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How to power a high current / low voltage circuit through wall outlets? I'm trying to wrap my head around this basic problem but I'm stuck here: I'm using 4 of these LED strips and I'm not sure how to safely power them. They are rated at 18W/M, so 4 strips * 5m * 18W = 360W with an input voltage of 5VDC, meaning they would draw 72A if they all were lit at the same time. Let's say I only turn on 75% of them at once, the circuit would then draw 54A. My understanding of wall outlets is probably wrong, but I read that they are rated at 15 or 20A. Does that mean that I cannot use the same outlet to power this circuit? If not, can I use outlets on the same circuit or will I have to use separate lines from the circuit breaker? Or any other solution that I'm not aware of? Optional question: Power converters that can output that much current are usually big, so I'm considering setting up 4 power converters in parallel all coming from the same power supply, is there any other potential issue I should be aware of if I choose this route? <Q> You need 360W @ <S> 5V. 200W 5V are fairly common, I would consider 2 of these. <S> https://www.amazon.com/SHNITPWR-Converter-Adapter-Transformer-WS2812B/dp/B07TSKK4FR/ <S> Don't put them in parallel, run 2 strips on each. <S> The input <S> current is not the same as the output current . <S> The input power is more directly related to the output power , the output power is less due to inefficiency. <S> If your power supply is 80% efficient, your input power would be 360W/80% = 450W. <S> At 110 VAC this would be only 4 amps on your household outlet. <A> I suspect it's a misprint (or a deliberate mis-specification). <S> Should 1.8W per meter. <S> My reasoning is as follows: <S> For the 5m roll, the connecting wires would not support 18A @5V <S> they are just 1-2A capacity. <S> 60 LED's per meter @ <S> 18W means each LED is dissipating 300mW which would be OK on a PCB with some copper, but unless the tape was stuck down on a metal surface, things would get too hot. <S> So maybe do a sanity check from other suppliers of similar items, and/or find one with a current rating. <A> Wall outlets use higher voltage <S> so 10A at 110VAC is 1.1 kilowatts. <S> You just need a 72A 5V power supply which is only 360 watts. <A> It's always tricky to guess at skill level of the asker. <S> But I'm concerned there are several skill areas where you're going to need a lot of help, no offense. <S> We use a unit called "Watts" to measure useful power irrespective of voltage. <S> Watts = <S> volts x amps. <S> Your load is 72A x 5V = 360W. <S> The common receptacle is good for 120V <S> x 12A (continuous) = <S> 1440W. <S> So it is simply a matter of power conversion. <S> As you notice, when volts go down, amps go up in proportion. <S> One thing that nobody's talking about is voltage drop . <S> This is where your design falls apart. <S> The curse of individually addressible LEDs is their need to run on a very low voltage, and being long linear strips. <S> Voltage drop is caused by Ohm's Law: <S> Vdrop = <S> A <S> * R <S> The resistance is decided by how thick the wires are. <S> But you are hit by a double whammy when voltage is very low: Current is very high so voltage drop is very high: <S> And, that voltage drop is taking a much bigger bite out of your total voltage. <S> Let's consider your 360W going through the 120V mains versus your 5V strips. <S> Let's say the wires have a resistance of 0.05 ohms. <S> Mains: <S> 360W is 3 amps @ <S> 130V. <S> A*R = <S> 3 <S> *0.05 <S> = 0.15 volts drop. <S> So 120V becomes 119.85 volts. <S> Drop is 0.125%. <S> 5V: 360W is 72 amps @ 5 V. <S> A*R = 72*0.05 <S> = 3.6 volts drop. <S> So 5V becomes 1.4 volts. <S> Drop is 72%. <S> Whoops! <S> See, delivering 5 volts over the same wire simply does not work. <S> The wires must be much larger. <S> The built-in wires in these LED strips is not thick enough to do the job for 5 metres, let alone 20. <S> They were never imagined to be used in a long string; you're thought to cut the string at the designated spot and use it for arrays. <S> So you will need to feed the strips at more frequent intervals than every 5 metres. <S> And you will need fairly thick feed wires if the LEDs are linear. <S> So there are some real challenges with your setup.
| Or use several smaller 5V supplies. Right off the bat, a 72 amp supply will require mighty big copper, or at least mighty big aluminum.
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How does this opamp circuit act as "good" voltage reference? If I understand correctly, we are using MC1043 because we don't trust Vcc to be stable. I get that the output of MC1043 is 2.5V which is extremely stable. But I don't get how the output of opamp will be stable. Because this opamp is getting supply from VCC and VEE which are not necessarily stable. What am I missing here? <Q> The supply voltage rejection of the LM348 is typ. <S> 96dB, meaning any ripple on Vcc will only be translated to the output of the OpAmp by a factor of \$10^{-\frac{96}{20} <S> } = 15.8 ppm \$ (parts per million). <S> The output voltage of the LM348 will be determined almost solely by the stable voltage of the reference source. <S> But, as Peter Green has pointed out in the comments, the PSRR is typically given for 50 / 60 Hz. <S> At higher frequencies the rejection will be not as good, the circuit designer has to take care to filter out any high frequency noise on the supply rails. <S> Effective decoupling is important with every good circuit design. <S> Additional: What might have a higher impact than the supply voltage are other parameters of the OpAmp. <S> For example the input offset voltage of the LM348 has a maximum value of 6mV. <S> With the amplification of \$A_v <S> = 4\$ <S> this could result in the output beeing off up to 24mV. <S> But even that is typically not a problem: These high precision sources are all about low drift over time and temperature. <S> As long as the error added by the OpAmp is static it can easily be calibrated for. <S> Important is, that it does not change. <A> What you are missing is that the op-amp's output is, ideally, controlled purely by negative feedback. <S> The output voltage is given by the equation in your diagram and you will notice that neither V CC nor V EE are factors in the equation. <S> Figure 1. <S> Internals of the ancient 741 opamp. <S> Source: Wikipedia . <S> If the output deviates from the desired value the negative feedback will increase the difference in voltages between the inverting and non-inverting inputs. <S> This difference will be multiplied by the open-loop gain of the op-amp adjusting the current through Q14 or Q20 as appropriate to pull the output (and hence the feedback) to the desired position. <S> In the real world supply noise can disturb the op-amp and for this reason we add decoupling capacitors close to the supply terminals. <A> At high frequencies, all opamps have no ability to reject the variations in VDD and VEE. <S> Thus its your task to use RC low pass filters on all the power pins. <S> Given the ESR of capacitors, I'd filter like this: a) <S> raw, trashy spikey VDD into b) 10 ohm in series, into 0.1uf shunting to ground into c) <S> 10 ohm in series, into 10uf shunting to ground <S> then d) into your opamp. <S> =============================== <S> Why use the resistors? <S> to isolate your capacitors from the surge current demands of other loads on those rails; and to give excellent low frequency rolloff; and to dampen, to prevent ringing.
| The output voltage of an OpAmp is independent from the supply voltage to a very high degree.
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RFID (13.56MHz) GND plane over antenna I dissasembled multiple RFID devices to study their design. The following reader attracted my attention: This is a four layer PCB, the anntenna seems to be 4-turn symetric and is layed out on the two internal layers of the PCB. You can see the cross-overs right next to the print "REV:B...". The thick solid plane on top (which is also on bottom) is NOT the antenna, it is connected to ground, and has has small vias on the edges of the plane. The reader antenna and parametrics of this reader are very similar to another one, but the reading distance with this reader is much better than with others, yet having the same NFC chip from NXP What I really wonder is the fact that there is a solid copper ground plane over the antenna. Usually RFID design guides recommend to not route any signals or ground near antennas and keep antennas away from large copper surfaces or metal plates. It is very interesting, that this cheap generic reader from China, is way better than genuine other readers with the same size. The following questions are unclear to me: Why is this ground plane on the sntenna?2.) Does the plane influence the reading distance (positively,) and if yes, why? In which case would such a design would be needed? <Q> 13.56MHz RFID uses magnetic coupling between reader and credential. <S> This coupling is used to power the credential. <S> To create communication between the two, the magnetic field is required to be modulated. <S> This will either be full amplitude modulation, type A, or a percentage of the overall field as is present in type B. <S> This modulation is easy for the reader as this is creating the field. <S> For the credential, it has to dump some power in order to load the field, reducing the amplitude output by the credential. <S> Most credentials these days are type B - the field can be varied quicker with type B. <S> To achieve this coupling, the two magnetic circuits need to be tuned. <S> The frequency of tuning and Q factor of each indicator needs to be determined by the designer. <S> Generally most cards will have a resonance around 13.56MHz but best performance from the readers is likely to be when the reader coil is not tuned exactly to 13.56MHz <S> but around 50 to 80KHz off. <S> The mutual inductance present when the cards are within range will create two smaller resonant peaks, detuning them. <S> You often find better range when this is the case. <S> I have seen designs like this in the past, they can help direct the magnetic field in front if the reader instead of both in front and behind. <S> It leads to an increase in range of reading the cards but not by a factor of two. <S> I cannot remember how it effects the Q of the circuit. <S> It is not necessary to do this in a design. <S> If you don't have the right tools, it's harder to implement and can take multiple iterations to get right. <S> The increase of range is likely to be smaller than worth the hassle. <S> To implement a reader coil, use 4 turns about the size of a credit card. <S> Different sizes can be used but the coupling factor reduces for smaller coils, and for larger then the card needs to be held more towards the edges where the Hfield is stronger. <S> You'll need to measure the inductance of the coil, from there you can estimate the parallel capacitance. <S> If you have access to a VNA then you can fine tune it <A> I came across some similar R/W myself and had the same question, then I found this document: https://eccel.co.uk/wp-content/downloads/antenna_1356.pdf <S> The ground plane's purpose is to reduce EMC emission, it seems. <S> I have no idea in terms of how much it affects the reading distance. <A> You might be looking at a patch antenna: <S> A patch antenna is a type of radio antenna with a low profile, which can be mounted on a flat surface. <S> It consists of a flat rectangular sheet or "patch" of metal, mounted over a larger sheet of metal called a ground plane. <S> https://en.wikipedia.org/wiki/Patch_antenna <S> They are quite common in UHF designs for RFID, due to their directional gain. <S> Used in many other applications too, e.g. RF motion detectors (for automatic doors), directional communication links etc... http://fab.cba.mit.edu/classes/862.06/students/alki/GA.html <A> The theory says GND next to your loop is harmful due to the eddy currents created which circulate in loops in such a way to create a magnetic field that opposes the intentional field generated from your loop. <S> https://en.wikipedia.org/wiki/Eddy_current <S> Note the shape, the GND plane is not a complete loop, there is a gap where the Rev is printed in silk. <S> That means eddys can only circulate in little loops in the plane but not around the entire loop. <S> The added GND plane capacitance factors into the resonance which should be tuned to get the required 13.56MHz. <S> I.e. you've already done the detuning yourself, so to first order the antenna is unaffected by metal housing. <S> This doesn't prevent additional losses from eddys due to surrounding metal but mostly preserves the resonance cause you've defined the capacitance with this ground pour. <S> As others have noted it helps with EMC as well.
| I think it may be that adding the ground plane yourself shields the antenna from the rest of the housing.
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LEDs have a low glow "noise" Interim conclusion edit at bottom. I'm driving a stack of 5mm LEDs off 24volts as in the diagram below. I have four stacks of these, two have 5 white LEDs and share their base drive, one has 7 red, and one has 7 yellow and each of those is driven off a different output. What I'm noticing is that they glow very very gently, and flicker as they do so (almost like an imitation of a candle-flame, which might be cute if I wanted it). Unfortunately, it's critical to the purpose of this design that these LEDs be totally dark when off. I notice that the LEDs are not all glowing the same amount, some seem to be more alive than others (though which ones is repeatable). I guess this is because they have slightly different characteristics, and whatever current is running down that stack is intermittently enough to make the most sensitive glow while still doing nothing to the others? Originally, the resistor and capacitor from the base to ground were not in the circuit. I first added the capacitor because I could see some high frequency (in the MHz range) noise on my scope, but that made no difference, then I added the pull-down resistor on a whim. That too made no difference. The scope shows the voltage at the bases of the transistors is essentially zero, and the noise peaks are perhaps 15 mV. So the question is, what's causing the glowing, and how do I fix it. Is it perhaps a leakage current/noise thing in the transistor? They're 2N2222. Again, what would be a good course of action. Do I need to use a FET instead of a BJT? I went with a BJT so I could control the current through the LEDs even when a stack of them might have had significantly different forward voltages, which differences would be potentially exacerbated by having more of them. (3v3 at the base minus the 0.7 forward BE junction puts 2v6 on the 150R and should give me a pretty predictable 17 mA through the LEDs, and the worst case voltage on the transistor still results in a pretty modest power dissipation. Well, that was the plan!) One thing I know I don't know is what is the lowest current level that will make an LED emit some level of light. Edits: These include a little more info, and answers to some of the questions so far raised: The flicker is so faint that it is undetectable unless in a darkroom (but this matters to me!) The transistors' bases are driven directly of the digital output of a NodeMCU It still flickers with a hard ground on the base and the CPU removed I can't put R1 between the collector and the diode, as that would lose my current control There's no mains noise on the base lines. Noise is very random,looks very white (to the naked eye), and around 15 mV peak. Putting the scope on base and then the emitter both show essentially0v (suggesting "low" leakage through that 150R) There is between 2v and 3v2 on the various LEDs in the stack. I think this tells me that it really is a leakage current issue. I had hoped to see the collector sitting at 24 V (supply rail) but it's not, it's about +10 V. Edit 2: Interim conclusion.So, based on extensive and very generous help in the answers and comments, I believe this really is leakage. I have experimented a little with the idea of adding resistors across the diodes which helped some, and it's quite possible that more experimentation might have found a good solution. The circuit proposed by MicroservicesOnDDD looks fascinating, but to be honest, it's more complex than I was hoping, and I admit I've not tried it (though I might get to it at some point). The two things that I've done at this point are: I added an additional LED (making six in the white stack, and eight in the red and orange stacks). This made a distinct improvement. In a burst of "crude, but effective", I added a simple relay and am now turning the 24 volt power for these LEDs off entirely when they're not wanted. The current relay is 240 volt capable, which is simply what I had in stock and so has higher current draw than it likely needs. However, it does, of course, entirely suppress the glow. So for now at least, this is my solution. Not terribly elegant, but provided the "click" that it makes when turning on isn't a problem, absolutely effective. simulate this circuit – Schematic created using CircuitLab <Q> This is likely not due to a flaw in the driver, or in the signal source, sinceonly some flicker. <S> Because LED's are not absolutely identical, the voltage across each in a series string will not be equal for DC due to differing leakage currents, and for AC this would be exacerbated by differing capacitance wihin each device and from device to ground. <S> Therefore, a signal not providing enough voltage to light all could cause a few to glow feebly. <S> The easiest fix is to put equalizing resistors across each LED. <S> perhaps 100 kΩ as a first try. <S> 10 nF caps across them might also be needed, if the resistors aren't enough to make the LED strings completely dark when nominally off . <A> In my setup, the glow starts when the voltage is 26 volts (could easily vary depending on the transistor). <S> Capacitive coupling would have a similar effect. <S> One 10k resistor across all 5 LEDs removes the glow in my setup. <S> This works up to 31 V, the highest that my power supply will go. <S> The 10k resistor will steal a few milliamps, you could try a higher value resistor if that is an issue. <S> This is basically what DrMoishe suggested, but only one resistor may be necessary. <A> Here is an LTSpice XVII circuit which (at least in simulation) does what you want: <S> When the 3.3 voltage source is connected (representing a HIGH on your microcontroller pin) there is 17mA running through your 5-LED string. <S> When the 3.3 voltage source is disconnected (by deleting the little bridge inside the circle) the current running throught D1 is reportedly in the femto-amp range (40 femtoAmps). <S> U1 is a voltage reference used to implement a constant current sink at NPN1 using R8 and R10 as a voltage divider and current/voltage delivery to NPN1. <S> This is what causes the 17mA to flow. <S> R9 is a pull-down for when the circuit is off. <S> When the circuit is off, R5 pulls up PNP3 and R7 pulls up PNP4, turning them off except for current leakage. <S> Also, R6 pulls down NPN2, turning it off, which turns off PNP3. <S> Effectively, the whole circuit is off except for leakage currents. <S> Finally, the 10Meg resistor at R14 is a big enough drain that it takes almost all of the leakage current (in the picoAmp magnitude in simulation). <S> Depending on R14, which I switched back and forth from 1Meg to 10Meg, the current remaining in D1 ranges (in simulation) from +100fA to -100fA. Placing a 10Meg potentiometer at R14 should allow you to make the LEDs completely dark. <S> Not really sure if D6 is necessary. <S> And I'm not sure if the LEDs will glow with the current going backwards <S> (it's such a low amount at femto-Amps that it should not cause any damage). <S> I did not build this circuit to verify it.
| Based on my testing (see comments), the glow can be caused by transistor leakage.
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Can you identify this diode? (vintage audio amplifier) I am restoring a Japanese Teisco mixing amplifier from the 1970s. It's a solid state 5-channel mixer with 16ohm speaker outs and a line out.Amongst other problems, I have discovered a couple of the bridge rectifier diodes have shorted out. Here is a photo of the diode bridge: Now i am trying to find a suitable replacement for these diodes as they have no markings.I cannot find any circuit diagrams or info on the unit at all.The diodes and the PCB both have no markings.If we can't identify this component ~ Is there a common equivalent used for typical solid state amps from this time? Surely.The filter capacitor is 2200uf 63V.Would a 1N4007 do the job? Thank you in advance, Looking forward to a reply. <Q> Those were a very common diode at one time. <S> My very strong recollection is that they were rated at one Amp. <S> 1N4005 should replace them just fine. <S> Be sure to replace all four diodes. <A> Put in the highest peak-repetitive forward current diodes you can. <S> And install a small fan. <S> The only way for heat to exit is by the PCB foil, and there is very little of that. <A> 1A is a bit wimpy for an audio amplifier; 1A into 8 ohms is only 8W. I would concur with Mattman that these were probably 3A ( <S> A Google image search for "3A diode" shows several in this case shape (merely illustrating my confirmation bias; "1A diode" shows some too, but fewer I think) <S> A BY228 Fast Diode 3A 1000V <S> SF5408 <S> SOD64 3W1000V <S> (image from BYW95A link above) <S> Fast recovery is not needed here <S> but now we have the package style <S> SOD64 <S> allowing us to search a supplier for SOD64 , in this case giving 27 results; mostly 3A rectifiers. <S> So you can even keep the original look if you are so inclined.
| After eliminating avalanche diodes and other fluff, these show several possible 3A rectifier diodes : BYW 95A Rectifier diode , SOD64, 200 V, 3 1N5403 rings a bell, or 5405 for higher voltage) in that case shape. 75NS Ultra-high Speed Fast Recovery Diode
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Convert kWh to watts per m² I am from a computer science background, I need some help with some electrical engineering problem. Electricity consumption for a home in the last 56 days is 7561 kWh. There are two floors. The combined area is 300 square meters. Can I use this information to get watts per m² for this house? <Q> No, the units are not compatible in any way. <S> Unless you have an area. <A> If the relevant area you had in mind is 1m 2 , then that's 5.6 kW <S> /m 2 . <S> If you were thinking of 1km 2 , then it's 5.6 mW/m 2 . <S> Did you have an area in mind? <S> (edit) <S> Now that the OP has let on that the area is 300 m 2 , we can divide that 5600 W by 300 to get about 18.7 W/m 2 <S> (/edit) <A> Now that you've posted all the required information, an answer is possible. <S> 56 days is 1344 hours (56 days <S> X 24 hours per day.) <S> You have 7561 kilowatt hours of energy consumed. <S> Divide the energy by the time to get average power. <S> That's <S> 7561 kWh/ 1344 hours gives an average of 5.62 kW. Divide 5.62 kW by 300 square meters and you come up with 18 watts per square meter. <S> That's an average of the power consumed by every square meter of your house over 56 days. <S> I don't know how much good it really does to know that number. <S> Is it for some kind of home energy efficiency rating or something? <A> The beauty of SI is how you can easily balance units and convert between domains. <S> So <S> 7561 kWh ... <S> kWh = <S> = <S> J/s <S> * s == <S> kg⋅m2⋅s−2 <S> While you cannot express it in Watts per m² (without doing other funky units around it), <S> you can express it in kg m² per s² <S> but then you could say that kg⋅m2⋅s−2 is equivalent to N.m = <S> = <S> Pa⋅m3 <S> so a kWh could be express as Newton-metres are Pascal per unit volume
| 7561 kWh per 56 days is about 135 kWh per day, or 5.6 kWh per hour, more simply written as 5.6 kW.
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Why the triangle reference wave is used in PWM for sine modulation? I'm implementing PWM for a three-phase inverter. The documentation on the subject, especially the technical implementations, is very good but as far as I am able to understand none of the one I read talks about why the reference signal is a triangle or a sawtooth. In one document I found that this type of modulation is the oldest in the game (PWM concerned) so maybe is a thing of old technology that becomes standard? <Q> The "triangle reference" method that has some history as a standard or popular method was more formally known as the "triangle interception method. <S> " It was based on generating switching commands based on the interception of a triangle wave with a sine wave as shown below. <S> It was not too difficult to implement and it provided reasonably good results for variable frequency drives (VFDs). <S> Results were improved by changing the ratio of the triangle wave frequency to the sine wave frequency over the range of the operating frequencies. <S> After the invention of large scale integrated circuits and microprocessors, more sophisticated PWM strategies became preferred. <S> Image from Zubek, Abbondanti & Nordby <S> “Pulsewidth Modulated Inverter Motor Drives with Improved Modulation <S> ” IEEE Trans. <S> Ind. Appl. <S> , Nov./Dec. 1975 <A> You don't need to have a triangle-wave or a PWM for that matter to generate a sine wave. <S> You could use hysteresis modulation which does not require a PWM. <S> However, the switching frequency is variable with a hysteresis modulation and it might require more computing power. <S> If you need a PWM, a triangle wave is the way to go, a sawtooth wave would cause problems for H-Bridges and 3-phase inverters/rectifiers. <S> The top transistors or the bottom transistors would all stop conducting at the same time. <S> While with a triangle wave, the top and bottom transistor of different branches will not start conducting at the same time except if the duty cycle is the same for all branches. <S> Edit : In the case of 3-phase rectifier/inverter using, here's the comparison between sawtooth PWM and triangle PWM. <S> As you can see, with the sawtooth, your transition from the zero-vector "000" to the zero-vector "111" , something you don't see with the triangle wave. <S> With the triangle wave, transistors do not switch at the same time <S> Another reason why a triangle wave is better. <S> It is easier to sample signals. <S> I added arrows on the graph to highlight the sampling instants. <S> With the sawtooth wave, no matter where you sample, there could be a transistor switching. <S> Edit 2 <S> : The last image represents the PWM for a space-vector modulation from 0 to 60 degrees with 1 zero-sequence vector "000". <S> You can see easily that only one transistor switch at a time. <S> The PWM frequency used is 2400 Hz. <S> In real-life, 2400 Hz is a bit low, it's more usual to see 10 kHz and more for PWMs. <S> You can also see Va, Vb and Vc <A> For sinewave PWM generation (or audio applications like Class-D amplifiers) triangle waves are preferred over sawtooth, because of their symmetry. <S> This gives 2-sided PWM where both the rising edge and falling edge are modified as the duty cycle changes. <S> This means that the centre of each PWM pulse stays in the same position in a PWM period, whereas with a sawtooth waveform (single sided PWM), the centre of the pulse moves as the duty cycle changes, which adds distortion to the reconstructed sinewave. <A> Think of it numerically/digitally instead of with waveforms: You would need a period timer. <S> What does a period timer that counts up or down at the same rate look like? <S> If it wraps around to zero at the end, it looks like a sawtooth. <S> If it reverses direction once it reaches the top or bottom, it looks like a sawtooth or triangle wave. <S> If you graphed a digital timer it would look like a staircase or ziggurat. <S> But if the timer is analog it looks like a sawtooth or triangle. <A> Sawtooth / triangular reference waveforms can be trivially generated using widely available hardware counters (timers), with minimum CPU involvement. <S> Signals generated by such modulation become asymptotically precise as PWM frequency increases, so unless you can't increase the PWM frequency to be 20-50 times higher than the signal frequency, there will be no improvement from a different reference waveform. <S> This is the reason they are so widely used.
| With the triangle wave, you can set the ADC trigger when the triangle wave is at 0 or 100%, because at these points you are assured that no transistors will switch.
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Can I prevent a 9V LED flickering when dimmed to around 30%? I bought an overly bright 9 volt, 3 watt LED to light up a bird box which has a webcam in it. The LED module is this one: https://www.amazon.co.uk/gp/product/B07CWSRF6N However when I dim it using a simple 12V dimmer to 50% or less, it starts to make the video feed flicker due to the strobing LED's interaction with the camera frame-rate. It's being driven by a spare 9V DC power supply previous used for a broadband router. Can I fit a capacitor across the LED to help smooth it? If so, what kind of value would I need? Or am I better just to fit a big load resistor in front and remove the dimmer? Thanks! EDIT: Here's a pic of the dimmer innards: <Q> There is no inductor on the dimmer PCB <S> so it is almost certainly driving the LED directly with the PWM signal. <S> If you attempt to filter the output just with a capacitor it will probably cause the LED to run at almost full intensity even if adjusted for a lower value. <S> The capacitor will charge during the on period and then power the LED during the off time. <S> It will also cause very high peak currents in the dimmer that could potentially damage it. <S> To avoid flicker you would need to provide the LED with DC. <S> Your big resistor suggestion would do just that. <A> You don't need a dimmer if you don't intend to modify the luminousity frequently. <S> What you need is to set the right voltage so that it will draw the right amount of current. <S> If you put 9V to it you will draw the maximum 300mA. <S> If you want 50% less, you must draw 150mA. <S> Idealy <S> you could use a constant current power supply giving 150mA and 1.5W (between 1 to 3W) <S> (or half than 3W). <S> If you can't find such supply, try with a 6V supply. <S> Ideally an adjustable power suply so that you can regulate manually. <S> but also, can be hard to find at this voltage. <S> It's not sure which voltage will give you half than 300mA, but I can say it will be much more than half the indicated voltage for 300mA. <S> Can be 7 or even 8 volts. <S> Only way to know is to try. <A> Smoothing capacitors can be calculated using the following formula: C = I⋅Δt/ <S> ΔU <S> where C = capacity of the capacitor in μF <S> I = <S> Charge current in mA <S> Δt = half-period in ms ΔU = ripple voltage in V <S> However, in this case adding a smoothing cap might actually make the problem worse. <S> It is important to note that the intensity of an LED is dependent on current and time it is switched on, but not voltage (so long as it's above the forward voltage). <S> If is is not running hot at full power, I recommend you try to reduce the brightness using a mechanical shutter such as a translucent film to block some of the light. <A> If the LED is strobing when you reduce the duty cycle, it suggests that the dimmer in question is using some sort of (poorly) low-pass filtered PWM, or the LED's drawn current is higher than the dimmer's rated output current. <S> As you already suggested, adding a capacitor will reduce the cut-off frequency and hold the output voltage a bit longer until the next cycle. <S> As long as you add any capacitor which is rated to the required voltage and ripple current, it should improve the behavior. <S> If you want to select a capacitor based on the PWM frequency, you would need to open the dimmer and measure it, and add an additional capacitance which leads to a cut-off frequency of alround 10 times smaller than the PWM's frequency.
| Adding a cap might smooth the voltage of the PWM to a level below the minimum forward voltage, and the LED will no longer light up.
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How can I know a signal's bandwidth before sampling? Shannon's theorem tells us the minimum sampling frequency based on the bandwidth of the signal which has to be sampled. But I wonder, how can I know the signal's bandwidth before sampling it? It is generally assumed that one knows in advance the continuous time dependence of the signal (that is, f(t)), but I don't understand how this information can be available without a measurement, which in turn requires a sampling rate to be chosen. <Q> It depends on the purpose of the sampling. <S> If you need to sample a signal accurately - such as recording music where distortion is undesirable, you know the signal's bandwidth at the ADC, because you already ran it through a low pass filter. <S> This also works with slices of bandwidth above the Nyquist frequency or even above the sample rate, provided <S> the slice has been selected via a narrowband filter. <S> For example the output of a narrow 10.7 MHz IF filter may be sampled at 10 MHz, and will appear as signals around 0.7 MHz. <S> This is known as undersampling. <S> If you need to detect any signal activity but you're most interested in the presence or absence of signals, or the envelope, or total energy, so that accuracy is less important, then just sample the signal, and be aware that aliasing introduces ambiguity in the frequency of any components. <S> Comparing two sampled channels, one via a low pass filter and one without, will alert you to activity outside the Nyquist frequency rate; if you need to do something about that, then you can take further steps (such as, bandpass filters ahead of the ADC to identify which spectral image is active, or changing the sample rate). <A> This is where engineering judgment and/or experience plays a role. <S> The definition of "sampling at a super-high rate" can mean different things for different people or applications. <S> There are also cost and feasibility tradeoffs to consider as well. <S> The engineering part comes in where you have to evaluate what you think the type of bandwidth may be <S> and then give that a go. <S> Research into unknown things very often takes trial and error methods and successive refinement of your process. <A> All possible phenomena that you can measure have a maximum bandwidth, if nothing else the recording device (microphone, photodiode, etc) is a low pass filter. <A> Before sampling you will have to run it trough a low pass filter in order to prevent aliasing. <S> You should not sample a signal with higher frequency components then you can sample. <S> Since you'd have no clue what you have sampled and how to reproduce or present it. <S> If you do not know the bandwidth of the source, then what you trying to sample anyway?
| You can always look at the specs of your measurement device and sample sufficiently to cover it's bandwidth.
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In an equation like i(t) = CdV/dt, is "V" the cause and "i" the effect? In mechanics, almost all the time, force is the cause and acceleration the effect. If you apply a net force on an object, an acceleration can be observed. However this seems not that straightforward in electronics in relation to voltage and current. I used to think that voltage is the cause and current is the effect. But then I came across solarcells, current sources etc and immediately realized my lack of imagination. For each of the 3 passive components, is there a preference to think one quantity as cause and the other as effect or it is completely arbitrary? Mechanics: If force is applied, velocity is changed . If velocity is changed(accelerating electrons), a force(emi) is produced. Resistor: If voltage is applied, current is produced. If current is applied, voltage is produced.Capacitor: If current is applied, voltage is *changed*. If voltage is changed, a current is produced. Inductor: If voltage is applied, current is *changed*. If current is changed, voltage is produced. <Q> It's not so simple even in mechanics. <S> Imagine you are a nail, with a hammer coming fast towards you. <S> In order to stop the hammer in the space available, it has to be accelerated (decelerated with a change of sign). <S> While undergoing that rapid acceleration, it applies a large force to the nail. <S> The problem is what you mean by 'cause'. <S> Usually, we think of effect following cause, that cause never follows effect. <S> The wayward football broke the neighbour's window. <S> It is totally reasonable to argue that the broken window did not cause the football to be moving. <S> However, in electronics, and mechanics, where the behaviour of accelerations, forces, voltages and currents are conveniently described by equations of the form dI/dt = V/L, and f = ma, the two go hand in hand. <S> We can say the voltage across the coil caused the current to change, if that's convenient at the time, for instance when 'charging' the inductor of a boost DC-DC converter. <S> We can also say that the changing current caused a large voltage across the coil, when 'discharging' the inductor of the same converter into the higher output voltage. <S> Both descriptions are correct. <S> It is not reasonable to argue that one did not cause the other. <S> I've had long arguments about whether a BJT transistor is voltage controlled or current controlled. <S> It is of course both, it's nonsense to argue that the current flow doesn't cause the voltage difference or vice versa , but concentrating on one or the other may be more useful choice at any given time. <S> Similarly, the large force applied to the hammer head by the nail <S> caused it to slow down rapidly. <A> I don't think that the "cause and effect" consideration is particularly helpful. <S> In fact many physical properties have "duality", i.e. they have "partners" that behave the same way if you swap the variables. <S> For example capacitors and inductors are dual with respect to voltage and current. <S> At the same time mass and springs are dual with respect to force and velocity. <S> It's actually quite helpful to analyze mechanical or acoustic systems using an electrical analogy and using electrical circuit analysis tools & methods. <S> These are called Mechanical Electrical Analogies. <S> It's actually straight forward to determine whether to equate Force with Current or Voltage. <S> Hence two different analogies are used, depending on the problem. <S> The "Mobility Analogy" (Force equivalent to current) and "Impedance Analogy" (Force equivalent to voltage). <S> That's simply so that the product is the power. <S> Voltage time current is power and so is Force times velocity. <S> This way you get full symmetry in the physical laws. <S> $$F = <S> m <S> \cdot \frac{\partial v}{\partial t}, v = C \cdot \frac{\partial <S> F}{\partial t}$$ <S> where F is force, v is velocity, m is mass and C is the compliance . <S> Good reading: <S> https://en.wikipedia.org/wiki/Mechanical%E2%80%93electrical_analogies <A> Why not ask a similar question about a more basic equation, Ohm's law? <S> Voltage or Current? <S> The fact is that some equations do not imply cause on their own. <S> They are simply relations between two things. <S> Much like how 1 meter is 1000 millimeters, neither causes the other. <S> They both exist within the same space and simply can be found via that relation. <S> It is natural for humans to think of force as a 'cause' <S> because that's what we do... <S> we exert forces on things. <S> However, a force exerted on a mass (without other intervening forces) can be expressed as acceleration. <S> It's simply a relation between a force and a mass. <S> Much in the same way, your question is a restatement of an old debate that does not really serve as more than philosophical discussion. <S> Voltage is a measurement of a difference in electrical potential between two points. <S> It is defined as the amount of work per unit charge required to move a charge over a given distance. <S> Current is a measurement of the flow of charge within a given area. <S> It is defined commonly as a flow of one Coloumb of charge per second. <S> Voltage can be measured without current (equipotential lines in a electric field, as an example) and current has been defined quite a few ways in the past without relation to voltage. <S> Within the context of circuits, they are simply normally guaranteed relations , rather than causes and effect. <S> Even your equation above is a relation and has a form that expresses voltage across a capacitor with respect to the current. <S> $$v(t) <S> = \frac{1}{C}\int_{t_0}^{t_1} i(t) <S> dt <S> +\ v(t_0)$$
| If the voltage is equal to the current times the resistance, then which one produces the other... The dual variable to Force is always Velocity (not acceleration or displacement).
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What is the point of this resistor network? What is the purpose of this resistor network? Why not just use one resistor? or connect the cathode of RA93 diode directly to the mosfet drain (this mosfet is high side of the h-bridge). It seems to me that pin 30 of the relay is connected to pin 87 through diode and the resistor network.I just am not able to figure out why? The diode is probably to preventing the circuit in case someone plugs in the battery terminals backwards, is this right? Thanks. simulate this circuit – Schematic created using CircuitLab simulate this circuit Alternate schematic (by @Transistor) emphasising bulk capacitor and relay bypass of soft-start. Purpose of D3 is unclear. Operator to review. <Q> Your network seems to consist of a series / parallel combination of eight 27 Ω resistors that is equivalent to a single 54 Ω resistor. <S> The significant thing is that these are surface mount and will be placed by machine. <S> Presumably eight are required for power handling. <S> The circuit could have used one 54 Ω through-hole 1 W resistor but this would require an extra component and, probably, hand installation and soldering. <S> With pick and place equipment this could be avoided for this component (although not for the caps, relay and MOSFETs) and possibly there were some savings by using resistors which had been bought in bulk for elsewhere in the circuit. <S> Another reason one might see series combinations is because of high voltage, such as dimmer-circuits, etc. <S> The components may have a maximum voltage rating considerably less than the circuit voltage and this is solved by series connection of several resistors but that is not the problem in your application as it is low-voltage battery powered. <A> Since this is a low-voltage circuit, the reason is almost certainly to reduce the power dissipation in each resistor . <S> The maximum power dissipation in the network is probably higher than the rated maximum dissipation for a single resistor, so they spread the power among 8 resistors...each resistor is only required to dissipate 1/8 of the total power. <A> <A> This looks like a soft-start circuit. <S> Incidentally, you've drawn the relay symbol the wrong way around. <S> The markings on top clearly denote terminals 87 and 30 as being connected to the switched contacts, but you've drawn the diagram with them connected to the coil. <S> I'll assume they really are the switched contacts. <S> When the relay is open, power flows to the H-bridge (and the inductor driven through it) only through the resistor network. <S> This is commonly done to reduce the inrush current when starting a motor; once the motor is running, the relay closes and power flows directly. <S> Startup is the highest current draw condition, so the resistor bank has to be able to handle it. <S> The relay is rated for as much as 70 amps! <S> So a resistor network is used to distribute the power dissipation across more than one individual resistor. <S> These surface-mount devices are very small and can't handle much power by themselves, but as a team that capability is multiplied. <A> D3 is your standard back emf absorbing diode that stops collapsing magnetic field in coil from generating a transient and blowing fets. <A> They use several resistors for power dissipation reasons but the resistors are effectively shunting the relay forcing a certain level of current from the "H" bridge in order to trigger the relay. <S> If the current that's required to operate the inductor L1 is too high, then the relay will trip. <S> There is no frequency indicated so the actual load from inductor L1 is not known. <S> However, if the load is too high or the frequency is too low the current will then be forced across the resistor network and the relay. <S> This looks like a garage door opener overload circuit. <A> The resistores form a Current Dump! <S> And as they are only Surface mount types can't individually shunt the Power so hence the Parallel in Series configuration. <S> This is a very common way of Dumping Current and or Power in the Professional World.
| Using resistor network instead of a single resistor is to increase power dissipation or maximal rated voltage of the single resistor.
