Power Supply Load Tester - Page 1
Forum section: electronics / projects

author: Chryseus
author_rank: Regular Contributor
timestamp: March 22, 2013, 11:05:44 pm
content: Hey guys I came up with this design for an active load for testing power supplies and was wondering what you think of it, I've designed it for a maximum load of 300W (30A @ 10V).The output is set by the non-inverting input to OA1, SW1 shorts the input so it's used as a on/off load switch (I suppose it could be in series also), OA2 amplifies the voltage drop across the current shunt R4 to provide a proportional voltage output, OA3 basically just buffers a fraction of the input voltage for voltage sense (D1 provides a little over voltage protection), the main pass transistors are two 2N3055 in parallel driven by two IRF530 (I'd probably use IRF540) with a worse case current from the +15V being about 4A.One thing I do want to add in are thermal cutouts since this thing will be generating a good amount of heat, as far as calibration is concerned I will probably do that mostly in software.I'm pretty sure it will work as expected but I'm quite interested if anyone can suggest ways to improve it, just need to get some more parts before I do some real testing.

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 11:19:44 pm
content: Definitely some things that could be changed, though I'll take my time with a response so it's not stupid. Just wanted to point out one thing - I was trying to figure out what the heck you were doing with the long-tailed pair, until I realized it's not a long-tailed pair, it's just Q1 and Q3 in parallel. Might want to rearrange those to make the arrangement more clear.

author: kxenos
author_rank: Frequent Contributor
timestamp: March 22, 2013, 11:28:39 pm
content: Here are my comment with a glance: D1 is shorted. It needs to connect to ground. Plus putting both Qs in the same loop will probably make it oscillate. It's better to have 1 op-amp per transistor. Also probably you will need some capacitance in these FB loops.

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 11:32:04 pm
content: I'll just list a couple quick ones, first:If you use MOSFETs as the actual load transistors, you won't need that Darlington-esque configuration.The clamping arrangement with OA3 and D1 is not good. You'll be directly shorting the output of OA3, which it won't like very much, and with that much current through a 5.1V Zener, there is no way you are going to clamp the voltage to 5V anyway. Use a lower Zener (4.7V, maybe) with a series resistor to limit clamp current (connect the negative feedback to the op amp after the series resistor to counteract any leakage through the diode)What on earth is Q5 doing? I'm not sure what's going on there, but I do know one thing - the voltage put out by OA2 is going to be subtracted by Q5's VBE, which will be highly variable, so it's kind of pointless to put it through the ADC.What is the point of C2?If C4 is supposed to be filtering the DAC input, there should be a resistor between the DAC and C4 to form a low-pass filter. Otherwise, you're just asking the DAC to drive a capacitive load, which it may or may not like, but it won't do you any good.What is the point of C1? It is highly likely to cause instability.Even without C1 this is likely to be unstable. You will need at least some capacitance between the OA1's output and inverting input.The only time you need an LC filter on the power supply is if you're trying to do it without a regulator at all, which you're clearly not. Even then, a 1 mH inductor isn't going to help much at mains frequencies. The inductors they use for that are as big as transformers.The large value of R18 (LED resistor) suggests a relatively high input voltage, so those voltage regulators are going to get very hot. (That, or the voltage isnothigh and the LED will be very dim.)The +15V is fused at 5A? What kind of regulator is that? Have you looked at the power and temperature calculations for that regulator putting out 5A? Even with just 17V in, it's going to be dissipating 10W. (My suggestion is not to use a bigger regulator with a better heatsink, but to design this not to need that much current. I'm also building a load and it can draw 10A while running off a 9V battery.)

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 11:32:48 pm
content: Quote from: kxenos on March 22, 2013, 11:28:39 pmOops - didn't notice that one! Thought it was connected to ground. (Note I still commented on it - even if it were grounded, it's not OK.)

