Why do I keep killing these MOSFETS? - Page 1
Forum section: electronics / beginners

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 12, 2017, 10:26:37 pm
content: Hi,The circuit in the attached image should be a simple constant current load, but I'm having some unexpected problems.1. I keep killing one of the MOSFETS. I know when this happens when all of a sudden the current jumps to over 12 amps. For what it's worth, I measured about 1kgate to source and 2.5ksource to drain on the last failed one. The MOSFETS are mounted on a relatively large heatsink with a 12V DC fan. The heatsink gets hot, but not so much that I can't touch it. The MOSFETS are mounted uninsulated, but I don't think that should matter since the drains are tied together anyway.2. The current tops out at about 3.3A, corresponding to about 9V on the inverting inputs (forcing the output voltage to the rail since the non-inverting input voltage is greater). By my understanding it should top out at over 10A, since R1-4 in parallel, while hot, measures about 3.5and the Rdson of the MOSFETS is .04, low enough to be trivial by comparison. I've since noticed that my supply is putting out 36V instead of 50V, probably a bad filter cap or something, but even then I figure I should be getting about 10A max.3. The op-amp seems pretty unstable. This is the least of my concerns right now, since it seems to be working well enough that I can adjust the current in the circuit with the pot (up the 3.3A, at least), it just fluctuates a lot. I'm wondering if this has something to do with the fact that I'm using a breadboard for the op-amp part of the circuit? I don't think this is a defective op-amp, unless I have a handful of defective LM324s.I suspect there may be a simple explanation for all this that will make me hang my head in shame once revealed, but for now I'm stumped (this is the beginner's forum, after all...). I'm also wondering if I shouldn't rebuilt this circuit with BJTs, since in my experience they are less fragile than MOSFETS.Anyway, any input is appreciated. Thanks.Edit: updated with revised circuit

author: pelule
author_rank: Frequent Contributor
timestamp: March 12, 2017, 10:43:17 pm
content: As two MOSFETs are never exactly the same (just near the same, when produced at the same lot), you need to align them with a smal resistor. See the International Rectifier AppNotes:http://www.infineon.com/dgdl/para.pdf?fileId=5546d462533600a401535744b4583f79

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 12, 2017, 11:15:23 pm
content: Thanks for the reply, but I'm not quite sure I follow you. I get what you're saying, that I can't assume the MOSFETS to be identical, but I'm not sure how to align them with a resistor (I'm still working through that linked doc, but some of it's a bit over my head I think). Anyway, I was trying to account for mismatch by driving them with different opamp outputs. I now realize that this didn't matter since I had the inverting inputs tied together, and have modified my design accordingly. Is this what you were talking about?That all being said, does mismatch even matter that much here, given what the MOSFETS I'm using are supposed to be able to handle? They're rated for 50A continuous and 300W power dissipation. Like I said, the circuit is maxing out at 3.3A with a 36V power source, so that's about 110W dissipation from the MOSFETS (once you account for what's dissipated from the power resistors).

author: rstofer
author_rank: Super Contributor
timestamp: March 12, 2017, 11:28:58 pm
content: What's your voltage between the gate and source.Ideally, your MOSFETs would turn fully on and the dissipation would be Rds*Id which should be a very small number.I seriously doubt that you can get up to 10 Vgs and get there quickly.I would look at a MOSFET Gate Driver - they are designed to drive the gate hard enough (amps, not milliamps) to overcome the capacitance in a small amount of time and will provide sufficient Vgs to ensure that the device is fully on.

author: Benta
author_rank: Super Contributor
timestamp: March 12, 2017, 11:36:30 pm
content: I'm just wondering what on earth induced you to connect two opamps like this? Neither of them has a defined feedback path, they're competing and perhaps happily oscillating at a high frequency.1: remove one of the opamps and drive both FETs from one. Keep the gate resistors and connect both to the (now single) opamp output.2: limit the voltage from the pot by adding a series resistor, so the the input voltage to the opamp cannot be higher than +15 V - 2.5 V = 12.5 V

author: The Soulman
author_rank: Super Contributor
timestamp: March 12, 2017, 11:42:56 pm
content: Quote from: rstofer on March 12, 2017, 11:28:58 pmYes in a switching application, this a linear dummy load.Load sharing resistors for such a application are typically around 0,1 Ohm to keep size and power dissipation minimal.