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What's the difference between a CPU core and a CPU/GPU thread? I've heard the term core and thread being used synonymously a lot of times. In Minecraft, if you press F3 to show coordinates, it tells you how many threads you have in your CPU but uses the term "core" for them. I know some things about threads and cores, but they are almost indistinguishable. Both can run programs and have memory, but there are typically more threads and the threads can also appear in GPUs. Is there any difference in architecture between cores and threads? <Q> Please read this Wikipedia page . <S> Most CPU cores can only process one instruction at the same time. <S> Some CPU cores can process more (usually two) instructions at the same time. <S> Intel calls this hyperthreading , to the operating system (Windows or Linux, etc.) <S> and therefore the application software (for example: Minecraft) <S> such a CPU looks like it has <S> two (logical) cores since it can process two instructions at the same time. <S> At least, that's what it looks like to the software (operating system and application). <S> A thread is a "string of instructions" that are processed on one logical CPU core. <S> So a single-core but hyperthreading CPU can process two instructions at the same time so it can handle two threads at the same time as well. <S> To complicate things, the software switches (in time) between many threads so usually there are a lot more threads active than there are logical CPU cores present. <S> But that's OK, if processed quickly enough, you wouldn't notice this as a user. <S> This is because most threads (tasks) can wait a little bit for their turn to use the CPU. <S> Only for a quick response a thread might be assigned a higher priority so that it gets preference over other threads. <A> Think of one-handed and two-handed people as cores, and each hand as a thread. <S> In that case, adding an extra person always means you are always adding the capability of doing at least one extra task simultaneously. <S> But if the people are one-handed, then each person can only handle one task. <S> But if they have two hands then a single person can handle two tasks at the same time, albeit at lower effectiveness (for example, some tasks might requires all the attention of one person <S> and <S> but it might be cheaper to squeeze extra usage out of the same core <S> than to outright add another core). <A> CPU threads don't have all functional units that make a whole CPU (or CPU core). <S> Typically they don't have their own L1 cache but share it with the other threads on the same core. <S> They may also lack their own ALU but have to share it with the other threads on the same core. <S> If you have a multicore CPU, it's still useful to give each core two threads <S> so there could be one copy operation and one ALU operation executed at the same time per core. <S> More than two threads aren't too useful. <S> CPU threads are most useful if the software runs multiple threads of related functions and their data in parallel, so all required data can be accessed through the L1 cache. <S> They are less useful when there are context switches. <A> Core is physical processor. <S> Multi-threading is capability to run multiple threads on a single core, thus multiple threads have to share resource available by the core. <S> If thread occupies all resource of the core, another thread cannot run on the same core.
| A CPU core is a unit that can perform instructions. Another core always means another thread is available with which to do work, but an extra thread doesn't necessarily mean an extra core is present.
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What the problem of my freewheeling diode 1N5822? Hey, guys! I have a DC motor with a voltage of 12V and a power of 550W, using a freewheeling diode 1N5822, but now this diode is often burned. Check the information to see that its maximum instantaneous current reaches 80A, thus I think it can work properly in the circuit. Why this happened? What should this margin be? Is it related to reverse recovery time? <Q> Adding to the previous answers, without knowing the characteristics of your motor and its transient behavior, it is hard to assess your problem. <S> In any case, your diode is way under dimensioned. <S> You are probably running into a thermal problem. <S> Not only the motor's rated power, but also its inductance plays a role in the selection of the freewheeling diode. <S> Consider the following simulation, where the inductive load is swept from 1uH to 100mH. <S> Here you can see the power dissipation of the diode. <S> Although the average power does not change much, take a look how the inductance affects the energy that this diode has to handle in the next picture: <A> The 1N5822 datasheet says the diode is rated at only 3.0 A. <S> It also says: Peak forward surge current, 8.3 ms single half sine-wave superimposed on rated load, <S> I FSM = 80 A. <S> This is an absolute maximum and not where you want to be operating. <S> Your circuit will probably not have an 8.3 ms half sine-wave current profile. <S> You have given no motor current or run-down time to allow further calculation. <S> Remember also that on switch off the motor will have inertia and will act as a generator with the diode acting as a short-circuit until all the energy is burned up in the motor resistance (and mechanical friction). <S> I suspect that a simple stop-timing measurement will confirm that you are exceeding the diode's rating considerably. <A> This diode is rated for 3A. It can sustain a short peak of 80A, but your application needs a diode that has rating of approx. <S> 50A.
| Upon turning off the motor, the diode creates a short-circuit with this inductive load and should be able to dissipate all this power without being stressed for too long.
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Power Density parameter in Ultra Low Power (ULP) devices I'm doing a bit of research on ULP processors. In a few papers (e.g. "Energy-Efficient Near-Threshold Parallel Computing: The PULPv2 Cluster" ) I found a figure of merit called "power density" and defined as Pdens = uW/MHz, so basically it is energy. The thing is I don't understand what it means, I'm not sure when this would be useful either. From my understanding: A circuit consuming 10 W @ 25 MHz has a Pdens1 = 10/25 W/MHz = 0.4 W/MHz. If it was running @ 50 MHz but still consuming 10 W then it would be Pdens2 = 0.2 W/MHz. To me, Pdens2 is better than Pdens1 since the goal is to minimize power consumption. Any hints? Or good introductions books/article to ULP which better explains figure of merit like this and Energy Efficiency (MOPS/mw), for example. <Q> The higher the frequency, the more the power dissipation will be. <S> All else equal, a processor that consumes 10 W at 25 MHz will need 20 W at 50 MHz <S> *. <S> The power density (I'm not familiar with this name for it, but I'll assume you have it correct) is roughly constant for a given processor, so if you have a power density of 0.4 W/MHz, you can predict how much power it will consume for different clock speeds. <S> It helps you to compare the relative performance of different processors; if one processor you're considering using is rated to consume 10 W at 20 MHz, and another one consumes 20 W at 50 MHz, but you only need to clock the processor at 10 MHz, you'd have to do the math yourself to find which one would consume more power at that clock speed. <S> But if instead they were rated at 0.5 W/MHz and 0.4 W/MHz, you can clearly see that no matter the clock speed, the second processor will consume less power than the first. <S> *note: <S> It's not exactly linear, but it's a good approximation. <S> At very low frequencies especially, though, don't expect the relation to hold. <S> Actually, don't expect your processor to work ; most processors have a minimum frequency below which they won't work. <A> In a few papers I found a figure of merith called power density and defined as Pdens = uW/MHz, so basically it is energy. <S> "Power density" in this case is just energy required to do one cycle of computation. <S> If you multiplied by how much work per cycle the processor did, you would have computational efficiency. <S> The thing is I don't understand what it means, I'm not sure when this would be useful either. <S> It is an easy to measure parameter that is roughly proportional to efficiency. <S> It isn't the same as efficiency itself, but measuring that would be hard since how much work per cycle a processor can do depends on the specific thing it is doing. <S> Having an easy to measure figure of merit is very useful. <S> To me, Pdens2 is better than Pdens1 since the goal is to minimize power consumption. <S> Any hints? <S> That is correct. <S> The processor that is more efficient is better according to this metric. <A> The lower the number, the better is power efficiency.
| This figure is used to compare power efficiency of different processors.
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Is it safe to touch 8 volts? Let me say first that i'm a noob in electricity. I measured the voltage from my diy microwave-transformer-welder and it was 8.I have no idea how much amps run through it but the fact that it creates lots of sparks and can create welds on metal may suggest that it is a high amount of amps.Is it safe to touch this or do I stick to my gloves? Edit: I added an image of my setup, that 230V AC is 16A. <Q> In fact most authorities round the world reckon that somewhere in the 40 V to 60 V range is safe to touch. <S> 8 V is high enough to send very large currents through metal wires. <S> The heat generated by this could burn you or start fires. <S> A re-purposed microwave oven transformer is heavy enough to break toes, if dropped. <A> I guess you'd better get worried about this bare 230 Vac wires you can easily toch or make a short circuit... <S> If you consider yourself a noob, be advised by those big red letters. <S> There are there for a reason. <A> I seem to recall that anything greater than 20ma thru you can be fatal. <S> At 8 volts, a resistance of 400 ohms would allow 20ma to flow (I=E/R). <S> Dry skin should be much higher than 400 ohms. <S> If, however, your skin is wet or you puncture the epidermis & reach the dermis, you can get a significantly lower resistance. <S> Science experiment: touch a 9 volt battery to your dry fingers and then your tongue. <S> The former should give about 4ua while the latter about 100ua. <S> A lot depends on your body chemistry. <S> Caution is a good habit to establish.
| 8 V is too low to send lethal currents through you, unless you implant electrodes under your skin near your heart (don't do that).
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Is the so called kTC noise, that is inherent in an RC filter, dependent on bandwidth? An RC filter has intrinsic noise called the kTC noise which is given by the equation $$v_\text{noise} = \sqrt{\frac{kT}{C}}.$$ Does this equation calculate a noise density with units of volts per root hertz? Or does this equation calculate an RMS noise voltage that is not bandwidth dependent? <Q> The KTC noise is actually the total RMS noise power at the output of the low pass RC filter. <S> So, actually the total noise power does depend on the bandwidth. <S> Interestingly, for the case of RC filter although the resistor is the only noise source present in the system, the total noise power is independent of the resistor value, as seen from the noise power expression, <S> $$v_n^2 = \frac{kT}{C}$$ This is because although increasing the resistor value increases its noise PSD (= 4kTR) but the effective bandwidth ( \$\frac{1}{4RC}\$ ) goes down proportionally. <S> $$v_{n}^2 = <S> 4kTR <S> \cdot \frac{1}{4RC}$$ <S> This can also be understood from the principles of thermodynamics. <S> Since the origin of Johnson noise is due to random motions of electrons inside the resistor, the kinetic theory of gases is applicable for it. <S> And the equipartition theorem says that in thermal equilibrium, energy is shared equally between each state of the system and the shared energy is given by \$E <S> = \frac{1}{2}kT\$ . <S> For the RC system, the only energy state is the energy stored in the capacitor which is given by: $$E = \frac{1}{2}CV^2$$ <S> Equating <S> the two terms gives: $$\frac{1}{2}CV^2 = <S> \frac{1}{2}kT$$ <S> $$ <S> \implies V^2 = <S> \frac{kT}{C}$$ <S> This is exactly the noise power derived from circuit equations but is derived much easily from laws of thermodynamics. <A> the ktC noise formula simply happens when you insert the noise bandwidth of an RC filter, B=1/(4RC), into the formula for voltage square over a resistor. <S> The R cancels out. <S> So, no, there's correctly no bandwidth in that formula, because the bandwidth is "hidden" in the C. <S> The wikipedia article on Johnson-Nyquist Noise actually containst this explanation, and a lot more stuff that you might want to know about kTC noise, considering you seem to be coming at this from a non-intuitive, formularistic side. <S> It's usually easier to understand what's happening than to have all the right formulas exactly right :) <A> Ultimately it is because the noise (and signal) is counted in electrons, and the electrons jiggle about a bit (technical term) and their speed of jiggling depends on temperature. <S> So the count of the number of electrons in the capacitor changes constantly. <S> The noise (rms) on the number counted is simply the square root of the number (same as a repeated random coin toss). <S> The magic value 'k' is the conversion factor. <S> The bandwidth is in cycles (in & out) per second, and the Temperature gives the 'how far/how many' move in and out of the capacitance. <S> The square root part annoys every one because it doesn't follow the normal electrical rules. <S> Rather its because electrons are quanta (quantum mechanics). <S> As explained by others the formulas can be swapped around to pretend that the resistors are noisy, while I find it easier to simply count the electrons in the capacitor. <S> The answer is the same. <S> The Physics is the same. <S> Very useful to know if you do photonics.
| The noise power is the product of the noise power spectral density (PSD) and the bandwidth (more accurately "effective bandwidth") of the system.
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How to calculate \$R_{\text{in}}\$ in this circuit? I know for an ideal inverting op-amp, the equation for finding gain would be \$\frac{-R_1}{R_{\text{in}}}\$ , but how can I find \$R_{\text{in}}\$ in this circuit? <Q> You can do a thevenin equivalent of vin, 2R and the first R so that it looks like this: <S> Ux is the voltage divider output of R and 2R and Rx is the parallel combination of 2R and R. $$ <S> U_x = <S> v_{in}\cdot\frac{R}{R+2R}=v_{in}\cdot\frac{1}{3}$$ <S> And for Rx $$ <S> Rx = <S> \frac{R\cdot 2R}{R+2R} = <S> \frac{2R}{3}$$ <S> So <S> $$R_{in} = <S> Rx + R = <S> \frac{5R}{3}$$ <S> Finally $$V_{out} = <S> U_{x}\frac{R1}{R_{in}} = <S> v_{in}\cdot\frac{1}{3}\cdot\frac{3}{5R}\cdot <S> -R1 <S> =v_{in} = -\frac{R1}{5R}v_{in}$$ <S> So you can say that $$R_{in} = <S> 5R$$ <A> First, measure the Thevenin resistance. <S> Here, Thevenin resistance = (2R || R) + R = <S> (5R/3) <S> So,Thevenin voltage = <S> (R/(R+2R)).Vin = Vin/3. <S> Now your circuit is reduced into: <A> 1) Assuming your question relates to Rin being the load seem by the left voltage source, Rin will be 2.5 <S> * R. <S> This answer requires an ideal opamp, with infinite gain, which forces the voltage to zero at (-) input of the opamp. <S> 2) how to chose a value for Rin? <S> very low values will get hot, and likely overload both the voltage source and the opamp (and the opamp power supply) but the random thermal KT noise will be low, which your system design may require. <S> For thinking purposes, I'd start with 1Kohm resistors, evaluating the heating and the random electron noise, then adjusting up or down from there.
| Then, calculate the Thevenin voltage.
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Drawing 2.0 A from 9 V battery for 5 seconds I'm building a launch controller for a model rocket, and I've worked out that I need to draw 2 A for no more than 5 seconds. Given most commercial 9 V batteries can give me ~500 mAh, would a 9 V alkaline smoke alarm battery give me what I need? <Q> An Estes rocket engine squib looks about like this: <S> (The above picture comes from a document I wrote on the topic.) <S> Estes provides very clear all-fire specifications for their squibs: <S> \$\frac12\:\text{J}\$ within \$50\:\text{ms}\$ into a resistance that is spec'd at \$\frac23\:\Omega\$ . <S> This implies an average of \$\sqrt{\frac{\left[\frac{\frac12\:\text{J}}{50\:\text{ms}}\right]}{\frac23\:\Omega}}\approx 3.9\:\text{A}\$ . <S> As a \$9\:\text{V}\$ <S> alkaline battery will have about 3 times that resistance, internal to the battery. <S> This means that the supply voltage must be \$V <S> \approx 10.4\:\text{V}\$ . <S> If you go review the charts at this site, Discharge tests of 9 Volt transistor radio style batteries , I think you'll find that even at just \$1\:\text{A}\$ , a fresh \$9\:\text{V}\$ battery might deliver about \$6.5\:\text{V}\$ for a short time. <S> (That chart on that web page is consistent with the figure of about \$2\:\Omega\$ for the internal battery resistance, fresh.) <S> If you were to attempt to draw four times that much, you can be pretty sure that it won't meet the all-fire specification. <S> Not even close. <S> If you do the calcs, you can perhaps consider adding a \$22\:\text{mF}\$ capacitor with a voltage rating of over \$10\:\text{V}\$ , placed in parallel to the \$9\:\text{V}\$ battery. <S> That may work. <S> But they do cost some money and they aren't small. <A> Back when I was into model rockets I had a commercially made launch controller. <S> That used 4 D cells in series. <S> These have a much higher capacity than the small 9 volt cells do. <S> Push the button and the igniter in the rocket motor would burst into flames the instant I pushed the button. <S> I have no idea how much current it actually drew. <S> The igniters were destroyed in the process so holding down the button even for several seconds didn't matter as they stopped drawing current during the launch. <S> If I remember correctly the ones I bought only needed 3 volts. <S> I can't check my launch controller, tossed it decades ago. <A> Its hard to say without knowing the details of the fuses you are trying to ignite. <S> Here are a few battery choices. <S> AA or AAA batteries will have lower output impedance than 9V batteries. <S> Taking the Duracell MN1604 as a typical example of an alkaline 9V battery . <S> The internal resistance is about 2.3 ohms at full charge. <S> At 2A discharge the output voltage would be 9V - 2.3 ohms <S> * 2A = 4.4V. <S> https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2016/03/MN1604_US_CT1.pdf <S> Taking the Energizer NH22NBP as a typical example of a NiMh 9V battery . <S> The internal resistance is rated at 1 ohm. <S> This is considerably less than the alkaline version. <S> At 2A discharge the output voltage would be 8.4V - 1 ohms <S> * 2A = <S> 6.4V. <S> https://data.energizer.com/pdfs/nh22-175.pdf <S> Even better yet use a AA batteries. <S> Taking the Duracell MN1500 as a typical example of an Alkaline AA battery. <S> The cell voltage is 1.5V and the internal resistance is 0.071 ohms. <S> A stack of six alkaline AA batteries in series would have an output voltage of 6 * 1.5V = 9V, and an output impedance of 6 <S> * 0.071 ohms <S> = 0.426 ohms. <S> With six batteries in series and a 2A discharge rate the output voltage would be 9V - 2A <S> * 0.426 <S> ohms = <S> 8.148V. <S> https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2020/02/14085905/MN15US11191.pdf <S> Taking the Panasonic HHR150AA as a typical example of a rechargeable NiMh AA battery. <S> The cell voltage is 1.2V and the internal resistance is 0.02 ohms. <S> A stack of eight NiMh AA batteries in series would have an output voltage of 8 * 1.2V = 9.6V, and an output impedance of 8 <S> * 0.02 ohms = <S> 0.16 ohms. <S> With eight batteries in series and a 2A discharge rate the output voltage would be 9.6V - 2A * 0.16 ohms = <S> 9.28V. <S> https://b2b-api.panasonic.eu/file_stream/pids/fileversion/3515 <A> There are several "9V" batteries. <S> If you really want an alkaline primary battery, you could use a PP6 instead of the familiar PP3. <S> It has about twice the capacity, but uses the same clips.
| In general rechargeable NiMh batteries will have much lower output impedance than alkaline.
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How does oversampling improve the measurement precision? Regarding improving the precision of a measurement, I have heard that a measurement precision can be improved by filtering or oversampling. When I think of filtering I can make sense because filtering reduces outlier measurement points and noise amplitude so it is easy to see that it reduces the spread around the mean value of our measurement. Less fluctuation results as narrower standard deviation hence better precision. But I couldn't come to a similar qualitative understanding regarding oversampling resulting better precision. How could this be explained? <Q> If you had two identical recordings of a piece of music both with their own background noise (assumed to be non correlated) and you added numerically those two recordings together, the music amplitude would increase by 6 dB (assuming you got things exactly aligned) <S> but, because the noises are uncorrelated the resultant noise amplitude would be only 3 dB larger. <S> This is because uncorrelated noises add like this: - $$N_T <S> = \sqrt{N_1 <S> ^ <S> 2 <S> + <S> N_2^ <S> 2}$$ <S> That's a signal-to-noise improvement of 3 dB. <S> In effect, the scenario above is the same as sampling a noisy signal twice. <S> You get a 3 dB improvement in SNR. <S> If you sampled 4 times you'd get a 6 dB improvement in SNR. <S> If you sampled 16 times you'd get a 12 dB improvement in SNR. <S> Another name for this is dithering . <A> Oversampling provides more measuring points allowing averging over a higher number of samples to improve precision. <S> Elliot's answer is correct that oversampling helps in the presence of suitable noise <S> but I disagree with the word "only". <S> As a counter-example, assume a 1 LSB peak-peak sawtooth waveform, period = final sample rate, is added to the input signal. <S> Then a signal exceeding quantisation interval N by 0.25 LSBs will generate value N until the sawtooth reaches 0.75 LSB, and then generate N+1 for the final 25% of samples. <S> Average enough samples (N * 0.75) <S> + <S> (N+1 * 0.25) <S> and you will see N + 0.25 at he final sample rate. <S> (I have shipped product that does this, it works). <S> It should be clear that other distributions including noise can produce similar increase in resolution. <S> A further approach is to subtract the residual error after quantization, in a feedback loop, from the input signal. <S> This is often done with low-bit esp. <S> 1 bit quantizers and high oversampling factors. <S> With the right design of feedback loop it can arbitrarily reduce noise level within a specific bandwidth of interest, at the expense of noise outside that band (which is later filtered out during the decimation process). <S> This "noise shaping" is common in higher order sigma-delta converters. <A> This can be explained with the following analogy. <S> Suppose you are doing an experiment where you measure the time period of a pendulum using a stop watch. <S> Assume the watch has a resolution of 1s, implying that your individual measurements cannot be more accurate than 1s. <S> So, for instance, if you measure a time interval of 9s then, you will be sure that the real time period lies between 8s and 10s i.e., \$9 \pm 1s\$ . <S> To ensure that your measurement are correct, you repeat the experiment multiple times. <S> Suppose you make 10 measurements as follows: <S> 9s, 7s, 11s, 8s, 10s, 9s, 11s, 12s, 9s, 11s. <S> You would expect that the true value of the time period will be given by the mean of these measurements, which is 9.7s. <S> But should we round this number to 10s since the accuracy of our instrument is only 1s and fraction of a second cannot be measured? <S> It turns out actually that the uncertainty in the mean goes down as \$\frac{1}{\sqrt{N}}\$ with the number of measurements. <S> So, after averaging 10 measurements your uncertainty goes down to \$\frac{1}{\sqrt{10}}.1s = 0.3s\$ . <S> Thus, your measurement result should indeed be: \$9.7 \pm 0.3s\$ , implying better accuracy (or resolution) compared to the individual measurements. <S> Same principle holds for oversampling. <S> When you oversample a signal, you sample it at a very fast rate, much higher than the rate at which the signal is changing. <S> Since the signal is not changing much in the consecutive samples, oversampling is identical to taking multiple measurements of the signal at a particular time. <S> Thus an oversampling ratio (OSR) of 4 implies making 4 measurements for the signal value. <S> The following low pass filtering stage is nothing but an averaging stage, which takes the average of all the extra samples from the previous stage, thus average of 4 samples in case OSR is 4. <S> As explained above, for OSR of 4 the uncertainty (given by the input noise standard deviation or square root of the noise power input to the ADC) will go down by \$\frac{1}{\sqrt{4}} = 0.5\$ . <S> Thus, your SNR increases 4 times and you gain an additional bit of accuracy. <A> Oversampling only helps improve precision if your measurements are subject to randomly distributed, zero-mean noise. <S> I have heard (but can not provide a citation) that random noise may be intentionally introduced into a system so that oversampling can be done. <S> It's important to note that precision is not the same as accuracy. <S> You won't necessarily increase the accuracy of a measurement by oversampling. <S> Any systematic errors and uncertainty will remain.
| If the noise is limiting the effective precision of your measurements then averaging multiple samples will improve the precision of your results.
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Are the 3 resistors in parallel or in series? Are \$R_1\$ , \$R_3\$ , and \$R_2\$ in series? or are \$R_1\$ and \$R_2\$ in parallel? I'm new to all this. Here's the circuit: <Q> Well, \$\text{R}_3\$ , \$\text{R}_4\$ , \$\text{R}_5\$ and \$\text{R}_6\$ are in parallel and they are in series with \$\text{R}_1\$ and \$\text{R}_2\$ . <S> simulate this circuit – <S> Schematic created using CircuitLab <S> So the total resistance is: $$\text{R}_\text{total}=\text{R}_1+\text{R}_2+\frac{1}{\frac{1}{\text{R}_3}+\frac{1}{\text{R}_4}+\frac{1}{\text{R}_5}+\frac{1}{\text{R}_6}}$$ <A> Just for a definitional answer: <S> Two resistors are in series if only one terminal of resistor A is connected to a terminal of resistor B (and their common point isn't connected to anything else). <S> Just because it's worth saying: <S> it doesn't matter how you draw them , it only matters how you connect them . <S> As for solving these kinds of problems, you can always replace two parallel resistors with another resistor of value 1/ <S> ( 1/Ra + 1/Rb ), regardless of what other connection there might be. <S> (Series resistors, on the other hand, can only be replaced by a single resistor if nothing else is connected to their common point.) <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Resistors R3, R4, R5 and R6 are connected in parallel. <S> And this circuit is connected in series to R1 and R2. <A> Actually R3,R4,R5,R6 are in parallel combination as the voltage across the resistors is same and the combination and R1,R2 are in series as the current will be same through them
| Two resistors are in parallel if each terminal of resistor A is connected to the terminals of resistor B
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A conflicting result in the diode and capacitor circuit Consider following circuit: simulate this circuit – Schematic created using CircuitLab The question is determining output voltage.D1 is an ideal diode with zero threshold voltage. Certainly when input goes from zero to peak voltage the diode is off but the problem is after that time. Both on and off assumptions for the diode is valid when $$T/4\le t\le T/2$$ If we assume diode is off then $$V_D = 0 - V_{out} = -V_1\lt 0$$ because $$T/4\le t\le T/2 \implies 0\le V_1\le V_{peak}$$ On the other hand assuming diode is on leads to $$ i_D = -C_1V_1' \gt 0 \implies V_1'\lt 0$$ which is true because $$V_1' = V_p\omega \cos{\omega t} \ \ \ \ and \ \ \ T/4\le t\le T/2$$ What is my mistake here? According to my book during entire positive half cycle diode is off but I don't know what's the problem with on state. <Q> It's assumed that \$i_D=−C_1V_1^′\$ , meaning that the current flows from the cathode end to the anode end of the diode. <S> Putting it simply, the \$i_D\$ arrow in your drawing is flipped. <S> This assumption in itself is OK. <S> It means that the calculated current will have an opposite direction. <S> The problem starts though with the second part of your assumption <S> , i.e. \$i_D>0\$ : This opposes the assumption that you just made (negative \$i_D\$ ). <S> You cannot have both assumptions at the same time, because they just contradict each other. <S> The correct assumption in this case would be <S> $$i_D=−C_1V_1^′ < 0 <S> \rightarrow V_1^′>0$$ <S> OR if you flip the \$i_D\$ arrow, you would have <S> $$i_D=C_1V_1^′>0 \rightarrow V_1^′>0$$ <S> which is the same. <S> Considering the derivative of the voltage across the capacitor: $$ <S> V_1^′=V_p\omega \cos(\omega t)$$ <S> Selecting a few time points, a frequency of \$100Hz\$ <S> and \$C_1=1\mu <S> F\$ <S> (considering the original arrow direction) yields: $$i_D(t <S> =T/4)=-C_1\cdot V_p\omega <S> \cos(\omega t)=-1\mu <S> F\cdot <S> 1V\cdot <S> 2\pi 100Hz \cos(\frac{\pi}{2})=0A$$ $$i_D(t=3T/8)=444\mu A$$ <S> $$i_D(t= <S> T/2)=628\mu A$$ <S> This can be double checked via simulation Simulating the on-state assumption (diode with no forward voltage and negligible off-resistance - otherwise no current could flow): As you can see from the waveform, the current values match the calculated ones. <S> In this case, the current is indeed flowing from the cathode to the anode, meaning that the assumption is correct. <A> The diode on/off trial method may not always work if there is a capacitor in the circuit, it could conduct for part of each cycle. <S> But, for now, lets assume that it will work. <S> First, solve for steady-state only, don't worry about startup. <S> At steady-state, the diode must be off, else the capacitor would charge to infinity (there is no other path for the diode current). <S> Next, recognize that the cap is a high-pass filter. <S> Since there is no resistor (infinitely large), the cutoff frequency is infinitely low, it will pass everything but DC. <S> There is no current in the diode at steady-state (previously concluded), but the diode also won't let the output voltage go below zero. <S> High-pass filter, voltage never below zero <S> : conclusion: sine, with bottom of sine at zero. <S> If you must know what happens for the first cycle(s), you need to analyze in pieces. <S> For 0 ≤ t ≤ T/2, diode is off. <S> For T/2 ≤ t ≤ 3T/4, diode is on, cap charges to Vpeak. <S> For 3T/4 ≤ t, diode is off. <S> Here is the answer using the simulator: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I modified the diode parameters to make it close to perfect. <S> You can see what I did in this answer. <S> Simple circuit transfer function and output graph <A> When you assume the diode states than you need to ensure that there are no contradictions in the resulting circuit configuration. <S> By just looking at the I-V characteristics of diode it seems like the ON state is possible but the resulting circuit does not follow KVL. <S> For the current direction you assumed, the voltage drop across the loop is \$V_1 - V_c > 0\$ , since \$V_c < 0\$ . <S> There are no contradictions associated with the OFF state and hence it is the correct state. <S> EDIT <S> The circuit diagram on the left shows you the original circuit which is transformed to circuit on the right if diode is assumed to be on. <S> For diode to be ON, the current has to flow in the indicated direction. <S> This implies the capacitor plate connected to node <S> \$V_{out}\$ will be positively charged by the current. <S> Consequently, the voltage drop across the capacitor \$V_c\$ , as indicated below, will be positive (I have reversed the polarity of \$V_c\$ from from what you indicated). <S> Now, apply KVL in the A - B - \$V_{out}\$ - D loop: $$0 + V_s + <S> V_c + V_d = 0 <S> \implies <S> V_s <S> + <S> V_c = 0$$ .This is a contradiction since both <S> \$V_s\$ and <S> \$V_c\$ <S> are positive as explained before. <S> But even before solving for these equations, you can see that if you have the circuit on the right the current will flow from the voltage source towards the ground not the other way around, which you indicated. <S> Since diode does not allow such a current, it will be turned off. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> It's quite simple: the assumption the diode is in forward mode is incorrect. <S> Certainly when input goes from zero to peak voltage the diode is off Correct. <S> No current flows through the capacitor and diode. <S> Therefore \$V_{out}(t <S> ) = V_1(t)\$ for \$ 0 <S> \le t\le <S> T/4 \$ . <S> Both on and off assumptions for the diode is valid when $$T/4\le t\le T/2$$ <S> This statement is incorrect. <S> As we saw above, at \$t = <S> T/4\$ <S> applies: <S> \$V_{out}=V_1(T/4) = V_{peak} \$ . <S> Therefore \$V_D=0-V_{peak}\$ , so it is, or better, it stays reversed biased. <S> In other words, only "the off assumption for the diode" is true. <S> Note that $$ <S> i_D = -C\frac{d \Big(V_1(t)-Vout(t) <S> \Big)}{dt} = 0 <S> \neq <S> -C_1V_1'$$ <S> Note that when \$V_{out}(T/4)=0\$ were true, then the remainder of your elaboration is true (I inserted a part): <S> $$ i_D = -C\frac{d \Big(V_1(t)-Vout(t) <S> \Big)}{dt} = <S> -C_1V_1' \gt 0 \implies V_1'\lt 0$$ <S> which is true because <S> $$V_1' = <S> V_p\omega <S> \cos{\omega <S> t <S> } \ <S> \ <S> \ \ and \ <S> \ <S> \ <S> T/4\le t\le T/2$$ <A> Your problem is that is some inappropriate cases, you used V1 as the voltage across the cap. <S> In fact, isn't V1, it's V1-Vout <A> Vout will be a sinewave with its negative going troughs peaking at 0V and its positive going peaks peaking at 2*root2*V1 where V1 is the RMS value of the voltage source. <S> The +ve and -ve signs on the cap are the wrong way around.
| I think the problem here is your on-state assumption, because it contradicts itself See, according to the following statement: $$i_D=−C_1V_1^′>0$$
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How to generate clock in 7555 timer? I am trying to simulate ICM7555 timer on proteus schematics. However, after following the datasheet for Astable operation, I don't see any results on the analog analyzer. Can anyone tell me what I am doing wrong? PS - new to electrical engineering also, I think that the problem can be in the simulation of proteus since it was saying before that simulation model is missing. I downloaded this footprint recently from the internet. <Q> Reset pin, pin 4 should be tied to V+ <A> Pin 4 (reset) needs to be tied high, especially for a CMOS part. <S> Pin 5 should have a bypass cap to reduce noise on the internal reference. <A> The reason this doesn't work is because there is no simulator model. <S> You can see in the error <S> log that none of the pins exist. <S> You said you downloded this part recently from the internet. <S> This is fine, but when downloading parts from the internet, it doesn't include the SPICE model. <S> If you want this to work, you have 2 options. <S> You can either use one of the 555 models already built in to proteus, or you can edit this part so it will simulate. <S> The question is why did you download this model when Proteus has 555 timers built in? <S> The ICM7555 is pretty much idential to the LM555, <S> NE555 and so on. <S> They are the same part, but from different manufacturers, so there isn't any need to have the specific part number for simulation. <S> So, your 2 options are: Option <S> 1 - Use a different part that has the SPICE model for your simulations. <S> Option 2 - Edit <S> this current part to make it simulate. <S> For option 2, it is a bit long winded, but you will need to right-click the part, and go to properties. <S> Tick the box that says 'Attach hierarchy module' then click OK. <S> You then right-click again and select 'Goto Child Sheet'. <S> In here you can edit how the part works. <S> You then go to the left hand menu and select the 'Device Terminals'. <S> Now you simply place the inputs/outputs and attach them to the relevant pins and rename them to their names on the schematic part. <S> You can then exit the child sheet by right clicking anywhere and selecting 'Exit to Parent Sheet'. <S> As you can see, option 1 is much, much easier.
| So you just need to find a pre-existing 555 timer in the parts list and place it in there.
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What exactly does a resistor's tolerance rating mean? I thought that I had this pretty much figured out over 30 years ago during my initial schooling, but various answers posted to a support forum of a major semiconductor manufacturer(name withheld to protect the possibly innocent) by staff support engineers (and no countering posts) have caused me to second guess myself. There are other questions concerning resistor tolerance, but none even mention the problem I talk about below. EDIT: This is concerning the statements that the chip makers support engineers make concerning why there are differences between measurement runs. The phrasing used makes it sound like resistors change randomly within their tolerance rating. When nobody called them on this explanation(which was given over a year ago, which is why I did not post in the thread.), given by two different people, is when I came asking. The chip in question is a type of ADC. The excitement signal is give or take a 125Khz, 2Vpp sine wave with a max current around 30mA. I always thought a resistor's tolerance rating was the permissible difference between its actual value and its posted value. Meaning if we took a group of 100 ohm resistors they could vary as follows: A 5% resistor could have any value that was -5 or +5 ohms from 100. A 1% resistor could have any value that was -1 or +1 ohms from 100 And a .1% resistor could have any value that was -.1 or +.1 ohms from 100 But once it was manufactured, and discounting abuse its value would not change, due to the tolerance rating. I'm not talking about changes due to temperature, that is based on its TCR. E Nor am I talking about aging, these are measurements taken within an hour(or less) of each other. What caused this question is that more than once in answering a question they stated that the various different values between measurement runs could be due to a resistors "tolerance". As if a resistor's value changes randomly within its tolerance rating, something I have never heard before. If they had stated it might be due to resistance changes due to temperature, I would have thought nothing of it. Or if the poster had been someone other than a tagged support engineer. And there were no posts questioning these conclusions, which messed with my head even more. Maybe I slept through that part of class and it's such a common thing no one bothers to talk about it, but I'm pretty sure that a resistor's value does not change if there are no outside forces(e.g. temp) influencing it. Am I correct, or have I some how managed to misunderstand how a resistor works for a LONG time? Or does tolerance have a meaning I somehow just have not encountered before? I do have medical problems, but I'm fairly sure that they have not degraded my thought processes that much(if they have, I need to stop talking). <Q> A decent manufacturer will specify pretty clearly in the datasheet what is meant. <S> For example, here's the relevant table from one vendor I've used: <S> As you expected, the resistor tolerance is the limit on the resistance at 25 C, and the variation over temperature is covered by a separate TCR specification. <S> But remember the resistance can also change due to other environmental factors, such as prolonged operation at high or low temperature, operation at high voltage, mechanical stress, etc. <S> If any of these stresses is applied to your device you could see its value vary from day to day, with the value at 25 C remaining within the specified tolerance limits. <A> ... <S> the various different values between measurement runs could be due to a resistors "tolerance". <S> As if a resistors value changes randomly within it's tolerance rating ... <S> The resistance of a resistor should not change within minutes (unless you strained it with a lot of of heat or something). <S> They are probably referring to different effects (which I am not even sure can create that fast variations). <S> That is, if you measure the resistor a few times in the same day all the readings should give you very close readings, and the only differences will be due to equipment precision, calibration or temperature variation. <A> First of all it's necessary to understand the need for multiple tolerance bands for a component. <S> For the supplier (component manufacturer), it would be a means to reduce inventory by categorising and differential pricing. <S> The supplier would aim for a nominal value and would need to get out-of-range components (caused by process variations) within range, by trimming. <S> Likewise trimming process variations would necessitate checking & sorting into various tolerance groups. <S> For the user (product manufacturer), it would enable choice of narrow- or wide-tolerance parts based on the criticality or otherwise of the nominal value to the application. <S> It goes without saying that narrow-tolerance components would be priced much higher than wide-tolerance ones. <S> Out of spec components/products could be the result of machine-setting not centred on the tolerance band and repeatability issues in the manufacturing as well as test & measurement processes.
| Tolerance does mean the variation from the specified value, the one in the product description, and the actual one, which you can measure.
