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How to create rectangular holes in ARES? I bought this three phase rectifier from local store and it has rectangular connectors, Now I need to make its package on the ARES, but I cant figure out how to create rectangular holes in the package. <Q> Forget rectangular holes! <S> This kind of part is usually bolted to a heat sink, "dead bug style" (legs pointing into the air). <S> The terminals do not go through a PCB; it is not a through hole part. <S> The terminals accept sliding "push on terminal" connectors. <S> ( source of picture) <A> To create a SLOT in ARES(PCB part of Proteus) requires 3 things. <S> First is the ability of your PCB Fabricator to route out slots, and to know his minimum routing tool size. <S> The jokes about rectangular drills are not very funny for someone asking for help. <S> Second, [C]reate a "PAD" in DILPAD mode. <S> Make a name [ie SLT-100x40DP], then click OK. <S> Next window you make the PAD dimension the outside dimensions of the copper for your slot. <S> The drill hole should be .005" greater than the preferred or minimum routing tool of your PCB fabber. <S> Smallest I see looking around is 0.5mm - .020", but you have to find out Third, [C]reate a "PAD" in PADSTACK mode. <S> Make a name, different than your DILPAD name [ie SLT-100x40], then select the DILPAD you created [ie SLT-100x40DP] in the Initial Style drop down box Under STYLE NAME select SLOTTED. <S> In the SLOT box enter your slot dimensions [w=20th h=100th] and your PCB Fabber's slotting tool dimension[t=0.5mm]. <S> Allow for copper plating thickness and round corners in deciding to the dimensions of your slot. <S> On the right you can see what other layers you may wish to connect to, if you never do multi-layer boards, this info doesn't matter. <S> You can always edit this later. <S> Click OK. <S> iIn <S> the list of available PAD STACKS your new plated slot [ie SLT-100x40] is available to use like any other hole or pad making a new package. <S> Edit as needed. <S> Finished slot info is placed on the Mechanical layer so be sure to include this layer in your Gerber .zip <S> Slots are handy, specially in high frequency applications- such as for a PCB mount antenna connector. <S> Besides saving board space, solder has more capacitance than copper. <S> A big hole could cause your project to function poorly. <S> A ton of solder in a big hole is not an ideal solution. <S> However, that said above, slots do cost more than holes, and lots of the less expensive PCB proto places won't allow slots. <S> So, find out the cost of a slot vs. a hole in your application, then make your choice. <A> To answer the actual question: "How do I create rectangular holes?" <S> : In the PCB programs I've used (mostly Protel/Altium) there is no built-in way to specify rectangular holes,as few board shops have rectangular drills. <S> If you want non-round holes or other special milling on a board, you will have to discuss the matter with the board shop to determine if they can do what you want, and how they want it specified. <S> Since you will be making this board yourself, using home-brew techniques, you just have to produce artwork that you will understand.
| (I've seen suggestions to mark the ends of a slot with a non-drillable hole size (perhaps .001"), so the board shop will have to consult you or a readme file to find out what you really want.)
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Why is electric heating so expensive? The loss of electricity in transmission and distribution is only 6% , so why is it that electric heating is so inefficient? I would imagine that the generators at power plants are far more efficient than any engine you can put it in my house for heating, and I also assume there is very little energy loss from going to electrical energy to thermal energy (if not, then where does the energy go?). Where am I going wrong in my thought process? <Q> The efficiency of converting electricity to heat is 100% (although sometimes some of the heat ends up in places you don't want it). <S> However, the conversion of fuel to heat, the conversion of heat to mechanical energy and the conversion of mechanical energy to electricity are far from 100% efficient. <S> Look up Carnot cycle for some insight into the fundamental physical limitations. <S> This is why it is always more efficient to convert fuel directly into heat in your home. <A> Your thought processes are fine as far as they go. <S> What you are overlooking is the inefficiency of converting fuel to electricity. <S> Dave Tweed mentioned it, but look at https://www.nema.org/Products/Documents/TDEnergyEff.pdf for an overview. <S> Yes, a central generating plant will do better than a home generator, but better is not nearly good enough. <S> Burning oil for heat will give efficiencies as high as 98%, although this takes some doing. <S> 80 - 90% is more typical. <S> A conventional coal-fired electrical generating station will run in the neighborhood of 40%. <S> Add on another 6 - 8% transmission loss, and a good general number for relative efficiencies is 2:1 in favor of non-electric heating. <A> For the same reason it would be uneconomical to burn fine furniture in your fireplace rather than raw logs. <S> Electricity is a more useful and expensive form of energy than heat, and you lose the majority of the energy in the fuel turning it into electricity in a thermal plant (even with a large well-designed plant, about half to two thirds is lost!). <S> When you burn natural gas directly, most of the heating energy goes into your house (in our case, with a high-efficiency furnace, the lost heat is so low that a small plastic tube is used as the "chimney", so you know most of the heat goes into the heat exchanger). <S> You are correct, however, that centralized large installations have efficiencies. <S> Your own standby electric generator would not be as efficient as the thermal power station. <S> Crowded downtown areas sometimes have central heating plants. <S> For example, in Toronto there is a plant that supplies about 0.6GW of steam to 140 buildings downtown. <A> If you're looking at cost efficiency as well as energy efficiency, an additional factor is that a significant fraction of your electric bill is for the maintenance and operation of the grid instead of power generation. <S> Depending on where you live those costs may be broken out separately on your bill: on mine they collectively make up about 4 of the 12 cents I pay per kWH of (coal generated) power. <A> I believe there are two questions conflated here. <S> One is thermodynamical (Carnot) efficiency and the other is cost in real money. <S> I believe that the heat pump (mentioned in a few answers already) can be more Carnot efficient (even with the coal-generated electricity) than the direct combustion. <S> There is a nice book that discusses these issues, "Without the Hot Air" by David MacKay. <S> http://www.withouthotair.com/
| There are economies of scale associated with central plants (transportation of fuel is cheaper, for instance, since there is no need for a local distribution network to get the fuel to individual homes), but these are not nearly enough to overcome the basic problem of generator inefficiency.
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Paralleling power supplies for higher current I am looking at using two Cincon ( CFM80S240 and CFM60S240 ) power supplies in parallel to achieve a higher current output. Is this possible? If so, can I just hook them up in parallel, or is there some specific way of doing this to prevent one from "fighting" the other? Both power supplies are rated for 24V with 1% ripple and noise, +/- 1% voltage accuracy, +/- .5% line regulation, and +/- 1% load regulation. Ideally, this would be possible since they would both be the same voltage. But, due to tolerances in manufacturing, there is no guarantee that they would be identical voltage outputs. I wasn't sure if I should also hook up a capacitor in parallel with them to help stabilize the output voltage. <Q> Also, in general, it would not be allowable to parallel two different types of supplies as you have. <S> You may be able to use ballast resistors to allow paralleling, but they would have to be calculated, would waste power, and would degrade regulation to your load. <A> I think you can parallel them by connecting two diodes in series with the supplies like in the following figure, it works almost exactly the same way a rectifier does. <S> The diodes take care of the small offsets in the supplies' voltages and prevent them from shorting. <A> I'm not an expert, but you may want to look into 'battery isolator' circuits that are used in charging car batteries with an alternator.
| There are ways that you can configure your circuit with diodes or something similar so that the voltages don't 'fight'. Unless the documentation specifically says this is allowable, you should not parallel supplies (and if it is allowed, be sure to follow all the related recommendations from the manufacturer).
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Is there some kind of jumper for banana plugs? I have a three-phase cable harness (low voltage) for connecting to a motor. I have it connectorized so I can easily plug/unplug the motor. I would like to insert a short inline cable so I can measure the current on any of the three phases. Here's what I have so far, to measure 1 phase: I've added a pair of banana plugs to connect the white phase to my DMM. I would like to do the same for the red and black phases, but I'm only going to be measuring one at a time, and I was wondering if there is any kind of double-ended banana jack so I could just plug the two halves of each unused phase together so that it conducts current and it doesn't have an exposed conductor. I couldn't find any. Any suggestions? (The red/white/black connectors are Anderson Power Products connectors, by the way.) <Q> Found one from Pomona: they're called banana plug splices. <S> Pomona 1829 <S> (single) Pomona 1823 <S> (dual) <A> You could easily make one from a length of 5/32" ID copper tubing with heat shrink over it or, failing that, http://www.amazonsupply.com/power-probe-pnls029-female-adapter/dp/B002YKIJ8O <A> Four pairs of Quick Disconnects (insulated) would work. <S> Male on one side, Female on the other for each pair. <A> Of course this relies on your multimeter being able to take proper 4mm shrouded connectors, most will, but some of the bottom of the barrel ones will only take partially shrouded or unshrouded connectors.
| Then a set of Quick Disconnects to Banana Jacks, one with a male and one with the female quick disconnects. Any radioshack or auto store carries them, just make sure you get the right size for your cable gauge/voltage/current capacity. For this application I tend to use 4mm stackable shrouded connectors, since they are "stackable" you can easily plug a pair of them together, and since they are shrouded you don't have to worry about the male part touching something.
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Using 16bit SDRAM with 8bit bus? I have a design with an 8bit SDRAM which I want to replace with a 16bit one,the problem is I don't have much pins left to connect all 16bit data bus, is it possible to access the whole memory using an 8bit data bus ? The ram ( IS42S16800E ) is organized as 4096 rows x512 columns x16 bits per bank, using 9 bits column, I'm only able to access half the ram , this is acceptable if I have to, however, I tried using 10 bits for column address, and it worked, ran a memory test and I can read/write to every location, but I'm not sure how/why this works, if I understand correctly, it should only use 9 bits for the column address ? so does this depend on the organization of this specific ram ? or is this how all SDRAM work internally ? Update: I fixed my ram test (Thanks Dave Tweed) and this time it fails, writing to address base+0 overlaps with base+512 which means the 10th bit is ignored and it wraps around to 0. If I use 9 bits for the column address I can still access half the ram. Here's my ram test: uint8_t pattern = 0xAA;uint8_t antipattern = 0x55;uint32_t mem_size = (16*1024*1024);uint8_t * const mem_base = (uint8_t*)0xC0000000;/* Test data bus */for (uint8_t i=1; i; i<<=1) { *mem_base = i; if (*mem_base != i) { BREAK(); }}/* Test address bus */for (uint32_t i=1; i<mem_size; i<<=1) { mem_base[i] = pattern; if (mem_base[i] != pattern) { BREAK(); }}/* Check for aliasing (overlaping addresses) */mem_base[0] = antipattern;for (uint32_t i=1; i<mem_size; i<<=1) { if (mem_base[i] != pattern) { printf("address bus overlap %p\n", &mem_base[i]); BREAK(); }}/* Test all ram locations */for (uint32_t i=0; i<mem_size; i++) { mem_base[i] = pattern; if (mem_base[i] != pattern) { BREAK(); }} <Q> If you have a device which puts out 16 bits, and you want to use all of its storage, then in theory what you can is: in the reading direction, take the 16 bit output obtained from your address lines A1.. <S> AN and pass it through a bank of 2x1 multiplexers, to go from 16 to 8 lines. <S> Your A0 line to drive these multiplexers, so that in effect your A0 line selects which of the two 8 bit halves of the 16 bits are passed to your data bus. <S> in the writing direction, you need some buffers which drive each of your 8 lines to one of two possible data pins on the ROM based on address A0. <S> These two directions can be combined into one using bidirectional multiplexers. <S> For instance, the ON Semiconductor 74FST3257 mux/demux/bus switch provides four 2x1 bidirectional muxes. <S> With a pair of these, you could route 8 points on one side into a choice of two different sets of 8 points on the opposite side, based on the A0 line. <S> This is complicated enough that it should send you looking for an 8 bit memory part in the first place. <A> 4096 rows is 12 bits of address space, 512 columns is 9 bits of address space, and 4 banks is 2 bits of address space so, added all up, that's 23 bits, which is 8 388 608 addresses. <S> The data sheet describes the IS42S16800E as an 8 meg X 16 RAM, so with your 9 bits of column addresses you should be able to access the entire array. <A> Not quite sure what the problem is? <S> I will just quote the parts from the datasheet that mention/explain the behavior: <S> Each 33,554,432-bit bank is organized as 4,096 rows by 512 columns by 16 bits or 4,096 rows by 1,024 columns by 8 bits. <S> Physical size of the RAM is fixed, so if your data is half the size you can store twice as much - or to be more accurate: You can store twice as many 8-bit words as you can store 16-bit words. <S> Inputs A0-A9 (x8); A0-A8 (x16) provides the starting column location. <S> Which means you use 10 address pins in 8-bit mode (x8) and 9 address pins in 16-bit mode (x16). <S> So what you tried is actually defined in the datasheet and absolutely fine. <S> Other SDRAMs don't have that option and if you use them with a 8-bit databus you are not using the full memory. <S> I guess it depends on how exactly their internal address decoders work.
| Some SDRAMs have the option to work in a 8-bit or 16-bit mode which means for those devices you can always use the full RAM.
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How to increase torque and decrease RPM of a DC motor? The motor is an old one from a cassete player of some sort. I am powering the motor with a 9 volt battery, but it's RPM is too high for the desired purpose. I have used gears to reduce RPM but it is still too high. I don't think I can increase the gear ratio anymore. Can you suggest a way to decrease the RPM further and increase the torque (if possible). Please suggest a circuit to accomplish it. <Q> I shall make the assumption the machine you have is a brushedDC machine. <S> Torque ~= <S> current <S> (Kt being the constant of proportionality) <S> Speed ~= voltage. <S> (Ke being the constant of proportionality) <S> In actual fact this is loosely accurate for all machine types but rather than being directly proportional it becomes a bit more complex & usually involves frequency. <S> ( see <S> How do DC motors work with respect to current, and what consequence is the current through them? ) <S> To reduce the speed the simplest method is to reduce the voltage be it by replacing the 9V battery by a... 8V battery or via PWM. <S> To increase the torque ... that a bit harder. <A> Rewind it with more turns of thinner wire. <S> For a given voltage, speed * turns will be a constant, so quadrupling the turns would give 1/4 the speed. <S> To fit those turns in the same space, you would need to halve the wire diameter. <S> Gearing is easier : here will be ways of increasing the gearing : the simplest way to get a large speed reduction is using worm gears. <S> But you haven't told us anything about how much speed reduction you need. <A> To decrease the RPM at the same torque, you want to power it from a "stiff" voltage source that is lower. <S> PWM is one way to achieve that with minimal losses. <S> To increase torque beyond what you get with 0\$\Omega\$ source impedance, you can go negative on the source impedance or use feedback to actively adjust the voltage to maintain speed (more complex and you need to get the control loop right). <S> I think you could do simple negative source resistance (also called IR compensation) with an op-amp driving the reference terminal of a linear regulator, which might be appropriate for a small motor.
| All you can do, electrically (as you can do it mechanically via gears), is to reduce the source impedance to maximise the current that can be delivered for a given stator impedance.
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What are some of the segmented displays with the lowest power consumption? I have to select a low-power display for a solar-powered digital clock . I have looked into numerous technologies, including e-ink. However, I was still unable to compare between them due to the lack of online information regarding average power consumption. I am aware of the existence of solar-powered pocket calculators, which use some kind of 7-segment displays, so I know that it should be possible to build a time-tracking device that is fully powered by light. I can accept refresh rates as low as 1 frame per minute, but preferably I would like to have 1 frame per second. The most stringent specification is power which, ideally, should be 100% supplied by a small solar panel. Does anyone have a good suggestion for a good low-power display part that seems suitable for these specifications? <Q> Have a look at old Sharp LCDs from calculators. <S> You won't find lower power 7 segment LCDs. <A> It sounds like you want a LCD display. <S> These can be very low power, and is why they are used in watches and many battery operated handheld devices. <S> The driving electronics for a LCD will take some power too, so you really have to look at the whole system. <S> For a do-it-yourself low power clock, I'd probably start by looking at some of the very low power PICs with LCD drivers built in. <S> The trick to lowest power will be very carefully written firmware that uses the LCD driver effectively and keeps the processor in sleep mode to the exent possible. <S> For you PIC, you want to use the timer 1 oscillator driving a 32,768 Hz watch crystal. <S> Managing the power and storing it to keep the clock running when the sun isn't shining will also be a major design issue. <S> A supercap is probably necessary, with a separate circuit that charges up the cap and only runs when there is power from the solar panel available. <A> Lowest power consumption for a constantly changing display like a clock is probably a low-voltage LCD display with static drive (that means one common and many segment wires- one per segment), and of minimal area (obvious trade-off here that you'd like to be big enough to read). <S> Low voltage also limits your ambient temperature range, but static displays are much better than multiplexed ones for temperature change to begin with. <S> Alternatively, the off-the-shelf e-ink displays claim a typical current during the change of state (which might last 0.5-2 seconds) of 500nA/cm^2. <S> The static LCD displays draw continuous power that is related to the area, the square of the voltage, and the refresh frequency (usually something like 50Hz). <S> Multiplexed LCD displays require a voltage divider to get the different voltage levels so they tend to draw more power at a system level. <S> You can drive static LCD displays directly with a microcontroller if it has enough pins (you have to generate symmetric waveforms that have no more than bout 50mV of DC component or the display can be damaged by electrochemical action, or <S> some low power micros such as PIC and MSP430 have built-in LCD controllers (just write a bit to the register to turn a segment on). <S> You can drive e-ink displays at 5V directly from a suitable microcontroller (not a 3.3V type), or to full contrast they need 15V and a high-voltage driver. <S> They also require DC balance on the drive but that simply amounts to pulsing the segments 'off' for as long as pulsing them 'on', and keeping track of the current state so you don't write them on more than once without turning them off first. <S> You will also need n+1 pins to drive an n-segment e-ink display. <S> Chances are either type of display would work for you and would not represent the largest current draw in your system. <S> Static LCDs are more familiar and would change faster from 5V <S> so I think I'd tend that way for a clock display. <A> The segmented LCD VI-602-DP-RC-S and the E-Ink display SC005221 <S> both fit your needs. <S> You'll have to carefully manage power consumption for the microcontroller that drives them, but they are both capable of running off a solar cell.
| Consider using a microcontroller with a built-in LCD driver, as that will probably be lower power than anything you can do by simply twiddling port pins.
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Why are there so many resistors in a typical schematic? I know that some resistors cause pull-up or pull-down or work as a voltage divider and/or current divider . But sometimes I cannot understand their function in some simple circuits: Take this motor driver (L9110) for an example . Why are there 2x4.7k resistors? The L9110 is rated to be able to recieve 2.5v-12v according to the datasheet Or take this zener diode voltage regulator , The 40Ω resistor, ohm's law says that V = I x R, increasing the resistance lowers the current. But why is that 40Ω resistor needed at all? Why not let it flow straight through (~0Ω)? Or lastly this RS232 level shifter , I understand the 10kΩ resistors. But why is there a 4.7kΩ resistor between RX(Device) and TX(Device)? (Should they even be connected?) <Q> All three circuits display examples of pull up, pull down, voltage dividers, and current dividers: R1 and R2 <S> are pull up resistors. <S> You have to have both because you have two switches which can be in a different state (one high, one low). <S> The 40Ω resistor is the top half of the voltage divider . <S> The zener is the bottom half of the voltage divider. <S> The zener can be thought of as automatically adjusting its resistance so that the voltage will always be 6 volts. <S> Without the 40Ω resistor, the top wire would be at 10V, and the zener would blow up trying to bring that wire's voltage down to 6V. <S> If the 10V supply was current limited to a current less than the zener's capacity, then the zener would pull the wire down to 6V, putting the power supply into current limiting mode (rather than voltage regulating), and the circuit would work fine. <S> Since the supply is a voltage regulator, though, then you need the 40Ω resistor so the zener can do its job without blowing up, and without getting a current limited power supply. <S> R3 is a pull down resistor. <S> Ignore RX(MCU) and R2 for now, they just tap the TX(DEVICE) <S> line. <S> D1 and C1 form a negative power supply. <S> RS-232 technically requires -12V for signaling. <S> The TX(DEVICE) <S> line will go to -12 occasionally, and the diode and capacitor store that charge so that the TX(MCU) line can use it without building a special -12V power supply into the circuit. <S> It has some limitations, but for those RS-232 devices that demand obedience to older RS-232 specifications, it can work well. <S> R3 <S> therefore is a pull down - when TX(MCU) isn't high, <S> then RX(DEVICE) will see a low current negative voltage. <S> If the device uses <S> -12V <S> on its TX line, then the RX line will reflect the device's adherence to the RS-232 specification. <A> In the case of the Zener regulator, the supply is 10 volts, and the Zener will do its best to limit the voltage across itself to 6 volts. <S> If the 40 ohm resistor wasn't there, a very large current would flow as the Zener tried to do its thing, and the magic smoke would be released from many parts. <S> If I've done the math right, the 40 ohm resistor will carry 100 mA to drop the 4 volts from the 10 volt supply to the 6 volt Zener diode. <S> Since the 100 ohm load resistor has 6 volts (controlled by the Zener diode) across it, it will pass 60 mA, and the zener will pass the remaining 40 mA from the 40 ohm resistor. <A> 1 and 3 have the same reason. <S> It's a current limiting Pull Up or Pull Down. <S> In 1, the pull up resistors are there for 2 reasons. <S> 1, to set the default state of the pins to Logic High, and 2, because if they were not, pressing the button would cause a short of VCC/5V directly to Ground. <S> Bad stuff happens when you do that. <S> In 3 it is a pull down. <S> Notice that RX(device) <S> is connected above the resistor but below the transistor. <S> When the transistor is off, the Pull Down (through the capacitor) brings the line to Logic Low. <S> When the transistor is on, the RX(Device) line is pulled up to Logic High through the transistor, which is a lower impedance path to a Voltage Level (Essentially a diode) than the resistor is. <S> Without the resistor, enabling the transistor would create a mostly direct path from 5V to Ground, again, a bad thing. <A> the short answer is that an open input needs some finite current, to assure the voltage. <S> If the current is zero, then the voltage is not determined. <A> All of those are for current limiting. <S> Ohm's law states that V=IR , so if you make R equal zero, then for a fixed V you get infinite <S> I and your part blows up (because P=IV ).
| So the pullup and pulldown resistors, bias the pin to some fixed voltage.
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Combing two DC voltages, one high volt low current the other low voltage high current I am a new member and have found lots of useful information on this forum whilst learning about electronics. I am wondering if I connect a low voltage high current source to a high voltage low current source, if I might get a high voltage source that could use the current from the low voltage high current source. I have previously joined similar low voltages through two independent bridge rectifiers and then placed them in series across a filtering capacitor which I could manipulate to a higher voltage (x2 volts) out or a higher current out (volts x 1). This worked well. But before I try combining a high voltage such as say 500VDC at 100ma with a low voltage say 1VDC at say 50amps, I would like an expert opinion of the possible outcome, given that each supply will be isolated. <Q> What you are proposing is not feasible based on power considerations. <S> To use your example, a 500 VDC source at 100 ma can provide a maximum of 50 watts. <S> A 1 VDC source at 50 amps can also provide a maximum of 50 watts. <S> Let's say you could somehow borrow 1 amp from the 1 VDC source and give it to the 500 VDC source. <S> Now you would have a 500 VDC source providing 1.1 amps or 550 watts from 2 sources that are limited to a total of only 100 watts. <A> You would have to use a switching converter to step up the low voltage source to a higher voltage, and a load sensing and balancing circuit to ensure sharing. <S> So you could get 500V at less than 200mA from 500V @ 100mA and 1V @ <S> 50A. <S> (The 1V to 500V conversion would be fairly inefficient.) <S> But as Barry noted you can't just create power or energy where it doesn't exist. <A> You can do exactly what you describe, but not with the results you expect. <S> You could connect the two sources to a common supply bus, isolated from each other by a pair of diodes. <S> The higher voltage source would dominate, providing high voltage to the bus. <S> If your load draws more current than the high-voltage source, the high voltage will droop further and further until its voltage equals that of the low-voltage source. <S> At that point, the much higher current capacity of the lower-voltage source would dominate, preventing much lower droop on the bus. <S> If you graphed voltage versus loading on this bus, you'd see a very abrupt elbow at the low-voltage source's voltage. <S> You should be aware that this approach would be pretty brutal to the higher-voltage source.
| You can in theory combine the two sources to provide a single output with an output power equal to the sum of the power available from each of the sources, minus conversion losses and overhead.
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12V LED strips in series with a 24V power supply? I am purchasing 50x - 1Meter / 12V / 72 LED /Rigid SMD 5730 strips directly from china. I don't know the LED manufacturer, nor the resistor being used. (yes, this is a gamble) The strip is one that could be cut at every 3rd led. So, it seems that every 3 LEDs is wired in series, and there are 24 groups of 3 wired in parallel. Each strip is specked out at 12V/18W. I would like to take 2 strips, and wire them in series with a 24V 18W power supply. (By series, I do not mean chaining them in parallel and calling it series like the billion articles polluting my google searches) My question is, can I wire two 12V strips in series using a 24V power supply? It seems like I can, but am worried I am overlooking something that I don't even know to ask, or I do not have enough information without the LED spec sheet. Thanks! <Q> Well, yes and no. <S> On the one hand, connecting 2 in series ought to work. <S> On the other hand, you will need 24 volts at 36 watts. <S> Since each string takes 12 volts at 18 watts, its current must be 1.5 amps. <S> Connecting them in series will still take 1.5 amps, but the total power will be 1.5 x 24, or 36. <A> This is an old post, but i just wanted to contribute my observation. <S> If half of the leds on the 24V would have burned out on the left side (see image). <S> The result would be catastrophic because the remainder leds on the burned side would receive an increase in voltage (about 16 V) and on the right side the voltage would drop to about 8V. <S> In conclusion, as the led burn out the voltage increases on the same parallel side and if your LEDs are rated at 12V it would be just a matter of time until all lights in the same parallel burn out due to high voltage. <S> That is why it is recommended to always connect led strips in parallel. <A> This is of course correct and a real problem as long as you have only a few LEDs on each side. <S> However, if you connect 2 X 2 meters with 2 X 120 LEDs, a burnt unit (3 LEDs and a resistor) will increase the voltage only about 0,6 Volt on the other side. <S> This will not do any harm especially if you turn down the power supply a little bit (if the power supply can be adjusted). <S> Furthermore you can balance the systen anew by cutting of 3 units as soon as 2-3 units are burnt on the other side. <S> Sorry, no native speaker of English.
| Both your parallel and series would work, however that is if we assume that none of the LEDs would ever burn out.
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Battery Ampere-hour rating vs Battery Amps (not an experienced user) I have almost no experience whatsoever with the technicals of electronics so this is probably a very easy question for someone who does. Do you know if a battery with a 200Ah rating can put out 200A for one hour or are there limitations? According to this website's third paragraph ( Battery Ratings - Chapter 11 - Batteries And Power Systems ), you can. For example, an average automotive battery might have a capacity of about 70 amp-hours, specified at a current of 3.5 amps. This means that the amount of time this battery could continuously supply a current of 3.5 amps to a load would be 20 hours (70 amp-hours / 3.5 amps). But let’s suppose that a lower-resistance load were connected to that battery, drawing 70 amps continuously. Our amp-hour equation tells us that the battery should hold out for exactly 1 hour (70 amp-hours / 70 amps), but this might not be true in real life. With higher currents, the battery will dissipate more heat across its internal resistance, which has the effect of altering the chemical reactions taking place within. Chances are, the battery would fully discharge some time before the calculated time of 1 hour under this greater load. But wouldn't that mean you could hook up a 200Ah battery and ask it to put out 12000A for one minute or 720000A for a second? That seems very unrealistic, lol. I'm trying to find the proper kind of off grid battery that can power a microwave through a 3000 watt power inverter. The microwave needs 1800 watts and the battery needs to be 12 volts so that would mean I need about 150 amps, I'm wondering if a battery with a 200Ah rating could do it? <Q> Your hunch that batteries have a current limitation is correct. <S> In general, it's hard to tell the current rating [A] from capacity [A·h]. <S> You have to look it up in the datasheet. <S> For example: coin cells with 500mAh capacity may have only 3mA max current. <S> Another <S> (opposite) example: automotive starter battery with 40Ah capacity may have 500A max current. <S> Lead-acid batteries are interesting in this respect, because there are two distinct types. <S> Starter lead-acid batteries are designed specifically to deliver high peak current for a short period of time. <S> Deep discharge, however, dramatically shortens the life of a starter battery. <S> So, it's not suited for routine operation at high depths of discharge. <S> Your typical starter battery in the automobile works at very shallow depth of discharge. <S> Deep cycle lead-acid batteries are designed (as name suggests) to discharge further. <S> But they can not provide as much instantaneous current. <S> Here's an example datasheet for a deep cycle battery. <S> Have a look at the nominal capacity on p.1. <S> Notice that capacity depends on discharge current (i.e. the rate of discharge). <S> - Depth of Discharge Starter Battery <S> Deep-cycle Battery100% 12–15 cycles 150–200 cycles50% 100–120 cycles 400–500 cycles30% <S> 130–150 cycles <S> 1,000 and more cycles ( Source. ) <S> p.s. <S> If you want to read-up, here's an excellent web site on batteries - Battery University . <A> No, not necessarily. <S> Amp-hours tells you how much energy is in the battery, the amps tells you how quickly that energy can be delivered (For the most part, for a fixed voltage). <S> Your suspicion is well founded, because the Amp-hour doesn't tell the story on the batteries wattage capability. <S> Car batteries post their peak amperage because that is important when starting a vehicle. <S> For your purposes, you'll want to find a deep cycle battery. <S> Significantly discharging a car battery pretty much ruins it. <S> Deep cycle batteries are designed to be discharged to a much greater degree, down to as low as 20%, and those are rated such that they can deliver the rated current continuously. <S> You may be able to find a suitable battery, depending on how long you are going to run your microwave. <S> A decent 20+ amp rated deep cycle should still be able to unload 150 amps for a minute or two. <S> Don't run the battery full blast non-stop though, it will overheat. <S> Shop around for batteries for boats, RVs, etc to find good batteries. <A> In the answers above no mention was made regarding battery temperature environment. <S> If the battery will be outside, possibly in a camper, to be used at temperatures below freezing, for a cup of hot coffee, cocoa, etc., the capacity drops dramatically. <S> At 20 below F a car battery will deliver a fraction of its 77 degree capacity. <S> At 40 below F is will be down to a mere blip. <S> Auto service calls to jump start a dead battery in -40 degree F temperatures years ago would start with an 18V jump & if that did not work would go to 24V just to obtain a decent cranking speed to start the car. <S> All this is dependent on the overall condition of the battery in the beginning. <S> Older heavily used batteries will deliver still less capacity. <S> http://www.batteryuniversity.com/learn/article/discharging_at_high_and_low_temperatures <S> and http://alternative-energy.6pie.com/batteries/temperature-effects-on-batteries.php provide some good information on the topic. <A> The best explanation I have read or know of is as follows, and makes the assumption that you do not want to run your battery to less then 50%, as deep cycle batteries do not last long is you drain them to more then 50%. <S> So the quick answer for you is as follows <S> :Take the amp hour rating of your battery and half it. <S> (200ah/2= 100ah)This means your required load can't exceed 100A. <S> Your required load is 150A. (1800w/12v= 150A) <S> So without doing to much fancy maths but assuming a linear relationship, the quick answer is that a 200ah battery will not last long supplying 1800w load at 12v. <S> Working backwards, you would need a minimum of a 300ah battery to run your 1800w load. <S> (150A x 2)
| A lot depends on the design of the battery.
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MAX3232 overheating/burnt after connecting to PC I've got a MAX3232 Serial-to-TTL RS232 board which I am currently using in factory assembly lines. I don't have its picture, but here's a similar product: This is the circuit I've managed to figure out of the board(I don't have the original schematic): I'm using the RS-232 board to send streaming data from an electronic Weight scale to PC, where the data is grabbed by my software for display. The problem right now, is that sometimes the board stops sending data. I used HyperTerminal to check if the weight scale is sending data or not. When it doesn't, I've tried disconnecting and reconnection the serial port connection, unplugging the serial cable and plugging again, turning off the weight scale and even the PC on/off again and such, sometimes with no results. Sometimes the weight scale (WS) stops sending data. When I left if turned off for a while, and turns it on again, the WS sends data again. I suspect the MAX3232 IC got hot, because I can see the glue I put on the board looking like it's melted off the IC. When it cools down it is operational again. I've used an oscilloscope to check on burnt ICs - these ones clearly don't have output on pin 14 (refer to schematic), with voltage level nowhere near +- 7V. I've searched on the internet, and also found other people with issues regarding their IC (well this is actually MAX232) heating up when connecting to PC: MAX232 overheat problem MAX232 Overheat So far I've done some countermeasures: Replace the burnt/old IC with new MAX3232 ICs from element14. Since the boards were very cheap and made in China the IC were probably of low quality. Connected the unused CMOS input (pin 10) to GND with 2.2K Ohm resistor. Added 20 Ohm resistor in series to pin 16 (VCC). I'm afraid that these measures wont be enough. Maybe I need to isolate my connection. Does anyone have any idea what's going on, and know what to do? <Q> Just solved this overheating problem on a batch of max232 chips using a couple of resistors. <S> See the input pins 10 and 11 http://www.maximintegrated.com/en/products/interface/transceivers/MAX232.html and click the picture to enlarge. <S> There are supposed to be 400k resistors internally but measuring the resistance to pin 16, I got 40 megohms on this batch. <S> So these pins will be floating and picking up RF and oscillating at a high frequency which in turn will be drawing a lot of current and making the chip hot. <S> I used a couple of 390k resistors as pullups, but anything from 10k to 400k would probably do. <S> Might be worth adding 5K pulldowns on pins 13 and 8 too. <A> I have used these 10 x 15mm converter modules from AliExpress for more than 3 years now, probably around one hundred of them. <S> Just recently, on my last 20 or so circuits, I stumbled on this fire hazard issue. <S> I used all the advice I could gather from different forums and I want to share my full recipe in the order I tried them because I reached a really stable behavior which doesn't consume much current and is also protected somewhat: <S> First, I tried 10k <S> pull-up resistors (on some forums they say that unused TTL pins should be grounded, but I measured some UART pins and for me it makes more sense to pull them up). <S> This increased the stability a bit, but it did not solve it completely. <S> Next, in addition, I limited the current that module consumes by putting that 20 ohm resistor in series with VCC. <S> Here the behavior was the same as at the previous step, with the change that when the chip would latch up it would get warm, instead of very hot. <S> So I left it like this, considering it a minor improvement. <S> Next I added 100 ohm resistor on the RS232 pins. <S> Personally I did not notice a change, but there is at least another person on some forum saying that he could see improvements. <S> I tend to believe this, (I'm sorry I cannot mention him. <S> I'm bad with ID's after reading all the posts in a 3 google results page) <S> it will serve as a current limiting protection... <S> Next, I added a 47uF electrolytic cap on the power pins in between the 20 ohm resistor and VCC. <S> Again same story, somebody else stated that it fixed him. <S> Of course not mine. <S> I left it in circuit. <S> Next I added a 20 ohm resistor in series on the TTL pins. <S> No change, left it there. <S> May serve as a protection, not to draw too much current from the micro-controller pin. <S> I moved the power rail from 5V to 3.3V, keeping all the parts mention above. <S> This totally solved the issue for me. <S> At this point I had a procedure developed to make the thing latch. <S> I was powering the converter first, and then connecting RS232 line then disconnecting it and also removing the micro-controller link. <S> Now I tried all the things I could <S> and I cannot get it to heat up again. <A> I had this issue too. <S> Was using 1uf tantalum polarized capacitors. <S> If I unplugged the power while leaving serial attached, then reattaching power, pin 6, the negative voltage source, would go positive ~1v. <S> For whatever reason, adding a couple .1uf <S> ceramic capacitors for decoupling, in addition to what it called for, fixed the issue. <S> It was VERY reproducible. <S> Now I can't reproduce it. <S> Any EE able to explain? <S> Seems insane. <A> For my application I couldn't lower the power rail to 3.3V as suggested by one of the posters, as I was powering a device using the RTS/DTR pins which needed a higher voltage than the chip was producing at 3.3V. <S> I ended up using 10K pull-ups on the inputs plus 50 ohm in series with VCC and that solved the problem for me, while keeping the output voltage high enough for my application. <S> I probably could have also used a diode on VCC to drop the power rail to something around 4.0V, and that would probably have worked. <A> Don't use a 20 Ω resistor in the V CC line. <S> Place two (BAT85) <S> Schottky diodes as explained in AN218 from EXAR.If still RS232 latch-up add a 1k series resistor in the RS232 input line. <S> No more latch-ups.
| The chip would heat up and stop working.