author: Benta
author_rank: Super Contributor
timestamp: March 12, 2017, 11:45:32 pm
content: Quote from: rstofer on March 12, 2017, 11:28:58 pmThe MOSFETs are operating linearly. VGS is probably around 3...4 V.Gate drivers are for switching, but not all MOSFET applications are hard switch types. Just sayin'

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 12, 2017, 11:50:13 pm
content: Quote from: Benta on March 12, 2017, 11:36:30 pmMy idea was to drive each FET from its own opamp because there is a full 2V range that the threshold voltage could fall in, according to the datasheet, so I didn't want one to be above Vt and the other not, etc. I noticed earlier that my design was wrong in that their inverting inputs were tied together. My modified design as attached to my second post has each of them with their own feedback loop to accomplish this.

author: Benta
author_rank: Super Contributor
timestamp: March 13, 2017, 12:01:32 am
content: Quote from: CaptainNomihodai on March 12, 2017, 11:50:13 pmSorry, didn't see your second schematic before I posted last. Yes, that solves the feedback problem.Did you test it yet?

author: raspberrypi
author_rank: Frequent Contributor
timestamp: March 13, 2017, 12:10:42 am
content: Dont you want to tie pin 1 and 2, and 13 and 14 together with a resistor in series? Every op amp I hae seen has this set up to keep it balanced.

author: David Hess
author_rank: Super Contributor
timestamp: March 13, 2017, 12:33:13 am
content: The MOSFETs are not sharing current at all; only one is being used at at time.

author: AndersJ
author_rank: Frequent Contributor
timestamp: March 13, 2017, 12:44:01 am
content: The two feedback lines are crossed.The op-amps have no feedback.

author: rstofer
author_rank: Super Contributor
timestamp: March 13, 2017, 01:47:05 am
content: Having the load resistors in the bottom side means the source voltage is increasing as the load increases. As a result, the gate voltage needs to go even higher to get Vgs into range.Ordinary MOSFETs take a LOT of gate current to switch in small time. Small time is required to keep the device out of its active region which is where all the heating occurs. Sometimes logic level MOSFETs will work with lower gate voltages but when you see specs like Vgs=10V at 28A versus 3V at some low number of mA, you know you have to drive the gate hard and high.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 13, 2017, 04:00:37 am
content: This is not how you parallel MOSFETs in linear operation.LM324 is also a particularly poor choice.Tim

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 13, 2017, 04:29:46 am
content: Quote from: T3sl4co1l on March 13, 2017, 04:00:37 amCan you (or anyone) elaborate on this? Please note that I have fixed the error in the original schematic (now removed) where the feedback lines were shorted together. Also what's wrong with an LM324? I mean, all the opamp has to do here is match the voltage across the resistors, it doesn't need to be particularly fast or anything, right? I picked it because it could run single supply.Quote from: rstofer on March 13, 2017, 01:47:05 amWell, since this is a dummy load, Iwantmy MOSFETS in the active region. That being said, I think your first sentence answers my question as to why I'm topping out at 3.3A... I was failing to account for those resistors pulling the source voltage up and assuming that my Vgs was the gate-ground voltage.Quote from: David Hess on March 13, 2017, 12:33:13 amI'm pretty sure this is false. I don't see how that could be the case, and my measurements indicate otherwise. Even if it were true, though, these FETs are rated for much more than I'm giving them.So, question #2 is answered, but it's still not clear as to why my FETs keep dying.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 13, 2017, 05:19:59 am
content: Quote from: AndersJ on March 13, 2017, 12:44:01 amyup and further investigation the logic reveals a nightmare. when a mosfet flows too much current, it will turn off the other mosfet. this is race condition not sharing, who is infront will further push its contender to the back, making him alone win the race = burnt. or maybe a rescue condition, you shutdown let me burnt. except when the the first one died in open/disconnect mode, the other will follows. but if the burnt one got shorted, the other will be spared.