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How to detect distilled water? I'm trying to create a circuit that detects water, however, I'm looking for a very accurate circuit that can even detect distilled water. My current circuit uses an Arduino that has a pin constantly set to HIGH , it triggers an event when water touches two wires that shorts the circuit to ground, creating a LOW signal. Unfortunately, this only works for non-distilled water. I tested this circuit with water from my kitchen tap and it failed, which is surprising because my city uses well water and I didn't think it could be pure enough to not conduct electricity. Is there a technique or design I can use to detect water, regardless of its purity? <Q> I like Russell's suggestion of optical methods. <S> It's fairly independent of water purity. <S> Couple of drawbacks : permittivity is temperature dependent (88 at 0C, 80 at 20C, 55 at 100C) <S> so precision (e.g. using it to estimate depth, or the proportion of the plates submerged in water) would need some knowledge of temperature to compensate the error. <S> Cannot easily distinguish water from other high permittivity liquids, although alcohols and hydrocarbons appear to have about half the permittivity of water <S> A bit more complex than a simple resistive measurement. <A> Here is the type of sensor that @RussellMcMahon mentions (photo from here ) <S> Other methods include capacitance, heat loss, and a simple mechanical float switch (usually a floating magnet operates a sealed reed capsule). <S> When the tablet dissolves, the contacts close. <A> Your statement when water touches two wires that shorts the circuit to ground suggests that you have a misunderstanding of how that works. <S> A "short" would absolutely pull the pin to ground, but what you have there is a low resistance, not a short. <S> In addition, the Arduino, with HIGH and LOW signals suggests you are reading a digital input, so you only see that pin going low when presented with low enough resistances. <S> Your tap water isn't presenting a low enough resistance to trigger the circuit. <S> You need to do an analog measurement so you can get a "number" rather than a yes/no result. <S> Now, when you go down that path, you will discover a number of other challenges, starting with the electrolysis and corrosion effects of running a current through your sensor. <S> Other answers can address this and other methods.
| Another approach is a capacitance meter : water has relative permittivity of about 80, therefore measuring the capacitance between two insulated plates a fixed distance apart will show if there is some high permittivity liquid between them. There is also the ancient one-shot flood detection method of putting an aspirin tablet in a clothespin.
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Why does AC input have a parallel resistor and capacitor? Well, I am trying figure out a situation with images and a circuit diagram. I am checking all part looks normal, but if I insert this board on a machine the resistor is exploding and one leg of resistor is breaking off. Actually I don’t understand why a resistor and capacitor was inserted parallel to the input of the AC power supply. Additionally, the resistor is exploding, but the fuse is solid yet. simulate this circuit – Schematic created using CircuitLab New schematic as a JPEG image: For the circuit, I didn't find a varistor symbol, so I use an adjustable resistance symbol in circuit instead Resistance: 470 kilohm <Q> The HTCC capacitor is used for filtering and interference suppression of high frequencies as well as the L1/L2 is used for common mode rejection. <S> The resistor is used to discharge a possible load at the HTCC when disconnecting the power supply from main. <S> This HTCC capacitor is also a self-healing type, also typically marked with X2 or Y2 types. <S> They can withstand pulses up to 5 kV.In <S> the schematics the L1 and L2 should be shown as inductively coupled for common mode rejection <S> , they must be on the same magnetic core. <A> <A> The capacitor directly across L-N is an 'X' capacitor and is used to reduce RF conducted emissions. <S> Without R1 this capacitor could hold a high voltage so that touching the input terminals may give an electric shock. <S> R1 is chosen to be small enough to discharge the capacitor in reasonable time but large enough to keep power dissipation down to an acceptable level. <S> This design has a 470k resistor for this a 470R resistor would dissipate excessive power and probably explode.
| The 470k resistance is there to discharge any stored charges in the filter capacitors so that the device or disconnected mains plug does not give a surprising shock if touched.
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Capacitance and voltage rating Lets say I have 2 capacitors, one rated 100 µF, 10 V and the other 100 µF, 300 V. Now I charge them both with 5 V for 1 min (or until they reach 5 V, fully charged) and connect them to a same load, for example an LED one at a time. Will they both light that led up for same amount of time? Please, no tank examples... <Q> Presuming that the capacitors are electrolytic, are in good working condition and their capacitance values are identical (since electrolytic capacitors have a tolerance range of -20% to +80%) they will test identically. <S> The working voltage of an electrolytic capacitor is the maximum voltage which should never be exceeded. <S> At the same time, using a 300V electrolytic capacitor in a 10V circuit is not a good idea. <S> In an electrolytic capacitor, the dielectric is an insulating oxide layer formed on its aluminium foil anode by electrolytic action when a positive 'forming' voltage is applied to it. <S> The working voltage is decided by the thickness of the oxide layer. <S> During an electrolytic capacitors lifetime, the oxide layer would be maintained only if the working voltage is between 25% and 90% of the forming voltage. <S> If not, the dielectric thickness would reduce leading to capacitor failure. <A> Will they both light that led up for same amount of time? <S> The capacitors voltage rating is just telling you not to exceed that voltage; it doesn't tell you anything about how it functions or stores energy i.e. its a stress rating value. <S> Please, no tank examples... <S> Sorry, I couldn't help myself. <S> This one is called Deborah. <A> It depends on the dielectric material. <S> Some dielectric materials have very non-linear characteristics meaning that \$C(V)\$ instead of \$C\$ being constant. <S> These non-linear characteristics are used by some snubber circuits to reduce losses. <A> Both capacitors have the same capacitance C, I suppose these capacitors are standard electrolytic ones. <S> The capacity C of each component is nearly constant within the voltage range. <S> The voltage rating printed on the capacitor is the maximum voltage you may charge with. <S> The electric charge Q of each capacitor is (after full charging) <S> Q = <S> C <S> * U that is for each capacitor: Q = <S> 100 µF * 5 V = 0.5 mAs, or 0.5 mC (milli-Coulomb).When connecting <S> a (or both) capacitor(s) <S> to a load, be careful to limit the current e.g. with a resistor. <S> Because when you directly connect the 5 Volts to a diode (LED) <S> a high current will flow for a short time at 5V through the diode, limited only due to the diode characteristic curve and due to internal resistors of the capacitor, leads and transition. <S> This high current could destroy your LED. <S> The voltage (and the current) will both go exponentially towards zero.
| They might destroy the LED due to the sudden inrush of current into an unprotected LED but, assuming the LED survives this (or is protected against it) then both capacitors will illuminate the LED equally for a short period of time.
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Does power delivery over ethernet always have to use isolated DC/DC converters? I am examining the possibilities for building a cheap ethernet device that would be powered over ethernet using PoE technology. I came across this article which shows the generic block diagrams: Looking around, I always find the DC-DC converters used in PoE applications to be isolated. I am trying to understand whether that is always necessary. Assuming that the PoE RJ45 cable would be the only cable connected to my device, is there really any reason for me to isolate from the input voltage? This is also assuming that the data lines are already isolated, so the only potentials without isolation is the power lines. Edit: To clarify more: I still want the TX and RX lines to stay isolated via transformers. I just want to skip the DC-DC part isolation (just graetz bridge, bulk capacitor and a DC-DC converter without isolation.) I don't mind being not compliant with the standard here. As for the shielding, I would leave it disconnected. <Q> If you want to call it PoE, you need to stick to <S> IEEE 802.3af-2003 or IEEE 802.3at-2009. <S> These standards call for isolation. <S> This answer could end here: <S> This is also assuming that the data lines are already isolated, so the only potentials without isolation is the power lines. <S> You also will have a shield in your ethernet cable. <S> On which end are you going to isolate that? <S> Both? <S> Not much of a sensible shield then. <S> Also, when doing ethernet, you need the magnetics (==transformers), anyway, so get one with center taps on the cable side – no extra effort, basically every ethernet transformer does, anyway. <S> So, without a DC/DC converter that isolates the regulated voltage from the voltage on the line, you'd tie your device's data logic to a fluctuating, noisy voltage that's not guaranteed to be at any fixed relation to the data signals. <S> So, really, don't. <A> I don't have the spec in front of me <S> but I've done a number of PoE designs. <S> As I recall the PoE isolation spec is to meet safety requirements and avoid ground loops. <S> If your device takes the 48 Vdc and uses it within a fully enclosed, non-conductive enclosure, with no other cables, then the 1500 Vac power isolation (typically done with a flyback circuit) isn't needed. <S> As noted, the RX and TX data pairs will need their own PoE rated pulse transformer since the ethernet transceiver also expects it. <S> Don't forget the PoE power can come in either on the TX / RX pairs or on the other two spare pairs, so to be compliant <S> you'll need two bridges. <A> Not mentioned elsewhere it is even quite possible that two different devices are powered over longer cables in the same building and then could share grounding through some common signalling cables. <S> The use of small diameter wires for POE and 48V drive implies that there will be voltage drops across the cables. <S> Such drop is proportional to the current draw of the target device and dissimilar devices would see voltage drop differences that are rather different. <S> The only way that you can mitigate these voltage drop differences is by using isolated DC->DC converters.
| no, you cannot build a PoE device without isolation.
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For SR timing diagrams, does a bubble on the output of Q' indicate that it is the same as Q? I understand that with flip flops when there is a bubble on the clock it means the clock cycle is in terms of the falling edge, but when there is a bubble on the output Q' isn't this inverting Q twice, bringing it back to just Q? As in, "not Q' = Q"? I don't understand why this would be beneficial because now won't both output timing diagrams be the exact same? <Q> A bubble on an input or output means 'inverted'. <S> That is why an inverter is shown as a buffer symbol (no logic state change) with a bubble on its output (then inverted). <S> The '>' on an input means that is it positive-edge triggered. <S> It will react only to a LOW-to-HIGH transition. <S> An input with a bubble followed by '>' is negative-edge triggered. <S> The logic gate inverts the input then looks for a positive edge transition. <S> Therefore the gate input will react only to a HIGH-to-LOW transition. <S> To your example... <S> You have a negative-edge triggered SR flip-flop with complementary outputs. <S> Your Q output will obey the Set/Reset (SR) function of the gate. <S> Your Q' output will always be driven with the inverse of the Q output. <S> Its name is has an apostrophe to denote that it isn't Q. <S> The symbol shows that it is inverted. <S> The name and symbol are indicating the single inversion in different places. <A> Note that many EDA tools have gotten away from using the "bubble" to denote an active low input or output on a schematic symbol, and have gone with an ANSI/IEEE standard that uses a skewed triangle. <S> Here is an example of LVDS differential driver and receiver symbols: <A> The 'bubble' is the symbol. <S> The Q' is the pin designation.
| Q' is the inverse of Q .
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Why does current only choose path of least resistance sometimes Say we have the following setup simulate this circuit – Schematic created using CircuitLab Input is pulled high and current flows through it as long as the switch is open. As soon as it closes current switches to the path of least resistance and goes through the switch to ground. Imagine a similar setup with two paths with different resistances simulate this circuit In this case current will flow through both paths, and will not simply go through the path of least resistance. <Q> 'Path of Least Resistance' is a phrase that only really applies where you have alternative paths for, for instance, a walker who chooses to go through a gap in the wall next to the gate, rather than open the gate. <S> Wikipedia says this: In physics, the "path of least resistance" is a heuristic from folk physics that can sometimes, in very simple situations, describe approximately what happens. <S> In electronics, the current always divides between several paths in the inverse ratio of their resistances (or impedances if it's AC). <S> In your first example, the switch when closed has orders of magnitude less resistance than the input. <S> The current division ratio will be so close to 100% to the switch, that we engineers approximate it to 'all through the switch'. <S> In your second example, it's a reasonable ratio. <S> Note that the input current doesn't 'divide' as such, looking at all the paths and deciding how to split. <S> The node where the paths split from has a specific voltage V. Current flows through each path I= <S> V/R, V is the same for each path, hence the current is in the inverse ratio of resistances. <S> The input current to that node is then the sum of all the output currents. <S> Looking at the comments about water channels, and thinking about the origins of the phrase, it's also often used for lightning. <S> Who hasn't been told, when perhaps visiting a tower with a 'lightning conductor' fixed to the outside, that it takes the path of least resistance. <S> Just like water eroding its own channel, ionisation in the air creates its own low resistance channel. <S> This happens to such a degree that lightning may well not strike the most obvious high point in the local area, but proceed by ionisation steps to create an alternative low resistance path to ground through which the main strike then occurs. <A> Why does current only choose path of least resistance sometimes <S> It doesn't, it travels all paths inversely proportionally to the resistance of the path. <S> Input is pulled high and current flows through it as long as the switch is open. <S> As soon as it closes current switches to the path of least resistance and goes through the switch to ground. <S> In your first example, when the switch is open: Current will try to flow from 3.3V, through R1, then through both the switch and the input. <S> Yes, I know, sounds odd. <S> But the thing is, although the switch is open, it is still a resistance, just an infinitely high one, so the current will be infinitely low (zero in the ideal case, but not necessarily zero in the real world). <S> Depending on what the input is, noticable current might flow through that, or if that is open-circuit, an infinitely small or zero current will flow. <S> When the switch closes: Its resistance drops dramatically to a very low value. <S> Now more current can flow from 3.3V through R1 through the switch to GND. <S> A noticable amount now - around 0.33mA. <S> Again depending on what the input is, current will still try to flow through that. <S> Whether it is infinitely small, or some measureable value, who knows. <A> Meet conductance Conductance is a unit you never think of, even though you almost work with it all the time. <S> Conductance is 1 / resistance. <S> The unit is the siemens . <S> Yeah, like the service equipment. <S> (Some people call it the "mho", but I think those people work for Square D :) <S> You have a 10 ohm resistor. <S> What is its conductance? <S> 1/10 siemens. <S> "Path of least resistance" is wrong. <S> It's used by business managers and motivational speakers because it's nicer to say than "I don't have to outrun the bear". <S> The actual rule is <S> Current flows on all paths simultaneously, in proportion to their conductivity Pardon the graphic. <S> The schematic editor won't support this. <S> Current splits 10:1 down those two paths. <S> In parallel paths, conductance adds . <S> So the total conductance here is 0.11 S. Easy peasy, when you think in conductance! <A> In the first case, when you close the switch you are connecting vin directly to ground. <S> If you had Vin coming through some other circuit, you would notice a current coming through that element too . <S> In your circuit when the switch is open , depending on Vin the current might flow through the resistor ( Say if Vin is 3.3v and the switch is open no current is going to flow ). <S> If Resistance is small current is going to be more through it <S> ( which is the case of the second schematic).
| Current will flow through all resistors which have a potential difference across them.
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Why does adding an emitter resistor to a common emitter amp reduce its gain? According to the Wikipedia page for common emitter amplifiers, the addition of an emitter resistor decreased the gain of the circuit. However, to me it appears that if the resistor was omitted, then V_out would be approximately equal to the ground mode (when the transistor is conducting). But when the resistor is present, V_out would be equal to the drop across the resistor. Why then does the resistor reduce the gain? Please note I am very new to transistor circuits, and to electronics in general. <Q> Gain is the output voltage change divided by the input voltage change. <S> In the small signal hybrid pi model <S> the collector current changes proportional to the change in base voltage (relative to the emitter) <S> so there is a voltage change at the collector proportional to the load resistance (parallel with ro). <S> That is because the emitter current change causes the emitter voltage to change in the same direction as the change in base voltage, partially counteracting the effect. <A> It's useful to think of a transistor operating in three modes, cut off, linear, and saturated. <S> When cut off, the collector current is zero. <S> When saturated, the base current is high, and VCE is small. <S> These are the two modes when using a transistor as a switch. <S> You appear to be thinking of the saturated mode in your question. <S> Wikipedia is talking about the linear, or amplifying mode, where the collector current is beta times the base current, and VCE is a few volts. <A> In a common-emitter amplifier, think of the transistor as a variable resistor controlled by the base current ( \$i_b\$ ): simulate this circuit – <S> Schematic created using CircuitLab <S> The output voltage (and thus the voltage gain) is determined by the voltage across the variable resistor (i.e. \$V_{ce}\$ ). <S> Now put a resistor between the low-end of the variable resistor and the GND: <S> simulate this circuit For the same VCC <S> , the voltage across the variable resistor drops due to the increased total resistance. <S> So the output voltage (determined by the voltage across the variable resistor) drops and so does the gain. <A> The emitter resistor causes negative feedback . <S> This can be shown using a block diagram based on the classical formulas for describing the input-output relations. <S> It is a well-known fact that negative feedback reduces the gain factor. <S> This is - in most cases (opamp) - a desired result of negative feedback. <S> Note that the output current Ic is determined by the input voltage Vbe via the transconductance gm= <S> d(Ic)/d(Vbe) . <S> (In the block diagram I have used the commonly used simplification <S> Ic= <S> Ie; all voltages and currents are small-signal quantities and not DC values). <S> The gain formula shows how the resistor RE reduces the gain. <S> More than that, you can see that the simplification Vout <S> /Vin=RC/RE (as mentioned in one answer) is applicable for RE>>1/gm only. <S> Comment 1 : <S> Because the signal output voltage across the collector resistor RC is referenced to common ground - and NOT to the supply voltage - there is a sign inversion between input and output voltages which is not contained in the block diagram (the block RC should habe a negative sign). <S> Comment 2: <S> It is easy to explain in words how feedback reduces gain. <S> An increase of the input voltage Vb (at the base node) causes an increase in Ie which in turn also increases the voltage across RE. <S> Hence, the emitter voltage Ve goes up and reduces Vbe (if compared with Ve=fixed at ground for RE=0). <S> Hence, the Vbe increase is smaller than the input signal (Vb) <S> increase - and the Ic increase is remarkably smaller if compared with the case RE=0. <S> (I like to add, that the block diagram and the corresponding explanation of the feedback effect proove again that the BJT is a voltage-controlled device Ic=f[exp(Vbe/Vt)] .) <A> Consider the following circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In this circuit Q1, R1, R2, RC, RE, Cin and Cout represent real components. <S> re represents the intrinsic impedance of Q1. <S> \$ re <S> \approx \dfrac{25}{I_C <S> } \$ <S> ohms if \$ I_C \$ is in mA at room temperature. <S> The usual design procedure is to pick a quiescent \$ I_C \$ and choose RC such that <S> \$ V_C \$ is half Vdc at this current to maximise voltage swing at the output. <S> Now if we assume \$ h_{fe} \$ is large <S> Then \$ I_C \approx <S> I_E \$ <S> So the gain is \$ <S> \dfrac{RC}{RE + re} <S> \approx <S> \dfrac{RC}{RE} \$ <S> if re is small relative RE. <S> I usually play with RC and RE to get nice standard values. <S> Since we know the quiescent \$ I_E \$ <S> we know <S> \$ <S> V_E \$ <S> and we want <S> \$ V_B \$ <S> to be approximately 0.6V higher than this. <S> knowing <S> \$ I_B = <S> \dfrac{I_C}{h_{fe}} <S> \$ <S> we can choose R1 and R2, The current in R1 should be large compared with <S> \$ I_B \$ <S> so it does not vary. <S> I usually go with ten times \$ I_B <S> \$ with maximum \$ h_{fe} \$ as a rule of thumb. <S> Finally we can choose Cin and Cout to have a low impedance at the minimum frequency of interest.
| If you add an emitter resistor, the base-emitter voltage does not change as much so the current change is less and therefore the gain is reduced.
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Do SMD mount bases exist? I'll soon need to test and work with SMD components, but these components are very small, about 2.5 mm x 2.5 mm (DFN package for example). I had a simple question that I can't answer despite my searches. Do bases for placing SMD components on exist? What I mean by that, is a component (SMD or Through-Hole) easier to solder on a PCB where I could place and remove a SMD component. If yes, what is the name of such a component? <Q> You still have to solder the SMD part to the board once. <S> The Chinese ones work fine. <S> Photo from Aliexpress. <S> This is a great way to play with new parts, test tiny things like 0402 LEDs and so on. <S> If you want to completely avoid soldering to the SMT part, there are sockets however they are huge and expensive and generally designed for applications such as programming SOT-23 MCUs. <S> Not really a viable option. <S> Another option is to buy an evaluation board, if one is available from the chip manufacturer, or a 3rd party module such as those from Adafruit. <S> This may be the most practical approach to deal with a BGA package in the early prototype stage. <S> And sometimes the part you want to use is available in a more "friendly" package as an option, so <S> you can use that and change to the unfriendly type for production. <A> Even if you may find compatible boards as shown in Spehro Pefhany's answer, it's very difficult to solder DFN or QFN packages manually. <S> Even on an assembly line it's not easy. <S> You need to apply solder paste in very tiny but exact amounts with very high precision. <S> Then heat with a hot air solder gun. <S> Succeeding or not is a question of luck. <S> The biggest problem is that you can't check the connections as the pins are hidden beneath the ic and not accessible to your multimeter probe. <S> So debugging your circuit will be painfull as you will never know whether your schematic is wrong or the ic is not properly connected. <S> If possible use a package with visible pins (with the largest spread between pins available). <S> If not possible, try to find a demo board (Development Board or Kit) for this ic. <S> Sometimes they are expensive but will save you a lot of time. <A> The answer is, for some SMD packages, yes. <S> The component you are looking for is called a socket. <S> Digikey, for instance has an enormous number of sockets available, including those for BGA, PLCC and QFN packages. <S> Many of these are designed to allow through-hole soldering to the board. <A> In Canada there's a company that'll blow your socks off. <S> Proto Advantage (proto-advantage.com/store/index.php) <S> They will even order and mount an ic from digikey on the adapter of your choice. <S> Mind you <S> it is costly <S> but if you absolutely need to breadboard they are THE source! <A> Search for surfboards at digikey.com <S> Part <S> number MK-9000CA-ND <S> Also Aries makes adapters where you solder the smt device <S> and it becomes a dip package. <S> Search for Aries Adapters on digikey.com
| There are breakout boards that allow you to solder the SMD part and then use it in something like a solderless breadboard or a lash-up by soldering to the large holes or adding headers to plug it in. You can probably find ones through the normal distribution channels as well.
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What is input capacitance of pins in IC and what effect does it have? What is the purpose and the definition of the input capacitance parameter in the datasheet. I have tried to research about this topic but was only able to find the input capacitance parameter with respect to operational amplifiers only. Even after those research, I was not able to understand the concept of input capacitance with respect to this buffer IC. Can someone help me with my following queries : What is the definition of input capacitance of a pin in an IC? How is it measured in the first place and is it applicable to all pins in the IC? I am sure, different pins of the IC must have different circuits connected inside. So, the input capacitance of those pins might be different. Is the input capacitance parameter relevant only to supply/signal pins? Why should we care about this parameter of the IC while we design our circuit and how is it relevant? <Q> 1) <S> What is the definition of input capacitance of a pin in an IC? <S> It's the capacitance from that pin to ground or 0 volts. <S> 2) <S> How is it measured in the first place and is it applicable to all Pins in the IC? <S> It's probably not measured on production items; rather it is essentially always that value unless there is a significant production anomaly on the die. <S> It's applicable to signal inputs. <S> 3) <S> And Why should we care about this parameter of the IC while we design our circuit and how is relevant? <S> It's a very relevant parameter given that a signal with significant source resistance will have difficulty keeping its rise and <S> fall times short/fast with too much input capacitance. <A> This means, if you apply a logic HIGH to that input pin your source has to supply sufficient charge to charge up the capacitance to the desired voltage level. <S> It might be measured directly with something like an LCR meter at the produced item or it might even be just a calculation from the design layout of the chip. <S> Depending on your source impedance it takes time to charge up the input to a specific voltage level. <S> If you know the source impedance as well as the input capacitance you can calculate the exact behaviour of transient events. <A> The AC drive capability and switching speed of most digital ICs is specified with a specific DC and AC load. <S> So if you want to use this part at its rated speed, you need to know the total capacitance that its driving - interconnects and components. <S> The input capacitance spec of a part helps you with that analysis. <S> Here's an example from the datasheet for an AC/ACT240. <S> Note that the AC characteristics are specified with a load capacitance Cl of 50 pF. <A> In simple terms: $$ I = <S> C \frac{\Delta <S> V}{\Delta t}.$$ <S> So if you need to change the voltage \$(\Delta <S> V)\$ within a certain time <S> \$(\Delta <S> t)\$ <S> you can calculate the current \$(I)\$ required. <A> Input signal pins capacitance will include: 1) leadframe capacitance between the metal piecees of the package 2) input capacitance of the bipolar and/or CMOS transistor; for low-noise-amplifiers, this may be picofards. <S> 3) ESD structures designed to rapidly ( < 1 nanosecond) turn on, for voltages only moderately larger than the VDD; <S> ESD structures can be the major shunting path for high frequency energy, which ruins DATAEYES. <S> I know of one guy who killed his wife because of the stresses of collapsing revenues, when the ESD diodes were ruining the bit-error-rate of a data-recovery IC, and the historic IC buyers were rejecting the new generation ICs. <S> summary --- I assume 3pf for input capacitance, whether or not is LNA now <S> ---- what is the VDD capacitance? <S> On one IC I designed, I plunked down 10pf caps on each end of a place-and-route logic block, in addition to all the Pactive-to-Nsub and the gate-to-bulk parasitics---thus the total >> 200pf.
| The input capacitance is the capacitance a signal source sees, when you connect the source to that IC pin.
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What happens to a high power inductive motor if you decrease the RPM to half the rated in terms of current, torque and heating? and what is minimum possible RPM ratio for a high power motor <Q> Induction motors are available with power ratings ranging from about 1/500 Hp (1.5 W) to 30,000 Hp (22MW). <S> It is difficult to say just where in that range the motor in question may be. <S> It is likely that the question is about a low to medium power motor rather than a motor that is more than a meter in diameter. <S> The motor current will not exceed the rated full-load current. <S> The VFD input current will be proportional to the mechanical power output of the motor, so it will decline as the motor speed declines. <S> The mechanical power output of the motor is proportional to load torque multiplied by speed. <S> There will be some harmonic current content in the VFD input current. <S> That will increase the current somewhat. <S> The heat lost in the motor will decline a little bit as speed declines, but the ability of the motor to dissipate losses will decline due to the reduction of the speed of the motor's self-cooling fan. <S> That will limit the time that it is safe to operate the motor at rated torque and any given speed. <S> At half speed, most motors will probably be able to operate at rated torque continuously. <S> If the motor is cooled by a separate blower that is selected and furnished as part of the motor by the manufacturer, it is possible to get a motor that operates continuously at or near zero speed. <A> An induction motor, controlled by a variable frequency drive, may be run at 50 % of rated speed applying 50% of the rated voltage at 50% of the rated frequency to obtain 50% of the rated HP and 50% of the rated torque. <S> Up to the rated speed V/f would be constant. <S> Above rated speed V would be held at rated voltage with only the frequency being increased for higher speeds. <S> The torque would naturally reduce. <S> Up to the rated speed the V/f ratio is intended to care of motor cooling. <S> However it is not advisable to go below 20% of rated speed without external cooling. <A> In loose terms for all motors $$ speed \propto voltage $$ <S> $$ torque \propto current.$$ <S> Thus: if you half the speed you must half the voltage; <S> you can ensure the same current by ensuring the same torque. <S> The \$power = voltage \cdot <S> current\$ . <S> Watch out for \$I^2R\$ losses though. <S> The minimum is a difficult question. <S> Testing might be the answer here. <S> Overspeeding can destroy the squirrel-cage. <A> The question is difficult to answer. <S> Holding frequency constant you can only change speed by changing the poles (nearly impossible) or changing the slip due to overload. <S> Using an inverter you can decrease the speed with nearly no effects to current and torque. <S> But reducing motor speed also reduces the fan speed, so machine could overheat. <S> Calculating torque depends on different parameters and is hard to do. <S> Simplyfied you can say it depends on current and current depends on load.
| If the motor is controlled by a reasonably good variable frequency drive (VFD) it can be operated down to practically zero speed at 100% rated torque.
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Finding the resistance of a resistor without multimeter So this is a trivial but frustrating issue I’ve come across. I have two 5 band resistors and I am having a hard time differentiating the tolerance band. I don’t have a multimeter on me and I don’t know if these are 1k ( Brown black black brown brown) or 110ohms ( Brown brown black black brown). Am I being completely oblivious to something? Edit: I do know that the tolerance band is supposed to be at the end with the largest band spacing, but it looks equidistant to me. <Q> Use 9 V block and check if the resistor becomes hot. <S> P = <S> U 2 <S> /R. <S> So your 110 ohms resistor should burn close to 1 watt. <S> (Voltage of new battery will be > 9 V.) <A> Make a bridge. <S> Feed it with low level audio signal and use a speaker as the detector. <S> Headphones would be better than a speaker due higher sensitivity. <S> If R1/R2 = <S> X/R4 the sound vanishes. <S> Then X = R4 <S> * (R1/R2) <S> In Wheatstone's bridge R1 and R2 are a potentiometer, R1/R2 is seen visually from the relative position of the slider. <S> Originally DC voltage and sensitive galvanometer were used instead of audio and speaker. <A> These are 1k resistors, both black stripes are closer together than the two brown stripes on the bottom. <S> The end with thicker and further apart stripes is the end with the multiplier stripe and the tolerance stripe. <S> This is the rule with 5-band resistors, but due to how hard it is to tell them apart it is best to mark every resistor set with its value and measure using a multimeter. <S> If you are calculating from the band colors, remember that resistor values most likely follow a value series, like E6 or E12. <S> 110 is not part of the E6, E12 or even E24 series, but 1k is. <A> You can build a circuit like simulate this circuit – Schematic created using CircuitLab <S> Where RQ and C1 build an RC circuit. <S> Take a watch and measure how long it takes for the LED to turn on. <S> Compare with other known resistors. <A> 5 band resistors have an extra (digit) band (3th) before the multiplier band (4th). <S> Although a bit hard to spot (I never tried it myself), as described in this question <S> you can start reading from the thicker band's side. <S> So in your case it should be $$110\ \Omega \ 1\%$$ <A> Start reading from the end of the resistor where the color bands are closer together. <S> In the picture, that appears to be brown-black-black-brown, brown. <S> Or 1-0-0-1, 1. <S> Which is 100 * 10^1, 1%. <S> 1000 Ohms, 1%. <S> When in doubt, get a meter. <S> That’s a very handy thing to have for kit building where it is difficult to distinguish colors or determine which are the thick and thin bands on resistors. <S> Yes, you can build yourself a Wheatstone bridge, but do you want to spend five minutes each time you check a resistor? <A> 1k resistor. <S> The broader band is tolerance, the spacing has nothing to do with it. <S> When my older eyes can't see, I use my phone camera as well to magnify. <S> That picture you took is perfectly clear. <S> That is the way to do it for over 40's! <A> Hence it is not hard to break the tie. <S> Do you have a small incandescent lamp (by which I mean "electronic component-sized" lamp)? <S> If so, by hooking it to, say, a 9 V battery in series with one or the other resistor will tell you right away, as the lamp will light much brighter when the series resistor is the 110 Ω one, as compared to the 1000 Ω one (it might not be visible at all with the latter at least in normal lighting, but should be with the former.). <S> This is, I'd say, a better idea than trying to get it "hot", because you don't want to risk cooking one or the other resistor. <S> An incandescent lamp is a resistor that is designed to get hot :)
| Assuming the resistors are not bad in some way and the only two values you are concerned about are 110 vs 1000, the first thing to note is that there is a lot of difference - a full order of magnitude - there. For casual measurements, I have a digital meter I bought for under $10 at a hardware store.
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Covid-19: can a UV-C source be improvised from a standard mercury vapor lamp? I suggest it because the usual professional UV-C sources are in short supply and very costly. A common 175 W mercury lamp designed for illumination includes the quartz 'lamp' tube itself, inside an outer bulb designed to block UV-C (< 300 nm) radiation. If the outer bulb were removed, this kind of lamp would seem to be able to provide UV-C to destroy a virus. I don't have a UV-C capable spectrometer nor a power meter filtered for UV-C only, so I can't test/do the experiment myself. Maybe someone else has such gear. A document on this is here, and it includes references. I do not suggest that this be done except by those who understand the hazards of the experiment. I believe that it's better to throw an idea on the table in front of engineers and technicians, than to hide ideas that may be helpful or to post untried projects on hack sites and prepper forums where they may do more harm than good. Known hazards include: UV radiation, high voltage, high temperature, broken glass. If someone has the gear to try it, maybe it could help the situation. The document is here , somewat hidden by obscurity. ADDED 4/3/ 2020: I should have stated that my goal was primarily to disinfect masks for re-use. I didn't want to limit the scope, being more interested in spectrum and power per square area at given distance. I too, worked with an e-prom eraser, but from a repair perspective. The original UV tube was broken and unavailable. I put in two UV-C germicidal lamps. It worked perfectly and saved the customer money. Some masks can be treated by heat or chemicals, and some may contain materials that are ruined quickly or are not properly disinfected by those and other methods. The type and construction of mask will dictate the best way to disinfect it for maximum re-use. Treating every surface of an arbitrarily shaped object with sufficient UV intensity for sufficient time is a challenge. It's not impossible. Two UV lamps could be used, or a rotisserie arrangement, using an alligator clip to hold a mask, or any other arrangement, may be worthwhile. We can use our imaginations. The risk of transmission from handling objects is significant and supported by expert opinion of an experienced doctor, James Robb, MD FCAP , here . There was, and will not be, any suggestion whatsoever that anyone get some Covid-19 Coronavirus and irradiate it, then try to prove it’s worked. There are experts in the field who know how much UV-C intensity per area and irradiation time is required to inactivate the virus. There is such information for other viruses online. While there are no studies specifically on Covid-19, Ozone has been shown to kill the SARS Coronavirus. Ozone is unstable and wants to revert to O2 and lose the third Oxygen atom. If the third atom of Oxygen does not find another Oxygen atom to combine with, it can break the virus so that it can't function. While Ozone can damage some materials, it does disinfect them. There are several references, to scholarly work, in online articles to support the statement made in the comment. Link1 to references and Link2 to references At least one paper states that atomic Oxygen can recombine in a time scale of a few milliseconds. Our senses and the above references tell us that it can be produced in copious enough amounts to be effective in spite of this. Solutions don't have to be clever, they only have to be workable. A Dallas, Texas mattress-cleaning company is already using its production line to sterilize masks for hospitals. It seems obvious but it's extending the lifetime of disposable masks from 1 day to > a week. What the UV (and Ozone comment) information may mean for the public, who may soon be told to wear masks or face coverings outdoors, that that a mask could be safely used for a very long time. Ozone generation is another topic. I clarify that I am not posting here for any pecuniary purpose. I have only one old set of PPE . Unfortunately it is only effective on October 31. A bit of levity keeps me motivated. <Q> Yes, but only for killing viruses in the near vicinity. <S> Not to use it when live people, animals are nearby. <S> Also it would accelerate ageing of colours, plastics, wood,...Not a very clever solution, but if desperately needed could be a way to go. <A> You mentioned the cost. <S> The UV filters for residential well-water systems are reasonable. <S> I assume they'd work for viruses too. <S> As said above, it is, of course, dangerous to living tissue, so modifications and use would have to be done properly. <A> If i get the question correctly:Those lamps have the proper wavelength according the Wikipedia : <S> And for covid19 from 250 to 285 nm (UVC ) could be fine. <S> also You needs to have minimum energy which could be like this: <S> Headgear-mers virus is one of RNA viruses and are categorized together should dose between 15–400 Ws/m2 . <S> Examples of well-known RNA virus doses are 110 Ws poliovirus/m2 , Newcastle disease 15 Ws/m2 SARS 226 <S> Ws/m2 . <S> The assessment measured in the same region as high SARS can calculate real-time inactivity like to achieve 90% [3, 9]. <S> The LED UVC have more life time like this : <S> Also here <S> you can see about UVC bulbs: <S> For example, mercury lamps have been used to disinfect surfaces and liquids for decades, and the bulbs are only about $100. <S> However, they are 25,000 times less intense than a Xenon bulb and <S> the disinfection process can take hours, making them impractical for hospital use. <S> LEDs could also provide cheaper UV light, but they are also far less intense than Xenon bulbs, according to Hart [5, 11] (Fig. 8). <S> Also consider there Smart Dosage could be fine which is described here : <S> Built Patent SmartDosage, together with the patented box Balance and <S> PowerBoost technology allow iris 3200 m UV lighting system of disinfection for automatically measuring conditions the room environment, such as the size of the room, temperature and humidity in order to determine in real time the appropriate dose, time, number and power of the lamp is required for complete disinfection of all while providing maximum power permitted in the United States for the production of germicidal UVC energy. <S> I think it could be used also, This book is good for <S> this propose.
| There are cheaper ways to generate disinfecting UV light.
