input to a DC/DC converter, output is 3.3V. the 24V is also passed to an opto-coupler (diode side) while the secondary side (NPN) has a 3.3V-made from the 24V. So the opto-coupler is used as a level shifter. I thought about separating the grounds for the opto-coupler. The input 24V will have a gnd1, while the 3.3V will have a gnd2 to raise the noise immunity. But, the 3.3V is transformed from the same 24V passed to the input of the opto-coupler (which has a different ground).In this situation, will this give good results in terms of EMI, SI, point of view?Thanks all! <Q> In that case, there is probably no advantage to splitting them at the opto-coupler, and also probably no reason to use an opto-coupler at all. <S> On the other hand, if the DC-DC converter is isolated and that isolation is needed, then you would use opto-couplers (or other isolation devices) for all signals that cross the boundary. <A> In this situation, will this give good results in terms of EMI, SI, point of view? <S> Separating grounds doesn't always do that, I can't go into great detail without seeing the application. <S> If you split the ground planes you can start making antennas especially around a DC to DC converter. <S> If you need isolation and you have a transformer, I could see reason for spliting the ground plane, if there isn't a good reason then in most cases your better off with a solid one. <A> If for example an Arduino of 3.3V is used at the 3.3V end, then connect the output GND of the optocoupler to the GND of the Arduino. <S> Then the output signal of the optocoupler is relative to the Arduino GND, and that is how it should be. <S> Wires with ground currents can cause voltage drops, and noise. <S> When the ground current can choose between two wires, it is hard to predict which one will be used. <S> It might not be the thickest wire, but the wire with the best magnetic coupling to the positive current going in the other direction. <S> For high frequency currents, it is even more complex.
| High frequency currents (any current for that matter) takes the path of lowest impedance, splitting the ground plane usually increases impedance for currents to return to the source. If the DC-DC converter is not isolated, then the grounds of the 24 V and 3.3 V logic are connected. It is hard to say without knowing your application and requirements what you should do. It is better to keep them seperated, to avoid unpredictable ground currents.
|
Why does noise-figure of components become less significant further down the chain in a cascaded network? I understand mathematically why this happens, but I can't intuitively get my head round this idea. To clarify this is the cascaded noise figure equation: <Q> Think of the gains from the inputs of each stage to the output. <S> This is assuming a chain of amplifier stages. <S> The input of the first stage gets multiplied by the gain of all stages together. <S> The input of each further stage only by the gain of the remaining stages. <S> For example, you have a microphone amplifier with overall gain of 2000, where the first stage has a voltage gain of 10 and produces a low impedance output. <S> The signal out of that stage (and into the remaining stages) is now less susceptible to noise for two reasons: <S> The impedance is lower. <S> Capacitively coupled noise has high impedance because it's in series with a relatively small capacitor, usually well under 100 pF. <S> The impedance of the signal forms a voltage divider with this capacitor. <S> The lower the impedance of the signal, the more the same capacitively coupled noise is attenuated getting to that signal. <S> There is less gain from this point to the output. <S> In this example, the gain is 10x less than what the raw microphone signal undergoes. <S> In this example, noise after the first stage is only amplified by 200 to the output. <A> This only applies where the nominal signal level increases from stage to stage. <S> This is usually the case where noise figure is often quoted, like in a receiver. <S> In a signal generator, where an attenuator is often part of the signal chain, noise figure becomes more significant later in the chain. <S> Noise figure is only one part of Dynamic Range , the real limitation to the performance of signal processing systems. <S> Dynamic Range is given by the maximum signal level less the noise figure. <S> When noise figure is quoted by itself, we are usually making some assumptions about the signal levels. <S> These assumptions may be valid, but sometimes are not. <S> For the front end of a sensitive receiver, if we make the assumption that the signal level is small, then noise figure is the only thing we need to worry about. <S> As the signal level increases stage by stage, the 'headroom' between the signal and the added noise increases, so we can tolerate more added noise in later stages which have more signal. <A> Here is a noise plot from Signal Chain Explorer. <S> The 3 OpAmps have gain of 10,10,10. <S> The gain-set resistors, and the internal Rnoise_density, are scaled up by 100. <S> Notice the noise contributions are exactly equal. <S> Total is 9.4 uV <S> Here is the schematic <S> simulate this circuit <S> – Schematic created using CircuitLab <S> Here is unequally weighted lownoise design: still gain of 10x per stage, but resistors are scaled up by 10: 10 Ohm, 100 Ohm, 1Kohm. <S> Total noise is 6.4 uV. <S> Thus that stage gain is your friend, your degree-of-freedom. <S> Time to run through a low-noise design, where the FIRST STAGE DOMINATES.Assume need 1nanoVolt/RtHz noise density (60 ohms Rnoise).Allocate <S> 40 ohms total to the first stage of gain. <S> [we'll use 20 Ohms ReferredToInput for 2nd stage, 1 Ohm RTI for 3rd]And set up that first stage for gain of 5. <S> Note <S> the 2nd stage has 1/25 the impact on the front-end noise. <S> You can have 20 Ohms * 25 <S> == <S> 500 ohms of total noise generation in that 2nd stage. <S> Set up that 2nd stage for gain of 10. <S> Note <S> the 3rd stage has 1/100 <S> *1/25 = 1/2,500 <S> the impact on the 1rst stage. <S> We used up our front-end noise budget: 60 ohms = 40 + 20.Design the 3rd stage for total noise resistors of 1,000 ohms. <S> Those scale down by 1,000/2,500 or less than 1 ohm at the first stage. <S> By the way, you can use the free tool Signal Chain Explorer to tinker withthese effects. <S> Find SCE at robustcircuitdesign.com and let us know how you use it. <S> Another example: Set up the Signal Chain as 1) <S> some opamp, with Rnoise of 30 ohms, Rg of 8 ohms and Rfb of 32 ohms <S> (gain=5) <S> 2) some opamp, with Rnoise of 400 ohms, Rg of 90 ohms and Rfb of 909 ohms(gain=10) <S> 3) <S> some opamp, with Rnoise of 900 ohms, Rg of 90 ohms and Rfb of 909 ohm(gain=10) <S> You may wonder why the LowPassFilter on output node. <S> Necessary for identical noise contributions, because the first stage high-frequency noise is filtered differently from stage #3, unless we artificially constrain the high-freq noise.
| This is because noise injected after the first stage isn't amplified by that stage.
|
Difference between bit rate and baud rate and its origins? Everyone seems to have different definitions everywhere I look. According to my lecturer: \$ R_{bit} = \frac{bits}{time} \$ \$ R_{baud} = \frac{data}{time} \$ According to manufacturers : \$ R_{bit} = \frac{data}{time} \$ \$ R_{baud} = \frac{bits}{time} \$ Which is the correct one and why? Feel free to give the origins of why it is defined as such too. Related question: link . <Q> Baud rate is the rate of individual bit times or slots for symbols . <S> Not all slots necessarily carry data bits, and in some protocols, a slot can carry multiple bits. <S> Imagine, for example, four voltage levels used to indicate two bits at a time. <S> Bit rate is the rate at which the actual data bits get transferred. <S> This can be less than the baud rate because some bit time slots are used for protocol overhead. <S> It can also be more than the baud rate in advanced protocols that carry more than one bit per symbol. <S> For example, consider the common RS-232 protocol. <S> Let's say we're using 9600 baud, 8 data bits, one stop bit, and no parity bit. <S> One transmitted "character" looks like this: <S> Since the baud rate is 9600 bits/second, each time slot is 1/9600 seconds = <S> 104 µs long. <S> The character consists of a start bit, 8 data bits, and a stop bit, for a total of 10 bit time slots. <S> The whole character therefore takes 1.04 ms to transmit. <S> However, only 8 actual data bits are transmitted during this time. <S> The effective bit rate is therefore (8 bits)/(1.04 ms <S> ) = 7680 bits/second. <S> If this were a different protocol that, for example, used four voltage levels to indicate two bits at a time with the baud rate held the same, then there would be 16 bits transferred each character. <S> That would make the bit rate 15,360 bits/second, actually higher than the baud rate. <A> The line bit rate is the number of bits per second being moved. <S> The baud rate is the number of symbols per second (Baud is named after Emile Baudot ) <S> The line rate and information rate can be different due to line coding <S> An example of line coding is QAM ; QAM64 encodes 6 bits per symbol ( \$ 64\ = <S> \ <S> 2^6\$ ) <S> , so the baud rate would be the \$ \frac {line bit rate} { <S> 6}\$ <S> As a (very contrived) example we might see something like this: Base rate = 64000 bits per second - this is the data rate Line coded using standard framing on a 32 bit basis adding 1 framing bit per word: this adds 2000 framing bits, so the line rate is now 66,000 bits per second. <S> Now we perform <S> QAM16 (encodes 4 bits per symbol), so <S> the baud rate (or symbol rate) = <S> 16.5kBaud <S> Another way that the line bit rate and data rate may be different is where we need to stuff bits in the bitstream, such as SDLC . <S> The SDLC framing symbol is 01111110 (0x7E) and is used for both the start and end of frame; clearly we don't want data fields to be a frame symbol and erroneously flag a start or end of a frame which would render the link useless. <S> To prevent this, if a sequence of 5 '1' bits are detected within the payload section of the frame (which the transmit source knows about), a zero is inserted into the bit stream to prevent a premature end of frame symbol. <S> The overhead on the channel is not deterministic, incidentally. <A> Baud rate refers to the number of "slots" per second. <S> With most forms of serial communication the data in each slot is a one or a zero. <S> But one could, eg, transmit a voltage indicating a value between zero and three, for four (vs two) possible values per slot. <S> With four values per slot one could transmit data twice as fast as with regular "binary" mode data. <S> This sort of encoding was used in the early days of telegraph (when all sorts of weird strategies were tried), but is hardly ever done anymore for communications of any distance. <S> However, multi-level encoding is still sometimes done inside computer integrated circuits, to reduce the number of wires required.
| The data bit rate is the number of information bits being moved per second.
|
DC power supply by RF signal through a coax cable I want to power a circuit which is only connected by a coax cable. Now I was wondering whether it is possible to convert a RF signal sent over a coax cable into a DC signal to save any bias Tees. The rf signal should be in the ISM band. I found rf to dc converters used for energy harvesting (e.g. http://www.powercastco.com/wp-content/uploads/2016/12/P2110B-Datasheet-Rev-3.pdf ) but the output power does not fit my needs. I need at least 2W (better more). Or maybe is there a chance with a classical rectifier from a power supply unit to get a nice dc from a rf (I am concerned because of the high frequency of approx 800-900 MHz)? Thanks in advanced. Any suggestions would be greatly appreciated. <Q> Converting an RF signal into DC sounds nonsensical. <S> Bias tees are a simple and practical way to connect RF and DC to a cable, with no significant problems resulting from their use. <A> You surely do better by superposing the signal and the supply DC. <S> The combination and the separation is so much simpler . <S> It needs only a couple of LC filters. <S> Your idea would be useful, if the signal was DC. <S> But it's UHF, so use DC as the supply power and combine & separate it by using low pass filters (=one L, one C) <S> Additionally the losses will be remarkably lower for DC. <A> It sounds like you are wanting "phantom power". <S> With phantom power you can simultaneously send a signal and provide DC power over a cable. <S> Think about the low noise amplifier in a satellite dish - the TV box feeds electronics in the dish with phantom power and, over the same cable, receives the amplified signal coming back.
| You can put DC onto a coax cable, this is how many systems power their remote pre-amplifiers.
|
Difference between bit rate and baud rate? Everyone seems to have different definitions everywhere I look. According to my lecturer: $R_{bit} = \frac{bits}{time} $ $R_{baud} = \frac{data}{time} $ According to manufacturers : $R_{bit} = \frac{data}{time} $ $R_{baud} = \frac{bits}{time} $ Which is the correct one? <Q> Bit rate is the number of the information bits per second. <S> To see the difference, consider a transmission line capable of having 4 different states, say 0 , 1 , 2 and 3V . <S> And the end equipment is able to detect each one of them. <S> In this case each symbol would be one of these four states. <S> And each symbol will carry two bits of information as opposed to the simple binary (2-state) line carrying a single bit at a time. <S> Now consider both, the 4-state line and the 2-state line are capable to switch the states in the same rate, which is making both of them to have the same baud rate . <S> But the 4-state line will transmit twice as many bits as the 2-state one, having the bit-rate higher two times. <A> This really belongs somewhere else. <S> Baud is the symbol rate : the number of symbols per second. <S> Bit rate is the number of bits per second . <S> If your symbols are bits <S> then baud and bit rate are the same. <S> If your symbols are equivalent to more than one bit of information (which they often are for later models of modem for instance, and do in many other cases) then the bit rate is higher than the baud. <S> A search on Wikipedia for 'baud' would have found this. <A> Bit rate describes the information transmitted. <S> Baud rate describes the number of "signal transitions" (symbols) used to transmit that information. <S> Depending on the transmission scheme used, your baud rate can be lower or higher than the bit rate. <S> Things are further complicated by the fact that most data transmission includes (significant) overhead - frame information, start/stop bits, error correction etc. <S> A nice tutorial covering much of this in detail can be found at http://m.electronicdesign.com/communications/what-s-difference-between-bit-rate-and-baud-rate
| Baud rate is the number of symbols per second.
|
Won't the ferrite bead (L3) will restrict sudden power requirement of bluetooth module (U4) I know the circuit is well tested by STm but, I think, C15 should not be able to provide enough instantaneous power.Does noise suppress by this ferrite bead (cross-over frequency 12MHz) will be same for other circuits if U4 is used? <Q> They are telling you to put a ferrite bead between the power connection of this module and the power connection to the rest of the board. <S> Perhaps they know the module is susceptible to power supply noise of a certain frequency, but they may also be trying to protect the rest of the circuit from RF being fed back thru the power connection. <S> In any case, C15 is a decoupling capacitor that they probably tell you to place physically close to the module, right across its power and ground pins. <S> C15 lowers the impedance of the local power net at high frequencies, so a little extra impedance from the power feed doesn't matter. <S> These two impedances are in parallel from the module's point of view. <S> As long as either of them is low, the power into the module will have low impedance and all will be well. <S> Calculate the impedance of C15 at the RF frequency, and you will see it is substantially lower than that of the ferrite bead. <S> The RF frequency is also high enough that a larger cap would probably have higher impedance at the RF frequency. <S> There should be no harm in paralleling C15 with a 100 nF cap for better impedance at lower frequencies. <S> However, mount C15 close exactly like they say, then add the 100 nF cap as a extra across C15 without compromising the placement of C15 itself. <A> If you have a concern that the 0.6 Ohm 300 mA ferrite bead will restrict instantaneous current draw requirements of the RF module then the correct thing to look toward is the capacitance value and equivalent series resistance (ESR) of the capacitor C15. <S> This capacitor should be close to the load point (RF module U4) as possible (both the + side and the GND side) and is intended to be the source of the instantaneous current requirements of the module. <S> If the 10nF value of such capacitor is not enough to keep the RF_MODULE_VDD quiet at a nice DC level then the correct approach is to increase its value and/or lower its ESR. <S> Some designers choose to use multiple capacitors at the C15 location, larger valued ones to provide more high speed current leveling and smaller valued ones to lower the overall parallel equivalent ESR value of the node. <S> The power distribution may very well be a longer wiring network and RF existing there is more likely to result in RF emissions problems. <S> With proper design and circuit layout of the RF_MODULE_VDD node the connections will be very short to the C15 capacitor (or its equivalent) and the possibility or RF emissions is greatly reduced. <A> you are right, the capacitor should be large enough to sustain current demand. <S> two reasons for that; normally there are decoupling capacitors with every component forming a large parallel capacitor network in the entire pcb between the power supply and the components. <S> but when an inductor is introduced it effectively reduces the impact of the rest of decoupling caps and so the capacitance between the ferrite beed and the ic/module needs to be adequate. <S> an inductor inherently acts as a resistance to current variations forming a high impedance between supply and the module. <S> there is this nice article ( https://www.murata.com/~/media/webrenewal/support/library/catalog/products/emc/emifil/c39e.ashx )where it mentions in detail about decoupling and filtering. <S> regarding to your specific problem, the formula for necessary capacitance between FB/inductor and IC is; C > <S> L/sqr(Z) where L is the inductance and <S> Z is the powerline impedance needed by the IC. <S> it can be calculated by; Z = delta V / delta <S> I delta <S> V is the allowable voltage variations (mostly in 10-100mV) <S> and delta <S> I is the change in current during operation which depends on the IC.
| As noted in the comments section of the question, the major purpose of the ferrite bead is to isolate the power source and power distribution network from the high frequency current changes that may exist on the RF_MODULE_VDD node.
|
What is the function of the zener diodes used in this setup? I have been studying some electronics schematics to get background information on a power generator we use in my work. In one of these schematics I ran into the schematic I drew below. As seen here the Zener diodes are in opposite direction of each other. simulate this circuit – Schematic created using CircuitLab My first impression was that I was looking at a (1)Transient-voltage-suppression diode that would catch voltage spikes. (1) https://en.wikipedia.org/wiki/Transient-voltage-suppression_diode But in electronics this is indicated with the below symbol: simulate this circuit As far as my knowledge goes on electronics (I am not an experienced electrician) zener diodes (or diodes in general) only conducts current in one direction. So I am curious to know what the function is of the diodes in the schematic from my work. Cause in my opinion they could be removed since they do not add any value to the circuit. Since if there is a current spike, it wont be caught. None of my college knew the function of these and I haven’t been able to find any information or example on google or this site. I am hoping someone can explain it to me. So my question is: What is the function of the zener diodes used in this setup? <Q> zener diodes (or diodes in general) only conducts current in one direction <S> That is only partly true, Zener diodes will also conduct when the reverse voltage exceeds a certain value. <S> Here that value is 2.4 V (they are 2.4 V zeners). <S> Add to that the forward voltage of the other Zener (about 0.7 V) and the two Zeners in series will start to conduct when the voltage across both exceeds 3.1 V. <S> When that voltage is negative (-3.1 V) <S> the same will happen. <S> These diodes limit the input voltage difference to the opamp. <S> Too much voltage difference will destroy the opamp. <A> zener diodes ... only conducts current in one direction. <S> That is, forward biased, they typically behave like a regular silicon diode, and have a forward voltage of ~0.7 V. <S> However, when reverse biased, they have a well defined breakdown voltage (2.4 V in your schematic above), and a reverse bias above their breakdown voltage causes current to flow. <S> Therefore, if the voltage across the D3/D4 pair exceeds approximately ±3.1 V (2.4 V + 0.7 V), they will "clamp" and prevent the voltage from increasing further. <S> So yes, this acts as a transient voltage suppression technique. <A> Two ziner diodes opposing each other, is working as a voltage/current regulator.ziner dioxde <S> A will regulators current and Voltage that goes through B. <S> The same goes for B to A. <S> The Aim of this to gives a regulated voltage/current to the Capacitor to Avoid break.
| On the contrary, the defining behaviour of Zener diodes is that they conduct in both directions, albeit with different forward voltages.
|
I’d like the 12v rocker switch LED to illuminate when the switch is turn OFF I've exhausted a google search and am hopeful expertise on this board can help me with my objective. My Arcade has LED lighted buttons when the cabinet is on. The LED button lights are powered by a dedicated 12v 3amp power source. My objective is to install a 12v rocker switch to allow me to turn on/off the power to these LED lights. HOWEVER, my goal is to have the light on the rocker switch illuminate when the switch is in the OFF position, and NOT illuminated when the switch is in the ON position. I would greatly appreciate guidance if this can be done. If so, does this require a "special" switch, or is it as simple as reversing any of the wires. If it requires a special rocker switch, would you have a URL link where I may purchase one. My google search for "12v LED illuminated when switch is OFF" comes up empty. Thanks in advance, <Q> Here is one example: http://www.mouser.com/ProductDetail/NKK-Switches/MLW3012-12-RC-1A/ <S> In this case the common is 2, switch toggles between 1 and 3, and lamp is on separate L+ and L- pins. <S> So you will want to connect common to 12V supply, then pin 1 will go to your lights, while pin 3 will go switch's light (L+). <S> L- goes to ground. <S> If you cannot find ON-ON switches in the style you like, it will be possible to use regular single pole (SPST / ON-OFF) switch by "shorting" back light when the switch is on. <S> However, this method will waste power, so I only recommend it if you cannot find ON-ON switches you like. <S> Let me know if you are interested in that. <A> What you want to do is very easy if you use a SPDT rocker switch. <S> Many will be available that way. <S> I was going to post a schematic, but I get a message that imgur rejected the image for some reason. <S> Connect the center of the switch to the 12 V power. <S> Then either side will be powered, depending on how the switch is thrown. <S> The one that is powered when the switch is in the OFF position has a LED and resistor in series. <S> The other is the switched power to run your load. <S> Schematic added Imgur seems to be up again, so here is the schematic I tried to post yesterday: <S> Again, one of the two branches will be powered at any one time, but not both. <S> Rocker switches are easy to find in SPDT configurations. <S> In fact, that's probably the most common. <A> Why reinvent the wheel? <S> LED wiring can be independent of switching function. <S> Latching Push button Red Led ON/OFF <S> Good up to 12V <S> , 3A. LED on all the time or only when switch is on ( <S> depending how you wire it) <S> So instead of wiring LED for NO (Normally Open), we wire it for NC (Normally Closed). <S> LED gets power when switch is OFF. <S> Turn it ON, LED goes off. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> simulate this circuit – Schematic created using CircuitLab <S> A simple possibility <A> I've used this circuit so many countless times before... <S> The diode on the left is to prevent the base current from driving the load. <S> However this poses a problem as due to the voltage drop <S> it's <S> voltage level is not high enough to pull the transistor up and completely shut it off, so we have the other diode in place to set the required voltage to shut off the transistor to something lower than what the diode set the value on the left.
| The easiest way is to use dual-throw switch (also known as DPDT, SPDT, or ON-ON) with a independent light input.