author: MK14
author_rank: Super Contributor
timestamp: March 13, 2017, 05:25:03 am
content: How well heatsinked are the Mosfets ?The 300 Watt rating, is done by keeping the Heatsink (with zero thermal resistance between case and heatsink, in practice it won't be zero!) at a constant 25 deg C, while soaking up 300 Watts of heat. I'd be interested in knowing how you are doing this ?Hint: It can only cope with a lot less than 300 Watts in practice.As others are trying to explain to you. You seem to have the op-amps the wrong way round, in that they are controlling the wrong/opposite Mosfet.I.e. Each Op-amp is sensing the current and then controlling the mosfet, connected to theOTHERpair of 10 Ohm resistors.This could be fixed by swapping the output connections around (pins 1 and 14).Possible causes of Mosfet destruction could be if they are overheating or if the gate voltage is (even for a tiny amount of time, like a microsecond), exceeding around 20 volts.There are much better op-amps, than the LM324.

author: Paul Moir
author_rank: Frequent Contributor
timestamp: March 13, 2017, 05:53:30 am
content: QuoteThis is a myth I have personally suffered which must die. There is no difference between a "single supply opamp" and any other opamp. The claim comes from that the LM324 could operate within ~1.5V of V+ and right down to about V- so could be useful with a +5V/0V supply, while the ancient dinosaurs it was competing with like the LM741 couldn't operate within something like 5V of V+orV-. Otherwise same thing. Of course with today's low voltage designs, most modern opamps are "rail-to-rail" or near enough that single supply operation is no problem at all.

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 13, 2017, 06:21:05 am
content: Quote from: MK14 on March 13, 2017, 05:25:03 amI've got them on a relatively large heatsink with a fan. It gets warm but not too hot to touch. Really good to know that the 300W figure is unrealistic, though.And, oops, those feedback lines are indeed crossed. I think that was just a mistake on the schematic, but I'll check the physical circuit tomorrow.Regarding gate voltage exceeding ~20V as a means of failure, is there any way, in this case, that that could happen since the opamp driving the gate is powered by only 15V? I don't see how, but then again I didn't notice that those feedback lines were crossed either...