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Why would the output voltage of a 5V SMPS (USB charger) increase with load? I've been using a USB voltage/current measurement device to evaluate power consumption of USB devices and performance of USB chargers and batteries. Most of the USB power supplies I've used are switched-mode power supplies which have a no-load voltage just over 5V and sag by varying amounts under load; however, I've noticed that one of my USB chargers, a Samsung "Travel Adapter" (5V, 0.7A rating) supplied with a feature phone, exhibits a significant voltage increase under load, until the supply's limit is reached. Here's what I'm getting: Amps Volts Error0.00 5.11 0.020.03 5.14 0.020.06 5.17 0.020.12 5.20 0.020.17 5.23 0.020.25 5.28 0.020.31 5.32 0.020.35 5.35 0.020.41 5.39 0.020.46 5.41 0.020.50 5.42 0.020.56 5.45 0.020.61 5.49 0.020.64 5.51 0.020.67 5.53 0.020.70 5.54 0.020.73 5.55 0.020.82 5.60 0.020.83 5.59 0.020.85 5.07 0.050.89 3.90 0.150.93 3.70 0.25 Notice that the voltage rises as the load increases up to 0.82A. Trying to pull more power than that causes the voltage to fall off a cliff as the limit of the power supply is exceeded. I find this behavior to be bizarre because no other SMPS I've worked with outputs increased voltage under load. What kind of SMPS design would cause the voltage to increase under load, and how would be it different from more typical designs? What advantages, if any, would this design carry? <Q> Sometimes a negative load regulation term is introduced on purpose in order to compensate for IR drop in the cable between the regulator and load. <S> This is common in automotive or industrial systems where the cable is long and bulky, thick cabling is undesirable. <S> If this regulator was designed for a specific phone, it is also possible that it's making up for internal drops in phone's power circuitry. <S> For example, if an LDO is used to generate a 5V internal rail, it's dropout voltage will go up as it's load current increases. <S> The phone designer could compensate for that and keep the LDO out of dropout by increasing the input voltage as a function of load... <A> Some cheap chargers don't use a regulator on the output side, monitoring the output voltage, and feeding back this information to the primary via an optocoupler. <S> Instead, they regulate based on what the "see" on the auxiliary winding on the primary side, saving the cost for the optocoupler. <S> On the aux winding of a flyback converter, you get a very approximate information about the voltage on the secondary winding, and sometimes, ringing and spikes dominate the picture instead of the theoretically ideal reflection of the output voltage. <S> If this is the case, the error actually would occur at light loads (because this is where the ringing might be severe), but is compensated for by decreasing the overall regulation such that the voltage is somewhat within the specification over the entire load range. <S> The details would depend on what the snubbers are optimized for, for example. <S> Cheapo USB chargers are sometimes reduced to an absolute minimum of components because the cost pressure is dramatic. <S> Muntzing is anything but an old-fashioned trick of the trade. <S> Searching for "primary side regulation in flyback converters" might lead to some background info like this article . <A> That's a really cool feature. <S> The 0.5V/ <S> A rise compensates for a 0.5 ohm cable drop. <S> Sure, 0.5 ohms sounds like a lot until you consider #28 wires, which are 65 milliohms per foot. <S> If you use a 3' USB cable made with #28 power wiring, this compensation is just about right. <S> Or, a 10' cable with #24. <A> The ratings never mean maximum value. <S> Ratings are based on some standard test conditions. <S> Also if you see the V - I curve, this seems more or less similar to a battery charging cycle except that V & I are reversed. <S> Usually in a typical battery charging cycle, the voltage rises up initially during which the current remains a constant. <S> After the battery's OCV (Open Circuit voltage) reaches a set point, the voltage remains a constant but the current reduces. <S> After some amount of time, the charging terminates. <S> But in your case, it is exactly opposite to a typical charging cycle. <S> Assuming you have reversed the order of the data, then there isn't anything wrong with the charger. <S> It is working fine. <S> You may find more details to the battery charging process here
| It could be that under mid- to high-load conditions, the ringing on the aux winding becomes less and the regulator increases the power just because it receives a decreasing ratio of the output voltage.
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Which is good method for step down 50Hz, 230V AC Which is low cost and safe for step down 50Hz 230V AC into 12V? Transformer or reactance or Regulator IC ? simulate this circuit – Schematic created using CircuitLab <Q> Use a transformer for the front end if you don't want to kill yourself. <A> Which is low cost and safe for step down 50Hz 230V AC into 12V? <S> The key word in the question above is "safe" and this means using a method that inherently provides isolation. <S> If you want to transfer some reasonable amount of power (rather than using an optical method that can transfer maybe 1 milli watt), you are forced down the road of using a transformer. <S> This is how I read "safe" in this context. <S> In this context I assume that safe means isolation so that anyone touching the reduced-voltage output is not going to receive an electric shock. <S> Other safe methods involve providing sufficient insulation around everything that connects to the circuit - under these circumstances you don't need to use a transformer BUT be prepared, when building the thing to get your fingers scorched by electricity. <S> If using a transformer means more cost then this price is worth paying for reasons of safety and prevention of litigation. <A> From my experience, one of the common ways of stepping down 230V-AC(50 Hz) is by using a step down (230V to 12V) transformer followed by an LM7812 IC to output 12 Volts DC. <S> Between the transformer and the LM7812 IC, you can also use some filtering capacitors. <S> If your application supports, you may just use a 9V (DC) battery avoiding the use of transformer. <S> If i am wrong , feedback's would be appreciated.
| A non-transformer method to reduce voltage can provide a fair amount of power (such as when using a switching buck regulator type of circuit) but it is still connected to live mains and can electrocute if not handled properly. And note that, you can use 9V battery, only if your working on some low power applications, or else the battery will drain out soon.
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Difference between capacitors of same value but different sizes I would like to know why some capacitors have the same value (capacitance) but their sizes are different? What is different between those capacitors? <Q> They will probably have different dielectric, meaning different working temperature and tolerance. <S> See table here: http://en.wikipedia.org/wiki/Ceramic_capacitor#Class_2_ceramic_capacitors <S> Also, bigger capacitors will usually have higher voltage rating, they cool down better. <S> It also might be age (caps get smaller with years) or manufacturing capabilities. <S> For example of the latter: if you were to buy strictly "Made in Russia" parts, you'd have to tolerate with much larger packages for the same thing, say, Murata makes. <S> Sometimes (or even usually) there is no real difference, so you can choose depending on the size itself: if you solder by hand, bigger size can be an advantage. <S> I also remember reading one interesting app-note, focusing on Capacitance as a function of DC Voltage. <S> Generally, physically smaller caps "degrade" more. <S> You can find it here: http://www.maximintegrated.com/en/app-notes/index.mvp/id/5527 <A> Voltage ratings, ESR/Q value, aging, temperature stability, price, packaging convenience for automated pick & place, etc. <A> Dielectric. <S> Dielectric is the material used between the plates of a capacitor. <A> For a given (fixed) set of constraints: Manufacturer , Manufacturing technology , Dielectric type , Target application , i.e.: decoupling, general purpose, high-frequency or power line filtering, <S> Mounting style , i.e.: SMD, through-hole or chassis, Capacitance value , The only feature that requires increasing the size of a capacitor is its voltage rating . <S> Reasoning the other way around, You can trade off a smaller voltage rating of the capacitors in your design for a smaller package size (assuming the set of constraints above). <A> The capacitance C between two plates of area A, separated by distance D, having a dielectric with relative permittivity Er is... C = <S> Er <S> * E0 <S> * A <S> / d <S> Where E0 is the permittivity of free space. <S> If the plates each have thickness t <S> then the volume V of such a capacitor is ... V = <S> A <S> * <S> (d + 2 * t) <S> Ceramic capacitors are made of many very thin layers of alternating metal and dielectric stacked together. <S> If a ceramic capacitor has N plates then it has a total volume V of... <S> V = <S> A <S> * <S> N <S> * t <S> + A <S> * (N-1) * d <S> Each dielectric will have a different Er value. <S> For example, X7R may have Er of 3300 but may drift 15% over temperature. <S> Whereas NP0 may have Er = 120 <S> but may drift only 30ppm/C. <S> http://www.ferro.com/Our+Products/ColorsGlass/Electronic/Multilayer+Materials/Dielectric+Formulations+for+Ceramic+Capacitors.htm <S> Therefore the total volume of the capacitor depends on what dielectric is used and how thick we make the electrode plates. <S> If you want the capacitor to handle more current or have lower ESR then the thickness of the metal layers needs to be increased. <S> The breakdown voltage of a dielectric layer is proportional to the thickness of the layer. <S> Therefore making thicker layers may create capacitors with larger voltage ratings. <S> The choice of dielectric involves a trade between how much capacitance we need in a given area, how much capacitance drift we can tolerate vs. temperature, and the required voltage rating.
| The plate size and material and dielectric materials have varying characteristics that make for the different sizes and voltages ratings.
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Is Back EMF relevant in dynamos, where there is electric generation? While doing some research about dynamos, I came across a situation wherein there was a question about the validity of my dynamo design. The counter to the design I suggested was that there would be back EMF generated by the design, but the design I have under consideration is that of a dynamo, where we generate electricity. I understand Back EMF is used in cases of motors, and not dynamos, is this the correct picture? The design of the dynamo is briefly described as follows: an outer rotating magnet that has a groove in which a fixed stator can fit in closely, of course, there is an air gap, and so on, as with standard dynamo design, but I did not see how anything like Back EMF, could really destroy the design under consideration, is this the correct? <Q> There is a back-emf even in dynamos that can be considered. <S> A perfect dynamo will produce a terminal voltage related to the speed that it was spinning and this also is the same for an imperfect dynamo. <S> However, under load, the imperfect dynamo has an extra factor to take into consideration; sure, as the electrical load increases, the dynamo shaft is harder to keep turning at the same speed (conservation of energy) but the current taken from the dynamo also flows through a few percent of copper wire turns that are not generating a voltage - this is called leakage inductance (also applies to transformer windings) and the current will produce a back-emf in those turns that reduces the forward emf due to the perfect dynamo effect. <S> Maybe this is what the OP is considering? <A> Whenever there is a field is built up in a coil, it eventually has to collapse. <S> This collapse results in the magnetic field being converted to "power". <S> This "power' will flow into every conductor connected to the coil. <S> Diodes are typically used to block this "power" from going where it shouldn't. <S> Read up on inductors. <S> There are well documented ways of dealing with the issues you appear to have. <S> Also, I am blind to the design you speak of, as I cannot see it. <A> It's the same thing; but that's the whole point of a dynamo... <S> To expand on this a little : any machine rotating a coil in a magnetic field will act as a voltage source across that coil, and that voltage is dependent on the speed. <S> (Also on the magnetic field and the construction, number of turns on the coil etc, but the speed is easiest to vary). <S> Now imagine there is also a voltage on the machine's terminals, and consider the current flowing between these voltage sources. <S> This current will be the difference between the two voltages, divided by the coil's resistance. <S> If the external voltage is greater, the current flows into the machine <S> Then we call it a motor. <S> Current times voltage dropped across the resistance is wasted heat; current times "back EMF" is work done by the motor. <S> If the external voltage is less, then current flows out of the machine, and we call it a generator. <S> Now current times "back EMF" is work done by you turning the shaft, and again that dropped across the resistance is wasted heat. <S> The rest (current * external voltage) is the power output of the generator.
| In a dynamo it's not called "back EMF" - just EMF!
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How to read this basic schematic? I need to build a circuit where light is converted to sound. The light falls on a CDS cell and the current is passed through the NPN transistor to amplify it. I know this is a very simple schematic, but what would be the real life model of this? Why is the resistor going to the ground? <Q> That is a really crappy circuit since it depends on the gain of the transistor being just right. <S> Real transistor gains vary widely, even within the same production lot. <S> Competently designed circuits work from the minimum guaranteed gain of the transistor to at least 10 time that, preferably to infinite gain. <S> I would start with the transistor stage setting its own DC bias so that the output voltage is near the middle of its range. <S> The signal from the LDR would be AC coupled into that stage. <S> The LDR would have a pulldown at least so that it and the pulldown produce a voltage signal by themselves. <S> For extra credit, come up with a auto-biasing scheme that keeps the output near the middle of its range regardless of average ambient light level. <A> It is a simple non-inverting transistor amplifier. <S> The resistor to ground is what allows it to be non-inverting. <S> When the transistor is off (the CDS cell not conducting), the Headphone output at the emitter of the Transistor is pulled to ground level through the 47Ω resistor. <S> When the transistor is on, the headphone output gets pulled up to 3V, with the resistor limiting how much current goes through the transistor, and preventing a direct short. <A> This is known as an emitter follower. <S> There is no voltage gain in the circuit, this acts as a impedance buffer i.e. it can drive heavier loads that the CDS cell could on <S> it's own. <S> Light on the CDS cell generates current which goes into the base of the transistor, this current is multiplied by the transistor (this gain is known as Hfe). <S> This current when forced through the resistor generates a voltage which is your output signal. <S> Your Hfe (from some old data sheets) varies but is at least 40 (and can be as high as 100). <A> The real world application varies upon the application and the type of the signal. <S> If linearity is not a concern, then the above circuit diagram can be improved to remove the 47(ohm) resistor and the circuit will act as a digital circuit, with almost no sound when there is no light. <S> On the other hand if linearity is a concern for analog applications, then the above is suitable to provide analog output specially if the amplitude of the light is to be varied on both sides of the normal value (i.e. if the value changes both above and below the dc value). <S> Just check up non-inverting transistor amplifiers for more information for the circuit. <S> A lot depends upon resistor values of the CDS cell and the resistance of the headphones. <S> Moreover, you might want to connect a capacitor between the output and the headphone to dc couple the output.
| This would be the signal fed into the amplifier stage thru a coupling capacitor.
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how does current go to infinity in an ideal LC circuit at resonance? Consider an ideal LC series circuit, excited by a sinusoidal source at resonant frequency. Assume zero initial conditions, for capacitor voltage and inductor current. simulate this circuit – Schematic created using CircuitLab 1) Can anyone explain how does the current amplitude go to infinity, at resonance? I understand the impedance goes to zero,which explains the steady state current as infinite.But can you explain the same by simple circuit analysis?I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite? After all for a simple L only circuit, the voltage across inductor is Vsinωt = L di/dt and it is not getting infinite 2)Now at resonance what happens internally, from an energy perspective?Is there any transfer of energy between inductor and capacitor OR Source is throughout supplying energy to inductor and capacitor? 3)Will the nature of current differ if we have initial conditions? <Q> I understand the impedance goes to zero,which explains the steady state current as infinite. <S> But can you explain the same by simple circuit <S> analysis?I mean the voltage across inductor is <S> Vsinωt - Vc = <S> L di/dt, so based on derivative being positive, the current increases. <S> But what makes the current go to infinite? <S> Yes, the net impedance falls to zero in a series resonant circuit when R is zero and, the voltage across each of L and C also rises to infinity <S> - it couldn't be anything else or you would not get infinite current. <S> That's the simple circuit analysis - trying to equate \$V_L\$ to the input voltage <S> minus \$V_C\$ doesn't really help other than remind you that the two voltages (across L and C) are oppositely phased <S> i.e. they are 180 degrees apart. <S> Should you think of the driving source as a constant current (just to prevent infinities and keep things sensible) <S> that current appears to conduct thru a dead-short-circuit at resonance and that current will produce an equal finite voltage magnitude across each element determined by the components impedance (equal at resonance). <S> The current will reach infinity non-instantaneously because resonance is at a pure sinewave and a pure sinewave has to have been in existence a very long time ago! <S> Energy transfer is exactly the same as when not exactly at resonance or with a little bit of resistance except the numbers are infinity (not helpful to analyse in my book). <S> There are initial conditions - <S> the application of a sinewave via a switch is an initial condition but, if you mean a dc voltage across the capacitor then this will continue to be present (on average) because the energy contained in it has nowhere to be dissipated (neither the perfect voltage source nor inductor) can do this. <S> In a real circuit this DC "energy" will be burnt in the non-zero resistance of the wires. <A> Can anyone explain how does the current amplitude go to infinity, at resonance? <S> The brief answer is that the AC steady state current isn't infinite but, rather, the circuit has no AC steady state solution . <S> Recall that one of the assumptions justifying AC (phasor) analysis is that the circuit is in AC steady state , i.e., that all transients have decayed. <S> For the circuit given, the time domain solution for the current is proportional to $$i(t) \propto t \cos\left( \frac{t}{\sqrt{LC}}\right),\, t \ge 0$$ <S> Note that this solution has no sinusoidal steady state <S> - the amplitude does not approach a constant as \$t <S> \rightarrow \infty\$ so this solution has no phasor representation and, thus, we should not be surprised that applying phasor analysis to this problem produces an undefined division by zero result. <S> Given the solution for the current, one can solve for the voltages across the inductor and capacitor as well as the energy stored in each as a function of time. <S> Different initial conditions will have different initial energies but will not affect the main result that the amplitude of the current will grow without bound once the switch is closed. <S> Can you please list the steps that lead to the derivation of the expression i(t)∝ tcos(t/√LC),t≥0 ? <S> By KVL, we have $$ <S> v_S = v_L + v_C = L\frac{di}{dt} <S> + \frac{1}{C}\int_0^ti(\tau)d\tau$$ <S> (we assume zero initial voltage across the capacitor). <S> Differentiating both sides with respect to time and dividing through by \$L\$ yields $$\frac{d^2i}{dt^2} + \frac{1}{LC}i = <S> \frac{1}{L}\frac{dv_S}{dt}$$ <S> Assuming <S> \$v_S = V\cos\omega_0 <S> t\$ yields the following non-homogeneous 2nd order <S> ODE: $$\frac{d^2i}{dt^2} <S> + <S> \frac{1}{LC}i = -\frac{\omega_0 V}{L}\sin \omega_0 t $$ <S> where <S> $$\omega_0 = \frac{1}{\sqrt{LC <S> }} $$ Assume a solution of the form $$i(t) = <S> t\left(A <S> \cos \omega_0 t <S> + <S> B \sin \omega_0 <S> t \right) $$ Substitute this \$i(t)\$ into the ODE to find $$A = \frac{V}{2L}, B = 0 $$ <A> I mean the voltage across inductor is <S> Vsinωt - Vc = <S> L di/dt, so based on derivative being positive, the current increases. <S> But what makes the current go to infinite? <S> After all for a simple L only circuit, the voltage across inductor is Vsinωt = <S> L di/dt <S> and it is not getting infinite <S> In an L only circuit, the voltage across inductor is $$ V\sin(\omega t) = <S> L \dfrac{di}{dt}$$This means current is either increasing or decreasing (as per sine wave). <S> In the case of resonance, voltage across inductor is $$ V\sin(\omega t <S> ) - V_c$$ <S> Here the key point is that the source voltage and capacitor voltage need not be of same polarity. <S> Infact, they will be almost 90 degree out of phase, which makes the difference more than that of a simple 'L' circuit. <S> This difference becomes more and more as time evolves. <S> It looks like capacitor stores energy from source and adds it with source in next cycle to increase the current more. <S> Increased current means more energy storage in inductor, which later supplies back to capacitor which results in more charging(more voltage) <S> and the process will reinforce each other.
| The amplitude of the current starts at zero and grows linearly with time once the switch is closed but for any value of time \$t\$, the current is finite , i.e., the current is never infinite.
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Lamp flickers if dimmed up to an extent I used circuit shown in this instructable to make an AC dimmer but everything goes fine if not dimmed much but then it starts flickering because of voltage fluctuation. http://www.instructables.com/id/Arduino-controlled-light-dimmer-The-circuit/?ALLSTEPS EDIT - embedding circuit <Q> A few things to try- and take extreme care as the author suggests- <S> no shortcuts. <S> short out the indicator LED in case the MOC is not getting enough drive. <S> I don't think this us your problem, since your flicker varies with the dimming, but as he mentions it it's worth a try. <S> make sure your load is 'simple' like an incandescent lamp. <S> Get it working first with that before trying CCFLs or LED lamps. <S> If you're using an incandescent bulb, try doubling the wattage of the load and see if it improves. <S> The 1K series resistor is very high-- <S> you could try reducing that to 1/2 or 1/4 and see if it improves. <S> (I would have put the load on MT2 rather than MT1 to minimize the dv/dt at the opto coupler, but I doubt that's related) <A> <A> I struggled same. <S> Please check your interrupt it should be at rising edge not at falling.1k ohm and 220 ohm are perfect <S> do not change them.you can swap mt1 and mt2 if triac is not firing at all.
| Assuming you are using an incandescent bulb, flickering can be caused by only triggering the triac on the one half cycle...might need to check your code.
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Powering 12v led strip from 16v dc I want to power up about 2 meters of led strip from a 16.5v dc battery (4s lipo). I know there are dc-dc converters, but I rather avoid using them due to weight (rc plane) The lights will not be used very often, probably they will stay on for about 30 minutes. I have read somewhere that I could plug few diodes in series with the strip, could this be a good idea? Are there any precautions to put in place (do diodes overheat)? I am basically looking for a solution that works, not for the best practice, performance or top reliability... Thank you! <Q> The easiest way, then, is to find out how much current the LED strip draws with 12 volts across it, and then to connect 7 silicon (not Schottky) diodes rated for that current or greater (greater is better) in series with the LED strip and the LIPO. <S> Depending on how much current they have to handle, they could get pretty warm, so you might want to place them somewhere where air can blow over them to cool them. <S> If you use 1N4001 or 1N4002 diodes, figure about a 0.8 volt drop each with 750mA through them, so you'll need about 5 or 6 in series between the LIPO and the LED strip, like this: <S> +--[DIODE>]--[DIODE>]--[DIODE>]--[DIODE>]--[DIODE>]--[DIODE>]--+ +| |+ [LIPO] [LED STRIP] <S> | <S> | <S> +--------------------------------------------------------------+ <A> Common led strips take about ~18mA at 12V per segment (typically 3 LED). <S> They are used in Automotive 12V rails, which is 12V nominal , but can go to 14V regularly, with larger spikes. <S> If you look at a segment, you will notice it has three LEDs and one resistor. <S> $$ <S> I = <S> \frac{V_s - <S> V_f}{R}$$ <S> Assuming a white or blue led with ~3.2V forward voltage drop, which is typically paired with a 120Ω resistor: $$\approx <S> 18mA = <S> \frac{12V - (\approx 3.2V \times 3)}{120Ω}$$ <S> At 14V, you get: $$\approx <S> 36mA = \frac{14V - <S> (\approx 3.2V \times 3)}{120Ω}$$ <S> This isn't exact. <S> The higher the source voltage and current goes, and <S> the LEDs forward voltage drop does change a bit. <S> At ~36mA the LEDs are being over driven, shortening their life some, so you won't get 1000~5000 hours that they should give at 20mA. <S> But its normal to overdrive them. <S> And they will be brighter to boot, so you might even need less LED segments. <S> But now lets change the source voltage to 16V: $$\approx <S> 53mA = \frac{16V <S> - (\approx 3.2V \times 3)}{120Ω}$$ <S> 53mA continuous is almost 300% of the typical recommended current of 20mA. <S> But at 53mA, the forward voltage drop goes up as well, making the math tricky. <S> Solution: As others have mentioned, you can use Silicon Diodes to drop the source voltage. <S> You can use from 3 to 9 (Dropping 2.1V to 6.3V), depending on how bright you want the leds, how much current the strip should pull, your batteries voltage range (A 4S Lipo is 14V Nominal, 16.5V <S> at max safe charge, 11.5V <S> at lowest safe discharged voltage). <S> Best thing to do, is take one or two segments of the led strip and TEST THEM. <S> Connect one directly at the charged 16V and see how long it lasts or how hot they run. <S> Look at the resistor and do the math. <S> Then start adding Silicon Diodes and see the difference. <A> I know there are dc-dc converters, but I rather avoid using them due to weight (rc plane) <S> Although you say that you don't care for such a solution I would like to add it as an option. <S> A switching step down regulator module such as LM2596 weights just 11.5 grams (costs about a dollar from ebay). <S> The benefits of such a regulator are : better output regulation (constant voltage at the output irrelevant of load changes) compared to the passive method (diodes) high efficiency which will generate less heat and waste less energy from the battery and you get the short circuit protection as a bonus
| Based on the color of the led and ohms law, you can figure out it's current at any given voltage. The life of the leds will be much shorter, and they will warm up considerably.
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CPU heat-sinks for cooling Transistor? Can I use CPU heat-sinks for cooling transistors? CPU heat-sinks are low volume and they have been made to cooling a sensitive hot Chip and if the datasheet has said that it requires to hold on the transistor less than 150 C,I thought maybe it will be possible and good way instead of massive heat-sinks. the CPU heat-sinks are always available in the markets But maybe finding the heatsinks with special shape and enough size are slightly difficult.Are these fans and heat-sinks suitable and normative for transistor?If so, How should I choose heat-sink?(for example:for one 100 watt transistor Is the 100 Watt CPU heat-sinks suitable and enough?) <Q> To appropriately decide on a heat sink you have to consider these parameters from your datasheet: From what I've read but never tried, you can even figure out necessary parameters using a circuit theory analog of sorts. <S> Check this this link. <A> The simple answer is yes. <S> I do this, as CPU heatsinks are cheap and big. <S> But, to get a proper thermal coupling between the semiconductor package and the heatsink, you may need to drill and tap a hole in the heatsink to screw them together, as CPU heatsinks tend to clip down to the motherboard. <S> Also, a couple of holes to bolt it to some support are probably essential. <S> As to the thermal conductivity, CPU heatsinks are cheap enough to massively overspecify, but then I am not trying to miniaturise anything. <A> Yes and no. <S> Yes, because ultimately all heatsinks serve same purpose: move heat from an element into surrounding air. <S> No, because a power transistor works in a different environment than a CPU. <S> First, the mounting doesn't match, both the transistor to the heatsink and the heatsink to the PCB. <S> Second, a transistor heatsink usually operates passively at relatively high temperature (often over 100 C), while a CPU heatsink is designed to keep the temperature as close to ambient as possible at the cost of noise and dust accumulation from the fan. <S> So you can't remove a fan from narrow-finned CPU cooler, because in passive mode it won't work as good as a dedicated transistor heatsink with large and sparse fins. <S> If you keep the fan, it becomes a point of failure, because CPU can handle fan failure gracefully by throttling down, while your circuit is unlikely to have temperature sensors and shutdown-when-overheating procedure. <S> For prototyping you can overcome some of those limitations (eg. by drilling new mounting holes), and just ignore the others (eg. fan failure and dust accumulation), but if you want to have a design that'll keep running reliably, you'd better buy (or design) a heatsink dedicated for the job.
| Sizing heat sinks is actually a function of the thermal characteristics of your device. As long as it can move enough power, it's fine.
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Driving EL wire with Supertex HV850 family drivers I was considering driving EL wire with Supertex'es ICs or some equivalent ICs from another manufacturer. The ICs are designed to drive EL panels not EL wire. Would this still work? Supertex'es ICs are rated for "lamp sizes" up to 42 square inch. My wire is 2.3 mm in diameter and 20 meters long. I'm going to assume 2.3 mm is the diameter including the outer protective sleeve and the actual diameter is probably somewhere around 2 mm. Looking up the formula for a cylinder's surface area I determined that my EL wire has surface area of about 2513 cm² (about 390 square inch). That's way too much for those ICs. According to this calculation I could drive up to 2 meters of EL wire with these ICs. But is this calculation applicable to EL wire at all? <Q> The area is not really what loads the driver- <S> what matters is the total capacitance (and the voltage, and frequency). <S> The wires have a spiral wire on the outside which doesn't have as much capacitance per unit area as a flat panel. <S> The HV816 claims to be able to drive 150nF, which they say is about 42in^2 of EL backlight. <S> From this link, they claim the capacitance of their EL wire is about 1nF/foot, so 20m should be about 66nF, in which case the Supertex HV816 should be able to handle the capacitive load, and if the voltage and frequency are right, it should be okay. <S> Check the specs on the EL wire you are using (or measure it). <A> The area of a EL display/lamp/wire is just a byproduct the Capacitance per square inch, which you can find on the EL Wire's datasheet. <S> The Supertex HV816 is a high voltage Electroluminescent (EL) lamp driver designed for driving a lamp capacitance of up to 150nF, or an area of approximately 42 square inches. <S> The other two factors are the voltage, and the frequency. <S> The HV816 has a frequency output of 100Hz to 1kHz. <S> EL wire is typically stated to require a higher frequency than EL panels (1~2kHz vs 300~500kHz). <S> The HV816 also has a voltage output of 360VPP, which again, EL wire is typically stated to require a higher voltage than EL panels (300VPP vs 150VPP). <S> As to why that's required, I have yet to get a solid answer on my question on that topic. <A> also it is worth mentioning that EL Wire also has a maximum VPP which isn't much more than 300V, I have had some fail at 339V.The thinner <S> the dielectric ie angel hair the larger the chance of a failure. <S> General rule, if it looks "patchy" then the frequency is much too low. <S> I have run it up to 7 kHz and although it was crazy bright <S> the wire didn't last long (less than an hour before a catastrophic failure) <S> Some folks run this directly from the mains, <S> NEVER DO THIS!!!!!always use an inverter even a simple one, a good circuit current limits input voltage and then increases frequency to around 500Hz.some even have a trimpot so you can set this parameter as it affects wire color and brightness on the green/blue wire. <S> Have also used the HV850, it needs input voltage regulation to work properly ie 4.2V <S> and if you try and run it directly from a lithium cell typically there is still useful power left well before the chip shuts down through under-voltage. <S> It does work well for backlights though with my own modification of a frequency feedback using CdS photoresistor so the backlight is brighter during the day when needed. <A> also worth adding a series resistor. <S> From experimentation 33K seems to work and limits drive current to a sensible level.
| The main consideration in driving an Electroluminescent display or wire is the EL Capacitance, Frequency, and Voltage.
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Can I solder a wire between two leads on a PCB to make up for a trace that isn't wide enough? I have a PCB that has 2 traces on one corner who's width only supports a ~10a load. I need support for a 20a load. Actually, it really only needs to support a 15amp load but I'm trying to build in some healthy breathing room. Can I just solder a wire between the two leads in parallel with the trace? The terminal block and the relay both support 20amps and these are the only parts touched by the load. I know it's messy but I figure it's not as messy as having to wire all of this up off the PCB. <Q> Yes. <S> Fixing production errors by using jumper wire is a time honored tradition by those blessed with hindsight. <S> For Power rails or low speed GPIO <S> (Think Push Buttons inputs or LED outputs), a simple wire would work. <S> As a precaution, you could cut the trace, and carry the full load on the jumper you are adding. <S> If the jumper fails, the load would try to go over the 10A trace, and can cause issues (i.e. fire). <S> High Speed Signaling or Protocols with tight tolerances, RF signal traces, Traces that require uninterrupted ground/vcc planes under them, matched impedance, etc, require significantly more thought into it. <S> Then again, some have wider tolerances than stated. <S> There are many hobbyist projects/hacks that add high speed USB ports to a board with a CPU that supports it, using nothing but regular wires and protoboard, not to spec. <A> That is one option if your design can handle the additional (capacitive/inductive) impedance created by the wire. <A> I would recommend cleaning the trace and end connections with sandpaper, laying clean copper wire on top of the trace between the terminals, then cover the top of the wire with solder, from one terminal to the other, making sure the solder flows between the wire and the trace.
| Another option if the trace has no solder mask over it is to build up a large amount of solder along the trace in order to increase its ampacity.
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Very low-quiescent (<2mA), 5V, simple "speaker" driver How can I drive a 30ohm speaker from a 5V microprocessor in a simple, very low quiescent-current (<2mA) manner? The "speaker" is actually a coil, not a speaker - and I'm trying to induce a magnetic field, like in an audio hearing loop. It's driven from a PIC microprocessor, primarily with brief, digital (rail-to-rail) pulses, but occasional DAC/analogue signals. It also needs to have separate gain control (that I can control from the PIC - there are plenty of spare legs) and (ideally) disable the whole thing entirely. I have a 5V, 200mA single rail supply. The coil is 30ohms. Ideally I'd like to be able drop up to 150mA through the coil in short bursts. Most important thing: low quiescent current, even when it's enabled. Preferably <2mA. Important things: simplicity, compactness, and low-cost Unimportant things: fidelity/distortion, temperature stability I've looked at audio amps (which seem over-complicated, and Quiescent too high), standard op-amps (but they seem to have low current source/supply) and various BJT options (some kind of emitter follower?) but this analogue stuff is a little beyond me. <Q> If you want to get 150mA with a 30 ohm coil (the resistance will increase with temperature) <S> you'll need no more than a couple ohms switch resistance. <S> One way to get that would be to use two complimentary pairs of MOSFETs in an H-bridge configuration. <S> For example, the DMHC3025LSD contains all four transistors and is less than 0.2 ohm for the total high+low switch resistance. <S> To get a fairly clean 1kHz you'd need something like 25kHz-50kHz minimum PWM frequency, so you'd probably need a gate driver for the MOSFETs rather than being able to drive them directly from a microcontroller PWM. <S> If you drive them both high or both low, the current consumption will be minimal (microamperes). <S> For an analog type output you would drive them with complimentary PWM waveforms to get bipolar output. <S> It might be difficult to get much fidelity unless you're using close to full amplitude. <S> If the inductance of the coil isn't enough to smooth the waveform at the PWM frequency you could add an LC filter, but that will affect the 'digital' waveforms too. <A> What frequency, turn on/off time? <S> If it's slow (~1ms) you could use a SSR (solid state relay)For higher switching speeds you could look into analog switches. <S> Here's an early hit on a digikey search. <S> http://www.ti.com/lit/ds/symlink/ts5a4624.pdf <S> But there are thousands to choose from. <S> Gain control could be PWM, or if not take the dive into analog electronics, it's fun. <A> <A> After some further experimentation, I've decided to use a single BJT emitter follower, biased into forward-active to allow analogue input. <S> However, I've added an extra line to a standard IO on the PIC. <S> When set as an output, this will serve to drive digital signals, and when set as an input, it will revert to analogue. <S> Gain will be controlled by a digital pot in place of (Re) <S> As far as I can tell this serves all my requirements - output fast sharp digital as well as clean analogue, low quiescent, and very simple/low-cost. <S> simulate this circuit – <S> Schematic created using CircuitLab
| I would think that any of the smaller H-bridge drivers for small motors would be pretty much ideal for the application as you describe it. You can drive the coil with pulses of either polarity, and you can use "class-D" or delta-sigma techniques to create analog waveforms — the inductance of the coil itself will provide much of the required filtering.
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What happens if SINR is < 1? What happens if the SINR of a signal is less than 1 in a (wireless) network? Is that practical at all to have a SINR < 1? Using the Shannon capacity formula, it seems that we must be able to have SINR < 1, but we get a rate < 1 bits/sec/Hz. <Q> Yes, it is practical. <S> For example, if you have an ASK signal with modulation depth 60%: <S> > <S> > am = <S> [ (ones(1,100) * 0.2) <S> (ones(1,200 <S> ) * 0.8) <S> (ones(1,200 <S> ) * 0.2) <S> (ones(1,100) * 0.8)]; Using a low-pass filter > <S> > d = <S> fdesign.lowpass(0.01, 0.02, 0.01, 100) <S> ;>> <S> hd = <S> design(d)hd = <S> FilterStructure: ' <S> Direct-Form FIR' Arithmetic: 'double' Numerator: [1x946 double <S> ] PersistentMemory: <S> false you can reduce the output signal bandwidth in order to be nice to the neighbors: > <S> > <S> amlp = filter2(hd. <S> Numerator, sig); <S> The recipient gets a noisy signal > <S> > <S> amno = amlp + 2*rand(1,600) <S> - 0.5 <S> ; and also uses a lowpass filter to reconstruct it: >> amre = filter2(hd. <S> Numerator, amno) - mean(amno) + 0.5; <S> This signal is sufficiently similar to the original signal that you can decide between 0 and 1 bits, but you need a rather narrow filter here in order to remove the noise -- in my case, 1% of the sampling rate (that is the .01 <S> above). <S> Note that we're only interested in the signal at the points 50, 150, 250, ..., 550, i.e. the middle of each symbol. <S> In order to be able to reconstruct that signal, I had to use rather long symbols (100 samples). <S> That is, with 100 Hz sampling rate, which would allow me to express frequencies up to 50 Hz, I can only transfer 1 bit per second. <A> There are plenty of examples where SINR (or SNR) is less than 1 and still quite usable. <S> The signal on an antenna (that you want to receive) maybe only 1 microvolt but all the other signals you don't want (from 50Hz upwards) totalize a few millivolts - radios manage to overcome this problem by filtering. <S> Even after filtering at a particular frequency for a wanted frequency, the result may still be SNR<1 <S> but, if the "system" is a direct sequence spread spectrum (DSSS) transmission that doesn't matter because the sum of all the different frequencies transmitted at means that noise cancels. <A> See, for instance, <S> http://www.gpssource.com/faqs/15 which discusses the SNR of GPS signals (typically about -26 db, or ~ <S> 1/400.
| Not only is it practical, it is commonly done.
