Oscillations in BJT (active load project) - Page 1
Forum section: electronics / projects

author: tfm
author_rank: Contributor
timestamp: May 28, 2018, 02:35:36 am
content: Hello,I am set on building a programmable active DC load for the lab at our University. User can choose constant current, constant voltage or constant power. It needs to withstand up to 200V and dissipate 1kW.After some research, I decided to try a first version of it paralleling a couple of power darlington BJTs. Also tried the "active ballasting" found on tAOE3 (the art of electronics), pg 113, Fig 2.82:(I hope I am not infringing any terms here by posting the picture. Please let me know otherwise.)So the circuit below is to be used when constant current is selected, here Vset sets 1A. In the final version, Vset will come from a DAC. Here are just the bare bones of the passing elements + active ballasting:BJTs Xmjh11022 are actually MJH11022. The model came from OnSemi:http://www.onsemi.com/PowerSolutions/supportDoc.do?type=models&category=809Here is the circuit response:I am attaching the Ltspice sim file + lib.It gets worse if current sink I1 sets a higher current.It gets better if load R5 is higher.My guess is that this behavior suggests parasitic oscillations coming from passing elements Q2 and Q4 Ceb, and its Miller effect. Also seems weird that because the way Q4 and Q3 are connected, the capacitance Ceb * (hfe +1), from Q4 should equal Ccb from Q3, since they are shorted. That sounds odd...But I can't figure out precisely what is the cause of the oscillation, specially since there is no real world parasitics (traces/leads inductances, etc) in the simulation.A few notes I think are worth mentioning:1) I have the same results if just one side of the circuit is used (ie, if I delete everything to the left of I1)2) It doesn't matter if I use an op amp in a closed loop to set the current as I did, or if I just replace Vin for a simple DC supply. The oscillation is the same.3) I know that the full specs (160VDC, 1kW) required will release the magic smoke of the passing elements in a moment. These requirements do not comply with its SOA. If I go with the MJH11022, I will probably have to set 4 of them in series, then 4 of those sets again in parallel to go safe. I am just sticking to a simple simulation now.4) I know I can also use MOSFETs, but we are short on power FETs and we have are many power BJTs available, so we are sticking with them. Ironic is that MOSFETs are known to have stability issues, not BJTs5) I couldn't tame the oscillations with resistors at Q1/Q3 base, or adding caps at any place i could think of.6) I did have a clean current, no oscillations, by adding small 33 ohms resistors at Q1 and Q3 emitters (between emitter and I1), but that failed if raise the current, say from 1 to 5 amps.7) If the darlington Q4 and Q2 are replaced for a MJ15003 (single, non darlington bjt) it won't help either, oscillations start just at 1.1 amp, for the same 1 ohm R5 load.8 ) I don't want to increase the emitter degeneration resistors (of the main passing elements), since the whole purpose of the auxiliary Q1 and Q3 transistors + current sink I1 is to ensure equal current sharing and avoid thermal runaway using lower ballasting resistors. I want to dissipate as much power as possible using the passing elements, not passive components.I could use some help to really understand what is happening to this circuit and how to fix it!

author: David Hess
author_rank: Super Contributor
timestamp: May 28, 2018, 12:15:09 pm
content: This does not answer your question but that is an interesting circuit idea. Unfortunately it has the same problem; variation in the Vbe between Q4 through Q6 creates the same imbalance that they were trying to correct. Those transistors need to be matched which practically means operational amplifiers should be used instead as the most economical way to get matched transistors. Driving each output transistors with its own operational amplifier with feedback from its emitter resistor is usually how it is done.I cannot tell what the frequency of oscillation is or the units. Knowing them would help.Q4-Q6 operate with a Vce down to 0.6 volts which should be fine for the BC337 but some transistors will have a problem with this.Does the simulation oscillate with Q4-Q6 removed? They may have too much transconductance in which case adding emitter resistors will help.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: May 28, 2018, 12:33:22 pm
content: Quote from: tfm on May 28, 2018, 02:35:36 amYou need tp look at the FBSOA for these transistors:You will be limited to about 40W dissipation by the second breakdown limits.Regards,Jay_Diddy_B