author: BrianHG
author_rank: Super Contributor
timestamp: March 13, 2017, 06:25:41 am
content: Doesn't the LM324 latch up if either of the inputs ever reach the rails, ie 0v, or 15v.I would use an opamp which supports rail-rail inputs at least in this circuit, or, add series resistors feeding the op-amp inputs, namely the feedback - input, to prevent internal latchup.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 13, 2017, 06:34:39 am
content: Quote from: CaptainNomihodai on March 13, 2017, 06:21:05 ampossible if you have inductive path to the gate. when opamp is slewing off hard, your gate path become boost converter. put fast acting diodes from gate to 15V and from gate to gnd to show some respect to the mosfet.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 13, 2017, 08:20:13 am
content: Quote from: CaptainNomihodai on March 13, 2017, 04:29:46 amCertainly,1. You need separate source resistors. Otherwise there is no way to sense the currents of the individual transistors. (This has been fixed.)2. Feedback needs to be to each respective op-amp. (This is a new bug that has been introduced.) As shown, you have a bistable flip-flop, which will hog as much current as possible through one transistor.3. Your transistors are fuckoff massive, as far as most anything in this circuit is concerned, except for power dissipation. (This is the price to pay for using ancient switching transistors in linear applications.)The "size" of a transistor refers to the gate capacitance (which the op-amp output pin must drive), its current capacity and Rds(on), the V*I capacity of its SOA, etc.Modern transistors are about 5x smaller on gate capacitance, for the same V*I capacity (of course, they're about that many times lower in power dissipation, too).If you've learned op-amps in school, I'm sorry to break it, but you've been lied to. There is absolutely no such thing as an ideal voltage source. No op-amp has one attached to its output pin. This isn't a terrible approximation at DC, but when things start moving, it's a preposterous error. A reminder that: a circuit is always moving, and even if your intent is to have it doing absolutely nothing, sitting statically at DC, it still must reach that operating point first, and if it spins out of control on its way there, that's your problem. There is no such thing as an "intent sensing circuit".The circuit does precisely what it does, no matter what you wanted it to do...Since op-amps are nonideal, we must consider the effect of the transistor's gate capacitance on its output pin. Capacitive loading is almost always a bad idea, and LM324 falls into the set of op-amps where this is true. Most likely, this circuit will oscillate, just from capacitive loading, without any other help from the transistor and load.To prevent oscillation, we must compensate the amplifier. The first step is to add a series gate resistor Rg (which you have -- a good start). The next is to add a series resistor R to the feedback path (i.e., from shunt resistor to -in). Finally, a C or Rz+C is connected between op-amp output, and -in. The R*C time constant must be on the order of the Rg*Ciss time constant, and Rz (on the order of R) can be used to adjust damping.It looks like this, except with an optional resistor in series with C1, and with the junction between R2 and R3 being the MOSFET (in which case, C3 represents its gate capacitance, which won't exactly be going to ground, but the shunt resistors are on the small side, so it's close).Measure damping by making step changes to the reference potentiometer (most often, by replacing it with a square wave generator), and observing the stability of the output current (rise time, undershoot / overshoot, ringing).Note that stability will vary with what kind of load is attached, as well as the current setting.4. LM324 is bad. Just, in general. It's okay for slowly varying servo loops, and poor quality audio applications, but not much beyond that. It has much more noise than most op-amps, mediocre input offset, slow response, and worst of all, a class B output stage. What does that mean? The exact output voltage isn't quite equal to the voltage "intended" by the op-amp's inner workings (namely, the voltage on its internal gain node), but has some "slop", just the same way a shaft, that fits loosely in its bearing, has some slop against rotation. It can be locked to a position or angle at one end, yet is still free to move some at the other end.The op-amp's input side eventually sees the error, and corrects for it, but this takes time. The output is the time-integral of the input, so over a short period of time (a couple microseconds), you can change the output by a quite noticeable amount (fractional volt), with very little current (indeed, this effect is strongest around 0mA output), until the op-amp corrects the error (10s or 100s of us later).By the way, lie #2: if you were taught op-amps are infinite gain, they aren't. They have high gain at DC, but having high gain at AC would be not only absurd, but physically unrealizable, and impossible to use (if not other things, like violating causality itself!). No, an op-amp is an integrator: its output changes over time, based on an input. An integrator has infinite gain at DC (ideally), so this works identically at DC, but the gain drops smoothly with frequency, eventually dropping below 1 at a well-defined frequency (beyond which it isn't useful; for the LM324, this is a couple of MHz).One more rub: the output pin, besides having nonzero impedance, and a really messed up impedance that depends on output current -- it can only source so much current, around 10mA. And when it's delivering a large current, it's not able to pull all the way up (only maybe 2-3V below +V), or down (maybe 1V above GND). This fundamentally limits how fast the gate voltage can be changed. This also provides a rationale for choosing Rg: it should be on the order of Rg = (+V) / (Iout(pk)). Or about 1kohm in this case. (That beefy >10nF gate will therefore have a time constant in the >10us range, which honestly isn't terrible considering that's only a little slower than the poor op-amp can go, anyway.)Now, with enough Rg to allow the output pin voltage to move freely, and an awareness of the limitations of the LM324, and external compensation to stabilize it, you have a hope of getting useful operation out of the circuit.Tim

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 13, 2017, 10:01:10 am
content: Quote from: T3sl4co1l on March 13, 2017, 08:20:13 amto be precise, LM324 short circuit and source current spec is 40mA.to the OP, your gate resistor is 100 ohm, at some points you are burdening the opamp since you theoritically try to pull 15V / 100 = 150mA from it way beyond its spec. dont be surprised if your LM324 will burn as well soon. set the burden current half of spec, 20mA or 30mA if you are too desperate for current juice, that makes Rgate = 15 / 0.02 = 750 ohm. btw forget the LM324 bashing, people have made usefull stuffs out of even an UA741 chip. LM324 is certainly perfectly fine for you application, given that you correct the surrounding details explained such as feedback path., and especially the feedback path.