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Meaning of temperature coefficient for resistors Is temperature coefficient of a resistor about the temperature of the resistor or ambient temperature or are they the same thing? What I mean to ask is that: Lets say I have a resistor and I have no data sheet about it and I want to find the temperature coefficient. Can I use the ambient temperature in calculations? Is the resistor temperature same as the ambient temperature after a while or is that always much higher? <Q> The temperature coefficient specification ostensibly provides a limit for the change in the resistance of a resistor from its nominal value at temperature \$T_0\$ to another temperature <S> T. <S> In fact it's usually defined using a box method at temperature extremes and does not really guarantee the slope of the temperature-resistance curve as you might assume. <S> Ideally, a maximum temperature coefficient of, say, 10ppm/°C would mean that if our 1.00K resistor measures 1.0015K at 25°C and the temperature changes to 35°C then the value should somewhere between: \$1.0015K + 1.0015K (35°C - 25°C) <S> 10 <S> ^6 <S> \cdot <S> 10ppm/ <S> ° <S> C \$ and <S> \$1.0015K - 1.0015 <S> K (35°C - 25°C) <S> 10 <S> ^6 <S> \cdot <S> 10ppm/ <S> ° <S> C\$ Or 1001.5\$\Omega\$ <S> +/- <S> 0.1005\$\Omega\$ <S> It doesn't matter why the temperature changes- ambient, self heating, nearby components, or some combination. <S> If you are trying to measure the actual temperature coefficient of a resistor, you can measure the resistance at two widely separated temperatures, using low enough current that self-heating is minimal (note that it cancels out to a first order if you allow it to settle out- also pulsed current can be used and the measurement made before the temperature changes much) and calculate the tempco as: Temperature coefficient = <S> \$\frac{R_X - <S> R_0}{R_0(T_X-T_0)}\cdot <S> 10^6 <S> ppm/ <S> ° <S> C\$ <S> If the tempco is large you might want to use the average resistance rather than \$R_0\$, but it shouldn't matter much in most cases. <S> Edit: Regarding the situation you mention - 0.2% change for a change in power dissipation of about 100mW.. you need a better resistor and probably a larger one that won't heat as much for a given dissipation. <S> Consider a 1206 249 ohm Susumu resistor. <S> P/N: RG3216P-2490-B-T1. <S> Tempco is +/-25ppm/°C <S> , it will increase by 15-30° <S> C at 20mA <S> depending on layout (see the link). <S> That should represent a change in resistance of about 375-750ppm or maybe 3-5x better than you are measuring. <S> If you need even higher accuracy, you could use a bigger resistor, several smaller ones distributed with copper around, or use a smaller value resistor and amplify the signal so it doesn't get as hot. <S> You could also use a Z-foil style resistor such as Y1630250R000T9R that has only 0.2ppm/K tempco, but they are pretty expensive (>$10 each). <A> It's not about the temperature of the resistor, its about the change in the resistance as the temperature changes, <S> so it's the ratio of the fractional change in resistance to the change in temperature at a certain temperature. <S> In order to measure it, you need to measure the resistance at some particular temperature and then increase the temperature(may be by passing a current through it) and measure the resistance again, then put values in the formula <S> and you'll get your coefficient of resistance. <S> Here's more about it. <A> The resistor temperature can and will differ from ambient temperature, especially if the resistor is dissipating power. <S> You might need fairly sensitive gear and wide temperature differences to measure the temperature coefficient
| The temperature coefficient is GENERALLY a number reflecting how much you can expect the resistance to change when the temperature of the resistor changes.
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Name of 4-conductor twinned headphone wire I would like to get more serious about repairing headphones, as well as making my own extension cords ( as the ones you can buy tend to be pretty low quality, and I am tired of breaking). The kind of cable I am looking for was the primary kind of cable used for headphones a while ago. [1] It is composed of four conductors. Sheathed in the insulation in pairs, with pairs of insulation bonded. You can split the insulation down the middle and and have two pairs of insulted wired ( each pair having two conductors )go to each speaker separately. What is this cable called? ( A link to some such cable would be useful. ) [1] It seems that modern headphones mostly use three conductor cable. Or it seems that way to me. Possibly because I use behind the neck type headphones. <Q> Wow, I thought that would be an easy search. <S> Then it wasn't. <S> Then I was really curious. <S> refers to two or more electrical conductors held together by an insulating jacket that can be easily separated simply by pulling apart. <S> Searching "double-pair" or "dual-pair" seems to bring up the cable you are referring to. <S> Belden comes up a lot as the manufacturer. <S> Example catalog (ctrl-f "zip"): http://www.belden.com/products/catalogs/mastercatalog/brilliance/upload/19Broadcast_Cables.pdf <S> An example of a store listing: http://www.interstatewire.com/double-pair-cable-1802b-z4b1000.html <S> Though a lot of the examples I'm seeing are shielded, twisted pairs. <S> I'm not sure where the cheap, old headphone cables are hiding. <A> Years ago I used to use a cable similar to what you are describing. <S> We called it Microphone cable and it had a rubber EPDM outer jacket. <S> The individual conductors were also insulated with Rubber EPDM and the individual conductors were bonded to each other as you describe. <S> The closest <S> I could come in a search has Polyethylene insulation: http://www.alliedelec.com/search/productdetail.aspx?SKU=70005399 <S> To get a list of selections, go onto any electronic supply website and search for Microphone or Musical Instrument cable. <A> I do not have a source of that twin-ZIP cord in bulk format <S> but I can tell you where to get it. <S> I regularly repair professional headphones. <S> AKG is one supplier who makes dual-ear headphones used in broadcast and recording. <S> The cords for the headphones is a wear item and replacements are readily available. <S> Same for most other suppliers of professional headphones / headsets. <S> Wear items are readily available as replacement parts. <S> Although these replacement parts may cost more than other sources, the headset wire is designed for continuous, heavy-duty use and it lasts a long time. <S> Much better quality than inexpensive consumer stuff.
| Looks like the cable you are referring to is called Dual-Pair Zip-Cord From wikipedia , zip cord: Sennheiser, Sony, Beyer-Dynamic, AKG are some of the headphones that I work with regularly. You may want to invest in a bunch of heat shrink and a good heat gun for your uneven lengths.
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How do I keep my constant current source constant? So I am trying to created a current source that will be used to drain a battery at specific time intervals for a specified duration. simulate this circuit – Schematic created using CircuitLab Here is the circuit that I am attempting to use, pulses from the square wave source will trigger the transistor allowing the current to flow and thus draining the battery. This needs to be implemented three times for my design but the first two (which are 10uA and 10mA sources) work fine and do not vary much with changes in the battery voltage, however I am struggling a bit with my final current source which needs to draw 100mA+ from the battery.The issue is not with getting the correct current, but in fact getting the current to remain constant with changes in the battery voltage, that is the voltage at the collector of the transistor. I was advised that I could possibly need a current buffer before the base pin of the transistor and I implemented this but it still does not yield a steady and constant current source!This is the circuit for the 'current buffer', not really sure what it is but that is what I was told to do: simulate this circuit As you can see all it is a pnp ahead of the previously show circuit. Again my issue is not with this, I just need a method of keeping the current very constant, ie. No significant change until the battery voltage has dropped < 1V or something like that! Any help would be greatly appreciated! <Q> Try this: simulate this circuit – Schematic created using CircuitLab <S> This circuit is capable of ~1% accuracy with good layout. <S> If the battery is removed, the op-amp will current limit into the base, which should not hurt it. <S> If you don't like that, or if you use another op-amp that doesn't current limit as well, add a small base resistor such as 100 ohms. <A> Drwaring 100 mA from a battery at 1V requires that the load resistor takes much less than 1V I would aim for ~ <S> 0.25 V, so I would lower the emitter resistor to ~ 2R2. <S> Your circuit has no overall feedback, I would suggest that you add an opamp. <S> This circuit from http://en.wikipedia.org/wiki/Nullor is the classic. <S> Divide your switching voltage down to 2R2 <S> * 100 mA = 0.22 V before applying it to the + input of the opamp. <S> In any case you will need a separate supply for the opamp (not the battery). <S> PS the 100 Ohm bases resistors in your circuits should not be there: you are using the transistors as voltage followers (common collector), hence the impedance at the base is (very) high and hence such a resistor is generally useless <A> You have been getting some very bad advice. <S> First of all, your top circuit depends on the fact that the voltage across R1 needs to be constant, and for this to be possible, the base voltage of Q1 needs to be held constant. <S> In other words, the base needs to be connected to a "stiff" voltage source — one with low source impedance. <S> R2-R4 control the source impedance, and R2 is actually completely counterproductive here, since all it does is raise the source impedance. <S> In your second circuit, you have Q1 in a common-base configuration, which has no current gain at all, only voltage gain. <S> This doesn't help at all. <S> It does help slightly that R2-R4 have lower values, but R4 is still completely useless. <S> If you give us some actual performance numbers that you are trying to achieve, such as the nominal output current and the tolerance you can live with, we can offer better circuits.
| You will need an opmap with an output range and common mode that includes ground, or use a negative supply.
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Will 74HC595 as LED sink damage LED when pin goes high? If used as a LED sink, what happens if a 74HC595 pin goes high? Might the LED blow? I expect that when a pin is low, current will sink and the LED will light. What about the opposite condition? When a the pin goes high, might there be risk of blowing LED's due to reverse flow? Why or why not? simulate this circuit – Schematic created using CircuitLab <Q> The answer lies in the reverse voltage rating of the LED chosen. <S> Common through-hole LEDs typically are rated at 5 Volts reverse, some LEDs higher. <S> This means that for up to 5 Volts applied in reverse bias, the LED will behave as a regular diode, blocking current flow. <S> In the schematic shown, the PWM signal is 0 to 5 Volts. <S> Thus, even when the PWM signal is low, and the latch output is high, the maximum reverse voltage the LED is exposed to is less than 5 Volts. <S> So the LEDs will be fine. <A> Being a diode , your LEDs will not suffer any damage, provided that the voltage being applied is below is breakdown voltage. <S> +5V is generally safe for diodes commonly used. <A> Almost all LEDs are rated for at least 5V reverse voltage. <S> In actual practice, I've never seen one that broke down even at 12V. <S> (Pulsed) reverse bias is a normal operational condition for multiplexed LED displays.
| Beyond this voltage, reverse breakdown might occur, potentially destroying the LED if the voltage is high enough.
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What is the difference between registers, flip flops and latches? I want the answer to the very basic level. I know what they mean individually, but what I am looking for is connection between them. <Q> Flip-flops are single bit devices with two stable states. <S> The outputs are typically Q and \$\mathsf{\small \overline{\text{Q}}} \$ . <S> There are several kinds, here are probably the most common: <S> SR flops have two inputs, S (Set) and R (Reset). <S> As they name implies, asserting either of these either sets or resets the flip-flop. <S> After the input is de-asserted, the flip-flip retains it state. <S> The inputs are usually negated, \$\mathsf{\small \overline{\text{S}}} \$ and \$\mathsf{\small \overline{\text{R}}} \$ , i.e. 0 is the asserted state. <S> J-K flip-flops are similar to SR flip-flops, (with J being Set and K being Reset), but they have a third property -- if both J and K are set, then the flip-flop toggles. <S> D flips have a clock input, and when this clock rises (or falls, depending on the type), the D input is clocked into the flip-flop. <S> Most D-flops also have the S and R inputs of a SR flip-flop. <S> Latches are the same as a flip-flop. <S> Several latches can be combined in parallel to form a register. <S> There will be inputs for each bit plus a clock. <S> An 8-bit register used inside a microcontroller would hold a single byte. <S> A 16-bit register would hold an address ranging from 0 to 65535 etc. <S> Often a register will have a common reset lead. <S> Latches can also be combined in series to form a shift register, in this case there is a single input, and the output of one latch is fed into the input of the next on the rising or falling edge of a clock. <S> The final output of the register can be used to feed into the input of another register. <S> The shift register can also have individual inputs for each latch, so the register is both parallel and serial at the same time. <S> This might be used for a register inside an ALU (arithmetic logic unit), that can do both arithmetic and logic operations, as well as shifting. <A> In the simplest of terms Latches are the smallest building blocks of a memory Flip flops are created using latches. <A> In the patois where I work, a custom logic circuit design group, the term register and flip flop are used interchangeably. <S> We would very much disagree with equating a latch with a flip flop however. <S> To us a flip flop is two latches in series with the one at the input being the master latch and the one at the output being the slave latch. <S> Though data does shift from the master to the slave we do not call a flip flop a shift register but in a strict sense it is a one-bit shift register. <S> We start using the term, shift register, when we put more than one flip flop in series. <S> It is more common for us to call flip flops registers when we have several, either in series or in parallel and we use outputs as a parallel bus of a data. <S> So this is using the term, register, as a memory cache for holding data. <S> In the attached drawing I am not using standard latch symbols or showing a clock signal as the question is not asking how a shift register works. <S> The drawing shows four latches arranged as two flip flops in series as a 2-bit shift register with two parallel outputs. <A> There are various types of flip flop/latch and terminology can be a bit varied. <S> The simplest device is the un-clocked R/S flip flop. <S> There are to inputs, a "set" line which when triggered makes the output go high and a "Reset" line that when triggered makes the output go low. <S> If both inputs are inactive then the flip-flop retains it's previous state. <S> If both inputs are active the flip-flop is driven into an invalid state. <S> Next simplest is the transparent latch. <S> Rather than set and reset inputs <S> there is a data input and an enable input. <S> When the enable input is active data is passed from input to output. <S> When the enable input is inactive <S> Then we have the edge triggered designs. <S> Clocked flip flops are triggered by a clock edge. <S> The value of the output after the clock depends on the inputs before the clock edge. <S> The big advantage here is that we can chain the devices and signals will move from stage to stage as clock edges come in. <S> There are a few types. <S> You can have a clocked version of the RS flip flop that only updates when a clock edge comes in. <S> You can have a JK flip flop which is similar but togles on every clock edge when both inputs are active. <S> You can have a T flip flop that toggles unconditionally on every clock edge <S> and you can have a D flip flip with a single input that is propagated to the output on every clock edge. <S> In some cases clocked flip flops also have an "asynchronous reset", "asynchronous set" or even "asynchronous load" function which allows the output to be changed independent of the clock. <S> Normally when people say "latch" they mean "transparent latch" but they may sometimes mean "D type flip flop". <S> Normally when people say "register" they mean a D type flip flop.
| A bit holds binary data - 0 or 1, similar to a flip flop, so you may use the term flip-flop and register interchangeably.
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Why resistor values are non-linear? What makes sense in 470 ohm? I never figured that out. Why do we have 220 ohm and 470 ohm, rather than 100 ohm, 200 ohm, 300 ohm? Any particular reason? maybe it's just legacy from old manufacturing process. <Q> Resistors are arranged in series with N values per decade. <S> Standard series are eg E12 with 12 values per decade, E24 with 24 ... <S> E96 ... <S> The values area arranged so the RATIO between adjacent values is about the same, so that $$ R_{n+1} = <S> k <S> R_n$$ Follow this thought to its logical conclusion and you see that for an EXX series, eg E12 <S> so XX = 12, then for 12 equally space ratio resistors per decade <S> the difference k is the xx'th root of 10. <S> $$ <S> k = 10^{(1/XX)}$$ so eg for the E12 series with 12 values between 1 and 10 Ohms or 10k and 100k, then $$K= 10^{1/12} <S> \approx 1.212 <S> $$ <S> So to get 12 values from 1 to 10 the values <S> would be $$1.212^0 <S> = 1.000 \\ <S> 1.212^1 = 1.212 \\1.212^2 <S> = 1.479 \\1.212^ <S> 3 = 1.780 \\... <S> $$ <S> Yielding E12 value: <S> 1 <S> 1.212 <S> 1.2 2 <S> 1.468944 <S> 1.5 3 <S> 1.780360128 <S> 1.8 4 <S> 2.157796475 <S> 2.2 5 <S> 2.615249328 <S> 2.7 6 <S> 3.169682185 <S> 3.3 7 <S> 3.841654809 <S> 3.9 8 <S> 4.656085628 <S> 4.7 9 <S> 5.643175781 <S> 5.6 10 <S> 6.839529047 <S> 6.8 <S> 11 <S> 8.289509205 <S> 8.2 12 <S> 10.04688516 <S> 10.0 <S> There is slight disagreement with the std value in a few cases. <S> eg <S> 3.169... -> <S> 3.3 8.28945 ... - <S> > 8.2 but the basis is clear. <S> The same method applies for other series <S> so eg <S> the E96 series values are spaced by $$k = <S> 10^{(1/96)}$$ <S> An extremely useful consequence of this attribute is that if two resistors N positions apart in a series have a certain ratio then ALL resistor pairs <S> N positions apart have the same ratio. <S> eg <S> to get a 1.5: 1 ratio we can use 1.5K and 1K. <S> 1.5K : <S> 1K = 1.5:1 <S> BUT in the series 1, 1.2, 1.5, 1.8 - the values 1.0 and 1.5 are two positions apart. <S> So, any two resistors 2 positions apart will have a ratio of about 1.5:1 in value. <S> so eg 2.7 / 1.8 = 1.5, and 6.8 / 4.7 <S> = 1.447 ~= <S> 1.5 etc. <A> In electronics, international standard IEC 60063 defines preferred number series for resistors, capacitors, inductors and Zener diodes. <S> It works similarly to the Renard series, except that it subdivides the interval from 1 to 10 into 6, 12, 24, etc. <S> steps. <S> These subdivisions ensure that when some arbitrary value is replaced with the nearest preferred number , the maximum relative error will be on the order of 20%, 10%, 5%, etc. <S> Copied from wiki E series <A> Resistors are built such that the increments between values make sense for the range of resistance in question. <S> For example, for low value resistors you might want 1 ohm and 2 ohm values, but for a 1 Meg resistor you wouldn't want 1,000,002 ohms to be the next value. <S> In order to make this work out, the decades are divided into a reasonable number of values per decade to acheive a given accuracy. <S> So for example for E96 (96 values per decade, 1% tolerance) each value is given by 10^(n/96) rounded to 2 decimal places and multiplied by the appropriate power of 10 for the decade in question. <S> (Where n is an integer representing the nth resistor in the series.)
| Resistor values are arranged so that adjacent values are in an (approximately) constant ratio.
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I'm looking for the simplest way to produce a 1-bit high resolution (1080p or so) DVI signal It doesn't need to be high performance, I just need to draw a bit mask. I'll be driving it from a micro controller, arduino or perhaps a BeagleBone Black. The application is to produce structured light patterns with a off-the-shelf projector. <Q> The Raspberry Pi 4 has two HDMI ports. <S> Use one for your UI and the other for your structured light generation. <S> https://www.raspberrypi.org/products/raspberry-pi-4-model-b/ <A> <A> As far as I remember, 1080p requires a 139MHz clock rate [1] . <S> Good luck getting that with an arduino or any bog-standard mcu, not to mention signal integrity issues if you don't design this properly. <S> A Beaglebone may be up to the task, I have no experience with them, but then you are probably looking more at software piping stuff out through HDMI. <S> This is much the same as using a Raspberry Pi or other PC-related stuff, where the programming probably will be easier than writing mcu-specific firmware to do this. <S> If you want a challenge, you could try using an FPGA on a devboard, for example the ZedBoard [2] . <S> This is a more expensive way (both in time and money) of doing it, though. <S> Good luck, whatever you end up doing! :)
| I'd go to the Raspberry Pi for the HDMI output instead of the Beaglebone black, and just use an HDMI to DVI converter, like http://www.amazon.com/HDMI-Adapter-Plated-Connector-Female/dp/B000FUVNX8
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How durable is the insulation on magnet wire? I'm planning a light sculpture. It will consist of a large number of RGB LEDs mounted on a tree-like galvanized steel frame. Each LED has four wires -- GND, +5V, data-in and data-out. For aesthetic reasons, I would like to use magnet wire to connect the LEDs, intertwined with the steel frame. So if one looked closely at the sculpture, one would see a twisted bundle of thick steel wire (non-conducting!) and several strands of conducting magnet wire. I'm worried that the thin insulation on the magnet wire might wear off and cause a short circuit, either with the steel wire frame or with other lengths of magnet wire. How much jostling/rubbing/abrasion can magnet wire take before the insulation is compromised? Once the sculpture is complete it will not be handled very much, but during assembly there could be a fair amount of jostling. [edit] I know there is no way to give a precise answer. I was looking for something like "I've done project XYZ which subjected magnet wire to a certain amount of abrasion, and based on that experience I would guess this project will [or will not] work." Also, I should have specified that I plan to use magnet wire with multiple coatings. <Q> It all depends on what insulation is used. <S> There are lots of different kinds of insulation and manufacturers make them with different thicknesses (single build, double build, triple build, etc.). <S> Most data sheets show the results of an scrape abrasion test. <S> In this test, a sample of the magnet wire is secured firmly in place and a needle is scraped along the surface of the magnet wire. <S> The datasheet usually records the number of scrapes the insulation can withstand until the needle breaks through the insulation and shorts the wire. <S> So that spec on the data sheets will give you a good relative feel for how different magnet wire types compare to each other. <S> But it doesn't really answer your question of how well magnet wire will work for what you are doing. <S> Here are a couple of thoughts. <S> I think you'd be surprised how tough a good quality magnet wire can be. <S> A popular type in the motor industry is a heavy build MW-35. <S> It holds up pretty well and should be able to withstand being assembled. <S> Just try not to use any sharp/pointed tools that could scrape or puncture the insulation. <S> Also keep in mind that for many, many magnet wire applications like motors and transformers, the magnet wire gets coated with a varnish or epoxy to give it more long term protection against rubbing, vibration, etc. <S> That doesn't mean your application would need varnish but magnet wire <S> is very often used that way. <A> There are two wire types that are ideal for your application. <S> The first and best is called "heavy formvar" and is the type of wire that is used in the automatic winding machines that make motor stators and armature windings. <S> This wire is available from many suppliers and costs around $50 a pound for #20 and up to $150 a pound for #44. <S> (smaller diameters are much more expensive)The second choice would be "polytheremaleze" which is also a very durable wire that is used in high temperature applications. <S> It has a nylon coating for insulation. <S> (think of how tough a tywrap is!) <S> This wire is also widely available from many manufacturers and distributors. <S> (google magnet wire) <S> It is approximately the same cost as formvar. <S> Another possibility is to use a wire called "ltiz" wire, which is made up of multiple strands of fine wire. <S> This is substantially more flexible before breaking than a solid wire of the same gauge. <S> Litz wire is available with both formvar and polythermaleze insulation, as well as "soldereze" insulation, which is MUCH easier to use because it disintegrates at the temperature of hot solder. <S> Thus you can solder it without having to strip it first. <S> Stripping the end of a wire is a serious problem with formvar and polythermaleze because of their toughness. <A> There are all sorts of magnet wire. <S> Cheap stuff from radio shack will scratch off easily. <S> If you go to a wire supplier (MWS?) <S> you can get triple (or maybe even quadruple?) <S> layers. <S> And lots of different materials.
| I think if you pick a good quality magnet wire and are careful not to nick or scrape it during assembly, you should be okay.
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Low power serial bluetooth module with Arduino I plan to use the HC-06 serial bluetooth module with an Arduino Leonardo. To my knowledge, the Arduino's serial pins output at 5V with 40mA of current. The HC-06's datasheet says it will accept input at 3V - 4.2V with 20mA - 40mA of current. I'm more of a programmer than an electrical engineer, but this tells me I will need a resistor between the Arduino's TX pin and the HC-06's RX pin. From my calculations I will need a resistor with 82.5 Ohms of resistance: $$r = \frac{V}{I} = \frac{3.3V}{0.04A} = 82.5 \Omega$$ Is this the correct way to solve the problem? Online I've seen people solve the exact same issue using a voltage divider with a 10kOhm resistor leading from the Arduino's TX pin to the HC-06's RX pin and a 20kOhm resistor leading from the HC-06's RX pin to ground. So, what's the correct way to solve this problem? If a voltage divider is the solution, why and how does it work? Or why does simply putting in a 82.5 Ohm resistor not work? EDIT: For simplicity sake, let me reiterate my question from a more general perspective. The HC-06 module requires a power supply voltage of 3.3V (which can come from the Arduino's 3.3V pin) and will also only accept logic inputs at 3.3V. So, what is an easy way to achieve this voltage reduction for the serial wire? <Q> There are two main sets of characteristics quoted for pins, absolute maximum ratings and typical ratings. <S> The HC-06's datasheet says it will accept input at 3V - 4.2V with 20mA - 40mA <S> Those are absolute ratings. <S> The pin is rated to handle 3V to 4.2V and 20mA - 40mA current. <S> In a normal circuit, digital input pins will draw very little current, as they are CMOS logic gates. <S> Without the pull-up or pull-down resistor enabled, the pin has very high input impedance. <S> Your proposed one resistor circuit looks like this: simulate this circuit – <S> Schematic created using CircuitLab <S> You effectively have a voltage divider with the other resistance being the input impedance of the HC-06 input pin. <S> So the voltage on the RX pin with 5V on the Arduino TX pin would be: $$ V_{RX} = 5V \frac{R_{really\,really\,big}}{R_{line} + R_{really\,really\,big}} \approx 5V$$ <S> This is outside the maximum voltage rating. <S> Solution <S> The resistors in the divider should be high enough not to violate the Arduino maximum current output and much less than the input impedance of the input pin. <S> They also must be low enough that the input capacitance of the pin doesn't 'smear' the signal too much (slew rate). <S> You can think of the capacitance as resisting the change in voltage, so sharp inputs start to get rounded off: (image taken from http://www.johnloomis.org ) <S> For you quoted figures we would then have: $$ V_{RX <S> } = 5V \frac{\left(\frac{1}{20 <S> K} + \frac{1}{ R_{really\,really\,big}}\right)^{-1}}{10K + \left(\frac{1}{20K} + \frac{1}{ R_{really\,really\,big}}\right)^{-1}} \approx 5V \frac{20K}{10K + 20K} <S> \approx 3.3V$$ <S> In Arduino land, people would by a logic level shifter board for this purpose. <S> One common one has a voltage divider for 5V TX to 3.3V RX, and a transistor for 3.3V TX to 5V RX. <S> Others have transistors both ways, so that the voltages can be different from 5V and 3.3V as using a voltage divider the ratio is fixed, and to effectively have a very low output impedance on the TX pins and thus avoid slew rate problems. <A> While a resistor divider will push the voltage down to proper levels, it is inefficient for serial communications or any quickly changing signal. <S> Resistors not only change the relationship between voltage and current, they also add a longer slew rate to the changing signal, meaning it will actually take longer for the signal to rise up to its HIGH logic level. <S> This will pose a problem with UART. <S> A better solution is to use a logic level converter, particularly one that is made to handle communication protocols. <S> You will give this device a 3.3V and a 5V as well as the signals as indicated by the documentation. <S> The Leonardo has a 3.3V pin <S> I believe so you should be fine. <S> Good luck! <A> The simplest solution is a resistive divider to lower voltage from 5V to 3.3 V, but you can't raise 3.3 V to 5V without a transistor. <S> Probably, there will be no problem, because Arduino will sense the 3.3 V input as HIGH input, and you can get a working circuit with only resistive divider on the Arduino TX - HC-06 RX line. <S> The best solution would be a level shifter. <S> You can build transistor level shifters like https://electronics.stackexchange.com/a/107388/40609 . <S> Or simpler, use the 74LVC245 IC. <S> If a voltage divider is the solution, why and how does it work? <S> Or why does simply putting in a 82.5 Ohm resistor not work? <S> You should read Voltage Divider vs. Resistor in Series
| There might be a pull-up resistor, that connects the pin internally to Vcc, or a pull-down resistor, that connects the pin internally to GND, which will increase the current draw. The solution is to use some kind of level shifter.
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Slowest transmission speed supported by TOSLINK TX and RX? I'm writing an Arduino communication library that uses differential Manchester coding (see: http://forum.arduino.cc/index.php?topic=257985.0 ). I'm wondering how to communicate between two Arduinos that are far enough apart that connecting their power supply grounds is not practical, but without resorting to a wireless system. In that case, a physical communication system that does not need to pass current is needed. I thought of using Toslink fiber optic modules that are typically used for digital audio signals. The only concern that I have is that the bit rate that my Arduino library (and the Arduino hardware itself) will be able to support is not more than about 20k bps. Digital signals typically carried by Toslink have a bit rate starting at about 1M bps. Since I am not familiar with fiber optic hardware I am concerned that the typical digital audio implementation (e.g. the TX & RX hardware and associated passive components) would not support a relatively glacial data rate of 20k bps. I could just purchase some Toslink modules and try it out, but I thought that before I buy any parts I would make an attempt to learn about the hardware and ask if there are any inherent pitfalls in sending data at kHz (or slower) rates over Toslink. So, is this feasible? Are there any obvious problems that will come up? <Q> The answer is: it depends! <S> It seems that most or all Toslink transmitters have bandwidth right down to DC. <S> On the other hand, most Toslink receivers require a minimum of 100kHz for the modulation frequency for the part to function. <S> That's too bad, because Toslink is a very practical medium for electrically isolated communication and the TX or RX hardware, as well as the fiber optic cables, are pretty inexpensive. <S> I did find one RX by Toshiba that is specified for operation down to DC, the TORX1952. <S> Here is a link to the part at Mouser (The US supplier) and the datasheet: Toshiba TORX1952(6M,F) at Mouser.com DATASHEET <S> It's not exactly cheap at $10 each for QTY 1 <S> but it should work for low speeds. . <A> I'd say you can go as slow as you want. <S> Communication in an optic fiber is achieved pulsing a light, usually of a "single" wavelength, and detecting the pulses on the other side. <S> Speed is limited by the speed of the emitter and the receiver, and of course by the lowpass response of the fiber. <S> There is no low limit to bandwidth though, this sort of connection even allows DC: if you leave the emitter on the receiver would detect a steady ON. <S> It might be that if you buy a module it can include some sort of circuitry that may or may not limit the badwidth on the low side. <S> If you just buy the connector with the emitter/receiver indside and hook it directly to the micro you are good to go. <S> If you really need to avoid current flow optic fiber is a great idea, I'd like to know why you have such a specification because maybe a differential/twisted pair is suitable too. <S> You can insulate the devices using optocouplers and call it a day. <A> Driver chips for the two standards are readily available as well. <S> Both standards work very well to support multi-MB data rates and below. <S> They solve the problem by providing a differential pair of signals, which means that the voltage and current transmitted on the positive side is always matched by the negative side, such that there is no net current from one side of the interface to the other. <A> The simplest solution would be to use signaling via current rather than voltage. <S> In current loop signaling, you signal a one by sending 5ma (for example) around a loop. <S> On the other end, the loop is directly fed through the LED side of an optocoupler and not electrically connected to the reciever at all. <S> There is no common ground, it is immune to interference and since the current is the same around the entire loop it is independent of the length of the cable, no voltage drop. <S> This is exactly the method MIDI uses to avoid ground loops in musical equipment. <S> The only bad part is that it is single duplex <S> so you need two loops with 4 wires total for bidirectional traffic <S> but it is dead simple to implement. <S> There are a variety of MIDI interface circuits online that you can directly use.
| LVDS and RS-422 signaling standards work very well in this type of scenario.
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Can the power source for a USB device be different from the data source? I was wondering if the data source (USB Host) for a USB device has to also deliver the power 5V circuit e.g. data and power need to have a common ground. Or can I connect the data lines to the data source and have a separate power supply for the 5V line? <Q> As Chris pointed out in the comments, you can have self-powered devices. <S> However, even with self-powered devices you need to have a common ground to ensure that the signal levels on the data lines are within spec. <A> However, with USB OTG (on-the-go), the presence/absence of Vusb (the power voltage) is an important part of the protocol for suspending device activity and doing the host negotiation. <S> It needs to be connected even if both devices are self-powered. <A> You certainly can supply power from elsewhere - bench power, another USB port, whatever. <S> As long as it shares a ground but does NOT back-power the USB data host. <S> I have a spliced USB cable on my desk for this kind of thing. <S> Trim insulation, cut red wire, add connectors to taste.
| With "plain old" USB, it doesn't matter a whole lot where the power for the device comes from if it doesn't draw power from the host.
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Is it possible to make a complete analogy between electricity and something more graspable? (e.g. fluids) A similar question has been asked here , but mine is more general. Here is the background: it seems to me that electricity is not so intuitive as it may sound. For example, many (unskilled, as I am) people think an electric plug is like a (water) tap: it delivers what it can, and stop at a maximum flow. But it is not possible to stop the water from flowing, by putting too large a bucket under a tap, whereas if you plug a washing machine you could break the current! Improving the understanding with analogy seems to be an efficient way to learn (not talking about scholar stuff) about electricity and "feel" it (not too intensely, though). Water seems to be a good candidate for these analogies: current/water flow, electron/molecule, tension/height difference, resistance/decrease of the pipe section, capacitor/volume with membrane. I also heard about an analogy between electricity/water for transistor. My question is: is it possible to define a global analogy, which would: (i) allow to better understand any electrical circuit [ideally, a complete transcription would be possible], (ii) be exhaustive (i.e. work for any electronic/electrical component). Other examples include allowing to understand what would happen if the frequency of the electricity changed, or if a power plant produced energy in excess, etc. I insist: I'm not an electrical engineering. I'm just curious :) <Q> Using analogies to fluid dynamics (air/water) is useful for some purposes for the complete novice, but it also can lead to confusion. <S> If someone really wants to understand electricity, they need to dispense with the fluid analogies and just learn how electrical fields and charges work. <S> There really isn't a global analogy. <S> I find it useful to refer to William J. Beaty's articles about electrical misconceptions . <S> Also to the web site All About Circuits . <S> And finally, the more in-depth reference book, The Art of Electronics by Paul Horowitz and Winfield Hill. <A> JYelton answer is very good in my opinion, but I will add alternative answer. <S> Is not just about moving electrons. <S> Electricity is complex and "contains" many diffrent phenomena magnetic fields electric fields <S> ionic reactions <S> chemical reactions <S> There is no complete analogy to electricity in "macro" world. <S> You can read some very old books (1900-1950) about physics, chemistry and electricity. <S> These old books are suprisingly good for beginners and most of them don't require modern engineer knowledge. <A> Falstad physics applets: <S> Circuits with visible electricity <S> If we use the water analogy, we find many interesting things that are hard to understand ...about water! <S> For example, water flows slowly through a hose, yet the hydraulic energy flows incredibly fast. <S> Also the hydraulic energy is counter-intuitive: it can easily flow backwards relative to the water flow. <S> Or, water can be moving back and forth in a pipe, while the energy is racing along in a single direction. <S> The water company doesn't sell us any water, they just offer an expensive pumping service, and all the hoses were pre-filled with water before we started. <S> (After all, water isn't a form of energy ...and neither is electricity of course: electric currents aren't flows of energy, amps aren't watts.) <S> Also, in a closed water-filled loop (a hydraulic circuit,) the start-up transients are fairly hard to grasp. <S> So are the energy reflections from joints and T-connections, resonance phenomena for "AC water," and energy loss from the hoses as radiated sound (which is analogous to antennas and energy lost as radio waves.) <S> Also, water is invisible. <S> Well, actually it's much like electricity: if you have a water current in a transparent pipe, you see nothing, it just looks like a static rod of solid glass. <S> It could be moving slowly, or flowing fast, or wiggling back and forth, but without dirt or bubbles we have no clue. <S> In this it's just like electricity: <S> electric currents are invisible, while electricity itself is easily seen, the mobile electrons within metals are silvery in color. <S> Metals are reflective like mirrors because any dense "fluid" of electrons is reflective like a mirror. <S> Heh, so we should start using glass pipes full of liquid mercury as an improved fluid analogy for electric circuits: <S> generators are liquid-mercury pumps, resistors are pipes full of mercury and compressed sand, frictional pipes which get hot whenever there's an Hg current. <S> The silvery Hg stuff always goes slowly in a circle, while the hydraulic energy races from place to place in the pipes at extreme speed, yet when we're looking at transparent Hg-filled tubes, we see nothing happening, just some thin silvery rods with a transparent plastic coating. <A> One analogy I liked to use for understanding current is a train. <S> Think of the electron as connected carts on a train. <S> If I push one cart (even just a little) this first cart will move since they are connected. <S> Then parallel tracks or larger carts may describe other properties. <A> The fluid-analogy is flawed in some ways, e.g. due to allowing open circuits and the fact that the fluid molecules actively move through the tube while the electrons in the conductor actually merely nudge their neighbours, who in turn nudge their neighbours and so on, which propagates much faster than the actual electron movement speed (which is slow ). <S> So in my opinion a better, though not trivial, analogy would be a (closed) track of magnetic wheels: <S> <=============================== <S> > connected to start <S> S S S <S> S ....... <S> | | <S> | <S> | <S> N <S> N <S> N <S> N <S> ....... <S> <=============================== <S> > <S> If you push the one on the left side, it will repel the next one and slow down (and maybe move slightly back, but not entirely to its origin), while the next one accelerates until it repels the second-to-next one and so on. <S> The fact that the wheels will slightly move towards the direction of "conductance" explains why the "circuit" must be closed, otherwise you'll run out of wheels <S> magnets in the track itself can simulate resistance a junction plate, the switch of which is equipped with a magnet as well, can act as a transistor <S> (I haven't yet come up with an analogy for a capacitor though)
| For better understanding how electricity electricity works you have to learn some chemistry, physics and try to build some picture of electric and magnetic fields in your mind. Electricity is unique and there is no complete analogy between electricity and something else.