author: capt bullshot
author_rank: Super Contributor
timestamp: May 28, 2018, 01:40:06 pm
content: The oscillation is a result of U1 (the error amp, providing a large gain) beeing faster than your boost stage (your output transistors and current sharing circuit).It's a common and simple rule of thumb that the error amplifier (here U1) must be slower than the loop - you may want to look up circuits providing either current or voltage boost to an OpAmp in the literature - I believe both The Art of Electronics and Jim Willams' App notes have some examples and notes.Modifying the circuit as shown brings you stability.And yes, you'd need to make R2/R4 large enough so that Q1 / Q3 Base to Emitter voltage variations have small enough influence to the current sharing. You'd also want to keep Q1 and Q3 at the same temperature by mounting them in near proximity.

author: David Hess
author_rank: Super Contributor
timestamp: May 28, 2018, 02:02:46 pm
content: I assumed that "2) It doesn't matter if I use an op amp in a closed loop to set the current as I did, or if I just replace Vin for a simple DC supply. The oscillation is the same." meant that he measured the output at the top of R5 so I ignored U1 but you unintentionally brought up another point here.U1 is an OP37 which is an OP27 decompensated for a closed loop gain of 5 or greater. Adding a feedback capacitor between the output and inverting input will just make things worse and likely create a local oscillation around the OP37 no matter what the simulation says. I have run into this before with the OP37 before I knew any better and some specific values of feedback capacitor sort of worked but the control loop and even the OP37 in isolation was never robustly stable like it should be. Use a unity gain stable operational amplifier.An OP27 would be more suitable but still too fast so it will require external compensation.Note that an OP37 can be externally compensated for unity gain stability if the circuit configuration provides for a minimum *noise gain* of 5 but this is seldom a good idea; increasing the noise gain to provide stability at low closed loop gains defeats the purpose of using a low noise operational amplifier.http://www.analog.com/en/analog-dialogue/articles/techniques-to-avoid-instability-capacitive-loading.htmlhttp://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-25.html

author: capt bullshot
author_rank: Super Contributor
timestamp: May 28, 2018, 02:19:26 pm
content: @David Hess:You're right with your concerns about the OP37. I wouldn't use that Amp here also.As I tend to put a generic OpAmp into such simulations, I didn't look at that detail. The simulation is to see if it can work in general, but for the real circuit there's still left optimization to do on the real prototype, this is my approach to simulation. This includes component choice.@tfmI forgot to mention, this brings stability together with the 33R resistors in Q1 and Q3's emitters. Having no emitter resistors, the circuit still oscillates, as it does with the control loop removed. I messed up the schematic a bit while searching for the oscillations cause, so I forgot about these.So we've got two sources of oscillation here:The output stage itself, which needs the 33R resistors in Q1 and Q3's emittersand the error amplifier, which needs compensation as shown, and as David mentioned, you should better use a unity gain stable amplifier.P.S.Don't connect a real OpAmp as shown in your original circuit (without series resistors to it's inputs). Especially these OP27 style amplifiers have protection diodes across their inputs, which will start conducting at more than 0.6V voltage difference between + and - input. With low impedance sources as in your circuit, the amplifier will be destroyed at first turn on due to the high current through these diodes. This is why I added R11. Anyway, R11 is also required for C1 to have an noticable effect.I also increased R7 to 100R, just because this is the "rule of thumb in this context" value for an emitter followers base resistor - there's been a nice RAQ about this recently:http://www.analog.com/media/en/analog-dialogue/raqs/raq-issue-151.pdf

author: SiliconWizard
author_rank: Super Contributor
timestamp: May 28, 2018, 02:28:16 pm
content: Try replacing the opamp with an AD8510 for instance. It will behave.