author: amspire
author_rank: Super Contributor
timestamp: March 22, 2013, 11:56:29 pm
content: c4757, sound comments. I will just add a bit extra.Quote from: c4757p on March 22, 2013, 11:32:04 pmYou will not get 150W from a 2N3055, and you will get a big current error due to the large base current that will be included in the current measurement but will not go through the load.Using MOSFETs rated for Safe Operating Area in the DC linear mode is much better, but as discussed many times before, the best way to have multiple mosfets in an active load is to have a seperate regulator circuit for each MOSFET.Transistors are easier to parallel (with an emitter resistor in each) and you could drive with a MOSFET connected from the transistor common base connection to the common collector connection. This arrangement will mean the current in the load is not identical to the current in the sense resistor.What is the minimum load voltage you need?By the way, you will need a powerful sense resistor - at least 10W, but it will get hot, and that could cause it to drift. You might be better off going for lower offset opamps and a sense resistor of 1mohm.QuoteDefinitely. Just get rid of itQuoteYes - it has to go.QuoteThis current source regulator arrangement without any frequency is usually unstable except it can be stable for very low sense resistor values. With 7.5mOhms, you might be lucky, but otherwise you will need to add a 10K resistor from the regulator opamp input to the sense resistor, a capacitor from OA1 inverting input to output (maybe 1nF ?), and a 100 ohm to 1K resistor from the output to each IRF510 is a good idea. Not sure what the 10K on the opamp output does - delete it.QuoteOnce you divert the power transistors' base currents back to the load as I suggested above, the 15V rail power consumption will drop from many amps to down in the 10's of mA. Definitely make sure the LED gets a decent amount of current - several mAs at least.If you are going to add fuses, put them before the regulator not after. Fuses are resistors and so they degrade the regulation of the supply rails.

author: c4757p
author_rank: Super Contributor
timestamp: March 23, 2013, 12:17:06 am
content: Couple more - then I'll stop, before you start to feel like I'm a vulture diving in to pick your circuit apartQuote from: amspire on March 22, 2013, 11:56:29 pmQuick comment about that - I used to use 7-10mA as a rule of thumb for indicator LEDs, but they have gotten much more efficient over the years, and now I find that 10mA through most typical red LEDs is annoyingly bright. I tend to use 3 or 4 mA now. (Seems like many people have forgotten this rise in efficiency, since indicator LEDs on consumer products have been getting blindingly bright lately...)Regarding the Zener diode (more schematic pickiness, really) - don't just label it with the part number, put the Zener voltage as well. I was mildly annoyed to have to dig through a datasheet to see that it was a 5.1V diode. It can also be useful to label values of other parts/systems that are not immediately evident (operating points, op amp/voltage divider gains,the power supply voltage, etc.)Be careful when designing in fuses. Carelessness quite often leads to your semiconductors protecting the fuses, rather than the other way around, and I suspect that is going to happen here. It's just reckless to fuse a 30A load at 50A, unless it's so massively overengineered that it really can tolerate 50A.At 30A that current sense resistor will be dissipating 6.75W. Can it handle it? What about temperature rise and temperature coefficient?Read that again - the 7.5 milliohm resistor will be dissipating 6.75W. 7.5 milliohm is not an unreasonable resistance for awire. You're going to need some massively heavy wires and massively thick PCB traces. (At such a high current, I'd probably try not to run the current through the PCB at all. Transistors and sense resistor can both be external.)

author: Chryseus
author_rank: Regular Contributor
timestamp: March 23, 2013, 02:09:07 pm
content: Quote from: c4757p on March 22, 2013, 11:19:44 pmI will certainly do that, this is not the final version of the schematic.Quote from: kxenos on March 22, 2013, 11:28:39 pmYour right about D1, I will be removing that and probably doing something else for over-voltage protection, in any case there is quite a wide margin for up to 60V input.I'm no expert with op-amps but why would 1 op-amp driving two transistors be more unstable than 1 op-amp per transistor ?Quote from: c4757p on March 22, 2013, 11:32:04 pmI plan to use only MOSFET pass transistors in the final schematic, I use the BJTs because I have a bunch of 2N3055 laying around.I did not really think D1 throughQ5 was just intended as a buffer (emitter-follower) although I can't really see a point of it now.I originally put C2 to filter noise, looking at it now I think putting it in parallel with R11 would be a much better idea.C4 is just for de-bouncing the switch, I will probably change that to be part of the digital board in any case.C1 I have no idea why I put there, I will move it between the output and inverting input of OA1 as you suggested and knock down the value a bit.As for the CLC filter on the supply I assumed that would decrease ripple, a quick simulation certainly supports that but if you have any other reasons why it's a bad idea besides cost I'd like to know.R18 is high because I find most LEDs are way too bright for my tastes above a few mA, the peak input voltage is expected to be around 19V from a 15VAC transformer, I will move the LED to the 5V in any case.As for the 15V and 5V I'm still not fully decided on those, power dissipation isn't really a huge issue as they will be mounted to fairly good heatsinks, perhaps I will design a buck converter instead at a later stage.Quote from: amspire on March 22, 2013, 11:56:29 pmI'd did not really think the current error through, as said before I will be going to MOSFETs in my final design.As I asked before what is the advantage of using op-amps for each transistor rather than just one?The minimum load voltage I expect to be a few volt, granted I will probably not get 300W at such low voltages without exceeding the transistor current ratings but I don't really mind that.I was planning to use a large power sense resistor, although using low offset op-amps also seems like a good idea, it really depends what is easier for me to get and price.Quote from: amspire on March 22, 2013, 11:56:29 pmI'm not that experienced with op-amps in terms of stability, could you suggest any reading material on the subject?What is the purpose of adding series resistors to MOSFETs, I was under the impression they draw very little current, the only thing I can think of is it would increase the turn on time.The 10k on the output is to ensure the MOSFETs turn off, this is not needed ?I'm not sure what you mean by a 10k resistor on the op-amp input to the sense resistor.Diverting the transistor base current to the load ? you mean using the load to drive the transistor base current ? wouldn't that increase power draw from the load ?As said before I prefer my LEDs to be dim.The fuses before the regulator is an excellent point I will be sure to change it.Quote from: c4757p on March 23, 2013, 12:17:06 amBetter you guys point out the problems than have a bad circuit!Quote from: c4757p on March 23, 2013, 12:17:06 amAs mentioned before this is only a draft schematic, the final one will be of very high quality as is usual with most of my schematics.The 50A fuse is a bit extreme I agree, maybe I can find something decent just over 30A, I don't expect it ever to blow unless there is a totally catastrophic failure, I plan to add thermal cutouts as well.I need to take a closer look at what current shunt resistors are available, I was planning on using a large one in any case with thick wiring (8awg maybe?) to avoid voltage drop, only low current stuff and the digital control board will be on PCB, maybe the power supply as well.Thanks for the excellent feedback!