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Why is my microcontroller passing a current through it when not connected? MCU: ATTiny13 I noticed this after trying to debug why pushing my switch (connected via R2, a 507kOhm pulldown resistor) makes the LED dimmer while depressed. The switch was powered by the same supply line as the Vcc input to the microcontroller. Upon disconnecting the Vcc input (Pin 8), I noticed that the LED was still lit when the switch was depressed. If I removed a connection from the ground pin 4, the LED still lit up, but less brighter. The circuit below represents what I observed. The switch is removed to simplify the problem: Why does this happen, and how can I stop it? It is interfering with the output when the button is depressed. Here is a picture of the circuit on a breadboard. The supply line (5V is the red wire, Ground is black): <Q> Inputs of many modern CMOS devices have ESD protection diodes from the I <S> /O pins to the supply rails, which hope to divert transient overvoltages to the supply before they cause damage. <S> A side effect of this is that the chip can, at least to a degree be powered through an I/O pin, once the pin rises enough against the (unsupplied) supply to forward bias the diode. <S> Even in technologies without explicit protection diodes, it could happen to a degree, though often resulted in very unreliable operation (classic mistake - forget to power a chip and see it "sort of" work <S> - I did it myself with an SPI flash this past January that somehow never got a ground, and would provide expected responses right up until I tried to write flash locations). <S> Generally you do not want to power a chip this way <S> - it is outside the absolute maximum ratings, and the protection diode may not be sized to carry the full operating current. <S> You do see it at times though, both in intentional experiments, such as an RF-powered ATTiny RFID tag emulator experiment, or accidentally in cases such as trying to measure power consumption of a sleeping MCU and having it actually draw power from your serial debug port rather than the supply you are trying to measure. <A> The datasheet for your device has this table: <S> In this table, VCC means the voltage applied to the VCC pin, not the net in your circuit which you have labelled VCC. <S> Since you have not applied any voltage to the VCC pin, you must not apply more than 0.5V to any other pin. <S> Your PB4 connection is violating this rule. <A> Current flows thru internal clamping diodes. <S> Internal circuitry (simplified) looks like this: <S> In this Atmel document (random application note containing information about clamping diodes) <S> you can read: To protect the device from voltages above VCC and below GND, <S> the AVR has internal clamping diodes on the I/O pins (see Figure 1). <S> The diodes are connected from the pins to VCC and GND and keep all input signals within the AVR’s operating voltage (see Figure 2). <S> Any voltage higher than VCC + 0.5V will be forced down to VCC + 0.5V <S> (0.5V is the voltage drop over the diode) and any voltage below GND - 0.5V will be forced up to GND - 0.5V. <A> Just a addendum to Chris Strattons correct answer. <S> You are indeed powering the device through the protection diodes. <S> There are several ways of providing ESD protection and all of them involve using diodes on the pins to connect to a rail inside. <S> So your conduction path and internal power of your chip will be at least 1 diode drop below the supplied "power" on the pin. <S> You can test this out by measuring the the Vcc pin <S> it will be about 0.7V lower than Vcc. <S> When you disconnect the ground, you are pulling less current through the protection diodes and shifting the operating point of the voltage supplied to the LED. <S> But by a little bit, so this may not account for the difference. <S> The protection diodes are designed to handle Amps of current during a ESD strike so they are fine with this little trickle. <S> Your danger in operating the chip this way is that you could induce latch up, but most chips are designed to be able to NOT latch up in these conditions so that isn't a concern so much. <S> But a possibility. <S> Another danger will depend upon what the exact nature of the ESD protection on chip. <S> If it is a clamping style and you have a high dV/dT event <S> then the clamp might fire and short out the supply. <S> But this is also unlikely. <A> BUT the important thing is that you are doing something which is "illegal" and completely outside specification and anything can happen and you should not be surprised if it does. <S> Your circuit diagram is WRONG. <S> This is the REAL circuit diagram that you are using: <S> This is a hardware version of "garbage in, garbage out". <S> If you do something random and get a random result you should be happy - the universe is working as you'd expect. <S> More later maybe ...
| You are probably powering the device through its protection diodes
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Which of these marks signifies Pin 1 on the STM32F (LQFP64)? I'm using the STM32F105, in a LQFP64 package. It has two circular markings, on opposing corners of the IC. This picture is for a different ST ARM, but the markings and silkscreen are similar: The larger mark in the upper-right of the picture has a flat bottom, whereas the smaller mark in the lower-left is slightly concave. This package doesn't have an exposed pad, so there are no clipped corners underneath. The datasheet only mentions one mark: and I haven't found any other package documentation that mentions it. Does anyone have a clue for me? Update: The project is completed and the answers are correct. Pin 1 is the smaller, concave mark in the lower-left . Don't forget to upvote the good answers! Interestingly, I've since purchased another batch of these chips and they only have the one mark. Thanks, all! <Q> This is a comment - converted to an answer. <S> If there is still some doubt in your mind, use a continuity tester to make sure all the grounds are connected, since these will likely have unique locations you should be able to verify which of the pins are connected together. <S> For example pins 27,28 and 49,50 are a unique grouping that doesn't occur on other sides. <S> Next step would be to use a diode checker and to test the power pins against group. <S> Use a little bit of correction fluid to the package body to mark your pin with a white (or yellow etc.) spot. <A> I have designed a board for the STM32F103R, like you. <S> Tested and "proven" now. <S> Anyway, if the text is right-side-up, pin 1 is in the lower left. <S> The other answers all mention things that could be considered a "hint", Luckily those hints all point in the same direction and happen to be correct. <A> Going by this image of the LQFP100 version of this processor, the text orientation gives the PIN1 location away (the lower left dot seems to be the pin1 identificator on your chip). <S> Of course, this deduction might be wrong, but I think it's far more likely that it isn't. <A> The smaller (smallest? <S> in some cases) circle is always pin 1. <S> In your figure, that would be the bottom-left pin.
| It's very very likely that since this is a processor that the ground is common throughout the die.
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Pictures of different adapters, trying to understand how they work (PLC to PC communication) I purchased the blue adapter in the picture as a USB/RS232 converter. It takes in the RJ12 cable from my PLC, and converts it to a USB input. Now when I try to use an RJ12-DB9 Female and then connect that to a DB9-USB which then connects to my computer (via USBO and then try to communicate to my PLC using this method (also seen in the picture), I can't seem to get my program to work. What am I missing here? I'm sure all of my configuration settings are working, so what could be going on here? Basically the blue adapter works with my program, but the combination of the 2 adapters in the lower part of the picture do not work and I'm wondering why. I understand there are some electronics in the blue adapter (HIN202 for example), but these chips just seem to "assist" in maintaining the RS232 communication process, rather than actually provide infrastructure for it to work. Edit: Sorry guys....find picture attached and thank you for your comments, hopefully this pic confirms or provides better insight into my situation. <Q> 3 possible issues to always be aware of when using RS232 to USB converters. <S> (in order of most common to least.) <S> Converter driver comm port number chosen matches comm number in PLCcommunication software. <S> Power from USB port enough to drive alldevices connect. <S> Communication watchdog timer when converting frommultiple protocols. <S> 1st example: see <S> http://bin95.com/Industrial-Training-Videos/AB-PLC-DH485-RS232-USB.htm <S> 2nd and 3rd example: if you connect USB to computer, RS232 to RS232 to RJ485 converter box (pic box), the computer's USB port will not have enough power to power circuits in RS232 to RJ485 converter box. <S> And/or communication during multiple conversion gets slowed down too much giving comm error on PLC side. <S> Solution <S> : Get a converter cable specifically designed to convert USB to RJ485 instead of connecting two different converters together. <A> In my answer to your other question , I linked to the pinout required for your RJ12 cable. <S> Check that the cable you are using matches the pinout required. <S> Getting USB-Serial adapters to work with the correct drivers can be really tricky. <S> I've seen some cheap adapters with poorly written drivers not work at all. <A> Many PLCs use custom wiring on multipin connectors (dsub, etc). <S> Your usb-serial must be aware of any specific changes required for that device (usually the PLC provider wants to sell you there own special converter, that will also include special drivers).
| Test the USB-Serial adapter with something other than your PLC to confirm that it is working correctly.
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Good method to remove tenting on vias after assembly for debugging? This is a question about modifying an assembled board to make it easier to debug. In a perfect world, I would have left the bottoms of these vias un-tented. Whoops. I have two BGA parts on a PCB, with a 32 bit wide bus with a few control lines and a clock line connecting them. Almost all of these traces are only on internal layers. There looks like a signal integrity problem on the bus. The good news is that the board only has parts on one side, and all the vias are through-all. Can anyone suggest good ways to remove the tenting over the vias so they could be probed with an oscilloscope, or at least be able to solder 30 gauge wire to them? The connection needs to be pretty good – the bus runs at about 100 MHz. There are also some traces on the bottom layer, which I need to avoid damaging. My current ideas: use x-acto knife to scrape off solder mask, first with a side to side scraping motion and then swivel the blade in a circular fashion. Use fine grit sand paper and remove the solder mask from the local area (covering several other vias, and maybe a few traces) This article has a little more information too: http://www.edn.com/electronics-blogs/signal-integrity/4327009/Scrape-it-How-to-probe-a-microstrip-trace-with-no-accessible-test-points-or-vias However I am hoping someone has a method that they consider a "joy to use." I don’t want to just dive in; this prototype costs more than my car and I just have the one. <Q> The accepted answer to this EE.SE question: How to remove solder mask? <S> is to use a fibreglass brush pencil. <A> I think it would be best to get control of the brush by chucking it in a drill press (or mill if you have one) and use a heavy solid block of material (a cast iron angle bracket if you have one) and double-face tape the board to the angle bracket. <S> There might not be enough room for that, and it might not work. <S> When I've had to do this (thankfully only for one or two places at a time) I've always used a common #2 X-Acto blade, but I'm always worried the blade will snap since you're applying force sideways to get it to scrape- safety glasses are a must. <S> Practice on a blank board of some other (cheap type) <S> first, then practice on a blank copy of your real prototype as solder masks seem to vary a lot- <S> this one might not be completely cured and you might be pleasantly surprised at how it comes off if it's 'hot off the press'. <S> Speaking of 'hot', it might be worth trying preheating the board moderately (say to 100°C) to soften it a bit, of course that will cure it more. <A> Certain more aggressive flux removers (e.g. TechSpray 1631-16S) <S> will, with sufficient time and agitation, also remove solder mask. <S> If very thin traces are involved, this might be safer than using a fiberglass or brass brush.
| You might want to try a small brass wire brush (practice on a scrap board first).
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Definition of Full-scale error I was looking for the concept of "Full-scale error" (related with laboratory measurement terminology) on the web without success. I would appreciate your help. <Q> For example, an analog voltmeter with a range of zero to 100 volts specified as having an accuracy of +/-1% of full scale can be off by plus or minus 1 volt when there's between zero and 100 volts across the meter. <S> So, with the meter reading 1 volt, the actual input could be anything between zero and two volts, and with the meter reading 100 volts the actual input could be anywhere between 99 and 101 volts. <A> The definition varies a bit, but often it must be added to a 'zero' error, and unless otherwise specified, applies only under reference conditions- warmed up for 1 hour, ambient temperature 20°C + <S> /-1 <S> °C or whatever. <S> Temperature changes to scale and zero, and aging often must be added on top of that. <S> In the case of thermocouple instruments, the cold-junction compensation is added on top (and may be specified as a ratio such as 20:1, meaning a 20°C change in ambient temperature could cause an additional 1°C error). <S> An error specification of 1% F.S. means that if the scale is 4mA to 20mA, that error will not exceed +/-1.6uA (1% of the full scale deflection, which is 20mA-4mA = 16mA). <S> If they're complete d*cks <S> it could mean +/-2uA error, as I said, definitions vary. <S> So if the measurement is at mid-scale (12mA) <S> then that error could be +/-2 <S> % of reading. <A> As an example, an error of 10 mV in a full-scale deflection of 2V would be a "full-scale error" of 0.5%. <S> In other words you compare the error with the limits of measurement of the device and express that as a percentage.
| An error expressed as a percentage of full scale means that any measurement made will fall between the limits given by that quantity.
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RC high pass Circuit Step Input why C Initially acts as SS In an RC high pass circuit, when you apply a step input, the capacitor doesn't have any drop initially. My professor's explanation for this was that the capacitor doesn't change voltages instantaneously. He presented the formula \$ V= \frac1C\int_0^T idt\$. Could someone explain to me why there is no drop across the capacitor? I didn't understand the explanation given by my professor. <Q> From the professor's explanation you know that the voltage across the capacitor is proportional to the integral of current wrt time, so for the voltage to change instantly, the current would have to be infinite. <S> Unless you have one if those imaginary circuits with no resistance or inductance in a loop, that can't happen. <S> In your case you have the R of the RC, so the current is limited to input voltage divided by R. <A> Can someone explain to me why there is no drop across the Capacitor <S> For an ideal capacitor, the current through is proportional to the time rate of change of voltage across $$ <S> i_C = <S> C \frac{dv_C}{dt}$$ <S> The faster the voltage <S> changes , the larger the magnitude of the current. <S> A voltage 'jump', an instantaneous change in voltage, would imply an infinitely large current. <S> Thus, the voltage across the capacitor must be continuous (no 'jumps') <S> for the current through to have a value. <S> If the input voltage to the RC circuit is an ideal step, the voltage across the capacitor, the instant before the input voltage changes, is zero. <S> Since the voltage across the capacitor must be continuous, the instant after the input voltage changes, the voltage across the capacitor must be zero (no jumps). <S> However, the current through the capacitor can instantaneously change. <S> Before the input voltage changes, there is zero current. <S> The instant after the input voltage changes, the capacitor current is maximum and then decays exponentially to zero. <A>
| Since an instantaneous step contains all frequencies, and the reactance of a capacitor decreases as frequency increases, at the instant of the step the reactance of the cap will be zero and it'll look like a short circuit.
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Difference effect filter in before and after amplifier Is there any difference between low pass filter that put before amplifier and after amplifier? Which one is good for PC audio amplifier? <Q> When you amplify something, you are increasing the power. <S> If you look at the cost of components, it will cost you less to filter a "pre-amped" signal. <S> Also, why would you want to amplify the "dirty" parts of a signal? <S> Clean it up, then amplify. <A> The answer is <S> yes in ideal circuit theory and the approximation to it that we call "The Real World". <S> Consider an ideal RC low-pass filter. <S> The open-circuit output voltage, due to an input voltage \$V_s\$ <S> is $$V_{o} = <S> V_s <S> \frac{1}{1 + <S> j <S> \omega RC}$$ <S> However, if one connects a load impedance \$Z_L\$ to the output of this filter, the loaded output voltage <S> is $$V_{ol} = <S> V_s \frac{Z_L}{R + <S> Z_L}\frac{1}{1 + j\omega <S> \left( R||Z_L\right)C}$$ <S> Thus, the DC gain and frequency response of our low-pass filter depends on the load \$Z_L\$ <S> However, if there is a good amplifier (high input impedance, low output impedance) between the low-pass filter and the load, the DC gain and frequency response will effectively be independent of \$Z_L\$. <S> There are other reasons you might want the filter before the amplifier rather than after. <S> For example: With the low-pass filter before the amplifier, the amplifier is notrequired to needlessly amplify frequencies well above the cornerfrequency. <S> If the low-pass filter is after the amplifier, the power delivered tothe load must pass through the filter. <S> and which one is good for PC audio amplifier? <S> Without more details of your application, it isn't clear what the 'good' is. <A> Ideally there is no difference. <S> Luckily we live in the real world, so math has its own limits. <S> If you are interested only in the low portion of the spectrum of your signal, then why would you amplify it all? <S> Amplifying costs power, and that's something you generally want to save. <S> Processing a signal before it gets amplified is not the golden rule, there are tons of situations where the opposite is true, but I've got the feeling that you want to build a set of loudspeakers <S> thus your input is line signal, while your output is a somewhat powerful signal meant to drive the speakers. <S> If that's the case you will need a LP filter on your input unless you want any kind of signal messing around in your circuit. <S> If you don't use a LP and your amplifier has a somewhat wide band it can amplify some high frequency signal which you can't even hear, wasting a lot of power... <S> For nothing.
| Moreover a filter after an amplifier should handle more powerful (aka amplified) signals, so it will need "bigger", more expensive components. If the filter is just a mathematical thing that you put in a signal processing chain, in this case its position it's irrelevant.
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Why the neutral of the UPS or inverter or generator is advised to be earthed/grounded? In my company we had few earthing related issues. When an consultant was brought in and apart from solving the issue, a suggestion was made to ground the neutral of the UPS and diesel generators that we have. The situation prevented me to seek clarification from the consultant, but I could not help myself pondering why? I Googled and it seems few other results also suggest the same but the pros and cons are not listed anywhere. I thought this would be the right place to ask and hence this question. <Q> I am not sure about UPSs, but a diesel generator is built to generate a voltage between its two (four) output terminals. <S> Whether these terminals voltage with respect to earth is high or low is not defined, and can't be. <S> The generator per se is not connected to earth, so its neutral terminal might be some volts above (or below) earth as well as some (tens of) tens of volts above it. <S> This might lead to tons of problems especially when you connect a correctly earthed apparatus to a generator powered one: their grounds are not at the same voltage <S> so a spark might occur on connection, much like static on windy days. <S> This should not be harmful for people but unfortunately electronics is quite more delicate and can suffer static discharges a lot. <S> Tying neutral to earth prevents this problem, and eliminates the risk for operators too. <S> Please note that this might not occur always since the generator chassis can be connected to the neutral, and the generator is usually placed on ground so a neutral-earth connection usually already exists. <S> That's a poor connection, but sometimes is enough to prevent problems. <S> I am guessing that some UPSs may have the output fully insulated from the AC line input, <S> if that's the case tying the neutral to ground is a good idea. <S> If the output is not insulated I'd say that tying the neutral to ground is a bad idea since you are basically connecting the AC neutral line to ground, and that's something that happens in distribution cabins but should not happen in the end user wiring. <A> For the UPS application > If the Neutral - Earth is to be linked, use of a Delta Star (3 Phase to 3 Phase + Neutral) transformer is recommended at the output of the UPS. <A> It is probably done to avoid high output voltages out of the ups. <S> The UPS takes the neutral as the reference point and works out the voltage on the output. <S> The trouble is the neutral on changeover between the mains and the generator can "disappear" so the ups loses its reference to zero volts and <S> the output increases and fries the connected loads. <S> Expensive! <S> So I suggest that the consultant linked the input to earth to tie it down to zero volts and control the output during changeover.
| It is not approved (in the UK) to bond Neutral to Earth more than once within an installation without galvanic isolation in order to prevent earth currents.
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The various terms for voltage I am trying to get a handle on some electrical circuit diagrams I've been encountering as I try to get back into electronics and building electronic gadgets. I've encountered a number of terms for voltage, and I am not sure what they all mean, so I was hoping someone could explain them to me. Here they are, along with what I believe they mean (if I think I know): Vpp (Voltage peak power?) Vp Vcc Vc Vr Vdd Vi Va Vt V0 or 0V (Negative or common "terminal", source of lowest potential?) There are also the common Vin and Vout designations, which are pretty self explanatory. I've also seen positive and negative voltage indicators in AC circuits, but that is also self explanatory. Are some of these simply context specific, with meanings only valid within a given circuit? Are they all commonly used? <Q> As suggested by other answers, most of this is pretty arbitrary anyway <S> Vpp : peak-to-peak voltage (for AC waveforms), historically this would also be used for the programming voltage for EEPROM or flash memory (particularly those devices that did not generate their own programming voltage on-chip). <S> May also be used for pull-up voltages (which could also be Vpu ). <S> Vcc <S> : positive power supply for many IC's, traditionally this referred to BJT based ICs, the 'cc' referring to the collectors of the integrated transistors. <S> Often this was matched with a negative supply, Vee ('ee' referring to the emitters of the transistors). <S> Vc <S> : Collector voltage for a BJT, <S> similarly Ve , <S> Vb may also be used for the emitter and base <S> while Vs , Vd and Vg may be used for the source, drain and gate of FETs. <S> Vr <S> : reverse voltage, particularly when referenced to diodes. <S> You may also encounter <S> Vz used to indicate a zener voltage. <S> Vf is used to indicate the forward voltage drop of the diode. <S> Vdd : positive power supply for many IC's, traditionally this referred to FET (NMOS, PMOS, CMOS) based ICs, the 'dd' referring to the drains of the integrated FETs. <S> Often this was matched with a negative supply, <S> Vss <S> ('ss' referring to the sources of the FETs). <S> Vi : input voltage. <S> Va <S> : used to indicate an internal analogue voltage point. <S> Vt <S> : May be used for the Thevenin equivalent voltage, or as suggested by WhatRoughBeast the threshold voltage (for a comparator or similar for example), or the termination voltage (also known as Vtt in the case of DDR=type memories). <S> Vo <S> : Output voltage for op-amps and the like. <S> 0V : <S> Zero-volts, not to be confused with Vo , refers to the system ground. <S> Also, Vhsys : may be used for the hysteresis voltage of a comparator type circuit. <A> Most of that notation is going to be arbitrary. <S> However, Vdd and Vcc are commonly references to your negative supply and positive supply, respectively (some devices require such configuration, like an op-amp for example). <S> Vpp is actually Voltage peak-to-peak, which has relevancy when speaking about time-varying voltage signals or output voltage ripple (like pulse signals or sinusoidals). <S> Vo is commonly read as Voltage-Out and is typically the voltage of interest if labeled as such. <A> sherrelbec's explanations look OK to me. <S> Of the remaining, Vp usually refers to a peak voltage <S> Vr is commonly a reverse voltage used, for instance, to indicate a threshold or breakdown voltage <S> Vi might equally be an input voltage, or the voltage appearing on a current sensor <S> Vt can equally be a threshold voltage or the voltage produced by a termination network <A> I've seen Vcc defined as the collector supply voltage, and Vc as the actual voltage at the collector (corresponding definiitions for Vss/Vs, Vdd/Vd, etc.)
| Vp : peak voltage (again, for AC) referenced to system ground, or 0V .
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Can I use + charge 3 LiPo cells in series with 3 separate charge controllers? I'm looking at this circuity from SparkFun . It's an mppt + LiPo charger, which is exactly what I need, but I need several cells. I'm actually fairly content to just build 3 of these, and wire each in series, but will that cause any balancing / other issues? I'm assuming as long as each charge controller is grounded correctly (i.e. one's ground is the next one's VCC), it should be fine, but want to confirm <Q> it's own PV panel. <S> See below for cautions. <S> The batteries can be connected ground1 = system ground, V+1-ground2, V+2-ground3, V+3= <S> Vout <S> PV panels should be nominally identical and batteries 9cells) should be nominally identical. <S> With this arrangement, charging may occur at slightly different rates due variations in panels, insolation variations (which should be minor), slightly different contamination levels on panels etc. <S> This will lead to one cell being charged before the others and one "coming last". <S> If the differences are as small as they should be <S> this will not matter. <S> If there is more than enough sun <S> the others will catch up in the same day. <S> If panels are at say 90, 85, 75% at the start of a day then as long as discharge is not more than 75% of max possible all will be well. <S> If say cutoff point was 3V/cell thgen monitoring cell voltage individually and stopping discharge when any one cell dropped to 3V would be easy and safe. <S> Rechaarging would then satrt at about 15%, 10% and 0% capacity. <S> On recharge, when the 1st cell reached the CC/CV knee its charge rate would slow (if the charger was able to deliver I_chg_max) and the others would start to catch up. <S> As long as on average there is enough sun to sometimes deliver full charge between discharges the system will self balance, as long as V_cell_moin is not allowed to drop below 3V or whatever lower target voltage is set. <S> Modern electronics makes per-cell monitoring cheap and easy and there is little point in omitting it in systems where imbalance is likely. <S> Thusly: <A> You can't just wire the power inputs in series, because the chargers won't share the voltage evenly. <S> You would need to feed each charger with its own isolated power supply. <S> If you are charging a 3 cell pack then it might be easier to use a single 3 cell charger, then put a balancer circuit across each cell to even out the pack voltages. <S> Once a Lipo pack has been balanced it doesn't take much to keep it that way, so the balancers should only have to handle a few hundred milliamps at most. <A> Multiple Lipo battery is a bit dangerous without special extra circuit balance cells (keep them as same charge level) as unbalance increase fire risk. <S> One popular practice is to use single cell, as example in cell phone ext batt pack, inexpensive voltage boost chip boost 3.7V to 5V. Low cost small break out board boost 5 to 35 volts. <S> Example of one 12V booster, some has high V and A too <S> , http://www.coolcomponents.co.uk/adjustable-4-12v-step-up-voltage-regulator-u3v50alv.html
| Wherever imbalance may occur - whether n this system or others, discharging without individual cell monitoring risks discharging one or more series cell to below its safe minimum level. Yes, this will work PROVIDED that each cell has Each PV panel has to be connected to Vin and ground of its related charger and NOT to the other PV panels.
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Batteries with same voltage but different number of plates Why do some batteries have the same voltage with a different amount of plates? What is the specific purpose of that? E.g. 12 V with 11 plates and 12 V with 13 plates. <Q> There seems to be some temptation here on EESE to assume "plates" refer to or equal cells. <S> I believe this is mistaken. <S> In effect, each cell is a number of cells in parallel, though sharing the same bath of electrolyte. <S> Six of these cells in series then form the 12v battery. <S> Likely, the 11 or 13 plate designs are two currently or recently common tradeoffs in the parallelism vs. size or thickness. <S> I'd expect we'd need a battery designer to explain the advantages or disadvantages of each. <S> Edit: in fact this is the case. <S> Automative-, motorcyle- etc style batteries often have right on their data plate the number of plates per cell , and 11 is a very common number. <A> In principle, a battery is made up of "cells" (which I think is what you mean by "plates") - and the word battery comes from that <S> (originally a battery is a "a usually large group of things, people, ideas that work together" - so by definition a "battery" ought to be more than one cell although it is now used for a single cell too...) <S> The cell has a "nominal voltage" that depends on the chemistry (material of the anode and cathode) and temperature. <S> However, as soon as a current starts to flow, the voltage drops due to internal resistance. <S> When you measure the voltage on a "12 V" car battery, you can easily measure 13 V or even 14 V. <S> I believe the key to your answer is understanding that a battery voltage can be specified "under load" <S> - it is more meaningful to know you can get a battery to deliver 20 A at 12 V than to know it measures 12 V when nothing is connected. <S> So if you want a small battery capable of (short term) <S> high peak loads while maintaining the nominal voltage, you can increase the number of cells. <S> When you do so, you may need to put in place a regulator that prevents the low-load voltage from exceeding the capabilities of the circuit under load... <A> Different types of cells inherently have different voltages due to their different chemistry. <S> It can therefore take different numbers of cells to get to about the same voltage when using different types of cells. <S> For example, lead-acid cells like in a car battery produce around 2 V, typical AA alkaline cells about 1.5 V, lithium coin cells usually around 3 V, etc. <S> Your 12 V car battery contains 6 lead-acid cells, although it would take 8 AA alkaline cells in series to make the same voltage. <A> The number of plates is the plates per cell . <S> As others have said, the cell voltage is fixed by chemistry. <S> The number of plates per cell will always be an odd number- a 13-plate battery will have seven negative plates and six positive plates per cell. <S> Ref. <S> Battery Reference Book <S> By Thomas P J Crompton Available cranking amperes, and life are affected by the shape and number of plates per cell (deep discharge capability seems to be more affected by the free space under the plates). <S> I would think that more plates = higher cranking amperes, but more costly and probably shorter life, all other things being equal. <S> The first part is confirmed by this reference. <S> Edit: As Olin points out, the references I pointed to and this particular terminology specifically refer to lead-acid batteries. <S> It's been consistent since the 1920s, when battery cases were made from glass jars encased in hardwood, and "battery repairmen" would dismantle batteries and replace the plates (at some risk to themselves ).
| Cutaway images of automotive 12v lead acid batteries I am finding seem to show that each nominal cell consists not of a single pair of plates, but rather of a stack of many interleaved plates, alternately connected to the cell's positive or negative terminal. In order to achieve voltage under load, one can increase the number of cells (increasing the no-load voltage), or lower the internal resistance (typically, this requires a larger battery).
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Difference between applying voltage, and voltage across? I'm confused about the two terms, when voltage is applied and across a certain element in the circuit. <Q> What Ignacio said is the core of the answer, I hope I can help you out going a bit deeper. <S> Generally the only distinction between "applied voltage" and "voltage across" is how you are dealing with voltage itself: <S> you apply a voltage to a bipole taking a voltage source and putting it in parallel with the dipole. <S> you usually measure a voltage across some dipole, putting a voltmeter in parallel with it. <S> That's to answer your question. <S> Now what if you apply a voltage generator? <S> What would the voltage across it? <S> The answer is: there is no answer. <S> That is a limitation of the model we are using. <S> Ignazio makes the useful example of a diode: you apply 5V but across it there's only something like 0.7V <S> : that's because your voltage source has an internal resistance where the remaining 4.3V drops. <S> Remember that most of the times when you apply a voltage to a dipole, the voltage across it will be exactly what you are applying. <S> addendum <S> Since this is at the top now, and I've read some others very good answers, and since the question is very basic <S> I'd like to add two words about potential , a word that every answer uses. <S> A potential is a scalar field associated with a vector field. <S> This vector field must be conservative for the potential to exist, and for the electric field this is true only for electrostatic fields. <S> When things start moving around no potential can be defined. <S> I don't want to be the fussy physicist but a professor once throw a chalk at me for this imprecision (he was quite precise) so since this might be seen from young students I though this should be pointed out. <A> A voltage is always across two nodes, it is the difference between the electric potentials of these two nodes. <S> They are strictly speaking always applied by something, but we speak about applying a voltage across two nodes when we set the potentials of those two nodes by connecting them to the outputs of a voltage source, which role is to make sure the voltage across those is fixed to a known value. <S> The voltage of a node is often a shorthand for the potential of that node with respect to the ground of the circuit (which, as a reminder, is only a node which has been arbitrarily associated to a 0V value). <S> Electric potential is often compared to height in the liquid analogy where the flow of water is electric current and rocks along its path, resistance. <S> Reminder: <S> a node is a uniquely defined point of interest in the circuit (a pin, the intersection of several branches etc.). <A> The phrase "voltage across a circuit element" means precisely the potential difference between the terminals of the circuit element. <S> One can measure this voltage with a meter. <S> The phrase "voltage applied to a circuit element" is less precise <S> but I believe it means that one is driving the circuit element with a voltage source of some type and that the voltage across is, more less, fixed by this source. <S> The opposite of this would be the "voltage supplied by a circuit element" which would imply that the voltage across is generated by the circuit element, e.g., a battery, a charged capacitor, etc. <A> applied voltage means the voltage which is given by us to the component. <S> voltage across means the voltage which is reduced by the component due to the internal resistance of the component <A> Applied voltage to a component is the actual voltage given to the component. <S> Whereas voltage across a component is the voltage drop/voltage dissipated by the component. <S> It is always between two points as it is only the difference between those two points that produce an electromotive force. <S> Now, applied voltage to a component and the voltage across a component may or may not be the same value. <S> If you are applying 5 volts to a single resistor circuit, that resistor will get all of the 5 volts as voltage is conserved within a loop (KVL). <S> If you now have 2 resistors in series of equal value, each resistor now gets 2.5 volts of the total 5 volts. <S> Technically, in the latter case, a total of 2.5 volts is applied to the last resistor, and thus the voltage across it is 2.5 volts. <S> However, there exists non-voltage driven components, meaning they're driven by current instead. <S> A diode of any sort is a good example, where you may apply 5 volts to it, but the actual voltage drop may be something around 0.5 volts. <S> In this case, the rest of the voltage is sent back to the source and power will be dissipated by the internal resistance of the source. <A> The right way to say it would be apply voltage across something - it sounds more precise. <S> The voltage is a difference in potential between two points. <S> So, when they say apply voltage , they omit the word across , assuming you know between/across which two points. <S> Commonly, this phrase is used to say 'Apply voltage to a circuit', meaning provide power to a circuit, since you know where to connect two wires. <S> Similarly, you can say apply voltage across the circuit, but it might sound somewhat redundant, but more precise. <S> This phrase is rather used to specifically tell between/across which points the voltage should be applied or measured. <S> In both cases it is meant the say the same thing, but might be misunderstood.
| In both cases, voltage means a difference in electrical potential between two points. The two wordings though does not mean the same thing at all.
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Is there a difference in heat produced on a conductor between AC and DC at the same amperage and voltage? Is there a difference in amount of heat produced on a conductor between AC and DC at the same amperage and voltage? For example if you ran 50A at 480V through a conductor would the heat produced be the same for AC (60hz) and DC? <Q> Yes... assuming for AC you mean 480V <S> RMS , not peak-to-peak . <S> The power (in watts) should be: 480V * 50A <S> = 24kW <S> However if you are measuring 480V AC peak-to-peak, the RMS voltage will be ~340V, in which case the power would be less: 340V <S> * 50A = 17kW <S> Power is calculated using RMS so that it can be compared to DC in a useful way. <A> Yes. <S> DC current will travel through the entirety of the wire -- AC current will tend to move on the outside of the conductor (see Skin Effect ). <S> This will cause slightly more ohmic heating in AC than in DC. <S> That's why AC wire is usually stranded while DC wire is usually solid -- more strands = more surface area. <A> If the AC load is reactive instead of purely resistive, then the currents which experience ohmic (conversion to heat) losses in the transmission line include not only the real current which does work, but also the reactive currents. <S> Because the losses are tangible, industrial customers are billed for reactive power as well as real; for a large facility it is worthwhile to install banks of power-factor correcting capacitors to balance the inductive reactance of motor loads with opposing capacitive reactance, keeping the power on the transmission line mostly real.
| AC voltage and current is continuously variable in a sine wave.