|
Analog differential line driver and receiver circuit I want to do a "simple" task of transmitting 0-5v analog voltage from a potentiometer to a microcontroller 15ft away in a very noisy environment.the available supply is 5V only (no negative voltage). My idea is to use something like the digital line Driver/Receiver (AM26C31/AM26C32) but made for analog signals . Is this possible ? Which circuit could achieve this? Does the signal have to be scaled down to 1/2 then use a 2.5V voltage reference, in order to avoid using a -5v negative supply? EDIT : according to answers by NEIL and OLIN , This is how i understand the circuit should be at the sender side, is this the schematic correct ? <Q> No potentiometer with moving parts has any significant high-frequency output. <S> So, just filter the 'received' voltage. <S> The more important issue may be the groundconnection, the '0' volt reference level. <S> Are there multiple ground connections in the system? <S> A ground loop will pick up any local magnetic field fluctuations. <S> Best connection scheme for a three-terminal potentiometer uses shielded three-wire cable, with the shield connecting to the metal shaft and case of the potentiometer, and grounded in ONE place. <S> The three wires connectto the potentiometer terminals, of course. <S> The wiper contact of a potentiometer can be very reliable and low resistance with careful choice of materials. <S> The least expensive potentiometers (carbon film) are not as good as those with ceramic and conductive-plastic elements. <A> The two lines leaving your potentiometer would be the wiper, and the ground reference. <S> It would be a good idea to connect a largish C across wiper and ground pot terminals, to equalise the drive impedance, at frequencies where you will pick up noise at least. <S> Run the two wires twisted, to minimise their inductive area. <S> At the receiver, use a differential or instrumentation amplifier to measure the voltage difference between the two lines, and refer it to the local ground. <S> If you insist on a differential transmitter as well, then a dual rail2rail (as you only have 5v) opamp, with one inverting about mid-rail, one non-inverting, will do fine. <A> There is a creature called differential amplifier. <S> For example, THS4552. <S> It will be a good driver, while instrumentation amplifier would be a good receiver. <S> Just remember to keep good CMRR. <S> Avoid resistors and capacitors unless they are symmetrical for both lines. <A> I can propose you an external ADC with LVDS (low volatage differential signal) <S> interface, e.g. AMC1305L25 from TI. <S> Thus you only need LVDS receivers at your uC to use the SPI interface for communication. <A> Differential voltage <S> Yes, it is possible to do what you ask, and it will help with noise immunity. <S> On the sending side, you invert the voltage about 2.5 V with unity gain. <S> You probably want to buffer the other signal too so that both lines have the same impedance. <S> On the receiving side, use a off the shelf differential amplifier to recover the single-ended signal. <S> A better way <S> However , you obviously have power available at the sensor since you talk about it producing a 0-5 V signal as apposed to just a resistance. <S> If you're really worried about noise, convert to digital at the sensor and send the digital value. <S> That will have much more noise immunity. <S> It can also be isolated relatively easily. <S> If you've got that much noise, it's probably a good idea to not tie the ground together from points 15 feet apart. <S> Since the sensor is a potentiometer, the bandwidth is obviously low and the resolution requirement not very high. <S> A small micro with a 12 bit A <S> /D won't be much more expensive or larger than the circuitry to send a differential analog signal. <S> This is a really easy thing to do. <S> Send the data out the micro using the UART, and have that create a current signal that drives the LED of a opto-isolator at the receiving end. <S> Again, the bandwidth is low, so a slow baud rate like 9600 is sufficient and should give very good noise immunity. <S> Sending two bytes per reading results in 480 samples/second. <S> This is very easy to do and isn't pushing any limits. <A> The two tools used to transmit analog signals in fairly noisy environments are current loops (4-20mA is the most popular in industry) and galvanic isolation. <S> Combine those with shielding and most applications are covered. <S> If the situation is pathologically bad, you can consider an isolated supply, digitizing and using an optical fiber. <S> However, chances are in a moderately noisy situation you'd be just fine with a shielded cable to the pot and a low pass filter at the receiving end. <A> Your initial drawing showed twin-lead (+ & - wires) with a shield. <S> How much noise do you measure, on the right hand side of the shielded cable?And <S> what SNR <S> do <S> you need?Here is Signal Chain Explorer, with Sensor/pot producing 1 voltPP, into 5 meters of twinlead (diffpair wires), into 15Kohm, into 1uF+1nH ESL+1ohm ESR, into 12-bit ADC. <S> Here are the "analysis details", showing the resistor+realistic CAP+ESR+ESL reduces the wiring interference down adequately, with remaining errors being trash injection directly into ADC Vin.
| You could use a differential transmitter, but really, only a differential receiver is needed. That can be done very simply with a opamp.
|
Differences between flip-flops and gates Whatever I searched the web I found things about differences between latches and flip-flops. I'm so new in electronics, my question is what is the difference between gates and flip-flops? <Q> A flip-flop is a type of logic circuit. <S> It is made up of gates. <S> Flip-flops are generally used to store information while a gate only knows about present inputs. <A> A gate performs a logic function (AND, OR, NAND, NOR). <S> Its output always represents the current state of its inputs. <S> Flip-flops are bistable storage elements. <S> Their outputs represent the results of the inputs at some previous time. <A> Both latches and flops are made of more basic gates. <S> These are NAND's, <S> NOR's, Inverterters, Transmission gate, Tri-state elements, and possibly more depending on technology node. <S> But both a latch and a flip flop would still be considered a logic gate (but not a single stage logic gate). <S> A flip flop is made of two back to back latches with opposite phase clocks, in a master-slave topology. <S> This forms a lock-and-dam system where only on one active edge does the output of the FF change states. <S> Alternatively a latch is open during some transparency window <S> w.r.t the clock. <S> This means that the latches are typically open (transparent) during an entire half-cycle (when clock is either high or low). <S> See other posts which go into a bit more detail with circuits: <S> What is a flip flop? <S> Difference between latch and flip-flop? <A> The basic difference between Gates and Flip-Flops is much like the difference between DNA and humans.... <S> one is defined by and subsequently built entirely from the other. <S> Gates are the fundamental building blocks of all logic solutions and are inherently asynchronous but can be used to build synchronous (clocked) logic networks. <S> If you know little to nothing about digital logic and really want to learn enough to be dangerous I'd start with some basic design learning. <S> The problem today is that an introduction to logic is invariably on a very rapid progression from SSI (your typical SSI gate like an xx7400) to an HDL <S> (Hardware Descriptor Language, the language used to describe logic) and rapidly becomes quite complex. <S> To start learning with SSI, MSI and LSI (the older and smaller logic blocks) you really need to be using a book that's 8-10 years old. <S> Some texts do try to modernize like Roth&Kinney Fundamentals of Logic Design but becomes somewhat harder to read in the process. <S> (you can find this text online with a Google search "fundamentals of logic design 7th edition pdf") <S> My personal recommendation is books like Wakerly's Digital Design Principles and Practice which are focused on HDL as a solution, but lead you through SSI, MSI, PAL/PLD, CPLD and eventually to FPGA. <S> It's an excellent learning resource matched to Xilinx FPGA solutions and directly to their development platform. <S> Again your can search online and find copies of the book. <S> IMO for anyone trying to learn digital design these days you need to rapidly get to using PLD/CPLD devices. <S> They are still the only remaining opportunity to learn with DIP chips. <S> In a few more years these may be gone forever, but today you can still get a decent number of 22V10 <S> based PLD's in which you can build conventional gate based logic solutions. <A> An ideal gate and an ideal flip flop encapsulate two fundamentally different behaviors. <S> An ideal gate has an output which will go high or low instantaneously based upon the states of the inputs. <S> An ideal flip flop has an output which will remain at a steady value except on a rising, falling, or any edge is received on the clock input, whereupon it will instantaneously capture the state <S> the data input had an infinitesimal amount of time before the clock edge, and will start outputting the new state an infinitesimal amount of time later. <S> Ideal gates provide no means of creating any sort of delay, since their outputs respond immediately to changes in their inputs. <S> It is possible to build latching circuits out of ideal gates if the conditions necessary for latching will always be held for a finite amount of time. <S> Building a flip flop out of ideal gates, however, is not possible because a flip flop requires an element that can delay the change in the output until after it has sampled the input and/or delay any change in the input so that it will capture the state the input had before a clock edge. <S> When working with real parts, the distinction between flip flops and gates is less rigid than with ideal parts, because it often takes real gates' outputs a somewhat-predictable amount of time to change in response to their inputs. <S> This makes it possible to use gates to construct delays of the sort needed to make a flip flop work. <S> Still, I think the distinction is a useful one because a design that uses flip flops and gates can be made to work when using any combination of parts which are at least as good as they promise to be, while a design which tries to build flip flops out of gates may fail if some gates are faster than promised but other gates are not.
| Said another way, a flip-flop is a group of gates arranged such that they have memory of previous inputs.
|
New PSU will not Start - Only flashes when the switch is turned on I purchased a shiny new Tier 4 power supply, a RaidMax 500W Cobra Power. I have dismantled it to use as a power supply for my mini-cloud project. I am porting the +12V rail to power 4X Intel NUC 5i5MYBE blades through a 12V to 18V 285W step-up DC transformer, the 5V rail to power my managed 5-port gigabit network switches and thrift store wireless router using 5V to 12V step up transformer modules, 3X 500GB USB3 hard drives for Cinder storage on the 3 compute blades, 3X Raspberry Pi 3B units (Swift Object Store w/WD Labs 333GB PiDrive kits,) and 1X Raspberry Pi 3B MaaS controller. I have connected the brown "sense" wire to the 3.3V rail inside the PSU case and shortened the leed, although I should probably keep it full length and attach it to a full length 3.3V leed, while I do not think this is the source of my issue. I have connected +12V Power indicator and +12V power-on indicator leads to colored 3mm LEDs as indicators, have wired the green power leed to a switch to turn the supply off and on, and have attempted turning the unit on and off with and without 10W dummy load resistors on the +12V or +5V rails. I have attempted the following resistor combinations: 1X 10ohm2X 10ohm parallel and serial (5ohm/20ohm)1X 50ohm2X 50 ohm parallel (25ohm) The power supply power indicator LED stays lit when the unit is plugged in and stays on well after power is removed as it drains the capacitors. The power-on LED flashes when i hit the switch, as do the connected components, but does not stay on when I turn it on. I am not sure what I am doing wrong, but am running out of things to try. I am not confident that connecting the sense wire to the +5V rail is a good idea, but that is the only other thing I can think of to try. <Q> PC power supplies are connected to the motherboard - not only for giving the power, but also for getting some control signals. <A> Turning on and off over and over again is in most of the cases a sign for a short circuit. <S> I've no idea what the brown "sense" wire should be. <S> To turn the PSU on connect Pin 5 of the ATX connector to GND. <S> Here is a pinout of the ATX plug Update : In your picture is a purple wire connected to a black wire. <S> The purple wire is the +5V Stand by supply, which is present even if the PSU is not turned on. <S> Do NOT connect this to GND! <A> Logic loads tend to be more linear driving buck regulators for the CPU than boost regulators. <S> The reasons are difficult to explain but ATX power supplies generally only regulate the main 5V supply and the others are regulated by tight ratio coupling of the transformer ratios with rectifier diodes. <S> There are limits to cross-load regulation errors and dynamic load regulation errors than are suitable for a Motherboard but not suitable for loads using Step-up voltage or boost regulators. <S> While it boosts the voltage it also boosts the primary load current and this may cause an over current situation. <S> In order to debug, each step up regulator you have may need to be modified to have an inrush current limiter (see ICL's at Digikey) inserted in series which can be inexpensive solution. <S> There may be other instabilities created with the dynamic pulsed loads of boost regulators that trigger over-voltage. <S> Debugging this problem without suitable tools such as a scope with debugging skills may be frustrating but necessary.
| Check that you are not turned your PSU off by leaving some of the motherboard signals unnoticed or into the wrong logic state.
|
What is best way to generate PWM to drive LEDs without 555 and microcontroller? I want to drive my LED circuit at 40-60% duty cycle to reduce the power consumption. How can I do this without using any microcontroller? Is there any switching IC to do this? Details:= Input voltage to the LED panel is 5 volts. Every LED has a series resistor. And then they are connected in parallel. 4 LEDs: 3 volts, 20 mA. EDIT: I don't want to use many components. Something like op-amp ICs; we can just select two resistors based on the gain requirements to amplify the signal. <Q> Why not just increase the current-limiting resistor values? <S> No additional components required. <A> You can use just transistor or two and some resistors/capacitors to adjust the duty cycle and frequency <S> https://www.google.cz/search?q=transistor+oscillator <S> eventually you can send your signal on/off as a power source of such oscilators. <S> Alternativelly you can restrict the current with something like https://www.google.cz/search?q=transistor+current+source <A> And then use this to drive the gate of an N-channel MOSFET or base of an NPN to switch. <S> If you want just a single Schmitt-trigger inverter in a small size, just Google "TinyLogic" (Fairchild's series) or "Little Logic" (TI's series). <S> Schematic is from http://electronics-course.com/schmitt-trigger-oscillator which has a good explanation of how this simple oscillator works. <A> There are many ways to generate own . <S> Without knowing what you consider better or best, it is hard to suggest. <S> I would use anpamp (like audio or headphone amplifiers) if I were you.
| A simple circuit would be to use a Schmitt-trigger oscillator to generate a square wave with 1 resistor and 1 capacitor to get 50% duty cycle.
|
Is it possible to combine two fuses together (with different current rating) to make a bigger fuse? Let's say I need a 45A fuse.However, I have only a 40A and a 5A fuse in my pocket.If I put them together in parallel, will it act as single 45A fuse ? (so it will blow up if current is superior to 45A). My guess is the 5A fuse might blow up before reaching 45A, because the resistance ratio of each fuse might not be the same as the ratio of their current rating (5/40). Because of that, there might be more than 5A going trought the 5A fuse before reaching 45A. <Q> Fuses are safety devices and they are specified and characterized to be used "standalone". <S> Although in theory you could characterize a fuse so that you could determine what happens when you put two in parallel, no vendor will do that, because it is something no sane designer would want to do. <S> This means that, by doing that, you put at risk the equipment (and possibly its user) <S> the fuse is intended to protect. <S> Bottom line: <S> DON'T <S> DO THAT! <S> BTW: <S> a fuse is not simply characterized by its current rating, but by lots of other parameters, which vary among manufacturers. <S> Think only of the "reaction speed". <S> What do you expect if putting a 5A(T) <S> (slow) fuse in parallel with a 40A(FF) <S> (ultrafast) fuse? <S> Which will blow first? <S> Very messy and unreliable behavior is ensured! <A> It's done all the time with large medium voltage fuses, but they are matched sets of equal size (1/2 the rating) and tested together as a set. <S> You cannot do this on your own. <A> The other comment is wrong. <S> The current won't divide evenly. <S> Yes, they should break when the current reaches 45a but some fuses are fast blowing, and some slow, AND you need to blow 2 fuses (create enough heat to do so), so I would say that the fuse reaction time would rise quite a bit compared to a normal 45a <S> I still don't recommend to parallel fuses, unless they're the same, and even then, it's still better to get one big one. <S> As for the comment saying "the current will divide evenly and blow the smaller fuse", think of a thick 0gauge wire running from point A to point B, running a high amperage, for example 80 amps. <S> Then get the thinnest wire you can find, for example a 30 gauge wire, and run it from point A to point B in parallel to the 0 gauge wire. <S> Will the thin wire blow? <A> Don't Do This. <S> Let's say you put two fuses in parallel. <S> You would have some really unpredictable results. <S> The reason is we don't know the resistance of the fuses (which determines how the currents would be balanced between each one) and even if we knew what they are, it most likely won't work. <S> Here's why: Your aim is to make the circuit blow at 45A <S> so let's say your circuit's current is <S> 40A. One fuse is rated at 40A and the other is at 5A. Assuming they both have roughly the same resistance thus current splits evenly between them. <S> Thus you'll have 20A for each fuse. <S> That means the 20A going through the 5A fuse will blow. <S> Then with that 20A having nowhere to go, it then goes through the 40A fuse thus having the full 40A going through the 40A fuse. <S> That fuse then blows. <S> So by putting these 2 fuses in parallel, congrats you just have a 40A fuse with a slower reaction time. <S> Overall just buy a 45A fuse. <S> Don't mess around by combining fuses to get a different rating. <A> The question is how will the currents split. <S> That depends on the resistance of the fuses. <S> If the 5A fuse had 8 times the resistance of the 40A fuse then all would be good. <S> the two fuses would move into overload at the same time. <S> So the question is will that be the case in practice. <S> I strongly suspect the answer is probablly not <S> but there are two opposing affects in play. <S> Lets assume the two fuses are of the same size and type. <S> On the one hand the fuse wire itself in the 5A fuse will probably have more than 8 times the resistance of that in the 40A fuse due to the way current carrying capacity scales with wire size. <S> On the other hand there will likely be contact resistances. <S> These will likely be similar for the two fuses (and therefore make up a smaller proportion of total resistance on the 5A fuse) but contact resistances can also be quite unpredictable. <S> The combination may carry a bit more than the 40A fuse alone but probablly not the full 12.5% more (note that the current ratings of fuses are the normal operating current not the current at which they are guaranteed to blow).
| Therefore, putting two fuses in parallel will go against the very purpose of a fuse, i.e. having predictable behavior under the specified conditions. No, most of the current would go through the thicker wire, as it has the least resistance. In summary the overall result is likely to be unpredictable.
|
What is making my DC/DC converter's capacitor to blow up? I'm having some capacitors blown up and I am not sure what's the cause of this. It is definitely NOT OVERVOLTAGE and NOT in WRONG POLARIZATION . Let me introduce the scenario: I have designed a double cascaded Boost converter using this scheme: Vout can be obtained from: \$\ Vout=Vin/(1-D_\max)^2\$ where \$D_ \max\$ is the maximum duty cycle. I want to step-up an input voltage of 12V into a 100V output voltage. My load is 100Ω , hence it would be dissipating 100W. If I consider no losses (I know I'm being TOO idealist, calm down), the input voltage source will deliver 8.33A We can split the circuit into two stages, the first stage's ouput is the second stage's input. Here comes my problem: C1 is blowing up when the voltage accross it reaches aproximately 30V. C1 is rated for 350V and it's a 22uF electrolytic capacitor (radial) 10x12.5mm. I am totally sure the polarization is right. The second stage's input current should (ideally) be around 3.33A (in order to keep the 100W with 30V for this stage). I know the current might be higher, but it's a good aproximation for this purpose. The switching frequency is 100Khz . For some reason the cap blows up and I don't really know why. Of course that when this happens the cap (dead) is hot. May it be an effect of the ESR? This cap has a 0.15 Dissipation Factor at 1kHz. \$|X_c|= 1/(2*pi*100Khz*22uF) =0.07234Ω \$ So \$ESR=0.15*0.07234= 0.01Ω\$ (DF would also increase for a higher frequency) for C1. Since L2 is pretty large, I would expect C1 to deliver a pretty constant current equal to the second stange's input current (3.33A) so the power dissipated in ESR is supposed to be around: \$3.33A^2 * 0.01Ω = 0.11W\$ Can this make it too hot and explode? I doubt it.... Additional information: L1 is about 1mHy L2 is about 2mHy D1 is a schottky 45V diode I tried two different capacitors: 160V 22uF that blown up, and then I tried the 350V 22uF which also blown up. Measuring the current in the cap would be difficult due to PCB layout Both the first and second MOSFET has a small snubber RC network. I don't think it could cause any problem in C1. I am waiting for your ideas! EDIT n°1=L1 is pretty large, ripple is only 1% of the rated input current (let's say 100W/12V = 8.33A) so que can assume it's almost like a constant current at the input of stage 1. For stage 2 inductor current ripple is less than 5%, we can also think it's a constant current). When MOSFET 1 is turned ON, around 8.33A goes through it, but when it's turned off, that current (we said "practically constant") would go through D1. We can say current in the capacitor would be \$ I_{D1} - I_{L2} \$ . Then we finally find that the peak current in C1 must be in the order of \$ 8.33A - 3.33A = 5A\$. Pretty much current! and it would dissipate \$5A^2 *0.01Ω = 0.25W \$ ... but looks not so much power dissipated in the ESR. As someone said, I might also consider the internal inductance of the cap, but i think this wouldn't be a cause of power dissipation (we know inductors store energy but don't make it into heat) Anyways, despite of the calculation above was very simplified and it might be a little higher power dissipated, I still wonder if it's enough to make it boil and explode! <Q> The peak ripple current for C1 is approximately I(out)/D where D= duty cycle. <S> If the Duty cycle is say 50% at your 30 V output <S> then the ripple for C1 is 3.3/0.5 = 6.6 A. <S> As the duty cycle is reduced this gets worse. <S> If the duty cycle was 10% = 0.1 <S> then the current peak is 33 A. <S> If you then use your ESR value the power dissipated is about 0.4 W, much higher than you previously calculated. <S> If I look at 160 V capacitors on Mouser <S> (I'm assuming you are using Al Electrolytics) <S> then I see nothing generally available that could sustain the peak currents you need. <S> I'd suggest you use TI's Webench to work through a design and then look at the selected components. <S> You will notice on many of the designs they use very low ESR capacitors and often have two or even three in parallel. <S> For example they use Panasonic polymer caps often in the designs and they have very high ripple current ratings at very high frequencies. <A> Your capacitors may have quite big internal inductance - too much for 100 kHz pulses. <S> You should connect some smaller non-electrolytic capacitors in parallel with them until the oscilloscope shows that the voltage limits are not exceeded. <S> BTW. <S> the current rushes as pulses from the inductors as soon as the fets turn off. <S> The start of the current pulse is very sharp - as sharp as how fast the fets can turn off. <S> If the switching frequency is 100 kHz, the capacitors really should handle several MHz properly. <S> NOTE: <S> low inductance electrolytes for SMPS applications are developed but they cost some real money, not pennies as the ordinary models. <S> Late addition: All your output power is at first stored in the capacitors - no direct way from the input to the output. <S> The inductance causes it to localize more at the near ends of the inside plate roll. <A> I bet on power created by ripple currents. <S> Your capacitor has some ESR. <S> Pulsed current of your magnitude may leave there like ten-twenty watts quite easily. <S> So... Put several in parallel, with lowest possible ESR/ESL <A> Cap Max ESR Ω Max RMS ripple (uF) <S> VDC PART <S> # 120Hz <S> (mA <S> ) 120Hz,105C DxL (mm)--- <S> ---- <S> ------------ <S> --------- <S> ---------- ---------22 <S> 160 <S> 226CKE160MLN <S> 11.3094 <S> 92 <S> 10x12.5 <S> C*ESR= <S> Ts=22uF*11.3 Ω = 250us , f(bw) <S> =0.35/ <S> Ts = 5.6kHz <S> which is the fastest rate of charging it can handle and reach full charge voltage. <S> f switch = 100kHz PWM <S> variable D <S> thus as 100kHz it will appear as a lossy resistor only at 11.3 Ω with losses of \$Pc=I^2ESR\$ and a rated ripple current of 92mA <S> the device can only handle 1.03W at max temp of 105C or a rise of 85C above room temp 20C. Now to choose a 22uF cap, you want to follow the App Note recommendation and choose a low ESR cap and not a general purpose electrolytic ( G.P. e-cap) <S> What they don't tell you in school, ( and I have commented many times on this site) is that an G.P. e-cap has a ESR*C <S> >= 100 us while a low ESR cap < 10us and best case < 1us. <S> This is what you need when choosing a switch period < 10us. <S> Now it is not hard to sort Digikey or Mouser databases by ESR or search in other ways for ultra low ESR. <S> You might also want to read the MSDS datasheets of e-caps for toxic material exposure when they blow up. <S> The App Note advises you to expect under INDUCTOR SELECTION that A good estimation for the inductor ripple current is 20% to 40% of the output current. <S> E-Caps are rated in several ways. <S> D.F. @120Hz <S> ( for small line bridge rectifier use)max ripple currentESR (typ.) not aged after 10 yrs ! <S> It is important to remember that Caps are usually charged by dumping current pulses then discharged slowly between pulses, so the duty cycle determines the ratio of Peak/Avg current. <S> If the ripple voltage is 10% then the pk/avg current ratio is 10/1. <S> If the energy dissipation is the power dissipation in each pulse times the pulse repetition rate. <S> No problem as 100Hz and 1000x worse at 100kHz. <S> Hence the result of not understanding subtle advice in the App Note ... is a Chinese fire-cracker. <S> Refs from OP in comments that should have been in Question <S> http://www.mouser.com/ds/2/88/ckh_cke-611140.pdf <S> Cap specs <S> http://www.ti.com/lit/an/slva372c/slva372c.pdf <S> Boost Reg Design App Note
| As suggested in several othe comments - the sheer dissipation in your capacitors may cause some boiling.