author: schmitt trigger
author_rank: Super Contributor
timestamp: May 28, 2018, 03:52:36 pm
content: "Safety devices hinder evolution"Could not agree more. It prevents the natural application of Darwin's law.

author: tfm
author_rank: Contributor
timestamp: May 28, 2018, 05:42:51 pm
content: You are right, I gave it a quick look, that is why I mentioned:QuoteBut now I see that I've overlooked the real numbers. For that part, it will probably take 6 sets of four in series. Thanks for bringing it up!

author: tfm
author_rank: Contributor
timestamp: May 28, 2018, 06:17:00 pm
content: Quote from: capt bullshot on May 28, 2018, 02:19:26 pmI see, that is a nice approach. But I had issues with that in the past, precisely for what is happening now again: an ideal op amp will hide potential failures that otherwise may be caught at the simulation stage.QuoteI understand. Great. Could you elaborate (or point a direction) about the output stage instability cause - that requires the 33R resistors?I took a look at the HF response of BJT and HF impedance gyration, trying to understand better the oscillations on such a simple circuit - particularly Dennis Feucht articles:http://audioworkshop.org/downloads/AMPLIFIERS_OSCILLATION_BJT_CIRCUITS.pdfhttps://www.edn.com/Home/PrintView?contentItemId=4401129Acording to these articles, being main Q2/Q4 emitter followers, a small cap across base resistors R1/R3 should make it better, but they actually make things worse (without the 33R resistors).What is the cause of the oscillations (just considering the output stage), and why the emitter degeneration resistors fix them?QuoteGreat tip! Thanks.@David Hess: thank you for your replies. Very instructive.

author: tfm
author_rank: Contributor
timestamp: May 28, 2018, 06:48:52 pm
content: Quote from: David Hess on May 28, 2018, 12:15:09 pmIf I remove the op amp, then it does not oscillate with Q4 and Q6 removed.However, with Q4/Q6 present (and no op amp, just direct drive) it oscillates at ~7MHz:QuoteWhat was bugging me is that I've seen in more than 2 references (one of them tAOE3) that this current sharing trick was extensively used by some old HP/Agilent power supplies. It is elegant, I'd like to make it work in the real world.

author: capt bullshot
author_rank: Super Contributor
timestamp: May 28, 2018, 06:57:52 pm
content: OK, I can't give you all the theory, I'm rather a guy of practice. I'm used to solve complex differential equations in real time and watching the result on a a visualization device together with a mixture of experience, rules of thumb and a portion of gut sense. (vulgo a prototype circuit, a scope or other suitable instruments, and a good mixture of educated guesses, gut sense and some experimentation, combined with a bit of simulation and rough calculation).So here's my approach:From theory I know (yes, I've got the complete professional EE education with a Diploma, but I forgot most of the details of theory), circuits tend to oscillate if there's too much gain or too less phase margin. Your output stage is expected to work with simple BJTs from my experience (and the AoE circuits usually are expected to work, since they result from practice), but using a darlington tends to mess up stuff. So there's too much gain here, and darlingtons tend to be slower, so your phase margin is also gone. You'd need to reduce the gain, that's what the emitter resistor does. Some oscillator circuits don't look much different from that output stage, just some reactive components (L/C) added. L/C parasitics are everywhere in the real world.Any emitter follower has a tendency to oscillate, that is why you add a base resistor. Bypassing the base resistor with an capacitor make the resistor pointless for higher frequencies and the oscillation starts over again (that's what you've seen with caps across R1/R3).If you carefully watch your simulation with the emitter resistors set to a low value (say 1R), you see the current in the 0.05R resistors is anti-phase. So this thing acts somehow as a differential stage going into oscillation. You've got oscillation, then you'd reduce gain (add emitter resistors, these reduce the differential gain) or compensate (add collector to base capacitors, but ensure there's high enough impedance in the base driving circuit, add a base resistor if necessary).Opening the loop (as you did by replacing U1's output by a voltage source) is a good way to see where oscillations start from. A more advanced way would be to use a step (or square wave) source to see the step response of your circuit, that reveals otherwise hidden tendencies to oscillate.