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Is there a clever way to prevent a window comparator from oscillating? Here is my schematic It's basically a crude battery level detector (I am very aware that this isn't the best way to guage battery life, I'm not here for that) it's not crucial that it is super accurate. Problem When the battery voltage is near 4.05V or 3.45V the output of U9 oscillates and two LEDs are on at the same time, LED1 and LED2 or LED2 and LED3 respectively. Datasheet here It is necessary that only 1 LED is on at a time. Solution I have found one solution but I am asking a question here to see if there is a more elegant or clever one (hopefully). Board space is ridiculously important here and my solution really doesn't fit well despite being only 4 components. The biggest problem is I want to be able to rework it into existing boards, and my solution is ugly and time consuming (4 cuts, 4 jumps). I made U9 into two schmitt triggers with 100mV of hysteresis. Yes it works, but it makes the schematic ugly and adds 4 more components. Hopes My thoughts were maybe there is an off the shelf comparator that has built-in hysteresis (don't know how to search for this on digikey) or maybe you clever guys on EESE have a better solution. <Q> Sorry, but hysteresis is the way to go, and in this case it will take 3 extra resistors, not 4. <S> In order for the low-voltage hysteresis to work, C25 has to go, and you can take advantage of the Thevenin equivalent resistance in the R15/R16 divider to provide the input resistance you need. <S> And that said, your circuit has a couple of things that make my brain itch, so I'll deliver some unwanted analysis. <S> 1) <S> 2) <S> High output for the comparator is only claimed to be 2.4 volts (typ) for a 4 mA load. <S> 3) <S> As a result of the previous, if U10 is a standard CMOS part with 2.5 volt input threshold, it may not work properly. <S> It probably will, since at 2.4 volts your LED won't be drawing 4 mA. 4) <S> You can get rid of U11 altogether. <S> With a worst-case bias current of 600 pA, the offset due to a 1 M input resistance is only .6 mV, compared to a comparator input offset of 5 mV. 5) <S> You don't need the 22 pF caps on the output of the comparators. <S> If anything, they'll fight the hysteresis. <S> They'll lose, of course, but still... 6) <S> You probably don't need C31, since the input capacitance of Q2 is 27 pF (typ), but <S> a 10 to 100 ohm series gate resistor would be a good idea just on general principles. <S> 7) <S> Your voltage set point networks could be done with 3 resistors rather than 4. <S> 8) C38 can go, since VBAT is already filtered by C22. <S> Other than that <S> , nothing jumps out at me. <S> I do like to see someone as paranoid about decoupling as I am. <A> You're probably going to hate me, but I'll suggest using a microcontroller with an ADC converter. <S> Just have it wake up every now and then and set the output state. <S> The LTC1042 window comparator is an old-school solution (I think I designed one in back in the '90s) but its kind of expensive. <S> but they might go back and forth). <S> Neither of those suggestions are helpful with rework, but for a relatively small quantity of boards the cut and jumps doesn't sound too terrible (especially if you get someone else to do it for you). <S> The <S> Of course you're free to divide down the input to make more of that. <A> So you handle the window comparison and hysteresis in firmware. <S> The microcontroller has a multiple channel 10-bit ADC and plenty of I/O lines to drive the LED's. <S> The outputs can drive 25 mA <S> so you won't need external drivers. <S> Since it has an internal oscillator, all you need to add is a decoupling cap. <S> Costs about $1 in single quantities, and 70 cents in 100's. <A> Depending why and what you want the comparator for, you may find what you need from one of the hundreds of battery control/charger/power management IC, readily make and designed to handle great variety of real-life situations, spanning wide range of battery chemistry/design/capacity. <S> There are different IC for LiPo, NiMH, Lead Acid, etc. <S> Some are multiple modes.
| LTC6702 does have built-in hysteresis, but only 4mV. It is a window comparator that you can set the sample rate on, so both LEDs won't be on at once ( The 4.05 set point violates common mode limits by .25 to .5 volts, depending on your paranoia about operating temperature. The LEDs may not be as bright as you expect. If you want to reduce your parts count (and space requirements) drastically, I would replace all of your circuitry with a small microcontroller, such as the PIC16LF1613 which comes in a tiny 16 pin, 4 mm square QFN package (plus several other choices).
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What is the effect of the rated current for a breaker when a 230V 6A rated circuit breaker is used in a 24 V DC circuit? I want to use a normal household breaker for connecting to a 24V circuit. The input supply is a 230V AC to 24V DC SMPS. The output of the DC converter is connected to a load (rated 180 milliamps @ 24 V) in parallel with a battery bank having two batteries of 7.2 Ah each. If I use a normal AC MCB, and if there is a short circuit in the above circuit whereby the battery supplies a higher current, will the MCB trip? <Q> However, will that protect what you think it will protect? <S> That is, if a short occurs, will the wiring between the short and the breaker be sufficient to carry the current required to trip the breaker? <S> Or will the wires simply burn? <S> This arises because with a load rated at only 180mA, the wiring to and within the load circuitry is likely to be quite fine. <A> Right, AC MCB should not be used for DC application. <S> There are many factors, at first the design of the surge/spark current suppressed material. <S> DC is a continuous level current while AC has a point when the level goes to 'zero' which help the surge current suppression, hence DC breakers are usually more rigid and have a better suppression mechanism. <S> Other than that AC current has a skin effect on the contactor, while DC current has no such effect. <S> You may see lot of DC breaker manufacturers, one of them I recommend is E-T-A, you may visit their website here: http://www.e-t-a.com/ <S> You may find numerous type of DC breakers depending on your application. <A> A standard electromagnetic or bimetallic breaker should work fine on 24V DC. <S> Here's an example which is rated for 6A at both 250V AC and 48V DC. <S> ETA - 106-P30 6A - CIRCUIT BREAKER, 6A <S> This breaker has a resistance of 0.05 Ohms or less, so provided your power supply and/or battery can deliver at least 0.6V for 20 seconds at 12A (or 5 seconds at 30A etc.) <S> it will trip reliably. <S> A good 24V 7Ah SLA battery should be able to maintain ~20V for several minutes at 30A.
| A household circuit breaker (of relatively low amp rating) can be tripped by the levels of current that the batteries are capable of.
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How to generate Red light from a Blue-Dominated Spectrum I have a white LED light source, the spectrum of which looks as follows --- want to filter out red light from this white LED source. But as you can see the spectrum dwindles in the red zone. And therefore, when I put red filter (absorptive filter) in front of my white LED light source, the brightness suddenly drops down to abysmal values. Is there a way in which I could use this spectrum and get red lights without compromising on the brightness? I thought about interference and then I came across dichroic filters. But I'm not sure if that'll do it. I also came across these filters called LEE LED filters which are filters specifically meant for LEDs. But I do not understand the scientific principle they work on. I would be grateful if you can follow the link and explain to me the scientific principle. From the spectrum shown in the link, these filters can give you more than 85% red. But the overall transmission is just 10.1% of the original intensity. So it's kinda bleh. Or maybe I don't understand properly. A red LED light source would be just perfect for me, but a high wattage (50W-100W) red LED light source is hard to come across in the market. Let me know if there are other options you guys can think of, for producing cool red LED light. Right now I use a tungsten lamp with RED absorptive filter, but there is so much heat (infrared) in it that my houseflies get fried up. [ I work in an insect flight lab and we use red light for high speed videography, since houseflies dont see well in the red zone]. So I want to switch to LEDs. <Q> Your "white" LED is actually a very bright blue LED combined with a phosphor that absorbs some of the blue light and re-emits it as the broad yellow peak you see in your spectrum. <S> While it would be theoretically possible to substitute a different phosphor that emits the red light you seek, I don't know of any commercial products that do this. <S> However, very bright red LEDs are readily available. <S> They're used as the strobe lights on emergency vehicles, such as ambulances, fire engines, police cars (depending on your jurisdiction), etc. <A> You can't add light with a filter that isn't there in your spectrum to start with. <S> Operating theatre lights often use pure red LEDs to augment the spectrum of 'white' LEDs. <S> I would imagine that you need good colour rendering at the blood end of the spectrum in that environment. <A> Given the fact that one can easily buy off-the-shelf, 50 to 150w, LED flood lamp, with heat sink and drive electronics, but they typically comes with white LED. <S> One may consider simply replacing the central coin sized LED module and make corresponding small change in drive V and A. <S> Either do it yourself or <S> if, in (small) batch quantity, manufacturer can do it for you. <S> For example, I am using this 10W LED, which is 9 LED die, 3 series in one set and 3 set in parallel. <S> It is 350mA, 10V. The constant current driver board is adjustable up to max. <S> 5A. <S> The RED LED module shown (which I do not have on hands, but, same principle as the 10W I have) is 10 LED by 10 LED, each die is one 1W LED, module is about 32V and 3.5A. <S> The example spec. <S> is white LED (which is actually blue then changed to white) and voltage is slightly different from red but is same range 3.x to 4.x V Commercial 100W driver with CE, 85 to 250V AC in, out DC 20 to 38V, 3A <A> I'm afraid you understand it perfectly, although you don't much like it. <S> Once your light is generated, there really isn't anything you can do with the unwanted frequencies except filter them out. <S> There just aren't any good ways to convert one frequency to another. <S> (DPSS lasers are one exception, but not applicable here). <S> As a general rule, incandescents produce about 10% of their power as visible light, and virtually all the rest as IR, which is why your red-filtered light is very hot - the red filter is not eliminating the IR. <S> LEDs provide considerably better efficiency, but as you've noticed, not much of it is red. <S> With not much there to begin with, any red-filtered version will be, in your words, abysmal. <S> There are two types. <S> The most common is used in cameras and other sensors, and has the problem that they won't handle a lot of heat. <S> The other variety is used in things like LCD projectors, and they are designed to take heat, although cooling design is also important. <S> One source is http://www.mecanusa.com/polarizer/ColdFilter.htm , but you'll notice that these puppies can be expensive. <S> As to LEE LED filters, a general, non-technical outline is here http://www.leefilters.com/lighting/led-02.html#led-explain <S> but it's actually pretty simple. <S> Regular filters work on incandescent light, with a standard black-body intensity curve. <S> As you may have noticed, your LED intensity has this big fat spike in the blue. <A> An longpass optical filter with a cutoff at 480nm would do the job. <S> I mean a "real" optical filter, like interferometric filters with very sharp transition. <S> They might cost 10th of $ For the brightness, this is the best solution. <S> Because the filter is quasi ideal, almost all the energy in the red part would pass through it. <S> If this is not bright enough, that means there is not enough power into the emitted light at the first place. <S> No way to boost this. <S> Except adding LEDs or focusing the output light to the ROI.
| Dichroic filters are used because they have relatively sharp cutoff characteristics and so can produce narrow passbands which provided "purer" colors, but they don't actually convert light from one frequency to another. As poster states, A very bright RED LED lamp (~50W) with a lamp tripod or something would be perfect for me. What you need, I think, are IR blocking filters, or cold filters, for your incandescents. The LEE LED filters just have what it takes to compensate for this spike.
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Outputting a clock signal from an FPGA Referring to question here: Click here , I'd like to use the 16 channel LED driver to run my 7-segment displays. I'm using a Spartan 6 LX9 FPGA to implement a 16-bit microprocessor that will take care of it all. The development board I'm using is pretty barebone and has an onboard 100MHz clock, or perhaps I should say oscillator. What I was planning to do is slow it down to perhaps a few megahertz, then output it via a GPIO to the LED drivers and synchronize it that way with a separate verilog module. My question is, do I have to take special care when trying to output a clock through a GPIO? Perhaps use a particular pin? I'm given to understand that the clock input is supposed to be through a predefined pin. The clock could be slower, as it doesn't really need to be too fast. Here's the online user manual to the development board (pretty basic): Click here . EDIT: More information: I'm concerned about clock skew. The chip I'm looking at writes at every positive clock edge. What I plan to do is to divide up the clock and send it out to the chip, while the FPGA also uses the same signal to synchronize and send data to the chip serially. Timing seems to be pretty tight. A slow enough clock might be the answer though. <Q> It doesn't matter if it's balanced clock or arbitrary data. <S> Also, output at units of MHz shouldn't give you any trouble. <S> In sequential logic, there is a trend to use as little of clocks as possible. <S> This is because clocks are very often the most important signals, as many others depend on them. <S> I'll try to depict here the issue of propagation delay, as projected to the thing called clock skew . <S> Imagine you have two clocks, one derived from the other: __/--\__/--\__/--\__/-- <S> <- original clock___/-----\_____/-----\ <S> <- clock divided by 2 Note that the clock below is delayed by 1 character (that would be one flip-flop delay, which is minuscule.) <S> It's just greatly exaggerated in the ASCII art. <S> But, in a black-boxed FPGA design you don't have to worry about that. <S> It's all taken care of for you. <S> All those worst-case timing issues are taken into account in the form of maximum clock frequency, that the design can run at. <A> The Spartan 6 family does not have dedicated clock outputs that connect directly to a clock network. <S> For a slow clock, such as you are suggesting, that probably doesn't even matter, because jitter and rise/fall time differences for I/ <S> O pins are slower than the clock. <S> For fast clocks, the recommendation I've found is to use an I/ <S> O pin in ODDR mode to make the compiler aware that both rising and falling edges on the signal matter, and let it handle the rest automatically. <A> What you are proposing to do, basically, is to take the FPGA 100 MHz clock, run it through a few flip-flops to reduce its frequency, and then output the reduced frequency. <S> Your proposed output is not, in FPGA terms, a clock. <S> It's just another registered output. <S> So go ahead and run it out through a GPIO pin. <S> You are, I think, confusing this with the need to run an FPGA clock into the chip via a special clock input pin. <S> In this case, the pin is connected to a dedicated set of drivers within the chip which distribute the clock to all the internal points which need it (and which provide considerable capacitive load). <A> There are specific requirements for inputs that are clocks because of the way the clock is distributed throughout the FPGA fabric. <S> However in your case, you are just outputting a divided clock on a GPIO pin <S> so there are no specific requirements other than those that apply to the pin <S> anyway <S> - ie: it's an output with certain current limits etc. <S> Here's a snippet from some code I wrote - it's in VHDL rather than Verilog though: <S> my_clk_div8: process (clk, i_enable_n)begin --- <S> On async reset if (i_enable_n = '1') <S> then clk125 <= '0'; clka <S> <= '0'; clkb <= '0'; clkc <= '0'; --- <S> On clock edge. <S> Implement 4 flipflops with last output inverted --- and sent to first input. <S> Divide 100MHz by 8 to get 12.5MHz elsif (rising_edge(clk)) <S> then clka <S> <= <S> not clk125; clkb <= <S> clka; <S> clkc <S> <= <S> clkb; clk125 <= clkc; end if;end process my_clk_div8;
| You can use any GPIO pin you want for output.
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How to get more current from 555 timer? My power supply is filtered regulated 5 volt 500 mA . I am using a 555 in monostable mode to switch the motor on for a certain amount of time after the 555 is trigerred. (The 555 is being trigerred by a counter circuit). But the output current from the 555 is too low. How can I use a transistor like 2N3055 to get full 500mA current? What other ways are there to achieve the same thing? Will this work? <Q> An emitter follower will have a voltage in the emitter about 0.6v lower than the voltage at the base. <S> It will work if you don't have a problem with the reduced voltage level. <S> Note that a 555 with 5v supply can have an output as low as 3v depending on the output current. <S> The alternative is to use a transistor or mosfet as a low side switch (switching the ground side of the load) <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The mosfet will be a better solution (more efficient) because it has a lower voltage drop across it (drain-source), and it doesn't need a constant current in the gate (for static operation) as the transistor does. <S> Just select a logic level Nmosfet so that it can turn fully on with low resistance <A> As a high side driver, the 555 is only rated for something like 100 mA, and it takes a good sized output voltage hit at that current, so it'll never work directly as a motor driver. <S> The emitter follower you've shown is better, but you can't drive it into saturation so you'll still lose about a volt across the transistor when it's on, starving your motor somewhat. <S> If you use a P channel MOSFET with a reasonable Rds(on) for the high side driver then you'll get nearly all of the 5V supply across the motor when the MOSFET turns on, plus it'll dissipate very little power. <S> The circuit shown below assumes you're stuck with a high side driver, and it'll work with a bipolar 555. <S> The LTspice circuit list is here if you want to play with the circuit, and if you want to switch to a low side driver get rid of Q1B and replace R6 with the motor and D2. <S> Just as an aside, the FDS4559 is a pair of about 3 amp MOSFETs in the same package, one P channel and one N channel for less than a dollar in onesies at DigiKey. <A> If possible, it would be better to put the load on the "high side" of the transistor — between the +5V supply and the collector, with the emitter tied directly to ground. <S> You'll also need a resistor in series with the transistor's base to limit the current. <S> 100Ω would give you a base current of 50 mA, which should be plenty. <S> For even lower loss, substitute an N-channel MOSFET for the NPN transistor. <A> Will this work? <S> Nope. <S> Not even close. <S> Take a look at the NE555 data sheet p.5 "High level output voltage". <S> Although you are using an extra transistor to reduce the load on the 555, it demonstrates that the output is not good at sourcing current. <S> Add the base-emitter drop in the transistor (~.7 volts), and you are not certain at all of getting a good motor drive. <S> This, on the other hand, will: simulate this circuit – Schematic created using CircuitLab <S> Depending on how much the motor draws, R6 may need to be decreased. <S> This will certainly work to 150 mA. Note that you can simplify the circuit if you use a PNP for the output stage. <S> Then Q3 goes away, and the collector of Q2 becomes the motor +, with motor - at ground. <S> R4 then controls the amount of drive current and would definitely need reduction.
| Using the transistor as an emitter follower will work, but you'll only get about 4.3 V at the output, because of the 0.7V drop of the transistor's B-E junction.
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What are the big plugs for that you find on high demand devices in North America (120V-Areas)? Talking about AMEA socket system and 120V. High power demand devices like A/Cs do have bigger plugs with some switches on them. What exactly is their purpose? Seems like a fuse but isn't the socket cirquit fused against overload? <Q> If you mean these types of plugs, they are to indicate what type of current the device needs/is rated for. <S> Image from Wikipedia <A> If this is what you're talking about, It's called an "IEC power entry module" and it's fused internally and supplied with internal switches in order to switch between various mains voltages, <A> High power demand devices like A <S> /Cs do have bigger plugs with some switches on them. <S> It's not really the "high power" which suggests their presence, but rather the possibility of a current leakage path developing.
| It sounds like you are talking about the plugs with a built in Ground Fault Interruptor designed to automatically disconnect the circuit if current flow from the "hot" line is diverted anywhere other than the "neutral".
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How does voltage drop work? I've been working with electronics for a while now, and I understand how to work with/account for voltage drop. But I'm still mystified as to how it works, particularly with components in series. Say I have two small incandescent lamps in series hooked up to a power supply. Through what physical mechanism is the voltage spread evenly over both lamps, resulting in both lamps running dim, instead of the first lamp consuming all the power and running at full brightness and the second running dark? EDIT: It seems that people were getting hung up on the fact that I was using LEDs in my example and were focusing on whether the LEDs would run at all with lower-than-rated voltage. This wasn't my intention - I am focused on the phenomenon that both drop an equal amount of voltage instead of the lead LED dropping more voltage than the second LED. As a result I changed the example to incandescent lamps which show a closer-to-linear relation between voltage and brightness. <Q> Through what physical mechanism is the voltage spread evenly over both lamps, resulting in both lamps running dim, instead of the first lamp consuming all the power and running at full brightness and the second running dark? <S> Ohm's law. <S> Both lamps, presumably having been manufactured similarly, have similar resistances. <S> Given that both have the same amount of current running through them (KCL), they will have similar voltage drops across them. <A> Think of each component as a two terminals black box. <S> This component will have a current that flows from one terminal to the other, and a voltage that you can measure across these terminals. <S> What characterizes the component is the V-I characteristic , i.e. the function that given a certain voltage across the component allows you to calculate the current that flows into it, and vice versa. <S> For example, for a resistor the V-I function is the known Ohm's law:$$V= <S> I\cdot <S> R$$ <S> The physical mechanism that leads to a particular V-I characteristics is not a single mechanism but varies wildly between components, and would be quite a huge topic to be treated in a single answer. <S> Let's take your LEDs: they obey an exponential law, meaning that:$$I_{LED}=I_0(e^{\frac{V_{LED}}{V_T}}-1)$$where \$V_T\approx 26\$mV is the thermal voltage and \$I_0\$ is a parameter of your LED. <S> Since you say that your LEDs require 3.3V, assuming that forward current would be 20mA you can calculate \$I_0\approx 1.5\cdot10^{-57}\$A from the above expression. <S> Now you connect two LEDs in series, each of them will have its own current and voltage, so how can you solve the situation? <S> You throw in the equations coming from the circuit:$$V_{LED1}+V_{LED2}=3.3V\\I_{LED1}=I_{LED2}\\I_{LED1}=f_1(V_{LED1})\\I_{LED2}=f_2(V_{LED2})$$ <S> Now you can happily solve this system (note that it is not linear). <S> It's quite easy because:$$f_1(V_x)=f_2(V_x)=f(V_x)==I_0(e^{\frac{V_x}{V_T}}-1)$$ <S> So you can write almost immediatly \$V_{LED1}=V_{LED2}\$ that leads to \$V_{LED1}=V_{LED2}=\frac{3.3V}{2}=1.65V\$. <S> Please note that all of the above holds if the IV characteristic is true. <S> There are some components that have "piece defined" characteristics, so that after finding a result you have to check that the characteristic you've chosen at the beginning is the right one. <S> The answer to your question basically is: math, that's the reason. <A> If the LEDs are rated at 3.3 volts then more likely neither will be seen to glow at all - LEDs need to have a "certain amount" of voltage across them before they appear to be glowing. <S> On the other hand if you were "forcing" the rated current through them both you would have to be developing about 6.6 volts across the two in series - one LED couldn't arbitrarily decide to be conducting full current with very little volt drop allowing the other one to be at full brightness. <S> So, turn it round - for one LED to be fully on <S> it has to be taking the rated current which has to mean that rated current <S> is flowing thru the other hence it must also be at full brightness. <A> Your all LEDs are will not work. <S> Because your LED needs 3.3V. <S> If you are connected two LEDs in series. <S> Current remain constant. <S> But Voltage will change (Divide - if you are use same LEDs) . <A> In series circuit - voltage splits proportionally to resistance. <S> In addition - bulb resistance is variable - depends on filament temperature. <S> Higher temperature is higher resistance. <S> When you have two bulbs with diffrent power - one with smaller power will get more voltage initially and heat up fast. <S> Resistance will rise. <S> Bulb with higher power will heat up to lower temperature and resistance will be much smaller than in normal conditions.
| You have to use identical or very similar bulbs with same power to split voltage evenly.
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Why is voltage higher at positive wire instead of negative wire in this example? If electrons travel from "-" to "+" and voltage decrease going through resistor why is voltage higher at positive wire instead of negative wire? <Q> An unfortunate convention. <S> Conventional Current assumes that current flows out of the positive terminal of a source. <S> When Benjamin Franklin was writing up his discovery, at the time without any detailed knowledge as to how it was actually occurring, he had a choice what is to be labeled positive and what is to be labeled negative. <S> image credit <S> http://xkcd.com/567/ <A> Electrons have negative charge, and hence where they come from has negative relative voltage (since energy here is positive). <A> Voltage is the difference in potential between two points, so to say a given point has a given voltage is meaningless unless you also specify the point you're measuring against. <S> If you're measuring against ground, with the negative terminal of the power supply (let's call it a battery) grounded as in your sketch, <S> then by definition the negative side of the battery has zero voltage because you are measuring the potential between ground and ground. <S> There isn't any. <S> But the positive terminal has a voltage of one volt. <S> If you were to remove the ground from the negative terminal and ground the positive terminal instead, the positive terminal would have zero volts (again, with respect to ground) and the negative terminal would have (negative) one volt--the higher voltage.
| Electron Flow is what actually happens and electrons flow out of the negative terminal of a source
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GPS Antenna Selector: How to electronically select one antenna from multiple available antennas? I want to connect multiple GPS antennas to a single GPS module and select one of them based on some logic. Basically, I want the module to receive signals from only the selected antenna. Is there an inexpensive/generic/common IC that can act as a Demultiplexer for GPS RF signal (without distorting it too much)? <Q> I believe the component you are looking for is called an RF switch. <S> They are used for example for switching the TX path of bluetooth and wifi chips to a single antenna. <S> RF switches are quite inexpensive (relative term, I know), and come for many purposes. <S> The terminology for RF switch classifications is similar to regular switches, i.e. SPDT means single-pole-dual-throw, i.e. a single common signal can be routed to two locations. <S> Here's a relatively generic one which works over a relatively large frequency band (incl. <S> GPS) <S> http://fi.mouser.com/ProductDetail/Skyworks-Solutions-Inc/SKY13270-92LF/?qs=sGAEpiMZZMtsfndvJ9ArQ1GAoWUJ3yIM3lKzNTG0W6Y%3d <S> Datasheet: http://www.mouser.com/ds/2/472/200128G-23362.pdf <S> Please note that RF switches are nowadays often minuscule components and can be difficult to solder manually for the unexperienced. <S> Here's a more generic article about RF switches by digikey: http://www.digikey.com/en/articles/techzone/2012/aug/rf-switches-simplify-multi-antenna-systems <A> These are available from several manufacturers in various form factors. <S> As for the control logic, you can use a micro controller. <S> Take into account that GPS signals reaching Earth surface have roughly a -120dBm power level, your switching circuit will add losses to this figure. <S> An alternative approach would be to use active antennas and enable/disable their LNA. <A> I'd suggest using an RF relay, which will allow you easily keep the signal path shielded, which is a major factor when dealing with signal levels this low. <S> The only way I'd use an IC here would be after an amplification stage. <A> At GPS frequencies, you could hand-build PIN diodes into switches. <S> On, with a few mA passing through, it's an ohm. <S> Off, with 5v reverse across it, it's 1pF or thereabouts. <S> With leaded PIN diodes, with care you could actually string them in space between the tabs on the backs of connectors, or between the ends of co-ax cables, there's no need to get everything into and out of a board. <S> Because GPS is spread spectrum, you don't need super isolation to make the receiver see one rather than another antenna. <S> Model what you get with 1pF off and 1 ohm on in a 50ohm system at that frequency.
| You need an RF mux or RF switches (DPST, DP3T etc).
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Nitrile and touch screens. What is the distance required from your finger to the screen to make it work? I am working for a safety company and we need to certify how we can use some nitrile gloves with touch screen, I read that your finger needs to be at .5mm to the touch screen to make it work. So depending on the thickness of the glove this could work or not. Any expert on this? <Q> Even heavy duty nitrile gloves are a handful of mils thick ( <S> e.g., here ), so you're unlikely to reach 20 mil for most gloves (roughly 0.5 mm). <S> Some do seem to peak out at 14mil, though, so you need to be a bit careful. <S> This paper <S> talks about designing for glove use. <S> Techniques involve careful tuning of the capacitive field by tweaking the distance between touch screen plates, increasing the power output of the driving electrode, and minimizing cover plate thickness. <S> As to how to certify, that's probably tougher. <S> Microsoft offers a standard at this site that you might find a suitable starting point. <S> You might find it easier just to develop a suitable specification that satisfies your customers, developing a test, and meeting it. <S> Another option to consider would be working with a touch screen vendor to see what their standards are and whether they have a product that verifiably suits your use. <A> I don't claim to be an expert on capacitive touch screens, however note that Nitrile rubber has a dielectric constant of about 2.5, which effectively reduces the thickness of the rubber by that ratio compared to air. <S> Typical thicknesses of Nitrile gloves are 0.004" and 0.0063" (0.1~0.16 <S> mm) <S> so even the thickest ones are only equivalent to about 10% of what you say is the allowable gap in air. <S> Things like cotton gloves would be considerably more problematic. <S> See, for example this parallel plate capacitor calculator. <S> The way the device is held may affect the functioning of the touch screen, especially if an external power supply is attached. <A> About the technology: I guess you are speaking about capacitive touch screen, but in your question you did not specify it. <S> About the touch sensor, I say it depends because I've worked in a company developing the sensor chip making the multi touch detection, tracking, and reporting to the OS. <S> It clearly depends on the screen itself, physical properties like mutual capacitance of the diamonds, and on the drive signals, and the overall SNR. <S> Plus it also depends on the performance of the image processing engine. <S> Some vendors would propose a hovering feature, since they could achieve a more sensitive sensor. <S> The touch detection specifications are given by Microsoft for MS certification process. <S> I guess it's the same for other OS providers. <S> For your case, I suggest you make the gloves behave like a finger, from the electrical point of view, at the contact of the touch screen. <S> Forget the thickness of the gloves, and add a conductive material on the finger tip, that makes it success for all touch screens on the market.
| It depends on the touch screen technology, and on the touch sensor.
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Large Wires on a Perfboard I am working on a project on Perfboard which involves controlling a motor that draws about 10A, and I am using 18AWG stranded wire to connect the MOSFET on board to the battery and motor which are off board. The problem is that the wires are too large to fit in the holes, and if I cut some of the strands short so that the wire fits in the hole, the other strands work their way out of the insulation and are in danger touch nearby components. What I'm doing now is laying the wire flat across a few holes and soldering it surface mount, which seems to work better, but I'm worried about pulling the copper right off the board. The oval shapes on the boards above also seem to be useful for this purpose as there's more copper to attach to. What's the best way to solder power wires to a piece of protoboard? How do you handle strain relief? <Q> Insert and solder pins, tin the wire ends, and wrap it round the pins a couple of time before soldering. <A> Use wires which will fit in the holes and short enough of them together (solder them together in parallel) to carry the current you need, then strain relief them by bundling them into cables and then using short pieces of wire to twist-tie them together through pairs of unused holes. <A> Drill out holes for a big terminal strip- <S> don't attach the wires directly to the perf board. <S> Use thick solid wire for the power connections to the MOSFET .. <S> And keep them short. <A> You can crimp a pin or blade to the end of the wire, then solder that to the copper side of the board (or do the same operations in the opposite order). <S> The pin is easy to solder to the surface, and the crimp gives a non-brittle connection to the wire.
| For strain relief, put the wires against the board and secure them with pieces of solid wire inserted in a couple of adjacent holes and twisted a few times on the other side, to keep the stranded wires in position.
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220V AC to 12V DC transformerless and main isolated converter? I am trying to build a 220/12V AC/DC max 100mA converter without transformer. I tried so much and burned some circuits. at the end I decide to write here. I searched but couldn't find and answered question here. Could you please help me to achieve that. Thanks <Q> What you are trying to do is VERY DANGEROUS, mainly because you do not know enough to know what the risks are. <S> Good, safe, reliable, certified safe power supplies are cheap and plentiful and available in "wall wart" and "power brick" styles very cheaply requiring only a DC jack on your device. <S> If it is just a home project you can probably find an old power supply around the house you can use. <S> If you MUST build a mains power supply into your device there are MANY things you must know & take account of in the design & layout & manufacture of the device. <S> But these cannot be relevant because you would be INSANE to produce & sell such a thing without knowing a LOT more than you do now & having a professional certify the electrical safety before you were able to sell it to anyone without a high risk of someone being injured or killed - and if you are in a country where such a thing is likely, you will also be sued for damages by the injured party & probably taken to court by the authorities for selling an unsafe product. <A> It sounds like you want a Capacitive Power Supply: <S> http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf <S> However, such supplies are not isolated. <S> The transformer is the only real form of isolation you can get, unless you want something silly like a mains powered light shining on a solar panel. <A> You can use SR086 or SR087. <S> these are excellent transformerless IC for generating the needed voltage with additional 3.3V and 5V regulated voltage to drive the Controller IC's from the mains respectively. <S> The mains can be from 90V to 270V AC. <S> check out the below datasheet. <S> make sure you use the proper rating of the components. <S> Its available at Microchip if you want in bulk. <S> Galvanic isolation is not provided. <S> make sure you add a circuit for shock prevention from the user. <S> If you are using it in a concealed container and not available for the user to touch directly then you can go ahead and use it. <S> has around 55% conversion efficiency. <S> drawbacks : it won't work if you connect transformer and wont work from the inverter with modified sine wave O/p. <S> http://kazus.ru/nuke/users_files/03032008/7568180.pdf <A> What you are trying to do is QUITE DANGEROUS. <S> Are you aware that that kind of systems is not providing any insulation between mains and output <S> and so what you name 0V or +12V could possibly be at full 230 V mains potential? <S> Please clarify what are you about to feed by this circuit. <S> Thx
| If you want to build a capacitive powers supply I suggest you get an isolated mains transformer to protect yourself.
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Identifying open switches in a serial circuit I have an equipment that includes a circuit made of a bunch of NO switches arranged in serial. This circuit includes a LED that turns on when all the switches are closed. When one or more switches are open I need to identify which ones are open. The following image depicts the existing circuit: simulate this circuit – Schematic created using CircuitLab My first approach was to put resistors in parallel with each one of the switches and measure the voltage with an ADC input. This kinda works but I feel it will not be very reliable so I'm looking for a different way to do it. Can anyone suggest a better way? <Q> O pin. <S> Then set one pin to Input, the other to Output. <S> Send a short pulse by raising the output from 0 to 1 and back to 0 quickly. <S> Have the Input pin detect whether it received that pulse. <S> If the switch is closed the pulse will be seen through the two capacitors. <S> Then just sequentially run through each of the switches performing the same operation described using the two <S> I/O pins on each side of it. <S> This doesn't give galvanic isolation, but it does eliminate any issue of messing with your original signal. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Resistors across the open switches is probably the easiest method for you. <S> Eg: <S> Sony Walkman CD players with remote control on the headphones. <S> Older cellphones used the same technique for their earbuds. <A> One way is a chain of one or more input shift registers, such as CD4021B. <S> Three IO lines would be required for a CD4021B (or similar chip) to read the inputs, but you would be able to read the state of an unlimited number of switches if you chain the shift registers together. <S> The chips are about 50 cents in single chip quantities. <S> Depending on your implementation, it may be possible to share two of those three lines (clock and data) with other components.
| If it's an open switch, no pulse will be seen. If you're using a uC that has enough IO, you can go ahead and wire in a small capacitor to each junction you're interested in connecting to one I/ You need to pick the resistor values so that your a/d readings are spread apart, preferably spread apart equally. MANY consumer wired remote controls work this way.
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How to connect IR sensor ( Infrared detector/receiver )? Would you give me a hand ? I have an IR sensor with 3 pins. It looks like the one in the picture, and I did not find any datasheet for it on the internet. actually, I have 3 questions : 1- I want to identify its pins ? (Where can I put the positive, ground ...etc. ?) 2- How to connect it with a visible LED so that when the IR sensor receive IR Rays, the LED lights up ? 3- Should I transmit IR Rays with a specific frequency ? or just connect IR LED to the power only ? <Q> I have an IR receiver that looks like yours. <S> The one I have is equivalent to this TSOP38238 from Adafruit :: This IR detector demodulates a modulated IR at 38kHz and provides a digital decoded signal at the output. <S> Assuming it's the same one; here's the pinout: <S> The DATASHEET is available here (TSOP38238). <S> Head over to Adafruit for all the details including how to connect to a visible LED: <A> if can power it up do a continuity test find in the 3 pins which is ground after you found the ground mark it if your able to or remember it and find another pin with a voltage of 3.3 volts and that it u found the pin out hope i help someone <A> It looks very much like a TSOP382xx from Vishay, available at Adafruit: https://www.sparkfun.com/products/10266 You can find a datasheet there also. <S> The carrier frequency depends on which model you have, the last two digits in the part number. <S> As you can see from the block diagram above pin 1 is your output (far left on your picture). <S> The receiver sinks pin 1 when the correct carrier is detected. <S> But it can only sink a maximum of 5mA <S> so you must limit the diode current if you hook one up directly to the output. <S> I would use a buffer stage of some sorts. <S> Hint: you could probably make one yourself with a PNP transistor.
| I have a method to find the pin outs ,my method is by luck though but is quite easy if you got the sensor out of a PCB of an old electronic if its able to power up.
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ICEO, ICBO physical interpretation in BJT I am stuck with the equation: I CEO = (1+β) I CBO My questions are: What is the physical interpretation of this equation? Will both the currents exist simultaneously? Are any of the reverse currents getting amplified? <Q> 1/ <S> I CBO is reverse leakage current going from the Collector to the Base. <S> This current is then amplified by β to produce additional Collector current, thus the "1+β" term. <S> 2/ <S> Both currents exist simultaneously, but I CBO is included in I CEO . <S> 3/ <A> ICBO = <S> Reverse Leakage <S> Current between Collector and Base while Emitter is Open . <S> (IE=0) <S> ICEO <S> = <S> Reverse Leakage <S> Current between Collector and Emitter while Base is Open . <S> (IB=0) <S> IC= <S> BIB+(B+1)ICBOif(IB=0){IC=(B+1)ICBO=ICEO} <S> These are two different Leakage Currents for two different ways of Active Biasing of a BJT Transistor. <A> ICEO and ICBO exist/or measured when you connect the transistor in two different configuration(common base and common emmiter respectively) <S> so there is no question of asking whether both exist simultaneously(answer <S> is no).Consider you have <S> transistor say T1 to be measured. <S> You measure both Icbo and Iceo when you connect T1 in two different config as mentioned above. <S> So now when you want to know what the value of Icbo <S> wrt measured Iceo Iceo=(1+beta)Icbo will be the expression for solving. <S> In a nutshell to calculate either Icbo or Iceo you need to know to measure just either of Iceo or Icbo for which the expression you mentioned helps. <S> Hope you get it! <S> Cheers <A> they are leakage currents ,be so small that they are often neglected β = <S> (IC-ICEO)/IBalpha=(IC <S> -ICBO)IE <S> these two equations are used when leakage is not neglected <S> * <S> but <S> * β=IC/IB alpha= IC/IE <S> these commonly used <A> I think the confusion with leakage currents, and other parameters, occurs because many textbooks don't explicitly state that the open terminal (b in the case of Iceo) and (e in the case of Iceo) is left 'open circuit' to allow the measurement of the parameter being measured. <S> This is also true for h-parameters etc. <S> where terminals are either shorted out or left open to effectively 'freeze' parameters in order to measure another parameter under known and benign (for the measurement) conditions. <S> The equation relating the relationship between the parameters (Iceo and Icbo) apply when the device is in a correctly biased circuit. <S> It is feasible for anyone to breadboard a transistor circuit and measure these parameters but one requires accurate meters capable of measuring very small currents. <S> An easier way is to simulate the circuits using LTspice or similar free simulation software. <S> Of course this last suggestion of simulation is simply to help the student understand how these parameters are measured and used.
| The reverse current I CBO is amplified, just like external Base current would be.