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Pull-up/ pull-down resistors in CMOS gates Can someone help me with understanding why the next two circuits have those outputs?My initial schematic contains only one inverter, and the signal is 5V for output high state, and 0V for low.Now, using one grounded resistor (that I think is pull-down resistor), my output values are 4V in HIGH, and 0V in LOW. Why does it work like that, and is it really a pull-down resistor as i think? The second one is and its output states are 5V for HIGH and 1V for LOW level. Why is it like that and is that a pull-up resistor?Thank you for your help! <Q> The first circuit has a pull-down resistor. <S> the second circuit has a pull-up resistor. <S> The low output voltage only goes down to 1 volt due to the internal resistance of the output stage of the inverter and the current drawn by the resistor. <A> Yes, pull down and pull-up <S> but not sure why the inverter needs them since the CD4049 is a push-pull output. <S> Could just be for a safe power-off state. <S> The inverters have voltage droop when trying to sink or source too much current due to high output resistance. <A> In first schematic, when output is HIGH, R1 and pmos pullup form a voltage divider. <S> Increase R1 and you will see the output voltage increase.
| The high output voltage only reaches 4 volts due to the internal resistance of the output stage of the inverter and the current drawn by the pull-down resistor.
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Switch between 2 voltage dividers depending on input voltage I have a power managent IC that has features of Under and Over Voltage Protection through the EN and OVP pins. These pins are enabled by manipulating the voltages on the pin via a 3-resistor voltage divider. Now my circuit takes in two configurations for the Li-ion battery for power, either at 3.7 volts (using 1 18650) or at 7.4 volts (using 2 18350). This makes it tricky for the power management IC. It is possible to use two of the PMIC for the different configuration making sure that only one is always active. But it take a lot more components and not much more GPIO pins for monitoring. I was wondering if there is any way to switch between two different voltage dividers depending on the input voltage. Without using timers, I do not want to keep adjusting it. The input voltage share the same line, but if the voltage is let's say lower than 5V, it will use the voltage divider on the left and if it's higher than 6V it will use the divider on the right. <Q> You could use a 4053 multiplexer with a comparator and voltage reference for selecting. <S> Be sure to carefully analyze this idea to see if it makes sense in all possible situations. <S> The reference has to work from the lowest possible voltage with one cell, so you'll have to divide the battery voltage down. <S> A better solution might be to add a couple of jumper blocks or a DIP switch to change from 1 to two cells. <S> You could use a single chain of 5 resistors. <A> TS12A12511 <S> This circuit is not finished though since the voltage divider for enabling/disabling the switches is proving quire hard, below 5V the IN pin must read <0.8V and above 6V the IN pin must read <S> >5V <S> EDIT: <S> I think I found the way to do it, is to use a power supervisor IC: <S> As to if I actually use this on my circuit I will still asses it. <A> maybe you can get the ratios you need by only adjusting the upper resistor in the divider. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> here the switch could be facilitated by a shorting link on the battery connector, or a switch in the battery compartment
| I had an idea of using a analog switch SPDT IC to switch the pins, I could only find one where it can more than 5V
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Can I safely use a 6 A, 600 V diode in place of a 6 A, 400 V diode? Inexperienced electrical person here, but trying to save significant cost to repair an oven. The oven I have has a MR754 diode off the main convection heating element, that has failed. Unfortunately, since it is part of the wiring, the manufacturer only offers up the entire wiring harness as a replacement part at $500+. Pending all other things checking out with a multimeter, I would like to simply splice in a replacement diode (~$2 part). However I cannot locate an MR754 in my area and it's multiple business days to get one delivered via online order. Local vendors do have an MR756 available. Both are 6 amp current, but the available replacement would be 600 V vs 400 V. Is this safe to use in this application? What is the risk, should I try this? Do I put the other components in the oven (i.e. element or control board) at risk? <Q> Yes, higher voltage rating is fine. <S> It just means more safety margin against mains transients (which is good). <S> The only difference between the -4 and the -6 is the voltage rating (see datasheet ). <A> Generally as the comments and answers already state it's ok to use a diode of higher voltage rating when the amp rating is sufficient. <S> It would even be ok to have a slightly higher amp rating normally. <S> In your case however i see no fuse in the wiring diagram and it looks like the diode could have blown due to a defective heating element. <S> A short circuit within the heating coil will cause the current to go up and this will as a result destroy the diode when it's going over 4 amps. <S> You might see your replacement diode fail very soon otherwise. <A> replying instead of answering due to lacking reputation <S> The question has been answered, but Stian Yttervik asked if there are any risks at all. <S> In almost all cases, no. <S> Most diodes will be at around 0.7 to 1 volt of forward voltage, (almost) regardless of current and breakdown voltage. <S> However, higher voltage ratings often come with a higher forward voltage drop and (at equal current into the load) higher power losses. <S> If the circuit used a very-low-drop Schottky diode with a voltage drop of maybe 0.5 volts and you replace with a high-voltage diode with a voltage drop of 1.5 volts, then the replacement diode will become much hotter than the original diode since it has to dissipate (three times the voltage) <S> * (same current). <S> This might become a problem inside a tight enclosure. <S> In addition, some small-signal circuits might rely that a diode has a certain voltage drop. <S> In that case you should also try to match the forward voltage, or at least the diode family (replace Schottky with Schottky, silicon with silicon, high voltage with high voltage). <S> And finally, please don't replace a Zener diode with another Zener diode with higher breakdown voltage - since for Zener diodes the breakdown voltage is the most important parameter!
| Since it is more likely for a heater coil to wear off and fail than the diode you should check the heating element first.
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Wiring a transformer with two primary windings I'm building a circuit that involves the use of multiple transformers. I've mainly used transformers that had one primary and one or two secondary windings. However, this time, the transformer I'm working with has two primary windings and two secondary windings, like in the diagram below: simulate this circuit – Schematic created using CircuitLab What is the proper way to wire such a transformer? Do both primary sides have to be wired to something in order to work? In my simulation, I'm only using one primary winding and one secondary as a bridge-rectifier, so the other one is not used at all. If I decide to make S1 and S2 center-tapped, will I have to use P2 as well? EDIT: More info, I'm working with 115 Vrms. This is a signal transformer. If I have to give it a model, it's the 14A-56-28 <Q> First of all your diagram is showing 2 separate transformers. <S> If it is one transformer there should be only one magnetic core like in picture below. <S> There is no 2 primary winding transformer model in CircuitLab. <S> So if you really need it you should create it. <S> A) <S> "What is the proper way to wire such a transformer? <S> " You can wire in any order. <S> Good practice is to wire all winding in same direction to avoid phase reversal. <S> B) <S> "If I decide to make S1 and S2 center-tapped, will I have to use P2 as well? <S> " <S> No. <S> Just leave P2 ends not connected. <A> You should connect the two primaries in parallel for 120 V input, and in series for 240 V (assuming the transformer is designed for 120/240V input). <S> You must observe the polarity of the primary windings when connecting them - the datasheet should show the correct connections. <A> Likewise, with the secondaries. <S> Connected <S> thus,the secondary current at double the secondary voltage would be half of that at the rated secondary voltage. <S> In parallel connection, identical ends would go together ('dot' to 'dot' and 'no dot' to 'no dot'). <S> In series connection, opposite ends would go together ('dot' to 'no dot'). <S> In practice, numbered terminals would be wired as per the wiring chart provided. <S> All windings would need to be used to utilise the transformer to its maximum VA rating.
| A transformer with 2 identical primaries and 2 identical secondaries would be wired with the primaries in parallel for rated voltage and in series for double that.
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MW versus MWh is one total capacity and the other maximum output? I still don't completely understand MW versus MWh, I have been working on a Wikipedia article for Tesla's "megapack" battery which are measured in both. I assumed one was an indication of how much power it could dump at any one time and the other the total amount of power stored. For example one the applications for a backup powersupply for trainlines where it needs to dump a huge amount of power of a relatively short amount of time so as to get trains to platforms during a blackout. So I assumed 7MWh was the amount it could pump out in any one time and 4.2MW was the amount of power it stored in total, am I wrong? I want to put a simple explanation in the article. https://en.wikipedia.org/wiki/Tesla_Megapack#Examples_of_installed_Powerpack_and_Megapack_systems <Q> MW is power, the rate of doing work. <S> MWh is energy, how much total work can be done. <S> To use a water analogy, MWh is the size of the water tank, MW is the size of the pipe going to the tank. <S> Generally, larger tanks will have larger pipes, but that's not always the case. <S> A small tank with a large pipe can fill and empty fast. <S> This is what you get with high power LiPos, that fly drones for only 5 minutes. <S> A big tank with a small pipe will take a long time to fill or empty. <S> If you want to shift a train to a platform, then you need a lot of power, high MW rating, but not for long, small MWh rating. <A> Say you were pulling a wagon up a hill using your battery. <S> It takes energy to do this <S> (the unit for energy is a joule). <S> So it takes some number of joules to get to the top. <S> Both can get you up the hill, but one can get the job done twice as fast. <S> Your 7MWH rating means that the battery will be out of energy after the product of multiplying the power times the time is running equals 7MWH. <S> In your example it would theoretically run out of energy after running 1 hour, 40 minutes (7/4.2). <S> So 7 MWH is how much energy (also termed "capacity") <S> the battery contains. <S> MWH another unit of energy and can be directly converted back to joules. <S> In summary, two batteries with the same MWH rating will go the same distance up the hill before running out of juice. <S> One with a higher MW rating will get you there faster. <A> MW is MegaWatt and MWh is MegaWatt hour. <S> Taking the Brisbane example... <S> 2.2 MWh is the installed battery capacity which can energise the installed 1.1 MW load for 2 hours.
| MW is a unit of power and MWh is a unit of energy. A watt is a joule per second so a battery capable of 4.2 MW means that it can produce joules at rate twice that of one with a 2.1 MW rating.
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Why do we use capacitors in parallel with DC motors? Below is the circuit diagram of an L293D motor driver IC driving 2 12V DC motors. What I don't understand is the use of the capacitors marked 104 in parallel with the motors. <Q> Generally a 0.01~0.1uF capacitor is wired across brushed DC motors to reduce radio frequency EMI caused by arcing between the brushes and commutator. <S> Sometimes two capacitors are wired in series, with the center connection going to the case to 'ground' it at RF frequencies. <S> For best effect the capacitor(s) should be placed on or inside the motor. <S> In this case a capacitor has been included on the driver board. <S> This makes it less effective at higher frequencies because the wires from the board to the motor will still be able to radiate EMI. <S> Still it's better than nothing, and may prevent misoperation due to interference from an unsuppressed motor getting into the driver and input wiring. <A> DC motors use brushes on the commutators. <S> These spark due to inductance of the coils as they switch from segment to segment. <S> Figure 1. <S> Communtators and brushes. <S> Image source: eReplacmentParts . <S> The capacitor shunts (or "absorbs") the high frequency spikes from the commutation and prevents damage to the driver chips. <S> A further factor not clear from the schematic is that snubber diodes should be used to prevent inductive kick-back from the motor's inductance causing damage to the driver's output transistors. <S> While the diodes protect the driver, the capacitors take the "edges" off the current spikes and help reduce EMI, etc. <S> Figure 2. <S> The L293 datasheet shows that snubber diodes should be connected across the motor unless the D version is used. <S> The left side of the schematic shows a H-bridge connected motor and the right side shows alternate configurations for single-direction motors - one connected to GND and the other to V CC . <A> Your question is probably only answerable through context; you're right, instead of C2 and C6, everyone would expect flyback diodes, forming a ground path for voltage spikes that happen upon switching off an output. <S> Maybe these capacitors are supposed to fulfill the same role? <S> Generally, you'd avoid having capacitance here – it's hard enough on the output drivers to reverse the output voltage, no need to burn an extra bit of energy from a capacitor to heat them up! <S> Now, honestly, seeing that this is a motor driver using the often abused 1986-designed L293D, the layout quality of the schematic is more than questionable, <S> the schematic happily mixes capacitor values and SMD capacitor designators, this clearly has not been designed by someone with a lot of experience or attention to detail. <S> Therefore, the answer to your question might simply be that they don't fulfill any useful role here, but have been erroneously included and shouldn't be there.
| Maybe these capacitors were necessary for EMI reasons, as they absorb high-frequency noise, e.g. from a mechanically commutated DC motor.
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Determine if a heatsink is required (Darlington TIP120) I'm planning to switch 1A load at 5V with a TIP120. Looking at it's power derating graph I see that at 5W case can be around 140C, but how can I tell if it will get hotter than that? In the datasheet I see that thermal resistance junction to ambient is 62.5C/W so does this mean that it will dissipate 5 times that = 312.5C. That seems like a bit too much for such load. If power dissipated is VCE * Ic (VCE is 4V for TIP120 at 5V collector voltage, right?), then this means 4 * 62.5 = 250C which is still a lot. I'm getting a feeling that this can't be right. Also if I dissipate 4W and still provide 5W for my load, then that's almost 50% efficiency, thats horrible. I'm probably off on my calculations, I could really use some help.. <Q> I believe you need to calculate the collector-emitter voltage before multiplying by (current) and (Thermal Resistance, Junction−to−Ambient) <A> No, 312.5C is the temperature that the TIP120 would reach if the heat pathway was only through the thermal resistance from the case to air. <S> You also need to consider other exit pathways for heat, like the Junction to case which is 1.92C/W <S> this means 5W*1.92C/W = 9.6C above ambient, which seems more reasonable. <S> However Junction to Case means that the part temperature will be ~10C above the PCB temp, and that can be difficult to determine because then you have to calculate copper thermal resistances and estimate how much of the heat can go back to air ect. <S> Do yourself a favor and slap a heatsink on the TIP120. <S> 1-3W you might be able to get away with, 5W will probably require a heatsink. <S> Your application will be happier also as temp swings show up in signals. <S> In the past I've even used small pieces of aluminum or coins and taped them to parts with a small amount of thermal paste, you'd be surprised at how much a little more surface area will increase the heat dissipation of a part. <A> In the TO-220 package, you can dissipate around \$2W\$ without the need for a heatsink. <S> $$P = <S> \frac{150°C - 25 <S> °C}{62.5 \:C/W} = <S> 2W$$ <S> But I suspect that the load will be connected between supply rail and TIP120 collector and also the BJT will be working as a switch (in saturation). <S> So in this case power dissipated will be equal to: $$P = I_C \times V_{CEsat} <S> \approx 1A \times 0.7V <S> = 0.7W$$ <S> And the TIP120 junction temperature will be <S> \$0.7W \cdot 62.5 <S> \:C/W = 44 <S> ° <S> C\$ above the ambient temperature. <S> If this is the case. <A> You want to refer to Vce(sat) <S> at 1A. <S> Which is around 0.8V typically, so you're looking at 800mW. <S> By my (conservative) standards that's a bit on the high side, but it might be okay for you. <S> A square inch or so of PCB copper would deal reduce it quite a bit. <S> But really, you want to use a MOSFET and not that darlington with the high voltage drop.
| 312C is well above the absolute maximum rating, so at 5W you'll need a heatsink.
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How to build a programmable sticky button I'm trying to build a button that can exist in two modes. Either it springs back after being pressed, or it stays pressed. I want to switch between these two modes using some complex logic, controlled by an Arduino or a RaspberryPi. The dimensions should be around 1'-by-1', and I'll have a 5-by-6 grid of these buttons to make a puzzle. I'm struggling to come up with a simple design for one such button. I thought of using a solenoid, but they have a duty cycle and use a lot of power in one of the two modes. Ideally, I would only need power to switch between the two states. That way, a player can play with the puzzle for a while, then leave it alone overnight and come back to playing the day after. I'd rather not keep the solenoids powered overnight. Another idea I had was to use two electromagnets to pull some kind of a latch back and forth, to jam the button in its pressed state. But that seems complicated and finicky. <Q> Someone has already invented this. <S> The circuit breakers used to simulate aircraft breakers in flight simulators can be released by a small current, but they are not cheap. <S> The device pictured sells for hundreds of dollars. <S> I see a relay mechanism used like a firearm sear , some contacts, and a spring-loaded plunger - you may be able to duplicate this at a lower cost <A> There are solenoids that can pull in instead of push out. <S> Ones with springs on them so that they default to out. <S> Taking one of these, and you can build a simple latch. <S> When the button is pressed down, the solenoid arm is pushed into a hole on the side of the button, locking it in place. <S> You can then just pulse the solenoid to unlatch and the button pops back up. <S> So no need to power it all the time. <S> For this you would need to fabricate a button cover and spring setup (hello 3d printers). <S> Or you could modify an arcade button type switch with a drill. <S> Alternatively you could make it electronic. <S> Instead of keeping the switch pressed in physically, have it light up an led. <S> This would mean you need to power it, but it will be much less current than a solenoid. <A> Maybe a small servo could be used to pull down the push button stem. <A> You can use a permanent magnet strong enough to hold the button down, but not strong enough to pull it down. <S> Put an electromagnet in series with it in the magnetic circuit, so that a pulse on the electromagnet will counteract the permanent magnet and let the button come up. <S> I guess the easiest way to arrange this would probably be to attach a permanent magnet to the bottom of solenoid, so that if you push the solenoid in, then the magnet will hold it down, but you can energize the solenoid to release it.
| You could get button switches with LEDs built in that you can control. How you do it would depend on the solenoid you use.
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Inrush current from DC motors causing Arduino to reset I've got an Arduino controlling a relay. The relay controls two small DC motors. All of these components are powered through a 9v power regulator. I need the power regulator because the power source is a 16 volt battery, which would fry the motors and arduino if I connected them directly. When the Arduino powers on the relay is initially open so the motors are powered off. The problem comes when the Arduino switches on the relay and motors start. The inrush current overloads the regulator and as a result, the Arduino loses voltage and resets. I have determined from experimentation that the regulator has enough ampacity to handle the DC motors at steady state, it's just the inrush that is the problem. I am very inexperienced in this area, so it's not clear to me how to solve this problem. Might I use a capacitor to provide power to the arduino during the inrush period? Or maybe a capacitor could be charged up and provide power to motors on startup? Or maybe capacitors aren't the answer? Using a voltmeter (on amp mode) it looked like the DC motors pulled about 1.5 amps on startup, and then went down to 0.5 amps on steady state. Whatever the answer, a detailed explanation of wiring would be helpful, as well as what numbers I need to consider when picking out a capacitor or whatever it is I need to buy. <Q> I'd be inclined, given the huge headroom (15 - 9 = 6 volts), to just use a 2nd regulator for the MCU. <S> Make sure the motor-regulator (dissipating 1.5 amps <S> * 6 = 9 watts) is well heatsinked. <S> Or use a switching_reg for the motors. <A> Measure voltages to see if it is the battery or the power regulator that cannot provide the 1.5A of startup current. <S> Then add a huge capacitor parallel with it. <S> The size of the capacitor depends on how long it takes the motor to reach its running speed. <A> This comes up a lot . <S> It's funny, I once went in backflips trying to find a way to share the reg, <S> and then I found out the regulator was a $4 part... <S> forehead slap <S> But what also gets people is voltage drop on the wires from the regulator or battery to the motors+Arduino. <S> Usually it's like this <S> (and this is about distribution, so 1 line represents 2 wires) /- <S> Arduino Batt -- <S> Reg --------------------------------------------- <S> < \- Motor <S> So you see the recipe for disaster. <S> Even if the reg can deliver the current, when the motors are at Locked Rotor Amperage on startup, that's kicking a big voltage drop down <S> whatever piddly 18 AWG wire the hobbyist happened to grab for that long run. <S> This means, <S> don't do this /- <S> Reg -- Arduino <S> NO <S> Batt ----------------------------------------------- <S> < NO \- Reg -- <S> Motor <S> NO <S> Do this /----------------------------------------------- <S> Reg -- <S> Arduino <S> Batt -< \----------------------------------------------- Reg -- Motor
| Yes, it may well be exactly what you say, in which case the answer is another regulator just for the Arduino.
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Thévenin with dependent source and unknowns? I have this problem where I need to find the Thévenin resistance. However I have a bit of a trouble of thinking how I should do it with a dependent source and that everything is unknown. How should I go about this problem? Thanks in advance! Edit: I'm trying to figure out the third equation that involves V1. Could some try to give a hint on how to manage that? <Q> There are several ways to determine the Thévenin equivalent circuit of a given electrical diagram. <S> In the present case, there is one fixed source and a voltage-controlled current source. <S> According to a paper written by Mr. Leach, it is possible to apply the superposition theorem and alternatively turn sources on and off, even the controlled one. <S> The trick, in this case, is to start by determining the control variable, \$V_1\$ in this exercise. <S> Here we go: <S> Once the control variable is obtained, then you can apply superposition again to obtain the Thévenin voltage: <S> Then, to determine the Thévenin resistance, short the \$V_A\$ source and install a test generator across the output from which you want to determine \$R_{th}\$ . <S> This is exactly as what is described following the fast analytical techniques to determine time constants. <S> Here, you can simplify the analysis because \$R_3\$ is in parallel with the intermediate result which itself is in series with \$R_2\$ . <S> The circuit to solve ends up being the simple below one: <S> Once this is done, a quick Mathcad sheet and a SPICE simulation tell you if this is good or not. <S> In SPICE, a .TF (transfer function) calculates the output impedance at the considered node, directly giving the small-signal resistance \$R_{th}\$ <S> you want: <S> Applying superposition to controlled sources represents a valid extension of the original theorem as this example shows. <A> One way to find the Thevenin equivalent circuit is to find the open-circuit voltage and the short-circuit current, then use Ohm's Law to find \$R_{TH}\$ . <S> If you make a significant effort to solve this on your own but have a specific problem or question, ask another question and we will try to help. <A> You might want to consider adding a test source to the open terminals. <S> Then dividing the voltage by current across those terminals. <S> That way you can determine the thevenin resistance.
| There are a number of general circuit analysis techniques you could use, such the node-voltage or mesh-current techniques.
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How can I monitor this AC circuit? I have a machine that has a circuit more complicated than this, but this to make things simple: The voltage source is 24 VAC coming from a PLC, and it has more than 100 × (bulb + resistor + switch). How can I insert one relay in this circuit to be able to perform a command outside this circuit when ONE or more switches is closed? simulate this circuit – Schematic created using CircuitLab My question is: Why this circuit didn't work? Relay model: RMI 45 Datasheet: http://www.farnell.com/datasheets/1430520.pdf Update: When one of the switches closes, the lamp don't go on and the relay keeps closing and opening frequently. <Q> for this aopplication you can use a current-sensing relay, as you have drawn. <S> this is a different type of relay to much more the more common potential relay. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> A diode dropping circuit consisting of D1 to D6 will drop about 2.1 V in both forward and reverse direction. <S> This is enough to power a bidirectional opto-isolator with 10 mA for most of each half-cycle. <S> The circuit will work with 100 mA (one lamp + resistor). <S> The circuit will work with 10 A (100 lamps + resistors). <S> Average current trough D1 to D6 will be 5 A with 100 lamps. <S> Diodes should be rated at ≥ 10 A. Diodes will each dissipate P = <S> VI = <S> 0.7 <S> × 5 <S> = 3.5 W when 100 lamps are on so they'll need adequate ventilation. <S> Q1 can be used to drive a logic circuit or a relay switching circuit. <S> (Note that it can only switch a few milliamps due to the limits of the opto-isolator's current transfer ratio.) <A> Use a current transformer. <S> Your load varies from a minimum of 24 V / 500 <S> Ω = 48 mA (one switch on, lamp burned out) to a maximum of 100 × lamp + resistor current. <S> If your lamps are 1.2W (480 Ω), then the maximum would be almost 10 A. <S> If you use a 200:1 current transformer, then the secondary current will range from 0.24 mA to 50 mA. <S> This is more than enough to activate a transistor that can operate your "common" relay. <S> Normally, you would "burden" a current transformer with a resistor in order to get a linear response. <S> In this case, we don't care about linearity at all, so we use a zener diode to limit the voltage instead. <S> A 5V zener will only be dissipating about 250 mW in the worst case (all lights on). <S> We take 0.1 mA from this 5V supply to drive a Darlington transistor pair. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is a variation of this circuit that eliminates the two bridge rectifiers. <S> The only catch is that you now need to make sure that the phasing of the transformer is correct. <S> Note that Q1 and Q2 are Darlington transistors internally. <S> simulate this circuit <A> Here's a solution using 108 diodes, a relay and an electrolytic capacitor. <S> A PCB would make the task less difficult.
| Another bridge rectifier allows us to control a relay that has a 24VAC coil (which matches your power source).
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Adding to the output of a DC/DC Boost Converter using the same power supply as the input? I've got a project that requires a variable output between 24-48V. This needs to be continuous-current, not PWM. Items that I have handy from my 'box of bits', and ready to work with are: an 0.6-36V Variable Boost Converter , and a 12V Regulated Switching Power Supply . Obviously, I'd prefer to use these rather than purchase a specialised supply if possible. There are two things that I know at this point, therefore: I can use the 12V supply as the input to the Boost Converter, but the output voltage won't be high enough. If I had a 12V battery, I could add this in series with the Boost Converter output, to get a variable output from 12.6-48V. The diagram below shows 'Point 2'. There is a 12V source in to the Boost Converter, and an additional 12V source in series with the output, leaving me with a variable 12.6-48V. This is great, except for the fact that I only have one 12V source. Which then brings me to my question, can I use the 12V supply in place of the battery, in series with the output from the Boost Converter? There are now effectively connecting wires going from the output of the boost converter back to the input, which makes me nervous! The image below shows the same circuit, but with the wires linked back to the original 12V source, rather than to a second one. I have built the closest circuit I could in Falstad to hopefully illustrate what I'm thinking. They don't have a Boost Converter component, though! Could this work as a solution? Is there a way to effectively split the 12V source into two separate supplies so I can enact the first diagram? Failing the above, do you have a hardware recommendation for a source that's directly variable through the appropriate range? <Q> Your 0.6V.... <S> 36V converter is said to be buck-boost converter in the linked advertisement. <S> The circuit doesn't need a transformer, a single inductor is enough. <S> It has advanced controller IC and the needed switching transistors and diodes for both buck and boost operating modes. <S> The controller selects the needed mode automatically. <S> Here's an example of the buck-boost idea. <S> In this case it's used for broad input voltage range. <S> https://www.analog.com/media/en/technical-documentation/data-sheets/8390fa.pdf <S> Boost converter actually adds in series with the input voltage extra DC which is generated as inductive pulses and stored to a capacitor. <S> So, the 12V DC input is already used in series with inductively generated DC. <S> It's no more availabe to be connected in series another time. <S> That would be a short circuit if we assume your linked converter really operates as common boost coverter when it should increase the voltage. <S> No datasheet = <S> the already purchased circuit must be reverse engineered. <S> But you can make an inverting switch mode converter which generates -12V for you. <S> Between its -12V output and the +36V output there's <S> 48V. <S> Unfortunately we do not know surely is the GND of the 36V converter at the output side same as the GND of the input side. <S> If it's not something unwanted can happen. <S> In addition the resulted 48V wouldn't be against GND, it would be +36V and -12V with GND not connected. <S> Another possibility is to use your battery idea. <S> Have a flyback converter which has fully isolated output side. <S> You wrote "No PWM, it must be continuous current" Unfortunately all dc to dc converters eat current pulses from the input supply voltage. <S> They can be smoothed with a LC low-pass filter. <S> Designing it isn't trivial. <S> Actually many converters have a LC input filter, but its smoothing operation is designed only to prevent radio interference. <S> You may need much more effective filtering. <A> Your question lacks a good understanding of power conversion with the conversion of voltage, current and impedance. <S> So a good answer begs a better question. <S> Always start with good specs like any battery or IC datasheet summary. <S> When you define a boost regulator the load referred to the source is in theory the same as a passive transformer. <S> with Zin/Zout= N^2 where voltage ratio = <S> N <S> (boost V reduces input Z) <S> If efficiency was ideal then the input impedance of the converter will be that average. <S> But since it is regulated by pulsed voltages of ramped inductive current the impedance will oscillate about this average. <S> Since your source will now be loaded by this reduced oscillating load impedance , its regulation will be greatly stressed by the impedance ratio of it's source to the transformed load with instabilities due to the real Z(t) ratios of source to load. <S> Your Falstad simulation is not defined properly in your question or simulation and has several errors. <S> All you have is a 3x VCVS using +12Vin to make an inverting boost converter to get 12- <S> -36V = <S> 48V. <S> The LED string characteristics were not defined properly and thus the simple LED was taking > <S> 1 kA !. <S> I won't give you an answer until you defined your question better with good input v. output specs for V,I & Z(t) or ESR or load regulation error. <S> I replaced it with a 48V 100W bulb <S> which with 4x voltage reduces R by 16. <S> e.g. that appears as 23 Ohm load steady-state, but the VCVS input transforms that load to the input as 1.44 Ohms=12V^2/100W. <S> Even though VCVS=3x +1 =4 <S> It still has a gain of 4x input voltage thus the input impedance is 4^ <S> 2 =16x (times) lower than the output load. <S> Thus for good stability your voltage source impedance needs to be <= <S> 2% of this load <S> and that's just an average and <S> not what you get with a DCDC boost regulator which has to store power in pulses in the choke then switch that power to the output. <S> thus Zin depends on peak/average current ratio. <S> I suggest you do a lot of reading on boost converters and use a very low impedance source such as a battery or a step down converter from AC. <A> You will most likely blow up your converter. <S> As seen above, the output of the buck-boost is connected to the negative battery terminal. <S> Most buck-most converters are using Cuk topology, which has the negative terminals of Input and Output shorted together, and your going to SHORT your battery through the converter. <A> Why not power the "0.6-36V Variable Boost Converter" with the additional battery and then add the "12V Regulated Switching Power Supply" in series to the output? <S> This should work, because there are no ground loops in contrast to the other suggestions mentioned so far.
| Buck-boost is a well known principle where the same circuit can produce with acceptable efficiency and peak input current as well higher and lower output voltages than the input voltage.
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Are optocouplers enough of a protection to sensitive electronics There are certain lines in my car which I want to detect when hot. My initial thought was to hook a voltage divider to drop from 12v (or ~14 rather) to 3.3 and feed it to my pi but I soon realized that I'm probably going to have a bad time due to fluctuations and stuff. Would it be enough of a safety precaution if instead I use an optocoupler such as 4n25 and feed the IRLED from the line I want to monitor via a 1.3k resistor to get about 10mA? Additionally would changes in the voltage of the electric system (and therefore the line I'm feeding the LED from) cause changes in the voltage of the transistor side of the 4n25? Or in other words does the brightness of the IRLED(?) affect the state of the transistor or is it stable once it reaches a certain threshold? <Q> Messiness on the input side of the opto could cause changes on the output side. <S> But if this is a digital system it is probably not much of a concern because digital systems only care if you are above or below a threshold. <S> If you are concerned, you could add a schmitt trigger on the output side for hysteresis. <S> Note that the power on both sides of the opto has to be isolated too for the opto to provide any real protection. <A> For the buck converter, only a varistor is enough. <S> For the LED see schematic below. <S> If R1 resistance is too high for the LED reduce it to 4.7k <S> but I think it should be plenty enough. <S> You can put two zenner in parallel for higher safety. <S> (or 3 or 4). <S> D4 protects from reverse voltage. <S> The varistor protects only for surges above 20V. Between 15 and 20V it only mitigates the surge <S> but it doesn't regulate the voltage with precision. <S> The clamping voltage is much higher than the working voltage. <S> Here 15V. The maximum voltage given by the battery. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I going to take a somewhat contrarian view here. <S> If you don't understand your environment - a car power system in this case, anything you come up with is more of an ad-hoc solution. <S> You can't properly engineer a solution to an ill-defined problem. <S> In you're case, you need to define what it is you're trying to protect against. <S> Things to look for are 1) steady state voltage ranges, 2) surges and sags, 3) spikes & transients (pretty large during a load dump), 4) noise, etc. <S> SAE ( think) publishes documents that contain this information, that you have to pay for, but there's probably a good deal of information on-line. <S> Once you have that information, and knowing the damage thresholds for the device you're trying to use - some flavor is Raspberry PI <S> I assume - you will be a position to properly design a solution for your problem.
| An optocoupler would provide quite a lot of protection from whackiness on the lines. If the voltage is above the "clamping voltage" (see parameters) it will short to ground and eliminate the current completely. Here it illustrates my comment. You must chose the varistor according to the working voltage. Starting point would be to do a little research to find out details of a vehicle's 12 V power system.
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Power that is consumed by diode? I’m wondering if I’m making the right approach when I’m trying to calculate the power consumed by the diode in the circuit. Here is the problem: So my approach is: Power loss by diode $$V_f \times I_f$$ where \$V_f\$ = Diode forward voltage drop. \$I_f\$ = The current flowing through the diode. So the first thing I did was to calculate the current $$I = V/R = 5/1000 = 5 mA$$ And then I use $$ V_f \times I_f= 2 \times 0.005 = 10 mW $$ Is this correct? I feel like there is more to the problem with the fact that it states ”red color light”. Should I check on a sheet or similar? <Q> I think that you have neglected the voltage drop across the diode which limits the current. <S> My solution is : $$\because V_{R_1}=V_0-V_f=5- <S> 2=3\,\text{V}$$ <S> $$\therefore I_f= <S> I_{R_1}=V_{R_1}/R_1=3/1000=3\,\text{mA}$$ <S> $$Power Loss = <S> V_f\times <S> I_f=2 \times0.003=6\,\text{mW} <S> $$ <A> You calculated the current wrong. <S> Apply KVL. <S> You will see that the voltage accross the resistor is actually 3 volts. <S> Other than that, your approach seems correct. <A> You just forgot the voltage drop across the resistor so... <S> Power loss by diode <S> $$V_f×I_f$$ where \$V_f\$ <S> = Diode forward voltage drop. <S> \$I_f\$ <S> = <S> So the first thing I did was to calculate the current through the resistor. <S> \$V_R= <S> V_0-V_f = <S> 5\,\text{V} - 2\,\text{V} = 3\,\text{V}\$ . <S> \$I= <S> V_R/R=3/1000=3\,\text{mA}\$ <S> And then I use $$V_f×I_f=2×0.003=6\,\text{mW}$$ <A> \$P_{diode} = <S> 2\,\text{V} \times 0.003\,\text{A} <S> = 6\,\text{mW}\$ . <S> \$P_{resistor} = 3\,\text{V} \times 0.003\,\text{A} = <S> 9\,\text{mW}\$ . <S> \$P_{total} = <S> 6\,\text{mW} <S> + 9\,\text{mW} = <S> 15\,\text{mW}\$ <S> consumed for supplying the diode. <S> The efficiency of the system is $$\frac{P_{diode}}{(P_{diode} + P_{resistor})} = <S> \frac{6\,\text{mW}}{(6+9)\,\text{mW}} <S> = 0.4 = 40\%.$$ Efficiency can be improved by using a supply with voltage closer to the diode voltage drop. <S> This will reduce the \$9\,\text{mW}\$ loss from heating the resistor. <A> \$V_R=V_0-V_f = 5 - 2 = 3\,\text{V}\$ . <S> \$I= <S> V_R/R=3/1000=3\,\text{mA}\$ . <S> \$V_f×I_f=2×0.003=6\,\text{mW}\$ . <A> \$I_F = <S> (V_O - V_F) / <S> R_1 = <S> 3 <S> / 1k = 3 mA\$ . <S> \$P_F = <S> V_F \times I_F = 2 × 3 = 6 mW\$ .
| The current flowing through the diode = current through the resistor.