author: MK14
author_rank: Super Contributor
timestamp: March 13, 2017, 10:29:44 am
content: Quote from: Mechatrommer on March 13, 2017, 10:01:10 amTo be even more precise, his 10mA is apparently actually bang on!If you are going to correct someone else, then it is best to get your facts right (I may regret saying that, at some point!).The 40mA on the datasheet seems to be a typical figure at room temperature.https://www.onsemi.com/pub/Collateral/LM324-D.PDF(I started using the TI datasheet, but it seemed to contain a technical error/mistake (as regards the output current), so used the onsemi one).But10mAis guaranteed as a minimum (and across the full temp range). (Or 20mA at room temp, but best to allow for the internal case temp to get quite warm, so the 10mA is still the better figure to use).There are a number of limitations with the LM324, which might make choosing a better op-amp, a good idea. Such as its limited voltage output swing, compared to more modern parts (e.g. full output rail to rail). They (modern/better/rail-to-rail-output) are still relatively cheap, so why not use them ?

author: BrianHG
author_rank: Super Contributor
timestamp: March 13, 2017, 11:32:55 am
content: Quote from: MK14 on March 13, 2017, 10:29:44 amHis power-up negative input will be at 0v as well as potentially his + input depending on pot setting, he also needs rail-rail input support, or at least an opamp which is guaranteed to not seize up under this condition.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 13, 2017, 11:46:48 am
content: Quote from: MK14 on March 13, 2017, 10:29:44 amso i guess correctness is to its own context. your linked datasheet is not much different to my reference here (regarding this matter)Quote from: MK14 on March 13, 2017, 10:29:44 amname the rail-rail opamp the same or close price point as LM324 or TL074/072/071, 30V++ supply range. i'll buy it everyday, as i'm clearing up my old stock.

author: MK14
author_rank: Super Contributor
timestamp: March 13, 2017, 11:46:58 am
content: Quote from: BrianHG on March 13, 2017, 11:32:55 amI was only trying to start the ball rolling (onto better op-amps), when I mentioned rail-to-rail outputs. The more additional specifications I mentioned, the more likely I would cause someone to disagree with some of it, so I avoided piling on the extra specs.Even when I choose an op-amp for my own projects, I can get into huge dilemmas as to which one to choose. With some of the specifications (potentially), you have to choose what one, two or three requirements will be met, well. Then compromise on the other specifications.E.g. Price vs functionality vs ease of use, etc etc.Sometimes even the lack of availability in some packaging formats, limits ones choices.

author: MK14
author_rank: Super Contributor
timestamp: March 13, 2017, 12:16:24 pm
content: Quote from: Mechatrommer on March 13, 2017, 11:46:48 amI tend to agree with you.If he/she (OP) was doing this for a production run, with tight cost targets and/or does not want to spend a dollar or few, on better op-amp(s). Then I agree with you.I also have been disappointed/surprised at how pricey the better op-amps can be. For hobbyist or one-offs (or very low volumes) it does not matter that much. But for any kind of serious production volume (and tight cost targets), it is a problem/issue.Also breadboard friendly DIP packages, seem to be largely forgotten in the later/latest op-amps as well, forcing the use of adapter boards, if breadboarding. Which also probably adversely hits some of the beginners on this forum.

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 13, 2017, 02:54:18 pm
content: Thanks for all the help everyone. I'll implement these suggestions when I get home from work today.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 13, 2017, 03:39:39 pm
content: Quote from: MK14 on March 13, 2017, 12:16:24 pmif it can be bought next door it doesnt matter, otherwise buying 1 piece from china or america land to here is not practical either even for hobby job. i tend to buy parts at least 10 pieces for each chip as many as i think i'll need to save shipping cost. cost can multiply dramatically for high end chips and is going to be problematic for poor hobby like me who has very limited portion of budget to invest in the hobby. edit: and we hobbiest need to buy excess parts to consider for blowing out during building and testing. blowing out $10 part is not fun.

