uA741 rail splitter with complementary Darlington emitter follower - Page 1
Forum section: electronics / beginners

author: wobbly
author_rank: Regular Contributor
timestamp: September 17, 2024, 07:29:23 pm
content: Not a problem-solving question this time - the circuit works nicely! In fact it works better than I thought it would...This is closely derived from a post I found on EE stack exchange earlier (damned if I can find it now, sorry). Built it on a new breadboard and used 4 discrete (TO-92) transistors and a uA741 op-amp. Supplied it with 10 Volts DC and it sprang into life and works very nicely. I'm greatly pleased.But this raises a question in my mind... the 741 is not a rail-to-rail op-amp. But yet it is still able to keep the "output" GND rail at a very steady and fairly accurate mid-point between the +/- 5V rails. Since the 741' outputs can get to within about 2 volts of it's power rails, how does this affect my circuit's performance, given that it looks to be working well under zero load?I pondered this for quite a few minutes and then noticed that the current draw from the bench PSU was very slowly increasing over time. Perhaps 1 mA every few seconds. Grabbed the thermal camera and noticed that the two BJTs feeding into the GND "output" were slowly warming up. Thermal runaway perhaps?This was all without any load on the output GND and power rails.Turned everything off and waited for it all to cool down to ambient again. Powered on again and then started to use a 100power resistor to poke each power rail (from GND) in turn. Both +5V and -5V rails seemed to keep regulating just fine, not even a twitch on the scope. Great! But this stress caused the BJTs to heat up a bit faster - not drastically so - but it was noticeable. Once they got to about 80°C they tended to keep warming up even after the load was removed (definitely thermal runaway).Again this is all without heatsinks. No harm done though.Is this heating of the Darlington BJTs caused by the fact that the 741 is not a rail-to-rail op-amp? If I were to use a better op-amp (suggestions welcome) would this heating be reduced?Obviously I'll be using bigger integrated Darlingtons in the final circuit and with a chunky heatsink. But I was just curious about the cause of the heating under no load.Insights welcome. Cheers.

author: g0fvt
author_rank: Newbie
timestamp: September 17, 2024, 07:45:22 pm
content: The thermal runaway is somewhat inherent and not related to the 741 itself, you may improve the thermal stability by epoxying the diodes to the transistors.

author: ledtester
author_rank: Super Contributor
timestamp: September 17, 2024, 08:06:01 pm
content: Quote from: edavid on September 17, 2024, 07:45:52 pmIn Falstad the m suffix stands for milli so perhaps the resistors are just modeling ESR.

author: wobbly
author_rank: Regular Contributor
timestamp: September 17, 2024, 08:17:24 pm
content: @ledtester, yes you're right. Falstad doesn't have ESR support yet, so you have to add a tiny resistor when using decoupling caps, else the simulation quits.

author: magic
author_rank: Super Contributor
timestamp: September 17, 2024, 08:18:11 pm
content: Not sure what makes you think this has anything to do with rail to rail. Both the inputs and the output are sitting at approximately ground, far away from each supply.Thermal runaway is generally prevented by addition of emitter resistors and thermal coupling between the power transistors and the bias diodes. In audio amplifiers, instead of diodes there is often a TO-126 transistor configured as Vbe-multiplier and mounted on the main heatsink.Quote from: edavid on September 17, 2024, 07:45:52 pmOf course it is, if the load varies between sourcing and sinking. During such transition there is a brief time when the output is effectively open circuit. But it's also true that capacitors should reduce the resulting glitches.