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Short Circuit Protection of High speed Data lines I need to protection data lines of a PIC32 micro-controller (SPI and parallel) running at ~10 MHz, from shorting to ground/Vcc or mis-wiring. I usually (in low speeds) use series resistors to limit current and it worked well, but of course this does not work now because of the speed. I am thinking to use line buffer/inverter (something like 74LVC04) this will save the MCU but the buffer will get damaged. I searched many logic families (ACT , HCT , LVC etc..), but none provide short circuit protection. I there a better solution? Edit After Comments: This is a development board for testing and validation, due to coding mistakes pins can be configured incorrectly . or mis-wired , connecting MCU output to target output instead of input. Edit #2: A PTC ( 0603L004 ) might be a solution however it is slow ! <Q> OK, in all honesty: then the purpose of the development board is to learn to avoid these mistakes. <S> The solutions to this particular problem (needing error-resistant high-speed off-board communications) usually involve transition to a specific far-reach bus. <S> Your "problem" has a "solution" that would make your original problem, talking to peripherals, so much more complex that it's not really a "solution" any more. <S> If your peripheral can destroy your controller, then take care to not wire it incorrectly. <S> That's typically not really hard, compared to other engineering pitfalls. <S> For example, you'd design an adapter board that has one side very clearly labeled "signal" and one side labeled "output", and if possible, two different connectors on each. <S> Problem solved. <A> You can infer this from the datasheet based on Vo(h) and Vo(l) for a given output current. <S> As an example, the PIC32 datasheet list Vo(l) as 0.36V at 6mA, which corresponds to an Rds(on) 60 ohms. <S> Shorting it to 3.3V would give 55mA, which is more than the datasheet allows (16mA), a pretty tight limit. <S> That said, a short on one pin won't necessarily damage the part. <S> The bigger problem often is ESD damage. <A> I am an electrical engineer. <S> I have worked on microprocessor-based designs of a lot of products. <S> I have occasionally miswired things (more times than I would like to admit). <S> Also, at some point we have to give our hardware to firmware engineers and they wire things up wrong all the time, too. <S> I have never seen a GPIO get fried due to being short-circuited to VCC or GND or to another output. <S> Obviously I don't recommend doing it on purpose, but I have done it more times than I can count, and had FW engineers do it wrong, too. <S> I don't think you need to worry about it unless you are going to expose the output to a voltage greater than VCC or less than GND. <S> The only thing I have done on occasion is used external logic to make sure that the FW team can't accidentally create a "forbidden state." <S> (Like turning on the high and low side of a bridge simultaneously to create shoot-through). <A> The problem with modern MCUs is the tiny size of the FETs, and the enormous local-heating temperature rise rates. <S> I recall computing dTemp/dTime of 1,000 degrees per microsecond for such (tiny) FETs, with many milliWatts dissipated in a square micron or two, and the large thermal resistance of cubes of silicon that are only 1 micron on a side. <S> ================================== <S> For over-temperature protection, any temp-sensor needs to be within 10 microns of the hot-spot. <S> If the hot-spot is deep submicron output FET, short channel for high drive, the FET can self-destruct before high temperature is detected. <S> Practical solution? <S> At high temperatures, the FET becomes less conductive. <S> Or detect the overtemperature and turn off the FETs (area expensive, to do that). <S> I recall stories of era-1995 MCUs that self-destructed when outputs shorted. <A> These work well at high speeds and are more reliable than for example optocouplers. <S> There are also specialized parts specifically intended to be used for SPI, I2C etc. <S> Though this solution is mainly used when you expect the secondary to be noisy and want to avoid EMI, voltage surges, ground currents and other nasty stuff from hitting your MCU directly. <A> Perhaps you can first configure as IO. <S> Them set the output briefly and read the result. <S> If it sits at 0 when you drive a 1 or 1 when you drive a 0 then refuse to engage the SPI. <S> I'm also an EE <S> but mostly I write software. <S> I agree with one of the previous commenters that said you should protect anything that leaves the board from ESD. <S> So TVS diodes should definitely be present. <S> If any of the pins in the connector have more than 5V <S> then you should protect your other pins from that higher voltage as well.
| Stuff that goes off board can benefit from additional TVS diodes to increase the ESD robustness. One way to protect is to galvanically isolate the bus from the MCU by using digital isolators, example . CMOS drivers have an inherent current limit set by the drain-source ‘on’ resistance of the output driver FETs, Rds(on). That's kind of a weak driver.
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Roadmap for learning Electrical Engineering I am just starting to learn electrical engineering, but I am confused about what I should do. Should I be reading about it, taking online classes, or just make projects? If possible can you provide me with a roadmap of what I should do in order to best learn about electrical engineering? I am really interested in making exciting projects related to the field. Any suggestions? <Q> Do both! <S> Do projects as well as reading literature. <S> You don't want to only sink into literature, or you will be bored and frustrated quite quickly <S> (well, most likely at least, it depends on what you are really interested in the most). <S> You also don't want to do only projects from the internet without any real explanation of theory, otherwise you will probably never understand the fundamentals behind your projects. <S> If you really want to learn all the basics I can recommend you the book The Art of Electronics . <S> It is quite impressive (in terms of number of pages) and written for courses at Havard, but that should not mislead you: <S> The book is fantastically written, very well explained and includes a lot of practical examples. <S> You won't get bored with theory. <S> In fact the book has gained a decent readership from the maker scene. <S> And for the projects aspect: The internet is full of little and big projects for every level of difficulty. <S> Analog Devices for example offers this set of tutorials with practial courses . <S> This should also cover a lot of the basics that would be explained in the above mentions book, but in less detail. <A> Build some bipolar ac_coupled multistage audio amplifiers. <S> I used several examples in my stackX answers. <S> And then learn about how transformers and generators and motors ad high power linescan interfere with the audio signals. <S> Then think about the fun involved. <A> Find out what you are most interested in: <S> ac electro-mechanics analogue <S> digital microprocessors <S> embedded programming radio frequency <S> radio amateur linear <S> switch-mode etc. <S> Then concentrate on that specific field and master it starting with basic principals. <S> The physics classes at Khan Academy has a sound scientific basis. <S> There may be many more.
| Just look around for tutorials and projects that you are interested in and start experimenting.
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How to reduce MOSFET heating in Buck Converter I have designed this Buck Converter which convert 60VDC to 12VDC at 10A. Switching Freq 100KHz . Facing MOSFET too much heating issue. MOSFET ON time and OFF time is decided by a UC3845B based circuit. MOSFET gate is biased with 2.2R resistor and pull down of 5.1K is there any way to reduce MOSFET heating?I've increased MOSFET rating to 110A 80V. Previously was 75V 75A but no succcess. Edit 1: Updated Schematic for better understanding. Edit 2: Previously tried this INFINEON MOSFET . Heating was less. Then used this ST MOSFET .Heating was more in ST MOSFET Hello, here is an update, can I use below circuit as bootstrap?here in place of 5V Input either I can use 60V directly or 12V from output. <Q> If it is confirmed the MOSFET needs a better drive then the best is to advise a suitable circuit. <S> A simple bootstrap circuit described by Monsieur Balogh in a TI application note can do the job nicely in a cost-sensitive application. <S> As noted in some of the comments, the UC384x was not really meant for hi-side drive - unless you make it entirely float and tie its GND pin to the MOSFET source and supply the IC via the rectified buck output - but <S> this little circuit does the job fine: <S> Below is the circuit I tested with component values: <S> This is excerpted from a seminar I taught in an APEC seminar in 2019. <A> MOSFET ON time <S> and OFF time is decided by a UC3845B based circuit <S> The UC3845B has a push-pull type output but, unless this chip is powered from a supply that is several volts higher than the 60 volt rail (as shown in your circuit) you won't efficiently drive the MOSFET in your buck regulator. <S> Given that the UC3845 is only rated at a maximum of 36 volts, you are likely driving the MOSFET very, very ineffectively and it will get very warm on load. <S> The gate voltage needs to exceed the main supply voltage by around 10 volts in order to get source and drain ohmically connected at a low value. <A> I’m guessing <S> the N-FET isn’t being turned on all the way. <S> It's being biased in linear mode and so has a substantial IR drop across it, which is being shed as heat. <S> You don't want that. <S> How bad is it? <S> Let's assume the FET Vgs threshold is about 4.5V. <S> Then when it's on: <S> Vd = <S> 60V <S> Vg = <S> 36V - 2V = <S> 34V (limited UC3845B output swing) <S> Vs = <S> (Vg - Vth) = <S> (34 - 4.5V <S> ) = 29.5V <S> Vds = <S> (Vd - Vs <S> ) = (60 - 29.5V <S> ) = 30.5V <S> If you're drawing, say, 1A from the supply, that approximately 30-35W you're dissipating in the FET, peak. <S> What keeps it from frying immediately is the stepping ratio that determines the on-time of the FET: <S> Vout <S> /Vin <S> * <S> W = 12/60 <S> * 35W = 7W <S> And that's at <S> only 1A. Clearly, this isn't workable. <S> What happens when you do that? <S> Here's the peak wattage shed in the FET: <S> Vd = <S> 60V, <S> Vg = 65V <S> Vs = <S> about 60V (transistor is fully on) <S> Vds = <S> (Vd - Vs <S> ) = 0, or close to it <S> So in theory almost no power gets shed in the FET. <S> In reality, this will be: Iout^2 <S> * Rds(on) <S> 10^2 <S> * 0.020 ohm = 2W peak at 10A output <S> With stepping ratio being 12/60, the FET is on about 20% of the time: 2W <S> * 12/60% = 0.4W <S> That's very manageable for a TO-220 FET. <S> How to do this? <S> You need a bootstrap circuit to generate the higher gate drive voltage (about 5-10V above Vin), and a gate driver that accepts that voltage to make the above-Vin gate signal. <S> The bootstrap voltage can (and usually is) generated from the inductor flyback via a diode and capacitor. <S> Problem is, the UC3845B is not a bootstrapped high-side driver. <S> It's really designed to be a low-side driver for a flyback topology. <S> Further, it's limited to +36V. <S> For both reasons it’s a poor choice for this application. <S> You could mess about with making a bootstrap + level shifter, but why? <S> Select a different device instead. <S> Example: this 75V input dcdc controller from TI <S> (bonus: <S> it’s synchronous <S> so your supply will be more efficient): http://www.ti.com/lit/ds/snvsai4/snvsai4.pdf <A> Pls reduce the resistance in Vin to maintain 15V across the IC UC38xx . <S> The MOSFET minimum required gate drive voltage is 10v <S> other wise it will be in ohmic region and generate heat as well as reduce power due to low contuctance <A> Normally these are called Tj (Junction temperature) and all related heat dissipation figures are shown together. <S> Also read https://en.wikipedia.org/wiki/Junction_temperature . <A> Designed output power = <S> 12V <S> x <S> 10A = 120W <S> High efficiency say 95%, heat dissipate = 120W x 5% = <S> 6W <S> You have to use a large heat sink with fan. <A> MOSFET heating was mainly due to Inductor. <S> Inductor Flyback or Back EMF was causing problem. <S> As said in Answer by @Verbal Kint sir, the ground of 3845B was completely floating as it was tied to source of MOSFET.Previously used PQ 32X20 core inductor with 0.4mm wire (don't know turns). <S> Temperature was upto 100 degree <S> C.Maybe logically more turns could cause more Back EMF or Flyback. <S> Now used a FeSiAl core with 1 mm wire gauge and 14 no of turns. <S> Noe, temperature is upto 60 degree <S> C.Maybe using a bootstrap circuit or a built in IC will do more effect. <S> But PCB was designed and needed to solve the issue badly. <S> Thanks everyone here for help. <S> Specially Verbal kint sir and hacktastical sir.
| This is a problem with source follower MOSFET configurations and they way around it is to use a proper "high-side MOSFET driver" chip. To fix it, the N-FET gate needs to be brought all the way above 60V, to at least 65V to 70V, to ensure the FET is fully turned on to its lowest Rds(on). You can check if a heat sink is needed by calculating (at least theoretically), by checking the numbers in the datasheet.
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Why are lithium batteries stored at 50% voltage and not a lower voltage? I understand that a higher voltage causes more degradation of the material in the cell, but I'm curious as to why the typical storage voltage is not lower than 50%. I've always assumed that the lower the voltage, the less material degradation, but that starting a storage cycle with a voltage too low runs the risk of dropping the voltage to a point where the cell cannot be recharged, but I haven't been able to find a definitive reference for this. <Q> I've always assumed that the lower the voltage, the less material degradation, but that starting a storage cycle with a voltage too low runs the risk of dropping the voltage to a point where the cell cannot be recharged, but I haven't been able to find a definitive reference for this. <S> A definitive answer is difficult because few studies have been done that are freely available, and the results from one particular cell type may not be applicable to others. <S> I found a study on Aging of Lithium-Ion Batteries in Electric Vehicles <S> which tested Panasonic NCR18650PD Li-ion cells. <S> The results show that storing at 3.45 V causes less degradation than 3.7 V, so your assumption is correct for normal operating voltages. <S> At 3.45 V the remaining capacity is very small (probably less than 1%), so your other assumption (that leaving too little charge may risk dropping into over-discharge) is also correct. <S> Another interesting thing they found was a big jump in degradation above 55% charge. <S> Storing at 50% charge was not much worse than lower voltages, but 60% and above was much worse. <S> This was attributed to changes in chemical composition of the cell at different charge states, so a cell with different chemistry may not react the same. <A> In short, lower is better, but there is a limit. <S> Voltages both too low (below 2.7V) and too high will damage Li-Ion cells, and they are best kept at "happy medium" levels. <S> Also, there is self-discharge (5% in 24h, then 1–2% per month, plus 3% for safety circuit if there is one) which all battery chemistries have, and higher level charge helps when storing a cell/battery for a longer period of time (including waiting on store shelves to be purchased). <S> According to "Battery University" ( https://batteryuniversity.com ), the lower the "depth of discharge", the longer the cell life, and a 50% "DoD" (between 30% and 80% charge level) gives the best compromise between useful capacity and lifetime. <S> Using the whole cell's capacity, you can get maybe 1000-2000 charge-discharge cycles, but using only 50% of its capacity you can get around 10000 cycles. <S> The same source does say the same thing you say here, namely that the higher voltages degrade a Li-Ion cell. <S> Furthermore, a voltage below 2.7V will cause irreversible damage to the cell and a reduction in its capacity, so that should NEVER be allowed to happen with a Li-Ion cell. <S> So, according to this information, the ideal charge (or SoC - state of charge) level is around 40-50% which is below the damaging "high voltage" and <S> at the same time gives plenty of "headroom" for self-discharge which would eventually bring the cell down to dangerously low levels. <S> References: " <S> BU-204: How do Lithium Batteries Work?" <S> https://batteryuniversity.com/learn/article/lithium_based_batteries "BU-409: Charging Lithium-ion" https://batteryuniversity.com/learn/article/charging_lithium_ion_batteries "BU-808: <S> How to Prolong Lithium-based Batteries" <S> https://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries (This is the most relevant link to your question) <A> A very quick and dirty high level summary is at above 3.8v the battery needs to "deflate", anything below 3.8, the battery needs to "inflate". <S> Basically stressing the cell, 3.8v or it's nominal voltage, is the happy medium where it can just rest. <S> Storing cells undercharged is not ideal but not a huge deal, storing them at charge in my experience causes huge degredation in cell quality, even if stored for more than a day or two. <S> This usually is reset after the next charge but does leave lasting damage. <S> I liion and lipo, aything below 2.7v results in permenant damage. <S> You typically want to stop at 3.3v. <S> Depending on your application, you would not want to go below 3v <S> since current draw at this low a voltage on a cell causes it to sag significantly more, potentially dropping it below 2.7v. <S> On a lipo, do notbring it down past 3v. <S> On Li ion, 2.7v. <A> Probably a lower voltage would be a bit better. <S> But the existing practice still preserves battery longevity pretty well. <S> Also, you need to leave room for self-discharge without getting too low. <S> The batteries may sit for some time after leaving the factory. <S> And once the cells have been assembled into a battery pack, there will be some type of protection circuit that consumes a small amount of current. <S> And if the battery pack is then installed into a device which will be shipped to a consumer, there may be additional standby current consumed by the device. <S> But the hope and expectation is that the device will at least power on when it gets to the consumer, even if it has been six months since the battery left the factory. <S> Marketing doesn't think it is a good out of box experience if the device is DOA and has to be charged before it can be used. <S> I should actually add that if it were up to marketing, the devices would be shipped fully charged. <S> They don't care about longevity issues. <S> The reason the batteries ship at 45% state-of-charge (SOC) is because that is the maximum that is allowed by the shipping regulations.
| I guess the batteries are less hazardous when they are at low/middle SOC.
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Why is the linear voltage regulator going hot although within its specification? I'm having an issue with my NCP718 voltage regulator overheating to the point of thermal shutdown. I have attached a photo of my schematic. I am inputing ~15-20v and outputting 4V from the reg. The reg is rated to 300mA and I am drawing about 50mA at idle according to my switch mode power supply. Even at such a low current draw the regulator gets too hot to touch and eventually shuts down. Any tips would be greatly appreciated, I am a little stuck right now! <Q> The reg is rated to 300mA <S> That's one of its many limits! <S> Not the only one. <S> You're thermally limited. <S> Let's see what that means. <S> Linear Voltage regulators work like this: They use an internal transistor as an "adjustable resistor", which they always adjust so that, given the current drawn at the output, the voltage dropped over the resistor is such that the desired output voltage is kept. <S> In other words: they're designed so that the difference in energy per charge (electron) in the voltage difference between in- and output is converted to heat . <S> That gives us the very simple formula of the power converted to heat: $$ <S> P_\text{linear reg., heat} = <S> \left(V_\text{out}-V_\text{in}\right)\cdot I_\text{out}\text.$$ <S> In your case: 11 to 16 V drop × 50 <S> mA current = <S> 550 mW to 800 mW idle power. <S> Go into your data sheet, look for "thermal specifications" or similar, look for "thermal resistance". <S> Look for your package variant. <S> If you've got a heat sink attached to some part of your chip, look for the resistance junction-to-that-part, add the thermal resistance of the heatsink, and multiply with the power converted to heat. <S> You sadly didn't specify the package variant, so I'm assuming SOT-23-5. <S> That's got a junction-to-ambient thermal resistance of 235° <S> C/W, so 550 mW will heat it up by more than 100°C. <S> Yowza! <S> So, clearly, this was a bit of a design mistake: <S> When dropping reliably high voltage differences, the way to go is usually a switch-mode regulator. <S> Linear regulators "burn" all the voltage difference between in- and output. <S> Switch-mode regulators just store energy and deliver at a different voltage than the input – which means you have a bit of losses here and there, but not "burn all the energy per electron that's in the input voltage-output voltage". <S> You didn't calculate how much power you'll be converting to heat in your linear voltage regulator, and thus didn't use one where you could have attached a sufficiently sized heat sink. <A> If you have an absolute top limit to the current that needs to be drawn, then you can drop some of the voltage with a series resistor before the regulator. <S> Size the resistor to still give you at least the minimum headroom at the LDO at the maximum supply current. <S> The LDO input capacitor must still be directly to the LDO for stability. <S> This is not an efficient solution, the best would be to use a switch mode DC-DC converter as the other answers state. <S> However, it is a practical workaround to get an existing board working with minimal changes. <A> With 20V input voltage and 4V output voltage <S> you have a 16V drop on the regulator. <S> With 50mA load current this is already a power dissipation of 800mW. <S> The small TSOT-23 package will really have problems cooling 800mW away and even the larger WDFN6 needs big groundplane beneath it. <S> If you reduce the input voltage to be only something like 9V for example the regulator has to drop only 5V, which reduces power dissipation to 250mW. <S> If you have to stick with this high input voltage it would be very reasonable to use a DC/DC converter, which can convert your voltage with high efficiency and that way it will stay cool. <A> You had to have a look of the total power consumption of your regulator: (20V - 4 V ) <S> * 50mA <S> = 0,8W <S> And this could be hot!!! <A> That regulator will need to drop at least 15 V - 4 <S> V = 11 V at 50 mA <S> that's already 0.55 W. <S> There are two variants of the NCP718, one in a WDFN6 package and one in a TSOT−23−5 package. <S> Which one are you using? <S> Table 3 in the datasheet tells us what the junction to air thermal resistances are, for the WDFN6 is is 65 °C/W <S> but for the TSOT−23−5 <S> it is 235 °C/W. <S> Normal practice is to provide a large copper area on the PCB, this copper area then works as a heatsink . <S> Dropping more than 11 V or more across a small regulator isn't such a good idea unless the current is really small (less than a few mA). <S> You can also add a zenerdiode or even a resistor to drop part of the voltage that appears at the input of the regulator. <S> I would try to make the voltage drop across the regulator around 2 to 3 V at 50 mA.
| You can either add a heatsink on top of the package or you should reduce the input voltage.
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Best way to track light source? I want to build a 1-D (azimuth), bidirectional optical tracker. The idea is to have two identical transceiver, mounted on DC servo-motor. The user can move the two servos on rails (reasonably fast), while the servos have to automatically adjust their angular position to face the other transceiver. The distance between the two transceiver should be at least 20 cm. The drawing below will certainly explains better what I mean: So far, I have thought about using 1-D PSDs (Position Sensitive Detectors like this one ) and LASER diode. PSDs are basically sensitive surfaces that have a direct relationship between the LASER dot position on the surface and the output current: To estimate the angle that the servo motor must rotate, we need two distances: the position of the LASER dot on the PSD ( x , given by the above equation), and the distance between the LASER and the PSD ( y ). With that, we can apply some basic trigonometric formula to find the needed angle: $$\theta=tan^{-1}\Big(\dfrac{x}{y}\Big)$$ The problem with this solution is that it is needed to have the distance y . I found two solution to estimate this distance: RSSI (Received Signal Strength Indicator) and ToF (Time of Flight). I read somewhere here that RSSI is not reliable enough to estimate distances. ToF would be much more reliable here. I found this chip which is an all-in-one ToF module that has a range accuracy of around 5 % (depending on several factors). The problem with this is that it increase the complexity of the system and since both sensors (PSD and ToF) work in the IR range, they might be reached by the other sensor's source and it would increase the error in the x and y direction. Is there another way to reliably estimate the distance between the LASER and the PSD?Has anyone any idea on how the system can be simplified while allowing for angular tracking between the two transceiver? <Q> You don't need the actual angle. <S> You just need to know whether the laser dot is centered, and which way to turn if it is not. <S> A simple servo algorithm will keep the detector — and anything attached to it — pointed at the laser. <S> When I first wrote this answer, I was assuming the existence of a lens in front of the PSD that turns the angle of arrival into dot position on the PSD. <S> Of course, that only works if the other laser is already pointed at us. <S> If there is no lens, then the feedback for our servo needs to come from the PSD on the other one, and the system is only useful after some other coarse positioning mechanism has both lasers already hitting both PSDs. <A> To estimate the angle that the servo motor must rotate, we need two distances: Put a lens in front of the PSD <S> and you don't need any distances. <S> The lens turns angle into position: Measure which position the beam hits when the angle is zero (perfectly aligned). <S> Call that voltage zero. <S> Rotate until the voltage is zero. <S> Done. <S> Your sensor above comes in a 12mm version. <S> If you select that, and combine with a 10mm focal length lens, your FOV for detecting the laser beam will be +/- <S> 30 degrees. <S> You can buy PSD kits that do this and include a closed loop controller and stepper motor, but your problem is so simple <S> I don't think you need one. <S> Light Source <S> Here is the spot of an 1mm LED from ~ 1 meter distance (almost 180 degree emission angle) behind a 4mm focal length lens (my phone camera). <S> The LED is sitting in sunlight. <S> The physical width of the spot on the sensor is only 30 microns (21pix*1.4um). <S> Scaled to 10mm focal length, this becomes 75 microns. <S> I could improve this just by using a small active area LED, but you get the idea. <S> You can focus an LED with very wide divergence angle to a very small (relative to your 12mm sensor) spot. <S> Probably much smaller than your electronics can detect. <S> Make the divergence angle larger than the physical range of angles the source can be rotated through and you will never lose sight of it. <A> You could do coarse, fine adjustment without knowing the exact distance. <S> Like solar panels do it. <S> For this you would need an additional omnidirectional light source - the sun. <S> But for my opinion is to be expected that this system will be unstable, because the dot position is a function of the angle of the both servos, not only one. <S> Further, my opinion would be better to expand the method of two omnidirectional light sources and kind of fringe and maybe fast low res imaging sensor, like a bee or a fly. <S> It would eliminate the possible unstability. <S> If you have an absolute encoder on the rails, then the solution becomes trivial.
| With some simple fresnel lens and pair of LDR you can move the servo at coarse speed and align the beam, then use you method to fine track the laser beam.
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Precision voltage clamping 0V to +5V with +/-12V input I'm looking for an opamp "voltage clamping" circuit with a +/-12V input and a 0/+5V output. I mean : When input < 0V = output stays at 0V When 0V < input < +5V = output 0 to +5V When input > +5V = output stays at +5V Thank you for your help ! <Q> Here is the textbook example: Note however, that each opamp is either in saturation (there's a huge differential voltage on inputs when no need for an opamp to clamp) or it tracks its inputs precisely (when it clamps). <S> Therefore, you'll need opamps that are able to switch fast and without ringing from saturation to tracking. <A> I assume here that you want the 0-5v signal to be processed by an A/D.If you want to protect the input then use a device such as the TLV6001 to buffer the input signals. <S> A simple circuit such as this will provide accurate translation of the 0-5v signal for any application. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The TLV6001 is rated for rail-rail operation and so allows the full range of MCU or A/D input ...in addition it is rated to carry 10mA in the input protection diodes. <S> This would allow the configuration above to withstand voltages of +/-200 V on the input resistor R2 (providing your resistors are rated for this voltage). <S> Since the TLV600 is powered by the MCU +5 <S> V supply it cannot produce an output voltage above 5 V or below 0 V, so the input is accurately clamped WITHOUT impacting the A/D range at all. <S> You must make sure that your MCU solution is always drawing a minimum current greater than your expected protection current (this is only an issue if you put things into a sleep state). <A> simulate this circuit – Schematic created using CircuitLab <S> This should work.
| If you really need to use an op amp, you could buffer the output.
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Sustained over-voltage protection for a 1 MHz digital logic line I am working on a board which is user-insertable into a DIP socket. Incorrect insertion is always a concern, and in this specific case insertion "shifted-by-one-pin" is especially worrying. What then happens is that a 3V3 digital logic output line, which normally goes from an FPGA to a TTL input in the socket, instead gets connected to 12V power supply. Once the power is on, this immediately fries the FPGA and renders the board useless. Given the geometry of the situation, purely mechanical misalignment prevention is going to be difficult to say the least, making it even more important to have a protection circuit in place. Now, I looked into Zeners, TVS diodes, signal diodes in series with the signal line, etc, but couldn't really find a solution that would combine these three items: Pass ~1 MHz digital signal reliably from the FPGA to the TTL input. Have the reaction time short enough to prevent FPGA from getting damaged (EPM1270, so basically anything above 4V is unacceptable). Be able to withstand sustained over-voltage for at least several seconds. Any suggestions where to look next? Update : I am choosing my answer, because it works for my case, but since it has its caveats, other responses may be better depending on circumstances. <Q> This is what I did: I inserted a 1SS404 diode in series with the signal. <S> Signal from BA_SAFE to BA doesn't get too attenuated, and the reverse recovery time is short (~3ns). <S> Everything appears to work fine, and if I unplug BA and connect 12V in its place, BA_SAFE remains unharmed. <S> Update : there is an important caveat here. <S> What happens when BA_SAFE switches to LOW heavily depends on the diode type, and more specifically on its reverse recovery time <S> (T rr ). <S> For example, with Schottky diodes that are usually specced at T rr < 1 ns the only way the charge between D5 and BA can drain is through R32. <S> If its resistance is too high, it can take quite some time, thus extending the falling edges and affecting the signal. <S> Even at the relatively low 1 MHz it turned out to be an issue for me, one which I couldn't sidestep by decreasing R32 resistance (because then the voltage at BA would be too low). <S> On the other hand, if the diode's T rr is too long (and there are some popular ones that need tens of microseconds ), the current drain from D5-BA segment will be so strong that its potential can go below ground: I measured around -400 mV for over 10 us with 1N4007. <S> As this was not something I was willing to expose BA to, I ended up picking up a diode with <S> T rr = <S> 50 <S> ns ( BAV19WS-7-F ), which in my case is enough to preserve sharp falling edges without any measurable negative voltages. <A> The simplest way is a TVS diode to GND to clamp the pin voltage with a series resistor to limit the current through the TVS. <S> The problem is that the resistor limits your bandwidth but 1MHz isn't too fast to begin with. <S> A smaller resistance will require a diode that can handle more power. <S> To limit the current in the diode to 10mA, you would need 870 Ohms which seems high. <S> If your diode can handle 20mA <S> then you can use 435 Ohms which seems more reasonable <S> Shouldn't be too difficult to test by just putting a resistor in series with your FPGA output and scoping the output to see if signal transition times are acceptable. <S> Do note the diode will add its own capacitance but you can put a forward biased diode if it has smaller capacitance in series with the clamp diode to reduce the overall capacitance of both diodes. <S> I only thought of this just now though so if you want to try this route you need to investigate more. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Small series depletion mode MOSFETs and TVS clamp. <S> You only need one MOSFET if it can never see negative voltage. <A> You might be able to use a zener, but because you don't know the available current, you'll want to put it in series with a resettable or replaceable fuse. <S> The choice will depend on how much resistance you can tolerate.
| You could also try adding a small capacitance in parallel with the series resistor to alleviate rise/fall times issues due to the resistor.
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To ground or not to ground: portable power supply So I'm creating a backpack system to carry around and shine a 100 watt HID lamp (using the arc tube for UVC mineral collecting.) This system currently includes a 12v lithium battery, a power inverter to 120vac, and a 120v ballast setup (including a 300vac cap) that drives the HID. The actual voltage output to the HID is 20-30vac. With all these different voltages, I'm a bit confused about ground. Should I have some sort of common ground in this system, or should every part be isolated? Again, all this will be sitting in a backpack while the light will be in hand. So also figuring out whether I should build a box containing these things out of aluminum or something non-conductive instead like wood. <Q> CONSIDER a system with Metal case grounded to 12V negative, with 12V positive well isolated from user contact making 12V user contact very low possibility. <S> Then float the AC supply. <S> Make it very hard to contact either AC leg. <S> In outdoor situations if you get good contact between user and ground and between one DC battery pole and ground and if user makes body contact with the other DC pole you CAN get strong user-muscle-lockup. <S> I know of it happening to a friend standing in water (flounder fishing, 12V LED light) - much less likely with your application. <S> Locking all 12V+ inside a full metal 12V- "full metal jacket" makes contact "impossible" [tm] <S> - keep it that way !!!. <S> Having AC floating allows one lead contact without harm. <S> The same would be true of the battery <S> BUT AC is much less liable to cause can't let go shock. <S> DO NOT common DC and AC sections for same reasons. <A> It really depends on if there are isolation transformers in the system, because a fault could go through you instead of back to the source. <S> AC doesn't really care where it goes, it will go to a lower potential, if that happens to be you, then you could be electrocuted by dangerous (above ~60V) voltages. <S> If you want to be safe would probably be best to have some kind of chassis ground around each component and all the grounds tied together. <A> The question of earthing would not arise, interconnection would be simpler and the backpack lighter, should you put together a modular system comprising of a battery and <S> an automotive 12V DC 100W HID ballast having cables with moulded connectors. <S> Standard automotive interconnection cables, with in-line fuses, would take care of input and output short-circuit protection. <S> There would be no question of isolation, with the battery having a plastic housing and only the ballast <S> a die-cast aluminium enclosure. <S> The system could be housed in a light-weight aluminium enclosure with the ballast mounted inside for improved cooling. <S> Should cooling not be an issue, a lightweight, flame-resistant and rugged polycarbonate enclosure could be considered.
| In consumer products it is common to have a chassis ground for faulting currents and a way to break the current (fuse or breaker) in the event of a fault.
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Is it possible to solder wires small 0.5mm to 1mm diameter vias? I need to reprogram the micro-controller on my keyboard. I am new to soldering and would probably buy some equipent and try out on some regular size through holes. I am wondering if it would be possible to solder wires to these tiny vias(looks to be 0.5 - 1 mm diameter.) Maybe it would be helpful with some high flowing solder? Any tips are appreciated. Thanks in advance. Below is a pic of what kind of hole I am talking about. <Q> These are pretty big vias. <S> But every via has a pad around it, 0.1-0.2mm around. <S> Use x-acto knife and carefully scratch out the solder mask around it. <S> Glue up a 2x5 0.1" header to the edge or your board for easy and robust access by your test leads. <S> Then solder thin magnet wires ( AWG34-36) from your selected vias to the header pins. <S> Use low-quality magnet wire, so you don't need to scratch ends - the lacquer will come off naturally under hot soldering iron. <S> That's it. <A> I am wondering if it would be possible to solder wires to these tiny vias <S> Yes, so long as the vias are not tented <S> it is possible. <S> Maybe it would be helpful with some high flowing solder? <S> Regular 60/40 <S> Tin Lead rosin-cored solder wire should work fine. <S> You don't need much, so I recommend using a small diameter, eg. <S> 0.7 <S> mm. <S> You also need a fine gauge hookup wire which fits in the hole. <S> I use plated 30AWG (0.25 mm) wire wrap wire. <S> Any tips are appreciated. <S> Use a temperature-controlled soldering iron. <S> Make sure the tip is clean and freshly 'tinned' with solder. <S> Strip <S> the end of the wire and put it though the via. <S> On the bottom side (where the end is poking out) apply solder and soldering iron tip between the wire and hole so they are all touching each other, until the solder melts and goes into the hole (should only take a second). <S> Cut off excess wire. <A> Use a thin iron tip, tin the via and the wire (wire wrap or magnet wire. <S> I would tear apart a high density ATA/IDE cable). <S> Once both are tinned, you can tack them pretty easily. <S> Once tacked they will be fragile, so tape or hot glue <S> the wire <S> somewhere <S> so it doesnt get pulled off.
| Then use a lot of rosin-based flux and regular lead-based solder to make the vias solderable.
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How to calibrate current transformer such as SCT-013 without having to actually generate the actual current? I have a current transformer like this one but is rated at lower Ampere rating (30A/1V). I would like verify how accurate its reading is. I will be using it read the current consumption on a 220V mains line Is there some way to verify its reading without actually having to generate the Ampere on a line. I do not own an AC bench supply that can generate that amount of power. Even the lowest rating that I can buy for this transformer is 10A that is still a lot of power to generate. I hope that going further after changing the burden resistor built in, I will be able to confirm the reading is accurate to at least 1%. All equipment I have access to is a DC bench power supply, multimeter, digital oscilloscope, and a function generator. Is there some way to validate its reading? <Q> All equipment i have access to is a DC bench power supply, Digital Oscilloscope, and a function generator. <S> Use your signal generator set at 50 Hz or 60 Hz and, through the primary hole wind ten turns (or more) but note the number. <S> Drive current through those ten turns and monitor that current using your multimeter. <S> If you can get 100 mA RMS through 20 turns, that's equivalent to 2 A RMS in a single primary wire. <S> If that isn't enough, wind more primary turns (say 100) or, build a little DC power amplifier to produce more current into the home-spun winding. <A> If you have only one multimeter, check the current with it first, and then measure the AC voltage from this transformer. <S> Since you don't mention a multimeter, you could use a very small value resistor on the neutral side (like 0.1Ω), measure the AC voltage across it, leave it in the circuit and measure the AC voltage from the current transformer. <S> Your AC voltage measurement would be done with the oscilloscope. <S> Just remember that the AC voltage RMS value is 0.707 times the peak voltage. <A> There are many sources in the average home ie heater, stove, water heater, many lights turned on... <S> your real problem seems to be that you need an excellent quality ampmeter which reads much better than 1% to be used as a reference. <S> Might I suggest you settle on the accuracy of a meter your budget will allow. <S> Please keep in mind frequency and <S> load impedance will affect linearity of the readings. <S> Modifying the resistor will only change calibration of output for one value, the other readings may become worse.
| The simplest thing I can think of is to place it over a wire going to your electric heater or an electric stove and verify the current with an AC ammeter.
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Two secondary flyback transformer coils in parallel This feels like a simple question, but my google game is letting me down: The below flyback transformer has two secondary coils that are meant to be connected in parallel, as per the below image, and a turns ratio of 1:1:1:0.5 (pri:sec:sec:aux). If I link the two coils in parallel, is the turns ratio still 1:1? Has it changed to 1:0.5 as though adding inductors in parallel? Is it 1:2 as though you doubled the output of a single coil? What would happen if I connected them in series instead? I'm new to designing any kind of power supply and am struggling with some of these concepts, although I'm familiar with the general idea of how a flyback converter works. <Q> What you said are two ways to reduce the leakage inductance in a flyback converter. <S> I often use this technique. <S> Connected in parallel - It will be 1:1, but the current will be double. <A> Think of two parallel wires as a single wire of twice the area. <S> The turns ratio is now clearly still 1:1. <S> Connected in series, the turns ratio is 1:2. <A> Has it changed to 1:0.5 as though adding inductors in parallel? <S> Two identical inductors ( \$L\$ ) that are not sharing a common magnetic field have a net inductance of \$L/2\$ when connected in parallel. <S> Inside a transformer, the two windings do share the same magnetic field hence they produce (near) identical terminal voltages and can be paralleled to increase current drive capability just like you can with paralleling identical batteries. <S> What would happen if I connected them in series instead? <S> Just like with batteries connected in series, the voltage doubles.
| Connected in series - It will be 1:2, you obtain double turns.