author: tfm
author_rank: Contributor
timestamp: May 28, 2018, 07:02:45 pm
content: Quote from: tfm on May 28, 2018, 02:35:36 amAnother question about this circuit. Horowitz/Hill say that this circuit works particularly well with MOSFET (also on Figure 3.117, pg. 214) and with power Darlington BJTs, because of their negligible base/gate current.But what is the caveat of using regular non Darlington BJTs ??

author: capt bullshot
author_rank: Super Contributor
timestamp: May 28, 2018, 07:13:30 pm
content: Don't have the AoE at hand ATM, so I cannot look it up now, I remember the circuit, but not the caveat you mention.I can imagine, the issue with non-darlingtons is the larger base current required by them (as a rule of thumb, the darlington may have an hfe of 200...2000, a power BJT rather in the ballpark of 10..100), so for a 10A output you'd have to supply maybe a total 0.2A base current, the current sharing steering transistors may want the same or double amount of collector current to operate properly, this results in a good amount of power dissipation in the steering transistors and a lot of output current the error amp has to provide, you won't get away with these small ones.

author: David Hess
author_rank: Super Contributor
timestamp: May 28, 2018, 09:04:45 pm
content: Quote from: tfm on May 28, 2018, 06:48:52 pmThat is about right for a Darlington transistor oscillation. The control loop for an operational amplifier in this situation would usually oscillate much slower.QuoteIt is a great idea as I mentioned but I think the reason HP was able to get away with it was by using matched transistors or possibly a transistor array. You can certainly do that but transistor arrays are more expensive and less available and separate matched transistors will not track as well with temperature; see my answer below. It may seem wasteful to use a 20 transistor integrated operational amplifier for each power transistor where a transistor array would be adequate for several but the former is a lot more economical than the later.Quote from: tfm on May 28, 2018, 07:02:45 pmThe necessary drive current is so much higher that it degrades the thermal tracking of Q4 through Q6. This might not matter if Q4 through Q6 are part of an integrated transistor array or hybrid which is likely what HP used.

author: jbb
author_rank: Super Contributor
timestamp: May 29, 2018, 12:40:40 am
content: At risk of being a killjoy, you have a 200V rating. Why not just use boring single BJTs with huge emitter resistors? Surely you don’t need a very low compliance voltage? (problem: emitter resistors must e low inductance.)Maybe a realistic solution is just to deploy 1 opamp per power transistor. It’s crude but you can divide and conquer.I’m also curious about how you intend to manage the series connection. Were you thinking about a cascode connection?Oh yes. With 2kW dissipation you’ll find that the heat sink itself has quite different temperatures along its length (with respect to airflow). This might make it harder to do the current sharing.

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 03:25:45 am
content: Quote from: capt bullshot on May 28, 2018, 07:13:30 pmQuoteI am afraid I am slow on this one. The base currents of the power BJTs have to be significantly higher compared to the darlington/FET counterparts. But that base current is summed at the power BJTs emitters and will add a small heating effect on the same power BJTs - small compared to the much higher current through them.Why do you say that "the current sharing steering transistors may want the same or double amount of collector current to operate properly"?If we consider the Figure 2.82 from tAoE, the sum of the Ie currents through the auxiliary BJTs is obviously 30mA, so if one power BJT has its Ic raised (because of thermal effects) compared to the others, then the voltage drop at its emitter resistor (50mohms) gets higher. Therefore auxiliary BJTs base voltage gets higher, then they get a bigger share of the 30mA and the drop on its 220R collector resistor gets higher (and their Vce lower). Since this 220R resistor is also the base resistor for the power BJT, its Vbe gets lower and so does its Ic current.If the above is right, then all the auxiliary BJTs do is compensate the thermal effect on any Ic (from power BJTs) differentially, compared to the others, dropping the voltage at the base of the power BJT that is ramping up its Ic/Ie current.The higher base current from the standard power BJT (non darlington) could make it more difficult for the auxiliary BJTs to regulate the base voltage of the power BJTs, but that isn't the case again, since the higher Ib would also set a somewhat higher drop at the 220R base resistor.I am sure I am missing the critical point here... would you care to show me?