author: pqass
author_rank: Super Contributor
timestamp: September 17, 2024, 08:35:53 pm
content: Darlingtons heating up is probably caused by both being slightly on; current flowing from positive to negative rail (not in/out of GND). This can occur when the double diode stack Vf is higher than the Vbe of the two emitters in the darlington.Power amps with this AB output stage typically have low/sub-ohm (0.3R) emitter resistors and use aVbe multiplierinstead of the 4 diodes (pot can be used dial an exact Vdrop and therefore minimum current through the darlingtons). See Q9 onthis amp in Figure 1. Your op amp output would feed one side of the Vbe multiplier and would hold that side at ~1.2V because that's what is needed to hold the output at 0V (via magic of negative feedback)But for your application, 1R emitter resistors with diodes physically touching the darlingtons (or heat sink) usually produces acceptable results.You don't even need an op amp; seeFigure 3.

author: langwadt
author_rank: Super Contributor
timestamp: September 17, 2024, 08:45:30 pm
content: Quote from: magic on September 17, 2024, 08:18:11 pmadd a resistor from the opamp output to the transistor emitters

author: wobbly
author_rank: Regular Contributor
timestamp: September 18, 2024, 05:01:18 pm
content: Quote from: edavid on September 17, 2024, 07:45:52 pmThanks for the tip about the diodes. That has indeed cured the thermal problem.This capacitor though, do you mean one capacitor to each +/- rail or something else?

author: wobbly
author_rank: Regular Contributor
timestamp: September 18, 2024, 06:27:33 pm
content: Quote from: edavid on September 18, 2024, 05:46:07 pmI did try it both waysOn the output side of the circuit, yes, it makes little difference to the stiffness of the rails (which I'm quite happy with). I guess the only difference is the total amount of capacitance available for decoupling (if that's the word?).Although I did try removing one of the two 10uF caps on the extreme left of the circuit, where the Vref is set. Doing so made that rail a bit more twitchy to current draw pulses.Feels peculiar to me to have only one cap to one of the rails on a bipolar supply though. Won't all the electrons fall out?Works nicely though.Many thanks.

author: Zero999
author_rank: Super Contributor
timestamp: September 19, 2024, 08:51:55 pm
content: What's the maximum current through the 0V reference point?Bear in mind the 0V point is arbitrary. It's just the point from where all voltages are measured. Here's an example showing the effect of moving the 0V point on two 9V batteries in series.It would be better to use pass transistors.

author: wobbly
author_rank: Regular Contributor
timestamp: September 20, 2024, 07:29:31 am
content: About 450mA.

author: mikerj
author_rank: Super Contributor
timestamp: September 20, 2024, 07:50:41 am
content: Quote from: wobbly on September 20, 2024, 07:29:31 amThen you likely have a problem; 450mA*5v = 2.25Watts, far above the capability of a typical TO-92 transistor. Which parts are you actually using?I strongly suspect you will need power transistors mounted on a heatsink to keep temperatures down to a reasonable value.

author: wobbly
author_rank: Regular Contributor
timestamp: September 20, 2024, 08:06:10 am
content: That's the plan, when I build the circuit properly I'll be using Tip120 and Tip125 (both TO-220 package) with separate heatsinks.

author: wobbly
author_rank: Regular Contributor
timestamp: September 20, 2024, 03:08:58 pm
content: Literally nothing. My only goal is to learn how this kind of circuit works. And I am learning, although it is a real challenge.I'll build the circuit and mess about with it for a bit and then re-configure it and then ultimately recover all the parts and make something better and hopefully with a purpose. I'm just having fun (I never thought I'd say this) building analog circuits and experimenting.Thanks to all of you for your guidance.

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 20, 2024, 03:37:41 pm
content: Having fun is the *best way* to learn electronics. Or any trade, for that matter.Now you have learned about the Vbe’s negative temperature coefficient and thermal runaway conditions. This issue has nagged electronic designers since the dawn of transistorized circuits in the early 1950s. Fortunately there are many proven techniques to prevent this.Side note; if you think modern Silicon transistors are bad, you should have seen the earlier Germanium transistors!

author: wobbly
author_rank: Regular Contributor
timestamp: September 20, 2024, 03:57:10 pm
content: @schmitt trigger,Yeah, fully agree. Fun = progress!Oddly enough I have been reading the Analog Devices "Op Amp Applications" book for the last couple of days. The History chapter discusses the temperature troubles with the germanium process. This book (though clearly awesome) is way above my level of understanding, so I'll be using it as more of a reference than a bedtime readRIP Bob Widlar.