|
Capacitor Gaining Voltage Over Time? I recently bought two 3300uf 100v capacitors and have connected them in parallel. I charge them up to 100v and discharge them. I then hook up a multimeter and notice the voltage going up very slowly, about .01 volts every 20-40 seconds. So I discharge the capacitors and the voltage goes back to zero. When I woke up this morning, I checked the capacitors and it had gone up to 5 volts! And I am able to power an LED with them. What is going on here? Edit: Thanks to Robert's comment in one of the answers, I think he's right. This is probably dielectric Absorption. <Q> What you've observed is called "dielectric absorbtion" or "recovery voltage phenomenon". <S> It's cause by kind of interia of the dipoles (ions) in the electrolyte while charging and discharging. <S> From wikipedia : Dielectric absorption is the name given to the effect by which a capacitor, that has been charged for a long time, discharges only incompletely when briefly discharged. <S> Although an ideal capacitor would remain at zero volts after being discharged, real capacitors will develop a small voltage from time-delayed dipole discharging, a phenomenon that is also called dielectric relaxation, "soakage", or "battery action". <S> For some dielectrics, such as many polymer films, the resulting voltage may be less than 1–2% of the original voltage, but it can be as much as 15% for electrolytic capacitors. <S> Further: <S> When the capacitor is discharging, the strength of the electric field is decreasing and the common orientation of the molecular dipoles is returning to an undirected state in a process of relaxation. <S> Due to the hysteresis, at the zero point of the electric field, a material-dependent number of molecular dipoles are still polarized along the field direction without a measurable voltage appearing at the terminals of the capacitor. <S> This is like an electrical remanence. <S> From a Mouser note 7 Recovery Voltage <S> Where a capacitor is once charged and discharged with both of the terminals short-circuiting and then left the terminals open for a while, a voltage across the capacitor spontaneously increases again. <S> This is called “recovery voltage phenomenon”. <S> The mechanism for this phenomenon can be interpreted as follows: <S> When charged with a voltage, the dielectric produces some electrical changes within, and then the inside of the dielectric is electrified with the opposite polarities (dielectric polarization). <S> The dielectric polarization occurs in both ways of proceeding rapidly and slowly. <S> (Fig. 28). <A> Dielectric absorption Equivalent Circuit 100V to 5V C1V1= <S> C2V2 <S> before = <S> after discharge after long time <S> Main Cap C1= 3300uF at V1=100V and <S> V2=5V <S> therefore C2 = <S> C1 <S> * V1/V2= <S> 66 <S> mF <S> equiv <S> dielectric absorption Capacitance <S> ".01 volts every 20-40 seconds" or 10mV/20s= <S> dV/dt thus Voltage rise on C2 at 100V and C1 at 0V <S> The discharge on C2 at V1=100V due to series ESR2 on absorption cap, C2 Ignoring leakage R for now, <S> V2/ESR2 = Ic2 = <S> Ic1 = <S> C1 <S> * dV1/dt or ESR2 = V1/C1 * dt/dV1 = <S> 100V/66mF <S> * 20s/10mV = <S> 3MΩ simulate this circuit – Schematic created using CircuitLab Note for old E-Caps each component value in the equivalent circuit <S> can be estimate by various tests. <S> Your test estimates the ESR2 <S> * C2 = T2 = 180ks C2/ <S> C1=20 <S> with as the absorption/cap ratio. <S> Bed Side notes <S> if dV/dt was 10mV/30s , can we estimate the minimum amount of sleep you had? <S> if 60% charge time of 5V is reached at a 10mV/30s rate this would take <S> 5V/10mV *30s <S> = 15ks = 4.17 h without knowing the dV/dt in the morning rate, we can only assume it was much lower such as 2T or 3T meaning 8 h or 12 h sleep or the ESR2 reduced overnight. <S> Parallel Leakage R values are known to reduce with aging and conditioning old large E caps using a large Series R will raise the leakage R values towards the original values in many cases. <S> This is a safe practice when dealing with big-old low ESR caps to prevent interlayer short circuits. <A> Different possible explanations: <S> Static electricity: it's perfectly normal that there's a potential difference between ground and e.g. clouds. <S> That can amount to quite a lot of V/m in "free air", but with very little charge behind it, ie. <S> trying to measure that is practically impossible. <S> However, if you have two electrodes stretched across that gradient, you'll likely be able to charge your caps. <S> RF harvesting: Everything acts as an antenna. <S> Now, that doesn't matter because RF induction is by definition AC and will cancel itself out, but if you happen to have some rusty/salty/... <S> conductor interface <S> , that can act as an accidental diode. <S> See: early crystal radios. <S> Residual charge <S> : I don't know your caps well, but maybe it's chemically sound to assume that shorting them doesn't eliminate all energy stored within, but only what is accessible quickly. <S> In that case, if you'd let caps shorted for longer, this effect wouldn't take place. <A> Actually the dielectric absorbtion story is ok, but complex To put it short:The electric field deforms the molecule structure just like some thick and soft cloth gets gets a dimple when you press it by your hand. <S> By the time that dimple vanishes and the cloth rises a bit upper again against the gravity. <S> The molecular deformations in the electrolyte and insulation layer reverse gradually and the ionic particles return to their original places. <S> That means a new electricity, because the charge distribution is changed.
| When a charged capacitor was discharged until the voltage across the capacitor disappears, and then being left the terminals open, the slow polarization will discharge within the capacitor and appear as recovery voltage.
|
What's the name of a connector usually used to connect a frontpanel to a PC mainboard? Could someone tell me the technical name of this connector? Visual description: ~2 mm thick (for one row) 6 ~ 8 mm long ~ 2 mm wide for each pin (e.g. 4 mm for 2 pins) usually black housing It may have 1, 2, 3 or more pins. Is there an upper limit? <Q> Different manufacturers have different names for them. <S> It's a housing, into which contacts crimped on the wires are inserted (they snap in). <S> Digikey uses this description: n Position Rectangular Housing Connector Receptacle Black 0.100" (2.54mm) where n is the number of positions and 0.1"/2.54 <S> mm is the pitch. <S> The largest I see in stock (single row) is an Amphenol 36 position one. <S> If you need more you can stack multiple connectors but that leaves things open to more incorrect ways to plug them in. <S> Since the manufacturer has to manufacture an injection mold cavity for each size, the limit for a single-piece molding will be determined by their profitability calculus. <A> That's a female pin header connector. <S> There's probably no real upper limit for their size, just usability. <A> There is however a somewhat generic term that has come into play for this type connector and wiring in the last few years. <S> Search for this term on sites like eBay and Amazon and <S> you will see what I refer to.
| Often referred to as Dupont jumper wire cables, "Dupont" connector housings or "Dupont" crimp pins etc. As others have stated there are a lot of different names for such type connectors from various manufacturers.
|
Turn off n-p-n BJT on non-zero voltage I have a controller that uses 0.2V voltage as logical zero and 0.4V as a logical one and want to turn a LED on/off depending on these two levels. I've built a cicrcuit like one below and tried to choose R1 value to turn the transistor off when control voltage is low, but it was eather always open and control voltage was only affecting LED's brightless, or (with big R1) it was always off. Maybe I'm missing something or some other circuit should be used? How to solve my problem and turn the LED on/off using 0.2/0.4V control voltage levels? <Q> Here is a circuit idea that can translate the 0.2V -> 0.4V signal to an on/off signal for the LED. <S> The two PNP transistors create a differential pair to act as a comparator. <S> The left side takes in the input signal and the right side compares to a reference of ~0.3V created via a voltage divider across a forward biased diode. <S> The load resistor (R6) on the right leg of the differential pair develops a voltage swing of near 0V to just over 1V which works well to switch the output NPN driver transistor on and off. <S> The circuit works in simulation but can likely be optimized to lower over all current draw on the power source. <S> As shown the LED current is 9mA and the full circuit current draw is ~32 to <S> 41mA. A sizeable chunk of that is going to heavily bias the diode. <A> Your basic problem is that a BJT needs about 600-700 mV B-E before it starts conducting meaningful current. <S> When the input signal is 200 mV, this puts 620 mV across B-E, which should cause some collector current to flow. <S> When the input is 400 mV, then there is only 420 mV B-E, which will cause very little collector current to flow. <S> This collector current is then amplified by Q2 and again by Q3, at which time it's strong enough to drive the LED. <S> There are some additional wrinkles to get around the fact that Q1 won't act as a nice on/off switch. <S> It will have some leakage when the input is 400 mV. R4 is intended to require some minimum Q1 collector current before Q2 turns on. <S> Another trick is hysteresis, which is what R5 provides. <S> This is positive feedback that turns Q1 on more when it's already on, and <S> less when its already off. <S> It provides snap action instead of a soft transition region where the LED fades on and off. <S> Adjust R1 to set the threshold between on and off. <S> Adjust R5 so that the hysteresis "dead zone" is maybe half the input voltage swing, which would be 100 mV. <S> I posted this circuit only to show one way that the B-E voltage threshold of BJTs can be overcome. <S> Another would be in the classic comparator configuration. <S> However, regardless of all these, I'd probably use a real comparator if actually faced with this problem. <S> It would be simpler and more reliable across temperature and voltage fluctuations. <A> Here is the kind of circuit that would usually be used- <S> it uses half of U1, an LM393 , which is an inexpensive dual comparator in an 8-pin package (available from multiple manufacturers): <S> simulate this circuit – <S> Schematic created using CircuitLab <A> How to solve my problem and turn the LED on/off using 0.2/0.4V control voltage levels? <S> two ways: amplify it to a more usable level. <S> use a comparator.
| One solution is to bias the transistor so that the 200 mV difference is right at the B-E voltage that makes a large difference in current: R1 and R2 form a voltage divider that creates about 820 mV from the 5 V supply.
|
comparison between AC and DC Generally speaking at same voltage level and same current which, current is more dangerous AC or DC or are they the same? <Q> If you are considering levels which are dangerous for human being. <S> I think it depends on what part of your body is in contact with conductor and whether you are touching ground(earth) or not. <A> From a purely electrocution point of view they are both the same since they will result in the same current flowing thru your body. <S> I've heard some claims that if your hand is clenched around a conductor electrocuting you, that you have a better chance of being able to let go with AC as apposed to DC. <S> Whether you actually get killed or hurt at any particular voltage depends on parameters of the situation, like how moist your skin is where the contacts are, the contact area, and where in your body the current flows. <S> These factors are much more variable than <S> any small advantage AC might have in being able to let go. <S> And note that the letting go part only matters if contact is such that the current is clenching muscles that then cause better contact. <S> Consider them both dangerous according to the voltage. <S> Don't get a false sense of security by thinking one is a little less dangerous than the other. <A> Electrocution is not the only danger. <S> If you start an arc, e.g. in a faulty switch, an AC arc may be self-extinguishing (during the zero crossing) whereas a DC arc will not. <S> You'll often see switches or relays rated for a given current, say 10A, for 240V AC ... or only 30V DC. <S> Apart from the fact that such an arc leaves your circuit powered despite the switch being OFF, it also dissipates power, generating heat, and potentially starting fires. <S> So, there are some additional dangers from even moderate (48V downwards) DC voltages. <S> There are further dangers specific to low voltages, because (for the same power) they involve higher currents, and large batteries can supply thousands of amps. <S> Never carry a metal ladder in a battery room. <A> One other thing to know is that AC can flow even when there is no apparent conduction path. <S> Thats because ac current can flow through capacitances, and high frequency currents flow better through capacitances than low frequency currents. <S> I know the question was probably intended for DC vs line frequency (commercially 50 or 60Hz), but this is definitely a concern with higher frequencies. <A> well, this guy survived and it was AC:
| In the end, both are dangerous at the same voltage. AC voltage adds some advantage technically as in every cycle it is going from positive to negative with zero in between.
|
accuracy shunt resistor beter than 0.1%? I want to make a circuit that can measure a current more accurate then 0.1%. For instance if I measure 7A, I want to be able to read 7.000 A. If I use a 0.1% shunt resistor I do not know if it is 7.007A or 6.993A it can be +/- 7mA. I did take a look at some big stores and the best shunts they sell are 0.1% accurate with a price about 5 or 6Eu a piece. How can I get a better accuracy? Can I even go further like 6 digits? What is possible? thank you for the help! <Q> You should expect to have to apply your own calibration to build instrumentation to this sort of level - you can't just expect to put a bunch of bits together and achieve fantastic precision/accuracy. <S> You will probably find some fairly simple slope-and-offset linear calibration is all you need for this case, though ammeter shunts can be tricky beasts. <S> For a start you need to make sure you are definitely measuring the voltage across the shunt, and not seeing a lot of ground lift happening elsewhere in the circuit. <S> Also think about the temperature coefficient of the shunt resistor - you might find the accuracy of your reading depends on how long the load's been turned on for... <A> In theory you can get some of following approaches to get better accuracy: <S> select from more shunts, th 0.1% accuracy means, that any of these is somewhere in 99.9%-100.1% range, but does not ensure where exactly. <S> By testing more of them you cen find by chance one, that is exactly the value you need (or any given close range <S> od it) <S> carefully combine more shunts together - if you use one wicht is 99.9% and another one, which is 100.1% (both half value) in serie <S> , you get 200%.0 of the half value (means 100% of wanted value). <S> Similarry you can combine them in parallel or make even more complicated net. <S> Depends of what values you have <S> , you cen get much close to what you want. <S> recalculate the result - if you have 99.9% and measure it preciselly, then you can simply divide the result with 99.9% to get exact value (either digitally, if you have something digital on the way, or redraw the analog scale to show the right values) - by the way you can create the scale on blank background from scratch based on your own calibration process You can combine all above methods to get better results with what you actually have. <S> Practically you will run to problems with stability, reliability and reproduceability - to have exact values you need exactly controlled environment as there is thermal change in all parameters, also radio noise can come to way, earth magnetic field (and all others), local gravity, etc, etc, so you cannot get infinite precision by changing just one piece of you equipement, you would need more stable environment for better precision, the more numbers, the (exponentially) more complicated and expensive stabilisation is needed. <A> The easiest approach to very high accuracy is to buy a fairly accurate and extremely stable resistor that is of roughly the correct value. <S> Then, use a calibration setup of known and acceptable error to provide a known current (or currents) and adjust your circuit so it reads more accurately. <S> This is best done digitally. <S> This will also cancel out initial error in your ADC voltage reference and any other linear error sources in your circuit. <S> Current measurement with ppm-level accuracy is possible but you will not achieve it on your first or second try unless you go out and spend big $$ on an instrument, and even then it will be transistory. <A> Beware of the temperature coefficient of copper. <S> Over 100 degree Centigrade change, the resistance will change 40%.Over 1 degree, that is 0.4%. <S> This becomes important in OpAmp circuits using non-inverting circuits, where the feedback current must exit to "Ground" and return to the power supply. <S> As current varies, because Vout varies, the localized ground foil changes temperature. <S> With the standard copper foil thickness --- 1 ounce of CU per square foot --- being 35 microns or 1.4 mils (0.0014 inches), and electrical resistance being 0.00050 (500 microOhms per square), with Thermal Resistance of 70 degree Cent per watt per square, you have a Thermal design issue. <S> Get a quadrille pad and draw lots of squares, thinking about current flows and heat flows. <S> Learn to think about thermal shorts and thermal opens & isothermal systems.
| The simplest calibration, if you are confident there is negligible offset voltage, is to multiply your ADC reading by a scale factor.
|
Relay does not turn ON I am having trouble turning ON a relay from the ESP8266 output pin. I have wired the relay as shown in the picture below. The only change is that the transistor is BC556. Other specs of components are:Relay 5V , running this on 5V power supply. Here are some more facts:- I have wired the exact same circuit on the bread board and it works perfect.- I am using GPIO2 as the output PIN and apart from driving this relay , it also has a pull up resistor of 10K connected to GPIO2. - The relay works fine sometimes but many times it either does not work or is very slow in reacting. For eg. when my output tuns LOW, it soemtimes takes several seconds to turn ON the relay.- I measured the voltage across the transistor Vce = 1.1 V which means the voltage across the relay pins is 3.9V. I thought this is the reason why the relay is not able to trigger but then I have exact measurements on my breadboard circuit and it works fine. I even tried switching the relays but I get the same result. - From BC556 datasheets it says saturation Vce is .09 - .25V but I get 1.1 voltage drop Is this normal? I tried BC 557 with the same result. I even varied the base resistor to decrease it till 100 ohms but the Vce improve by 0.1 V only. Any suggestion on what should I do to troubleshoot this further would be helpful. Thanks <Q> One thing that limits what you can do is that the ESP chip is 3.3V and your supply is 5V so you cannot use just a PNP transistor to do what you want. <S> Here is one way to accomplish the task without changing the logic (low = on). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Your circuit has two problems - the voltage across the relay coil will only go between about 4.3V and 1.1V- neither fully on nor fully off. <S> That is because it is an emitter follower and has a 0.6~0.7V offset <S> and it cannot amplify the 3.3V coming from the ESP chip. <S> The 1.1V is particularly dangerous since it is a grey zone of relay operation and some relays may stay pulled in (depending on temperature, whether they have been tapped, etc.) <S> and some may drop out. <S> The 4.3 means you are not getting full voltage on the coil so it might not pull in at very high temperatures. <S> With the circuit above, the relay voltage will go from about 4.9V to 0V, as it should. <A> The problem is that you are likely not fully turning the transistor on and off. <S> You mention in the comments that the relay is taking 13 mA when off. <S> This is obviously because of your setup. <S> One, the ESP is 3.3V logic. <S> Even with a 5V pull up, you create a voltage divider between 5V and 3.3V, so the transistor is still conducting. <S> You may also have fried the ESP, just saying. <S> Two, the 1 kohm resistor may not be enough to saturate the transistor. <S> It's 5 mA maybe, or less. <S> Three, with a PNP transistor, wiring the relay on the emitter, like <S> a low side driver is not recommended. <S> The emitter will never be a diode drop lower than its collector. <S> Solution, replace the PNP with a NPN. <S> Remove the pull up, and change your code to invert the logic (1 = on). <A> To be clear, here is your circuit being discussed <S> : You need a NPN transistor there, not a PNP. <S> The base should be wired as shown, the emitter to ground, and the collector to the relay. <S> Otherwise, you have to make sure the transistor can support the relay current. <S> Let's say the digital signal is 3.3 V when high. <S> Figure <S> the B-E drop will be 700 mV, so that leaves 2.6 V across the base resistor. <S> With the 1 kΩ resistor as shown, that will result in 2.6 mA of base current. <S> Verify that the digital output can source that. <S> Most can, <S> but you should check anyway. <S> Of course it can't be a open drain output if used this way since those can't source any current. <S> Let's say you use a transistor with a minimum guaranteed gain of 50. <S> That means it can support up to 130 mA collector current. <S> The relay must be rated for that or less. <A> In such a circuit supplying a low switching voltage usually turns on the transistor. <S> And supplying a high usually turns off the transistor. <S> However, consider the case of supplying the PNP transistor's base with a 3.3V high signal. <S> It might be expected the transistor is turned off. <S> But if we look closely at the question we see that the relay is run at 5V. Not at the same voltage as the micro processor is running at. <S> If 5V is supplied to the relay and 3.3V to the transistor's base, there could be as much as 5 - 3.3 or 1.7V difference between the emitter and base. <S> Enough to activate the transistor even when it might be expected the transistor is turned off. <S> It might be that the relay is never turned off. <S> So that it appears never to switch. <S> Perhaps even thought of as defective. <A> Normally, an NPN transistor is used, with the emitter grounded and collector to the relay. <S> That way, the transistor will be saturated when on, so the voltage across it will be about 0.2 volts. <S> That will invert the logic, so a High from the ESP8266 will turn on the relay. <A> What is the relay? <S> This is usually in the datasheets, but is often not so obvious as you would like (Guess who was had by this one....). <S> Look at the pinout diagram on the datasheet for a ' <S> +' by one terminal, they mean it, both Omron and TE are buggers for this. <S> Do respect the 'must operate' and 'must release' ratings from the datasheet, these are temperature sensitive and you can set yourself up for a lot of reliability pain if you ignore these.
| Quite often small low voltage relays have a polarised coil (It saves power by having a permanent magnet biasing the field), and will not switch if you connect the coil the wrong way around. Second solution, switch the PNP and relay around, and use a NPN as a PNP driver, because of the 3.3V logic of the ESP.
|
What happens if you feed 3.3V to a 5 to 3.3 voltage regulator? The PCB circuit I'm using expects 5V input voltage which it then converts to 3.3V using an AMS1117-3.3 voltage regulator. The power supply I'm using is 3.3V. What happens if I feed the 3.3V to the input of the 5 to 3.3 voltage regulator? The input of the voltage regulator has an easily accessible jumper header. Whereas bypassing the voltage regulator would require the use of bodge wires. Can I get by with just feeding the 3.3V into the regulator? What are possible drawbacks if it even works? What would happen when the PSU voltage drops below 3.3V (as I doubt it will provide a precise and constant 3.3V)? <Q> The critical parameter is drop-out voltage, which I'll call \$V_{do}\$, listed in your datasheet on page 3. <S> The output voltage will generally not be more than \$V_{in}-V_{do}\$. <S> (Obviously this breaks down when \$V_{in}\$ is very low. <S> If \$V_{in}<V_{do}\$, for example, all bets are off) For the AMS1117-3.3, the drop-out voltage is listed as 1.1 (typ) to 1.3 (max) volts. <S> That means to be sure of getting 3.3 V out, you need to provide at least 4.6 V in. <S> And if you provide only 3.3 V in, you shouldn't expect more than 2 V out. <S> In fact, that far below the nominal input voltage you shouldn't count on the listed drop-out voltage applying, so you can't even be sure of getting 2 V out. <A> Read the datasheet! <S> The dropout voltage is clearly specified as 1.3 V (page 3, fourth row, "Dropout Voltage"). <S> Since you are using the 3.3 V output version, the input voltage must be at least 4.6 V for any of the other specs in the datasheet to apply. <S> So the correct answer to what will happen when you supply 3.3 V in is <S> "You don't know" . <S> All other promises made in the datasheet are null and void since you are violating the minimum dropout voltage requirement. <S> Probably , the output will track the input minus some voltage, at least for input voltages that aren't "really low". <S> Many chips that are meant to run from 3.3 V need 3.0 to 3.6 V. <S> It is unlikely, and certainly not guaranteed, that the output voltage will be 3.0 V or more with only 3.3 V in. <S> Overall, bad idea . <S> Solder <S> a small wire across the input and output of the regulator. <S> Or, remove it altogether and jumper across its input to output. <S> Or, get a 5 V supply. <A> In the Electrical Characteristics table in the datasheet, there is a line for "Dropout Voltage", with a typical value of 1.1 volts. <S> That is the minimum voltage difference between input and output of the regulator at which the regulator is guaranteed to work. <S> So, for a 3.3 volt regulator, the input voltage must be above 4.4 volts for correct operation. <A> It will likely not pass through 3.3V, as the are internal components that will use up some of that voltage. <S> Even a simple diode drop will take up some of it. <S> The regulator may also oscillate. <S> But this is simple enough to test. <S> There would be little harm in attempting it and measuring the result yourself. <S> Or just avoid the issue and solder on a pass through wire. <S> 10 second job really.
| I would expect that if your input voltage is only 3 .3 volts, the regulator output will be around 2.3 volts.
|
Can a DC-DC converter destroy a SLA battery or wear out very quickly? I have 2 12V SLA batteries for recreation use in my camper. One very old (+10 years), another a bit less (unknown, maybe 5 years?). I noticed both of them were working fine in the beginning, but now the first one is completely used up and ready to recycle (5V after 5H charge!).But the second one is also more or less broken . For example it quickly discharge to 9V after half of the charging time on 60W only! Now, unless the first one was its due time and because I left it too long connected parallel to the second one (can that be?), I suspect that this DC-DC converter I use for my laptop is at cause. The DC-DC converter 95W and converts from 12V to 19.9V , but it pulls amps in a very alternate way. E.g. alternating between minimum 4A and maximum 9A. So, I'm wondering if this alternating current can quickly wear out an SLA or other negative effects? But my more general question remains priority:Can a (certain designed) DC-DC converter destroy or very quickly wear out an SLA? With very quickly I mean, in months or so, compared to the years left in it.(Since I'm going to buy a AGM battery as replacement and don't want to kill it.) Thank you , more experienced battery users! <Q> The switching nature of a boost regulator is typically not a problem for most batteries. <S> The concern here is the current pulled. <S> If it is in the acceptable range of your battery. <S> Or if you let the battery drain too much. Trying to pull AMPS from a dead battery will hurt it. <S> Even deep discharge batteries would be affected by this. <S> But also, those batteries are very old and you are mismatching them. <S> If you have them in parallel, one will attempt to charge from the other due to their different conditions. <S> So replace both with new and identical ones. <S> Additional concerns, continuous load current of a battery is not the peak, surge or pulse current. <S> The Datasheet for your battery will have both, as well as the expected capacity at any given load. <S> A car battery may have a 600 CCA rating, but this is over seconds and not minutes let alone hours. <A> A bigger factor is if you let the DC-DC converter draw the battery down until it is completely flat- that will kill the battery in short order. <S> It will do that even if you have nothing connected to the output of the DC-DC converter. <A> The second 50% is for emergencies only. <S> If you always take it down to 11v, then you'll only get a few hundred cycles <A> Ordinary SLAs are done often after three years of continuous use in 24/7 UPS power supplies. <S> The 10 year models are very expensive. <S> They can survive that long only if the charge-discharge cycle specifications are followed excactly. <S> As already commented, Your load is not a problem, but if you let it draw the battery flat, it probably stays flat finally because some of the cells start to get reverse current. <S> The "cabin" hints long pauses to exist between the periods of usage. <S> Check the available quidance. <S> An example: http://www.powerstream.com/SLA.htm
| Let your batteries stay unused as empty for months, they will wake up much more tired. If you want long life from an SLA, don't routinely discharge more than 50% of its rated capacity.