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RC filter on a clamping circuit I have designed a clamping circuit to protect pins of my microcontroller and I have to add a low pass RC circuit at the microcontroller end to filter out the high frequency noise. My circuit analysis is weak and I am not sure what part R1 and R21 would play in determining the frequency. Resistor R1 is used to bias the diodes in a region where their forward voltage is 0.2.So the node between R1 and R2 is maintained at 3.5 Volts.R21 is a pull down resistor because the input voltage is 12-20V or open circuit. <Q> You don't need R2 to form a low pass filter from the 12-20V input - R1 and R21 form an equivalent series R to the capacitor C3. <S> Filter cut-off (ignoring the effects of the diodes) is: - \$f_C = \dfrac{1}{2\pi R C}\$ where R is the parallel value of R2 and R21 <S> (6.667kohms in your circuit). <S> So, get rid of R2 and simplify things a bit then, tune the filter by selecting the capacitor value to give the cut-off frequency you desire. <A> Look at the steady state condition (take out the capacitor) <S> The node between R1 and R2 will be held at 3V5 due to the 0.2V drop across the diode (D2 pin 3,2). <S> R2 and R21 form a voltage divider with R2 having a 0.2V drop giving 3V3 at the output. <S> The purpose of R1 is to drop the input voltage and limit the input current to manageable values. <S> For a 12V input R1 will drop 8v5 (0.425mA), at 20V this drop will be 16V5 (.825mA) <S> The maximum value for R2 can be calculated (assuming there is no output current taken) <A> It's not really correct to talk of a cutoff frequency (in an analog sense) for this circuit, as it is asymmetrical, nonlinear (and <S> the output is digital anyway). <S> Here's (in simulation) <S> what the output looks like if you feed it a 16V 5kHz square wave (R2 = 510 ohms, C3 = 12nF). <S> If you use the simple \$f_c = \frac {1}{2\pi ((R1 + R2) <S> || R21)C}\$ with these values, you get a cutoff frequency of 1.97kHz. <S> Of course what actually gets through depends on the threshold(s) of the digital circuit- which will vary from unit-to-unit and with temperature. <S> If it's a Schmitt trigger input, the maximum frequency will be a lot lower and more unpredictable because of the typically poorly specified limits to the thresholds. <S> Usually with this kind of thing you're probably not trying to hit a certain frequency bang on, but merely pass (for sure) frequencies below a certain threshold, so only the order of magnitude is important. <S> If your maximum frequency is 200Hz, something that blocks signals above a few kHz is likely just fine.
| There will also be somewhat of an effect on the maximum frequency from the actual input voltage level.
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What is an appropiate way to isolate a DC signal? The question might sound kind of funny, but I'm unsure how else to word it. So... Pictures!! Currently, I have a setup like this: simulate this circuit – Schematic created using CircuitLab What I would like to have is something like this: simulate this circuit For the digital stuff, an optoisolator does the trick, and for AC signals an isolation transformer does the the trick and in each one of these devices, the input and output grounds are separated. What method, or what should I be looking into in order to isolate two DC system groundsfrom each other? (The DC signal can vary from mV to 10's of volts.) <Q> I can suggest two solutions. <S> If the isolation can be less than perfect, meaning many megohms in the ground path, use an op-amp in differential mode for the DC signal input side and drive the black box with the ground and output of the difference amp. <S> For total isolation you can use optical much like the digital. <S> The light (LED) brightness is controlled by an op-amp with an optical detector in the feedback. <S> The same kind of detector is used on the isolated side and a simple calibration can correlate the two detectors. <S> Check figure 3 here http://www.digikey.com/en/articles/techzone/2012/dec/linear-optical-isolation-for-safe-sensor-operation <S> and these are the high end units <S> http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=sbos129 <S> Transformers are also still used of course. <S> http://www.analog.com/static/importedfiles/tutorials/MT-071.pdf <A> There are "analog" optocouplers (e.g. <S> the <S> IL300 from Vishay) <S> that can be used to opto couple analog signals with high linearity. <S> Instead of just one photodiodes they contain two . <S> One is used for the isolated output and one for a feedback loop to ensure linear response: <A> Analog Devices and Texas Instruments (and probably others) make isolation amplifiers specifically for this sort of application. <S> TI ISO 122 and ISO 124 are a couple of choices. <A> VFC32 <S> The question is how much accuracy do you need?
| The conventional method is convert your slow-moving DC signal to a frequency - this will still be the same frequency no-matter what isolation system you use and converting back to DC from a frequency is also relatively straight-forward i.e. there are chips that do this such as the
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What is the purpose of star pcb? I'm fairly new to electronics, and I was recently looking at some LEDs listed for sale online. Some LEDs were mounted on "star pcb": From the picture, it looks like there are 6 connections. I'm pretty sure the LEDs just have 2 (an anode and a cathode). Can someone explain the purpose of the other connections, or what the star shape pcb is used for? I tried googling, but all I could seem to find were more star pcb items for sale. <Q> These are not usually ordinary PCBs, but rather metal-core PCBs. <S> Rather than a fiberglass-epoxy composite material like <S> FR-4 <S> as the core of the PCB, the core is made of aluminium or sometimes copper. <S> This has the advantage of far better heat dissipation which is a concern for high-power LEDs. <S> Cree provides some detailed information in their application note: Optimizing PCB Thermal Performance for Cree <S> ® XLamp ® LEDs . <S> A typical high-power LED looks like this: Making good thermal contact to the thermal pad <S> is tricky because it's so small. <S> The star PCBs make it a lot easier. <S> There are RGB LED modules which may have three LEDs on one die, and thus six connections are required. <S> It is cheaper to manufacture just one PCB that can work for many kinds of LED package. <A> I assume you are looking at a unit such as this: Digikey link . <S> The "circuit board" is actually a heat-sink heat-spreader to draw away the heat from the LED module in the middle. <S> As you can see in the photograph, half of the pads are labeled + (anode) and the other half are labeled - (cathode). <S> So yes, there are actually only two terminals. <S> [Edit: changed "heat-sink" to "heat-spreader", acknowledging Conner Wolf's correction.] <A> This star pcb board is also used for RGB LEDs <S> so it gives the flexibility of breaking out all of the anodes and cathodes of each individual LED. <S> Since one board will work for both RGB's and normal single color LEDs <S> it's cheaper to just design one board that works for them all. <S> Here 's an RGB LED with the star pcb.
| As for why these star PCBs have as many as six pads when an LED has only an anode and cathode, I suspect it's an issue of cost.
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What applications require isolated transformers? I am reading that the principal disadvantage of autotransformers is that unlike ordinary transformers, there is a direct physical connection between the primary and the secondary circuits, so the electrical isolation of the two sides is lost. What applications that use a regular transformer require the high and low windings to be electrically insulated from one another?Thanks! <Q> The problem with mains power is a dangerously high voltage that is referenced to ground. <S> That means that if you are standing with your bare feet on your floor and touch (the wrong) one of the mains conductors, you will get an electric shock. <S> These shocks can easily be lethal. <S> Now the problem with an auto-transformer is that you can never be sure that you attach the common wire to the neutral side of your mains supply. <S> Therefore depending on how you plug your auto-transformer into the wall socket, you may get a lethal zap when you accidentally touch its output, 50% chance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> By using a proper transformer, the secondary side is galvanic separated from mains power. <S> That means no matter which end of the secondary winding you touch, you won't feel a thing. <S> Only when the output voltage of the transformer is high (say > 40V) <S> then you will feel it, but only when you touch both wires at the same time. <S> Medical applications as mentioned in other answers are on the extreme end of why you want galvanic separation from the mains supply. <S> Medical applications require a fair amount more precautions than just a simple transformer. <S> These transformers typically have extremely low capacitive coupling too, but that is worth a question on its own. <A> The one that springs to mind is "any medical applications". <S> When electrical equipment is used in a medical setting (or any other setting where electrical safety is a priority), an isolation transformer is considered an essential (and legally required) part of the power supply chain. <S> This significantly reduces the risk of electrical shock. <S> Quoting from http://www.digikey.com/Web%20Export/Supplier%20Content/SignalTransformer_595/PDF/signal-transformer-pi-isolation-transformer.pdf <S> Adequate isolation between a power source and a user of electronic equipment ensures the safety of that equipment. <S> Given the high voltages that exist in modern electronic equipment, proper isolation protects an operator from contact with excessive electrical energy should a short circuit occur in the equipment. <S> Isolation transformers have represented a traditional solution for providing high isolation in electronic circuitry. <S> Even with the increased use of efficient, switched-mode power supplies (SMPS), isolation transformers can improve the overall isolation of an electronic design without severe penalties in added size, weight, and cost. <S> Isolation transformers offer an effective means of meeting the requirements of domestic and international safety standards for electronic equipment. <S> In the United States, for example, such standards are set by the Occupational Safety and Health Administration (OSHA), with product testing performed according to appointed laboratories, such as Underwriters Laboratories (UL). <S> Throughout Europe, safety standards are established by the International Electrotechnical Commission (IEC), with testing performed by the laboratories of individual member nations, such as the Verband Deutscher Electrotechniker (VDE) in Germany. <S> Such systems include medical diagnostic equipment, computer systems, and telecommunications equipment. <S> The systems may incorporate linear power supplies, SMPS, and sometimes a combination of both. <S> A single isolation transformer can help an electronic design meet all of its isolation requirements. <S> With proper system design, an isolation transformer can also help reduce the size and cost of the power-electronics components following it in a design. <A> Another widespread usage of Isolation Transformers is in marine applications and especially steel hulled vessels. <S> In this application there are other considerations apart from safety the greatest of which is the mitigation of galvanic errosion of underwater metal parts when connected to shore power. <S> Reference: <S> BS EN <S> ISO 13297:2012: Small craft. <S> Electrical systems. <S> Alternating current installations <A> Use a transformer to isolate, applied throughout the circuit in which interaction between parties is required, without offering a common path to power. <S> In other cases, is to couple a signal to a power line, such as PLC (Power Line Communications).
| Isolation transformers enable a variety of electronic systems to meet safety requirements. In medical applications, the patient connection to the main supply line through the ground wire, can be deadly, even for very low current levels.
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What is SPI Interrupt (SPIE) in SPI control register? I have been trying to understand SPI Interrupt i.e. what does it do and why is it used ? Reading about it on the web has not helped me. Can some one please explain it in a simple terms ? <Q> SPIE or "SPI Interrupt Enable" is used to enable the interrupt. <S> (Assuming you are in Master mode) <S> There are two ways of transferring data using SPI: 1.Polling: <S> Here you write data in SPI's data register (SPDR) and then poll on the flag for SPI Transmission complete (SPIF). <S> 2.Interrupt <S> based: <S> In this case, when you enable Interrupt for SPI, you write data in SPDR. <S> Controller will transmit the data and will raise an interrupt when transmission is completed. <S> A question may arise that why there is interrupt for transmission but not for the reception, I am tempted to answer <S> but It would be better if you figure that out! :-) <A> Assuming you are talking about AVR, the SPIE <S> (SPI Interrupt Enable) bit is to enable SPI related interrupts. <S> The number of interrupts an SPI peripheral can have will vary depending on the device, but if we take the Atmega8L as an example, it only has one interrupt. <S> Looking at the datasheet for what SPIE does, it says the following : <S> • Bit 7 <S> – SPIE: <S> SPI Interrupt Enable <S> This bit causes the SPI interrupt to be executed if SPIF bit in the SPSR Register is set and the if the global interrupt enable bit in SREG is set. <S> Basically when SPIE AND SPIF AND the global interrupt enable bit is set, you will get get an interrupt. <S> What's the interrupt ? <S> Well lets see what SPIF does. <S> In the datashet, it says the following: <S> • Bit 7 <S> – SPIF: <S> SPI Interrupt Flag <S> When a serial transfer is complete, the SPIF Flag is set. <S> If SS is an input and is driven low when the SPI is in Master mode, this will also set the SPIF Flag. <S> SPIF is cleared by hardware when executing the corresponding interrupt Handling Vector. <S> Alternatively, the SPIF bit is cleared by first reading the SPI Status Register with SPIF set, then accessing the SPI Data Register (SPDR). <S> The text here (I think) is pretty self explanatory, so I won't repeat it. <S> If you don't understand what a particular register or bit does, flip through the datasheet. <S> Sometimes you might have to jump from bit to bit in order to get the full picture of how it works or what its supposed to be doing. <A> In a typical application that uses SPI communications, the microcontroller needs to receive data at some point. <S> To receive data, the microcontroller has to read and write the SPI signals in a specific way as dictated by the protocol. <S> Now since you can't have the microcontroller constantly busy doing this, microcontrollers sometimes are equipped with a hardware module solely responsible for the SPI communications. <S> This allows the microcontroller to work on other stuff instead of handling SPI. <S> The interrupt is raised by the SPI module and it is used to notify the microcontroller that a byte (or more) was received and it is ready for processing.
| An interrupt is generated if SPIE in SPCR is set and global interrupts are enabled.
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How do capacitors help in smoothing potential difference from a full wave bridge rectifier? Let me establish some facts I have understood by myself: The polarity of resistor does not change, as the current from A.C source flowing through does not change direction. The D.C potential difference value fluctuates from zero to maximum. To reduce this fluctuation a smoothing capacitor is used. What I don't understand is: Is the equation \$t=RC\$ applicable to both charging and discharging of capacitor? Is the time constant the same in both cases? If so then why does the capacitor charge fully in time \$t\$ as the potential difference rises to maximum initially but discharge slowly as the potential difference falls? Its time constant for charging and discharging is different apparently! It's greater for discharging and less for charging. Why is this? Initially, as the capacitor charges won't it cause some of the current to be diverted towards itself and less current would flow through resistor, as they are arranged in parallel? <Q> The charge time of the capacitor is hardly affected by load resistance at all (except in extremes). <S> The charging of the capacitor is determined by the resistance of the forward conducting diodes in the bridge rectifier - this resistance is likely to be around 1 ohm or less. <S> The discharge time of the capacitor is determined only by the load resistor which is probably many ohms hence charge and discharge times will be different. <S> Here's a picture that shows an applied AC voltage and a single diode charging up a capacitor: - After a few cycles of AC the capacitor is starting to receive full charge. <S> Remember the picture is a half-wave rectifier so the negative excursions of the AC aren't contributing. <S> With a significant load resistance there will be a droop/decay on the capacitor voltage between conduction periods of the diode but, under normal circumstances the voltage will have a trend that is asymptotic with the peak AC voltage minus 1x diode volt drop of about 0.7 volts. <S> Here's a picture that shows the droop in the capacitor voltage between diode conduction periods. <S> This is a full-wave picture and assumes the diode is perfect: - <A> Yes, t = RC applies, but the diode(s) in the rectifier are nonlinear elements. <S> Their effective resistance changes with the polarity of the voltage across them. <S> In both cases, the resistance value you use in the equation is the parallel combination of the load resistor and the diodes. <S> However, when charging, the diodes are conducting — have a low resistance — and so the RC time constant has a low value that's completely dominated by the diodes. <S> When discharging, the diodes are not conducting (high resistance), and the RC time constant is dominated by the load resistance. <A> The time constant is a property of the circuit. <S> The time constant is (not) <S> the time needed for charging the capacitor (nor) the time needed to discharge the capacitor. <S> From the figure above, there are three time intervals that you should take care about, each is different than the other. <S> The time interval represented by the green line is the (charging time). <S> The time interval represented by the red line is the (discharging time).The time interval represented by the purple line is the (time constant). <S> In the charging process, it represents the time needed to charge the capacitor from zero to 63% of its capacity. <S> In the discharging process, it represent the time needed to discharge the capacitor from full to 37% of its capacity. <S> The time constant is the same in both processes.
| A more exact answer would consider the leakage inductance of the transformer feeding the bridge and the cable resistance feeding the transformer but, in general, the equivalent forward resistance of the diodes in the bridge ensure the cap charges quickly (compared to the discharge due to load resistance.
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How to keep circuit powered when battery disconnected for short time In an assembly, I'm using spring loaded pins to connect a battery to a circuit board. It's possible from dropping the assembly that power may become disconnected for milliseconds at a time from the shock of dropping it. I have a capacitor bank that holds a charge for 30+ms when the positive node of the battery is disconnected. The problem is I don't think the capacitors will have any affect if the negative battery terminal is disconnected, or both at the same time because there will not be a complete circuit. Is my logic flawed? Is there a better way to do this than using a capacitor bank? <Q> The battery and the capacitor should be connected in parallel. <S> It is better to solder the capacitor to the main circuit board if possible. <S> Considering the capacitor connection is always okay it will work just fine. <S> The beauty of parallel connection is that a damaged or disconnected or open branch does not affect other branches in parallel of the network. <A> If the capacitor bank is on the same side of the connector as the battery, then it will be disconnected at the same time and won't help. <S> A capacitor bank will draw a large inrush current when first connected to the battery, so you might want some kind of inrush limiter, like a small value power resistor (on the order 0.010 to 0.100 ohms), a ferrite, or both. <S> This protects both the battery and capacitors from currents which are too large, and keeps the input fuse from blowing unnecessarily. <S> The capacitors will charge a little more slowly. <S> The longer time the capacitors can smooth out battery disconnect, the longer it takes to charge them up in the first place. <S> Take a look to see if power input ramping up slowly doesn't have a detrimental effect on what's being powered. <S> Really this may be more of a mechanical engineering problem. <S> Whatever is holding the battery in place needs to keep it pressed against the spring contacts enough that electrical contact isn't broken once made. <S> Take a look inside a digital camera or cellphone with removable battery and see how they do it. <S> Usually sliding into place and then click to hold it there, with a button press to release. <S> If there's too little play it might be hard to slide into place, if there's too much play the battery contacts may bounce. <S> The benefit of getting the mechanical design right is that the large capacitor bank can be designed out, saving cost, and the product will run fine with the only large capacitors being the ones already in the voltage regulators (assuming you have some in the design). <A> You might want to look at this IC. <S> http://www.onsemi.com/PowerSolutions/product.do?id=CAT6500 <S> in case you can afford another small battery source. <S> You can set preference of the source regardless of their voltage in comparison to the other one.
| The capacitor bank will work fine if both positive & negative pins are on the circuit board side of the connector, so that it doesn't get disconnected when the battery does.
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How can one make an analog voltage memory circuit? I am looking for a circuit that can, on input, remember a certain voltage and output that voltage indefinitely even after the input has been taken away. The circuit should not change its output until a new input has been provided. I understand that such a circuit can be made by digitally sampling the input up to some arbitrary resolution, but I would like to know if a simple analog solution is possible. I would also like to keep this solution purely electronic, as I can also envisage a mechanical solution in which a feedback circuit mehanically controls a potentiometer. Finally, I would ideally not like the circuit to rely on the passive stability of any floating inputs. The circuit should be stable for at least hours. <Q> This is not a practical answer unless you happen to work for a company with the resources of, say, Intersil, but the technology exists to make this work. <S> Consider the ISL21080 type references which hold charge, hopefully for the life of the equipment they're installed into, based on a tiny capacitance isolated by quantum tunneling effects. <S> Provided they don't get too much in the way of X-rays etc. <S> they'll remain pretty stable for years . <S> See, for example, this application note . <S> I might add <S> this kind of thing gives me the willies. <A> EEPROM technology split into two branches in the early 1980s - one thin oxide (FLOTOX) with Intel and Seeq and the other thick oxide (Xicor). <S> In the early days, there were weaknesses to both routes. <S> Thin oxide leaked charge and thick oxide was inherently impossible to scale. <S> There were other issues, but they don't apply here. <S> Given the thick oxide didn't "leak" electrons, I asked the designers at Xicor about the theoretical resolution of a single thick oxide cell if we discounted the limitations of the sense amps, and they said it could approach 1ppm (roughly 20 bits). <S> Since I was also associated with LTC, which was one of the leaders in precision voltage references that were inherently power hungry, that led me to think a single EEPROM cell could be adapted to be a high precision and very low power voltage reference. <S> My longer term thinking was this technology could be developed further for use in AI and used in conjunction with n fan-out non-blocking, reconfigurable non-volatile multiplexers. <S> Fast forward about 15 years <S> - Xicor ended up developing such a device and then subsequently was acquired by Intersil. <S> Given the inability to scale, the longer term vision is probably not practical. <S> However, other technologies can enable the vision when combined with a software software reconfigurable mux. <A> This device does exist although it's not readily available in single unit quantities, its output amplifiers will get in the way <S> and it's very non-linear. <S> It is a Floating Gate MOSFET, used in Flash memory, EEPRom <S> and the ilk. <S> The programming charge can be variable though somewhat unpredictable as the FN tunnelling (Fowler Nordheim) will be variable across the die. <S> While non-linear it is a proportional effect <S> so you could imagine designing a circuit that linearized the programming effect (of Vth shift). <S> It will be stable over weeks to months <S> so it meets the requirements of hours that you say you'd need. <S> But a lot depends upon the specifications that you need, how much drift is acceptable etc. <S> Just to be clear here, I am talking about the individual device/transistor not the complete component as the support circuits of a Flash will prevent you from operating the cells in this way. <S> Here are 3 references from an EDN article talking about a company called GTronix which was acquired by National Semi (now TI). <S> Lee, BW, BJ Sheu, and H Yang, “Analog floating-gate synapses for general-purpose VLSI neural computation,” IEEE Transactions on Circuits and Systems, Volume 38, Issue 6, June 1991, pg 654. <S> Fujita, O, and Y Amemiya, “A floating-gate analog memory device for neural networks,” IEEE Transactions on Electron Devices, Volume 40, Issue 11, November 1993, pg 2029. <S> Smith, PD, M Kucic, and P Hasler, “Accurate programming of analog floating-gate arrays,” IEEE International Symposium on Circuits and Systems, Volume 5, May 2002, pg V-489. <S> THere is another class of device that is called a MNOS transistor (Metal Nitride Oxide Semiconductor) in which tehere are two dielectrics in the gate, one of which is Si3N4 which has a lot of traps. <S> This device operates very similarly to the flash cell above. <A> I left a comment and thought about it for a minute and will say with hopeful certainty that it does not exist - some drift away from the "sampled" voltage is not only likely but a certainty. <S> Resolution is critical it seems, (as implied in your question) and this is why I say it doesn't exist. <S> Noise is another factor that will reduce the fidelity of what you have sampled. <S> Even a digital system (with more than enough resolution) will be inaccurate in reproducing the voltage you apparently "stored". <S> Anything taken to limits will be a problem. <S> The potentiometer idea (suggested in the question) is also flawed because it relies on the reference voltage across its terminals being kept (or reproduced) - you cannot know how these things minutely drift <S> but, again, it's all down to accepting an error or rejecting that error. <A> Yes, the ISD chips work this way. <S> In fact an inventor in the 1990s claimed to have found a way to store an entire 1 hour movie in a 16MB analog memory chip. <S> The problem turned out to be (Yup you guessed it!) <S> voltage drift over time. <S> Sure the chip would store your movie just fine for a day, possibly two but even powered eventually it would degrade beyond use because the individual values stored could not be recovered without referring to the original file. <S> I actually looked into using this to store SSTV signals but ran into the same problem, conventional floppy disks or even VHS tape were far more reliable.
| For an ordinary application, digital is most likely the way to go.
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Circuit to switch about 5W of 12V using CMOS logic inputs using N-channel MOSFET? I want to switch some 12V landscape light LEDs, totaling about 5 watts, using MOSFETs, with the control signal provided by a digital output line from an Arduino microcontroller running at 5V. I thought I had some logic level N-Channel MOSFETs lying around, but apparently not. What I have is several 400N80Z and IRLB8721PBF power MOSFETS. I tried using the 400N80Z in m circuit, but the LEDs don't light fully. I have a variety of low to medium power NPN and PNP switching transistors on hand (BC33716, PN2222A, and 2N4401 NPNs, various others, plus a few TIP120 NPN power darlingtons. Clearly logic MOSFETs are the right tool for the job and I need to order some. In the meantime, though, can I rig up a circuit that will fully saturate one of my power MOSFETs with a 5V CMOS logic signal from an Arduino? I'm a software guy, and only passingly familiar with analog circuits. <Q> The FCPF400N80Z has a maximum gate threshold voltage (V GS(th) ) of 4.5V, which is marginal for logic operation, but the IRLB8721PbF has a maximum threshold voltage of 2.35V. <S> This should work decently well for even 3.3V logic provided you don't need to pass more than about 3A of current. <A> When being bitten by the threshold voltage, an option is introducing a simple voltage amplification stage using a NPN BJT. <S> You power your load from 12V, so you can use that voltage to pull the MOSFET's gate up to a level much higher than the logic level. <S> Basically what happens is the gate is pulled up to 12V unless the BJT is forcing the gate down to ground. <S> The first stage inverts the behavior of the whole power switch, but that should be easy to solve in software of the controller. <S> simulate this circuit – <S> Schematic created using CircuitLab Advantage of this circuit is that it barely loads the controller output, only about 0.5mA is drawn from the output pin. <S> Disadvantage is that you have to route the +12V to the output stage, just for pulling up the gate. <S> When you use a long lead from power supply to the buffer stage, you should decouple the power supply locally with a small (order magnitude) 100nF capacitor. <A> Note, in data sheet max. <S> 5A us 'infinite heat sink', refers to case temperature at 25 (or similar <S> ) degrees centigrade. <S> Practical heat sink give about half current rating. <S> 1 watt LED is about 300 to 350mA. <S> For 5 watts, one transistors with good size heat sink is enough. <S> Spread to two is better for added safety margin by keeping current to 1A. If individual LED is used, 2 or 3 LED (depends on color, voltage) can be connected in series, so the total current is kept low. <S> Connect Base with 300 ohms resistor to one of the digital output pin to drive about 10mA into the transistor. <S> E to ground. <S> C to LED, resistor (matching current rating of LED) to positive power. <S> To reduce current, rise efficency by reducing heat loss in current limiting resistor, multiple LED can be connected in series (one LED is 2 to 4V, depends on color). <S> Data sheet,figure 2, Vce Sat is 0.6 volt at 1A Ic, which is good for 2 to 3 LED in parallel (1W LED is 300 to 350mA). <S> As original poster mention 'right tool', transistor is suitable/usable <S> (have not check price for this one to see if it is 'best solution') and have been working well for decades in application like this one, at a few watts. <S> MOSFET has added advantage of small heat sink requirement, especially more significant in much higher power applications.
| Since you already have it on hand, you may use TIP120 NPN power darlingtons which is rated at 5A as Adafruit TIP120 data sheet .
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KVL loop current direction convention--should be same in all loops? Do we have to take the same sense of direction for loop current, like clockwise, in each loop when applying KVL to a circuit or can we take different directions in different loops? If we must take the same direction always, please explain why. <Q> No, they don't have to be in the same direction. <S> You just need to remember which direction you assumed because your final answer will tell you if the current is in the same direction as your assumed direction. <S> If you solve for current and you get a positive answer, it means your assumed current is in the same direction as the actual current. <S> If you solve for current and you get a negative answer, it means the actual current is in the opposite direction from what you assumed. <A> I assume this question is in the context of mesh analysis and, in particular, those cases where two loop currents circulate through one circuit element. <S> The important rule to remember is the passive sign convention . <S> Having chosen a reference polarity for the voltage across a circuit element, if a mesh current enters the positive labelled terminal, the contribution to the voltage is positive else it is negative. <S> For example, consider the following circuit: <S> For the chosen reference polarity for the voltage across \$R_2\$, we would write $$V_{R2} = <S> \left(I_1 - I_2 <S> \right)\cdot22 \mathrm <S> \Omega$$ since \$I_1\$ enters the positive labelled terminal whilst \$I_2\$ enters the negative labelled terminal. <S> Now, let \$I'_2\$ be a counterclockwise mesh current in Loop 2, i.e. <S> , \$I'_2 = <S> -I_2\$. <S> Then we would write $$V_{R2} = \left(I_1 <S> + I'_2 \right)\cdot22 \mathrm <S> \Omega$$ since both mesh currents enter the positive labelled terminal. <S> As an exercise, let's write the KVL equations for the two meshes for the mesh currents as drawn: For Loop 1: $$I_1 \cdot <S> 47 <S> \mathrm <S> \Omega + <S> \left(I_1 - I_2 <S> \right)\cdot22 \mathrm <S> \Omega - 10 <S> \mathrm V = <S> 0$$ and Loop 2: <S> $$I_2 <S> \cdot 82 <S> \mathrm <S> \Omega + <S> 5\mathrm V - \left(I_1 - I_2 <S> \right)\cdot22 \mathrm \Omega = <S> 0$$ <S> Now, in terms of \$I'_2\$, the equations become: $$I_1 \cdot 47 <S> \mathrm \Omega + \left(I_1 <S> + I'_2 \right)\cdot22 \mathrm <S> \Omega - 10 <S> \mathrm V = 0$$ <S> $$-I'_2 \cdot 82 <S> \mathrm <S> \Omega + <S> 5\mathrm V - \left(I_1 <S> + I'_2 \right)\cdot22 \mathrm \Omega = <S> 0$$ <S> In either case, we get the same system of equations. <S> $$69\cdot I_1 - 22 <S> \cdot <S> I_2 = <S> 10 = <S> 69\cdot <S> I_1 + <S> 22 <S> \cdot I'_2$$ <S> $$22\cdot I_1 - 104 <S> \cdot <S> I_2 = <S> 5 = <S> 22\cdot I_1 + <S> 104 \cdot I'_2$$ since <S> \$I'_2 = - I_2\$. <A> To begin the analysis by KVL is necessary to provide voltage references. <S> This is what should be kept constant throughout the analysis. <S> Voltage references, allow establishing the signs of each term in the equation of KVL. <S> Once raised the equilibrium equation, each term can be analyzed according to the corresponding voltage-current relationship. <S> In this step, the direction chosen for the current can change the sign of the term, according to Ohm's Law. <S> Definitely, no matter the direction assigned to the current of each loop, provided it is maintained throughout the analysis. <S> After obtaining the results, we must interpret the sign of them, according to the voltage and current references. <A> No. <S> KVL basically means that the closed loop integral around any loop for time-invariant electric fields must be zero. <S> Choose any direction. <S> You can traverse an element shared between multiple loops in different directions for each loop.
| One does not need to choose the same orientation (clockwise or counterclockwise) for each mesh.
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How does USB Type C handle reverse polarity / signal duplication The new USB Type C connector doesn't have a physical reverse polarity protection any more. You can plug it in any way you want on both ends, there is also no A and B end any more, it's all the same. So how does this new USB type handle that the polarity doesn't end up being reversed or signals being routed to the wrong point? Is there some sort of routing going on in the connector and the devices don't have to handle anything and can be sure the polarity is always correct? This is assuming that not half the signals in the cable are redundant. <Q> Below is the pinout for the receptacle: GND TX1+ TX1- <S> Vbus CC1 <S> D+ <S> D- SBU1 Vbus <S> RX2- <S> RX2+ <S> GND | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | |=+====+====+====+====+====+====+====+====+====+====+====+= | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | |GND RX1+ RX1- <S> Vbus SBU2 D- <S> D+ <S> CC2 Vbus TX2- TX2+ GND <S> You will note that all the pins are rotationally symmetric, so if you flip the connector, TX1+ connects to TX2 <S> +, TX1- connects to TX2-, etc. <S> and most importantly, Vbus and GND always match up. <S> The trick lies in the controller and cable -- the CC pins are used to detect orientation, at which point the controller routes appropriately: 2.3.2 <S> Plug Orientation/Cable Twist Detection <S> This allows for signal routing, if needed, within a DFP or UFP to be established for a successful connection. <S> As you might imagine, the cables are going to be a fair bit heftier due to the extra wires. <S> A minimum of 15 wires plus braid required for full-featured Type-C (i.e. USB 3.1 -- recommended 4-6mm outer diameter) <S> 10 wires plus braid for legacy <S> Type-C USB 3.0/3.1 cables (intended to connect to Type-A or Type-B on the other end -- recommended 3-5mm outer diameter) <S> For USB 2.0 or earlier, whether connecting to Type-C or a legacy type on the other end, the usual four wire configuration is permitted (recommended 2-4mm outer diameter) <S> Source: USB 3.1 Specification @ <S> usb.org -- specifically, the Universal Serial Bus Revision 3.1 Specification PDF available for download at the top of the page) <S> Also a great blog post explaining all the details about the Configuration Channel pin: <S> http://kevinzhengwork.blogspot.de/2014/09/usb-type-c-configuration-channel-cc-pin.html <S> Archive.org <S> (in case it goes offline) <A> Since the cables are passive and are meant to be backwards compatible, the signals are duplicated top and bottom. <S> This has the advantage of doubling the power pins and thus increasing current capacity. <A> The 2×12 (i.e. 24) pins are arranged in a way that inserting them in both ways will direct the electrical energy to the same path. <S> As Vladimir says geometrical symmetry. <S> Each of the pins has a clone pin on the other row of 12 pins.
| The USB Type-C plug can be inserted into a receptacle in either one of two orientations, therefore the CC pins enable a method for detecting plug orientation in order to determine which SuperSpeed USB data signal pairs are functionally connected through the cable.
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RJ11 socket in Eagle libraries? I am trying to add RJ11 (telephone wire) sockets to my PCB on Eagle but am having trouble finding it in the libraries. Is this a standard socket that is included or do I need to install some external library? If the latter, where can I find such a library? <Q> Sparkfun releases many of the parts used in their products in the form of the footprint and schematic symbol. <S> It seems like they have an RJ11 connector in their connections library. <S> You can download them from the linked github page. <S> As for more Eagle library resources: Adafruit also releases their eagle parts as well, although I'm not sure if they have an RJ11 connector. <S> I have published and maintain my own as well , although I take great care to make sure they are correct, they are still unverified so take them <S> w/ a grain of salt. <S> Generally if you can't find the part, you'd have to make it yourself. <S> There are many tutorials explaining how to make a part in Eagle. <A> The best way to deal with a missing component is to make your own. <S> Find a component that meets your spec. <S> Draw a schematic symbol for it. <S> Draw a PCB footprint. <S> Marry the two as a component. <S> Place the component you created in your design. <S> Relying on third party libraries is not good practice. <S> Unless you check every part, you cannot be certain that the pin mapping is correct, or the part will match the footprint. <A> I doubt there's such a thing as a generic rj11 socket. <S> https://github.com/sparkfun/SparkFun-Eagle-Libraries
| The SparkFun connectors library has a footprint + schematic for an rj11 socket that they sell.
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What is the easiest way to identify what leg of a 220V panel a 110v outlet is on? I need to connect two devices on the same leg so they can communicate. Basically I am trying to send my internet to my remote garage over the power line. I need to have them both on the same leg in order for it to work. I have a 220v 100A line from my house to the garage. I need to plug one device in my computer room outlet and one in my garage in an outlet wired to the same leg as the one in the house. How can I easily identify which is which? <Q> This is assuming North American 120:120 wiring. <S> Suppose you plug an extension cord into the outlet, then run it back to the spot in the house. <S> Also suppose you have a voltmeter rated for safe mains operation and know how to use it (if that's not true, get assistance from someone (friend, neighbor, licensed electrician, whatever) who does know how to do these things safely . <S> Measure between the two hot wires. <S> If they're on the same leg, the voltage will be very low. <S> If they are on opposite legs, it will be more like 240VAC. <A> Easiest w/ simple tools: <S> Short it out and see what breaker trips. <S> Bad Idea, don't do it. <S> I have done it and it'll freak you out. <S> It is safer to shut off circuits one by one until a light turns off or a loud radio shuts off. <S> Easiest w/ tools: <S> Fox/hound device. <S> Or, just run a capacitor across the 2 legs so that the signal can cross. <S> They sell Plug-In Phase Couplers that do just that plug in to the oven or dryer outlet (then the dryer or oven plug into them.) <A> Once you an absolutely sure mains is disconnected then you can go into your computer room and short the neutral and power together. <S> Then go to your garage and use your meter to test for any continuity between the given leg and neutral. <S> This is of course just a suggestion and is not something I have done before. <S> Please comment for better or worse.
| Be sure to disconnect mains and test it with a probe first since these voltage ranges can be deadly . I do not know a great deal about wirings in homes, but if the outlet is indeed fed from one of the single-phase sources in your garage then the first thing that comes to mind is a continuity test. Identify the 'hot' wire on each pair (should be 120VAC relative to earth).