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 03:40:53 am
content: QuoteI understand your point. And agree. But we have to do a sanity check. If that holds, and HP did use transistor arrays or matched ones, well then they could simply have used matched power BJTs, mount them together (which they did anyway) and use smaller passive ballasting resistors.Are matched power BJTs more difficult to get than small signal BJTs? I don't know.BTW, did they actually measure the auxiliary BJTs in case they didn't use the arrays? I don't think so, even in not-so-large production quantities. And (if) how come they did use the arrays, being more expensive? Those have always been more expensive, right?

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 05:04:01 am
content: Quote from: jbb on May 29, 2018, 12:40:40 amHi, jbb! Don't mind, your suggestion will never be a killjoy!There is no hard reason I won't go with a single BJT, really. But since there is no critical time constraint for doing it, I'd like to take a look at the options. And even if, in the end, I settle on the ol' good single BJTs, this whole exercise of looking into these details is fun to me!!Now, one thing I'd like to avoid is huge emitter resistors. It is kind of just ugly to me, because all that power (BTW, 1kW, not 2kW) will require a massive heatsink + convection. So I should put some effort to dissipate all the power on the part of the circuit that is meant to take it! Not on the resistors. Might be just silly...QuoteI am pretty sure the 1 opamp per BJT will work, and it is indeed realistic. There is no reason it shouldn't be done like that, from my perspective, really. I am just trying to make this suggestion from tAoE work too. Better? I think so,ifit doesn't need matched aux BJTs. I guess that is yet to be seen, despite David Hess solid points - please see my next post. Simpler? I guess it will be, after it is been done.QuoteWell, I am considering a cascode or just spreading the Vbe using plain resistors. I haven't tried on the bench with such a high 200V, but it should work:You know, if we go berserk, and have 20 of those in series, we'll still be wire-bound limited:With the advantage that it is in series, so the same current is guaranteed through everyone.QuoteYes, but eventually, if the active current sharing approach works, it should suffice to tune the different legs until thermal equilibrium is reached.

author: capt bullshot
author_rank: Super Contributor
timestamp: May 29, 2018, 05:14:09 am
content: Quote from: tfm on May 29, 2018, 03:25:45 amYou point at the critical point: The higher base current. In your idea and if every transistor and resistor was born equal, only a small collector current might be necessary for the steering transistors to regulate the current sharing. In practice, it isn't. The power BJT would need quite a high base current, that is set for the worst case lowest power transistors hfe plus headroom. For a quick and dirty calculation, this might be the ten- to twentyfold current than with a darlington. For proper current sharing, the steering transistors would have to take away current from the base for the worst case highest power BJT hfe plus headroom, which may be triple the worst case lowest hfe. From this, a worst case steering transistor collector current of double the maximum required power BJT base current results. This in turn results in quite high power dissipation for the steering transistors and all the related stuff like thermal tracking.You can get better figures if you select the transistors for hfe, but that's usually a rather expensive process, so the preferred way would be to design your circuit for large tolerances and take measures to keep critical things like power dissipation in the steering transistors low. Then you end up with darlingtons or MOSFETs. By making the power darlingtons emitter resistors larger, you'd reduce the requirements on Vbe and hfe matching of the transistors. So your circuit design will always be a compromise between cost (for selected components), precision (how perfect is the current sharing), and power dissipation (trade larger emitter resistors for cost reduction).I don't have any problem to figure HP and others could do the power supply circuitry without any matched components by taking the right compromise, but I'm not aware of a specific model or circuit. If someone has an example, I'd like to see it. Service manuals of that age often provide you with all schematics and detailed circuit descriptions, one can learn a lot from that.