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2024, 06:36:20 pm
content: Quote from: wobbly on September 20, 2024, 08:06:10 amAgain, using them as pass transistors is the answer. The circuit works by switching the pass transistor on, when the supply current to the op-amp exceeds a certain point. The resistors are higher values because a Darlington pair has two base-emitter drops, rather than one.EDIT:R1 & R2 changed from 270R to 10k.

author: magic
author_rank: Super Contributor
timestamp: September 20, 2024, 07:05:38 pm
content: You can have the exact same effect and save one 330Ω resistor. Place the other resistor between U1 output and the virtual ground and connect both darlingtons as emitter followers directly to U1 output(And the values of R1, R2 are a little too low for a 12V supply.)

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2024, 07:18:37 pm
content: Quote from: magic on September 20, 2024, 07:05:38 pmIs it as good though? The ground will bounce around more with emitter followers, because the output will have to slew as the current changes direction.QuoteThe values for R1 and R2 are a drawing error, which I've fixed.Thanks for pointing it out.

author: magic
author_rank: Super Contributor
timestamp: September 20, 2024, 07:31:51 pm
content: Hmm, that makes sense, AC response may be better at high frequencies.OTOH, I suppose the other configuration could oscillate in some unlucky cases, as it is basically two complementary feedback pairs.editAnd you will now have this new fascinating problem that the internal compensation cap of your chip is referenced to one of its supply pins, so to that 330Ω resistor. If this happens to be the rail providing your load current, and if said current is insufficient to turn on the external power transistor, the supply pin will see ripple exactly equal to ΔIload·330Ω and the ripple will be applied to the compensation cap. So once again, same effect as 330Ω in series with the output, for this particular case.

author: Zero999
author_rank: Super Contributor
timestamp: September 20, 2024, 08:06:27 pm
content: Quote from: magic on September 20, 2024, 07:31:51 pmMaybe capacitor across the IC would help? But then there will be a surge, when both transistors turn on, when the power is first applied. Perhaps emitter resistors would reduce the surge to a safe level.This configuration is widely used to increase the current drive of op-amps, so it can't be that bad.

author: magic
author_rank: Super Contributor
timestamp: September 20, 2024, 08:29:06 pm
content: I guess it just means that perfect AC linearity is not a concern for most people(Because I don't see a meaningful difference between this and the simple class B EF output stage. You have some current between OUT and Vs- (or Vs+), there is an internal cap effectively tracking these two pins, it shouldn't matter which side has a series resistor and sees the ripple voltage. The difference I do see is that this effect disappears during times when the opposite supply provides load current. And you could eliminate it completely by "cascoding" the supply pins, which will be done anyway by those applications that need a higher voltage swing too.)

author: Zero999
author_rank: Super Contributor
timestamp: September 21, 2024, 08:23:21 pm
content: Quote from: wobbly on September 20, 2024, 03:57:10 pmYesterday, I noticed a post which you deleted, when there was no need to, because you're right. We appear to have gone way too in depth for the beginner's section. I've had more time today to provide a more detailed explanation.It's not obvious how the pass transistor circuit works, unless one looks at what's going on inside the op-amp.Here's a schematic showing a hypothetical op-amp with external pass transistors (Q1 & Q2). Focus on the internal output transistors inside the op-amp (Qout1 and Out2). Don't worry about the other components for now. I can explain the rest if you're interested, but it's outside the scope of this thread for now.When a load is connected between +v and the op-amp's output, Qout2 passes current, which also flows out of the -Vs pin.When a load is connected from -v to the op-amp's output, current flows out of the +Vs pin, into Qout1 and out of the output pin.The +Vs and -Vs pins are connected to the supply rails via R3 and R4. They each drop a voltage, in proportion to the current. The +Vs pin is connected to Q1's base and the -Vs pin to Q2's base. If the voltage drop across R3 or R4 exceeds 0.6V, the base-emitter threshold voltage of the appropriate pass transistor will be exceeded, causing it to turn on and divert current around the op-amp's internal transistors.EDIT: Added emitter follower.