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Inductor selection for a switching regulator This is somewhat related to this question , but hopefully a bit broader. I'm selecting components to fill out an lm2734 . This regulator will take a 9-12V (11.1V nominal) input down to 5v @ 1A max. They provide guidance for selection on page 12 of the datasheet, and using the formulae there, I arrive at an inductance value of 4.9uH. My question : what do I want directionally here (size and cost notwithstanding, and assuming the output current and DCR are correct), an inductance value that's within some percentage of the one calculated above, or just a value that's higher? Should I just be using the largest inductor I can fit to minimize ripple current? To recheck my work, I looked at the sample circuits they provide in the DS, and calculated the inductance for them to compare to the inductor value they recommend. They seem to be taking inductor values as much as 2x greater than the formula suggests (which may also mean that I'm just doing it wrong). E.g. for a 12V in / 3.3V out @ 1A configuration, I get 4.3uH, and they recommend a 4.7uH inductor however, for an 18V in / 1.5V out @ 1A configuration, I get 3.1uH, and they recommend a 6.8uH part. <Q> These sorts of parts (i.e. parts with unspecified internal compensation) can be tricky because the size of the output inductor will determine the mode of operation (CCM or DCM) and the type of feedback compensation required, along with all of the other variables that have been touched on (core loss, saturation, etc.) <S> Because of the vagueness, there are no hard-and-fast rules here; you need to do some empirical designing. <S> I offer some rules-of-thumb: <S> Choose an inductance value that will make the ripple current somewhere around 20% of the maximum load current, and the maximum switch current (DC + ripple) at least 20% below the internal current limit of the part. <S> Many (most?) of the "off-the-shelf" DC/DC converter inductors on the market are designed with this sort of operation in mind (mostly CCM, low ripple, low ESR) <S> Choose (or design) <S> a part that provides the inductance and current rating in an acceptable package size Build the circuit and test under many corners (zero load, max load, overload, short circuit, input brownout, load transients, etc.) <S> to ensure that the converter doesn't go unstable <A> Should I just be using the largest inductor I can fit to minimize ripple current? <S> Remember that, as a general rule, as you increase your inductors package size, your inductors DCR goes up. <S> There are various other aspects that should go into your inductor selection, such as switching frequency, inductor core material, output capacitors, etc... <S> Some things to make sure apart from the proper inductance value are: <S> Make sure the inductor has a proper thermal current rating at the inductance you want Be extra careful about the saturation current rating the inductor. <S> Try to keep the DCR as low as possible. <S> Linears design tools, like LTpowerCAD II, will give you a general idea of what to look for in an inductor. <S> It uses similar equations used in the datasheets to give you these values <S> so you are on the right track. <S> What I would recommend as a next step is determine your most common operating Vin and then try and design, with other voltages in mind, around that. <S> Various inductor manufacturers have their own tools to help you find the proper inductor for your application. <S> I've always been a huge fan of Vishays inductors and they have a great tool to help you out . <S> In the end, you will need to do various testing on your power supply such as no-load, full load, in-between. <S> If you want to test it a lot, you could do a loop-feedback test on the FB pin to see how your compensation network holds up. <A> I can try to answer your question in a different way. <S> During the selection of inductor, you may have to choose the value that the manufacture suggests and verify it with your circuit calculations. <S> Also the inductor value you choose should have the saturation current more than the load current of your circuit. <S> So here in the application circuits which you are referring to you may have to check the output capacitor value also.
| The value of inductor here is also dependent on the capacitor you use at the output which together decide the switching frequency of the regulator.
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LM339: why is the output +5V The circuit diagram below is given in my book (written by Coughlin and Driscoll). simulate this circuit – Schematic created using CircuitLab Here it is said that, at the condition when \$ E_d=+2V\$, the output voltage will be \$V_o=+5V\$. But for an op amp, when the non-inverting voltage is higher than then inverting voltage, the output voltage will be \$+V_{sat}\$, which is approximately +15V. Why is the output for this Op Amp IC (LM339) +5 Volts? <Q> Well, the LM339 is a comparator, not an op-amp. <S> It is intended to compare the two input voltages and put out one of two levels depending on the result of the comparison. <S> It is not intended to be used with feedback, except possibly positive feedback for hysteresis. <S> The output is an open collector, and there is a pull-up to 5V, meaning that when the non-inverting input is higher than the inverting input the output transistor is OFF, and the output will go to whatever voltage <S> the pull-up (Rp) is connected to. <A> There is actually a contradiction in the LM339 datasheet: <S> The output consists of an open-drain NPN (pulldown or low-side) transistor. <S> vs. <S> When both IN– and IN+ are both within the common-mode range: – If IN– is higher than IN+ and the offset voltage, the output is low and the output transistor is sinking <S> current – If IN– is lower than IN+ and the offset voltage, the output is high impedance and the output transistor is not conducting <S> TI has confirmed the second case is correct for a similar part here: https://e2e.ti.com/support/amplifiers/etc_amplifiers__other_linear/f/18/t/631667?LM393-Contradiction-in-the-LM393-data-sheet- <A> According to the datasheet, The output consists of an open-drain NPN (pulldown or low-side) transistor. <S> The output NPN sinks current when the positive input voltage is higher than the negative input voltage and the offset voltage. <S> The VOL is resistive and scales with the output current. <S> Hence when Non-Inverting(+) is greater than the Inverting(-) terminal the output will be driven to Ground. <S> But when Non-Inverting (+) is less than Inverting(-), the output will be driven to 5V by the pull-up resistor.
| The output NPN sinks current when the positive input voltage is higher than the negative input voltage and the offset voltage.
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Solar panel and dc motor I want to run a 130V dc motor 2hp. I have 6 solar panel 120V dc out. I connected them in series to the motor directly and it runs, but when I check the voltage at the terminal when it is running it read 60V. Where is the rest of the voltage and why not all the volt is use it up? What can I do to make it run full power? Second question: Can I attach a dc-dc boot converter at the end terminal of the solar panel and increase the voltage directly without using anything else? What is wrong with that idea? thank you for your responses.I am trying to run a water pump. I have purchase a mppt unit that has 80v input max, and two lead batteries deep cycle, but I notice that I have to convert the voltage to ac then back to dc and make it 130v. how can I make this work, do I connect the panels in parallel and then charge batteries, have ac inverter and ac-dc inverter, I am not sure if this Is practical at all and how many batteries do I need, can I <Q> Wire your panels like so: <S> Wired <S> exactly as shown, you'll get <S> 18.5V <S> * 7 <S> = 129.5Vat 5.9A <S> * <S> 2 = 11.8A= 129.5V * 11.8A = 1528.1W <A> I'm assuming that you meant the 6 solar panels give 120VDC when they're all connected in series. <S> Not 120VDC each, and thus 720VDC in series. <S> Please edit your question to clarify. <S> Anyway, the primary piece of information that you're not considering is the rated power output of the solar panels. <S> Your motor is 2HP, which equates to about 1500 Watts. <S> The power capacity of your solar array needs to be at least 1500W. <S> A bit more, actually, to account for losses and less-than-peak sunlight. <S> My guess is the panels you have add up to far less than that. <S> This has the consequence of driving the voltage down as the panels are starved of charge carriers. <S> It's called a "brown out". <S> The same thing happens when to the power grid sometimes. <S> When the power plants can't produce enough energy to keep up with demand, the total voltage in the grid starts to drop, which makes everyone's lights dim. <S> The answer to your second question is yes. <S> Since you gave no more information about what you're trying to do, I can give no more information than that. <A> The solar you are using have what is known as a load curve. <S> This simply means that their output voltage and current follow a distinct relationship. <S> You don't say, but the 120V is probably the open circuit (OC) voltage of the panel. <S> Which means how much voltage it will generate with no load. <S> At OC the panel will also not generate very much power. <S> To get the most out of your panels you need to have a MPPT controller (MPPT = <S> Maximum Power point controller) which is essentially a fancy DC to DC convertor than allows the panel to operate at the point that allows it extra maximum power. <S> In short, you loading the panel too heavily. <S> In fact it would be a very rare circumstance that you could operate the panels at their maximal point by directly connecting the motor as they have very different characteristics. <A> The "where have all my volts gone" question is covered already. <S> "Can I attach a dc-dc boot converter at the end terminal of the solar panel and increase the voltage directly without using anything else? <S> What is wrong with that idea?" <S> In order for a DC-DC converter to work, it has to draw even MORE current from the source (your panels) in order to pump up the voltage. <S> Even in a perfect world, your net gain would be zero... <S> but the DC-DC converter will ALSO waste some power (heat loss, etc), so your realistic net gain will be a net LOSS. <S> Upshot: you'll have even LESS power to your motor than you have now. <S> If you're going to run that pump at the full rated 130VDC on photovoltaic, you're going to have to add more panels. <S> Given that you're now reading 60V at the motor, you'll need AT LEAST twice as many panels to get the motor up anywhere near in full sun with perpendicular insolation. <S> The batteries... assuming 12V lead-acid batteries, you'll need them to store and deliver a total of at least 125 amps per hour to run that pump, probably more than that because the pump motor won't be 100% efficient. <S> Count on twice that. <S> If you're using photovoltaics to CHARGE those batteries, then the photovoltaics will need to supply enough to run the pump motor PLUS enough to charge the batteries. <S> Your sum total power requirement will be more than twice as much as it takes just to run the pump, if the pump needs to run full-time. <S> If the pump is only operated 50% of the time, you could load-switch... charging the batteries while the pump is off, not charging while the pump is running. <S> So... if the pump needs to run 100% of the time, AND you need to charge the batteries, then you'll need at least FOUR times as many PV panels as you now have. <A> When the motor is running, the internal movement of the rotor creates a negative voltage according to the solar pannels polarity. <S> So your diference of voltage is going exactly where it should; to power the motor.
| The reason you're seeing 60V is because the motor is overloading the solar panels. As long as you motor is not damaged by running at a low voltage and it can do what you need then this setup may be sufficient and you can get away without any fancy controllers. The panels are giving the motor literally every electron they're generating from solar energy.
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Can a 5V regulator connect directly to a voltage divider to obtain a 2V output? In my circuit, a lm78l05 is used to provide +5V voltage to other parts. And now I need another +2V source to shift my input signal from +-2V to 0~4V as the ADC ad7682 limits the input signal range from 0~4V. Can I just connect the lm78l05 with a simple voltage divider(maybe made by ada4841)?(as the picture showed) If yes, any parameters I should consider carefully such as the opamp offset current/voltage or input resistance of opamp? If no, any good suggestion to achieve the goal? Thank you very much! simulate this circuit – Schematic created using CircuitLab <Q> This idea will work providing you don't need to supply more than (say) <S> 20 mA of current at 2V to other devices because it's likely that the op-amp wouldn't be able to do this - some op-amps will supply (say) 50mA such as the AD8605 (from memory) of course. <S> DC offset is an issue if you want to use an op-amp so check the data sheet. <A> This will function okay, particularly if you use good resistors, and don't need more than some % DC accuracy. <S> The 78L05 is not a very good regulator and its a worse reference so it's almost surely going to limit the overall DC system accuracy if not dealt with. <S> The 78L05 has an accuracy of +/-200mV and a typical temperature coefficient of -0.65mV/K. <S> Since the relatively high-end op-amp you're looking at has an offset of 300uV maximum (40uV typical) and a drift of typically 1uV <S> /K the reference errors will dominate (by 250:1 for offset and drift typically), even with the 2/5 divider. <S> The resistor tolerances will add some errors too. <S> In fact, I suggest you use the ADC reference 4.096V output as the primary reference source rather than your power rail and use precision resistors. <S> The particular op-amp you've chosen is not quite suitable for buffering the ADC reference output (only guaranteed to work to 4V input, not 4.096V). <S> It also has an enormous input bias current ( <S> 5.2uA maximum- enough to light up an LED) <S> so if DC accuracy is important to you, resistances have to be kept quite low (the internal ADC reference can only supply +/-300uA <S> and you had best keep loading to less than that). <S> The main advantage of this op-amp is low noise, but there are many better choices if you don't need such low noise (it's only a 16-bit ADC, <S> so without doing any calculations I suspect that is not a requirement even over the maximum 125kHz BW for that ADC). <S> Edit <S> : Also see EM's answer regarding the error in the level of the offset voltage. <S> The above comments still apply except the divider should be 1/5 from 5V or ~1/4 from 4.096V. <A> That won't work. <S> Vin Vout -2 <S> 6 <S> 0 <S> 4 <S> 2 2 <A> Just a suggestion... use a pair of metal-film resistors as a voltage divider between +5 and ground. <S> MFRs are a lot quieter than carbons. <S> R1 should be exactly 150% of R2.
| You could measure the voltage with one channel of your ADC and deal with it digitally, but power supply noise will again limit accuracy and will likely have some unfortunate statistical correlations that limit what can be done in the digital domain. Why not find an adjustable regulator that can do the job - most are able to supply at least 150mA and can be adjusted by a resistor divider from about 1 volt upwards.
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Why can current flow through the reverse biased base-collector junction (N-P junction) in a BJT with a forward biased base-emitter junction? If the base-emitter junction of a BJT is forward biased, then current can flow through the reverse biased base-collector junction (N-P junction). This disagrees with my understanding of the PN junction, as I thought electrons cannot flow from the P-side to the N-side of the reverse biased junction, since there is a depletion region between them. I understand why the current can flow through the forward biased base-emitter junction: the external voltage (positive connected to P-side, negative connected to N-side) creates an electric field from the N-side to the P-side, which cancels out the built in electric field caused by diffusion of carriers across the dissimilar materials. This collapses the depletion region. However, in the reverse biased base-collector junction the external voltage will support the built in potential and cause a larger electric field (from the N-side to the P-side), which will stop positive charges flowing from N to P, and stop negative charges flowing from P to N. But if you forward bias the base-emitter junction, and reverse bias the base-collector junction, electrons can still flow from the collector to the base, which is from P to N, which as I just explained in the previous paragraph should not be able to happen? So what allows the electrons to flow through the reverse biased PN junction, as in the case of the collector-base junction of a BJT? <Q> In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. <S> The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. <S> Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current. <A> Normal diode: <S> the n-doped side is full of free electrons, while the p-doped side is full of mobile conduction-band vacancies (holes.) <S> A forward bias will produce e-fields which force the two populations together, turning the diode "on." <S> A reverse-bias voltage will pull them away from each other, forming a potential barrier and turning the diode "off." <S> Transistor CB diode: <S> the p-doped base is full of free electrons! <S> That's backwards. <S> If they remain there for long, they'll quickly fall into holes, which ends up producing an EB base current. <S> However, if first they wander too close to the CB junction because of thermal diffusion, they'll be gripped by the large e-field within the depletion layer, and violently attracted into the n-doped collector region. <S> (Fast charge-carriers crossing a large potential, that heats up the transistor because of Ic x Vcb.) <S> (That example was for NPN. <S> For a PNP transistor, we reverse the carrier polarities, where free holes have been deposited in the n-doped base, and they tend to wander too close to the CB junction and get violently flung into the p-doped collector region.) <S> Conclusion: <S> diodes don't work normally if we somehow moved some of the n-side charge-carriers over into the p-side, or vice-versa. <S> The extremely thin base-region in BJTs allows this to happen. <S> In that case, a diode starts working backwards, and its reverse-bias voltage causes increased current, not decreased current. <S> Yet at the same time it also works normally, with the reversed junction preventing the free electrons of the Collector region from flooding into the Base region. <S> Notice that the out-of-place carriers can easily pass through the depletion layer. <S> The depletion layer was never actually an insulator; it doesn't pin charges in place like rubber or plastics do. <S> Instead it was more like a vacuum region, with the intense e-field repelling the usual carriers, keeping them on their proper side. <S> But when we have out-of-place charge carriers in the diode, that e-field in the reverse-biased depletion layer no longer behaves like a potential barrier. <S> Instead it becomes a particle-gun. <S> (Heh, see, the NPN transistors can be viewed as triode vacuum-tubes! <S> Depletion regions are the vacuum, and doped silicon forms the electrodes. <S> Of course the Base must be very thin, because it acts like a perforated grid electrode!) <A> The Base Collector Junction is Reverse Biased which means that no current flows from the Collect to the3 Base. <S> It doesn't mean current can't flow from the collector to the emitter. <S> When The Base Emitter P-N Junction is forward Biased it ceases to act like a diode and is just a conductor and since the N-Doped Collector region is filled with electrons and the Baser-Emitter Depletion region no longer exists the Current Flows From Collector to Emitter. <S> BUT there still is a reverse biased P-N junction between collector and Base <S> so no Current flows from the Collector to the Base. <A> The main cause of collector current in case of bjt is the size of base region. <S> The base region is very small in area in contrast to emitter and collector. <S> Taking the npn transistor, we simply connect the emitter side to 'x' volt, base to 'y' volt and collector to 'z'volt, where x < y < z must be strictly followed to forward bias the emitter-base region and reverse bias the collector- base region. <S> Since the emitter side is -ve with respect to the base, the electric field is established from base to emitter. <S> So the major charge carriers(electrons) move towards the base. <S> Now comes the core of transistor. <S> Once the electron enters the base (which is very thin and lightly doped), very few electrons(5 out of 100) get the opportunity to recombine with holes. <S> In addition to this, the electrons cant stay in base region for long time due to its small area. <S> When you(electron) come running from ground(emitter) towards home(collector), you cant stay on the gate(base) for long time. <S> You pass through the gate immediately. <S> The same thing happens with electron. <S> It passes to the collector-base depletion region immediately. <S> So the charge carriers are electrons in the depletion region. <S> The electric field established in this depletion region is directed from collector to base due to reverse bias. <S> So the electron again tends to move towards the collector giving rise to collector current. <S> If you simply search on web, you can see that the base is small in size like in the above one.
| In BJTs, charge-carriers are able to cross the reverse-biased CB junction because those charge-carriers have been placed in the "wrong" side.
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Sending audio over infrared? I am an undergraduate and new to electronics. I wanted to know if it is possible to send audio over an infrared LED? I plan to use a 555 timer to generate PWM output with carrier at around 44 Khz to satisfy the Nyquist rate. On the receiver side I would use a one or two stage active low pass filter to get the output. Is it feasible? What is the maximum range I can get? <Q> That was - early in the history of LEDs - pretty common, even without the PWM stage. <S> A very carefully aimed IRLED with a very narrow beam can transmit audio reliably over several miles in clear weather. <S> The greater the distance, though, the greater the likelihood of incidental noise from reflection, refraction, and air movement. <S> The old Popular Electronics magazine published several articles long ago about transmitting unmodulated audio with IRLEDs. <S> Back then, LED beam angles were nearly all pretty broad, so theirs required focusing/collimating lenses. <S> Today, you can easily find IR laser LEDs <S> so there's little need for collimating lenses except arguably at the photodetector end. <A> Some years ago, I built a very simple AM transmitter and receiver using a red laser pointer and CdS photodetector. <S> The practical range for that was around 100 meters at night. <S> It probably could have worked at a longer range, but aiming accurately was quite difficult even with a tripod and lens on the transmitting side. <S> The receiver had no optical components. <S> You might be intrigued by Alexander Bell's photophone invention which transmitted audio via modulated sunlight. <S> Your basic idea is sound, but if you're new to electronics, you might find that it's a bit easier to design and build a passive low pass filter, followed by an amplifier. <S> Also, if you plan to use this indoors, you might wish to consider optical filters to eliminate interference from fluorescent lighting. <S> For your proposed system, the key to maximizing range will be the precision of aiming and the optics used. <S> I'd recommend doing some experiments first with a simple system using, say, just a sine wave driving an IR LED or laser and the receiver at some distance away. <A> If you want accuracy and repeatability, then PWM is not my favourite way to do it (due to the emitter and receiver asymmetrical rise and fall times and their tendency to change with temperature). <S> I would use a voltage to frequency converter to drive the LED and a frequency to voltage converter to demodulate the receiver.
| Depending upon the angle of the IRLED, you can get a VERY long distance out of one... but aiming gets rapidly more difficult with longer distances.
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How to select ADC resolution, considering input signal SNR and digital signal processing? What I knew: The ADC resolution selection depends on the ADC inherent noise and input signal SNR. Our application: Very high precision measurement of signals ranging 1kHz to 10kHz. We want to get as accurate result as possible, even 24-bit result. The above statement, however, does not take digital signal processing after ADC sampling into consideration. Since DSP techniques such as FIR, FFT and etc. may reduce the noises greatly, is it possible that increasing ADC resolution may also improve the measurement result when taking DSP techniques into consideration? <Q> A general rule of thumb is that is you want something to not contribute to your noise budget, that it must be at least a factor of 10 higher SNR than the dominant noise source in your signal chain. <S> As an example, if you have a signal source that is at 300 :1 SNR, run your ADC at 3000:1 and for all intents and purposes you can ignore the ADC. <S> The only way to do this properly is to do a noise analysis. <S> Post processing ( <S> via in DSP for example) has the potential to extract out salient features from above the noise but you have to be careful. <S> You have to have sufficient bit depth <S> so you don't introduce rounding/truncation errors. <S> You have to ensure that you are conserving the nature of the noise (gaussian/poisson pdf) or else the noise floor may rise in an unpredictable way and may not be amenable to DSP techniques. <S> These sorts of steps (matched filters etc.) <S> typically at best can improve the SNR by factors of \$ \sqrt{N} \$ and often the processing cost (# of operations) often follows \$ N^2 \$ <S> so these sorts of steps often become rapidly very expensive. <S> But agains a proper analysis will show this. <S> I would caution you against assuming that a DSP technique will automatically reduce your noise. <S> It is very important that you lot at your noise sources via histogram analysis to ensure that the PDF (Probability Density Function) is amenable to processing. <S> I.e. <S> it appears <S> well behaved, Gaussian or Poisson, is not multivariate and is stationary <A> ADC resolution all depends on your application. <S> If your just trying to take readings from something like a RTD sensor, you will likely have no need for a high resolution device and can save cost. <S> One thing to take in consideration too is that typically higher resolution ADC's have slower sampling times. <S> So you get a more precise reading, but not as many as them in the specified time period. <S> If this was the case with audio, a higher resolution ADC would increase bandwidth, but decrease the bit rate. <S> Your other concerns can be addressed though other circuitry options like filters and such. <S> But like I said before, it is all in what your application actually is. <A> Having an ADC that has better resolution than the external signal noise is a waste of money. <S> Having an ADC that has poor resolution and hence high inherent SNR can be countered by increasing the sampling rate of your process and implementing software recovery of the signal through filtering. <S> If your signal bandwidth is known then a lot more can be done in algorithms to improve SNR by excluding noise <S> - this is basically what an FFT does and software radio. <S> As we don't know much about your application it seems pointless pouring out detail that doesn't apply. <A> To get 24 bit precision you probably want a ΣΔ-ADC (Sigma-Delta ADC). <S> Which is a special kind of 1-bit ADC (with noise-shaping) followed by filtering of the digital values. <S> The advantage of the ΣΔ-ADC is that with just one bit the quantization can't be nonlinear.
| Quite frankly, if your using something that is going to produce a wide value range, like sound or high frequency you will need a ADC with a higher resolution.
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Is Micro-processor completely Digital? I read about different parts of micro-processor like ALU, registers etc. all different digital parts. Are there any analog parts inside the processor? <Q> This is a complex question, because what actually makes a part "digital" can have multiple definitions. <S> Fundamentally, reality is analog (at least at the scales which most microprocessors operate at). <S> Therefore, you can make a coherent argument that there are not actually any digital microprocessors. <S> "Digital" is a theoretical mechanism for simplifying the expression of analog systems where the analog voltages therein are (as much as possible) constrained to two states, each of which represent a boolean value. <S> This simplification makes it much easier for our puny human brains to contemplate complex systems, and much easier for people to write software to evaluate the behaviour of said complex systems. <S> However, if you are asking if any components inside most microprocessors operate outside this simplified view, the answer is generally no. <S> Some microprocessors have integrated ADCs (analog-digital converters), which by definition must operate at least partially outside of the digital simplification. <S> Some microprocessors also have DACs (Digital-analog converters), which are much the same as ADCs. <S> Schmitt trigger <S> input buffers are also partially analog. <S> Basically, at this point, the question is more, assuming you're asking about whether components inside a MCU operate outside of the digital simplification, the question then becomes "How do you define a microprocessor"? <S> Fundamentally, the *CPU core( of almost all microprocessors is purely digital. <S> However, many, many microprocessors integrate on-die peripherals like the ones mentioned above that are very much "analog" devices, so you must ask if you are defining the entire integrated-circuit as the "microprocessor", or just the actual processing core, which may only be a small part of the actual processor's IC die. <A> I've reverse-engineered old microprocessor chips and there are some things that are more analog than digital. <S> For instance, many chips (e.g. Z-80 ) implement register storage with a pair of inverters connected in a loop. <S> To write a new value to the bit, a larger transistor "overpowers" the smaller inverter transistor, forcing the desired value into the bit. <S> It's not just 1's and 0's, but the relative currents that matter here. <S> Reading a register value can be more analog than digital. <S> For example, the 8085 uses a moderately complex differential amplifier to read register values. <S> Another analog circuit in some processors is the substrate bias generator, basically a charge pump to generate the desired substrate voltage without requiring another supply. <S> The 8008 microprocessor has an on-chip reset circuit that detects power-on. <S> This circuit is analog, using a diode drop to set the voltage thresholds. <S> (Most processors have a reset pin, but pins were scarce on the 8008 because Intel insisted on an 18-pin package.) <S> Pass-transistor latches are very common in microprocessors. <S> This type of latch depends on the charge stored on the capacitance of an unconnected gate. <S> You can think of it as digital, but it's more analog than regular gates. <S> The analog vs digital distinction becomes an issue when you're simulating a chip. <S> If you can simulate everything digitally <S> it's easier, but for some chips you need a more physics-based simulation that keeps track of the amount of charge moved around rather than just 1's and 0's. <S> An example is when a circuit has transistors of different sizes and you need to determine the "winner". <S> A modern example of analog circuitry in a microprocessor is Intel's hardware random number generation . <S> This circuit uses a metastable latch to generate bits from thermal noise. <S> (The resulting bits are genuinely random, not pseudo-random.) <S> An analog bias circuit ensures metastability. <A> Is Micro-processor completely Digital? <S> it depends on your definition of "digital": any digital signal is actually represented in analog form. <S> traditionally, we don't pay much attention to that as it has limited practical implications. <S> Are there any analog parts inside the processor? <S> Yes. <S> there are analog input circuitry; there are analog comparators; they are opamps; there are pgas, there are programmable current sources; there are adc, and they are dac, to list a few.
| Some microprocessors have analog comparators that can be configured to act upon input analog signals.
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Is there a GPIO IC that has a parallel interface? Sorry if this is an obvious question, I come from a software background and don't know the names for a lot of components. I am designing some custom hardware based around a W65C02S processor (a modern version of the classic MOS 6502). This processor ( data sheet here ) has only a 16 bit parallel bus for talking to other chips but to interface with modern peripherals I need GPIO functionality. Are there any ICs which would provide setting/reading a GPIO line state by communicating with the parallel pins on this chip? <Q> (The 6521 is patterned after the 6821 PIA <S> that was designed by Motorola to go with the 6800 microprocessor). <S> You may also see references to the 6520 and 6820; the entire family of 6820, 6821, 6520 and 6521 <S> PIA's are all pin compatible and were designed by Bill Mensch , the founder of WDC. <S> Mensch also was instrumental in the design of the original 6800 and 6502. <S> The 6521 has two 8-bit bi-directional I/O ports, automatic handshake control of data transfers, and two interrupt line outputs. <S> By adding some decoding to the address lines, creating individual chip selects, you can put as many 6521's on your system as needed. <S> The 6521 was the preferred way to add parallel ports to a 6502 processor 35 years ago, and still is. <S> Although the Apple ][ didn't have any on its main board, some of the I/ <S> O boards used it, such as the one below (it used a Motorola 6821 instead of the 6521, same functionality as mentioned earlier). <S> The Apple 1 had a 6820 on its main board. <S> The Commodore PET had two 6520's on its main board (used for the keyboard). <S> The W65C21 is available at Mouser in various packages. <S> You might also want to check out the W65C22 VIA (Versatile Interface Adapter) , which is like the 6521 (two parallel ports) <S> but in addition has two 16-bit timers and a shift register. <S> The 6522 was also used on the main board of the PET. <S> The W65C22 is available at Mouser . <A> One possible solution would be to convert the parallel bus to I2C, and then use any number of I2C to GPIO chips. <S> Perhaps some combination of: http://www.nxp.com/documents/data_sheet/PCA9564.pdf http://www.ti.com/lit/ds/symlink/tca6418e.pdf <A> Implement a memory bus interface and any peripherals you need (GPIO's, UART's, SPI, I2C, anything at all really) in an FPGA.
| The W65C21 PIA (Parallel Interface Adapter) , also made by Western Design Center (WDC) along with the W65C02S, is a parallel GPIO chip specifically designed to work with the address/data/clock interface of the 65xx series of chips.
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Cool White LED bulbs: Are they "full-spectrum"? I don't have much knowledge on LED lights and find the technology quite fascinating (yes, I know, it is not that new!) I have read that in order to create "white" light, LEDs actually need to emit light from all spectrums. My question(s): How do LED light bulbs (cool white or warm) generate white light? What would be the light spectrum for cool white LED light bulbs? What would be the light spectrum for warm white LED light bulbs? Thanks! P.S. This is my first question here. If it should be forwarded to another SE site, please let me know! P.P.S. I have done my research BEFORE asking but could not find a technically accurate answer. <Q> gbulmer puts you on the right track. <S> The phosphor takes roughly half of the light from the LED and converts it to a second frequency of light. <S> The two frequencies of light combine in our eyes and look to be some variation of white. <S> A power LED <S> I have emits yellow and purple to attain a "cool white". <S> Warm white has more red in it. <S> In short though, the color spectrum of White LEDs is generally horrible unless you get a really expensive one designed for full spectrum use. <S> In general, white LEDs consist of two-ish spikes of color in the spectrum with everything else very low in comparison. <S> The result of poor color spectrum is that some colors won't even be present even though it appears to us as white. <S> You need 3 separate bands of color to be capable of producing all of the variations in between. <S> With only two bands of color, you could be shining the light at something green and it'll come back looking dark grey. <A> To extend horta's answer, you might want to have a look at the CREE guide to LED color mixing . <S> As previously said, the two colors (blue and yellow) mix to create a white. <S> This is shown below on the CIE 1931 color space : <S> The mixed color (white) will be on a line between the two components (blue and yellow). <S> The ratio of the intensities of blue:yellow determines the final color. <S> Theoretically, you could achieve a white by mixing other colors (eg. <S> cyan and red). <S> One of the advantages of a blue and yellow mix is that many standard "color temperatures" can be achieved. <S> As you can see the blue-yellow line is quite close to the line of standard color tempratures (the " Planckian Locus ") <A> White LEDs are coated with phosphors that glow with the desired color temperature. <S> "Cool-white" LEDs (more blue) have a color temperature above about 5000K, while "warm-white" (less blue) ones have a color temperature below about 5000K. <A> Google image search for "light spectrum for cool white LED light bulbs" gave e.g. http://www.enkonn-lighting.com/product/dimmable-led-light-bulbs-ceramic-cool-white/ google image search for "philips light spectrum for cool white LED light bulbs" gave e.g. http://www.philipslumileds.com/technology/quality-white-light without further information <S> , I think the technique of searching the web, and looking at the images works. <S> So is this question is complete?
| For the most part, "White" LEDs are nothing more than a single color LED with a phosphor on them.
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How is current split with two parallel LED's? I have two LED's in parallel with different forward voltages, and want to know how much current flows through each of them. They have one series resistor connected before being split. Like so: As LED's don't follow Ohm's law, I'm not sure how to calculate the current through each LED. I thought I should treat the LED's like voltage sources and apply KVL loops, but I'm still stuck. <Q> If two LEDs with different forward voltages are connected as shown, then for idealized electronic parts , the LED with the higher V f will allow no current through it, and will not light up at all. <S> The LED with the lower V f will be the only one lit up. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> To understand this better, note that the voltmeter as shown above will read 2.4 Volts, the forward voltage of LED1, and that is insufficient to light up LED2. <S> To calculate current drawn from the battery (first diagram in the question), the voltage drop across the 100 Ohm resistor passing said current, must equal the difference between supply (5 Volts) and V f (2.4 Volts): $$I = \frac{V}{R} = <S> \frac{5.0 - 2.4}{100} = 0.026 <S> A = <S> 26 <S> mA$$ <S> LED1 will also thus have 26 mA flowing through it, and LED2 will have 0 mA . <S> When using real world components, the behavior is marginally different. <S> Note the V-I graph for <S> this 2.7 Volt blue LED : <S> Even though the datasheet indicates a forward voltage of 2.7 (typical) to 3.6 Volts, the actual current it will allow at 2.4 Volts, shown by the red line, is just under 1 mA going by the graph. <S> Of course, the graph is an approximation. <S> Even two LEDs from the same production batch will have slightly different actual V-I curves, with temperature variation adding yet another set of variables. <S> Be that as it may, this ~ 1 mA current through LED2 will reduce the current drawn by LED1 by approximately the same amount, if one were to simplify things somewhat. <S> The exact currents through the two LEDs can only be determined experimentally, due to the environmental and manufacturing variables affecting the various parts. <A> If the two LEDs were absolutely perfectly matched to each other then they could share the same resistor. <S> Because they are not perfectly matched in VI <S> characteristic one might appear a little brighter than the other because it will tend to hog more of the current. <S> To avoid this, it's usually deemed preferable to use a resistor for each LED but, despite this, some LEDs will just appear brighter but <S> (statistically) less so than if the LEDs all shared one resistor. <A> If the LEDs are of different colors, for example, they will have significantly different forward voltages, and the higher-votage (usually shorter-wavelength) LED will pretty much be off. <S> In general, you will need to solve a system of equations: $$V_{supply} = R_1(I_{led1}+I_{led2}) + V_{led}$$$$I_{led1} <S> = f_{1}(V_{led})$$$$I_{led2} = <S> f_{2}(V_{led})$$ <S> where f1 and f2 are functions expressing the respective IV characteristics of the LEDs. <S> You can find an approximate solution using the plots, or, if you're interested, you could use a mathematical model (e.g. see Diode modelling article in Wikipedia ) and find a symbolic solution or an approximate numerical one, using essentially the same successive approximation method as you would with the plots. <S> On a more practical note, you need to use separate ballast resistors if you want both LEDs to work. <S> You may also be able to feed LED1 (2.4V) off of LED2 through a small ballast resistor, especially if LED2 is a high-britghtness, high-current diode. <A> The "cheater" answer is to put a low ohm resistor (eg 1 ohm) in series with each led, measure the voltage across each and calculate the relative currents with good ol' Ohms law. <A> In the early 80's, I made an invention based on this "bad" idea - a 3-LED zero voltage indicator. <S> It is interesting to see how it operates. <S> The LED 1 was green (VF = 2.5 V) while LED 5 and LED 7 - red (VF = 1.5 V). <S> The base resistor 8 can be omitted; the ratio between the resistances 2 and 3 can be changed but their sum should be kept constant.
| Often the spec sheets will contain IV plots, you may get an idea by matching the voltage and adding the currents together and doing a few successive approximations.
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LED burned on Arduino I connected a green LED to pin 13 and digital GND on Arduino Uno R3. pinMode was set to OUTPUT and digitalwrite was set to HIGH . The LED glowed normally. However connecting same LED to the 5V and GND burnt it out. How? Did not pin 13 on HIGH also have voltage of +5V ? <Q> The microcontroller at the heart of the Arduino can handle about 40 mA of current per pin. <S> Connecting the LED (apparently without a current-limiting resistor) directly to 5V and ground will cause it to burn out, as it pulls more than its rated maximum current. <S> Pin 13 will also be 5V, but the current is being limited (not in a good way) by the microcontroller. <S> The LED will try to pull more, but the microcontroller can't source it, so the LED doesn't burn out (as quickly). <S> However, the microcontroller is not designed to be a current limiter, so connecting an LED directly to an output pin can cause damage over time. <S> How much current the LED tries to pull, depends on the LED. <S> Most standard LEDs operate with about 20 mA, and can handle shorter durations of higher current. <S> You need to use a current-limiting resistor with your LED, regardless of whether you use a digital pin or connect it directly to Vcc. <A> The digital pins of the Arduino can only source 40mA <S> , the 5V rail can source much more than that depending on supply. <S> See here . <S> When you attached 5V across the LED it likely tried to pull amps of current and instantly smoked. <S> See this relationship. <S> See how if you apply 5V, the current it would try to pull is exponentially higher than 40mA. <S> The LED you used was likely rated for 20mA continuous. <S> Poof <S> Additionally, if you put a voltmeter across the LED when connected to the digital output, you would not read 5V, likely around 2.2V if you used an LED with a similar curve to the generic one I posted. <S> This is due to the arduino limiting the current to 40mA. <A> The Arduino output pins do not have hard, low resistance, switches to connect to +5V or ground - they use transistors which have a significant resistance which will limit the current the pin can deliver. <S> When you connect the LED across the power supply directly, the power supply will deliver as much current as it can in an attempt to maintain 5 volts - this will destroy an LED.
| The LED blew up because you did not have a current limiting resistor.