|
Trouble deleting dimension in Eagle I want to get rid of the measurement label in the PCB (the double arrow showing the dimension) but I am not able to delete it. It is showing overlap errors with the bottom traces. How do I delete this? I am using Eagle 7.7. Things I have tried: Hit the Delete button and tried to delete the arrows as well as other parts. Hit the Ripup button and tried deleting at multiple points. <Q> The OP surely has solved this problem by now, but I found this question when faced with the same problem, so the answer will probably be helpful to somebody. <S> At the time of this writing the current version of Eagle is v8.3.2; the procedure may be different in other versions. <S> Where is the hot zone? <S> Ah, there's the rub: <S> it is unmarked and very tiny . <S> Random clicking in likely locations will generally miss it. <S> Fortunately, though small and invisible, its size and position are predictable. <S> The hot zone is 1 mm tall and 1 mm wide wide. <S> In the versions of Eagle I have used it is found positioned at the exact midpoint of the dimension line. <S> An additional hot zone is found at the spot where the user clicked to begin creating the dimension, but this hot spot is less easily located. <S> The OP has also encountered a second problem that is very common: if you naively add a measurement (dimension) line, it will by default likely be placed on either the top or bottom layer. <S> This is a bad idea because the lines will be treated as traces that may trigger DRC rules. <S> The OP encountered this when the DRC complained about trace overlap errors. <S> AutoDesk recommends placing measurements on layer 47 (Measures). <S> If you do so, you will not trigger DRC rules. <S> There is another common source of confusion when measurements are placed on the Top or Bottom layer: the default settings violate common DRC rules for line width and for font, so the DRC creates a hatched design covering the lines, arrowheads, and the text of the measurement. <S> Even if the actual dimension is then successfully deleted, the hatched coverage remains, making it appear as if the dimension line and arrows still remain. <S> This is the problem that user1155386 has noted in the comments above. <S> It is not possible to delete this phantom dimension line, but it will go away if the DRC is run again or if the file is closed and reopened. <S> I hope this answer will help somebody, because personally I found these problems somewhat frustrating and did not find any answers in the documentation or in other questions on this site or other sites. <A> In Eagle, to apply a tool on an element, you have to Select the tool from tool box. <S> Find the origin (the + mark) of the element. <S> Click on that origin. <S> but the problem is there is no visible origin of the dimensions. <S> As explained in previous answer , there are multiple zones on a dimension which act as the origin (pointed in green arrows below) <S> Select the desired tool from tool box <S> Right click on the selection Select <S> tool-name : <S> Group Done! <S> I also found that a dimension cannot be Move , Rotate , or Mirror alone in Eagle, but can be done if the dimension is stuck in a Group . <S> I'm on Autodesk Eagle 8.5.1 Linux. <A> A good idea is in the left panel, select group , then you can select the area that contains the dimension mark. <S> Then, you can use the delete icon to delete the group you marked. <S> It works to me.
| To delete a dimension, just choose the trash icon and then click somewhere in the "hot zone" for the dimension marker. If finding the origin is giving me a hard time, I just use the Group tool Select the entire area of the element isolating it using Group (if you can't isolate it, try disabling some unwanted layers through Layer Settings )
|
Low pass sallen key filter using single supply for audio transmission I have used TI filter tool design to make sallen key low pass filter, their are four op-amps stages. But filter tools has not calculated AC coupling capacitor that has to be included after every stage. We are amplifying audio signal hence both positive and negative cycle needs amplification. Otherwise the DC voltage that is being generated after every stage will make the amplification unsymmetrical. (Schematic is attached below) Do we have to manually insert AC coupling capacitor or their is provision in the filter design software which I am unaware of. Regards, Edit: After reading the comments, I have done changes in the circuit. Then I performed simulation in the Multisim. I am simulating only first stage of the circuit. And it is working according to the design. Can anyone please explain adding 100K ohm resistor between inverting and non inverting terminal is solving the problem. In my previous design I have made inverting amplifier using single supply, but V/2 (DC bias) was only given to non-inverting terminal of the op-amp. I was applying same logic for salley key configuration <Q> Ordinarily this is done by generating a pseudo-ground about halfway between ground and the power supply voltage. <S> For a low-pass filter which is intended to reject all AC signals and only pass the DC component this may not be necessary, or if the signal has an appropriate offset which allows operation on the AC component. <S> I've not bothered to analyze your filter sections, but if the low-frequency gain is greater than one, the cascaded gain may well give you trouble. <S> However, you cannot simply add blocking caps between stages unless you provide a pseudoground to reference each section's signal to. <S> Furthermore, since there is no DC path in the non-inverting branch of the filter, adding a blocking cap will cause the simulator to barf, and a real circuit will drift to one saturation level or the other. <S> So, no, you cannot simply add blocking caps. <S> Furthermore, any blocking cap will add a high-pass component to the filter, and so it will be characterized as a band-pass filter. <A> Your DC gain is only 2 per stage so only the input and output need AC coupling. <S> Choose C from = <S> 1/(2pi <S> * f *R) for the output load and input R values then DC bias for input. <S> e.g. 100K <A> The design you show there is not really large enough to see well. <S> I can see that each stage has <S> a gain of more than 1 so any DC at the input will be amplified. <S> In that case you will need isolate each stage with a coupling capacitor and provide DC bias for each stage. <S> You can probably reduce the number of components required by just putting a series capacitor at the input together with two resistors to provide half-rail biasing. <S> Then if you put a capacitor in series with the bottom of the gain defining resistors each stage can take the bias from the previous stage. <S> (I can't read the ref designator). <A> That's what Vcm is for. <S> Set it to mid rail. <S> Now either use it to reference your input and output, or, AC couple input and output. <S> With a decoupling capacitor, a factor of 10 is sufficient. <S> Set its RC -3dB frequency a factor of 10 lower than the lowest audio you want to pass through the filter. <A> Hmm, Low pass implies DC coupling. <S> In audio work that is usually not needed so you could add decoupling capacitors. <S> Given each stage has 2x gain <S> then you have maximum 16x input off-set error of first stage. <S> With interstage decoupling your amplified error would be 2x last stage off-set error only or none with a final output capacitor. <S> I would consider an input cap and one in the middle large enough to cover your desired low frequency cut-off given the feedback network impedance. <S> This would result in a maximum of 4x the off-set error and a capacitor free direct DC output coupling. <S> Selecting high quality op-amps and resistors might be another alternative to keep the 16x off-set error within workable limits across temperature and supply variations.
| When using single-supply op amps for AC amplification such as filters, it is necessary to provide an offset voltage, since a single-supply amp is incapable of providing a negative signal.
|
Is it possible to get accelerometer magnitude value lesser than 1G? I'm using lsm9ds1 accelerometer. I retrieved row data in mg for Y, Y, Z accelerometer axes every 10 ms and calculated the magnitude sqrt(x^2 + y^2 + z^2) but this magnitude is sometimes less than 1g. How can I get magnitude < 1g? Is that physically possible? magnitude in mg <Q> If you are looking at individual samples or short term averages then that's perfectly reasonable. <S> MEMS accelerometers have 3 important characteristics, a bias offset, a drift rate and a noise level. <S> The bias offset is as you'd expect from the name, the average reading at the 0 point. <S> The datasheet should be able to tell you the maximum this should be. <S> What is a little odd is that the value can be different every time you power the sensor up. <S> The drift rate is how fast the bias will change over time. <S> And the noise level is how much random noise is added to the current measurement, this is generally large enough that for any meaningful accuracy you are typically looking at having to average a few thousand measurements. <S> There is also a non-linearity in the output <S> but this is normally fairly constant and compensated for at the factory. <S> It's never perfect <S> but it's normally relatively small in comparison to the other errors. <S> Each of these parameters is different for each axis. <S> So yes, getting a value under 1 g is quite possible. <S> As is getting a value over 1 g. <S> In fact at times it's impressive when you manage to get anything usable out of the results. <A> For example, if it were in free fall, it would read 0 g. <A> I guess by "norm" you really mean magnitude. <S> That would be the square root of the sum of the squares of each component. <S> On the surface of the earth, the measured acceleration should only be 1 g when not accelerating. <S> There are also small differences in the earth's gravity due to altitude, what rock is beneath you, how magma is moving around near you, and latitude. <S> However, these differences are quite small. <S> Every accelerometer also has some offset and gain error. <S> That is what you are seeing at the right end of your graph. <S> It looks like this one is reading a couple percent low. <S> Note that this error can be orientation dependent. <S> There are really three separate sensors in a 3-axis unit, and each can have its own gain and offset error. <S> Usually you deal with this by calibrating each unit during production, then saving calibration factors in non-volatile memory. <S> You hold the sensor fixed in each of 6 axis-aligned orientations being down. <S> From those, you can determine the gain and offset to apply to each individual axis. <A> A typical accelerometer measures specific force , which is non-gravitational force acting on an object. <S> An accelerometer in free fall in a vacuum would measure 0g since there are no external forces acting upon the accelerometer, despite that fact that the gravitational acceleration would be 9.8 m/s^2. <S> It may help if you understand how exactly a MEMS accelerometer measures acceleration. <S> This blurb on Wikipedia should help . <S> Modern accelerometers are often small micro electro-mechanical systems (MEMS), and are indeed the simplest MEMS devices possible, consisting of little more than a cantilever beam with a proof mass (also known as seismic mass). <S> Damping results from the residual gas sealed in the device. <S> As long as the Q-factor is not too low, damping does not result in a lower sensitivity. <S> Under the influence of external accelerations the proof mass deflects from its neutral position. <S> This deflection is measured in an analog or digital manner. <S> Most commonly, the capacitance between a set of fixed beams and a set of beams attached to the proof mass is measured. <S> When your accelerometer is sitting undisturbed on a surface (and is properly calibrated), the magnitude of the acceleration should be precisely 1g. <S> Otherwise, it will output an acceleration equal to the sum of the external forces acting upon it. <A> Gravity isn't uniform across the earth, surprisingly. <S> But I think this may just be a calibration issue. <S> The datasheet says (p12) <S> "LA_TyOff |Linear acceleration typical zero-g level offset accuracy(2) <S> |FS = <S> ±8 <S> g |±90 <S> mg" <S> That (up to) 90mg offset value is consistent with your observation of values of 10-20mg below 1g. <S> Try measuring it at the foot of a large mountain; you may get a different result .
| A value less than 1 g is certainly possible if the accelerometer is being accelerated downward.
|
How to convert analog output from sensors to digital without microcontroller? Okay, I want to make this clear - I am from computer science background, and would require more explanation than an average electrical guy. The scenario, I have piezoelectric and LDR sensors, which give analog output of course. I plan to transmit the output from this using a serial 434Mhz RF Module, which will be received by a Raspberry Pi on the receiver side and processed further. Now, I am using HT12E encoder to encode 4 lines of parallel data, which according to my knowledge is 4 digital inputs, each of which I plan to use for 4 different sensors. I have setup the transmission part, and got the RF working. The question, can I convert the output from the sensors to digital output, without using a micro controller, and send it to HT12E to encode, if so, how can I achieve this? If I have to use a micro controller, what would you recommend, Arduino or RPi, portability is important for me, and as far as I know, RPi doesn't have inbuilt ADC and is much bigger? Ideally, I would like to use the first method without micro controller mainly for portability and power consumption reasons. Please ask me about anything if I haven't been clear. <Q> You want to use a microcontroller. <S> Microcontrollers can be tiny integrated circuits. <S> This is the right approach. <S> Arduino and Raspberry Pi's are not microcontrollers. <S> The Raspberry Pi is a single board computer, and the Arduino is a microcontroller development board, associated with a "wiring" development environment. <A> Before we had lots of small microcontrollers, we had voltage to frequency converters. <S> Perhaps that would work for you. <A> If you are connecting four sensors to four digital inputs, it seems fairly obvious that you only expect to get binary data (on/off, 1/0, active/inactive) from each one. <S> In that case, all you need is a quad comparator chip, with the threshold on each comparator set to an appropriate level for that sensor. <S> However, there are additional issues, such as how often you need to sample the data in order to capture all of the information <S> you're interested in. <S> You need to provide additional details about what that information might be.
| Especially when using a low-bandwidth radio link, it usually makes a lot of sense to use some local intelligence — i.e., a microcontroller — to convert the raw data into the information you're really interested in.
|
Is a very large capacitor the same as a short circuit? This is a theoretical question. Let's say that we have two black boxes, each with two terminals. One contain an infinity (or arbitrarily) large capacitor in series with a 1 ohm resistor and the other contains just a 1 ohm resistor. Is there a way to tell which is which using finite finite resources (e.g. upper bound on experiment time, upper bound on voltage sources, upper bound on equipment accuracy, etc)? My thinking is that since the capacitor is arbitrarily large, it will not get charge to a noticeable voltage in a given finite time and thus will be identical to a short (zero voltage regardless of finite current). <Q> Yes, your analysis is correct for a infinite capacitor. <S> However, anything less than that can be detected in arbitrarily short time. <S> The problem is that the size of the signal to notice the difference gets smaller as the time to run the experiment gets smaller. <S> Larger current makes the effect larger in the same amount of time. <S> Let's say your current is limited to 1 A and you have a 12 bit A <S> /D in a 3.3 V microcontroller. <S> Let's see how large a capacitor this could detect. <S> The voltage change of a cap as a result of some Amps for some seconds is: V = <S> A s <S> / F <S> Where A is the current in Amps, s is the time the current is applied in seconds, and F is the capacitance in Farads. <S> Flippping this around to solve for the capacitance yields: F = <S> A s <S> / V <S> The minimum voltage change we can detect is (3.3 V)/4095 = 806 <S> µV. <S> Plugging in our particulars, we get: F = <S> A s <S> / V = <S> (1 A)(1 s)/(806 µV) = 1.2 kF <S> That's a very large capacitor. <S> If you can supply 5 A and wait 2 seconds, then you can detect a 10x larger capacitor. <S> Or conversely, be able to measure 1.2 kF to 1 part in 10. <S> Yet another way to look at this is to apply a constant voltage for a fixed time, then see how much the open-circuit voltage went up afterwards. <S> The voltage on the capacitor will rise exponentially, asymptotically approaching the fixed voltage being applied. <S> Again let's say we can measure down to 1 part in 4095 of the applied voltage. <S> That comes out to 0.000244 time constants. <S> If that's how long 1 second is, then the time constant must be 4096 seconds. <S> With a 1 Ω resistor, that means the cap is 4.1 kF. Note that cheap $20 voltmeters can measure much smaller voltages than a 12 bit A <S> /D running from 3.3 V. <S> Basically, it takes a unrealistically large capacitor to not be detectable via rather simple means. <A> Yes, not only can you say that an infinite capacitor acts like a short, but a wire IS an infinite capacitor. <S> First, recall the formula for a parallel-plate capacitor: $$C=\frac{\epsilon A}{d}$$ Grab your nearest monomolecular blade (1 Å) and stretch it a bit <S> so it's even thinner. <S> Now, cut your solid wire into two pieces, creating parallel surfaces separated by the thickness of your wire, so the parallel plate capacitor formula applies. <S> Observe that as the thickness of the blade approaches zero, eventually it passes between the atoms of the wire without interacting with any. <S> So the continuous wire before and the capacitor after are physically indistinguishable. <S> Also$$\lim_{d \rightarrow 0 <S> } C \rightarrow \infty$$ <A> In your theoretical world of infinite capacitance, I suppose the step input would be indistinguishable between the R and the series RC. <S> That is only the case because of the infinite capacitance: anything less would gain charge slowly and start to block the DC. <S> This happens since when solving for the initial current through a capacitor you ignore the capacitor and I = V/R. Since the capacitor would never charge up at all, being infinitely large, this would remain the current. <A> You are correct that for an arbitrarily large capacitor (approaching infinite farads of capacitance) and for a finite limit on testing strategy you would not be able to tell the differences between the two black boxes. <S> If you begin to bring the capacitor value back into realistic and implementable sizes <S> (so you can at least carry these black boxes around) and remove any restrictions on your test time and equipment accuracy you would begin to see the current into the series RC box begin to diminish as the capacitor starts to charge. <A> If you are dealing with DC, a capacitor is always a open circuit. <S> If you are on transient domain (ie: calculating the circuit reaction to a key switching), the capacitor is an short until it is fully loaded. <S> Then it will work as an open circuit like the DC model. <S> If you are dealing with AC, a very large capacitor (a capacitor with theoretical infinite capacitance) is an short circuit. <S> So, answering your question, Feed the black boxes with a 1V ideal DC Voltage source and wait. <S> After a very long time (infinite time), the box with the resistor alone will be draining 1A while the one with the capacitor in series will be draining no current. <A> The way to tell is to apply your voltage for a small given time, then disconnect the voltage source and short the terminals through a current measuring device. <S> That and the whole... "I can't lift the infinite capacitor box" thing. <S> BTW: <S> All capacitors present themselves as almost a short when power is first applied to them. <S> This can introduce a troublesome inrush current when powering up a circuit-board or system if there is a large capacitor farm. <S> This is one of the reasons you often see a small inductor added to the in-line before the capacitors.
| Since the capacitor stores energy and the wire doesn't the one that indicates a current when shorted is the capacitor.
|
AtTiny85 analog / digital pin configuration order in setup Bug in compiler or what ever? Simplified circuitry there is 2 LEDs and 1 analog input.LED connected to pin PB1 lights very dim when output is high. It seems that output has been configured to high impedance mode.That happens if inside setup() analog input A1 (pin 7, PB2) is configured after digital output PB1.If order of pin configuration is changed everything is OK.Is that compiler bug? /* ____ D5 PB5 1|o |8 Vcc D3 PB3 2| |7 PB2 A1 Voltage measurement, potentiometer D4 PB4 3| |6 PB1 D1 --|>--| GND 4|____|5 PB0 D0 --|>--|*/byte led0 = PB0;byte led1 = PB1;byte potentiometer = A1;int voltage;/* It is important in which order pin configuration has been set.Wrong order causes PB1 to high impedance, low current, output.This must be compiler bug, I think.If port BP1 setup has been made using straight to register there is no problem in order.Any other analog input port and digital port combination works correctly. */void setup(){ pinMode(led0, OUTPUT); pinMode(led1, OUTPUT); // Doesn't work if this is before pinMode(potentiometer, INPUT); //DDRB &= ~(1 << DDB2); // Alternative working setup for analog input A1 pinMode(potentiometer, INPUT); //pinMode(led1, OUTPUT); // Works if this led1 setup is after pinMode(potentiometer, INPUT);}void analog_input(){ voltage = analogRead(potentiometer); if (voltage < 500) { digitalWrite(led0, HIGH); digitalWrite(led1, LOW); } else { digitalWrite(led0, LOW); digitalWrite(led1, HIGH); }}void loop(){ analog_input();} <Q> It should be noted that ATtiny is not supported by Arduino IDE out of the box. <S> There are several third-party implementations ( <S> e.g. 1 , 2 , 3 ) which provide necessary files (mostly pin definitions) for Arduino, and you didn't tell us which one you're using. <S> The error looks like the implementation you're using has screwed up pin mask definitions, since applying the right mask directly in your code works. <S> You should check the code you end up compiling (specifically the constants used in <S> pinMode() function) for obvious errors. <S> There are makefiles for Arduino <S> (e.g. arduino-mk ) which can be instructed to provide you with the <S> preprocessed <S> *.cpp <S> file, as seen by the compiler. <S> I always prefer looking into those, because they represent the real code, after all the tricks that Arduino IDE pulls out. <S> If doing that sounds too hard, I suggest you file a bug to the developer of the implementation you're using. <S> Or just pick a different implementation, which will probably have a different set of bugs. <S> Compiler bug seems unlikely, because compilers are usually rigorously tested before they are released, unlike amateur projects on github. <A> As noted in the documentation, you cannot use the A# defines for pinMode() - you will get unexpected behavior. <S> You must use the Arduino pin number, or the PIN_xx define. <S> PBx works on the tiny85, but may not work on all chips - the PBx defines come from the compiler, and are #defined to the number of the bit within the port registers that controls that pin. <S> Edit <S> : Misread the original problem. <A> . <S> Is that compiler bug? <S> At least three things are involved sand potentially more. <S> Your hardware. <S> The Arduino port to your hardware. <S> Thee compilers.
| Without ruling out the other factors, it is hard to pin the blame on thee compiler.
|
What kind of hall effect sensor IC should I use to convert a frequency measurement in a continuous voltage? I need to measure the angular speed of a DC motor. My idea is to fix a magnet on the spinning wheel and a hall effect sensor near to the wheel. What I need as output is a continuous voltage proportional to the frequency measured. If there is a practical way to achieve this using an IC could you please explain it to me? Or should I make the encoder myself? this second option feels a bit like re-inventing the wheel to be honest. <Q> If you trigger a one-shot multivibrator from the digital output of a Hall sensor and low-pass filter the digital output (and buffer if necessary) you will get a voltage proportional to RPM. <S> It is best if the multivibrator is operated from a well regulated supply voltage. <S> This works best <S> if you get many pulses in the time you expect the voltage to react to changes in angular velocity. <S> Alternatively, it's always possible to do it digitally, and turn the resulting number into a voltage using a DAC. <A> Look at the LM2907: Freq to Voltage Converter <S> This is the first device I found when I googled "frequency to voltage converter IC" I've never used one. <A> The simplest can be built from a bunch of resistors plus capacitors. <S> There are also chips for that. <S> Alternative, a simple MCU can do that.
| A voltage to frequency converter would work.