author: David Hess
author_rank: Super Contributor
timestamp: September 21, 2024, 10:56:11 pm
content: Instead of a complementary Darlington output stage, consider using a 4 transistor diamond buffer.

author: magic
author_rank: Super Contributor
timestamp: September 22, 2024, 07:35:25 am
content: Quote from: Zero999 on September 21, 2024, 08:23:21 pmHighly hypothetical, because it doesn't work(Qcm2 is forced into deep saturation, and even that may be not enough to turn off Qvamp.)Anyway, I agree that these are irrelevant details. It's the output stage that matters, in this case.

author: Zero999
author_rank: Super Contributor
timestamp: September 22, 2024, 09:27:14 am
content: Quote from: magic on September 22, 2024, 07:35:25 amI missed off an emitter follower. It'll work now. Thanks.

author: wobbly
author_rank: Regular Contributor
timestamp: September 22, 2024, 06:50:15 pm
content: Hi guys,Thank you for your continuing interest in my humble newbie project. I appreciate your guidance.I've been working on it a bit more and added a few bits of house keeping such as a PFET on the input power connection for reverse voltage protection and a fuse and a 10-turn pot for trimming the virtual GND point. I've also added an RC filter to inhibit a slight tendency to oscillate, which seems to have worked quite well. Circuit now looks more like the attachment below. Still a work in progress though.I'll build this with actual TO-220 Darlingtons on my breadboard and then re-do it on a small perfboard for the final unit.I also changed over to using Texas Instruments LF411 op-amps (mainly because H&H said they are OK).Honestly, having a circuit on my bench that a year ago would have bewildered me, but which I now (kind of) understand is a huge win!

author: Zero999
author_rank: Super Contributor
timestamp: September 22, 2024, 09:54:37 pm
content: That will work, but it has a couple of issues.When the current direction changes, the op-amp's output voltage will have to change by four VBEdrops (two for each Darlington pair), about 2.4V, very quickly, causing the output to become high impedance for a short length of time. The capacitors will help to absorb that, but then they also increase the risk of oscillation. Adding a resistor, say 330R between the op-amp's output and the 0V node will help to soften this to some extent. It might also be necessary to introduce some series resistance to the capacitors, to reduce the oscillation.I could simulate this, but it's really something you need to build and measure with an oscilloscope. A good way to test this is to add a MOSFET switching a load resistor to the output. The MOSFET should be turned on by a square wave, from a signal generator or perhaps a 555 circuit. The voltages on the +V an -V rails should change as little as possible and there shouldn't be any ringing (oscillation).

author: magic
author_rank: Super Contributor
timestamp: September 23, 2024, 05:33:54 am
content: Also, R5 and R6 are doing nothing obviously useful

author: wobbly
author_rank: Regular Contributor
timestamp: September 23, 2024, 07:39:52 pm
content: Quote from: magic on September 23, 2024, 05:33:54 amI don't know either. I have seen a lot of circuits of this general family and some of them have these resistors, and the values very wildly (I've seen as low as 150 Ohm). I removed the pair of 4K7s I was using and it made no difference to the circuit whatsoever, so you're right. Thanks for the tip.I've been building it on a bit of perfboard of suitable size (pic below). Not quite finished, still need to add the output caps but the thing does work nicely. Reverse protection PFET does its job and the virtual ground is stable with small loads. I'll try loading it harder tomorrow.I think I've soldered the potentiometer in backwards though but I can live with thatDon't you love it when a circuit looks like its own schematic?