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Microcontroller - potential damage to IO pin I'm revisiting a design I made. The circuit is powered by 24V with polarity protection. I placed a power loss detection before the polarity protection: In case the power pins were replaced, the 3v3 zener diode will have forward voltage of ~1V. will this pin go through R32 to the "unpowered microcontroller" and destroy its IO pin? The absolute maximum is -0.3V in the mcu: It's only 100uA limited by R32, but I can't be sure. Is there such detailed information about the IO pin? <Q> Usually the datasheets, in the absmax section, give an information like "-0.3 V or -10 mA, whichever occurs first", but unfortunately not in this case; therefore a proper answer is not possible, but I can give a bit of insight. <S> The -0.3V limit is there because for each pin there is a diode from ground to the pin, and from the pin to vdd. <S> With 10 kOhm in series, this is a non issue; I would expect almost any IC to be able to handle at least a few mA in the protection diodes, and here you have a 10 kOhm resistor in series. <S> You can even bump the resistor up to 100 kOhm, if I understand your circuit correctly. <S> To sum it up, with your series resistor you are not violating the absmax because the diode will conduct and block the voltage around -0.3 V, probably even less. <S> With 100 uA you can sleep safe. <A> According to the datasheet this is indeed a violation of the absolute maximum rating, with the caveat that the current not being specified. <S> It might be okay, but not guaranteed. <S> These are available at a wide variety of ‘trigger’ voltages and can be simpler/cheaper than doing your own comparator circuit. <A> Just move the voltage detection to after the polarity protection mosfet and your MCU won't see the negative voltage. <S> Also, for detecting voltage, use a resistor divider circuit that gives you 2.5V or whatever your threshold is at 15V in (so, 56 kOhm to 10 kOhm or so, or even 560k to 100k to save power). <S> Then parallel the zener with the lower leg for protection; far under 3.3V it won't conduct a lot and won't impact your UV detection significantly. <A> Insert (add) a diode after R32 for any sensing of voltages <S> this is a standard protectionThis offer reverse protection to input pins however any <S> I /O pins will have two diodes one to the Vcc in forward bias to take care inputs exceeding the VccOther to Gnd to take care of reverse voltages occurring on in put pi <A> As others have said the current limiting R32 means it will all be ok, but you could move the UVLO tap to the right of the FET and then it would definitely be better. <S> But ... <S> In your schematic you have a comment about the UVLO circuit being wrong and needing a comparator. <S> Take a look at shunt regulators such as the tlv431. <S> These three pin devices can be used as comparator with integrated reference and driven straight from pretty much any supply with only a series resistor provided you can protect against reverse polarity. <A> That will definitely overcurrent the cpu pin. <S> I just use a schottky diode right on the v+ input to prevent against reverse polarity. <S> FSV10100v might work depending on your current requirements. <S> You can then spin off a small separate circuit if the supply is reversed to turn on some user indication (led?). <A> Have a look for injection current in the data sheet. <S> Also fix the divider, <S> set it <S> so the zener doesn't conduct under normal circumstances. <S> Failing this ask their web site. <S> But I agree keep the current into the micro to less than 1-2 mA is usually good enough. <S> An alternative is to use a bas99 as clamp diodes. <S> There was a paper done a long time ago about their use as esd protection. <S> The other part of this is to remember what you are protecting.. <S> micro pin <S> so after the 10k series resistor. <S> Use the 10k as a buffer. <S> Or split the 10k in 2
| I would look into a small power monitor IC which can detect when the input voltage is about to get low and signal your MCU.
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Which method or component is best to regulate 42 to 3.3 V? I am using a 36 V battey pack (10s 2.5 A each cell 3.6V.) Its maximum voltage is 42 V. I am using a STM32 microcontroller whose input is 3.3 V. I am designing a PCB in which the power supply is drawn from the battery pack (42 V.) I would like to power the STM32 microcontroller from the battery pack power supply. I would like to regulate 42 to 3.3 V so that I can power my STM32 microcontroller. Which method is best to convert it either using regulator or buck converter based on the pcb designing? Battery Specification: 10s1p Battery pack nominal voltage: 36 V Battery pack maximum voltage: 42 V Battery pack current: 2500 mAh Microcontroller specfication: STM32F401RCT6 Supply voltage: 3.3 to 3.6 V Supply current: 100 to 160 mA Which method or component is best to regulate 42 V so that I can power my microcontroller with 3.3 V? <Q> I quite like the LT8631 (1 amp at 3.3 volts and <S> input voltage range up to 70 volts) <S> : - <S> Or maybe the LT8630: - Or possibly the LTC7138: <S> - Or choose your own buck converter using the Analog Device's selection tool . <A> A buck converter is the best approach A linear regulator circuit in the worst case is going to have to drop 42V down to 3.6V <S> = 38.4V. <S> At 160mA the power dissipated by the regulator will be 6.14W. <S> On the other hand a buck convertor will drop voltage to the required level with possibly 90% efficiency (less energy wasted as heat) and likely reduce the current consumption to 3.6/42*160mA = <S> 13.7mA allowing your battery to power the circuit 11.67 times as long. <S> https://www.monolithicpower.com/en/products/dc-dc-power-conversion/switching-regulators/step-down-buck/converters/vin-max-48v/mp2492.html or similar could do this. <A> The best method is to use a buck-topology switching regulator. <S> For example, Analog Devices LT3437 has an reference circuit/example for your application <A> A buck converter is the right solution, and many of the chips suggested in other answers are easy to use as the datasheets already have designs that you can just use. <S> But it does still require a PCB layout and other design work. <S> For example SRH05S3V3 can provide 3.3V at up to 500 mA from an input voltage anywhere between 9V and 72V. <S> It has three pins only: Vin, Vout, GND. <S> Even the input and output capacitors are optional, as the module has small capacitors integrated. <A> If you're making a bunch of them, in particular, and are concerned about cost, you might consider the XL Semi XL7015 , which costs only about 25 cents in 100's, <S> about 1/20 the cost of the LTC boutique parts. <S> Typical efficiency is only about 70% with 36V in and 160mA out vs 85% for the LTC part, so there is a cost in battery consumption (about 0.08W more loss). <S> There's also less voltage margin <S> and it's a physically larger TO-252-5 part. <S> On the plus side, it's capable of a lot more output current. <A> I think you can find off the shelf BECs for 10S - or even higher. <S> If you plan on building your own thingy, you must account for the voltage drop which appears with the battery's depletion. <S> BECs are specially built for this purpose, hence the name (battery eliminating circuit). <A> As your input battery pack is 36V then linear voltage regulator is out of question (although it is possible <S> due low consumption amperage). <S> With linear voltage regulator power dissipation would be (Vin-Vout)*Iout. <S> For example if consumption amperage would be 1A then (36-3.3)*1=32.7W of wasted power. <S> The best option would be a buck converter (switching power supply) which monitors voltage on output capacitor and recharges it when it will be required. <S> Efficiency of such power converters is quite high. <S> If you are looking at very inexpensive solution then check LM2596 -- datasheet . <S> Power consumption of MCU is very small and there plenty of other solutions on the market.
| The easiest solution is to use a premade buck converter module.
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What will happen if you supply a higher voltage to a buck converter? The title says enough, but let’s give a real life example too. Take one of those LM2596 35V boards. What will happen to it if it gets 50V on the input? Most likely it will burn, but will this voltage propagate to the load? Will it get damaged too? Is there any kind of protection? Does it make a difference if the over-voltage is a spike or continuous? <Q> It's in the datasheet: <S> Stresses beyond those listed under Absolute Maximum Ratings may cause permanent damage to the device. <S> These are stress ratings only, which do not imply functional operation of the device at these or any other conditions beyond those indicated under Recommended Operating Conditions. <S> Exposure to absolute-maximum-rated conditions for extended periods may affect device reliability. <S> If you check the functional block diagram, an overvoltage will most likely damage the transistors, failing short, so, shorting the IC's input to the output and present the input voltage across the catch diode, the output capacitor and the load. <S> Typically, the input capacitor is rated 1.5 to 2 times the intended input voltage, so, it most likely will survive. <S> This input capacitor may also helps shorting the overvoltage when it is a (short) spike. <S> In section 9.2.1.2.5 it is advised to use a low ESR capacitor for this reason. <S> A low ESR aluminum or tantalum bypass capacitor is required between the input pin and ground pin to prevent large voltage transients from appearing at the input. <S> This capacitor must be placed close to the IC using short leads. <A> As power semiconductors often fail short-circuit, there is every danger it can pass the full input voltage to the output. <S> After an overvoltage spike destroys the power switch, normal input voltage may be destructive to the output. <S> Sometimes you will see explicit protection, e.g. a crowbar or large zener across the output combined with a fuse on the input. <S> A cheap buck module may omit this... <A> Well, the LM2596 will burn out since its internals were not desigend to withstand much more than 35V being applied. <S> Since semiconductors normally fail short, this would either then pass the 50V to the input or short to GND. <S> I would not count on protection since why would you ever apply 50V to a 35V device?
| The catch diode may get damaged (when its rating is less than 50V), the output capacitor will likely blow/explode when its rating is too low.
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Can I somehow use a neon bulb as a switch? I’m trying to create a timing light. Due to the lockdown I can’t get any other parts. I have a neon bulb which does light up when I hold it next to the spark plug lead but it is too dim to be useful. Is there any way I could somehow use this as a switch so I could light up another LED? <Q> You can try this. <S> 27V is 3x 9V battery in series. <S> If the LED stays on after the neon triggers you may have to play with the voltage or R1 value. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Add in LEDs in series, usig two parallel LEDs(back-to-back) <S> so one LED conducts and the other protect the active LED from Vr <S> >-5V <S> Don't overdo the brightness and put a tube over the end so the wirebond doesn't fuse open but is visible in high ambient light. <S> I presume the neon tube has a current limiter. <A> You are seeing capacitive coupling from the 25,000 volts in the spark plug wire, to the neon bulb lead, thru the neon gas and out to your body that is leaning against the car fender. <S> You don't need the neon bulb. <S> Just tape one lead of a 1MegOhm resistor to the plug wire. <S> Then run the other resistor lead over to clamp diodes, to clamp at gnd and +3.3v. <S> Use that clamped voltage pulse to toggle your switch behavior. <S> how much current will you get? <S> assume the resistor lead is 1mm by 3cm, spaced 3mm from the spark plug wire, with plug wire relative dielectric constant of 5. <S> The capacitance is C = <S> Eo <S> * Er * area/distance <S> C = 5 <S> * 9e-12 farad/meter * area/distance (area/distance = <S> (1mm <S> * 3cm)/3mm == <S> 0.01er <S> C = <S> 45e-12 <S> farad.meter * 0.01 meter = <S> 45e-14 farad = <S> = <S> 0.5e-12 <S> farad C = <S> 0.5 picofarad Assume <S> the spark voltage rises 25,000 volts in 25 microseconds, This is 1 billion volts per second. <S> What is the current? <S> ignoring the 1Megohm, <S> I = <S> C <S> * dV/dT = <S> 0.5e-12 <S> * 1e+9 = 0.5e-3 = <S> 500 microamps. <S> which imposes 500 volts across the 1Megohm. <S> So use 1/2 watt, for the longer body <S> Now you place 10K ohm at the clamped node, and enjoy your 5 volt trigger pulse.
| The gas in the neon lamp is ionized by the high voltage which causes current to flow, a bit like an old-school thyratron or xenon flash tube.
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Does phone battery last longer if I charge often or less frequently? Which practice will help my Android phone lithium battery last longer? Charge often. Whenever phone battery drops to, say, 70%, charge it back to full 100%. Charge less frequently. Charge phone to full 100% after it has dropped to between 0% to 20%. An explanation for the choice would be helpful. <Q> The number of cycles and sitting too high or low is what matters. <S> The number of times you decide to plug in the battery doesn't mean anything. <S> First is worse not because you plug it in more (you're not charging it more). <S> It is worse because you are maintaining the voltage too high. <A> I'll disagree with DKNguyen even though his answers are usually good. <S> A major component of LiIon battery failure is mechanical cycling of the cell as it is charged/discharged. <S> (The LiFePO4 battery has a greater cycle life due to the solid Olivine structure which does not change shape/size as Li ions are passed to and fro. <S> So recharging from 70% charged is better. <S> Best of all is not to approach either extreme too closely. <S> About 20% to 80% is good but harder to manage. <S> Stopping when Vbat reaches 4.2V with no constant voltage mode is also good but also hard to achieve with a std charger. <A> Use your Li-ion battery in 20% - 80% SoC range at room temperature. <S> The performance of a battery is measured via its delivered capacity (Ah), which is the best indicator out of many other parameters, the capacity decides the run-time of the application in use. <S> The SoC range is not 0% - 100% above because a partial charge and discharge reduces stress on the battery material and prolongs its life. <S> Alos keeping the battery at < 10% for longer duration can lead to a faster capacity (Ah) drop. <S> As most consumers don't bother with these things and these things are hard to control, manufacturers use higher capacity rated batteries and claim lower than that to maintain user satisfaction and avoid warranty claims in some cases. <S> For more detailed explanation and tested data go to this link
| In this case the greater depth of cycle of the infrequent charge method places greater mechanical stress on the battery cell structure.
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What is meant by average power in AC circuits? I read that "The average power is the power averaged over a cycle or number of cycles." I know that the average value of power for both active and reactive power is a constant number. Buy I am confused that whether the cycle here means the complete cycles or cycles? If power is averageed not "over a cycle or number of cycles" but supposes it is averaged over a quatter of cycle or two third of a cycle, will we get the same exact constant for the average power which we get for the average power over a complete cycle or cycles? <Q> Look at these various scenarios for the power waveform obtained when various phase angles are involved: - The power waveform resulting from the multiplication of voltage and current sine waveforms is at twice the frequency of either voltage or current and is also sinusoidal. <S> Because it's at twice the frequency, if you averaged power over one half of the voltage or current period it still works out to be accurate. <S> However, you should be able to see that if it is averaged over a different period such as a quarter or two-thirds (of voltage/current period) then you can get a significant error. <S> But, for instance, if you began averaging at the trough of the power waveform and finished at the peak (equivalent to a quarter voltage or current period) then you would get the right answer. <A> Actually we consider only complete cycles. <S> If you consider half or quarter of cycle then the power that you obtain will be different. <S> Consider a sinusoidal power wave its average for one complete cycle is either a constant or zero but will have different values depending upon the quarter cycle or portion of cycle you take. <A> Average power is the average of all the instantaneous powers calculated by instantaneous V x I. <S> If the waveforms are periodic and you take the average over a time interval increasingly longer than the period, it approaches the average over one period because the error introduced by landing randomly somewhere in the last cycle is an increasingly smaller proportion of the summation. <S> If power is averageed not "over a cycle or number of cycles" but supposes it is averaged over a quatter of cycle or two third of a cycle, will we get the same exact constant for the average power which we get for the average power over a complete cycle or cycles? <S> You can't assume this due to phase shifts between the V and I. <S> The average over a single quarter, half, and full cycle of a full rectified sine wave are the same, but this has nothing to do with power. <A> In electronic electricity meters the energy is typically averaged over one second. <S> For each second: <S> the direction of the power is determined the active energy is added to +E or -E <S> the reactive energy is added to the correct quadrant: F1, F4 or F2, F3 <S> For each second, only one register of +E, <S> -E <S> and one of F1, F2, F3, F4 is active. <S> Only from one second to another the direction of active- and reactive- power (or energy) might change.
| The average is taken over some interval of time.
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What connector to use when using 5v 15A I recently purchased an APA102 LED light stip (60leds/m) and read that it requires 18watt/meter, I am using approximately 4.5 meters so will roughly need to provide 81w. I did, however, struggle to find a dual power supply that could provide this, so I settled on an RD-125A power supply that can provide up to 15A at 5v, and am planning on reducing the brightness of the LEDs to 75%, to account for this. I was then hoping to have a wire connecting the LED strip to a box containing the rest of the circuitry and the power supply. So I could detach the box, I wanted to have a connector that would allow me to plug the LED light strip into said box, and then from within the box, the connector would connect to the power supply and a raspberry pi. My first thought was to use a 4 pole DIN connector and socket, but I wasn't sure if it's safe/advisable to use one when using ~15A, and thus I was hoping to find out if This is safe If not, is there any other connector that would work <Q> TL;DR Distributing 5V is a bad idea. <S> I would highly suggest not trying to distribute 5V @ <S> 15A at all. <S> The voltage drop at 15A is going to be considerable unless you use very thick wire. <S> If you think about it, if your wiring loom and connectors had a total resistance of only ~0.3 Ω, your entire voltage would be dropped across it. <S> Even if it was only 0.1 Ω you are dropping a significant portion of your supply voltage. <S> Instead distribute a 24V supply. <S> This would drop your supply current down to around 3.2A or so, which is much more reasonable. <S> It will also be easier to find a supply capable of handling the current. <S> If your LED strips require 5V, then for each strip add a DC-DC convert to drop the supply down to 5V locally. <S> By having a converter per strip, the output current for any DC-DC is lower than having just one, and the distance over which you are tranferring it is vastly reduced. <A> I don't know this particular connector, but its data sheet could answer the maximum current question. <S> If you don't have one, at least give a link or something (like where you got it). <S> I think your main problem will be the wiring. <S> High current means high voltage drop, but you start only from 5V. <S> So to avoid this, you'll need to bring separate (thick) wire along with your LED strip, and feed it several places. <S> If you can find the datasheet, and your isn't big enough, I'd look around in RC modeling forums, they use low (11.1V) voltages with high currents (10s of Amps). <S> e.g. XT30, XT60 are reasonably cheap, have reverse polarity protection, and widely available. <A> "DIN connector" is a broad term. <S> Please edit your post to include a part number or vendor link <S> so we can evaluate the part you are considering. <S> Will you be plugging / unplugging things while power is on (hot plugging) or only when power is off? <S> Hot plugging is much harder on the contacts, and you can't make up the current capacity you need by tying together multiple pins in parallel. <S> Anderson is well known for their high-power DC connectors. <S> http://www.andersonpower.com <A> There's some similar looking connectors that are not DIN, which are rated for higer currents but not 15A <S> something intended for use with powerful batteries like like XT30 may be a bettter choice.
| You will struggle to find a connector with sufficient current rating that isn't excessively large and expensive. Use a higher voltage supply (e.g. 24V) with a local DC-DC convert for each strip. I'm not aware of any "DIN" connectors that age good for 15A, these connectors are signal connectors typically only good for about 1A.
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In below picture, is ammeter AC or DC? Is there direction for connection with load in series? If so what if I connect in wrong direction? This is front view of analog ammeter. There are 2 terminals with nut and bolt behind it but nothing marked neither plus nor minus. <Q> It's a moving iron meter hence this symbol: - <S> Taken from here <S> Moving iron meters are used for DC AND AC so, the meter needle will only move in a clockwise direction irrespective of current polarity. <S> If it used a magnet it would be polarized and with AC, the needle would barely move (because the needle would try and move clockwise and counter-clockwise at the AC frequency). <S> It's also worth noting that the symbol that looks like an upside-down "T" next to the moving iron symbol also has a meaning <S> as does the 5 sided star with a 2 inside: - Extra picture from here . <A> There are three clues: <S> The scale isn't linear so that implies that it is a moving iron meter. <S> The ≂ symbol indicates that it can handle DC and AC. <S> (The '-' indicates DC and the '~' AC. <S> The terminals are not labelled. <S> If polarity mattered they would be. <S> The meter can be wired either way. <S> The moving iron works on the principle that the iron place near the magnet attracts towards it. <S> The force of attraction depends on the strength of the magnetic field and not on its polarity. <S> See the linked article for a list of advantages and disadvantages of moving-coil meters. <S> They are favoured on cheap electrical devices such as 12 V battery chargers, etc., where they can be considered as a current indicator. <S> Calibration in many cases is going to be terrible. <A>
| The symbol, circled in red, indicates it's an AC/DC meter.
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The right way to have amplified speakers that mute when you plug in earphones I am designing a handheld device that has an I2S DAC, small speakers, amplifiers (~2 watts), and an earphone jack. I'd love advice on the best way to hook these together. My goal is high quality audio for a broad number of configurations, including different types of earphones, without going nuts on complexity and cost. Options I have explored: Have the DAC output to an earphone jack with internal switches, when the earphones are present audio plays through them directly, otherwise the signal passes through the earphone jack switches to the amplifiers and then speakers. This is what I've been using so far. It works, but is it really okay to drive earphones from the DAC output directly? The speaker amps are always active which is wasteful and pick up some noise. Also potentially noisy during insertion. The DAC feeds the amps directly, which connect to the earphones, or the speakers if earphones are not plugged in. Am I going to blow the earphones (or the user's eardrums) this way, or are earphones high enough impedance that the power delivered to them will be low? Will the brief shorting that occurs during insertion kill the amps? (I guess that depends on the amp) Implement insertion detection (I've seen a few ways to do this with some pros and cons) and route audio to either the speaker amplifier, or the earphones (with or without their own amplifier). Would I need to switch the signal, or would it be okay to route the signal to both in parallel and simply disable the speaker amps when earphone insertion is detected? Is there an integrated solution that can do this for me? Thanks! Relevant: Mutually exclusive headphone and speaker How does the phone detect if 3.5 mm jack circuit is closed? How (not) to leave class-D amplifier's input hanging <Q> Since the PAM8301AAF output is for an 8Ω speaker, the sum of R1 and R2 should be 8Ω. <S> The choice of R1 & R2 values would depend on the level of attenuation required. <S> 1W resistors would suffice. <S> The schematic shown is for one channel. <S> With the PAM8301AAF outputs to be kept isolated between themselves and ground, a standard headphone plug/jack cannot be used. <S> Separate mono headphone plugs/jacks, with separate cables for each headphone element, would be required for the isolation. <A> is it really okay to drive earphones from the DAC output directly? <S> Not really. <S> The <S> PCM5102 has a minimum rated load impedance of 1k Ω. <S> Most earphones are around 32 Ω, which is too much load for the DAC. <S> You should use a headphone amplifier to buffer the DAC signal. <S> are earphones high enough impedance that the power delivered to them will be low? <S> It will be lower due to the higher impedance, but possibly not low enough. <S> Headphone level can be reduced with an 'L' pad. <S> Another possible problem is that the PAM8301 's output is 'BTL' <S> (Bridge Tied Load) <S> so both sides of the speaker are driven and there is no common ground. <S> You could connect each earphone to one side only, through a capacitor to block DC voltage. <S> This would reduce the level by 6 dB, which might be enough make the earphone level acceptable. <S> However the PAM8301 is not specified for operation in this mode <S> so I don't know how how well it will work. <S> If you want high quality audio output via the headphone jack then it might be better to use a separate (analog) headphone amp which has lower distortion and noise, and no EMI concerns. <S> Whichever way you do it, stereo headphone jacks are available with auxiliary switches that can be used to switch between speakers and headphones and/or disable the speaker amp. <S> The PAM8301 has an active low 'Shutdown' input that just needs to be grounded to turn it off. <S> So the circuit could look like this:- simulate this circuit – <S> Schematic created using CircuitLab <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) <S> The PWM class-D amplifier outputs do not share either of their output pins with ground. <S> (b) <S> Typical 3-pole jack headphone wiring. <S> To answer your questions: Have the DAC output to an earphone jack with internal switches, when the earphones are present audio plays through them directly, otherwise the signal passes through the earphone jack switches to the amplifiers and then speakers. <S> This is what I've been using so far. <S> It works, but is it really okay to drive earphones from the DAC output directly? <S> The speaker amps are always active which is wasteful and pick up some noise. <S> Also potentially noisy during insertion. <S> This is the best solution if you are not overloading the DAC outputs. <S> The disconnected amplifier inputs should be shorted to ground while on headphones. <S> If the jack socket does not have suitable switches then try adding a 1 kΩ resistor from each amplifier input to ground. <S> The DAC feeds the amps directly, which connect to the earphones, or the speakers if earphones are not plugged in. <S> Am I going to blow the earphones (or the user's eardrums) this way, or are earphones high enough impedance that the power delivered to them will be low? <S> Will the brief shorting that occurs during insertion kill the amps? <S> (I guess that depends on the amp) <S> Yes, that would be a risk and attenuation would be required for protection of the user, protection of the headphones and to maintain an acceptable signal to noise ratio. <S> Implement insertion detection (I've seen a few ways to do this with some pros and cons) and route audio to either the speaker amplifier, or the earphones (with or without their own amplifier). <S> Would I need to switch the signal, or would it be okay to route the signal to both in parallel and simply disable the speaker amps when earphone insertion is detected? <S> Route to both should be fine. <S> Is there an integrated solution that can do this for me?
| The headphones may be connected to the amplifiers, through attenuators, with a DPDT switch to select speakers or headphones.
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Why is a transmitter dipole's ends the max/min voltage instead of the middle? If the AC voltage source is connected to the middle of the dipole, I'm confused on why the edges have the max/min voltage instead of the middle. In my mind, the middle of the dipole is closer to the voltage source, and there is even some resistance between the middle of the dipole and the end, which would imply a voltage drop, making the dipole's end a lower voltage than the middle. For a receiver, I understand why the ends have the max/min voltage (based on how the wave hits the dipole). However, I'm totally lost when it comes to a transmitter. Any help would be greatly appreciated. <Q> On a dipole there is a spread of current and voltage, in very simple words: at the end of the dipole, no current is built up b/c <S> it is an open end, so there the voltage rise to a maximum. <S> The impedance at the ends of the dipole is very high compared to the impedance at the feed point. <S> I think, the wikipedia Web site explains this quite well. <A> Current and voltage are not especially good to present what happens in an antenna. <S> The actual thing that happens in an antenna which is connected to a transmitter is that a radiowave comes along the feeding cable (it's in the space between the wires, in coaxial cable it's between the shield and the center wire). <S> In antenna the fields spread to wider area and if the antenna is properly designed a substantial part continues to the free space around the antenna as wanted. <S> The rest is dissipated in material losses or reflected back towards the transmitter. <S> The current which the fields of the wave induce to the metal cannot jump out of metal in usual power levels, so the end of the metal surface is a discontinuity which causes reflection. <S> From the theory of transmission lines we should remember that when a wave meets a strict "no current possible any more" border the electric fields of the arriving and reflecting waves are in every moment to the same direction at that border, so the sum field of the arriving and reflecting waves is as strong as possible with that power level. <S> With transmission lines we usually call the RMS value of the sum field "standing wave" <S> It's maximum is at the unconnected end of the line. <S> On the surface of an antenna less far from the feeding point the fields of the different direction propagating waves partially cancel each other (but do not disturb each other by any means) so the sum field can be weaker. <S> At the same time when waves propagate along the surface of the antenna they also lose energy to the space as radiated wave. <S> That radiation is attempted to be maximized in antenna designs, in transmission lines it's wanted to be kept zero. <S> Voltage is a coarse measure for electric field, all vector field structure is omitted, but the proportionality to electric field strength is still there. <S> Thus also voltage between the ends of a dipole antenna can be higher than at the feeding point. <A> If the AC voltage source is connected to the middle of the dipole, I'm confused on why the edges have the max/min voltage instead of the middle. <S> Basically, an antenna converts the operating impedance of the electrical feed-circuit (circa 50 ohms) to that of free space. <S> Given that free space has an impedance of 377 ohms ( \$120\pi\$ ), the voltage to current ratio at the extremes of the antenna have to be greater than at the electrical feed-point (circa 50 ohms).
| The key thing is that there's a propagating wave around the antenna and some reflection happens at the far end from the feeding point.
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how to select an LDO regulator to get 3.3 v 200 mA of output from 5v and 250 mA of Input? I would like to power my stm32 microcontroller the supply voltage is 3.3v and 160 mA of current. So i would like to regulate my buck controller output of 5v to 3.3v to power my microcontroller. My input to LDO regulator: input voltage : 5v input current : 250 mA Expected output: output voltage : 3.3v output current : 200 mA ( to power my STM32 microcontroller) Iam planning to use Texas instrument TPS730 LDO regulator to get 3.3 v as output. TPS730 power consuption: (Vin-Vout) *Iout = (5v-3.3v)*200 mA = 0.34 W In datasheet Maximum junction temperature is given as 150 degree C and Rthja as 225.1 degree C/W so 225.1 *0.34 = 76.54 (degree C) + 25 (room temperature) = 101.54 degree C so i think it is safe to use based on Power consuption. if iam wrong correct me whether it is safe or not. My doubts: i need an out put voltage of 3.3v and 200 mA of current but in datasheet it is given like What does this means . only if my output voltage is 5v then only i can able to get 200 mA as output current.If it is like this i cannot use this regulator because the maximum output voltage i need is 3.3 v so can i able to get 200 mA of output current with 3.3 v of output voltage if my understanding is wrong please correct me. or can i able to get 200 mA of current by having 3.3 v as output voltgae and What does this means i am confused.Is it telling like only if my input voltage is 5.5 v then only i can able to get 200 mA. if it is like this i cant use this regulator because my input voltage is 5v .Is it like that or can i use this regulator with 5v of input to get 200 mA of output Iam really confused after a large number of search i am asking this question in stack. Can i able to use this regulator by having 5v as input and i need 3.3 v output voltage and 200 mA of output current. Please find the Datasheet of Texas instruments TPS730 enter link description here In the above datasheet in Page 13 it has given the adjustable output voltage . Basically can I use this regulator based on my specifications ,If no is there any regulator based on my specs with decent power consuption. <Q> The regulator output current range is from 0 to 200mA. <S> So you can use it for a 200mA max load. <S> 160mA is within range of 0 and 200mA. <S> The regulator output voltage range can be set between Vfb and 5.5-Vdo. <S> Since 3.3V is included in the range, you can set it to output 3.3V. <A> Maximum recommended junction temperature is 125°C <S> so you're pretty marginal at 340mW, depending on the layout, number of layers, size of PCB and so on. <S> If the temperature inside the enclosure gets above 45-55°C it would be easy to exceed the 125°C. <S> Check that your 200mA is a real number. <S> If it's more like 100mA then you're probably fine. <S> The power dissipation of any linear regulator will be the same, plus a bit for internal use, so really your concern is the specifications and the package and how you are going to design the device to safely dissipate that amount of power. <S> Note that you don't actually require an "LDO" regulator, since 5V - 3.3V is 1.7V. An LM1117 or similar AMS1117-3.3 in SOT-223 or TO-252 package might be a better choice. <S> If you don't want to, or can't, do all the testing and simulation for a relatively accurate thermal design <S> it's better to be err on the safe side and use a bigger package with some copper on the tab. <S> As with any regulator of this type pay careful attention to the value and type <S> (eg. tantalum) of output capacitor. <S> The older LM1117 type can use a ceramic output capacitor safely if you add a series resistor. <A> Each line in the datasheet table is separate. <S> If your input voltage is between \$V_{FB}\$ and 5 V you are not violating the input voltage specification. <S> If your output current is between 0 and 200 mA <S> you are not violating the output current specification. <S> As another answer pointed out, there is also a thermal power consumption specification that you need to consider, and that might be harder to meet than these specs.
| Your previous question already covers that the regulator can have 3.3V output with 5V input quite easily because the difference is much larger than minimum required.
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Understanding the resistor method of slowing server fan speeds Researching how to compensate for the overly aggressive fan speed control in our HP DL280e Gen 8 server † , I have seen several recommendations for how to slow the system fans using a resistor in series with the 12v power input, for instance over on server fault . What I don't understand is how their suggested solutions work. 1W 10Ω resistor in series with a 12V 2.5A fan Naively using ohms law, suggests that at full speed this fan has a resistance of 4.8Ω so adding a 10Ω resistor in series should drop the overall current to 0.81A (9.7W rather than 30W) with 6.6W through / 8.1V across the resistor and 3.2W through / 3.9V across the fan. What I don't understand is how this works without blowing the resistor, putting 6W through a 1W resistor doesn't seem like a great idea. So I wonder what I'm missing. Also, although I can't find a data sheet for the V60E12BS2CB5-08 fan shown, I can't imagine it being happy with only 3.9V across it's 12v line. 10W 11Ω resistor network in series with a 12V 2.3A fan Again, ohms law suggests that at full speed this fan has a resistance of 5.2Ω so adding an 11Ω resistor in series should drop the overall current to 0.74A (8.9W rather than 28W) with 6W through / 8.1V across the resistor and 2.9W through / 3.852V across the fan. The power requirements are now within tolerance, but the spec sheet for the fan ( Delta PFR0612XHE ) says that the operation voltage is 7.0 to 13.2v, so it feels to me unlikely that it would work at 3.852V. 7W 3.7Ω resistor network in series with a 12V 2.3A fan Am I right in thinking that the ideal resistor for minimum speed (7V) with the Delta PFR0612XHE at 100% PWM would be 3.7ohm? At full speed with a 5.2Ω fan, a 3.7Ω resistor in series should, according to my naive assumptions, drop the overall current to 1.35A (16.1W rather than 28W) with 6.75W through / 4.7V across the resistor and 9.45W through / 7.0V across the fan. That's 34% of the original power to the fan. The power requirements and supply voltage would now both be within tolerance. Is my naive understanding even vaguely correct? What about driving at less than 100% PWM? So, if a 100% demand now results in a 34% power draw, I assume that it's power draw would scale proportionately with PWM demand. If the fan is driven at less than 100% PWM, it's effective resistance will go up as the current draw goes down. Re-running the above calculations, I get: Running the fan at 43%, i.e. 1A, it would appear to be 12Ω so adding a 3.7Ω resistor, the overall current would drop to 0.76A (9.2W rather than 28W) with 2W through / 2.8V across the resistor and 7W through / 9.2V across the fan. Thus a demand of 43% results in a 25% power draw. Finally the fan at 15%, i.e. 0.345A, it would appear to be 34.8Ω so adding a 3.7Ω resistor, the overall current would drop to 0.31A (3.8W rather than 28W) with 0.4W through / 1.2V across the resistor and 3.4W through / 10.8V across the fan. Thus a demand of 15% results in a 12% power draw. If this is correct, I assume this is the reason why I've seen complaints that the resistor method results in fans shutting down when the PWM demand gets too high (presumably as the rise in current drops the voltage across the fan to below the 7V lower threshold) and that going no higher than a 3.7Ω resistor would prevent this issue. Other research My specific fans Note that these fans don't have a rotation pulse out like consumer PWM fans, they only have a locked fan signal. This signal isn't even mentioned in the datasheet , but people have found that it is 0v when the fan is spinning, and non zero when not spinning (people ground it to bypass the failed fan detection). The fan takes a 3.3v inverted PWM in, so a constant 0v or 3.3v FG signal will fail safe, with the fans running at 100%, while increased duty cycle on the PWM will slow down the fan. Combined, these mean that the motherboard only knows if a fan is spinning or not, and it assumes that if it is asking for 50% and it is spinning, then it is spinning at 50%. Evidence that naive voltage divider calculations aren't useful. Most of the videos and forum posts about this suggest resistor values without giving any details of the fans they apply to or the precise effects they have. How to make a rack server quieter by putting resistors on the fans , by ElectronicsWizardry explicitly tests the voltages across the van though and mentions them in the the video. Starting with a 12V 1.54A fan, a 22Ω resistor resulted in 4.9v across the fan (but 150°C resistor *8') while 2x 51Ω resistors in parallel (25.5Ω) resulted in 4.7V measured across the fan. These are obviously much higher voltages than the naive voltage divider calculations suggest: 22ohm suggests 3.139V not 4.9V, while 25.5ohm suggests 2.8V not 4.7V. That means that the resiatances of the fan must also be much higher. Summary & specific questions I've done a lot of calculations to try and work out if this is a viable method to reduce the aggressiveness of our server fans, but are my calculations are even vaguely correct?* What am I missing?* And can I calculate the resistance required for a motor given it's specification, assuming it will sometimes need to run at 100% PWM? † At power up, iLO (the embeded management computer) runs the fans at 100% during self test (5x 68dB-A fans are really loud) and then drops them down to 45% PWM when idle (still too loud in an adjacent room). We know that it should be running the fans at around 15% with only two drives populated, but iLo intentionally ramps up fan speeds with non HP drives on Gen8 servers. <Q> The important aspect that you're missing is that fans are not resistors, so a 12V 2A fan will not sink 1A at 6V. <S> In fact, if the motor is able to achieve the target RPM at 6V and has a controller which generates the PWM duty cycle to achieve the target speed, the motor will still consume around 24 W, sinking 4A of current. <S> Are you sure the PWM duty cycle remains constant when a fan gets a series resistor? <S> The same would happen if the motor controller generates the actual duty cycle internally. <S> A motor with 7-13V operation voltage should be able to reach the specified RPM in the specified voltage range. <S> If the voltage goes below 7V, it will not immediately stall but gradually slow down. <S> At 3.5V most 12V fans will still rotate, albeit very slowly, and some may require a push to start. <S> It's really hard to say how much power a particular fan is getting with a series resistor without actual measurements (e.g. the actual voltage drop on the resistor). <S> Regarding 1W resistors dissipating 6W, it's indeed a bad idea, but if the resistor is getting its share of the airflow from the fan, it may well be able to dissipate 6 times its nominal power. <S> PS. <S> I had a very loud fan in an appliance once. <S> Getting one of those tunable DC-DC buck converters worked like a charm: there's no waste heat from it (running the fan at half the supply voltage results in ~10% of losses in a buck converter wrt motor power, vs. 100% losses on a resistor), the dependency between output voltage and RPM is much more linear than in case of a series resistor, there's no stall even at very low voltages, and it can be conveniently tuned with a screwdriver, no re-soldering required. <A> Motor speed is directly proportional to voltage. <S> Air flow is proportional to speed squared. <S> Mechanical power required to move the air is proportional to speed cubed. <S> If you want to reduce the air flow by 20%, 0.8 x full flow, reduce the voltage to the square root of 0.8 <S> = 0.89 <S> x 12 = 10.7 V. <S> The mechanical power required will be the full-speed power multiplied by 0.89 cubed, but the motor may only be 50% efficient, so estimate the electrical power at 22 watts. <S> The required current would be about 2 amps. <S> To reduce the voltage by 1.3 volts would require 0.65 ohms. <S> The power in the resistor would be 2.6 W, but it is not a good idea to run the resistor at its rated power, so use 5 watts. <S> Since this is all estimated, you might look for a 1 or 2 ohm 10 watt variable resistor. <A> The issue is that current taken by the fan won't be proportional to the voltage over it. <S> Therefore, the fan can't be assumed to be a pure resistance.
| If the fan has an RPM output, I would expect the speed controller to detect the slowdown and compensate by increasing the duty cycle, which would explain why larger that expected resistor values are needed.