|
How do flyback diodes (on an H-bridge) affect the motion of the motor? I'm designing an H-bridge and one of the features I'd like to have is to allow the motor to keep it's momentum even when the H-bridge is off. Obviously flyback diodes are used in H-bridge designs to allow the coils to discharge. I feel like I don't have a very intuitive understanding of this concept, but my hypothesis is that this will cause the motor to stop spinning (or at least slow down a bit until the voltage on one of the motor's terminals is no greater than the power supply +0.7V). Is this correct? What does this mean from a mechanical perspective? <Q> The snubber diodes only provide action in dissipating the energy "stored" directly in the windings of the motor. <S> Since the direction of current flow in the motor winding is in the same direction as when being driven, the dissipated energy will in fact continue rotation and not retard it (brake the motor). <S> If you consider the effect of a snubber diode on a relay for example, adding the diode makes the relay change state slower when power is removed. <S> You can control the time taken to dissipate the energy by allowing the flyback voltage to increase to some larger controlled level. <S> Here with a basic snubber: Here with a Zener to increase the snubber dissipative power to reduce the run-on effect: Notice here that the time taken to discharge the energy is much reduced. <S> The time taken to dissipate this energy becomes important where you want to reverse the voltage across the motor. <S> The time taken becomes the limit of your switch time: <S> The last element is the ability in an H-bridge to "brake" the motor by turning on either both high side or both low side switches at the same time to dissipate any inertia or momentum stored in your system. <A> Consider the image below. <S> These are the five legal states of a H-bridge. <S> There are two illegal states (will cause damage) and four that don't do anything. <S> All switches OFF - any current would freewheel by the supply & decay 1 & 2 ON - positive current can build up in the load 3 & 4 ON - negative current can build up in the load 1 & 3 ON - zero volt loop that minimises load current decay OR shorts winding. <S> 2 & 4 ON - zero volt loop that minimises load current decay OR shorts windings. <S> For each state, the current path is shown. <S> THe diodes are present to ensure there is always a path that the load inductors current can take. <S> There is however a massive difference between the freewheel paths of State1 & State{4,5} <S> In state1, the "natural" freewheel path, the safe & default path, the main DClink is in-circuit. <S> Thus to facilitate current flow, the load's voltage must be greater than this voltage. <S> This will cause the current to decay relatively quickly <S> and thus there will be some deceleration torque experienced at the shaft In state{4,5} a "zero volt loop" can be established across the load and thus the free-wheel conserves the the load inductors current. <S> Once the rotor has stopped however these schemes will facilitate a pseudo-locking mechanism <A> You have to distinguish between two different effects: The inductive energy stored in the motor coils, which generate "flyback" voltage pulses when you switch off current The back EMF of the motor, which (when the bridge is switched off) is simply the EMF of the motor acting as a generator. <S> The energy involved is normally relatively small and has little effect on a freewheeling motor. <S> (And if it does, the effect can be minimised using the snubber tricks in Jack's answer). <S> Back EMF - or generator EMF - is dependent on speed, and will generally be less than the incoming supply voltage. <S> Thus, if you simply switch off the bridge, letting the motor freewheel, the generated EMF is not high enough to turn on any of the diodes, and the motor will normally freewheel - with two important exceptions. <S> If you also remove the supply voltage from the bridge, then the generator will power the circuit through the diodes. <S> (One form of regenerative braking is to use a boost convertor to charge the battery while this is happening). <S> But as you don't want braking, you can avoid this by keeping the bridge powered even while all its transistors are off. <S> If the motor is being driven by its shaft - e.g. a car running downhill - faster than its top speed, the generated EMF can exceed the supply voltage, turning the diodes on. <S> This will increase the supply voltage and either charge the battery or possibly destroy other electronics on the supply.
| The flyback diodes simply return the inductive spikes to the supply instead of generating excessive voltage which could destroy the transistors.
|
Amplifier Repair - How can I find a replacement transformer? So I'm trying to repair an old Marantz 1040M amplifier, which after being run at high volumes on New Year's(!!) no longer powers on :( My electronics knowledge is fairly rudimentary, but I think I have identified the problem as being transformer related. The primary winding reads an open circuit on a multimeter, and there is no voltage across any of the secondary terminals when a voltage is applied to the primary. It has three primary terminals (0v, 210v and 240v for EU/UK use) and four secondary taps. I want to replace the transformer, however I can find no information online regarding its specifications. Searching for the part number yields zero useful search results. Some research has revealed that the 1040M is internally identical to the PM-200. I have found the relevant service manual, but that shows little clues as to the transformer specs - https://www.manualslib.com/manual/691537/Marantz-Pm200.html?page=8#manual I have contacted Marantz, who say that the company has changed hands so they have no access to specifications of older products! Both of the repair companies Marantz pointed me to are of no help either. The label on the transformer reads - TS1662002-0(ETP66P2E) and to the side of that - 6931M So basically the question is, can I repair this amp, or should I sell it for spares and write the whole thing off? Is there a way of somehow figuring out what voltages the board needs, just using a multimeter? Could I even crack the transformer open and repair the coil? I'm fairly sure the secondary winding is okay. Any advice much appreciated. Cheers, Oli <Q> Sometimes transformers have a thermal fuse underneath the black tape wrapped around the coils. <S> You could carefully remove the tape and replace the thermal fuse if there is one. <S> For example - http://www.electronicrepairguide.com/images/thermalfuseinside.jpg <A> I don't think there's any escaping that <S> but you might get lucky and find that the break is very near one lead or the other. <S> If not, dismantling that transformer may be difficult. <S> (But what is the worst thing that can happen? <S> It breaks. <S> And you are no worse off.) <S> But before you go and throw it out or attempt to tear apart the transformer you can do something else, first. <S> You already know the AC mains voltage. <S> So that's good. <S> What you need to do is to find out the AC secondary winding voltages. <S> Assuming those are still working okay <S> , you can go find a 6.3 VAC transformer and wire up its secondary to any of two wires that are still connected (test with an ohmmeter to work out all the leads that appear connected to each other.) <S> [If you can find a lower AC voltage secondary than 6.3 VAC then use that instead. <S> Up to a point, lower is less risky when testing like this.] <S> You can then measure the voltages at other pairs and, from that, work out the likely secondary voltages on the transformer. <S> If you do this and work out the required secondary voltages (through such testing) and already know the mains voltage, as well, then you can go look up a replacement. <S> (You'll need to find one that weighs about the same, or more, though.) <S> Or just find several such transformers, if you can't find one that matches up the same on all of the secondary voltages. <S> I'm not sure what else to suggest right now. <S> But it's late here. <S> If something else comes to mind, I'll add it. <A> Some transformer winding Houses offer a ' Detail Rewind service ' where they strip the core out and wind exactly the same windings with the same number of turns and wire sizes .This <S> can cost a little more and may be worth it <S> .Otherwise <S> you can look at the big electro caps and multiply the volts by say .7 and use lab supplies to slowly crank up the volts checking current .You <S> can now see if your amp still goes <S> .If <S> it does then its time to replace the power transformer .
| If the main power leads are no longer connected (you wrote, "the primary winding reads an open circuit on a multimeter") you will have to find where the wire break is located within the transformer and repair it.
|
How to work with small components I want to test this IC and therefor I need to implement a LC tank with really small components. Is there a trick how to properly solder those parts together? Especially for prototyping, where I might have to exchange parts from time to time. Can I solder them somehow on universal PCBs? I don't know the technical term for those kind of small components: inductor , capacitor <Q> That looks like a QFN package. <S> You really need a PC board for those. <S> The pads are too close together and have too little area for soldering wires to them to be feasible. <S> QFN packages are really meant for reflow soldering, not manual with a soldering iron. <S> If you are doing this professionally or this is a commercial product, get a hot air station. <S> If you really need to use a soldering iron, then extend the pads a good ways past the part, like 20 mils or more. <S> That way there is a place for the soldering iron tip to touch the pad and the pin of the part simultaneously. <A> If you are looking for a quick solution then you are far better off starting with a pre made development board. <S> http://www.mouser.com/ProductDetail/Texas-Instruments/FDC2114EVM/?qs=sGAEpiMZZMvNM%2fd3q5fCVzBrFaRB3C8F1tKLhQwKrCs%3d <S> You are likely to get non-working boards <S> the first few times you attempt this on your own. <S> If you do need to roll your own board to get ready for a commercial product then Sparkfun has some good tutorials to get you started. <S> https://www.sparkfun.com/tutorials/category/2 <A> You will find some difficulties soldering QFN Packages since the pad area is mostly under the component itself. <S> Usually, a reflow oven is used to mount those types of packages to a PCB. <S> The capacitor can be soldered easily but you need to let enough area for the solder to simultaneously heat the PCB pad and capacitor pin. <S> But in general, you will need to heat up the parts individually to be able to remove them one by one if you plan on removing them. <S> Again, you will need to make sure to not exceed the thermal capability of the part which is usually 125C-150C.
| For manual one-off use, a hot air soldering station is best.
|
Can I use a LED power supply for a general purpose I have one simple question: can I use a LED power supply for a general purpose, e.g., to supply some logical controllers etc. Are there any limitations of a LED power supply that I'm not aware of? For example, I would like to buy this power supply to supply a PLC. <Q> That looks to me like a standard, general purpose DC power supply, so it should be fine for your purpose. <A> As long as the output voltage and current match, you should be good to go. <S> Something to keep in mind, especially with cheap power supplies, is the output ripple and regulation <S> can be quite severe. <S> The power supply might say it's 24V, but due to poor design, the actual output is only 22V. Or the voltage might droop by a few volts. <S> At the end of a day, a power supply is a power supply. <A> Usually LED power supply is called one that keeps a specific amperage constant by varying it's voltage, which is probably something you don't want. <S> If you can't find it's data-sheet, then you should get a normal power supply.
| There are some power supplies designed specifically for use with LEDs that have a constant current output - the voltage automatically varies over some range to keep the current through the LEDs constant - that type of supply would not be suitable for other applications.
|
Using PNP as OFF switch I am a total novice but not unintelligent. I want to use a single solar panel to switch on some lights when it gets dark and also charge the batteries for the lights. I have previously successfully done this using a JFET, but have to have TWO solar panels, one to switch off the JFET and one to charge the batteries. I have read on numerous web sites that I can use a PNP transistor. As far as I can understand it they ALL say that a PNP is ON except when a base current is applied. Example circuits are shown. I have bought quite a few PNPs, different values. I have tried all of them in the suggested circuits. I have tried all of them with just a battery and a bulb. None of my PNPs are on with no base current, they all only switch on when I apply a current to the base. This is the opposite of what all these sites are saying. What am I not understanding? Connect battery +ve to PNP emitter, connect pnp collector to +ve of LED, connect -ve LED to -ve battery NO light.Connect -v of solar to pnp collector and +ve of solar to pnp base and LIGHT. This is the exact opposite of what I want. What am I doing wrong <Q> so for PNP emitter is + <S> ve and base is -ve so to forward bias connect emitter to +ve supply and base to ground with resistor (see attached image) <S> similarly for NPN connect emitter to GND and supply +ve voltage to base Note: <S> the explanation are purposely made simple for understanding purpose <A> PNP will work like this... <S> Connect battery +ve to PNP emitter, connect pnp collector to +ve of LED, connect -ve LED to -ve battery NO light. <S> Connect -v of solar to pnp collector and +ve of solar to pnp base and LIGHT. <S> This is the exact opposite of what I want. <S> Make bit more clear explanation on this. <A> I have read on numerous web sites that I can use a PNP transistor. <S> As far as I can understand it they ALL say that a PNP is ON except when a base current is applied. <S> Well, those websites are either wrong or you have misinterpreted what they say. <S> A PNP (or NPN transistor) is turned on (in various degrees) by the application of base-emitter current i.e. a current into the base or out of it. <S> None of my PNPs are on with no base current, they all only switch on when I apply a current to the base. <S> That's how they should work. <S> What am I doing wrong <S> Believing in crappy web sites sounds like your only crime! <S> I would name them if you can be bothered. <A> As you describe your PNP circuit (Vs->PNP_emitter->PNP_collector->LED_a->LED_k->0V) where Vs-0V is your supply, that makes sense. <S> You must draw current from the PNP base for it to conduct emitter-collector current, not drive it into the base as for NPN transistors. <S> Your base voltage must be taken below approx. <S> Vs-0.7 V, which allows for the diode drop between the emitter and base. <S> You will readily find plenty of text describing this on the Internet. <S> I suspect that your solar cell's output voltage is more than 0.7 V below your battery voltage so your solar cell is drawing current from the PNP base. <S> Try putting a 1 K resistor between your PNP base and the battery negative, or 0V as I called it. <S> The LED should light.
| To switch on any transistor PNP/NPN, you need to forward bias the base emitter junction
|
How to design an optocoupler circuit for raspberry pi I want to give sensor pulse to raspberry pi. The sensor signal comes from a running conveyor . The sensor gives 24v signal hence I used a optocoupler to convert the 24v to 3.3v . How should I design a best optocoupler suitable for PI. <Q> It seems your problem is that you have a 24 V signal that you want to get into a 3.3 V digital input. <S> Since the signal is coming from some industrial equipment, I agree that opto-isolating <S> it is a good idea. <S> So here is what you do: <S> Pick a opto-coupler. <S> Use whatever your company already has in stock. <S> If you don't already have something available, check out the FOD817 family. <S> They are cheap, widely available, and have good CTR. <S> They aren't for high speed applications, but should be instantaneous for a signal from a conveyor belt. <S> Read the datasheet. <S> Look at what kind of digital input is available. <S> If all you have is a regular digital input, then you have to supply your own pullup resistor. <S> 10 kΩ is usually a good value for such things. <S> Either way, find how much current the opto has to sink to pull the digital input low. <S> Divide the minimum required pulldown current by the minimum guaranteed CTR (current transfer ratio). <S> Now at least double that. <S> This is current <S> you want to drive the input of the opto with. <S> Check the specs for the 24 V signal and verify it can source the current computed in the previous step, with some margin. <S> I'll assume it can. <S> If it can't, you'll have to get more clever and the solution isn't as simple as hooking up a opto-isolator. <S> Look at the minimum guaranteed signal voltage (24 V nominal might only be 22 V min guaranteed, for example), and subtract off the maximum LED voltage from the opto datasheet. <S> This is the minimum voltage that will be across the resistor. <S> Use Ohm's law to calculate the maximum allowed resistor in series with the LED on the opto's input side. <S> Go one or two standard values down from that, or some value you already stock that is a bit lower than the calculated value. <S> Connect everything up. <S> Put the opto physically close to the digital input it drives. <S> In other words, the long pair of wires should be between the conveyor and the opto, not the opto and the digital board. <A> Does your sensor toggle between 0V and 24V with each impulse? <S> Are the GNDs connected? <S> You don't need an optocoupler then but can use a much simpler circuit: simulate this circuit – Schematic created using CircuitLab <S> How does this work? <S> Easy, the +24V from the input don't ever reach the Raspberry Pi GPIO port. <S> Only the 0V do. <S> Your impulses toggle between 0V and 24V, remember? <S> When the sensor outputs 24V, the diode is non-conducting and thus, the GPIO is pulled to 3.3V by the resistor. <S> When the sensor outputs 0V, the diode is conducting and the GPIO is pulled to 0.4V by the sensor. <S> Depending on your Raspberry Pi GPIO pullup setup, you may even leave out R1. <A> if the sensor generate high speed pulses use high speed optocouplers 6N137
| Just about any will do since you seem to have plenty of input voltage and presumably current available. The best would be one that has a internal pullup, because then you need not other parts than the opto itself on the digital side.
|
SimCom SIM5360 sharing SD card with uController I am designing a data logging system that uploads its data via 3G. I'm going to be sampling an I2C sensor at 10Hz, buffering this in local memory (or FRAM), and then dumping the contents of the buffer to an SD card periodically (every few seconds). Every 15 minutes or so, I want to upload the contents of the file to a web server using a 3G module (Sim5360). I understand that this module can be directly connected to an SD card, and can be instructed to upload files from the SD card. This will take processing effort away from my uC and will increase the upload speed that can be achieved, compared to streaming the data from the uC via the UART. In order for this to work, both the Sim5360 and the uC need to have access to the SD card, requiring a multi master SPI bus. Is this possible? As long as both devices have the ability to read and write from the SD card, I can ensure that there are no conflicts by simultaneous attempts to use the resource at the same time. Please can someone tell me if this is possible and if there is anything special I need to do to make this functional? Alternatively, if anyone has a better idea on how to achieve this, I'd love to hear it. <Q> It's not impossible, but you have to be very careful from a software point of view regarding the filesystem on the card. <S> Each device must flush buffers and unmount the card before "releasing" it to the other device. <A> It does not really matter for example if you have two slaves beside the MCU or a slave and a turned off other master. <S> So basically the turned off <S> Disconnecting the lines of the SIM5360 by a multiplexer controlled by the MCU is not recommended. <S> It would bring more risk into the system because due to a software bug the MCU might disconnect the SIM5360 while it reads/writes the SD card. <S> I think you should check two condition by the MCU before accessing the SD card. <S> Always check if the SIM5360 is properly, entirely turned off. <S> Check the status of the Slave Select pin of the SD card if it is in high logic state (assuming active low slave select pin). <S> If either of these conditions are false do not proceed. <A> Make Sim5360 "master" for the SD card. <S> So it can get dedicated access to the sd card when needed. <S> This can be achieved using some input pin on the uC that is controlled by Sim5360. <S> If uC is not able to write its buffered data to the sd card it just needs to buffer longer or throw some data away. <S> If the amount of data is reasonably small, it can even be stored in flash until it can be copied to sdcard.
| SIM5360 should not cause any trouble to the MCU given it properly unmounted the SD card before turning itself off, just as @pjc50 has mentioned.
|
Discrepancy on the ADC input impedance in the ATtiny datasheet 1..100k or 100M? I found a discrepancy in the ATtiny datasheet . While the section explaining the ADC shows a simplified schematic with a 1...100 kOhm series resistor, the characteristics table shows 100 MOhm. Figure 17-8. Analog Input Circuitry The ADC is optimized for analog signals with an output impedance of approximately 10 kΩ or less. [...] With that max. 100 kOhm series resistance in the sample/hold path it makes sense to me, that they say that the source impedance should be about 10 kOhm. But with the 100 MOhm impedance from the characteristics table I'd say requiring 10 kOhm source impedance is unnecessarily high. Table 21-8. ADC Characteristics, Single Ended Channels. T A = -40°C to +85°C ... Table 21-9. ADC Characteristics, Differential Channels (Unipolar Mode). T A = -40°C to +85°C ... Table 21-10. ADC Characteristics, Differential Channels (Bipolar Mode). T A = -40°C to +85°C ... What is the reason for that discrepancy? Or can I assume the 100 MOhm is a wrong value and should actually be 100 kOhm typical? <Q> The active value you first quote is when the ADC is sampling an input signal. <S> In effect, both are in parallel. <A> If you read further on, they explain why they recommend less than 10kohm source impedance. <S> It is so that the sampling capacitors can charge up in a negligible time. <S> If you use a higher source impedance, they will take longer to charge, and initial readings will be low. <S> The ADC input does have a >100Mohm input impedance at DC. <A> It's helpful to understand a bit about the data sheet generation process at a large silicon vendor. <S> The technical editor responsible for the data sheet typically gathers information from Design Engineering, Test Engineering, Applications Engineering, Development Tools Support, Marketing and possibly external early-adopter customers and consultants. <S> Electrical models in the data sheet are often simplified models of complex behavior, and the parameters in them are the results of many hours of measurements, calculations and production parameters gathered from different teams. <S> A lot of the information is typically "hot off the press" and the data sheet editor is under a lot of pressure to release the initial copy. <S> Before first release, the document is circulated for sign-off review to design, test, applications, devtools, marketing and finally, legal. <S> The point is that many disciplines provide input into the document, and good datasheet editors are miracle workers in how few errors slip through with so much data from so many sources. <S> Conclusion: Discrepancies happen all the time, and you should not be surprised by them (or blatant errors) in datasheets - just read the legal disclaimers in them. <S> Also, when you download a datasheet, always look for the errata on that datasheet and make sure you read it first. <S> When you do find a discrepancy you can not or do not want to solve , the vendor's support system can always be used to figure out the facts, but it typically takes time and can set back your design schedule. <S> That's why it's good practice to design with popular products that have been in the market for 1-2 years - other early adopters have found/fixed <S> most of the bugs in the data sheets for you!
| The 100 Mohm is a static value which pertains to leakage.
|
The concept of protection diodes across a fan/motor? I'm designing a circuit which includes a fan being controlled and I've seen on a few websites that I should include a protection diode across the fan (as shown in the diagram below). However, after searching and trying to think about it, I can't see what the purpose of the diode is? If it's in that direction it's not allowing any current to travel down that branch of the circuit anyway as it will always be reverse biased? Does anybody know/can explain if this is necessary/correct to use a diode in this way for a fan/motor circuit? simulate this circuit – Schematic created using CircuitLab <Q> The problem comes when you try to switch the load off. <S> You don't show a switch in your circuit, but if you had one say in the low side of the fan and you switched it off, <S> without the diode there would be no path for the inductive current except for the parasitic capacitance at the switch node. <S> Motors are inductive loads and an inductor can't change its current instantaneously. <S> V= <S> L*di/dt <S> so the V can spike extremely high if the current is forced to change quickly. <S> By putting the diode across the load you provide a path for the current to continue, with a small voltage across it to discharge the inductor slowly. <A> Does anybody know/can explain if this is necessary/correct to use a diode in this way for a fan/motor circuit? <S> you can think of a motor as an inductor. <S> all inductors have one important attribute: <S> current through an inductor cannot change suddenly, as just voltage across a capacitor cannot change suddenly. <S> so when your device switches off the motor / inductor, the current through it will want to continue to flow, create a huge voltage across your device, potentially killing it. <S> the diode provides a path for that current. <S> thus the name "free wheeling" diode. <A> It's a flyback diode. <S> When you disconnect the power from an inductive load, such as a motor, the load will induce a voltage to resist the change in current. <S> If nothing in the circuit allows the back EMF current to flow, the inductor will induce a voltage so high that something breaks down and allows that current to flow.
| The diode will allow the current to flow in the same direction and at the same strength as before the motor got disconnected and drain the inductor's magnetic field.
|
Is it possible to solder a capacitor to SMD board using regular iron? I need to replace the C219 of the following SM design: Is it possible to accomplish this task using regular soldering iron (used for through the hole design soldering) and some solder? Can I use a regular legged capacitor to replace the capacitor with, or does it have to be SMT? I have extra set of hands available, so holding it in place should not be an issue. Are there any precautions I should take as to not damage the board (very expensive)? There is some PSU wires next to the part (can be seen on the right) that apparently can't withstand a lot of heat. What temperature should I keep the iron at? <Q> Yes it is quite easy. <S> If you have no experience in this sort of thing, either get someone else to do it or practice on boards that are trash. <S> With that kind of part you can suck most of the solder off each pad with some fresh solder wick and then melt one lead and gently tilt the part away from th pad, freeing that end, then melt the other side <S> and it should come right off. <S> Clean the pads up with solder wick <S> (so the part will sit down flat), solder one lead then the other to center the part on the pads. <S> Strongly suggest getting a proper SMT part. <S> If you must hack something in, there are techniques to do it safely, but I doubt that is desirable. <S> Try about 320 <S> °C for the iron tip temperature. <S> Obviously you want to use a nice thin tapered tip that is small enough that it is suited to the job and avoid melting that wire you mentioned. <S> (both photo credits to Digikey) <A> Add solder to the pads. <S> Use a pair of tweezers to hold the capacitor and pull gently. <S> Heat one pad until the solder flows Switch to the other pad and heat it until the solder flows The capacitor should come up off of the board. <S> If it doesn't come off entirely, don't force it. <S> Repeat the heating of the pads until it comes off all the way. <S> Remove the excess solder with solder wick. <S> Be careful because this can easily pull the pads off the board. <S> Tin one pad Place the new part on <S> the pads - watch the polarity since you are replacing an electrolytic capacitor. <S> Push down on the capacitor (gently) with the tweezers. <S> Heat the tinned pad until the solder liquifies <S> The capacitor will pop down solidly onto both pads. <S> Remove the heat from the pad <S> Place the tip of the iron against the junction of the capacitor's free connection and the pad under it <S> Feed in fine solder <S> (I use 0.5mm solder) to the junction <S> The solder will melt and wick itself in between the capacitor and the pad Remove the solder wire from the junction <S> Remove the soldering iron from the junction <S> Solder cools <S> , job done <S> Then follows step 20: Cuss because the problem is really something else. <A> 1 - Yes, you can replace it using a regular soldering iron. <S> It has to be thin enough to fit in the space between the targeted component and the neighbooring ones. <S> I've seen technicians adding more tin to the pins, then heating them back and forth until the component moves out of its place. <S> If you have 2 irons, you could do it without this technique. <S> 2 - I do not recommend using a through-hole component on SMD pads, considering that board, according to you, is very expensive. <S> (I've seen pads come off boards because someone soldered wires on them for testing, then accidentally pulled on one of the wires) <S> 3 - Unless you don't really have a choice you can use the a TH component and secure it with some fixating material. <S> 4 - Iron temperature depends on the soldering material you're using, you have to reach a high enough temperature to melt it. <A> For removing, consider using chipquick. <S> It is a special type of solder you add to the pads to help remove the old SMD. <S> Clean all the chipquick off with wick. <S> You really do not need to have much solder on the pads when added the SMD. <S> You do need the pads to be tinned, and you need to have flux on the pad. <S> Then the solder will flow under. <S> Press the iron tip to the edge of the cap and add the solder. <S> It should flow. <S> Do one side at a time, for the first side you will need to hold cap in place with a tweezers. <S> You can buy this paste.