author: Ghydda
author_rank: Regular Contributor
timestamp: March 23, 2013, 09:13:51 pm
content: QuoteNo it would not as the base current will go into your sense resistor and thus will be a part of the load that the opamp will measure and account for.QuoteYou do not have a 60 volt input range due to the SOA of the transistors. You need to account for an over voltage situation during maximum current draw in the load, and thus the transistors cannot tolerate going anywhere near the rated collector-emitter voltage (or drain-source voltage if MOSFETs are used).QuoteThe problem will arise when the darn transistors heat up. Read up upon thermal runaway.Also if using MOSFETs in parallel from the same driver/opamp output you can in rare situations arrive at a condition where the MOSFETs will oscillate at many MHz due to the parasitic capacitance between drain and gate.QuoteThe reason why you should always use a gate resistor is that the gate is a small capacitor and very few opamps are comfortable driving a capacitive load - thus you ad some series resistance. And yes that will in fact lower the achievable bandwidth, but I'll bet the slooooooow LM324 will act as your real bandwidth limiting factor in any case, so why worry about it.

author: madires
author_rank: Super Contributor
timestamp: March 23, 2013, 10:25:30 pm
content: Quote from: Chryseus on March 23, 2013, 02:09:07 pmIf you use BJTs you should add an emitter resistor (something like 0.1 or 0.22) to each of them to help distributing the current. Otherwise one of the BJTs will take most of the pain and simply die. If it doesn't fail shorted (and blow the fuse) it will be followed by the remaining one.

author: Chryseus
author_rank: Regular Contributor
timestamp: March 23, 2013, 10:28:53 pm
content: Quote from: Ghydda on March 23, 2013, 09:13:51 pmAhh of course your right I didn't think about that, but since I'm going MOSFET only that no longer really matters.Quote from: Ghydda on March 23, 2013, 09:13:51 pmThe MOSFETs I'm going to be using are two IRFZ48V rated for 60V 72A 150W 12mohm each which should do the job, the controller is constantly going to be monitoring the voltage, current, power and temperature, I think 50V maximum input should be acceptable.Quote from: Ghydda on March 23, 2013, 09:13:51 pmIf I remember correctly MOSFETs are immune from thermal runaway.I've never seen an op-amp oscillate driving MOSFETs but I guess there is no harm in using separate op-amps, better safe than sorry.Quote from: Ghydda on March 23, 2013, 09:13:51 pmThat makes a lot of sense and as you said I don't really need the bandwidth.Quote from: madires on March 23, 2013, 10:25:30 pmYeah I've decided to go with MOSFETs as well in my own build, at this sort of current using a BJTs is a bit of a waste.Here is what I currently have:

author: hlavac
author_rank: Frequent Contributor
timestamp: March 23, 2013, 10:49:30 pm
content: 10k MOSFET gate resistors seem unnecessary high, 10 Ohm will stop the ringing on the gate without imposing unnecessary bandwidth limitations.

author: hlavac
author_rank: Frequent Contributor
timestamp: March 23, 2013, 10:53:02 pm
content: As for need for bandwidth, a good power supply tester will allow for step response transient testing, which requires quite fast changes in load...