author: Zero999
author_rank: Super Contributor
timestamp: September 23, 2024, 09:32:36 pm
content: Quote from: wobbly on September 23, 2024, 07:39:52 pmThe resistors are required, when there are diodes in series with the op-amp's output. This was the case for the schematic in your original post, when the resistors pulled the base towards either the negative or positive rail, with the op-amp pulling in the other direction, in order to turn the respective transistor off. In your most recent schematic, the op-amp turns the transistors on, so the resaistors are unnecessary.

author: wobbly
author_rank: Regular Contributor
timestamp: September 25, 2024, 06:44:12 pm
content: I'm going to call this little beginner project a success. At least by my low standards.If I understand correctly (I'm explaining this to myself more than anyone else)...When the load on each rail is equal, the circuit basically just keeps the virtual GND point steady, neither sourcing nor sinking current.In that case, the upstream PSU (e.g. my bench power supply) will provide all the power to both loads, and very little power will be dissipated in my Darlington circuit.But when one rail is subject to a greater load than the other, the op-amp compensates for this imbalance and begins to source or sink current accordingly and keeps VGND as well as it can.So I think the main limiting factor to how hard I can push this design is down to what load imbalances are likely to occur in use.Studying it with a DC load on one rail or the other and watching the Darlingtons' temperature carefully with a thermal camera, this design can sustain a load imbalance of about 2 Watts continuously. This looks quite symmetrical (+/-) and the small heat sink tops out at about 70°C at room temp after an hour or so.The VGND holds it's set point better at higher input voltages. With an input of 30V (the maximum), and a load of 130mA on one of the output rails, that rail only sags by about 40mV, which is awesome as far as I'm concerned!I put this together as a simple PSU project to power my future learning of op-amps and analog electronics in general, so I think it's unlikely that the rail load imbalance will get to anything like that kind of extreme (unless I connect something wrong - and that never happens obviously).Seriously though, sincere thanks to all of you for your kind advice. After being scared of analog electronics for so long, I'm finally loving it.Cheers!

author: Zero999
author_rank: Super Contributor
timestamp: September 25, 2024, 07:43:28 pm
content: Good, you've got it working so well?Have you done and transient/AC analysis/tests?

author: wobbly
author_rank: Regular Contributor
timestamp: September 25, 2024, 07:53:06 pm
content: No, I haven't. Don't know what those are. "Beginners" forum right?I haven't sent this up into LEO on a cubesat either, before you askJust kidding.I know what you mean though. The proof of this "design" (which I basically copied and pasted from 10 different sources on the internet) will come when I make some new circuits that take their power from it.I'm not there yet.

author: Zero999
author_rank: Super Contributor
timestamp: September 25, 2024, 09:01:23 pm
content: Quote from: wobbly on September 25, 2024, 07:53:06 pmThe idea is to test what will happen when the current through the 0V output changes.Here's an example of a test circuit.Normally the rail splitter sources 400mA into R2. When M1 is turned on, R1 is connected, which draws 800mA from the positive rail, with 400mA going through R2, the rail splitter has to now change from sourcing to sinking 400mA. Look at +V and -V with an oscilloscope, which can be set to AC coupled. The voltages should change as little as possible, without any oscillations.The resistors obviously need to be able to dissipate the power. You might want to start with higher values, then work downwards. Keep R1 =R2/2then it'll sink and source the same current.

author: Andy Chee
author_rank: Super Contributor
timestamp: September 25, 2024, 10:47:44 pm
content: FWIW, this amp design shares a similar topology to your rail splitter.Note D5,D6 which your circuit lacks the equivalent. These diodes add bias to the output transistors, thus minimising crossover distortion. This may or may not be important in your final application of your rail splitter.Also note C1,R1 which your circuit lacks the equivalent. These components form a low pass filter which rolls off the high frequency response of the amplifier. It is extremely difficult to get a circuit to respond equally from DC to GHz, so set yourself a realistic design goal, say 100kHz, then filter off everything above it.