| You can also you a paste of solder and flux that hold the cap in place before you touch it with the iron. Yes, you can do that.
|
Design of a single cell low voltage reverse battery protection circuit I have a design that will run off of a single cell 1.5V AA battery and I wanted to provide some reverse battery protection to it while also maximizing my run time. The classic way to do this is just with a FET. But then I was reading a TI app note where they note that when a battery is drained to 0.8V the voltage is not sufficient to fully turn on the FET and it will operate in the linear region. Therefor the RDSon is much higher and the voltage drop will not allow the regulator (which is placed after the protection) to turn on. Their solution in the app note is a regulator they have with a Vaux output that charges up to 2.5V before turning on the main output. This Vaux pin is used to pull up the gate of the FET thus turning it on. The part they use is a little too small for me and I was trying to figure out a way to get just this feature discretely or with some additional IC. Has anyone run into an application like this before or have any advice? I looked for some charge pumps but they stopped around 0.9V input where I want to get down to 0.7V (their circuit can do 0.5). I looked for some other regulators and I've considered using their regulator just for it's vaux capability :) Finally one AA battery is a hard requirement, I don't have a choice here. Here's the TI circuit for reference: <Q> Seems like a lot of hassle for just a reverse battery protection. <S> If the battery is reversed, a lot of current flows through the diode. <S> Many circuits (chips) can handle such a small reverse voltage but do check that. <S> The large current blows the fuse. <S> You could make the fuse a standard replaceable glass fuse or a polyfuse (these recover after some time). <S> Yes another solution is a mechanical solution. <S> I once worked on a product running on a single AAA cell. <S> It had a non-electronic reverse battery protection. <S> The positive battery contact was recessed slightly into the plastic case. <S> Therefore the flat negative terminal of a battery cannot make contact with it. <S> But the positive battery pole protrudes a little bit, just enough to make contact. <S> Have a look in some devices (like a remote control) which work on AA or AAA cells and you'll see what I mean. <A> The mosfet that you have shown is valid ,but the gate volts available is of course only the battery volts .Finding a mosfet that will give low on resistance at say 1 Volt will be hard,if not impossible. <S> Remember that the gate source voltage must be significantly above the threshold volts <S> .There <S> are devices that I have not used that have a gate source threshold of 600mV .They <S> may not give low enough on resistance for you on a single alkaline cell. <S> There are osc circuits that could be rectified to make a realistic gate voltage .I <S> have used a modified dynatron that starts on 700mV and can run down to 300mV on a TEG application. <S> There are other osc circuits like joule thief and the blocking osc that I have not tried but could work .At <S> start up the mosfet <S> has no gate volts so the body diode could waste <S> say <S> 600mV <S> .Placing a resistor across the DS of the revpol mosfet will assure low voltage start up .The <S> resistor value is low enough to provide good starting and high enough to keep revpol prospective currents safe . <A> You could mimic the functionality using analog /discrete components but my guess is that it is not going to be worth it. <S> First off, you may end up with lower efficiency overall. <S> If this is not a production design you could consider using a third party board like this one from SF .
| An alternative solution I can think of: simulate this circuit – Schematic created using CircuitLab For an evel lower diode voltage drop, use a Schottky diode but beware that these leak reverse current at higher temperatures.
|
How to know if a multimeter is defective? One of our service engineers asked me(or tested me) how to know if an analog multimeter is defective without measuring any current, voltage or resistance. Just by looking at the device itself. Do you have any thoughts about this? <Q> I'm not aware of any way to tell that a multimeter (digital or analog) is defective just by looking at it - other than obvious physical damage such as a broken case or obviously overheated connectors or other signs of overheating. <A> You can detect certain faults by looking at the needle of an analog meter. <S> The front spring if visible should unbroken and appear to be in good condition. <S> The needle should also not move much if the meter is moved gently from side to side and should return to 0 when you stop moving it. <A> You can set the meter to measure resistance, short the leads together and use the ohms adjustment to adjust the reading to full scale. <S> If the needle goes towards full scale you can tell that the meter movement is not damaged. <S> If the needle moves towards zero ohms <S> but you can't get it to zero ohms (full-scale for the V & mA) <S> the battery is likely about dead. <S> That is a little more than just looking at the device itself, but it doesn't require anything but the meter. <S> I can not remember the last time I used the ohms scale on the meter below, so the battery is nearly dead. <S> You should also zero the meter (infinity for ohms) without the leads shorted. <S> The zero screw should be near the center of the movement. <A> Meter movements are physically a bit fragile. <S> If the meter cannot be zero'd <S> (usually there is a screw located at the center of rotation of the pointer which adjusts the hairspring rotation) then it is damaged. <S> It's also possible to damage the meter bearings (aka jewels) if it is an older type with such bearing or the suspension (usually taut band - metal ribbons- in modern movements) or bend the pointer by physical or sometimes even electrical abuse. <S> If you move the meter around and the pointer moves but does not return to the original position (hysteresis) then it is sticky for some reason. <S> Sometimes these things can be repaired- for example if the hairspring turns are touching they can be carefully adjusted with electronic-grade tweezers). <S> So you can see certain common faults. <S> Of course there are many things that could be wrong with the meter that you would not be able to see visually from the outside. <S> Incidentally, there is a compromise in analog meters between sensitivity and ruggedness. <S> Ideally, for ruggedness, a meter movement would be some mA or more full scale, but for an analog multimeter (without internal amplifier such as an 'FET' type) <S> you want the meter to be as sensitive as possible. <S> A typical post-WWII meter was perhaps 20,000\$\Omega\$/V (you might recognize the units as the reciprocal of current) <S> so the full-scale meter movement current was 50\$\mu\$A.
| It should be on 0 and respond to the adjustment screw and should be centered with respect to the dial.
|
Using 11.1v for a 9.6v motor I have a DC motor that uses 9.6 volt batteries. Can I use 3 of my 3.7 volt lithium-ion polymer batteries (11.1 V) to power this motor? Somebody please help me. I have tried 2 of the LiPo (7.4v) and it could not give me the expected result. I am currently scared of burning my DC motor. I do not have access to 9.6v batteries. Also if there would be a possible circuit that can step down the voltage please do help me with it. Thanks a lot. <Q> Just put 2 silicon diodes in series between the 11.1V battery and the motor. <S> You will have 11.1 - 1.4 = 9.7V which should be fine. <S> Just pick 2 diodes that have enough current capacity for your motor. <A> i am currently scared of burning my dc motor. <S> most likely an non-issue. <S> motors aren't that sensitive to supply voltage. <S> if you want to be safe, read the motor's datasheet if you can find one. <S> also, dropping some diodes /leds, appropriately sized, could help as well. <A> As mentioned in Majenko's comment, a switch-mode buck regulator would work ok. <S> So would the diodes mentioned in previous answers; each ordinary silicon rectifier will drop around 0.6 - 0.8 volts, with the voltage drop increasing as current goes up. <S> The diodes solution wastes the dropped volts as heat. <S> The buck regulator would produce some heat, but typically will be more than 90% efficient. <S> As suggested by another comment, many motors accept wide ranges of voltage and typically are robust enough to survive wide ranges of load. <S> But if the load is heavy, having higher voltage and current capacity may allow the motor to burn out. <S> A better approach than most of the above would be to use an electronic speed control, an ESC. <S> These ordinarily use PWM signals to keep motor speed constant in the face of varying loads, by changing the duty cycle of pulsed DC delivered to the motor. <S> At slow speeds, motor power and performance can be controlled much more smoothly using PWM than by controlling voltage.
| If your motor won't have too heavy a load, it may be ok with the higher voltage.
|
Specifications for LED strips - power supplies + installing in automobiles There never seems to be any talk of using a "LED driver" when using LED light strips in automotive applications. From my research it seems that LED strips have no voltage regulation and only a voltage dropping resistor. This mean that the 14.2 typical peek "full alternator charging voltage" could cause some damage to these. Why does no one seem concerned? Also I seem to read a lot of double talk about "LED drivers". Shouldn't these just be common DC power supplies that supply a constant voltage of 12V with an amperage capacity that meets or exceeds that defined as required by the LED strip? And thirdly, why is there not much talk then of the amperage required by a string of "sticky tape" LED strips? Is it because the industry only talks in "LED driver speak" rather than "volt and amp speak"? ****** UPDATE AFTER ANSWER 1 BELOW: I guess what I am asking, and thanks for this detail by the way, regarding the common "roll of LEDs" on a spool that are made and shipped from China that so many people are selling now, regarding these, what I have read is that they use single resistors for each LED in the strip so something to regulate the voltage would be needed. Now a) is this true, that most of them use individual resistors and b) isn't that really just a constant 12V power supply with an amp rating greater than or equal to that required by the sum of the strips and c) doesn't this mean that the 14.2V commonly reached when an alternator is charging an auto battery at full capacity, that the strip is over powered and will burn out more promptly? Thanks again. <Q> Too much current and they blow out, so you need to limit that current to the amount that gets the desired brightness level (and is below the maximum). <S> You'd think you could control that current by changing the voltage, but it is very hard because a very small change in voltage results in a huge change in current. <S> Imagine trying to control the current flowing though a wire by changing the voltage across it - it would be very hard. <S> So there are two common ways to control the current though the LEDs.... <S> Use a current driver . <S> This is a device that constantly adjusts the voltage to the exact level to generate a specified current. <S> Keep in mind that this voltage is not fixed <S> so it is not like a typical voltage power supply. <S> Put a resistor in series with the LEDs. <S> Because the voltage drop across a resistor changes with the current, this accomplishes sort of the same thing- as the current <S> though the LEDs rises the voltage dropped across the resistor also goes up which lowers the voltage across the LEDs with lowers the current. <S> These two effects balance out and settle on a limited current level. <S> The resistor is much simpler, but less accurate (brightness changes with input voltage and temperature) and you also waste power in the resistor. <S> For small amounts of current, people typically pick the resistor, but as you get to larger currents the resistors get big and hot, so a current driver usually makes more sense. <A> After much research, I found this inexpensive solution. <S> I have ordered the parts and am going to give it a shot using a heavy duty heat sink. <S> The device is rated for 5A (12v <S> x 5a = 60w <S> ) and should handle (theoretically) <S> any of the 5 meter LED strip model calling for 60w or less for the strip to be powered (see second link which gives tables of types of LEDs and wattage needed). <S> http://www.eleccircuit.com/12v-5a-power-supply-regulator-with-lm1084it-12/ LED strip power requirements in watts <S> However in the super bright 5050 link, it would not handle more dense than 30/meter, assuming 5 meter length. <S> For that, a more elaborate supply might be needed. <S> http://www.ledlightsworld.com/page.html?id=38r <A> "12V" led strips tend to use a resistor value that aligns with the typical 12 to 14.5 volts of a automotive power system. <S> At 12V, they typically consume 17mA for a 5050 or 3268 package led. <S> 20mA is the typical max ok voltage of a low power led diode, and much lower than its absolute max. <S> This changes sightly with other led packages. <A> You are right with your concerns, but there is no easy solution. <S> Most LED strips are engineered for 12V. <S> They have three LEDs in series with one resistor to do the current limiting for the group of three. <S> For example for 3 blue LEDs with 3.3V voltage drop each (9.9V total) <S> the resistor gets 2.1V. <S> If you design for 20mA you need a 105 Ohm resistor. <S> When you feed the same LED strip now <S> 13.8V <S> , the voltage over the resistor raises to 3.9V and with the same 105 Ohm resistor the resulting current is now 37mA, almost twice the current at 12V. <S> You buy those thing mostly in places with small margins. <S> Those places want easy or no specs, to be beginner and dummy friendly. <S> Most of the customer buy sort of a modular kit, where the components fit together, without much knowledge. <S> So you get what you pay for, much much specs for not much money... <A> Amazon sells dc to dc buck converters that are small, perhaps 1 inch by 1&1/2 inches in size, and will regulate voltage down via a pot to your specs to limit current. <S> There are 8 dc to dc buck converters rated at 3 amps for 12$. I would not install led 5050, or any led strips in any automotive, RV, boat, etc, without a converter in the supply wires and they are dirt cheap. <S> It’s your labor when the leds begin to yellow and burn out quickly because they are overdriven and at anything above 12 volts you are overdriving them. <S> In a normal automotive environment I agree with a previous poster who said run them below nominal voltage by approximately 10-15% I run all mine at a regulated 11.2 volts.
| LEDs need a minimum voltage to work, but once they get that minimum voltage they "turn on" and then the amount of current flowing though them controls how bright they are. So, for automotive use you should use LED strips designed for 13.8V or use a voltage regulator for the LED strip to limit the voltage to 12V. One word about the unclear specifications. At the typical high end of 14 volts of a car on and charging via an alternator, the strip takes 20 mA per segment.
|
5 V from 12 V at 650 W? I have 220V AC to 12V DC with 130Amps output. Could I convert this 12V DC to 5V DC where 130Amps not change? <Q> Sure you can. <S> There are several options: <S> Get a resistor and a giant heat sink, if your load is constant. <S> Realise option 1 and 2 are more expensive and inferior to buying an off-to-shelf 230V to 5V power supply <S> Try to mess with your current power supplys regulator to get 5V out of it, and in the process get a nice feel <S> what 230V through your body feels like (hint: it's not great) <A> Yes, you can convert 12 V at up to 130 A to 5 V at 130 A. <S> But, at such a high power level (650 W), that should be a last resort at best. <S> You already have 220 V AC available to run the 12 V supply. <S> Get a 5 V supply that can also run from that 220 V AC. <S> Every time you convert electrical power from one voltage x current combination to another voltage x current combination, there is some loss due to the process not being 100% efficient. <S> At 650 W, even a 93% efficient converter will lose 49 W. <S> That's a lot of heat to get rid of. <S> The 12 V power supply certainly loses some power too, but dealing with the heat is already built into it. <S> Likewise, a 5 V power supply would also have this heat handling built in. <S> Converting from 220 V to 12 V, then converting again from 12 V to 5 V is going to waste more power, which means more heat to get rid of. <S> Also, while a 12 V to 5 V DC converter capable of 650 W is possible to make, this is not a beginner project. <S> For anyone that has to ask here, the answer is really "No, you can't". <S> Even if you did eventually get something to work that isn't dangerous to users and that doesn't regularly blow up parts, the result will have cost significantly more than a off the shelf power supply. <S> The relatively high cost of a 220 V AC to 5 V DC 650 W supply should tell you that this is not a simple problem. <S> Even finding such supply won't be easy. <S> One possibility is to look around where electro-plating equipment is sold. <S> Electro-plating requires such low voltages at high currents. <A> 130 amps!? <S> What are you working with? <S> That's industrial type of requirements. <S> At a 100℅ efficiency, Power out = <S> Power in. <S> If you step down the voltage by means of a very efficient DC DC converter, then the current has to go up, that's just the physical properties. <S> Now you could set up a current limiter so that the current doesn't exceed a specific max. <S> Also, your load is what determines how much current it really needs. <S> So even if a source can supply x amount of current, it doesn't mean it always pushes the max current to all loads.
| Get a Switiching regulator from 12V to 5V, at 650W, that's still going to need a heatsink.
|
If many EM signals exist in same time/place, when would interference occur? If electromagnetic waves can experience interference from same or different frequency transmitters in surrounding region; Then why do we not experience interference when many people use their mobile phones at once in the same place? This is the part I don't understand. Is it to do with phase? EDIT - Thank you everyone for your answer. I did read on Frequency Hopping, but my issue is in trying to "envision" what happens. I attached a pic of what's in my mind. If 2 Transmitters transmit waves... they "may" (and certainly will in far field) cross each other. Do the red circles not pose interference even if amplitude or frequency is different via spread spectrum? Or I'm missing something? Can the 2 waves ever end up being in Superposition even if they hopped frequency? <Q> If they were all on the same frequency, at the same time, and using the same modulation scheme, then yes, there would be interference. <S> But cellphones (and other services) use methods like multiple frequency channels, timeslots and "Code Division Multiplexing", also known as "Spread Spectrum". <S> These techniques allow many transmitters to share the same frequency band. <S> In general terms, phase has little to do with it. <A> Older cellphones used a band for the basestations to transmit out to the customers, and a 2nd band was used by all the customers transmitting back to the basestations. <S> These bands, in the 900MHz region, were 20 or 30 Megahertz wide, supporting hundreds of 30KHz narrow-band FM cell signals. <S> The basestations used expensive band-pass filters to keep their 50watt outgoing signal power OUT of the incoming millipico (femto) <S> watt basestation receiver circuits. <S> The receiver signal floor is computed at -174dBm + 10*log10(30KHz), or -174 <S> +45 = -129 dBm; with 0dBm being 1milliWatt into 50 Ohms or 0.632 vPP, the -129dBm floor is 0.632 / <S> [120dB + <S> 10dB -1dB]= <S> 0.632 / <S> [10^6 <S> * (3.16/1.12)] = 0.25microVoltsPP. <S> The 50 watt outgoing power, needed for the users the most distance away from the basestation, would easily INTERFER and probably destroy the receiver circuits located only a few feet away inside those little houses at the ground level, unless the bandpass filters were used. <A> The waves do not affect each other. <S> Then the radio receivers that use filtering might not be a able to keep them separate. <S> Different signals can interfere in electronic circuits also via saturation and <S> by that caused mixing (=radiotechnical mixer). <S> The signals can be in different frequencies, but one of them can by the sheer power clog the circuits <S> and that is one form of mixing. <S> The mobile phones do not transmit wildly all the time. <S> The base station conducts the orchestra and never two phones send simultaneously at the same frequency until they're in the operating circles of different base stations and enoughly far away. <S> The phones are designed to tolerate other phones nearby without any saturation. <S> That's not a design criteria for some cheap audio stuff <S> and that's you surely have noticed.
| Some electronic circuits that receive several different signals at the same time can suffer if the signals are on the same frequency or on too near frequencies.
|
Should SMD footprints be rounded? I was looking at the footprints in Altium's Atmel library, and I noticed that many (most?) of the pads had rounded rectangles. However, if you use Altium's own "IPC compliant footprint generator", by default the footprints are rectangular (not rounded). Is there a specific reason to use one of these over the other? It would seem that rounded pads would be easier to manufacture, and would make a more natural shape during reflow, but that's just complete speculation on my part. (On a related note, would rounded pads have to be made slightly larger than strictly rectangular pads?) <Q> Yes, SMD footprints should have rounded corners as per IPC-7351A Corner radius is 25% of the shorter side of the pad but not more than 0.25mm (10mil which is not exactly the same but close enough here) <S> Why?The corners do not add anything useful (no additional adhesion, no additional stability or conductivity). <S> But on reflow soldering the solder does not always flow into every corner possibly leaving copper exposed. <S> Additionally: it's better to have stencils with rounded corners. <S> Addition: <S> no, pads with reasonably rounded corners do not have to be bigger because the corners didn't add anything useful to begin with. <A> per IPC-7351B standard: <S> An exception to this rule occurs when the heel portion of the land pattern has to be trimmed due to the component body standoff being less than the paste mask stencil thickness or the heel having to be trimmed due to “Thermal Pad” interference. <S> In these two cases, the rectangular pad shape is preferred to compensate for the reduction in copper area of the land pattern pad length. <A> You can use rectangle pads and have rounded corners on your paste stencil, there can be issues with Gerber sizes due to pads being drawn not flashed. <S> Most use std. <S> rectangular pads. <S> Bradman175, you are <S> incorrect 90 degree corners have no real effect untill the signals reach the GHz...
| The only reason for pads with edges was that some tools did not support rounded edges. Also, the usage of oblong, or rounded, land pattern pads is considered advantageous for lead free soldering processes in comparison with rectangular pads, as the oblong shape provides for a pull of the solder on the pad.
|
What communication protocol does this pinout match? DATAP, DATAN, CLKP, CLKN, CLKSEL I am trying to build a laser diode driver circuit using a part from Analog Devices. (The datasheet is here for your reference: http://www.autex.spb.su/ad/ADN2848.pdf ) It appears that there are 4 data inputs: DATAP, DATAN, CLKP, and CLKN. There also appears to be a clock select (CLKSEL) pin. I am relatively new to this sort of thing, so I'm not exactly sure how to interpret the data sheet to figure out what communication protocol to use when inputting data. Does anyone know what the protocol is, or have any thoughts/advice if not? I also notice that the data sheet gives two values relevant to the data inputs: t setup and t hold. Are these relevant to the communication protocol that needs to be used? Thank you! <Q> There are differential inputs (CLKP is the positive phase, CLKN is the negative phase). <S> Differential signals are often used in high-speed circuits, for improved noise immunity and reduced signal swing. <S> The setup and hold values are the timing requirements between clock and data. <S> This has nothing to do with protocol, just when the data needs to be stable around the active clock edge. <S> This driver is totally protocol-independent. <S> It merely turns a serial bitstream into drive levels for the laser. <S> Protocol happens way upstream. <S> I suspect you've got some serious studying ahead before you can get this to work... <A> It looks like it's just a flip-flop with differential inputs. <S> Each signal is considered high when 'P' > 'N' and low otherwise. <S> The device updates 'data' on the rising edge of the clock. <S> The setup time is the minimum time period that data must be stable before that rising edge, and the hold time is the minimum time it must be kept stable after the edge. <S> CLKSEL allows you to bypass this. <S> If CLKSEL is high, clocking is used. <S> If it's low, it looks like data just goes directly to the diode. <A> The fact that the AC coupling is required means that you will need to use some line coding in all probability, 8b10 or such to avoid large DC shifts, do not do what SDI for example does and end up with pathologicals with long sequences of 1 or 0 <S> , it causes DC shifts in both the line RX and the ALC loop as well as making clock recovery a pain. <S> Apart from that the thing is completely protocol independent, it is up to you what bit patterns you send it. <A> I would usually expect to drive this type of device from an LVDS (Low Voltage Differential Signaling) output from a device such as an FPGA or ASIC for SONET etc Overview of LVDS .
| Well the part is clearly set up for AC coupled LVDS or maybe PECL logic, and obviously the line rate will have to satisfy the setup and hold times relative to the clock edge if the clocked mode is in use.