author: MK14
author_rank: Super Contributor
timestamp: March 13, 2017, 04:10:47 pm
content: Quote from: Mechatrommer on March 13, 2017, 03:39:39 pmI agree that the availability/affordability of components, varies between different hobbyists, how old they are and where they live (Country). Also how much of the hobbyists wealth, they are willing to commit to their electronics hobby.Going partly off-topic to reply to the post.The problem with many of the older, standard op-amps, such as the 741 and many others. Is that the microcontroller (used in many projects), may be running at a rather low voltage, such as 3.3VSome of the modern, rail to rail input and output op-amps will still work fine at 3.3V, for use around the A2D converter of the MCU.This makes hobbyists who expect to often do microcontroller projects (not all hobbyists), running at lower supply voltages, needing to stock up on such op-amps. E.g. For more complicated A2D circuitry.In the OP's case, without analyzing their circuit in too much detail. They need the op-amp to be able to climb above the 10 ohm resistors voltage drop, and have enough extra voltage to drive the (non-logic level) Mosfets. The LM324's may run out of output voltage headroom, as they are not rail to rail output, op-amps.As discussed in other posts, some of the other advantages of better op-amps, can also help in circuits like the OP's.But I also concede that if the circuit was better designed (as also already discussed). It would be able to work better, even with a LM324.My opinion is that the LM324 is still useful in some simple/limited circuit use situations. But as the op-amp circuit becomes more and more demanding/complicated. Better/modern op-amps, become increasingly important.I'm NOT brave enough to say on a public forum, that a LM324 with a clever enough circuit design CAN perform the OP's circuit function, for fear of being criticized too heavily.Proof/Example: All I said so far was a rail to rail output, would be good/useful, and I'm already getting lots of flak

author: David Hess
author_rank: Super Contributor
timestamp: March 13, 2017, 06:24:33 pm
content: Quote from: MK14 on March 13, 2017, 04:10:47 pmI am. The LM324/LM358 will work fine in this application although CaptainNomihodai does not give any performance requirements. Performance will be limited by its speed and drive capability and the fact that it is not a precision operational amplifier. At worst, some frequency compensation will be needed.1. Its input common mode range includes its negative supply.2. Its output stage is current limited3. Its output cannot sink significant current at the negative supply because of a base-emitter junction but the Vgs threshold of the MOSFETs makes this unnecessary.4. It only suffers from phase reversal if an input is pulled *below* the negative supply. I do not see that happening here.Before I knew better, I built a very similar circuit for testing inductors and ended up using an LM318. *That* is asking for trouble but it worked great.

author: bson
author_rank: Unknown
timestamp: March 13, 2017, 06:50:15 pm
content: I'd add gate pulldowns so the gate can't drift towards drain enough to exceed max Vgs and kill the MOSFET while the LM324 output is in the crossover region. (I'd also use an opamp that's biased to be active over its entire output range.)

author: David Hess
author_rank: Super Contributor
timestamp: March 13, 2017, 07:20:12 pm
content: Quote from: bson on March 13, 2017, 06:50:15 pmI agree with adding the pulldowns for safety reasons but if the LM324's output stage is in its crossover region, it will not be if the gate tries to drift anywhere.Take a look at the LM324 schematic and you can see the problem. Ignore the current limit transistor and there are 3 Vbe junctions in series with the bases of the upper Darlington and lower emitter follower tied directly to together at the class-A output of the previous stage. So the output can wander about 3 Vbe total or 1.8 volts, so what?Solve both problem if they are that with one resistor from the gate to ground. If the MOSFET's Vgs threshold is about 5 volts, then a 1k resistor between the gate and source will sink 5 milliamps protecting the MOSFET and forcing the LM324 output into class A operation. This also adds 5 milliamps of error on the output so use a larger resistor to ground if this is not acceptable or something more complicated.