author: Chryseus
author_rank: Regular Contributor
timestamp: March 23, 2013, 10:58:04 pm
content: Quote from: hlavac on March 23, 2013, 10:49:30 pmNow that you mention it does seem a bit high given the amount of capacitance, I'll decrease it to 100 ohm which should be more than enough.Quote from: hlavac on March 23, 2013, 10:53:02 pmI never thought about that, I think I will setup a small test circuit to look at the bandwidth response.

author: Ghydda
author_rank: Regular Contributor
timestamp: April 01, 2013, 10:50:26 am
content: Quote from: Chryseus on March 23, 2013, 10:28:53 pmYour original target was 30 volt / 10 amps.That gives 5 amps for each MOSFET. So far so good. Let's look the SOA for the IRFZ48V.First of all, IR doesn't even specify running this at DC - that is warning number 1.Second, the 10 msec line suggest maximum 7 amps for the 30 volt drain-source. You can bet you sweet heiny that it will not cut 5 amps at DC steady state. Based on my personal experience I would estimate the SOA at DC steady state is between 1 and 2 amps with 30 volt across drain-source.Third: these numbers are valid only if you keep the case of the transistor at a nice 25 degree C while burning your proposed 5A*30V = 150 watts into the transistor.Andthatalone requires an all but impossible heatsink solution.Quote from: Chryseus on March 23, 2013, 10:28:53 pmFirst, MOSFETs are indeed not prone to thermal runaway. But that statement is based on the presumption that the MOSFETs are driven in a saturated condition. And in your application you are not doing that. Therefore the thermal runaway risk applies and the gate drives should be separate for each MOSFET.Second, it's not the opamp that causes the oscillation. Now hang on a second because this is quite hardcore. When two MOSFETs are hardwired in parallel, drain to drain, gate to gate, source to source, you have a setup where the parasitic capacitance in the MOSFETs and the parasitic inductances in the wiring will cause the MOSFETs to interact with each other. If the right conditions are met the MOSFETs will act sort of like an astable multivibrator. The MOSFETs will each turn on an off very very rapidly (when attempting to turn on or off from the gate driver).To be fair, this rarely happens as the conditions for this phenomenon to arise needs to be just right. Chances are that you will not meet those particular conditions. But, when doing commercial designs this phenomenon needs very much to be dealt with.One obvious way is to separate the gate drives as you have done.Third, the new proposed design absolutely guaranties a "thermal runaway"-like condition. The two opamps each form a control loop ensuring the current in the sense resistor matches the input reference value.You cannot have two control loops measuring a common output and achieve current sharing between them. One of the MOSFETs will pass all the current and the other will pass nil. The solution to this is fortunately very simple; split the sense resistor into two; one for each control loop.

author: amspire
author_rank: Super Contributor
timestamp: April 01, 2013, 11:15:57 am
content: Quote from: Ghydda on April 01, 2013, 10:50:26 amAbsolutely correct. The current is only shared properly if the mosfets are saturated.The only mosfets that can be paralleled are the ones where the drain resistance dominates over the gate sensitivity. In general, these will be mosfets with a high gate turn-on voltage.You can hardly pick a less suitable mosfet for linear use or parallel use then the IRFZ48V. It has a low 2 to 4V gate sensitivity - that means that almost certainly, the channel resistance will give no assistance in current sharing - even within the cells within a single mosfet device. It has no DC safe operating spec. That means it should not be used in DC linear mode. The manufacturer never designed it for DC linear mode - it is optimized purely for switching.Any mosfet where gate sensitivity prodominates over channel resistance will have a problem thermal runaway.You will have to replace the IRFZ48V with a mosfet with a DC safe operating area specification, and you will have to drive each mosfet separately. You cannot parallel them, unless they are the very rare mosfets designed for parallel operation. (Some of the audio power amp mosfets). The mosfets designed for parallel operation will have a high RDS, and a large gate turn on voltage. They will not be mosfets ever used in a switching circuit.Richard.

author: fbs
author_rank: Newbie
timestamp: April 01, 2013, 11:29:40 am
content: Wouldnt it be better to use an opamp designed for driving capacitive loads, something like the mc34072?

author: mzzj
author_rank: Super Contributor
timestamp: April 01, 2013, 10:56:27 pm
content: Quote from: c4757p on March 22, 2013, 11:19:44 pmI am not sure if this was addressed already earlier but above like circuit is doomed to fail. Mosfet parameters vary wildly and the result is that one of them is hogging all the current and another half of the output stage is not doing anything.