|
Soldering a SMD sensor - options without etching PCB? I would like to break out this sensor . I don't have the equipment to design and etch my own PCB. Do most sensors come in standard dimensions, such that a pre-made break out PCB could be obtained? How would I go about finding such a break out PCB? The dimensions are listed in the datasheet (below), as well as a notation of 5-SMD, but I haven't been able to find an appropriate board. Am I approaching this wrong? It has no leads, but I assume the solder will flow under the pads just fine. Thanks. Update: I ended up doing both - the spacings were compatible with standard protoboard so initially I soldered to that. But ultimately I learned to use Eagle and got a board fabricated for about 20 bucks. Definitely recommended. <Q> Given the 0.1" multiple spacing, you could quite simply solder it to some veroboard/busboard (stripboard). <S> To do this you could place it against the solder side of the board, making sure to cut gaps in the strips below the IC, and then solder to the strips. <S> This shouldn't be too difficult as from the images in the datasheet <S> it appears that the pads extend up the sides of the IC. <S> Alternatively, you could place it on the component side of the board. <S> For this you would need to solder short lengths of wire between the pads on the IC and through the holes in the veroboard. <S> As a side note it is worth taking care in how you place the sensor. <S> Given that it is an accelerometer, placing it as square and parallel with the surface as possible is advised as it will make it easier to work out how X/Y/Z relate to the board. <A> Or, this sensor might sit on a pre-etched perfboard having 0.1" solder pads. <S> Solder wires to the pads after you solder the sensor to the pads. <A> Here is an example of the "cut islands in all-copper PCB and solder on" method (it's not beautiful.) <S> This one also has a recommended capacitor installed, so the "ground" island makes an L under the device to permit the capacitor (small brown item on bottom edge) to be installed as near the device as practical. <A> This sensor is a DFN (dual flat no-lead). <S> Pads are on bottom, and it seems there are also dimples on the sides that are tin-plated and connected to the pads. <S> Check out SparkFun or AdaFruit, they may have the breakout board with correct pad pitch. <S> The sensor has pads at 200mil apart <S> so 50-mil breakout board for SOIC chips should line up with pads quite well. <S> The challenge would be to find a proto board with correct spacing between rows. <S> If you don't have a prototyping board with pads at right spaces, the easiest is to 'dead-bug' the sensor. <S> The sensor is not a humidity sensor so it does not need fresh air and turning it upside down will only invert some motion axis. <S> Another thing you can try is to use a through-hole board, put 18-24 AWG solid wires like staples through the board in right places to match the pads, and then solder the sensor pads to these wires. <S> Wires will lift the chip up from the surface of the board so you can put solder under the chip with sharp soldering iron tip. <S> Surface tension will pull the solder between pads and wires, and the gap between pads is wide enough <S> so you don't have to worry about shorting anything. <S> If the dimples are plated, you can also use them to test continuity when soldering the wires. <A> Just glue the sensor upside down to something flat and solder thin wires to the pads. <S> I’ve done it even for more complex parts like this tiny bluetooth module (a PAN1326 with 0.6mm pads): <S> Unfortunately I don’t have a picture of the finished solder job <S> but you wouldn’t see much because it’s covered in hot glue.
| When I am desperate, I take an Xacto knife to a sheet of blank circuit board and cut as needed to give me a pattern where I can solder the device.
|
PCB Home Made Keep Inkjet Or Buy Laser My inkjet is about to run out of ink and I'm thinking of buying a laser printer to print the transparencies. General process: 2 prints of PCB cut aligned and glued to each other > Pr-Sensitised PCB exposure > development and etching. In general with the Inkjet I print several times on the same sheet and any offcut can then be attached to a paper A4 sheet for printing later on, minimal waste. My concern in buying a laser is the heat and the possible warping of the transparency especially running it through several times. Am I better of sticking with the inkjet due to the warping of the transparency that the laser can/could cause? I included a pic to show that all these three PCBs were done in multiple print runs with one A4 transparency sheet on an inkjet. I don't know but I don't think I can do the same with a Laser printer. Thanks all, but it seems lasers are to finicky. Right laser, sourcing and using the right transparency, then in future if I try to use the transfer method on non pre-sensitised boards it has to be the right toner for transfer. One mistake or fault and cost of repair could be a nightmare. For the cost of repair/parts of a laser I can buy a dozen cheap new multifunction ink-jets that will last me 5+ years. I may stick with the ink-jets, 100% success rate so far. It just occurred to me that with a laser and using the transfer method (translucent paper 'crowie') I could just use the laser to create a silk screen then iron it on the PCB. I may see if I can buy a cheap second hand laser. It would be nice to have a cct board with a silk screen of sorts. <Q> First, thanks to all replies. <S> Second, A tip I discovered. <S> I bought two second hand laser printers. <S> One, a multifunction for $40 that was <S> a never used spare in an accounting firm. <S> 8 years old but brand new and 90% toner in it (mono). <S> The second for $10 a colour laser with 65% toner in it. <S> Slight scratches on the far left of the drums but for printing A5 transperancies it's perfect. <S> Not two overlaped as with the inkjet. <S> I can also print a silk on transfer paper and then iron it on. <S> Minor draw back. <S> Can only run transperancy through once before they start to warp. <S> So planning is a must. <S> Cutting A4 into A5 risks only minimal. <S> Lasers are big and heavy. <S> But can't complain. <S> More than happy with the result. <S> Never buy another new printer again. <A> Old desktop pen plotters, such as the old HP 7475A, can work spectacularly well for this purpose, in conjunction with technical drawing pens, and some 3D printed adaptors. <S> Eagle CAD, being old, still has functions to export HPGL. <S> I've been able to go down to 0.18 mm trace width (~8 mil), and can succesfully do even small 0.5 mm pitch QFNs with relative ease. <S> The way I do it is by just plotting straight onto (freshly cleaned! <S> else the ink won't stick!) <S> copperclad board with ink, let it dry for a while, and etch. <S> The ink selection is somewhat critical, but cheap shellac-based drawing inks do the job rather well, such as the Koh-I-Noor ones. <S> If you have a 3D printer handy, using the heated bed as a sort of hot-plate to speed up the ink drying is a nice lifehack. <S> A minor downside is that the pens do require maintenance. <S> Another possible downside is that the plotting takes a few minutes on complex boards. <S> The upside is that the ink is cheap, the precision is pretty great, and there's no mucking around with exposure times and such. <A> You only need a black and white laser printer. <S> The cheapest laser printer costs about double that of the cheapest inkjet. <S> The cheapest inkjet cannot reliably work for printing masks, as sometimes there is a fine (0.05mm) gap between consecutive lines (i.e. solid areas are not solid but actually hatched, if looked under a microscope; and this passes UV). <S> Also, high quality expensive ink is required for inkjet, while lower quality cheaper ink gives better result with lasers. <S> Additionally, the laser ink will not dry up at the drop of a hat... <S> it is already dry to begin with! <S> Transparencies in a laser will not warp if the correct heat setting is used. <S> Any minor warp does not matter when pressed under a glass sheet. <S> I've got 0.1mm traces with not much effort. <S> Keep in mind that laser toner spreads about 0.05mm under the hot rollers, so clearance will increase by 0.05mm. <S> It may be possible to get 0.05mm traces if this spreading is accounted for, and the photoresist film is properly affixed and exposed. <S> Laser printed transparencies will have to be made darker by going over them with some fine pigment ink <S> (e.g. nano carbon ink from fluid ink refils, or nano carbon fountain pen ink), and then rubbing it off gently with dry soft tissue (2-3 rounds). <S> It easily becomes dark enough to block out the sun when seen with naked eyes. <S> For this, aftermarket lower quality inks may work better due to the lower wax content. <S> I use a slightly higher heat setting.
| Laser printing result on transperancy is better than Inkjet as it also only needs one print.
|
Adjusting LED currents same level for different voltage levels I have different voltage levels 1 V, 1.2 V, 1.8 V .. 5 V, 12 V and I want to connect a led each of them to see the voltages are clearly set. In this case, I want same brightness for each LED. I can do this by using simple resistor for each voltage level but low voltage sides (1 V, 1.2 V and 1.8 V) aren't efficient. So I thought maybe there is some voltage shifter that accepts different voltage levels as input, fixed voltage level for the output that I can use. Anyone know anything like that? or some other suggestions to handle this? Thank you <Q> Apparently you want to indicate that various voltages are present. <S> I'll assume the indication threshold doesn't need to be all that accurate, and that you mostly just want a quick way to see whether a power supply has come up at all. <S> I had a similar problem where I wanted to show quickly that the 24 V, 12 V, 5 V, 3.3 V, and 3.0 V supplies were up. <S> Here is a snippet of what I did: <S> The bottom rail is ground, but the ground symbol is cut off in this snippet of the schematic page. <S> The 24 V is used to derive all the remaining supplies from, so the LEDs are actually lit from the 24 V supply. <S> I used TL431s as voltage threshold detectors to turn on the LEDs. <S> There is a different resistor divider feeding the threshold input of each TL431, depending on the voltage of the particular supply being tested. <S> The resistors across the LEDs (R13 and R14 in the snippet shown) are to avoid the LEDs being dimly lit when they should be off. <S> A TL431 requires some current to operate, which would otherwise come thru the LED. <S> This method works for supplies down to 2.5 V, since that's the threshold built into the TL431. <S> It also requires a high enough voltage to run the LEDs from. <S> In this case I used green LEDs with a forward drop of about 2.1 V. For your supplies below 2.5 V, you can use a NPN transistor instead of <S> a TL431. <S> You can either just use a single base resistor to show that the supply has come to about 600 mV, or a resistor divider to raise the effective threshold. <S> In this case, you can get rid of the resistor across the LED since a bare transistor doesn't draw operating current when off. <S> The threshold for lighting the LED won't be as crisp and accurate as with a TL431, but may be good enough for your purposes. <S> Basically, you'd be using the B-E junction drop of the transistor as a voltage reference. <A> See simple bar graph display driver for what you want. <S> You can also go more fancy and use a microprocessor with an analog to digital converter. <A> Then you connect the LEDs all to the same voltage rail. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You could use low threshold devices for the the small voltages (<2V) and for the rails higher than 5V be sure to choose a transistor that is rated for the right gate voltage. <A> The smaller voltages would make it more difficult to light an LED with the same intensity using the method you talked about. <S> I'd probably use the different output voltages turn on N CH MOSFETS with very low Threshold voltages. <S> Vdd should be 12 volts. <S> Adjust your resistor values between the Cathode of LED and the Drain of MOSFET.
| If you just want the LEDs to show whether the voltage is present or not, you could use a mosfet to switch the LEDs. The resistor across the LED passes that current but at a voltage so low that the LED won't light visibly.
|
AC current sensing using micro-controller and suitable circuit I'm working on the project in which I need to measure AC current. The main problem I am facing is small current amplitude. I have to obtain the current waveform(I need complete waveform not just RMS value) of amplitude ranging from 100\$\mu\$A to 600mA and frequency ranging from 500Hz to 10KHz. But, I can't afford to have voltage drop more than 4V across sensing resistor. Do I have to use relays and set of different sensing resistors? Any other better alternatives to relays? <Q> If you have two ADC channels avalable you could run a scheme like this: simulate this circuit – Schematic created using CircuitLab <S> This avoids switching and allows you to have options on your dynamic range. <S> If you only have one ADC channel <S> you can get an analog mux and use it to select between the low and high gain and that way you don't have to do anything to your sensitive current signal. <A> Basically you have a 6000:1 difference, which means you need a clean 13 bits of ADC, which is doable but not completely trivial. <S> You'd need to keep the noise right down, because you'll have to have no more than a 6.7ohm resistor, which is only giving you 670uV at your low current end. <A> I'd suggest there is a ready made solution to your measurement problem front end. <S> Made for the hobby/professional tools market the uCurrent can measure down to pA. <S> The design uses a MAX4239 bridge/instrumentation amplifier, and I think could be easily modified or the design adapted to meet your needs. <S> There was a complete design description of the first version here . <S> Redesigning the burden resistors would appear to be a simple task with the only real problem being that of achieving the lowest contact resistance for your burden switches. <S> Getting higher output voltage could be achieved by raising the supply voltage. <S> If you want to auto range your input <S> I'd suggest that you should consider Au/Ag or Mercury wetted reed relays which have about the lowest contact resistance you can get. <S> Though I'd suggest you include one set of contacts that shorts out all your burden resistors while you select a range.
| Personally I think I'd switch a different shunt resistor in at the low end.
|
Can the capacity of a battery be determined from Volts, Amperes, and output of Joules for xyz amount of time? I'm trying to purchase a portable car battery charger (Noco Genius Boost) and I can't tell if it's possible to calculate the capacity in Ampere-hours or Watt-hours. The reason why I ask is because there is no indication of Ampere-hours or Watt-hours, or anything related to time. The product information in the details show: 12V, 1000A, and output of 7000 Joules per 3 seconds. Is the capacity possible to determine and if so, how did you calculate the math? <Q> Normally the capacity of battery is specified by "Watt-Hour" written on the battery or in the battery datasheet. <S> If this isn't specified, "Amp-hour" will be written on it. <S> You can calculate watt-hour using this: Watt-hour = <S> Voltage <S> x <S> Ampere hour (specified on the battery) <S> E.g Battery voltage = 12v , Amp-hour=1000, then watt-hour = 12 x 1000 => 12000 watts-hour. <S> what this means is that if the battery supply 12000 watts to your car/circuit, it will work for 1 hour (ideally though).1 watt = 1 Joule/sec; so 7000joules/3sec => 2334 watts. <S> It means that it can supply ~2.3KW to the load. <S> Now if you want to calculate the operating time of battery for a specific load, say, 200watt load, watt-hour/watt(load) = <S> > <S> 12000/200 <S> =>60 hours (theoretically) <A> Without knowing if that 3 second rating fully discharges the battery or if that is just the limit of the test rating, there's no guarantees on any calculations. <S> Your best bet is to Google high C rating lithium ion batteries with similar physical dimensions and weight to see what it's closest to. <A> No. <S> It is not possible to determine the capacity from the information given.
| Weigh the battery and compare the weight with batteries of similar weight and cost to get an estimate of the capacity.
|
Electromagnetic EM waves - How they coexist in same space and time without signal impact? I'm having difficulty understanding how 2 EM waves from 2 sources can co exist in same space and time, without changing one another from a Quantum Electrodynamics Perspective. If a charged source emits photons which excite electrons in Field Region X.... then how can another wave from a different source not impact the first EM wave due to an increase in Electron Excitation on ((same)) electrons in same Field Region X in same time. I didn't think different electrons were excited by different wave sources... otherwise, how can Electron State = Wave1 .... when another Force from Wave2 is exerted on it? <Q> The electrons will be excited by a superposition (sum) of the two EM waves. <S> There is no differentiation between waves from different sources. <S> Occasionally it is possible to run in to nonlinearities that depend on the overall amplitude of the EM waves in question, and in this case all of the waves would be affected. <A> You assume some electrons and also two waves in region X and wondered: "no interaction"? <S> The interaction can result to something that quite soon collapses returning the original waves that go on, only some extra time have been spent in X Other possiblity is an irreversible interaction. <S> Try to put a cellphone near some audio system,take a call and listen. <S> You very likely can hear the consequences of the irreversible interaction. <A> The carrier of an EM wave is a photon , <S> photons don't (most of the time) interact with each other. <S> If you delve into this question further its like asking why do triangles angles add up to 180 (in a normal geometric sense). <S> Or why did the universe come to be. <S> The answer is, they just are. <S> If you study the math behind it <S> you can find out what constitutes a wave and how waves pass right through each other unimpeded.
| Photons interact with electrons and other 'stuff' but can pass right through each other. The waves will pass through each other much like waves on a pond.
|
What's the best way to reduce the wattage on an audio amp output? I am building a simple audio circuit that will only be used for headphones. It is starting with a very low signal going into a passive volume and tone control board. The output of the volume/tone board goes into a small amp that has a 3.5 watt output. When the vol/tone board is turned to max it is too loud for headphones. I want to limit that 3.5 watts to maybe 3/4-1 watt so when the volume is turned to max it doesn't blow the headphones or my ears. What is the best and most reliable way to reduce the output of this small amp? Thank you <Q> I would recommend configuring the amplifier for a lower closed-loop gain. <S> The amplifier board you linked to is built around <S> a Diodes Inc. PAM8403. <S> You can pull the datasheet for that part to learn how to configure it's gain. <S> Unfortunately, the datasheet is not very clearly written. <S> It gives the closed-loop gain of the amplifier as: AVD = <S> 20*log [2*(RF/RI)] <S> However, "RF" is not identified on any schematic. <S> Based on the verbage, though, I think it is safe to assume that RF is the feedback resistor inside the IC, with a fixed value of 142kohm. <S> "RI", in the above equation, is the sum of the internal input resistor, which has a value of 18kohm, and any additional external resistor in series with the input to the chip (labeled RI on the front page application schematic). <S> What this means for you is that you probably want to add additional resistance in series with the input to the amplifier board. <S> If there is already an external RI on the little board that you have, then try replacing it with a larger value until you are satisfied with the volume range. <S> If there is no footprint for it on the PCB, you will have to find a way to kludge the resistor into the path - maybe splice it into the connection between the tone board and the amplifier board. <S> It is worth noting that your tone board is passive and, thus, has significant output impedance itself. <S> In this way, it is possible that the tone control will interact with the gain of the amplifier and vice-versa, altering the control law of the volume and/or tone potentiometers. <S> I won't offer a complete explanation here, but this is something to keep in mind if experimentation yields confusing results. <A> There are two ways to do this: First is to limit the input signal, perhaps by putting a resistive attenuator before the amplifier input. <S> For example, you can insert a series resistor at the input side of the volume control. <S> This is similar to not allowing the volume control to be turned up to full volume. <S> Using an attenuator at the amplifier input will increase your "noise floor", but probably not enough to be noticeable. <S> A series resistor will work, but then the attenuation will be largely dependent on the headphone impedance. <S> Many headphones are around 30 Ohms or so, but others may have an impedance of 8 Ohms. <S> You can reduce the headphone-impedance dependency by using a voltage divider (an "L network" attenuator). <S> Depending on the attenuator design you may have to use power resistors to dissipate the "thrown away" power. <S> The disadvantage with the output attenuator is that you will be wasting power, but this probably won't matter in your case. <S> A resistive output attenuator may also affect the frequency response of the headphones, probably in the low-bass region. <S> Again, this may not be enough to matter. <S> Or, you can use both techniques together. <S> I would probably choose the output attenuator if all other factors were equal. <S> [edit]: <S> I just looked at impedance ranges for modern headphones. <S> iPod-style phones are typically around 50 Ohms. <S> Others run up to 600 Ohms, and some run well under 50 Ohms. <S> With any of these headphones, the amplifier output power will be way under 2.5W, but this isn't the important factor. <S> This does mean that a output reduction that a resistive output attenuator will give you is headphone-dependent. <S> I would probably use a series resistor from the amp output (selected for the desired attenuation, but 220 Ohms will give you about a 20dB reduction), feeding a 47 Ohm resistor to ground, and connecting the headphone across the 47 Ohm resistor. <S> You can probably use 1/4W resistors. <A> Perhaps, you are using a turn-pot to adjust your headphone volume. <S> So, it is a simple voltage divider circuit at the opamp feedback which determines your gain. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Refer to the schematic above where the R2 (1K) is placed to restrict the gain to 10, in this circuit. <S> The calculations have to be done by you though, the values in this schematic are not to scale. <S> By the way our ears are logarithmic and thus, for audio applications instead of using a simple linear potentiometer, we use logarthimic one. <S> Do account for the same as well. <S> Your objective should be to place a cap on the maximum gain attainable by the circuit! <S> Refer to the logarthimic amplifiers and write down their equations to set the maximum gain. <S> You can make a logarthimic amplifier by using a transistor and a resistor. <S> If you can upload your schematic, it would be easier for us to debug it. <S> Happy prototyping!!
| The other way is to use a resistive attenuator at the amplifier output, before the headphone jack. I would suggest that you restrict your Gain using a fixed resistor such that you always have a set maximum gain which can't be exceeded!
|
Reverse piezoelectric effect in crystal We know what is piezoelectric effect and its operation.Whereas Reverse piezoelectric effect is just opposite that of the piezoelectric effect i.e, Electrical to mechanical and we get oscillations in the crystal. As i know the principle, now I'm eager to know its operation. If someone have designed the crystal using this, please share the same. <Q> No need to make inventions. <S> Piezoelectric speakers are still widely used as tweeters. <S> Piezoelectric (=crystal) earphones were popular in the past due their high sensitivity. <S> Ultrasonic devices use piezoelectric transmitters. <S> The high pitched buzzers have most often a piezoelectric sounder. <S> Nearly all clocks, computers and modern radio equipment have crystal oscillators. <S> They have a quarz crystal that mechanically vibrates due the piezoelectricity <S> Addendum: <S> The crystal is connected to an output of an amplifier or other circuit that produces the wanted AC voltage. <S> The oscillator is a special case. <S> The crystall is the feedback route between the input and output of an amplifier. <S> This makes the circuit to oscillate just like an audio PA system squeals when the microphone is too near the speaker. <S> The crystal makes the frequency stable and well defined. <S> The presented devices create mechanical vibrations fron the voltage changes. <S> The reverse of this is used in the following devices: <S> oscillator crystals (they act in both directions at the same time) crystal microphone ultrasonic receiver crystal pickup for a player for vinyle records acceleration sensor pressure sensor knock detector <A> If your looking for reverse piezo electric, look no further than your grill. <S> Piezoelectric spark generators use the reverse effect. <S> A mechanical shock from a spring releases energy to generate thousands of volts to generate sparks. <S> There are also microphones <S> that use this effect. <A> The motor produces waves running around on multiple crystals that rotate the object that is placed on top of them. <S> Youtube: <S> https://www.youtube.com/watch?v=uFZsH62ewYo Max
| A cool use of the (reverse) piezoelectric effect (electrical -> mechanical) is a piezoelectric motor.
|
Solving for steady state voltages (Op-Amp) simulate this circuit – Schematic created using CircuitLab so I have this Op-Amp problem. I need to find for the values of V1, V2 and Vout for t > 0. I have this solution but I'm not sure if I got it right. If not, what part did I have an error? It's not in the picture but the continuity variables V1 and V2 both have a value of 1 volt. Also, I don't know how to interchange the polarities but the positive terminal of the Op-Amp should be on top. <Q> No need for all that maths.... <S> A TL081 does not have much output drive and with only 10 ohms to ground the thing is going to struggle to move the output anywhere <S> , I would guess the thing will sit with the output pretty much at ground with the output stage hard in current limiting no matter what you do to the inputs. <S> Note also that your two schematics are NOT the same <S> , the one on paper has R2/R4 joined at Va, the other one does not, and the inverting/non inverting inputs are swapped between the two if you were to assume an ideal opamp instead of the 081 it would make a difference. <S> R4 can be replaced with a short circuit without changing anything, and Vout = <S> Vb <S> * (1 + R7/R6) <S> = <S> 2 <S> * Vb. <A> Your input 4exp(-t) diminishes to zero as the time goes on. <S> That's not possible, because the capacitors get finally discharged through the resistors. <S> ERRORS: <S> Seemingly you have not understood that V1 and V2 are the given initial voltages of the capacitors. <S> The nodal equations should be written with Va and Vb; the rightmost capacitor should have their difference as its voltage. <S> Vout should be written as amplified Vb. <S> You should solve the equations by using the Laplace transform - it has the places for the initial values as soon as you apply it to the derivatives. <S> You will get Va and Vb at first as their Laplace transforms, but the inverse should be easy. <A> Solving for steady state voltages <S> (Op-Amp) <S> I need to find for the values of V1, V2 and Vout for t > 0. <S> those are two different questions. <S> for the steady state solution <S> , Vout = 0 for ideal opamp; for the steady state solution, Vout needs to be measured empirically with a real opamp.
| The voltages of your capacitors V1 and V2 have in your final equations a limit 0,14V.