author: MrAl
author_rank: Super Contributor
timestamp: March 13, 2017, 07:49:52 pm
content: Quote from: CaptainNomihodai on March 12, 2017, 11:15:23 pmHello,Based on your reply in reply #2 your feedback arrangement is not that good.One op amp depends on the other op amp which then depends on the first op amp again which then again depends on the second op amp..etc. That is both unnecessary and harder to analyze intuitively.To fix this is really super easy. Just use the pot reference for one op amp ONLY, and use the output of one transistor as the reference for the other op amp, which then will be forced to output the same voltage within a tiny percent. The circuit is shown below. In this new circuit, the second op amp depends on the first circuit output but that's the end of the analysis, and it must output the same voltage across the second set of load resistors because it's reference is the first output and it's feedback is from the second set of load resistors where it has to regulate the voltage. In short it is a tracking regulator.Note this does not say anything else about the circuit, such as the incredibly high power dissipation in the two mosfets at some operating points.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 13, 2017, 07:52:57 pm
content: Simple fixes:1. 10mA minimum is a good point to start reasoning from. 20 or 40mA typical would also be fine. It's not like it's critical. Certainly nothing to get pedantic about. (I say this, and I'm one of the biggest pedants here!)Current is also limited by dV/dt, since I = C * dV/dt. 10mA into 10nF is 1V/us, a bit slower than the LM324 is, so the resistor will be useful (i.e., it will have a worst-case voltage drop near half the supply voltage, so it's doing something, and it's not too big, not too small).If the gate were, say, 1nF or less, it wouldn't be a big deal, because the LM324 can't move that fast in the first place. A 100 ohm resistor would be fine: mainly to avoid possible oscillation. (External compensation would still be recommended.)2. The LM324 will not blow up. It might get warm if it breaks into full oscillation. Even in a short, it's rated for "continuous" duty. You have to work to destroy one of these (or, as the phrase goes, "it takes a special kind of stupid"...).3. The transistor won't blow up. 15V is well within its rating.That said, a G-S zener diode (preferably a large one, like P6KE15.0A or SMAJ15A) will help keep G-S voltage within limits, even under surge conditions.That is, consider what happens when you connect a large 50 or 100V capacitor to the load terminals: the voltage rises within nanoseconds, limited only by series inductance and circuit damping. The peak current, during that transient, may be many amperes, and a fraction of it will be seen by the LM324 because the transistor has so much capacitance.Designing the circuit around transistor failure is also handy; in that case, you might make the series gate resistor, and source/shunt resistor(s) fusible types, and add a protection diode to the LM324 output pin, so it doesn't get damaged by transistor failure (i.e., suppose it fails D-G shorted, S open).4. If low supply current consumption is not necessary, then a pull-down resistor will stabilize the LM324's crossover distortion under quiescent conditions. Step changes in one direction will still be messy, but it will eventually settle out of it.This, by the way, is another handy feature of "single supply" amps: they are well behaved on VEE-referenced resistive loads. In this case, it keeps the high-side output transistor active, so it behaves like a simple emitter follower. Only when the output needs to change rapidly, will the low-side transistor be turnedTim

author: MrAl
author_rank: Super Contributor
timestamp: March 13, 2017, 07:56:37 pm
content: Quote from: T3sl4co1l on March 13, 2017, 07:52:57 pmHi,See the new circuit in the post just before yours. It's a good way to do it as long as the mosfets can handle the power.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 13, 2017, 08:10:27 pm
content: Quote from: MrAl on March 13, 2017, 07:56:37 pmYes, this works, though it's rather peculiar. A simpler shortcut is simply tying both +in's to the pot wiper: same reference, but not passed through a feedback loop.Tim

author: Audioguru
author_rank: Super Contributor
timestamp: March 13, 2017, 09:01:38 pm
content: The circuit is simply an opamp driving a Mosfet that is a source follower. There are two of these circuits, cross-coupled but they do not oscillate, instead one latches on and the other latches off.The one that is latched on dissipates about 72.5W in the Mosfet if the supply is the shown 50V so it overheats and is destroyed.The Mosfet can dissipate 300W when its case is held at or below 25 degrees C with liquid nitrogen or something else extremely cold.I show the voltages when power is applied and only one Mosfet becomes turned on. There is no input signal to switch the circuit so that the other Mosfet turns on and it turns off the one that was already turned on.