author: capt bullshot
author_rank: Super Contributor
timestamp: May 29, 2018, 05:20:37 am
content: Quote from: tfm on May 29, 2018, 05:04:01 amDoing so is kind of silly indeed. Power resistors can dissipate way more power than silicon devices for a given size, and dissipating power in a suitable resistor would be the preferred way for most designs.But I do understand your considerations, if you set yourself a goal to achieve, the way to get there can be way more entertaining and educative than just going for the standard way.BTW - if you just want to dissipate power in a resistor, you'd put it into the transistors collector, not the emitter. Makes things easier to design, since the parasitics of the resistor have less influence on the transistor. You'd still need an emitter resistor to operate the transistor as a current sink, but you'd design that resistor the usual way and for way much lower power dissipation than the collector resistor.

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 05:46:00 am
content: You know, that need for matched aux transistors is really bugging me. David has a point, but...On pg 113, tAoE mentions this approach, then again on pg 213 for FETs, and then they say at a footnote that "found this cute circuit trick used in some HP (...) E3610-series linear power supplies. It’s much simpler than using individual op-amps to bias each transistor, as some MOSFET manufacturers suggest".Matched stuff is not a cute trick, and is not much simpler than op amps, since once you replace them, you might be in trouble.Here is a thread where Hill advises someone about this approach, to be used in a real circuit:https://groups.google.com/forum/#!topic/sci.electronics.design/TfOGqGhjTT8And no one mentioned the need for matched aux transistors.Here is the schematics for the e3610a:https://sites.fas.harvard.edu/~phys191r/Bench_Notes/A1/agilent_e3610a.pdfOn pg 15 you can see the parts list. Q4,5 are the aux BJTs that diferentially regulate the power FETs: plain 2x 2n2222a. And the current sink is just a 100k trimpot connected to -12V. (BTW, Q4 and Q5 are drawn incorrectly: emitter and collector are inverted!)

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 05:56:21 am
content: Quote from: capt bullshot on May 29, 2018, 05:14:09 amOh, man. All this time I've been thinking about Vce at aux BJTs... and about the power BJT Vb. Oh man.Thanks.QuotePlease check my previous post.

author: tfm
author_rank: Contributor
timestamp: May 29, 2018, 06:01:59 am
content: Quote from: capt bullshot on May 29, 2018, 05:20:37 amWell, I am not so sure. If you consider the area or volume of, say, a 10W cement or woundwire resistor vs a 250W TO-247. Not to mention the clumsiness of a heatsink on the former.But I get your point too. Aesthetic is not a good reason to avoid a higher rated power resistor that could potentially simplify a lot.

author: capt bullshot
author_rank: Super Contributor
timestamp: May 29, 2018, 06:11:09 am
content: Quote from: tfm on May 29, 2018, 05:46:00 amThanks. If you look up all the component values for the different models, you'll find the current sharing is used only for the 60V/120V model. And here, the source resistors are quite large and have a high voltage drop, that easily compensates for component tolerances. The current sink can be tiny here, because the MOSFETs have zero static gate current. A resistor of large value across a relative high voltage can act as a poor current source, that's a common technique for differential amplifiers. Search for "long tailed pair" if you're interested.QuoteThat's their replacement for your 33R emitter resistors to cancel the oscillations. Quite neat, learned something new.Edit: that's wrong, after finishing reading that thread, it's just a wrong diagram. Transistors are connected as usual.But the whole power stage appears to be designed with tenfold headroom for everything ... Look at the large voltage drop across the output transistors, this leaves headroom for very low line voltages at the expense of high power dissipation. I've seen this in other gear too.

author: capt bullshot
author_rank: Super Contributor
timestamp: May 29, 2018, 06:17:48 am
content: Quote from: tfm on May 29, 2018, 06:01:59 amOne wouldn't mount the resistor on a heatsink in most cases. Resistors have way much higher surface temperatures allowed than your TO-247 transistors and do their dissipation by radiation and air convection. Of course, there are power resistors intended to mount on a heat sink, and these usually are larger size than your transistors. It's a matter of selecting the appropriate resistor. One can get open frame wire wound resistors that can dissipate 1kW in less volume than the heatsink required for the same amount of dissipation with silicon devices. One can also get transistors that can dissipate 1kW (you'd still have to be aware of the SOA). Most power resistors have way more overload tolerance than silicon.This is an example of a transistor that can dissipate 1kW:http://wunderkis.de/pwramp3/index.html