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What is common mode voltage exactly in current sense amplifier ? how to select a current sense amplifier based on common mode voltage? I understood the average difference between the input and output voltages is called common voltage .But still i have a doubt when it comes to current sense amplifier. i a m confused like is it the voltage drop across the shunt resistor .which we will feed as input to the current sense amplifier. i am using 36V 2500 mAh battery pack its maximum voltage is 42 v when the battery is fully charged and its minimum 30V when the battery is fully discharged. Iam using a 6.6 mohm shunt resistor it will give maximum voltage drop of 100 mv for max 15 Amps of current . is this voltage drop is called as common mode voltage. The current sense amplifier iam plannin to use is AD8207 whose input supply is 3.3v . so when i search in google about choosing the current sense amplifier i saw that when we use High side bidirectional current sensing amplifier the common mode voltage should be equal to the bus voltage Why. and the AD8207 current sense amplifier whose common mode voltage is equal to my nominal battery to load bus voltage that is 36v but it is not equal to my maximum bus voltage 42 v . so can i use AD8205 as current sensing amplifier in my circuit. What is bus voltage is it the nominal voltage of Battery pack given to the load or maximum voltage of battery pack given to the load. So according to my understanding the common mode voltage is the voltage drop between the shunt resistor and it is extremely low compared to the bus voltage and why we need a current sense amplifier whose common mode voltage is equal to the bus voltage . My battery to Load specification: Battery pack nominal voltage : 36vmaximum discharge voltage : 42vminimum discharge voltage : 30vNominal current : 2.5 Amaximum current that can be used in the bus (between battery to load): 15 Ashunt resistor value based on 100 mv dropout voltage : 6.6 mohm What is exactly a common mode voltage? and based on my bus voltage and my common mode voltage.Can i use AD8207 in my application. based on above battery to load specification can i use AD807 current sensing amplifier with 3.3 v of input voltage and 20 V/V of gain. Please find the attached datasheet of AD8207 : enter link description here <Q> For an amplifier, the input common mode voltage range is the range of inputs across which the amplifier will operate as advertised in the datasheet. <S> It is also important not to exceed the absolute maximum ratings . <S> Exceeding the common mode range means that the amplifier will not meet it's datasheet specifications and can have very odd effects such as output phase reversal (common in older JFET input amplifiers). <S> Some devices specify what will happen when operated outside the common mode range, but most do not; for a high side current sense amplifier that is powered from the same supply as being sensed requires a common mode range that goes to the positive power rail. <S> The AD8207 is specifically designed for this type of application; the common mode range exceeds the power rails (both positive and negative) by a significant margin, so you can use this device in your application if you power it from 5V ; <S> the maximum common mode continuous when powered from 3.3V is only 35V for proper operation which is lower than your application is sensing. <S> Trying to use this device to sense beyond 35V when powered from 3.3V will not damage the device (according to the absolute maximum ratings) but the operation of the part is not guaranteed . <A> Your understanding is backwards. <S> The common mode voltage is the average of the two inputs. <S> The differential voltage is the difference between them. <S> If you read Peter's detailed answer you will see that it may not be a good idea to use this amp unless you are able to power it from 5V instead of the 3V you show in your sketch. <S> Also I want to caution you that 15 Amps will cause a 6.6mOhm shunt to dissipate around 1.5W. <S> So if you will have 15 Amps, you may want to choose a slightly smaller shunt resistor, like 4mOhm or 2mOhm. <A> What is exactly a common mode voltage? <S> In your situation it is helped if we put numbers on your diagram: - The common mode voltage when 10 amps are flowing through the 6.6 milli ohm resistor is: - $$\dfrac{36 + 35.934}{2} = <S> 35.967 \text{ volts}$$ I understood the average difference between the input and outputvoltages is called common voltage <S> No, it's the average voltage of both inputs - it's got nothing to do with "average difference" - <S> that term makes no sense. <S> No, the specified common-mode input voltage should be greater than the maximum possible bus voltage (and preferably, by some margin). <S> So according to my understanding the common mode voltage is thevoltage drop between the shunt resistor and it is extremely lowcompared to the bus voltage and why we need a current sense amplifierwhose <S> common mode voltage is equal to the bus voltage <S> No, it's the voltage that is common to both inputs with respect to ground or 0 volts.
| when we use High side bidirectional current sensing amplifier thecommon mode voltage should be equal to the bus voltage
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Part information from schematic symbol I don't understand what this part is from my keyboard PCB design. Someone else worked on the design and I can't figure out what it is.Please help! Here's another image including other parts of the design. <Q> The actual connector depends on the footprint of P1. <S> Most programming connectors use set of header pins, either male or female. <S> Male is most likely. <S> As we already have a male and female, I'll add another type: a shrouded connector: <A> It is a 6-pin connector for the programmer. <S> Think of P for 'plug' or 'pin'. <S> Figure 1. <S> A 6-pin header connector. <S> Tip: Crop your photos. <A> It's a 6 pin connector. <S> From the text below it, it's for connecting a programming cable - <S> presumably this is some kind of custom keyboard where you can change the firmware. <S> Lacking any other information, I'd assume it is something like a pin header. <S> From the wikipedia page: It might not be a pin header, though. <S> There are all kinds of connectors out there. <S> If you don't have the PCB layout or a picture or drawing of the finished board, then you're kind of stuck. <A> It is a connector, but it doesn't have to be there. <S> It depends on whether you plan to use it and what kind of connector you plan on connecting to it. <S> You can simply leave 6 through-hole contacts on the board. <S> These contacts are used if you want to do some in-circuit programming. <A> By convention, a "P" designation on a schematic or wiring diagram indicates a plug. <S> A plug is by IEEE/ANSII conventions the movable part of a connection, or the more movable part. <S> The fixed/less movable part is designated by a "J", for jack. <S> So if you have a board that going to plug into a motherboard, the connector on the board is designed as P-xxx, and the connector on the motherboard is designated as J-yyy, where xxx and yyy are numbers. <S> Similarly, if you have a cable that plugs into a chassis or box (like an Ethernet cable plugging into your PC), the connector on the cable is a plug (P), while the connector on the PC is a jack (Jack).
| It is a connector, looking at the pinout most likely a programming connector.
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Mulitple switches connected to one pull-up resistor vs separate resistor for each I have 16 SPDT switches to be used as the address inputs to a UM61512A 64Kx8 CMOS static RAM module. I have designed the circuit to have separate 10k pull-up resistors to +5v when each switch is up and a direct to GND connection when the switch is down. As I was getting ready to solder it all up, it seemed redundant to have 16 pull-up resistors. Are there any issues with just using a single pull-up resistor and connecting all of the switches' up connections to it? Maybe connecting groups of four switches each to a single resistor - thereby only using four resistors? Are there any advantages/disadvantages to these methods? <Q> The decision to use a pullup resistor is more to do with the switches than with the input currents for your device. <S> Since you are using a CMOS memory, the input currents for both high and low are very very low (2uA max). <S> This means the maximum input current for all sixteen input to Vcc is about 32uA or less. <S> If you used a 1K Ohm pullup the voltage drop under worst case conditions would be about 32mV ..... <S> totally insignificant in defining the high voltage. <S> It terms of the switch however, there are other concerns. <S> If the switch is an overcenter break before make, you are safe, but if it's a make before break configuration <S> (like many small slide switches) there can be problems. <S> Direct use of VCC (without a pullup resistor as suggested in other answers) is not recommended practice, and could short VCC to ground in the worst case. <S> It is NEVER good practice to use VCC or in fact any voltage supply without defined current limiting as a pullup point. <S> My recommendation <S> : Use a single 1k Ohm pullup to VCC as your common pullup point for all sixteen switches. <A> In your configuration, that should be fine. <S> If it's CMOS, you may even be able to eliminate the last resistor and connect directly to +5V. <S> If you leave the resistor in, you should check the worst case bias currents to make sure it can support all the address lines being high simultaneously. <S> enough when the switch is open. <S> Those couldn't be consolidated like this, but you're using the switch to select between two levels rather than to override the pullup. <A> The only problems i can think of are 1: <S> if the pins being pulled up draw a significant amount of current then your voltage can drop when there are many connected to the pull-up, to determine the voltage drop just do n*I*R where n is the number of pins (always keep the worst case scenario in mind) <S> I is the current per pin and R is the pull-up resistor values. <S> If the voltage drops too much maybe you can put a smaller resistor 2: <S> Switching a switch (no pun intended) may generate a transient on the line. <S> this is easily solved with a decoupling capacitor in parallel. <S> So as far as you consider this i think you are good.
| Generally, individual pullups(pulldowns) are required for SPST switches or open-drain outputs where the resistor is required to pull it high(low)
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Calculating the resistor needed for a simple LED and battery circuit I'm having trouble learning the most basic circuit math! I've a 3V battery pack, two AAA, a standard plastic LED and a pack of resistors and I need to work out which resistor I need. I'm assuming the voltage drop of the LED is 2.0V; the datasheet only states a "forward voltage" of 2.0-2.4V and some wavelengths and lumens. The problem I have is that the math drop / current requires a value for current, but I don't know that yet. In the YouTube tutorials I've seen, I just gets written in as if its value was magically known (0.02A). My understanding is that the current is "drawn" by the loads on the circuit, which include the LED and maybe the resistor itself. So my confusion is that there's a circular dependency, or chicken-egg situation, I need to build the circuit to measure the current it draws in order to arrive at R and build the circuit. What's going on? Perhaps the datasheet or testing and measuring components in isolation breaks this deadlock? Perhaps I'm just tired. <Q> You said that the datasheet states a range of forward voltage. <S> The same line in the datasheet should also tell you the amount of current that was passing through the LED when they made the voltage measurement. <S> It may be called \$I_F\$ , the forward current, and that is the value your need to calculate the necessary resistor value. <A> The value of the resistor will determine the brightness of the LED. <S> The resistor limits the current drawn by the LED, preventing it from drawing too much current, and "popping". <S> Put in a 10k resistor, see how bright it is, and do the math with that. <S> This will give you a feel for what current values correspond to what brightness levels. <A> Given the limited data, we just have to make a few assumptions. <S> Almost all standard 5mm LEDs can take 20mA (0.02A), so let's pick that. <S> The supply is 3V. <S> The LED drops 2V. <S> That leaves 1V across the resistor. <S> Use the standard Ohms law equation V/I = R, gives 1/0.02 = 50 ohms. <S> If you haven't got a 50 ohm resistor, then 47 ohm is a common value, and close enough. <S> If the LED actually takes 2.4V, it won't be as bright as expected. <S> Some trial-and-error may be needed. <S> But LEDs will light at currents well below 20mA.
| Vary the resistor up and down, and do the math with each.
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Confusion with Ohm's law When I connect a cell with 100 Ω resistor. voltage across circuit is 1.4 V and current is 14 mA. That means total power of cell is 1.41 x 0.014 = around 20 mW. Right? Now if I connect two cells in series with same 100 Ω resister the volts across are 2.83 V and current is 27.6 mA If we multiply them we get around 78 mW. What I don't understand is if one cell has around 20 mW of power how come two cells have 78 mW instead of double of 20 mW = 40 mW maybe? <Q> This might help : \$P=\dfrac{U^2}{R}\$ <S> \$P_1=\dfrac{1.41 <S> ^ <S> 2}{100} \approx 20\,\text{mW}\$ <S> \$P_2=\dfrac{2.83^2}{100}\approx 80\,\text{mW}\$ <A> In this case, the power delivered to the resistor is not limited by the power available from the cell. <S> If you put two cells in series then you double the voltage. <S> If the resistance stays the same then Ohm's Law tells us that the current must also double. <S> So, if the voltage doubles and the current doubles, power goes up by a factor of 4. <A> We know that V= <S> IR (ohms law) .when <S> you added 2 cells in series then the voltage applied is doubled but the load connected(resistance) remained same. <S> So, the current also got doubled. <S> Now, power = V*I . <S> So, as now both current and voltage are doubled one would expect power to get four times. <S> This was what you obtained experimentally. <A> Extension/demonstration to @Marko Buršič answer: [1] <S> \$V={R}*{I}\$ <S> [2] \$P={V}*{I}\$ Calculating \${I}\$ from [1]: <S> [3] <S> \$I=\dfrac{V}{R}\$ <S> Then, replacing [3] in [2]: \$P={V}*\dfrac{V}{R}\$ <S> Resulting: <S> \$P = <S> \dfrac{V^2}{R} <S> \$ <S> As an addition, it could be interesting to consider the cell internal resistance which could cause a voltage drop in the cell terminals depending on the current it is sourcing <A> Applying Ohm's Law, you have rightly calculated that the power dissipated by a 100Ω resistor across a 1.4V cell is 19.6mW and that dissipated by it across a 2.8V cell is 78.4mW. <S> The power dissipated by a load cannot not be equated to the maximum power that can be delivered by the source across which it is connected. <S> Your confusion stems from that wrong assumption.
| The power is limited by two factors: the voltage provided by the cell the resistance of the load
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So why aren't supercapacitors used more often? I've recently found out that supercapacitors (C >= 1F or so) are actually a thing. I went searching for their advantages and disadvantages. They don't hold as much energy as a battery, but can endure many charge and discharge cycles. They have a lower inner resistance, so they can output a great current. Also they're not cheap, but they're not prohibitively expensive either. From what i gather, they could be a great solution for many problems, and yet, they are rare and I've never seen them around. Why is that? They seem so good. What's the catch? <Q> Low energy density: for a given amount of energy, they take up more space than almost all battery technologies <S> High leakage <S> : this document has some nice curves. <S> It looks like they've lost half the voltage after a week. <S> Bad discharge curves: normal batteries might give you 80% of the power before they've dropped 20% voltage. <S> Capacitors give out power evenly across their exponential curve. <S> So either you end up using a small amount of the capacity before the voltage drops too low for your application, or you need to mess around with boost converters on the output to keep it matched. <S> Edit: Peter Smith's link highlights some good applications. <A> The meaning of the term supercapacitor has changed over time. <S> In the 1980s, it referred to devices designed for very low current memory backup power and they had a relatively low output current capability (high ESR). <S> I remember one that was 4F, 5V in a relatively small package but had a current capability of perhaps 10mA. Newer devices have low (and sometimes really low) ESR and very high working current capability. <S> They typically also are low voltage (often 2.7V) so a practical application uses a series - parallel arrangement. <S> Take this family for example: <S> Capacitance up to 3000F, DC ESR nominally around 250 \$\mu \Omega\$ , 1 second peak current measured in thousands of amps but a limited temperature range to get datasheet operation. <S> In applications such as regenerative braking (becoming more common particularly in hybrid and all electric vehicles), where hundreds (or more) of amps may flow during the braking event, they act as a perfect energy store to later be used to recharge the onboard batteries. <S> The primary batteries could not take that as charge in direct form. <S> But, as noted by pjc50, there are downsides as well. <S> As with all technologies, they will be used where they best fit the requirement, and currently that is not really high. <S> I was evaluating them as a potential form of back up for power drop out in avionics (which was never finished due to lack of time and the fact that the temperature range was a bit suspect for that application) but that is a bit of a niche application (perhaps a few 10s of thousands of systems a year which is hardly the numbers manufacturers want to see). <A> Supercaps are really a compromise between a regular cap and a battery. <S> And, as every compromise, they can't fully match any of the alternatives. <S> If you can live with a battery it's better to use one. <S> 500000 recharge cycles on a typical supercap look impressive unless you check out the lifetime: 1500 hours. <S> That's right: if you keep it fully charged at 65°C, it will last a couple of months. <S> Not to mention the low energy density. <S> If you can live with an elco, you don't want a supercap either. <S> Not only you are free to chose any voltage you like, and keep the elco fully charged, but you also get lifetime and temperature ratings which are simply unavailable in supercaps. <S> For instance, try finding a supercap rated for > 100°C (typical requirement for engine compartment electronics). <S> So, you'd only use a supercap in applications where you can't put a battery and can't put an elco either. <S> Presumably, there aren't that many.
| So they are used, but until they start to approach battery densities they will be a niche product and even then the circuitry around them (to deal with the charge / discharge voltages) will have to become a lot less expensive. They're not widely used in cars yet because of the density and leakage issues, but for industrial equipment they're a good fit.
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what is causing this ring effect on my MOSFET testing circuit and how would I be able to calculate the MOSFET power losses? So I have the following circuit This is the data sheet for the MOSFET enter link description here Using equations, I worked out what my Rgate to be 2.36 ohms. where Rtest was a 470ohm ceramic block resistor (due to it getting quite hot when turned on)Connecting the oscillator, I see that I have this ring effect on the square wave as shown here: Just wondering what the cause of this ring effect is or maybe a little detail on what it is and how to fix it? I increased the resistor value to 47 ohms which does lower the effect but why? Also any ideas how I can calculate the Power loss of the MOSFET? <Q> The signal rise/fall times are so fast that either you have incorrect scope probing method with long ground clip, or there is too much stray inductance from the PCB wiring. <S> The series resistance slows down signal edges as it forms a RC filter with the FET gate capacitance. <S> If the probing is correct and the effect is real, then you need a snubber network to dampen the ringing. <A> Rtest is probably a wirewound resistor... <S> therefore an inductor. <S> I suspect this is the primary source of the ringing. <S> Oscilloscope probe calibration may be a contributing factor, as well as other long wires connected to the output. <A> on measuring the mosfet dissipation i would suggest you record the current with a current probe or a small series resistor on the source of the transistor and the voltage across DS, then integrate voltage*current for one cycle and multiply by frequencyin other words integrate the instantaneous power for one second <S> and you will get the energy dissipated in one second which is numerically equal to the average power
| As others have said, the ringing you see is most likely due to the probing you are using (for example having a crocodile clip to ground and forming a loop), if you are using wire wound resistors then they can also be guilty as they have a relatively high inductance, pcb layout can also be a culprit and if you are using a protoboard then i would look at that first and foremost.
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The Wien Bridge Oscillator The question is that: How could you make a Wien bridge tune to different frequencies? Since the frequency is depended on the value of the resistor and capacitor we should use a variable resistor and capacitor. But, I am not sure if this is the right approach. <Q> For a single frequency Wien Osc. <S> you need 2 resistors (of the same value) and 2 caps (of the same value). <S> Then have say 4 frequency bands. <S> Each band would be selected by switching in a pair of same value capacitors. <S> 8 capacitors in all. <S> This avoids the difficulty of trying to implement a dual gang capacitor. <S> Here's one I designed a while ago. <A> Here is the block diagram of the iconic HP 204C oscillator: <S> As you can see, it uses a ganged air variable capacitor as the primary tuning elements. <S> A pot is used to adjust the amplitude as controlled by the AGC (automatic gain control). <S> The complete schematic is in the manual which can be downloaded. <S> It uses BJTs and JFETs exclusively (no ICs at all). <S> Incidentally, my 204D, made approximately 50 years ago, still works fine. <A> decades ago, in university, I built a 100:1 tuning-range Wein using cadmium cells from Edmunton Scientific. <S> In a pack of 10, I found at least 2 <S> that tracked (as light varied from a 28 volt bulb) rather well. <S> To handle mistracking, another cell + another bulb was used to adjust the nominal_3:1 feedback gain.
| To vary the frequency over a large range you require a dual gang pot for the resistors so that you can vary their values together.
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Bit Stuffing on CAN Bus How a CAN transceiver detects the 5 successive high or low bits and inserting the opposite bit as stuff bit ?what is the logic behind the bit stuffing process?. I am sure that, it's not handled in CAN driver software side. <Q> The transceiver remains just a level converter, <A> The hardware looks at the transmitted or received bits and can count them. <S> Transmitting is easy, insert extra stuff bit when doing the line encoding. <S> Receiving more than 5 same bits before line decoding to remove stuffed bits results into receive error. <S> The point is to have a line code with transition at least after 5 same bits to keep the receive bit clock in sync with the data stream. <S> This allows for the devices to have cheap low precision systen clocks that have more tolerance. <S> If this was not implemented, it would be difficult to keep track of how many same bits have been received, unless the system clocks are very precise, which is more expensive. <S> Same thing why UART transmission has a start bit, to synchronize the reception of the next data, parity and stop bits. <A> This is not happening in transceiver <S> but in CAN module in microcontroller. <S> I suppose it counts bits, it's simple.
| Bit stuffing is on the interface, not the transceiver, like how there is also hardware in the interface to calculate the CRC and various other errors without ever having to touch the main processor, if it sees the successive bits, it inserts the inverted non data bit,
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Need help finding a solution to recharging old Li-ion batteries, maybe a power supply Sorry in advance if I say anything stupid, I am in no way an expert in the field. Everything I know is pretty much self taught through years of experience in refurbishing mobile phone. I deal with hundreds of phones that are sometimes decades old. Most of the batteries are still working, but I need some sort of solution in recharging them. I have reached the internet far and wide for a device to suit my need but at this point I'm not sure one exists so I need your superior expertise. So here is the situation. A lot of these batteries have been discharged for so long that they will no longer accept charge from the phone, appearing dead. But giving it a couple of seconds boost with a slightly higher than recommended voltage will bring it back to life and most of the times the battery will be working fine and hold good charge after. The problem is, I rely on primitive technology. I have an adjustable DC power supply that can be set between 3v-12v. When I try to resuscitate a battery, I set it to 9v and touch the appropriate pins to the battery a few times for a split second, and when the power supply's LED flickers a little it means the battery draws power, therefore is alive and can now be recharged. Another method, even more primitive, is this phone charger with a LED indicator on it. When idle is green, when the battery takes charge it goes red, when there is a fault in the battery it goes Orange, when the pins are shorted the LED goes off alltogether. What I need is some sort of power supply with an active display, maybe able to set voltage and amperage output, (there are many out there) but I also need to see the power it that it is drawn by the battery, maybe the recharge level, some sort of live feed from the battery as it is being recharged.I the closest thing I found are those Li-po chargers for RC models but not sure if those are safe for li-ion. I have a few of those cheap Chinese universal battery chargers but are very unreliable and I found only work on good batteries with some charge left in them. My budget is not very high, I found some industrial grade stuff for hundreds of ££ but for my small business, it would not be a smart investment.Any help or suggestion would be appreciated! <Q> Normal lithium safety still applies here, charge them somewhere that if it exploded into fire, it would not harm you or others, the proper way would be something to warn you if it became warm, but at a slow trickle charge, if it became warm, its already breaking down internally. <S> The lipo chargers for RC vehicles are a suitable option, but many have the same issue that they will not respond when flatter than they expect. <A> Your best bet might be finding similar size batteries and then taking the connectors off, yes <S> I know this is a pain, and rigging them onto the new battery. <S> Another way would be to find any new battery that fits and soldering wires to connect it. <S> I have done these type of things many times. <S> Recharging dead lithium cells rarely works and when it does "work" if you measure the battery its performance is terrible. <S> Also have tried this many times <A> I think I found some solution. <S> This adjustable DC unit that has charging / full / cc indicator. <S> Will give it a go. http://www.icstation.com/mobile/step-down-power-supply-module-display-constant-voltage-constant-current-adjustable-buck-converter-cvcc-adapter-p-13408.html
| As your working with what is likely single cell li-ion, and need to just make sure you get them to a chargable state, set your power supply to about 2.7V, and fit a 2.7K resistor in series, this will very slowly trickle charge it back up, If it does not increase in voltage after a long while, the battery leaks to a point it is not really usable.
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Are there any advantages to using rigid RF coaxial cables? While watching this Signal Path video , I came across this shot: I saw few rigid coaxial cables there and so I started wondering what might their advantages be. I could only find disadvantages: They're expensive Limited in available angles 45° or 90°. Otherwise they're not easy to bend compared to the other cable types Sold in sections so we may need to always join them Prone to corrosion The only advantage seem to be higher power capability. So, are there any advantages to using rigid RF coaxial cables? <Q> So, are there any advantages to using rigid RF coaxial cables? <S> First, the product you show might be semi-rigid rather than rigid cable. <S> In either case, the main advantage is lower loss, because of the solid outer conductor (as compared to the braided o.c. used in flexible cables). <S> In a production environment the rigidity of the cable is itself an advantage because it means the system will always be assembled with the cable arranged in the correct way, rather than however the assembly worker can make it fit. <S> The rigidity of the cable also improves phase stability. <S> Most of the disadvantages you cited can be mitigated, or don't even apply when manufacturing in volume. <S> Limited in available angles 45° or 90°. <S> Not true at all. <S> Sold in sections <S> so we may need to always join them Not true. <S> You can have these cables assembled at whatever length you like. <S> Certainly they can be fabricated up to at least a meter in length, which is plenty for the type of system you showed in your image. <S> Prone to corrosion <S> The outer surface is typically tinned copper or aluminum. <S> Either of these materials is likely to be present in numerous other locations in an electronic assembly, so these cables will be no more prone to corrosion than, for example, the surfaces of any PCBAs used in the product. <A> From a mechanical point of view, a rigid (or semi-rigid) cable has the advantage that they will support themselves. <S> That means that you don't have to worry about them getting pinched or stuck between things when they shift or move during assembly or transport. <S> As it doesn't move or flex or vibrate, the electrical length will be more stable over time than with flexible cables. <S> This can be important in applications where length matching or calibration is important, and is one of the reasons that test&measurement equipment is full of these rigid sections. <S> As others have said, you will have these manufactured by your suplier (or inhouse) to the exact shape and bends you need. <S> From an assembly point of view there is another advantage: they can result in fewer errors (but I don't think this is a significant factor when choosing for them). <S> As they are made to the exact size, they will usually only go in one way, and you can't by accident connect to the wrong port within the device. <S> I think they might also be cheaper at high frequencies, since the solid nature of the outer conductor is easier to manufacture than the very tight weaves needed for coaxial cables operating at many tens of GHz. <A> I expect the isolation is also better, and the crosstalk is lower. <S> Thus your ultimate system dynamic range will be better, which may be crucial for battlefield survival or to detect more planets around red dwarfs If you want a system that will NOT outgas from plastic cable coverings, the volatiles not being set free to condense on your billion dollars of optics, or condense on relay contacts, then semirigid or rigid is your friend. <S> Also the rigid may use very sparse spacers inside, allowing mostly air/vacuum as the dielectric and thus LOWER losses. <A> Lower passive intermodulation, phase stability, and better shielding. <S> Lower insertion loss, better temperature range, better dielectric and ... just everything. <S> But notice that your reference url shows a device in which two of the original semi-rigid cable assembly parts (W4 and W5) have been replaced with colorful flexible coax: flexible coax is available in better grades than it was when the 8445B was new.
| The solid o.c. also reduces radiation, which can be important if there are sensitive receivers nearby. A rigid/semi-rigid cable also has performance advantages. There are numerous shops that will bend these cables to whatever angles you like, following your drawings.
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Can I build a 4A circuit on cardboard? I'm planning to build a 12V 4A circuit, and I was originally going to put it on a strip board. However, the board I ordered came broken (and each piece would be a bit small for building a circuit), and I don't have time to order a new one. Additionally, I'm starting to realize that my circuit setup would be pretty difficult to set up keeping the strips in mind, and I would need to widen some of the holes to fit certain components. Since I have a decent amount of a 16 guage wire to link the components without a board, would it work to stick everything on a sheet of cardboard? Do I have to worry about the cardboard catching fire at a high wattage due to heat from the leads? <Q> Yes it would work and yes you might have to consider the possibility of the cardboard lighting up under high loads. <S> If you do it right you might be fine. <S> Maybe consider building a boardless circuit. <S> Look up "flywire circuits" and you will see examples. <S> It will take a little more skill but your results can be beautiful, last for years and be fireproof. <A> at-least temporarily, if it gets folded there could be a short-circuit. <S> FR2 which stripboard is usually made from is cardboard that has a fire-retardant resin soaked into it which also adds stiffness. <A> The alternative would be to use lug strips, mounted on an aluminium sheet/chassis, to interconnect the components (point-to-point wiring). <S> The aluminium sheet/chassis may also serve as a heatsink if necessary.
| if the leads are thick enough to not overheat then cardboard can work. If you use heavy solid wire for you ground you could even incorporate mounting loops in the wire.
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What is the purpose of the PCB jumpers in the attached image Can someone please tell me what may be the purpose of the jumpers in the attached image? Why not use a PCB trace instead? Figure 1 The PCB is a part of an old Videocassette recorder (VCR) <Q> It is a super-cheap single-sided paper-based phenolic PCB (probably punched rather than drilled holes, judging by how large they are). <S> The cost savings will easily pay for the salary of the engineers, and if you don't do it, they will save the cost of your salary by shunting you out onto the mean streets of Tokyo or Taipei and find someone who will. <S> Another issue from that time is that it's easier to get consistent quality of production with single-sided PCBs. <S> The chemical process of making plated-through holes is sensitive to contamination and easy to get wrong. <S> It only seems straightforward now that many low-cost fabs have mastered the process control. <S> In the 1990s it was not uncommon to see bad (intermittently bad on 100%-tested!) <S> plated-through holes and hair shorts. <A> The PCB is a single sided board. <S> A double sided board (with traces on the top and bottom) costs more to make. <S> Back when that board was new, a doubled sided <S> PCB cost so much more than a single sided board that it was cheaper to make a single sided board and install a bunch of jumpers to make connections that would have otherwise required a double sided board. <S> Today, double sided PCBs are cheap enough that it is rare to find a single sided board with jumpers. <S> It still happens, but mostly for really cheap junk. <A> As others have noted, the aim is ultra low cost of production. <S> It uses a low cost single sided "phenolic" PCB. <S> By using links it only needs copper on one side, and there is no need for via connections between the layers - which involve relatively expensive and time consuming plating processes processes. <S> The links are inserted by a machine that moves so fast it is a blur to the eye. <S> I don't know the precise insertion rate, but it's tens of links per second and sounds somewhat like an automatic weapon. <S> Here is a You Tube video of one in action - Jumper wire insertion of one at work - not quite as fast as I recall but still impressive. <S> Rate varies but sounds like maybe in the 10-20 links per second on faster runs. <S> Links are made from a wire roll on the fly - cut, shaped, 'hammered' into place & the ends cinched slightly on the copper side to retain in place prior to soldering. <A> Back when the VCR was made, it was cheaper to fabricate a single-sided PCB and use wire links, than to fabricate a multi-layer PCB. <A> Its a phenolic PCB, its an extremely cheap material to make these devices out of, If you wanted to produce say 1 million PCB's, plating VIA's to connect both sides of the board is a rather slow process, So to speed things up, and remove the need for those additional steps, they built the complexity into wire bending / placing machines that could shoot those jumpers into the PCB at a ridiculous speed, spooling up through the bottom, grabbed by a head on the top side, moving up to the length then moving to the other hole and feeding and bending the other end back down the other hole, at multiple per second, <S> At the end of all this, it has any additional components placed, and then fed through a solder bath. <S> This is also part of why the holes tend to be so wide, the lower quality of the material means the holes are not always in the exact right place after it relaxes from the punching step.
| It's to save money compared to a fancy double-sided PCB with drilled plated-through holes.
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Effects of PWM the source voltage for DC & BLDC motor? In DC or BLDC motor control..The standard is to PWM the H-Bridge switches to control the motor current ..Why not to PWM the source voltage (V-Motor) and just switch the H-Bridge switches (in lower frequency domain) as needed ..I know .. it may be not economical due to an extra switching element ..I just wanna know the effects of this way regardless the economics .. <Q> The question mentions DC & BLDC as if a single answer is expected for both. <S> However there are two completely different situations. <S> In the case of a brushed DC motor, you could control the speed by PWM of the source voltage and use the H-bridge just for direction control. <S> That would work, but there would be no electronic protection against excess current due to accelerating too quickly or overloading the motor. <S> You could ramp the voltage up to control accelerating current and implement some kind of electronic shut-off for overcorrect, but you would end up wanting an active current limit. <S> You could implement that with the source PWM, but that would amount to an armature voltage control with an overriding current loop. <S> The end result would be a control system with more power components, a control scheme as complicated or more complicated than usual and poor performance. <S> The case of the brushless motor, you would be feeding PWM DC to a six-step inverter and starting with an open-loop, three-phase permanent-magnet synchronous motor control. <S> You would have the same over-current protection problems as described tor the brushed motor and more difficulty in designing a remedy for the problem. <S> In addition, the performance would poor. <S> Here again, you have added complexity and subtracted performance. <S> Historical Note <S> The following block diagram shows how a PWM source voltage brushless motor controller (mostly for induction motors) was implemented about 50 years ago. <S> About 45 years ago, that design was obsolete. <A> Changing the supply voltage is the same as feeding in a lower voltage, say half the voltage means half the current, means 1/4 the energy turning the motor, <A> Why not to PWM the source voltage (V-Motor) and just switch the H-Bridge switches (in lower frequency domain) as needed <S> .. <S> I know .. <S> it may be not economical due to an extra switching element <S> You are right. <S> The main reason for not chopping the power supply separately is that it needs an extra switching element, which is not required if the H-bridge or brushless commutation controller has a suitable PWM input. <S> However sometimes it doesn't, and then chopping the power supply is another option. <S> This is often done on brushless fans, which have a very simple commutation system that cannot easily have PWM applied to it. <S> The PWM switch may be applied externally to a 2 wire fan, or inside the fan controller (which permits continuous tach output). <S> This pdf from Analog devices shows how it can be <S> done:- Low frequency PWM basically just turns the whole fan on and off too fast for it to speed up or slow down much. <S> This may only work at relatively high PWM ratios, is non-linear, and has poor efficiency due to the high surge current that occurs each time the fan is turned on. <S> With high frequency PWM <S> the effect is equivalent to the same average DC voltage, and speed is controlled linearly down to perhaps 10% of full speed. <S> The inductance of the motor coils smooths out the current <S> so efficiency is not impacted.
| Vs halving the PWM value for the motor at the same voltage, gives half the energy turning the motor, You also have things like beat frequencies, where the H bridge switching may not overlap cleanly with the supply, making things even uglier.
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can i cut a large strip board into a smaller one? I need a 17 hole by 8 hole strip board. I could only buy a much larger one. Can I cut the larger one to the size I need without destroying it? If so, how would I do that? <Q> Use a ruler or other straight edge to guide a craft knife. <S> Make many passes cutting a little bit at a time. <S> After the cut looks to be reasonably deep you can bend the board and it will break along the cut. <S> You can use sandpaper to clean up the edge. <S> If you are cutting across traces you might need to make a few knife passes on the back to ensure the traces are all cut before breaking the board. <A> It is a slow process, if you have a lot of them to cut, you might look for a quicker method. <S> It will create a lot of dust, best to do it in the garage or outside. <S> Depending on your skill level, you may need to use sandpaper to clean-up the cut. <S> https://www.amazon.com/Busboard-ST2-Traditional-Stripboard-Prototyping/dp/B00LLQFRAU <S> Edit: I have used the knife scoring method also. <S> My success rate was not 100%. <S> I use a dremel now. <A> I cut thousands of stripboards of many sizes with a fine-toothed hacksaw and a squaring holder. <S> But only one every day. <S> Some are made from fiberglass (Veroboard) and others are made from cheap stuff. <S> The cheap ones warp. <A> Score the copper side with a knife, then break it. <S> Works fine 99 times out of 100.
| I cut printed wiring prototype boards with a dremel tool and cutoff wheel.