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How many LEDs can you chain in series? Theoretically, on a 12v with a max of 10w power line, how many LED's could you chain in series before the LEDs further down the chain begin to dim? Where the LEDs have a forward voltage of 3.2 - 3.5 and a forward current of 20mA? I would be using a 470 ohm 1/2w resistor before the first LED. <Q> It's in series , not serial. <S> Assuming the LEDs are the same, those "further down the chain" aren't going to be any dimmer. <S> Since the LEDs are in series, they all have the same current through them, and will therefore have the same brightness. <S> To answer the question otherwise, do the math. <S> You have a 12 V supply and each LED might need up to 3.5 V. (12 V)/(3.5 V) <S> = 3.4 LEDs max. <S> Since you can only have whole LEDs, the answer is 3. <S> To calculate the current limiting resistor value: <S> You know each LED will drop at least 3.2 V, which means 3 of them will drop 9.6 V. <S> That leaves 2.4 V across the resistor. <S> You want the current to be 20 mA, so the optimum resistance for max brightness is ( <S> 2.4 V)/(20 mA) = 120 Ω. <A> Each one you add will make the chain dimmer. <S> If the forward voltage of all 3 was 3.5V <S> you'd have a current of 1.5/R, if the forward voltage of all 3 was 3.2V <S> you'd have a current of 2.4/R or about 60% more. <S> You do have to be be careful of this if you're running the LEDs close to their maximum capability. <S> Edit: If you need to run 21, say, you can run 7 parallel strings of 3 LEDs in series, so you need 7 resistors total. <S> Each resistor will be about 1.95V/ILED where ILED is the current <S> you want the LEDs to pass. <S> 470R would give you about 4mA, so you could run as many as 600 LEDs from a 10W supply (but 4mA may not be as bright as you'd like). <A> The answer should be the source voltage divided by the voltage drop of 1 LED. <S> So if your source is 12v, and your LEDs have a voltage drop of 3.5, 12/3.5v, gives you 3 LEDs. <S> If the the forward voltage drop of the LEDs was exactly 3 volts, you'd be able to drive 4 of them, and would not need a current limiting resistor. <S> If you have 3 3.3 volt LEDs, you need to figure out the amount of leftover voltage to calculate the current limiting resistor. <S> Thats 12-(3.3*3), or 2.1 volts. <S> Use 2.1 volts in your current calculation. <S> 2.1/.02= 105 ohms. <S> You want the next larger E24 resistor value, or 110 ohms. <S> If you assume your LEDs have a 3.2 volt drop, you'd use (12-(3.2 <S> *3))/.02 to calculate your current limiting resistor. <S> That gives 120, which is an exact match. <S> You'd use a 120 ohm resistor. <S> Edit: If you want to run 20 LEDs, arrange them in groups of 3 series LEDs, each with a 120 ohm resistor, and all the 3 LED chains wired in parallel. <S> Like this: simulate this circuit – <S> Schematic created using CircuitLab <S> P.S. <S> A 4.7k resistor would probably prevent even 1 LED from lighting. <S> It would only allow about 1.8 mA to flow through a single LED from a 12 volt supply, and with 3 in series you'd only get about 4.4 microamps through each LED. <A> You can chain as many as you have voltage for. <S> If your forward voltage is 3.2V thats 3.75 LEDs. <S> Since you obviously can only have whole LEDs, that would 3 plus your resistor. <S> If you are desiring a 20mA drive current though, that resistor is too large for even a single LED. <S> A 470ohm resistor flowing 20mA will see a 9.4V drop, leaving 2.6V for your LEDs...not enough for their forward voltage.
| You could run 3 (using a lower value resistor) in series. The resistance of the current limiting resistor will vary with how many LEDs you're driving in series.
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Why 2.4 GHz Band? Bluetooth, WiFi, Zigbee, Remote Controls, Alarms, Cordless Phones etc.. Why all of these protocols, devices, etc. use 2.4 GHz band instead of 3.14 GHz. What is so special about it? <Q> 2.4 GHz is one of the industrial, scientific and medical (ISM) radio bands . <S> ISM bands are unlicensed, which makes it easier to certify the equipment with FCC (or its counterparts in other countries). <S> However, what special about 2.4 GHz? <S> There is about a dozen ISM bands. <S> Some at higher frequency, others have lower frequency. <S> Not all ISM bands are international. <S> But 2.4 GHz is an international band. <S> update: <S> Microwave ovens also operate at 2.4 GHz, which is not a coincidence. <S> Short version in Q&A format: Q: <S> Why does so much wireless communication operate at 2.4 GHz band? <S> A: <S> Q: <S> Why is 2.4 GHz an unlicensed band? <S> A: <S> FCC has originally set aside this band for microwave heaters (cookers, ovens). <S> As a result, from the beginning, this band is polluted by the microwave ovens. <S> Q: <S> Why 2.4 GHz for microwave ovens? <S> Microwave ovens can work on pretty much any frequency between 1 and 20 GHz. <S> There's nothing special (like resonance), when it comes to absorption of microwaves by water at 2.4 GHz (see also here ). <S> A: <S> The frequency choice was based on a combination of empirical measurements of heat penetration for various foodstuffs, design considerations for the size of the magnetron, and frequency considerations for any resulting harmonic frequencies. <S> [These considerations were proposed by Raytheon and GE to FCC in 1946, when the decision about 2.4 GHz was made.] <S> The long versions can be found here . <S> [This link goes to Indiegogo, because this bit of historical research was crowd-funded.] <S> Also, this FCC document (54MB) from 1947 can be of interest. <S> Thanks, @Compro01 for finding this reference. <A> The "special" thing about 2.4GHz is that when spectrum was allocated for various needs in the 60's and 70's, no one wanted it, because it was thought that atmospheric water absorption made it useless. <A> It is 'special' since it does not go very far. <S> Strangely, this turns out to be an important advantage as many devices and people can use the same band in near by area without interference. <S> Tele density is the term used in phone industry as how many cordless phone per square mile. <S> Early generations (25 years ago) coreless phone use few MHz and tens of MHz and go too far. <S> Modern (now year 2014) cordless phone use GHz (some are not 2.4GHz) for short range and high tele density. <S> There is social vs technical story and dimension behind this. <S> My first job, 30 years ago, was first generation coreless phone using 1 MHz and 50MHz, work range a few miles, excellent for farmer and country size home. <S> Cell phone was just coming out at price 5% of a house, too costly for use in that social context so coreless fits social demand. <S> As more people use them, big interference, phone at times has 10 blinking LED showing searching for un-used channel as they are getting too crowded. <S> Then, move to higher frequency, 900MHz and the likes. <S> Then come Spread Spectrum. <S> It was time at IEEE tech conference that Spread Spectrum session were off limit to civilian. <S> That changed. <S> SS tech moved into consumer items, WLAN, GPS, Coreless phone, 3G Cell phone, Remote Control model, Bluetooth. <S> Next higher move to 2.4G did the trick of balancing social need, short range (BT is a few meters, WLAN tens of meters), spread spectrum, anti-interference, auto channel search (old RC modeler fly color flag on antenna to tell other to stay off their channel). <S> As other responder pointed out, cost did play a part. <S> My first 2.4GHz WLAN is 4 by 10 inch PC plug in card, at 2000 US dollar. <S> Now, we have finger nail USB plug at cost order of magnitude lower. <S> Also, SS and social demand at that times shaped the present situation as described by original poster, that many devices use 2.4GHz <A> When a frequency is chosen for widespread use, it is cheaper to use off-the-shelf parts in your design rather than having to start from scratch to use a particular frequency. <S> You can buy a ready-made transceiver that millions of devices use for a lower cost per unit than using a custom made transceiver. <S> http://en.wikipedia.org/wiki/ISM_band <S> and http://en.wikipedia.org/wiki/Electromagnetic_interference_at_2.4_GHz have some info regarding the frequency designations. <A> As others have said, it's an ISM band, and all of the other listed reasons are totally valid, <S> but I think another part of of the reason it's more popular than other ISM bands is that it is available in almost all countries whereas some ISM bands are only ISM in certain regions, and it is also fairly wide compared to other ISM bands. <S> As you go up in frequency the ISM bands get wider it seems. <S> In fact, 5GHz WiFi is getting more common all the time as 2.4 gets more crowded. <S> The 5.8 band has 150MHz whereas 2.4 only has 100MHz. <S> 5GHz can't go through walls quite as well <S> but they say it can go through smaller holes like under doors.
| 2.4GHz was 'special' since to does not go very far. Because it's an ISM band, and it's unlicensed, and it's international. Some of the reason is cost (both financial and power budget with distance), some is because what frequencies are reserved for other types of devices/communications and the interference caused by such deviations from those frequencies.
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What type of motor can provide smooth motion at low speed around 0.1 to 1 mm/s I am building a stage for tissue cutting and the motion of the stage should be very slow and smooth. The speed range is 0.1 to 1 mm/s. I got a stepper linear motor from PI (M230.10s), I test its speed at 0.1 mm/s and the chatter is 3 times of the setting value. I hope the motor can motor can move very smooth with chatter around 1%. Is it possible? <Q> One option, of course, would be to add gear/belt/leadscrew drive to the stepper so it can be run at high speed while the mechanism it drives operates at low speed. <S> This way, it'd act very much like a long lever. <S> A large ratio of motor to motion yields MUCH <S> smoother travel and higher torque. <A> Big vote for the 'obvious' suggestion by Nicolas D to use microstepping which will make the motor move smoothly, 16-microstep driver chips (Allegro A4988 for example) are readily available and there are some drivers which go much further (64 microsteps or more). <S> However, if you need small, smooth, accurate movements the best thing you can do is gear the system down so that you are not relying on the limits of motor quality for smoothness. <S> Moving 0.1 to 1mm/sec is not a very big range of speeds, you could probably gear the drive from the motor down to a ratio of 100:1 or lower (so 100x better accuracy). <S> If you know the maximum speed the motor can run, choose gearing that will move the device at your maximum speed (1mm/sec) at the maximum speed of the motor. <S> This will give you a far better range of control and increase the achievable accuracy. <S> You don't say what the gearing of your system is - as you state movement in mm <S> /sec <S> I assume you have some sort of lead-screw style mechanism. <S> Even quite basic CNC machines / 3D printers can achieve movement accuracy of 0.1mm whilst moving a lot faster than 1mm/sec, using very cheap stepper motors & drives. <S> Most of the inaccuracy ends up coming from the play in the drive system (usually lead screws) or flex in the machine body due to mechanical loads. <A> Maybe I'm missing the obvious here, but... <S> If you didn't drive your stepper with microsteps, you might want to try that first. <S> This allows the stepper to go smoothly in between "normal" steps. <A> A stepper motor can keep track of it's position by itself (1) without requiring a position sensor for feedback. <S> That's why steppers are popular. <S> On the other hand steps are not smooth, because... <S> well... because they are steps. <S> There are other types of motors that don't have steps: brushed DC motors , brushless DC motors . <S> With these types of motors, however, a feedback sensor is needed to know the position of the motor. <S> Another approach to your problem is to create a mechanical damper, which would help smooth out vibration. <S> (1) Unless the load torque is too high, which causes the stepper to skip steps. <S> Fortunately, that's not the concern of this thread. <A> I recommend a DC brushed gearmotor. <S> Brushless motors tend to have ripple around their commutation points which brushed motors don't have. <S> The gearing in a gear motor allows you to get drive ratios like 1000:1, which is important since it magnifies torque (which increases cutting power, so varying tissue densities aren't a problem) and reduces speed. <S> I also haven't seen brushless gearmotors, so you'd have to get a brushed gearmotor anyway. <S> Just be aware that gearmotors will have a little play (rotation before the gears lock) when switching directions. <S> More importantly, you want a good control system for it. <S> Ideally, you would use a servo driver (such as a Copley Controls or AMC) with encoder feedback. <S> This would allow you to easily do velocity control on your motor. <S> If that's not an option, then you would need a variable voltage supply with a decent current rating. <S> Maximum motor speed is directly proportional to the voltage across the motor, so a lower voltage means a lower speed. <S> If none of the above options would work, then you could look into building/buying a viscous damper for the motor. <S> It essentially be a fan blade, encased in an oil, attached to the motor shaft. <S> It creates a viscous drag on the shaft that dampens oscillations and prevents the motor from reaching high speeds. <A> For such a slow speed I would go with a standard brushed motor. <S> Cheap <S> Easy to control Smooth <S> Forget about the stepper, you won't get as smooth movement with it as with a brushed motor and anything above 10 uSteps is not accurate <S> so you will end up using an encoder anyway. <S> Get a high voltage brushed motor (24V at least) <S> so you can drive it very slowly with a large gear ratio and an encoder for feedback. <S> this motor will do the job with minimum damage to your wallet. <A> If I understand datasheet and your question then motor resolution is 50nm corresponding to 1 step. <S> You need drive 100um-1mm/sec, i.e. 2-20kHz, which is high speed at upper motor limit 1.5mm/sec. <S> So IMHO microstepping/gearbox is not solution (it helps for slow speeds but the cost is torque halving as you double micro stepping factor), you need proper ramping to achieve velocity without resonance (i.e. also some kind of chatter). <S> But I think that problem might be derivation. <S> If you calculate/derive at small time interval you see large jitter. <S> I'd look at raw position data. <S> Note: <S> But I don't understand what is supposed graph waveform, triangle with period depending on cutting travel ? <S> And how the position is measured (as number of microsteps or an external micrometer) ?
| I would suggest to begin with a brushed DC motor, because it's easier to drive, and it doesn't appear that application requires high performance characteristics of a brushless motor.
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Would having no power supply but a large input signal damage the amplifier? I am now learning about the power sequence of the circuit. And wonder that as every amplifier has its own limited input voltage (for example: ADA4004's input limit is V− < VIN < V+), would the amplifier be damaged if no power supply but a large signal (maybe +-2V) injected into the input pin of the amplifier? <Q> In general it is bad practice to apply input signals to a device before applying power. <S> When you do this, you can cause semiconductors to 'latch up'. <S> The semiconductor can act due to its physical build as a SCR (Silicon Controlled Rectifier) , which can cause it to short out its pins. <S> Subsequential application of power can cause excessive currents which destroy the device. <S> This does not only apply for single chips, but also for modules. <S> Recently chips have been designed with internal protection diodes, but I doubt you want to rely on them for this. <S> Modules that are hot swapable can most of the time be recognized by the fact that the power supply pins are slightly longer than signal pins, enforcing power to be applied before any signal. <A> With no power supply, the transistors in the IC would not do anything, so the input would be "ignored"... <S> and there would be no output. <S> It requires a power supply to open the switches to let the signal in. <A> OTHER devices are totally immune. <S> I'm not sure whether this qualifies as an "answer", because it doesn't specifically address the ADA4004 <S> ; I haven't seen that device's specs.
| Depending upon the amplifier or other device... SOME devices can be destroyed by applying any significant input voltage without first appying power.
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How to bring out internal signals of a lower module to a top module in VHDL? How can I bring out the internal signals of my VHDL source code to my testbench so that I can view them as waveforms? I use Active HDL. I would like to know if there is any tool independent method of achieving my objective. Any help is appreciated. I get this error now. My source code is entity SPI_DAC is Port ( -- inputs and oututs )end SPI_DAC;architecture Behavioral of SPI_DAC is --These are my internal signals signal ch1_byte_data_sent : STD_LOGIC_VECTOR(23 downto 0) := x"000000"; signal ch1_byte_cmd_sent : STD_LOGIC_VECTOR(23 downto 0) := x"000000"; --and a few other signalsbegin --functionality end Behavioral; My testbench code is entity tb_spi_dac isend tb_spi_dac;architecture behavioral of tb_spi_dac is component spi_dac port( --declaration, inputs and outputs ); end component;begin uut: spi_dac port map( --map ports ); --stimulus processend; <Q> What you are looking for is called external names (or hierarchical names) in VHDL. <S> They can be used to circumvent scope/hierarchy visibility. <S> The syntax is like the example below. <S> <<signal path_name : <S> std_logic_vector(7 downto 0)> <S> > <S> You can also access constants and variables with external names. <S> You have to change the type in the external type though. <S> You can use external names directly for read/write access. <S> However you should use aliases to improve readability. <S> alias <S> signal_name is <<signal path_name : <S> std_logic_vector(7 downto 0)>>; <S> The external name has to contain the path to the element you want to access. <S> The path can be either absolute or relative. <S> The individual elements in your path are separated by dots. <S> Note that you have to provide the labels of the instance/process/entity/... and not the name. <S> Absolute path start with a . <S> followed by the name of your top level. <S> For relative paths you can use ^ to move up in the hierarchy. <S> When using constants/signals from some package you can also use @ to move to a library. <S> An example for an absolute path is .tb_name.instance_label.sub_instance_label.signal_name <S> To access the same element from the testbench with relative names you can use instance_label.sub_instance_label.signal_name <S> When you want to access some testbench signal/constant from the sub_instance you can use ^.^.constant_name <S> To access some other constant in a config package located in the config library you can use @config.pkg_name.other_constant_name <S> You can use your simulators design/library browser like David pointed out to find the correct path name. <S> This feature was added in VHDL-2008 <S> so it should be supported by all tools that already have VHDL-2008 support (including ActiveHDL I think). <S> Most simulators do not use VHDL-2008 by default but provide a command line argument or configuration option to enable it. <A> How can I bring out the internal signals of my VHDL source code to my testbench so that I can view them as waveforms? <S> A test bench implies simulation - an entity without any ports is generally not synthesis eligible. <S> While I've never used Active-HDL I understand it has a design browser that should allow you to pick signals down in your hierarchy to display in your waveform <S> See Aldec's Compilation and Simulation video (5:02, min:sec). <S> And about now I get the impression the video may cause confusion, perhaps in this particular case. <S> At 2:22 from the end the video shows a do file (macro) which controls the simulation: <S> Where we see every signal in the top level of the design has been added to the waveform display with the wave command. <S> It should also be possible to specify a signal in anywhere in the design hierarchy. <S> The basic idea is a lot of simulators allow you to schedule signals (and some allow variables) to be collected for waveform display. <S> This short video simply doesn't show signals for subsidiary hierarchy levels. <S> (There's a lot stuffed in a short video presentation). <S> I would like to know if there is any tool independent method of achieving my objective. <S> As noted above your objective seems to be to view internal signals as waveforms. <S> Now for the bad news - there is no standardization for simulator features or interfaces, <S> scheduling signals for waveform dump, etc.. <S> These are all implementation defined. <S> It's pretty much guaranteed <S> The good news is they tend to copy concepts from each other such as do files, which you could generate programmatically for portability, using a common database describing functional verification on multiple implementation platforms, overcoming differences in syntax and semantics. <S> There'd likely also be differences in command line interface syntax for invoking the tools programmatically. <S> The idea of portability not embracing multiple GUI's gracefully. <A> Tools like xilinx has option to view internal signals. <S> Simple tool-independent method is declaring seperate output lines and connecting the internal signals to these lines. <A> If you have declared the signals in the testbench but are unable to see any output, you may have a problem in the instantiation statement where you instantiate the entity to be tested. <S> Check whether you instantiated it correctly, using an entity instantiation statement or component instantiation. <S> Make sure that the signals in your testbench are connected to the entity under test. <S> If that doesn't solve the problem, post the testbench code so that we can help.
| you can dump signals anywhere in a design hierarchy to a waveform viewer or waveform dump file with any implementation that simulates, the method for doing so is implementation defined.
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How can I make a safe Ni-MH charging station? I wanted to make a docking station for my robot. My idea was simple - robot drives in, batteries are disconnected, pins that I attach to my Ni-MH charger's electrodes are inserted into battery pack, and start charging. Like on this picture: But I've read on the internet, that chargers look at batteries temperature, to know when to stop. Is it going to work? Or are there any alternatives to dock my 4 Ni-MH battery pack? These are 4 batteries, 2400 mAh, connected in series, 1.2V to generate 4.8V. EDIT: I have also read about super fast charging, something like this: http://www.batteryspace.com/ch-v2880superfastnimhbatterychargerhomecarwithlcdmonitor.aspx . Will this work? <Q> Add connector for thermoresistor into charger dock. <S> Make thermoresistor go into contact with batteries when gobot goes to dock. <S> Install charger circuit into robot, not dock (by Dave Tweed). <S> Use delta-V charger (by Ignacio Vazquez-Abrams). <S> Use slow (0.1C 10 hour) charger. <A> You don't have to design a battery charger from scratch, there are special purpose ICs which handle the charging algorithm for you, like delta-V, timeout, etc as are mentioned in other answers. <S> Here are some places to start. <S> The links point at NiMH since you mention that, but these vendors also make ICs for lithium-ion batteries: Linear Technology , Maxim , Texas Instruments . <S> If you want to buy a working charger circuit board, you might want to switch from NiMH to lithium-ion or lithium-polymer for more choices. <S> Such as available from Adafruit . <A> From what I've read you measure the voltage from the battery to gauge charge, not temp. <S> Temp is a backup to avoid overheating and rupturing faulty cells. <S> In order to measure the voltage while charging you have to build a pulse charging circuit. <S> You feed the battery short, fairly high current pulses of DC, then measure the battery voltage between pulses. <S> When the battery reaches the target voltage, you stop. <S> Getting the algorithm right is tricky. <S> I believe that smart chargers taper off the charge rate as the battery gets close to full. <S> You should read up on smart battery chargers. <S> Also note that in order to do this right you should really measure and charge each cell separately. <S> Why not buy an inexpensive smart charger, dissect it, and move it's electronics inside your robot?
| You'd want to put the battery and charger board together in the robot so that the dock power pins would be a simple DC voltage input (unmodified output from wall adapter) rather than a battery charger. Use timed charger (require fully discharged battery). That way if there are differences in capacity or charge level between cells then you don't undercharge some cells (bad) and overcharge others (disastrous, can ruin cells.)
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Wire for project with 9 volt battery I'm working on a project that powers 20 LEDs with a 9 volt battery. What gauge of wire should I use? A couple of LEDs are in series with a 180 ohm resistor. Then each pair are in parallel. <Q> If this is one of this consumer 9 V batteries that are sortof rectangular with two round clips on one end, then any wire you have lying around is probably good enough. <S> Anything up to 28 guage should work fine for something of the size that fits on your bench. <S> A better answer requires a better spec. <A> Most of the top results of a search for "AWG wire current" reference the Handbook of Electronic Tables and Formulas, each giving the same numbers, presumably copied from the print edition. <S> If you use a different piece of wire from the battery to each R-LED-LED string, and if R=180 ohms and LED drops 3V each, that leaves resistor current I = <S> (9-3-3V)/180ohm = 33.3mA per string. <S> Ten of these strings in parallel (for 20 LEDs total) gives 333mA total from the battery. <S> The table offers a spec for 30 AWG of 142mA <S> "Maximum amps for power transmission" which "uses the 700 circular mils per amp rule, which is very very conservative." <S> Meanwhile Wikipedia gives a fusing current for 30 AWG copper wire as 10A for 30 seconds. <S> A 9V battery will not be able to source anything remotely like 10A, so melting the wire is a non-issue. <S> You could probably get away with 30AWG if it's all you have, or want a thin wire bundle for some reason. <S> Anything thicker would remove doubt about current carrying capacity and 26 AWG with table spec of 361mA max would already be above your total expected current. <S> Whether a 9V battery could actually deliver 330mA for a reasonable length of time is another matter. <S> There is wide variation in chemistry, construction, and thus capacity available. <S> Wikipedia says a lithium model has typical capacity of 1200mAh, giving you perhaps 3 hours runtime. <S> Lithium 9V will run $5-10 each at retail, which may be more expensive than you had in mind. <S> Alkaline will cost less, $3 each at retail, but also have less than half the capacity and also less current capability. <S> An Energizer or Duracell alkaline might only give 1 hour runtime, and a cheapo model with low current capability might be lucky to light it up at all. <S> If you want longer runtime, you'll need a physically larger battery which can store more energy. <A> It depends on how you connect the LEDs. <S> They can't all be connected in series, because that would exceed the 9V you have to work with. <S> You could connect them in parallel with appropriate current-limiting resistors on all of them, in which case (assuming 20mA for each LED) you would have 400mA total. <S> A combination of series-parallel would likely be best, but the highest current draw would be if they were all in parallel. <A> At 9V, you should use whatever wire is available to you, and easy to work with. <S> Based on your asking of the question, I suspect any gauge that is too small for the current is a gauge that you will probably have a hard time effectively working with. <S> Personally, my prototyping spool is 24 gauge solid wire, which is commonly available at RadioShack. <S> Why? <S> It's easy to work with, and (way) more than large enough for virtually any typical battery type application. <S> It's thick enough to stay bent in just about any position you want (for routing), and it's easy to solder. <S> As you get to and through the 30ish range, I find the wire much more difficult to work with, but maybe it's just what I am used to, or the types of projects I tend to do with it.
| Even a very thin 30 AWG wire can handle more than 400mA, so just about any wire you have handy will be sufficient. I sometimes use 24-28 AWG wire, found inside Cat5e network cable.
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What are the square holes for in most electronic project boxes? I'm new to the electronic scene and lately I been purchasing a lot of project boxes for the stuff I make, I've noticed in about 75% of the boxes I purchase they come with small square shaped holes on both the base and the lid, I assume these are for some kind of mounting or for pillars. I've looked online and on here and I can't find any details. Below is an example of one of the boxes I recently purchased of eBay. What is the technical term for these and what do they do? <Q> These are called bosses . <S> A boss is a post with a hole that can accept a screw. <S> If the boss is tall in relation to the diameter, it should be reinforced with gussets <S> (those shapes on the side that help keep the boss from snapping off at the base). <S> Sometimes the posts have molded-in or ultrasonically pressed in brass inserts that provide a thread. <S> Edit: Given the detailed photos the OP added, I'm thinking the square holes are intended to take square inserts, as mentioned above. <S> I'm not 100% certain of this, as most are round with teeth around the outside. <S> Making them square would have some advantages, but does not seem to be common outside of India (though perhaps that Chinese company is Indian owned). <S> For low cost, thread forming screws are used. <S> Note that they are not the same as self-tapping screws used for metal. <S> The thread form is quite different (much skinnier with sharper flutes) to cut threads into the plastic. <A> It seems obvious that square holes will have a somewhat wider tolerance for a range of screw diameters, which is handy for hobbyists who have an assortment of screws laying around. <S> The screw has to forge a thread for itself, and that means displacing plastic. <S> If the hole is round, then it works best when the screw has a fairly precise diameter in relation to that of the hole. <S> The inner diameter of the screw has to be narrower than the hole, because the inside of the thread (the "valley") provides space for the plastic displaced by the "crest" during thread formation. <S> If the plastic has nowhere to go, it is increasingly difficult to turn the screw. <S> The post could crack from the pressure. <S> Yet, the screw cannot be so thin that the outer thread diameter doesn't provide a solid grip; it will then easily strip the newly formed thread when tightened. <S> If the hole is square, then the screw thread's inner diameter can be exactly as wide as the the sides of that square, yet it will still fit without undue difficulty; the displaced plastic can flow toward the nearby empty corners of the square. <S> Yet, the screw can still be as narrow as what would be acceptable for a round hole of a diameter that can be inscribed in the same square. <A> EDIT: <S> As the OP pointed out he really meant SQUARE holes, so this answer isn't a match anymore. <S> Which brings up kind of a meta question: Is it considered good practice to remove an answer after it has been voted on but becomes somewhat irrelevant for the question? <S> I hesitated to write an answer for a pretty obvious thing <S> but here you go. <S> The hole with the red circle is indeed a PCB mounting hole, similar to this: <S> In most cases the holes are threaded (is this the right word?) <S> and you can put a screw in there. <S> The "pins" like those in the middle of your example casing either support the screw holes to hold the PCB in place or even snap into pillars coming the top of the casing. <S> Providing lots of mounting options is a big plus for a PCB casing, because you have more design choices when designing the PCB around it. <A> I'm not a native speaker, but you might call them mounting posts (or something alike).
| You can mount circuit boards on top of these holes, using short, self-tapping screws.
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How can you calculate base voltage of a transistor when there are biasing resistor and a resistor between 2 input terminal I just get to know about transistor a little. Here is my question As you can see that here the transistor is configured to be an amplifier in common emitter mode. R1 here is the bias resistor. What I don't know is how Vb can be calculated. Suppose the input voltage is known already but here the input voltage is affected by R1 and R2 that make Vb is different from input voltage. Also, R1 is connected between b and Vcc that also affect Vb. I learn how to calculate voltage, current and resistance (or impedance) of parallel and cascade circuit but this case look strange and KCL can't be apply here also (or it's just me who don't know how to apply it). Of course, in real life I can use voltage measure device to know its value but I am talking about math here. <Q> You must analize the Thevenin's equivalent circuit: where $$R_{th} = <S> \dfrac{R_1\cdot R_2}{R_1 + R_2}$$ and $$V_{th} = <S> \dfrac{V_{CC}}{R_1+R_2}\cdot R_2$$ then, the equivalent circuit where \$V_b\$ is \$V_{th}\$ and <S> \$R_b\$ is \$R_{th}\$ <S> Then, you can write KVL, for the input circuit: $$V_b = <S> i_b\cdot R_b <S> + <S> V_{BE} + i_b\cdot (\beta + 1) <S> \cdot R_E$$ <A> The simplest calculation is to assume the base current is 0, then Vb = <S> \$Vcc\frac{R2}{R1+R2}\$ <S> and Ve = <S> Vb - 0.7V. <S> Once you know Ve <S> you know the emitter current <S> Ie= <S> Ve/Re, and since the base current is assumed to be zero, you can find <S> Vc = <S> Vcc - Ie <S> * Rc. <S> That is valid if \$\frac{R1R2}{R1+R2}\$ << \$\beta <S> Re\$ <S> If you want more accuracy you have to solve this circuit: simulate this circuit – <S> Schematic created using CircuitLab <S> where <S> Rx = <S> \$\frac{R1 <S> R2}{R1+R2}\$ <S> V1 = <S> \$Vcc\frac{R2}{R1+R2}\$ <S> Ry = <S> Re \$\beta\$ <S> Which is easily solved as Vb = \$(\frac{V1}{Rx} + <S> \frac{V2}{Ry}) \frac{Rx Ry}{(Rx + <S> Ry)}\$ <S> Then Ve = <S> Vb - 0.7V <S> Ie = <S> Ve/ <S> Re <S> And Ic = <S> Ie (1-(1/\$\beta\$)) <S> so finally, <S> Vc = <S> Vcc <S> - Ie Rc <S> (as before) <A> You would really need to know the output resistance of the input voltage source to Vb. <S> Without it you can't really know for sure. <S> If an op-amp or a similar amplifier stage drives Vb, then the output impedance should be quite low. <S> If it's low, then it's as if you're applying a true voltage source to Vb. <S> In that case, all of the bias resistors really don't matter because the input voltage will drown out their effect. <S> Alternately, something like this is commonly driven by a capacitively coupled source. <S> This means that DC will be blocked and AC will pass through. <S> In that case, the bias resistors you see in your diagram are what will determine the DC set point of Vb is. <S> In that case, what the others post is correct. <S> If your input voltage source has an output impedance somewhere in between the two extreme cases I give (0 and infinite impedance) then you'll need to add that into the circuit and start crunching the numbers/equations.
| Some off the cuff rules that usually work so long as your bias resistors allow enough current flow generally means that you treat R1 and R2 as a voltage divider and find an approximate DC setpoint from that.
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Understanding load cell and instrumentation amplifier I want to build a weight scale with 0.1 gram resolution. But I have some problems in my initial stage. I am using a load cell with sensitivity of 1mV/V and excitation voltage of 3 V. With no weight, the load cell output shows -0.089 mV and with full load (500gms) it shows 2.895 mV. Is -0.089 mV the noise of my load cell? If so, how could I solve this problem? Why is my load cell mot showing an output of 3mV for 3 V of supply? I have connected the load cell to IN amp (PMI AMP04FP ) with a gain of 100, with no load, I get an output of 3.203 mV and with full load I am getting an output of 278 mV. Why is the IN amp showing an output of 3.203 mV? Is it the offset voltage? If so how could I solve the problem? Why am I getting 278mV instead of 300mV? I have gone through the basics. But couldn't get my mind clear because of a lot of information in a small time. I know that my questions are basic and need some reading. I would be glad if someone could explain me what is happening and how to proceed. I am willing to learn. So, any kind of help (links, books etc) is appreciated. <Q> Presumably your sensitivity of 3mV/V is with a 500 gramme load. <S> and given that there is a small offset of -0.089 mV and this rises thru zero mV to 2.895 mV, there is a change of 2.984 mV <S> - that's quite close to 3mV and, as always with mechanical-electrical devices you need to make a calibration adjustment. <S> In other words the sensitivity of the device is nominally 3 mV per volt and 2.984 mV represents a small (and acceptable) error of 0.53%. <S> An In-amp is also subject to errors both in gain and offset. <S> 3.203 mV offset represents an input offset of 0.032mV. <S> Realistically, the error from the in-amp is adding to the natural loadcell error of -0.089mV and making it +0.032mV <S> i.e. the in-amp input offset error is 0.057mV. <S> The device you have chosen has a maximum input offset error of 0.3mV at ambient temperature conditions so this error is well within specification for the device. <S> As for the output not being 300mV but 278mV this is usually due to the tolerance on the resistor that sets the gain. <S> It should be also noted that the gain accuracy for this device (assuming a perfect gain-set resistor) is +/-0.75%. <S> Other factors that can contribute to problems are layout, power supply levels, inaccuracies in your meter/measurement device etc... <A> This is your instrument amp? <S> http://www.analog.com/static/imported-files/data_sheets/AMP04.pdf <S> I found the comma's in your numbers a bit confusing. <S> I assume that 3,203 mV is what I would write as 3.203 mV. <S> So first there is some DC offset in the load cell. <S> That's not surprising. <S> (do you have a model number.) <S> and for the second there is also a DC offset in your amp. <S> The spec sheet says a max of 150 uV <S> (x 100 gain) = <S> 15mV. <S> You see about 3 mV <S> so that seem reasonable. <S> Then your full load number... why do you expect 300mV? <S> (sorry maybe this should be a comment?) <A> Assuming that full scale is 500 g <S> and you desire a 0.1 g resolution, I would go about as follows: <S> Determine the temperature range over which this should work. <S> Let us pretend it is 25°C to 35°C. <S> So choose a sensor which is rated for 500 g and has low tempco for zero and span in the 25°C to 35°C range. <S> Let this be 100 ppm for zero and 100 ppm for span. <S> The load sensors are never ideal and when you get a sensitivity of 1 mV per volt, it is just a nominal value. <S> It need not be exactly 1.000 mV per 1 V or 3.000 mV for 3 V at 500 g. <S> You need to adjust this offset and bring it to less than about 1 <S> uV. the sensitivity of the sensor is the change in output when load is varied. <S> So if you get, after amplification, 3.203 mV at 0 load and this changes to 278 mV at full load. <S> Without sounding to be too technical (such as is this for 0.0 g to 500.0 g change?) <S> , the difference is \$278-3= <S> 275 \text{ mV}\$ for a change in load of 500 mV. <S> This is called span and to get this to chnage to 300 mV, one has to trim the gain, make it a little higher. <S> In summary, do not expect the output of the sensor (or for that matter the output of an amplifier) to be 0 for 0 input (load). <S> Nor would one get 300.0 mV with a gain of 100, since the 3 mV sensitivity claim is a nominal one. <S> Then there is the drift of the amplifier and drift of "zero" and "span" of the sensor to be considered. <S> Best of luck.
| Generally these sensors also have an initial offset which combines with the offset of the op amp, and that is why you presumably get -0.089 mV when no load is placed on the system.
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