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: May 29, 2018, 06:52:50 am
content: Hi tfm and the group,I suggest that you look at this thread:https://www.eevblog.com/forum/projects/dynamic-electronic-load-project/In this thread I showed how I designed an electronic load that was stable. You should be able to apply the same techniques to your load. I suggest reading the whole thread, but you could start at reply 19.Regards,Jay_Diddy_B

author: David Hess
author_rank: Super Contributor
timestamp: May 29, 2018, 03:31:58 pm
content: Quote from: tfm on May 29, 2018, 03:25:45 amThe problem is thermal mismatch between Q4 though Q6 which will alter their individual Vbe at about -2mV/C.When the change in collector current of Q4 though Q6 necessary to match the voltages across the emitter resistors of Q1 through Q3 is small, the differential heating effects on Q4 through Q6 are also small and their Vbe change due to temperature effects will be small.If bipolar transistors are used for Q1 through Q3 instead of Darlington transistors, the collector current for Q4 through Q6 will be 10 to 100 times as great; it increases in direct proportion to the hfe of the Darlington's driver transistor which will be quite high. So the difference in power dissipation between Q4 through Q6 also increases by 10 to 100 times degrading their Vbe match due to temperature differences by the same amount. So the current sharing using bipolar transistors instead of Darlington transistors is much worse because it causes greater heating in Q4 through Q6. There are various ways to combat this including lowering the collector current of the transistors used to enforce matching or lowering the thermal resistance between them to lower the temperature differences which is what integration does. Using an integrated operational amplifier does both.

author: David Hess
author_rank: Super Contributor
timestamp: May 29, 2018, 03:36:49 pm
content: Quote from: tfm on May 29, 2018, 03:40:53 amAbsolutely in the sense that matched small signal transistors can be produced simply by making them part of an integrated circuit. Even a poor integrated circuit process and layout can achieve matching better than 5 millivolts and with some care in layout, better than 1 millivolt is easily feasible.Integrated matched power transistors are very rare and offhand I know of no examples. Some expensive high power integrated circuit processes support them.Some hybrid modules used matched power transistors with the hybrid construction enforcing better thermal tracking than discrete parts mounted to a common heat sink.QuoteMany manufacturers graded discrete transistors to produce matched sets and this is sometimes still done. Integrated transistors arrays are only more expensive if the required quantities are too low to justify their production. But integrated transistor arrays are not always suitable to the circuit due to other factors like parasitic coupling between the transistors or to the substrate so matched discrete parts are sometimes still used.

author: David Hess
author_rank: Super Contributor
timestamp: May 29, 2018, 03:40:15 pm
content: Quote from: tfm on May 29, 2018, 05:46:00 amThe voltage across the sense resistors reaches 3 amps * 0.6 ohms in the E3610 so 1.8 volts. Variation of Vbe in the 2N2222A transistors will be small compared to that no matter what their thermal tracking is so there is no need to match them. It also helps that either MOSFETs or Darlington power transistors were used so their collector currents are lower than if bipolar transistors were used.I was only addressing Hill's (?) comment that tracking would be worse with bipolar transistors. The increased power dissipation through the balancing transistors increases the mismatch because of thermal effects. This may be completely acceptable if taken into account.In the past, matched transistors or a transistor array would have been the most economical way to improve on this situation but today, using an integrated operational amplifier to enforce current sharing is by far the least expensive and highest performance solution.

author: iMo
author_rank: Super Contributor
timestamp: May 29, 2018, 11:51:14 pm
content: This works fine, no oscillation with I_R5=5.00A and Vc=250V.