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Why have a bottom mounted pad on a component and not use it? For my design, I am using the ICM-20689. The pinout for the chip is taken from the datasheet : Once I get to the bottom of the datasheet, I notice that there is a bottom-mounted pad that is not mentioned anywhere else in the datasheet. On the manufacturer's website, they have a link to SnapEDA for the PCB footprint and there is no bottom-mounted pad here either. So, is there a reason why they would manufacture the chip with a bottom-mounted pad that is not supposed to be used? <Q> A few reasons. <S> 1 | <S> these packages are standardized. <S> They come in big grids called lead frames, and the packaging factory will place the die-stack provided on the lead-frame and bond it out as directed. <S> If that packaging house already has a lot of QFN products with the bottom metal pad, it might be easier and cheaper for them to just stick to using that package even if there is no need for it. <S> (the package itself has this ground plane because other components might need it, see Niel_UK's answer) <S> 2 | <S> There might be a standardised footprint for these kinds of components. <S> Say TDK has a whole line of IMUs. <S> It could be interesting for them to have a lot of versions come in the same package with compatible pinouts. <S> Maybe a higher-end model has that bottom pad for ground because it needs it, and uses the other NC pins for communication. <S> Maybe an older version needed more supply rails. <S> If you as designer used that part, you might be able to just drop in this part into the previous part's footprint, not populate the voltage regulator components, and continue using the same board layout as you did for the older version, cutting down on your design time. <S> 3 <S> | Testing in the factory: The pads that are marked NC, along with the bottom pad, might be used for in-factory testing. <S> I know this is done with other types of components, where NC pads are actually bonded to the die and used in the factory to feed in calibration data to the IC. <S> The NC pads are then either just left to whatever, or the factory can disconnect them electrically after calibration (through on-chip fuses for example). <S> 4 <S> | <S> Finally, <S> but I am not sure about this: The chip in question is an IMU. <S> I would imagine it could use with higher mechanical stability when connected to the PCB, and perhaps for that reason the big bottom pad is used. <A> The 'must use' scenarios for a bottom pad needing soldering are for 1) high power applications for heat-sinking, 2) RF applications which need a solid electrical GND connection and 3) high current applications where the pad is one of the electrodes. <S> This device is low power, low frequency, low current, so meets none of those. <S> Why does the package have a large pad? <S> Bear in mind the way many ICs are made, with a metal frame that defines the pins, which are then separated later in the process. <S> Leaving the pad part of the frame in place may well be cheaper than removing it. <S> This is a very thin package, intended no doubt for space constrained application like smartphones. <S> Any insulation or mold material over the exposed pad on the bottom would make the package thicker. <S> I'm intrigued by their 10,000g shock specification. <S> If I was designing with it, I might be inclined to ask them whether soldering the pad down has any influence on that spec. <A> Usually the pad provides an electrical and thermal ground. <S> In this case it’s not necessary. <S> If the chip is applied on the bottom side and the SMT reflow is done in one step, chips with a small number of leads like this one can fall off. <S> The center pad allows adding a pattern to the board for extra solder paste to hold the chip onto the board during reflow. <S> When the extra paste melts, the added solder creates more total surface tension to hold the chip in place. <S> This is preferable to using adhesive as it saves a manufacturing step.
| Perhaps it's a standard package, and it would cost more to design a package without the exposed pad, than it costs to use it, if indeed it is more expensive than a package with no pad.
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How can a 4-bit CPU have 40 instructions? Currently, I am using Logisim(yes, still Logisim) to build a 4-bit variant of the 8-bit SAP-1 microcomputer. However, I ran into a problem with the instruction register. Let me explain. The SAP-1 has 5 instructions(LDA, ADD, SUB, OUT, HLT). These 5 instructions amount to 3 bits of data to read those RAM words. But that leaves only 1 bit for the address, and that means only 2 RAM addresses. To fix the problem, I tried to make the RAM words 6-bit instead(I don't know why I chose 6). But then I realized that having 6-bit RAM words would mean a 6-bit bus, and therefore a 6-bit CPU, which isn't good. Yet, 4-bit microprocessors carry on with upwards 40 instructions with kilobytes of memory. How do they do it, and how can I implement it into my 4-bit SAP-1? <Q> variable-length instructions <S> ---- suppose you want to jump with an offset of 550 fromthe current Program Counter? <S> uses at least 3 of the 4-bit nibbles, just for the JumpOffset. <S> Plus the opcode, which for Jump with Offset may need 2 more nibbles. <A> Also with the address problem, it doesn't mean that a 4 bit CPU can only handle 4 bit addresses, as an 8 bit CPU (like a Commodore 64 in the good old days) <S> can also address 64 KB of memory which is 16 bits. <S> It just means, it cannot handle it at once, so first 8 bits are loaded, and than another 8 bits. <S> So if your CPU needs to handle 1 GB, it is still possible with 4 bits, but you need many of those to acquire the final complete address within the 1 GB. <A> The SAP-1 has 5 instructions(LDA, ADD, SUB, OUT, HLT). <S> These 5 instructions amount to 3 bits of data to read those RAM words. <S> But that leaves only 1 bit for the address, and that means only 2 RAM addresses. <S> I think there are some serious misconceptions here. <S> Being "4 bit" <S> does not mean that your instructions are necessarily 4 bit, just as being 64 bit does not mean that your instructions are 64 bit. <S> Practical CPUs typically have either at least 16 bit instructions or a mechanism to make instructions variable length simply because a 4 bit instruction length doesn't let you do very much. <S> Yet, 4-bit microprocessors carry on with upwards 40 instructions with kilobytes of memory. <S> How do they do it, and how can I implement it into my 4-bit SAP-1? <S> First, lets pick a common definition of what "4 bit" means. <S> We'll say it means a system that has the general purpose integer registers all 4 bit. <S> It doesn't mean that all registers are only 4 bit, but we'll assume the basic ones are. <S> In that case, and with byte addressable memory, you would be limited to 16 bytes of RAM. <S> If you want more you have a few options: Make the address space larger, in which case you will probably need to make the registers larger, in which case you arguably no longer have a 4 bit processor <S> Make the address space larger, but introduce a specific address register that is larger than the general purpose registers that can hold larger addresses (e.g. you have 4 bit integer registers and a special 16 bit register for memory pointers). <S> Note that this may require you to be able to do <S> ALU operations on greater than 4 bit values. <S> Introduce a memory bank mechanism, where you have multiple 4 bit banks you can swap between, possibly by writing to a specific register, using a specific instruction, etc. <S> This effectively just encodes memory addresses in two (or more) register values.
| If you want to build a 4 bit CPU, you should look at what your CPU needs to do, and then pick an instruction size that makes sense for the application.
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Vias in a the layout of a CMOS Integrated Circuit Is there a downside to using maximum possible VIAs (adhering to DRC rules) to connect two very long metals (ex. Power rails in Mx and Mx+1) running in parallel at block level? I know that multiple VIAs are preferred over a single via for current sensitive connections and that redundant VIAs increase reliability, but, is there something called excess VIAs ? I see a lot of you sharing their thoughts on PCB layout whereas I was looking for answer in an CMOS IC chip layout. <Q> Excess would be the point where the board manufacturer replies saying it will cost extra to produce your PCB, as it costs them machine time drilling all those holes, <S> You can get pretty over the top with the number before they complain, (about 1200 on a 10x10cm 2 layer board from memory) <S> avoid the minimum size, the smaller drill sizes break more frequently, and have a slower depth drilled per second which takes longer, so instead of a 0.3mm hole in your via, make it a 0.5mm hole <S> and there will be much less friction, the larger the better if you don't need it tiny. <S> Other side effects on a circuit level is, well the hole is not as conductive as the trace, so you increase the resistance of any trace you shotgun spray with vias <S> , Other things can be a sort of tear along the dotted line problem, a rectangular PCB will normally curve when flexed evenly over the entire length, if you have rows of vias all in a line, that area will flex more than the rest of the PCB, and may be subjected to higher stress than the rest of the board, <A> Since you are asking about VLSI design rather than PCB design we don't have to talk about the cost of the vias. <S> However, in current VLSI manufacturing the horizontal distance between features on the same layer is comparable or less than the vertical distance between conducting layers. <S> This means that sidewall capacitance to adjacent structures becomes significant. <S> If there are not nearby signal wires then the fringing capacitance to the substrate will increase. <A> regarding n-silicon use of vias ---- in a linear setup of vias, clearly the vias eat up part of the metal width, thus reducing the current-carrying ability. <S> I suggest you use wide metal over wide metal, and have a line of vias down the middle of the metals, so the vias are fed with current from both sides and on each layer. <S> Summary: draw sketches of the current flows. <S> A long thin metal overlap with lots of vias in a line will cause massive current crowding. <S> Current crowding is NOT what you want. <A> There are usually rules regarding density in CMOS. <S> Because modern CMOS processes use CMP between metalization layers, they need an even density of metal across the entire die. <S> If your DRC rules do not contain any limit as to the maximum metal density, that is great. <S> Usually you do require to have a certain minimum pattern of no-metal, but since large metal traces are allowed, I imagine you need a whole lot of vias before you have too many of them.
| If you add many vias you are creating a large picket fence kind of structure, and there will be significant capacitance from the vias themselves to nearby wiring of other signals on the same metal layers.
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How to switch 1000w AC load with minimal heat? I am trying to heat a 1000w heating element with 120VAC. I am currently using a thermocouple and SSR to regulate the temperature, but the SSR is getting too hot at full power even with a heatsink. In open air, it's fine, but once it's in an enclosed, even with vents, there isn't enough cooling. I want to avoid moving parts so I didn't want to use a relay or a fan; is there an alternative? Edit: Since people mentioned zero-crossing, I am switching at the zero-cross points and while I say it was an SSR, it wasn't a prebuilt one, but one I made with a BTA-16 600B triac, 4n25, and moc3021. It was based on an AC dimmer module sold by RobotDyn since I wanted to originally use PSM to control heat output, but now I simply need the AC load to switch on or off and I'm not sure if there is a better way to do it. <Q> Alternative would be a over-specced SSR, based on the load, I presume your using something like a 10 to 25A one, so try a 60 or 80A one, the device is made to run at X temperature for that rating, and half the current means quarter the heating over the switch. <S> so a really big SSR will run with a 10A load at a much cooler temperature. <S> Edit: <S> Based on his new edit, Clearer what the issue is now, change out your opto-triac for a MOC3041, that way it will not trigger until it crosses through zero again, instead of monitoring it with the micro Edit 2: <S> Also the above still applies, a larger triac will reduce the dissipation, e.g. 30A triac datasheet in the same package as what your currently using would have half the dissipation. <A> If it is fine in the open air, and you want to avoid forced cooling, get a larger heatsink and a larger enclosure. <A> while I say it was an SSR, it wasn't a prebuilt one, but <S> one I made with a BTA-16 600B triac, 4n25, and <S> moc3021. <S> TRIACs have voltage drop, which is quite high no matter what their rating. <S> This is caused by the TRIAC needing to 'steal' some voltage to keep itself turned on. <S> At 120 V you need ~8.3 A to produce 1000 W. A typical TRIAC drops about 1.5 V at rated load (one rated for higher current should drop less at lower current, but it will still be significant). <S> Your TRIAC probably drops > 1 V at 8.3 A, producing > 8 W of power loss. <S> That will need a reasonably large heat sink with good airflow to keep it cool. <S> I'm not sure if there is a better way to do it. <S> A switch using high voltage power FETs could be more efficient, but you need two FETs wired back-to-back for AC operation, and the driver circuit must ensure that the FETs don't turn partially on when there is significant voltage across them. <S> High voltage high current FETs are also much more expensive than a TRIAC. <A> SSRs (thyristor based ones) have a loss that is about 1W/A. Overspecced ones will be a bit less, but generally in that neighborhood.
| So your options include using a mechanical relay (less loss and no heat sink but limited life and slower cycle time), getting a better heat sink (perhaps with fan), or using a higher mains voltage (using 240V with a corresponding heater would reduce the loss in the SSR to about half). You might want to consider a heatsink that can have its fins on the outside of the enclosure.
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Turn spacing for helical heating coil? I'm planning to develop a helical heating coil wound around a ceramic rod. The purpose of this coil is to radiate heat to its nearby surroundings. I want to know if there should be a spacing between each turn in the coil? If yes does the heat radiated depend on the spacing? If yes what should be the spacing inorder to attain maximum intensity of heat radiation, keeping all other parameters constant? I'm guessing, the lesser the spacing more are the turns in the coil hence more heat radiation, thus the spacing should be as little as possible (but not zero) inorder for maximum heat radiation. Am I right? EDIT: I'm interested in a high intensity heat to generate a high temperature in a small volume. Basically I just want to heat the surface just below the coil. The distance between the coil and the surface is not more than two millimetres. <Q> While I would be quick to jump on the "just simulate it" bandwagon for tasks like this, I'll take a shot Reflections off the rods, assuming the wire of the coil radiates fairly evenly about its circumference, having a wider gap between the coils means there is a larger proportion of paths from that circumference that can either freely escape, or reflect off the rod and miss another heating coil, so the heat is spread over a wider angle on average, meaning the wire should be slightly cooler. <S> Resistance, The smaller the spacing, the longer the coil is in total, you cannot trip the circuit its on, so the resistance has to be high enough to not do this, equally to get the maximum wattage out of a given supply voltage you want to be as close to the ideal resistance as possible, e.g. for a 240V, <S> 10A circuit, P = <S> I <S> I R, so 2400W / 10 <S> / 10 = 24 ohms, matching the resistance of your wire to reach this resistance would be ideal Thermal change in resistance, the larger the surface area <S> the same heat energy is spread across, the lower the temperature on that surface, so the more turns in your coil, the lower the temperature in the wire, and the lower the temperature change from ambient, the less increase in resistance when at operating tempeature, Based on all this, I would work to find your ideal resistance by knowing your supply voltage and current limits, find a wire thickness and number of turns that meets that at expected hot and cold temperatures, that fills as much of that surface as possible, increasing the spacing only if you need a wider angle for the heat to be spread over. <A> Define "heat radiation" for your purpose. <S> The total power radiated depends on the voltage and current. <S> The current depends on the coil resistance : increasing resistance reduces the current and total power. <S> So, the answer depends if you are interested in a high intensity heat to generate a high temperature in a small volume, or maximising heating at a lower temperature in a larger volume. <S> A good pottery kiln is an inefficient way to heat a large room. <S> I cannot tell which you want from the question. <A> You will want to maintain appropriate watt density (diameter-> surface area and wattage per unit length) on the resistance wire so that you get reasonable life at your operating temperature. <S> The quality of the contact with the ceramic rod will affect how much the wire overheats. <S> If the ceramic rod has grooves in it, it will be much better. <S> Of course you have to leave enough room to ensure the turns don't touch. <S> Sometimes you need the element helically coiled and then that coil coiled about the surface. <S> I really suggest you start with a known-good commercial product and modify from there. <S> Or, better yet, just buy a commercial heater if that makes sense. <S> You have not mentioned temperature range or wattage, but there is a very wide range of commercial heaters that are available with guaranteed specifications. <S> Exposed elements are not all that common in most, for good reasons, but necessary in a few types for severe cost constraints or extreme temperature reasons.
| Reducing turns spacing will increase the coil temperature and thus the radiated heat intensity, but the increased temperature will (with most coil materials) increase the coil resistance and thus reduce the overall power.
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Using a DB3 diode as a flyback for a 24vdc relay Can I use the DB3 diode as a flyback for a relay that works on 24v dc? [![enter image description here][3]][3] This is a new photo : [3]: https://i.stack.imgur.com/Snv9f.jpg <Q> It is a diac, a non-directional semiconductor switch that can be turned on when its breakover voltage is exceeded. <S> Breakover voltage of DB3 is 28V. Your unregulated power supply voltage could touch 34V. The DB3 would get triggered and short circuit the power supply, thereby damaging itself and/or the power supply in the process. <S> A 1N4001 would suffice as a freewheeling diode for this application. <A> You don't need a snubber diode. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> The bridge rectifier acts as a snubber for the relay. <S> When the AC switches off the relay current will continue to circulate through the diodes of the bridge rectifier. <S> Since you have clarified that there is a switch in the circuit but haven't given any details on it we can only suggest this circuit. <S> simulate this circuit Figure 2. <S> As @MarcusMüller has suggested, placing the switch in the AC side solves the problem. <A> https://www.st.com/resource/en/datasheet/db3.pdf Breakdown voltage 28V minimumRepetitive current limit: 2A, Also its a back to back device <S> , so there is no forward voltage, I would say this is a poor choice for a relay unless you needed the relay to close very quickly. <S> if your supply voltage gets a little too high, or your circuit re-powers <S> the relay before the breakdown has recovered, it will short out the supply to the relay may destroy itself from the excess amount of current flowing through it, Avoid if at all possible for this application.
| No, DB3 is not a diode.
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How does this MOSFET based motor driver designed work I have designed a motor driver using a MOSFET that takes a PWM signal from an Arduino and boosts it to operate the motor, but I'm not sure why it works. I placed the capacitor in parallel to block noise created by the motor and the diode to prevent back EMF, but I'm not sure how the MOSFET operates to boost the PWM signal or what the purpose of the resistors is. Any help is greatly appreciated, thank you! <Q> I'm not sure how the MOSFET operates to boost the PWM signal <S> The MOSFET is acting like an on-off switch changing from on to off as quickly as you switch your PWM waveform. <S> So, if the PWM is 50% duty cycle then the average voltage across the motor is 50% of 4.8 volts = 2.4 volts. <S> If the PWM was permanently a "1" then the motor voltage would be 4.8 volts. <S> I placed the capacitor in parallel to block noise <S> That doesn't work too well as a noise reducer because that capacitor is being switched on and off across the 4.8 volt supply at PWM speed. <S> This means that large current pulses will flow through the capacitor and from the power supply and generate EMI. <S> So, rather than acting as a noise reducer, it acts as a noise source. <S> If you really want to reduce noise properly then there should be an inductor between the MOSFET drain and the circuit that connects to it. <S> Then you would need to move the diode back to the MOSFET drain to prevent inductor back-emfs destroying the MOSFET. <S> what the purpose of the resistors is <S> The resistor from gate to source (usually around 10 kohm) ensures that the MOSFET turns off when the circuit is not connected to your Arduino. <S> The series resistor (usually around 100 ohm) is to reduce stress on the Arduino output pin that drives the large gate-source capacitance of the MOSFET. <S> Any help is greatly appreciated Noted! <A> Here it is driving the mosfet gate pin high and low, as its likely to be a logic level mosfet a voltage about about 3V will have it fully switched on and its resistance will likely be a fraction of an ohm between its source and drain pins R17 is a current limiting resistor in effect <S> , the gate of a mosfet is essentially a capacitor, so when the arduino's pin goes high or low it has to charge and discharge this capacitance, if that happens too fast it could damage the arduino or cause a big enough supply current spike that it may make it misbehave <S> R18 is there to pull the mosfet gate back to 0V, and thus open the switch, if your micro looses power, resets, or your codes not working quite right, this will turn the motor off, (assuming the PWM pin is not stuck high), <A> You need to gain a basic understanding of FETs. <S> Using them by following other people's circuits is a good start but a basic understanding will greatly add to your capabilities. <S> In this application the MOSFET acts as and electronic switch. <S> When PWM voltage greater than the FETs turn on voltage is applied to its gate the FET provides a a low resistance path between drain and source. <S> Here, when Q3 is "on" current flows from V+ (4.8V) via the motor then via the FET to ground. <S> The motor operates. <S> When PWM voltage is about zero the MOSFET is off and current cannot flow to ground through it. <S> If the motor is rotating a "back emf" of opposite polarity to supply voltage appears across the motor. <S> Diode D3 conducts and the motor "freewheels" until drive voltage is again applied. <S> R18 pulls the gate to ground and truns the FET off when there is no drive voltage. <S> Not needed IF drive is always applied - high or low. <S> R17 reduces gate charge current somewhat but its main task is usually to damp gate ringing voltages which can occur during switching.
| The Mosfet is acting like a switch, the PWM output of the arduino should not be loaded with more than about 10mA, so for a tiny current spike at every switch on / switch off change, the mosfet can potentially switch 10's of amps.
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Why are most consumer electronics power supply boards single sided and made of FR2 (Phenolic resin) Most consumer appliances/power adapters that I open up have an SMPS providing the power. This SMPS is layed out on a phenolic resin (FR2) board and is usually a one-sided layout. Can anyone explain why power supply boards are designed as such? My guess is that this board mostly contains through hole components, and is easier to wave solder when laid out on a single side. Since FR4 is used for multilayer boards/through hole plating and vias, it is cheaper to lay the power supply out on this type of board. <Q> Cost. <S> It is cheap, and it is good enough. <S> "The best is the enemy of the good enough" <A> Why are most consumer electronics power supply boards single sided and made of FR2 (Phenolic resin)? <S> Cost. <S> The FR2 will be a bit cheaper than glass fibre boards and may be a little easier to machine meaning less tooling cost. <S> Figure 1. <S> XT PC Power Supply PCB - Component Side. <S> Source . <S> Figure 2. <S> XT PC Power Supply PCB - Solder Side. <S> Source . <S> Note that some of the traces are rather circuitous to avoid crossover connections on the other side of the board. <S> Can anyone explain why power supply boards are designed as such? <S> The circuits aren't so complex and the board can be laid out without the need for wire jumper links. <S> There are none visible in Figure 1. <S> My guess is that this board mostly contains through hole components, and is easier to wave solder when laid out on a single side. <S> Double-sided boards are generally wave-soldered in the same way. <S> The traces on top facilitate jumping over other traces and the through hole vias connect the the two layers. <S> Since FR4 is used for multilayer boards/through hole plating and vias, it is cheaper to lay the power supply out on this type of board. <S> Yes. <S> You could get a feel for the cost saving by trying to price, say, 1000 PCBs on one of the online PCB manufacturing companies. <A> FR2 can be punched instead of drilled. <S> FR4 has to be drilled. <S> When you're making 100's of thousands, it's worthwhile to make one big punch and do a ton of boards in one pass. <S> Drilling is expensive and takes time. <S> So it's "cost" ;) <S> It's because of the materials they're made of. <S> FR2 is paper phenolic resin. <S> FR4 is fiberglass Single <S> sided boards are ALOT cheaper to work with 'cuz <S> you only have to process one side of them. <A> One reason for single sided boards in consumer electroniss is they are easily serviceable, and it is easy to service them during the warranty period instead of repairing the whole board. <A> The power supply boards having higher voltage, have higher risk of arcing, static and carbon formation making the circuit in-efficient and higher power losses and pose risk of short circuit damaging the psu and could be harmful.
| Single-sided board will be cheaper than double-sided due to reduced copper cost and processing steps.
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PCB design and manufacturing: How important is it to select actual vs. generic parts? PCB newbie here. I'm using Autodesk Fusion 360 to design a PCB that is a modified version of the Arduino Nano IOT 33. My questions are about selecting parts before sending to the PCB manufacturer: For generic parts like SMD resistors, capacitors, etc., can I just specify a generic resistor (see screenshot) with the appropriate resistor value, or is it important to specify an actual part number from a real supplier/vendor (like going on to Digikey or Mouser, etc. and specifying that exact part and serial number)? If it's necessary to specify exact parts, what if I pick a part that is in short supply with the manufacturer? Is there a way to pick parts that are likely to be in stock with the manufacturer already? I guess I'm not sure how it works- do PCB manufacturers order parts from suppliers, or do they typically have generic parts on hand already? My goal is to minimize the cost and lead time for my boards, and I'm ordering in a pretty low minimum quantity as I'm still prototyping right now. <Q> When I have boards built "outside" <S> I issue a full BoM (bill of material) and, if the vendor cannot fit a particular part, they call me (or email me) to ask for a replacement part number. <S> They usually have stock of other devices that will be good enough and, they will usually offer up these parts (and where they originally sourced them from) for my consideration. <S> As an added precaution, I always state what the techy details are on my BoM so typically a resistor will have stated: - footprint size resistor value tolerance <S> drift in PPM/degC voltage rating <S> I also codify "generic" components on my circuit diagram and this code also finds its way into the BoM: - Those component codes also get incorporated into component symbols (which some folk may find a little over-the-top): - I always get boards built with full traceability back to source <S> so that's something else that you might consider if quality is important (as it should be). <S> My goal is to minimize the cost and lead time for my boards <S> Your goal might be to do this but not at the expense of quality - keep hold of what parts are fitted if different to what you originally asked for. <S> Any potential vendor that says something like "don't worry, we know what you want" would be struck-off any vendor list in my opinion. <A> You can ship your own parts to them. <S> You may also specify parts where the assembler supplies their own parts for really generic components like caps and resistors. <S> They should have a bunch of these just on-hand. <S> Other times you specify parts and they order the parts themselves in a key turn solution. <S> But really, these are all very vendor specific issues that you NEED to discuss with them. <S> Certain circuits are pickier with certain "generic" components than others. <S> Surely, high frequency or RF boards can't just take any old SMD capacitor or resistor of the same value that will fit onto the pad. <A> As a rule, even for ‘generic’ parts, always assign at least one manufacturer part number - yes, even jellybean parts like resistors - to your approved vendor list (AVL) for your part. <S> This level of detail is essential for quality traceability, and if your company requires it, for RoHS and ISO audit. <S> Even if you don't care, you should be thinking this way when you interact with a contract manufacturer (CM), even a reputable one. <S> It shows them you're serious. <S> With the reference manufacturer part number in hand the CM can review your BOM and recommend alternates or substitutes as needed. <S> You would review these recommendations and either add or reject onto your AVL for that part. <S> This will make them easier to cross with lower cost parts later. <A> As others have alluded to, you should get in touch with your manufacturer as early as you possibly can. <S> As Andy has mentioned, you should get a feel for how they operate and they will be able to advise on what they do best. <S> Always make sure you are comfortable with what they are doing though - high quality looks expensive on a BOM, but low quality is lot more expensive in the end. <S> Some things to think about: Especially for small runs, it is likely they have generic components installed already and will offer them for free/minimal cost. <S> It would cost them more to change reels to run your ten resistors. <S> When you are far under the minimum order quantity for a component, they may have a suitable replacement or can bundle it with other orders. <S> They may have preferential rates from suppliers, but of course they will add their commission on top. <S> If you opt for them to handle sourcing, your contract should specify that they are responsible for checking <S> it's correct. <S> This might sound obvious, but the error rate in packing components is not zero. <S> If you have supplied them with incorrect components (for whatever reason), you will be on the hook for the rework costs (which are a lot higher than the original work costs..!). <A> Asking them to order and look for specific components will come at a cost if they don't have them in stock. <S> They count work hours even if they don't tell it explicitely. <S> You know what you are buying and what you need exactly. <S> You can decide yourself if a replacement is ok or not. <S> They can't because they will stick strictly to the parameters or part number provided. <S> It can cause useless loss of time and conversations to you and to them if they can't find exactly the thing in the BOM. <S> When you buy yourself, you can keep part of the components and build your own stock. <S> It can be useful to have the components separately. <S> You can check the cost yourself and have a better idea whether a high quality component is worth it or not. <S> Play your qulity/cost ratio, find the best deals, estimate for future bulk production etc. <S> To the question "can you take a generic replacement? <S> " I would answer yes as long as the important parameters are the same. <S> If you are not sure which parameters are important, then just find a replacement with all the same parameters. <S> It's not only true for passive components like resistors and capacitors, but also for pin to pin compatible ic's (still check parameters carefully).
| There are advantages of buying components yourself. When your create your AVLs for your parts, choose well-known vendors when you can that are easy to get from the usual places (Mouser, Digikey) with reasonable lead time, and, just as important, have complete datasheets. Manufacturers can order components for you but, from my experience, they prefer that you provide them.
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Reasons for constraining single-ended Op-Amp VCC when multiple power rails available If I am utilising an op-amp, say a LM321/LM324, say in single-supply operation as a buffer and have a multiple suitable power rails available then are there, assuming I do not expect to utilise the full output swing of the opamp for any of my rails and none of the rails exceed the maximum specification , any potential advantages to using a lower-voltage supply rail? <Q> If you have several choices on your VDD supply for the opamp, here are some thoughts: 1) use the cleanest VDD, and plan ahead to sustain the cleanliness. <S> This minimizes the use of area/cost required for "local battery" R+C low-pass-filters on the opamp's VDD. <S> The reasoning for the is the horrid PSRR of opamps at high frequencies. <S> 2) using an LDO output (if clean enough) instead of a higher-voltage switchreg with its spikes and ringing 3) use the same VDD as your ADC, to avoid use of protective clamps between opamps and ADC 4) if you have a huge amount of gain, as in magnetic-field-detection for robot-location, then using separate VDDs for the first few stages versus the final stages, is a way to reduce the likelihood of oscillation by sharing a common VDD. <S> 5) identify the best-regulated VDD, and use that, if the DC_STABILITY is crucial <A> any potential advantages to using a lower-voltage supply rail? <S> Yes, several. <S> The LM324 for example can take as much as 32 V. <S> But if the output goes to a device powered from eg. <S> 3.3 V then powering the op amp from a similar voltage could:- Prevent injection of current into the power supply of the receiving device (through its input protection diodes etc.), stopping it from malfunctioning or being damaged. <S> Greatly reduce power consumption in the op amp, reducing temperature drift and improving reliability. <S> The LM324 draws a maximum quiescent current of 1.2 mA at 5 V = 6 mW, but 3 mA at 30 V which equates to 90 mW. <S> That might not sound like much, but with a thermal resistance of 88 ° <S> C/W it would cause nearly 8 °C of temperature rise. <S> In a sensitive circuit the op amp could take several minutes to stabilize at the higher voltage. <S> If the device needs to operate at high temperature that 8 °C could push it over the limit. <S> But that's not all. <S> When the load draws significant current the power dissipation can be much higher. <S> For example driving a 150 Ω load to 3 V (drawing 20 mA, the guaranteed minimum output source current) would only dissipate 40 mW with a 5 V supply, but a massive 540 mW with a 30 V supply. <S> If the lower voltage is supplied through a regulator (or secondary regulator if the primary power source is also regulated) it could be more stable and make the op amp's power supply rejection ratio less critical. <S> Lower output voltage swing could reduce EMI caused by a strong signal that produces higher output than needed. <S> A 3.5V square wave is much 'quieter' than a 30 V square wave! <A> One main reason: interfacing your op-amp to other single-ended electronics like an ADC or a logic circuit (e.g., a comparator.) <S> This ensures the output swing of the amp doesn’t exceed the input range of the downstream device. <S> Likewise, if the upstream circuit is single ended, referring the op-amp to a midpoint reference (virtual ground) makes things easier sometimes.
| The lower voltage stress could improve reliability and reduce sensitivity to power supply spikes and stray currents due to surface contamination of the chip and PCB.
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Can I connect this meter to 220vac? Can I connect this analog meter directly to 220VAC without any additional components? And what is the meaning of writing on the right? <Q> The internal 150 kΩ resistor and diode suggest that it is indeed intended to be connected directly to the AC line, but the actual construction would not meet any modern safety standards! <S> The average value of a half-wave rectified sine wave is 0.45 times the RMS value, so at 220 Vrms, there will be about 0.66 mA flowing through the meter, which corresponds to 2/3 of full-scale on a 1-mA movement, consistent with the labeling on the face. <A> I suspect this meter will work (approximately) correctly when connected to your AC Mains supply. <S> The proper way to find out is to use something like a Variac and slowly increase the voltage applied while observing the meter. <S> I base this opinion on the fact that there is a rectifier plus resistor of value 150k in series with the meter movement. <S> I say "approximately" because these meters are notoriously inaccurate. <A> Actually, this device is a moving-coil ammeter (meter movement). <S> The resistor in series to it acts as a voltage-to-current converter . <S> So the combination of them acts as a DC voltmeter . <S> The bare movement has polarity; hence these "+" and "-" on the back. <S> Obviously, the resistor and diode are added later. <S> Note this is still a DC meter that has polarity (there is no blocking capacitor). <A> It's a 91L16 Series 40mm x 40 <S> mm moving coil AC voltmeter, with rectifier, intended for direct connection across mains. <S> It's accuracy class is +/- <S> 5% of full scale deflection. <S> All specifications are symbolically marked on the dial. <S> Here's a list of the symbols generally marked on panel meter dials. <A> The meter is "intended" to be an AC volt meter with up to 300 VAC input. <S> It's super cheap and nasty and should only be used in applications where users are either protected against any risk of shock or fire, if then. <S> HERE is one of many examples of the meter for sale. <S> They specify it as 300 VAC, as you'd hope :-) <S> Also for sale here and here where it is described as a "pointer type VU voltage panel meter moving coil structure AC voltmeter 91L16 <S> 40*40mm" :-) <S> :-(
| The diode in series makes it an AC voltmeter .
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Disconnect car battery with receiver I have no clue what I'm doing but I want to prevent anything down the line of my car battery draining it, so I was thinking of somehow disconnecting everything down line while it is off. Pretty sure I finally kinda sorta maybe understand amps volts, ampere-hours, and how to power and use a relay switch receiver. What I know for sure I don't understand is what to actually disconnect the batter with. All I could find were things that you manually use at the battery to disconnect the negative terminal. The battery is 12 volts and outputs up to 810 amps. Links: https://www.amazon.com/TalentCell-Rechargeable-12000mAh-Multi-led-indicator/dp/B01M7Z9Z1N/ref=sr_1_3?dchild=1&keywords=12v%2Bpower%2Bbank&qid=1587279333&sr=8-3&th=1 https://www.amazon.com/Solidremote-12V-Universal-2-Channel-Transmitters/dp/B01JGDV8UM/ref=sr_1_2?dchild=1&keywords=Remote+Control+Relay+Switch+receiver&qid=1587280585&s=hi&sr=1-2 https://www.autozone.com/batteries-starting-and-charging/battery/duralast-platinum-battery-35-agm-group-size-35-650-cca/377104_258029_25698 <Q> This seems like a waste of time and effort and probably introduces a reliability safety risk. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> General overview schematic of automotive power distribution. <S> and auxiliaries (lights, indicators, interior fan, etc.) are already controlled by the ignition keyswitch <S> so there is no need for additional isolation. <S> The always on power feeds things like the door lights and radio (which may need a password on power-up) <S> and these are really the only leakage paths that you need to be worried about. <S> All I could find were things that you manually use at the battery to disconnect the negative terminal. <S> Isolating the battery negative is the recommended practice for maintenance because it eliminates the chassis return path. <S> Consider what happens if you use a spanner to disconnect the positive terminal while chassis is still connected to negative: if the spanner touches any chassis metalwork while undoing the terminal a short-circuit will occur. <S> That can't happen while disconnecting the negative terminal and once done <S> it can't happen with the positive terminal. <S> If you were to proceed with your scheme you would only need to cut the auxiliaries positive feed. <S> The battery is 12 volts and outputs up to 810 amps. <S> There's the problem. <S> You need a huge solenoid and you've got to keep it on all the time so it will run hot. <S> You've also got to be very confident that it won't fail while running. <S> I suspect you are trying to solve an imaginary problem and can't think why you would want a solution that uses a remote control which introduces another level of complexity and potential random failure. <A> You want to disconnect the battery from the entire car, starter motor included?You'll need a relay that could withstand 800A in that case. <S> Like you, the mechanical cut-outs I've seen are like big screws <S> - Lots of surface area so they can pass lots of current. <S> Hard to get that much surface area in a relay. <A> You haven't mentioned whether you are using this car battery in your car (as its main battery) or outside of the car. <S> If inside the car, you can greatly reduce your control circuit current handling capability by leaving the cranking lead connected (as it is not using power until starting the car anyway); that is your largest load in the car and is best left alone. <S> You could also measure other wires (if there are more than 2) for any current while the car is off, but then you will leave the car's computer without power, and you could run into a few problems like not being able to start or unlock your car. <S> Either way, for very large currents (over 20-30 amps, and especially over 50-100 amps) you need to use a larger and more powerful relay with a significantly more power-hungry solenoid <S> OR you can use a few MOSFETS of a very low on resistance (Rds ON) in parallel, and these MOSFETS would require practically no power while being turned on or off, only a brief spike of current during the turn on and turn off. <S> In order to help you more effectively, you need to give us more information, photos of your setup, drawing of your intended wiring (diagram), etc.
| If outside of the car, you only have to worry about the maximum current being used, not about the maximum current capability of the battery. If we know more about the intended use and the way you want to wire everything up, we can help you better. The starter solenoid, ignition circuits (spark plugs, etc.)
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