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 14, 2017, 04:15:42 am
content: Here is the circuit updated based on suggestions I've received here. The Zeners are 14V because I have a bunch of them (and that's higher than the opamp will output anyway). R1 and R2 don't have values right now since I need to go through some of my parts bins to find low value resistors that can handle a fair amount of power (I'm currently paralleling a bunch of big power resistors to get about 2.5R, but it's pretty hideous and I'd like a lower value), and then the value of R3 depends on that.Since that 300W dissipation figure is apparently fictitious, I guess I'll have to parallel some more FETS in order to run this circuit up to 10A like I was hoping...I haven't actually built this new one yet, I'll update once I do.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 14, 2017, 05:37:17 am
content: if you use zener, i think diode to 15V is redundant. put the zener/diode as close as possible to the gate and drain pins to avoid inductance downstream. i suggest pulldown resistor on opamp +ve input since if the trimpot got damaged and wiper disconnect from resistive element, opamp +ve input will be floating and go whereever it like to. if it goes to gnd its ok, if it goes up then its not ok. a weak >100K (>10X the trimpot impedance) pulldown resistor should be ok. if you are me, i'll connect R6 and R7 to GND, not the high side of power resistor, value should be also greater than 10X the series R4,R5. adding capacitance C1 and C2 like that will only worsen the mosfet turn on and off criteria, i dont like that unless i have a strong driver.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 14, 2017, 07:07:45 am
content: Almost -- move R6/R7 to the right, in series between C1/ZD1, C2/ZD2. And move C1, C2 to before R4, R5.Also, for increased power dissipation on the cheap, connect a resistor in series with the load. If the load is always 50V and 10A, then a 5.0 ohm resistor will draw a constant current from a constant 50V supply; all the transistors need to do is be turned on, dissipating almost no power themselves.Any smaller amount of resistance increases the voltage compliance range (i.e., how low the voltage can go, while still being able to draw up to 10A), while still reducing the power dissipation in the transistors.Tim

author: jdraughn
author_rank: Regular Contributor
timestamp: March 14, 2017, 10:07:02 am
content: I skimmed the first few replies, but didn't see where anyone mentioned the pinouts of the mosfets. You show 2 going to gate, but when I googled the mosfets part number, it goes G,D,S, pin one being gate. Looks like your blowing them by putting 50v into the gate.

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 14, 2017, 04:08:55 pm
content: Quote from: jdraughn on March 14, 2017, 10:07:02 amThe pin numbers on the schematic are wrong (I guess the KiCAD library is wrong?). I have them wired according to the datasheet. Based on what people have said, I'm guessing that what kept killing the FETs was inductive kick on the gate due to the long wire between the gate and the pin and/or having my feedback lines crossed (which I never intended to do -- having them crossed on the schematic was just a mistake -- but they were plugged in and out of the breadboard enough times that it could very well be that they were crossed at least once).

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 15, 2017, 11:30:56 pm
content: New schematic. R9 and R10 are just to take some of the stress off of the FETS. It may be a while before I actually build these modifications since I need to find the proper resistors.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 16, 2017, 12:20:11 am
content: Perfect!

author: CaptainNomihodai
author_rank: Contributor
timestamp: March 17, 2017, 06:06:29 pm
content: Something occurs to me that makes me wonder if this circuit will work the way I intended, even with all these corrections....My intent here is that the FETs operate in the linear region, acting essentially as variable resistors and dissipating a lot of power. However, the linear region is when Vgs > Vt & Vds < (Vgs -Vt), correct? The problem I'm seeing is that there is no way that Vds < (Vgs - Vt), i.e., my supply is 50V, if I have, say 1R between source and ground and another 1R between power and drain, just to lessen the burden on the FETs a bit, that still leaves, at 5A per FET, a Vds of 40V forcing it into saturation, and things only get worse if I try to run at lower currents, since the voltage drop across the resistors will be less.Now that I've typed the above, I'm even more confused because my understanding of MOSFETs (which is obviously wrong, apparently) now dictates that a FET shouldn't be able to stay in saturation without an insanely huge current. My reasoning is that Rdson is generally close to zero, but saturation requires a relatively large Vds, so you would need a large current to create that voltage drop.I know I'm wrong, but where?

author: MK14
author_rank: Super Contributor
timestamp: March 17, 2017, 06:15:43 pm
content: One option is to put the circuit into (often free) simulators, such as LTSpice. Then you can play around with the circuit fairly quickly and try out stuff like that and see how it behaves. You can then learn more about how and why it works (or not), the way it does.Simulators are not necessarily 100% accurate, but it should give you an approximate idea on how the circuit will work out, WITHOUT blowing up Fets and things, while you are learning/experimenting etc.http://www.linear.com/designtools/software/#LTspice