|
How much amperage can pass through a current limiter I just completed an FAA required recurrent course for the aircraft I fly. The Power Point program & instructor teach that a wire with 75 amp rated current limiters at each end will provide 150 amps of load before one or both current limiters melts and opens the circuit. The use of the current limiters is a modification to replace a single 450 amp rated reverse current circuit breaker. Instead of one line with a 450 amp RCCB between the main battery and the main bus, three parallel lines are used with current limiters at each end of each line. The advantage of the three parallel lines is to not lose all power to the bus if only one or two of the three parallel lines opens. The explanation given is that total resistance in a series circuit is the sum of the individual resistances; therefore, six 75 amp current limiters (one at each end of the three parallel lines) will allow the same sustained current as the single 450 amp reverse current circuit breaker. I disagree and I've read that each current limiter is rated at 150 amps and each of the parallel lines will allow a sustained current of 150 amps. By analogy, if a line has current limiters at each end, one rated at 75 amps and one rated at 50 amps, the circuit will open with a sustained current in excess of 50 amps, not a total of 125 amps. <Q> I am not familiar with aeronautical current limiters, but it sounds like they are similar to fuses or circuit breakers. <S> Regardless of the details, any series connection through two current-limiting devices will limit at the lower of the two ratings. <S> The current-carrying capabilities will not add. <S> However, I just looked at this document ( FAA - Aircraft Electrical Systems ), and on page 9-59 they show a "main battery bus" and an "isolation bus", connected in series, with a current limiter at each end. <S> These two busses are connected together and fed at the junction. <S> This may look like a single series connection, but in this case the current feeding the junction can equal the sum of both the current limiter ratings. <S> I would make sure that you and your instructor agree on what kind of connection you are looking at. <A> I agree with you. <S> I'm not sure what these 'current limiters' are (fuses, resistors, some sort of active solid state device, etc.), <S> but if they are in series on the same physical wire, adding up their ratings will not increase the total current limit. <S> The effective current limit would be the lowest of them. <S> Unless they are resistive, in which case the current limit would actually drop. <S> E.g. for a 24VDC bus, a 1 Ohm resistor would limit the current to 24A max. <S> Putting two of them in series would limit the current to 12A. <S> I assume though these are more like fuses or breakers, in which case again, the effective limit would be the lower of them. <S> Of course if placed in parallel, then the limit would increase, but it's hard to predict the limit because they may not share the load evenly. <S> So in the case where three 150A circuits are expected to provide 450, well OK up until one of them trips (which might be at less than 450A if they do not share the current evenly), and as soon as that one trips the others will soon follow. <S> (What equipment is that? <S> 450A is a lot more than what I would see in the small planes I fly!) <A> Current limiters, fuses and PTC or resetable fuses usually have 2 current specs: rated current and tripping current. <S> Rated current is at which the limiter will operated normally without tripping. <S> The tripping current is the current at which the manufactures specs the limiter, fuse or PTC will trip in an specified amount of time. <S> So in your case 75A is the rated current for the limiter and the 150A is the tripping current. <S> Having redundancy in aeronautical system is common especially in their electrical system.
| So they are correct to say that the system will provide 450A before tripping and there's most likely 2 limiter on each end just as redundancy.
|
The meaning of Burst Mode Many times during reading about Power Supplies, I came to this word and couldn't find it's meaning? What does generally burst mode mean? What is the advantage of a power supply if it has burst mode. To be more specific in Buck converter. <Q> Normally with a buck regulator it continually switches a transistor on and off at a certain duty cycle to feed energy to the load. <S> When the load draws high current <S> the energy transferred is high and this results in a high duty cycle i.e. the transistor is on more than it is off. <S> When the load is very light and the input voltage is at the high end of its working range, you find that a very small duty cycle (less than 1%) is difficult to produce and so some regulators switch off the main process of continuous duty cycle and go "idle" for a while. <S> During this period of idle, the output voltage drops to a certain point and this triggers to process to restart until the voltage reaches a higher level. <S> Then the regulator goes idle again for a while. <S> This idle period is very much dependent on load current - if there is very little load current the idle period can be several milliseconds or more. <S> This is burst mode. <A> The word meaning of "burst mode" is equal to the meaning of "pulse skipping mode". <S> Typically, pulse is skipped from PWM control to maintain regulation at very light loads (i.e. 100uA ). <S> During the normal operation of DC-DC PWM, especially if you are in current mode control, you would use pulse skipping at a light load to improve the light load efficiency. <S> At a light load, the conduction losses, switching losses, and Iq losses will become significant. <S> Skipping pulses will therefor reduce switching losses. <S> In general, a DC-DC converter (buck or boost) will have two methods of operation: PWM or Pulse Skipping. <S> To pick between the two different modes, I would look at the efficiency curve in the datasheet and the load that you have. <S> Take a look at the graph below to get an idea of what you would need to look for. <A> Switchmode power supplies are generally efficient .Buck <S> converters are generally more efficient than isolated types <S> .These days the turn down ratio is often high .For example 12V in and 3V3 out .This <S> means that at full load a linear reg would be very lossey .However <S> at very light load the switchmode will draw more current than the linear reg <S> .This is because the fixed frequency buck converter has some fixed losses that in % terms are low at full load but dominate at very light load <S> are other terms like "Green mode" or" Ball cock " .There <S> are other ways of reducing standby power like frequency reduction and pulse skipping .
| .Burst mode is a term describing a scheme where the the converter is switched on and off during light load in order to reduce the standby power .There
|
Driving a CMOS 1.8V input with a CMOS 5V output I'm trying to drive a Telit UL865 UART 1.8V input with a ATMega32u4 UART 5.0V output. Both of these gates are CMOS. The datasheet for the Telit says that it has a 5K to 12K pullup on the input. I believe that I need a voltage divider to drop the 5 to 1.8, but I can't tell if the pullup resistor is on the gate side of the CMOS or the source/drain side. I imagine that if its on the gate side, the voltage divider will be thrown off by pull up resistor that is parallel to R1. Where would the pull up resistor be placed in a discrete schematic of a CMOS gate? <Q> The pullup is from the input to the 1.8V supply. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It would be better to use either a specific level shifter chip (the kind with two supply pins) or a transistor rather than a voltage divider. <S> The maximum '0' input level is only 350mV not accounting for any noise margin. <S> If you pick 200mV as the output voltage that is acceptable for the output you need less than 625\$\Omega\$ for the lower resistance, so a fair bit of current from the AVR, but probably okay <S> (1.1K + 620 maybe, but check how close to 5.0V the AVR output is when sourcing ~3mA). <A> so yes, it would affect your resistor divider design and you would want to account for that. <S> Ideally a level-shifter would be used, but I've done it with a resistor divider also and it should work OK. <S> The other downside to using resistors is that it will be less efficient. <S> If that's not a concern then OK. <S> So pick two resistors such that they produce a voltage less than 1.8 and above the max turn-on voltage for the input pin, when considering the internal pull-up. <S> When you account for the pull-up, you can't just consider it as parallel with the external top resistor though, because they would be tied to different rails. <S> Actually, if you ignore the internal resistor, as long as your external resistor divider produces a value less than 1.8V, it will still not exceed 1.8V when the internal pull-up is added. <S> Then you just need to be sure that the external resistors are low value enough that when the UART line is low, that the internal pull-up is brought low enough to be seen as a logic 0. <S> It complicates the design a little. <S> It's definitely better to use some level shifting logic. <A> The pull up resistor would be on the gate side of the CMOS gate. <S> Whenever a single-throw switch (or any other sort of gate output incapable of both sourcing and sinking current) is being used to drive a CMOS input, a resistor connected to either Vdd or ground may be used to provide a stable logic level for the state in which the driving device’s output is floating. <S> Here is a picture showing what I am explaining here:
| Be careful about baud rate, because you may not get full bandwidth with a resistor divider as you would with a level-shifter. If they are stating there is a pull-up, they mean it is right at the input pin,
|
Replace DPDT on-on-on switch with relays I want to control via Arduino a peltier cooler/heater that uses a DPDT on-on-on switch to set the voltage at the terminals of the peltier cell to +12 V, 0V or -12V. I have measured the voltages at the six terminals of the switch and the voltages in the three configurations are as shown in this figure: The terminals at the top and at the bottom of the switch are set to either 0V or 12V as in the figure (voltages written in black). The position of the switch forms the colored junctions that set the voltages in the central terminals. I would like to use an Arduino to control the voltages of the central terminals in a way analogous to the switch. I was planning on doing this by using relays, but I'm unsure how to do the wiring and if I need a 4-channel relay of if a 2-channel relay is sufficient. Do you have suggestions? Thank you. <Q> Breaking this down... controlling this via a micro needs to define 3 states: off (no voltage across the Peltier) <S> cool (0, +12) <S> heat (+12, 0) <S> I think this can be simplified as two GPIO bits, and two SPDT relays. <S> Why two relays? <S> SPDT types are common, inexpensive and reliable, <S> so it’s more convenient to model the on-off-on using a pair of these. <S> So here’s what that looks like <S> : simulate this circuit – Schematic created using CircuitLab <S> This works as follows: <S> COOL, HEAT low: (-) = <S> 12V, ( <S> + <S> ) = 12V <S> COOL <S> high: <S> (-) = 0V, (+) = <S> 12V <S> HEAT high: <S> (-) = 12V, ( <S> + <S> ) = 0V <S> COOL, <S> HEAT high: <S> (-) = 0V, (+) = <S> 0V (probably does no harm, but check) <S> The relay coils draw a bit of current, about 10 ~ <S> 30mA <S> depending on the type. <S> That's very small compared to what the Peltier draws, so not even significant. <A> Yes, that 2-relay module should work for you. <S> Keep in mind that one or both coils may be energized for long periods so it's not very efficient, but it should function fine. <A> Here's a scheme, using a diode matrix, push button switches, magnetic latch relays and a 'H' bridge. <S> Freewheeling diodes, for the relay coils, and LED indicators are not shown. <S> S1, S2 and S3 are pushbutton switches which select '+12V', '-12V' and 'Off'. <S> Being latch relays the coils are not kept continuously energised.
| Ultimately you would want a set of MOSFETS in an H-bridge arrangement, but using relays saves you the trouble of dealing with shoot-through, and is much easier to wire.
|
How does an antenna differentiate between different signal waves? I'm aware there are frequency filters, but at the electron and EM field level, if there are multiple waves directed at an antenna, how does it differentiate between 2 different waves and frequencies? If wave1 is captured by an antenna, how does the filter/antenna know that there is another wave2 and that the E field is related to a different wave? How does it know that those excited electrons are from wave1 and not a part of wave2 for example? <Q> The antenna can't tell the difference between signals at all. <S> Even if an antenna is broadly tuned to a band, there could be many different transmissions within that band. <S> It is only with tuning, filtering, mixing and finally decoding that the signals you are interested in can be differentiated from the rest. <S> For example, to receive simple AM broadcasts you could have a non-tuned antenna and feed the signals (all mixed up) into an RF amplifier and then into a mixer fed also by a carrier frequency of your choice (a local oscillator). <S> When you mix the carrier signal with the modulated signals from the antenna you get product and difference signals. <S> The difference signal turns out to be the audio you desire. <S> You could start reading here . <A> The resonant frequency of the antenna is determined by its physical dimensions. <S> So actually the antenna reacts to all the fields it is exposed to, but it reacts 'better' to those where it is resonant. <S> In a similar way, several guitar strings emit different sound waves when being played, each one proportional to its length, even if they are strummed with the same force. <A> All antenna has gain characteristic. <S> It can received any signals so long the signal is within the antenna frequency range. <S> With so many signals received by the antenna, the engineer select his wanted signals by tuning to the frequency, implementing a band-pass filter.... followed by analog digital conversion (can be optional)... <S> etc. <S> (This is a simplified description and the real system is far more complicated) <S> Above: <S> The antenna gain of OmniLOG 70600. <S> The antenna is supposed to be used from 680MHz - 6GHz, according to specs. <S> There are performance penalties outside the range. <A> Antenna is an electrical device which converts electric power into radio waves, and vice versa, you can imaging the effect on water when you put a stone in it, similar to this the electric power applied to the antenna produce a disturbance (electro magnetic wave) to the applied electric power which propagate in air, when this disturbance hits an antenna (receiver) it produce electric power depending on the strength of wave. <S> radio waves are prone to interference Now to interpret signal the receiver stage will have a wide band (the range of frequency the receiver should cover) amplifier which amplify the received signal <S> then we will have one or multiple tuned (band pass) filter which allow only the required frequency that is of interest, the next stage will be decoding of the message from the signal. <S> radio waves are prone to interference, frequency hopping, <S> FDM, TDM are few methods, techniques to address various issues that are dealt with radio waves as mode of communications, today there are various technologies available to make effective wireless communication, like WIFI, BLE, etc
| Most antennas work by being resonant at the frequency of interest.
|
Change all Off Sheet connectors to Ports I´m new to Altium Designer. It seems like AD can´t use Off Sheet Connectors for hierarchical design, so I want to change all Off Sheet Connectors to Ports. Is there an easy way to do this? Btw: There are a lot of Off Sheet Connectors in the design. How can I use NetLabels in the hierarchical design? Somehow only Ports appear as Sheet Entries. Thank you. <Q> I would beg to differ with the answers posted here, but Altium absolutely has the ability to do what you're trying to do. <S> Because people sometimes use different names than others for the same thing, I'm just going to show a basic example, not necessarily tailored to your question, but you should be able to modify this easily to suit your needs. <S> The key to the solution is the Smart Paste tool. <S> Lets say I have a bunch of ports <S> and I'd like to paste them onto nets of the same name. <S> Simply select and Copy. <S> Click Edit - <S> > <S> Smart Paste <S> Here I chose paste as net labels and wires: <S> The final result: As you can see from the second image, you can paste as Ports, Net Labels, even as Sheet entries where you can paste them onto a hierarchy. <S> This is very useful for connecting <S> say a connector block up to a new hierarchy with a large number of nets. <S> For example, from the same original Copied material, I pasted this onto a sheet symbol. <S> I trust you can modify this to suit your original question. <A> In Altium Designer ports are used to create sheet entries. <S> Net labels are used to connect wires and buses inside a sheet only. <S> The power supply symbols (GND, VDD etc) are common over multiple sheets, so you don't need to make ports for them (although you can if you want it for clarity). <S> I don't think you can easily change the off-sheet connectors into a port, so you need to place a new port and give it the right name. <S> Altium documentation states the use of off-sheet symbols only for a specific reason: Multiple sub-sheets may be referenced by a single sheet symbol. <S> Separate each filename by a semi-colon in the Filename field. <S> With the effective use of off-sheet connectors placed on the sub-sheets, you can effectively spread a section of your design over multiple sheets, treated as though they were one giant (flat) sheet. <S> Note however, that use of off-sheet connectors is only possible for sheets referenced by the same sheet symbol. <A> If my memory serves, off-sheet connectors were really only included in Altium Designer for backwards-compatibility with older versions of the software (Protel). <S> I generally use global net names instead. <S> I think you'd have to go through them one-by-one to replace them with ports. <S> I ran into this a few months ago at work -- an old design used off-sheet symbols but we needed to change them to ports for use in a hierarchical schematic. <S> We ended up having to go through and replacing them manually. <S> It would be nice if Altium would allow the "Object Kind" to be changed in the Schematic Inspector. <S> EDIT: <S> I tested Joel's answer and it works beautifully. <S> Please disregard this answer.
| As for changing off-sheet connectors to ports, I don't believe Altium currently has a good way to do that.
|
Pros/Cons/differences of High pass inductive vs capacitive filter A simple high pass filter can consist of a capacitor in series with a resistor or a resistor in series with an inductor. Why would I use one over the other? I also wonder the same thing about low pass filters. Do capacitive high pass filters have less energy usage or increased speed, or clearer filtering capabilities compared to inductive high pass filters? Thank you! <Q> You appear to be talking about these two variants: - Both are equally good but you will find that for low frequencies of cut-off (such as audio) the parallel output inductor becomes too physically big compared against a series capacitor. <S> Just try calculating a 100 Hz filter with a 1 kohm resistor. <S> The capacitor should have a reactance of 1000 ohms at 100 Hz hence, it has a value of 1.6 uF. <S> The inductor also has a reactance of 1000 ohms and therefore has a value of 1.6 henries and will be bulky in comparison, have an annoying self resonant frequency in the kHz region and will cost possible 100 times more than the capacitor. <S> If you are thinking about a much lower impedance then things change; if R is now 10 ohms, L would be 16 mH <S> and C would be 160 uF i.e. a less clear-cut differentiation. <S> Go do some math and look up inductors at distributors websites. <S> Exactly the same argument applies to low pass filters because the same components are used but the output for low-pass is taken across the "other" component: - <A> The two circuit configurations have the same transfer function, so can be made to behave identically as filters. <S> However, there is a major difference in the input and output impedances of the two networks. <S> The C-R configuration works by preventing the low frequency energy from a source from entering the filter while the R-L configuration shunts that energy away from the load. <S> If, for example, you had an power amplifier generating a broadband signal into a load, and you wanted to add a filter to select the high frequency components, the C-R configuration would be preferable. <S> This is because the C-R filter presents a high impedance to the amplifier for the frequencies you don't want at the output. <S> The R-L configuration, in contrast, presents a low impedance to these frequencies, forcing the amplifier to produce a signal that you are then just going to shunt to ground. <A> Exactly they are not in series, because the output is taken from the middle point. <S> No difference if the components are ideal. <S> This is due the losses and stray capacitance in the wire coils. <S> At low frequencies the inductors are bulky. <S> One C + one R filter is often too unselective. <S> For steeper frequency discrimination more complex circuits are needed. <S> They must have inductors and capacitors. <S> By adding more Rs and Cs one can get no advantage. <S> At low frequencies (sub 1 MHz) the need of inductors can be passed by using an opamp based active filter. <S> It has an opamp+Cs +Rs. <S> This all is proven in the mathematical circuit synthesis theory.
| One L + one R high pass filter is mathematically equivalent with one C + one R high pass filter. In practice the inductors degrade to unusable much faster than the capacitors as the frequency increases.
|
Connecting transistors in series I know we can connect two or more transistors in parallel to increasing the maximum switching current, but what about increasing the maximum voltage? Can we connect them in series? For example, connecting two 2N3055 to give 100V 5A rather than 50V 5A? <Q> A cascode amplifier is a kind of series connection where a common emitter stage drives a common base stage. <S> Wikipedia shows a variant with multiple common base stages: <A> Yes that is possible but less trivial than placing them in parallel. <S> Schematic created using CircuitLab Qsw does the actual switching <S> Qcasc is for dividing the voltage between the transistors <S> I have used this kind of circuits where I need to switch 5.5 V but are limited to using (on-chip) transistors which can only withstand 2.7 V. <A> can we connect them in series? <S> yes. <S> for example connecting two 2N3055 for gave 100V 5A rather than 50V 5A? <S> Yes, if the switching voltage is also correspondingly increased. <S> Otherwise, the top transistor takes most of the voltage drop. <S> Such devices do / did exist. <S> Google emitter-switched transistors. <S> ST made some.
| What you need is cascoding which is done like this: simulate this circuit –
|
Max. switching current in relay Maximum switching current of this relay at 250 V is 10 A. Why is the maximum switching current at 30 V also 10 A? I wonder why it can tolerate 10 A at 250 V but not 80 A in 30 V because the power is the same? P1 = P2 = 250 V x 10 A = 30 V x 83 A <Q> There is no 'power law' that applies to relay contacts. <S> When the contacts open carrying a current, an arc often forms in the gap between them, which damages the contacts. <S> AC current falls to zero 100 times per second (at 50Hz). <S> This helps to extinguish the arc, which mitigates some of the effect of having a higher voltage. <S> DC current does not have this feature, less voltage is needed to maintain an arc. <S> Current carrying, with closed contacts, is limited by heating in the contact area. <S> This will be a function of the contact resistance, which depends on contact area, pressure, material etc. <S> This joule heating is insensitive to whether the current is AC or DC. <S> Therefore it's not surprising to see the same current specification for AC and DC. <S> You will notice there are two AC specifications, 250V and 125V. <S> Obviously both cannot be 'right'. <S> They are ratings. <S> The point about a rating is that it's a test done under well defined conditions. <S> What that means is when the relay is tested with some specified circuit, for x thousand operations, some specified parameter, perhaps on resistance, will not have changed by more than y%. <S> If you want the detail of the test circuit and pass/fail conditions, then you need to track down the specification that this has been rated against. <S> But basically, more voltage means more energy available to heat contacts, and more voltage available to keep the arc going longer. <S> Different relays have different contact materials, pressures, masses, speed and distance of separation, and these will all have differing impacts on carrying current and breaking current and voltage. <S> Some relays may well have different rated currents depending on whether AC or DC is used, this one doesn't. <A> The contact area and heat sink capabilities determines the maximum current irrespective of voltage. <S> AC voltage has zero current assisting normal arc extinction where DC must rely on sufficient contact separation to extinguish the arc. <S> This is why maximum AC voltage is much higher than the maximum rated DC voltage. <S> Contacts can be used outside these parameters with snubber circuits but will be destroyed under very inductive loads, even at rated voltage and current. <A> Your calculations assume that the full voltage is dropped by the relay, which is certainly not the case, usually you have a very small voltage drop over the relay, as its contacts have a low resistance. <S> To understand the issue, it helps to look at the formula for the power dissipated in a resistance, which is: $$P = <S> I^2 R$$ <S> As you can see, it is not dependent on the voltage. <S> So if the dissipated power would be the limiting factor, only the current would influence that. <S> In this case power is probably the limiting factor (as it both specifies 10A for 230V AC and 10A for 125V AC). <S> As @Neil_UK <S> already stated, the reason for the different maximum voltage rating for AC and DC is because of the sparking behavior. <A> The ratings apply to the safety approvals above them. <S> You should not use the relay above 125VAC if you need UL approval. <S> The ratings are determined by a number of factors- one of which is heating (the current passes through a flexure that must not get too hot or it could fail). <S> The safety agencies also look askance on products that produce excessive amounts of smoke or burst into flames. <S> Contact life from erosion is another factor, contact gap (for inductive loads) and maximum surge current (for tungsten loads and motors). <S> Ratings are not determined by any single factor. <S> Chances are if you tried to put 80A through that relay the contacts would instantly weld shut , and then it would burn up soon afterward no matter what you did to the coil.
| The copper conductors in the relay have a non-zero resistance, at a certain current the relay might get too hot and plastic parts might melt.
|
Eliminating attenuation in summing box Full disclosure: when it comes to electronics, I'm the type that knows enough to be dangerous. And that's kinda why I'm here - I took an electronics course in high school in 2005, but it's been a while so I've forgotten a lot... I have two computers hooked up via a KVM switch, but I didn't like how audio was also exclusive when I switched inputs. I came across the Rane note " Why not Wye? " and built a balanced summing box so I could hear both computers at once. Works great, except that when both computers are on, the output volume attenuates regardless of whether or not either line has sound. I notice this a lot, especially when one computer sleeps and thus cuts the audio and suddenly the music on my running PC is louder. My circuit is the one on figure 4 of that article: Is there a way I can make sure the output volume is consistent regardless of whether or not both inputs are present? <Q> You can get most of the way to what you want with resistors, as long as you have enough spare level at the source, or gain following. <S> At the moment, your input resistors are small, and the load resistance is high. <S> This means that when one computer is low impedance, it is responsible for all the attenuation of the signal, with equal input resistors, that's bout <S> 6dB. <S> When it goes high impedance, the output level jumps up 6dB, which you find objectionable. <S> If the 'wye' had more attenuation built in, say 10k resistors to replace the 470s, and a 2k instead of the 20k, you would lose a lot of level. <S> This is why this method will only work if you have some gain in hand. <S> The attenuation to the signal when the other computer is on is due to the 10k resistor and the 2k, about 1.7k. <S> When the other computer goes high impedance, that attenuating resistor becomes 2k, and there is only a 1.5dB change in level. <S> The smaller you make the resistor that replaces the 2k, the less the level will change, but the more level you'll lose. <S> What a proper summing amplifier does is make that 2k resistor position effectively a short circuit, it presents a virtual ground to the summing junction, which totally isolates one signal source from the other, and provides gain so you don't lose level. <S> It's the better solution, but it requires amplifier and power source. <A> The resistor summing circuit that you are using is a lossey circuit. <S> What that means is that the resistors consume some power and end up reducing the net amount of voltage that can be delivered to the load. <S> When both computers are ON some signal from one computer can actually feed back into the resistors and output of the other computer. <S> This is what causes the sound level decrease that you noticed. <S> When the second computer is OFF or has gone to sleep the audio output driver from the chipset inside the computer goes to a high impedance and thus removes the extra current path back into the output. <S> This will block the DC current path and eliminate some of the signal level shifting. <S> You could start by trying a 1uF non-polarized capacitor. <S> (Note that a capacitor will be required on both the left and right signal paths of each input - so for your circuit total of 4 capacitors). <S> Another choice is to build or buy an audio mixer device that buffers the two audio outputs through opamp or transistor gain stages. <A> Buffer the inputs individually before mixing them together. <S> Use a jack that would short the summing resistor to ground if not used.
| One simple thing you can try is to add a capacitor in series with the series summing resistor on the signal output from both computers.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.