An interesting 3-transistor PSU from the old USSR "Radio" magazine - Page 1
Forum section: electronics / projects

author: Aramp
author_rank: Contributor
timestamp: September 22, 2024, 08:26:55 pm
content: I am posting an interesting construction that I have a question about.I was fixing a school lab power supply designed years ago that at some point underwent a mod resulting in total loss of capacity to withstand shorts and overcurrent (something you should expect to happen all the time when students try things at the lab). I could have replaced the PCB with something modern based on opamps. However, the space was rather small and required a relatively compact solution. In the search for one, I came across the attached schematic, which has current limitation, low output impedance due to the current shunt placed before the regulating transistor, and a separate very fast short prevention circuit using two diodes. I have modified the schematic to include modern and available transistors and diodes and made sure that replacements do not change any functions (as I also happen to have the old Soviet parts).The schematic generally does its job, but not ideally. The reaction to shorts works perfectly, as it is immediate. However, the current limitation is peculiar in that the actual limit depends significantly on the output voltage. The higher the voltage, the higher the current. That's not bad for the transistor, as the power transistor will have fewer problems passing higher currents when the voltage drop is low. However, that also puts higher requirements on the transformer power capacity and limits the usability of the power supply at lower voltages, as the current limit for below 1.5 volts is very low. Decreasing the resistance R4 can partially compensate for that, but there is a reasonable limit to how much it can be decreased.The theory of operation is straightforward in all but the involvement of the two diodes (D6 and D7) and how they control the current.Let me describe it the way I understand it. D6 is normally open, and D7 is closed. The transistor Q1 works normally, and its collector current only depends on the base current, which is defined by the potentiometer, as its collector negative potential is higher than the emitter potential in absolute value due to the voltage difference between the collector and emitter of Q3. When the load on the output exceeds a particular threshold (which depends on the value of R4), diode D6 closes, and thus, the current through Q1 gets defined by R4 and RV1. At the same time D7 opens and counters the current that is regulated by D1. The higher the current through Q3, the more current gets through D7, thus compensating for the increase of current at the output, potentially also defined by the increase of temperature of Q3. Thus D7 helps stabilize for the temperature drift. Without it, the increase in temperature of Q1 may have a runaway effect on the whole chain.When the output is shorted, the potential on the base of Q1 gets very close to the potential of the emitter (minus the drop at the diode), and Q1 closes, shutting down the output voltage. If a Schottky diode is used, that potential (and the output voltage) will be even smaller.Now, the question: is there a way to make the dependence of the current limit on the voltage drop less steep, thus allowing the construction to work at lower voltages without making things too complicated? If that is possible, this may turn out to be a very minimalist, safe, and functional PSU. Actually, D7 may be replaced by an LED to provide an indication of overcurrent (with some increase of the output potential when the output is shorted). As that's a secondary safety measure in addition to D6, even if LED malfunctions, the overcurrent protection will still work.

author: PGPG
author_rank: Frequent Contributor
timestamp: September 23, 2024, 12:19:40 am
content: This circuit has some minimum output voltage. To drive Q2 minimum voltage drop at RV1 is about 0.7V. The whole current to RV1 comes from R4. So to make Q2 conducting at R4 there will be 0.7*(3k3/500)=4.62V. Voltage at output will have to be (D6) 0.7V higher (in negative sense of higher) to ensure regulation so minimum output is 5.3V.Based on information "designed years ago" and seeing that it regulates at (-) supply side I suppose it was originally been done with germanium transistors so calculations would be different.What you see here is called foldback and is used to protect regulating transistor.

author: Aramp
author_rank: Contributor
timestamp: September 23, 2024, 06:05:47 am
content: Quote from: PGPG on September 23, 2024, 12:19:40 amThank you. I don't think your calculation is correct. The voltage drop on base and emitter junctions of transistors has no influence on Q3, as no base-emitter junctions are involved in the path from the ground to the base of Q3. All transistors are connected to the output through either the collector directly or through the emitter. As the emitter and base of Q3 are connected with 100 Ohm resistor, there is no drop on Q3 either.So in practice, the PSU produces voltages starting from almost 0. As I've mentioned, swapping the transistors for germanium did not change the behavior much. I didn't try to swap the power transistor (I could, but I don't think that will solve the issue I have mentioned), but I did swap the Q1 for germanium. Q2 was a silicon transistor in the original publication.In this configuration, foldback is provided by the shunt resistor and the base-emitter junction of Q2. As I said, that part is simple and is very common. However, this case is more complicated and involves the action of diodes. It introduces an interesting effect of dependence of the current limit on the voltage across the whole range, which is beneficial overall but would be nice to get under control. If there is any interest in discussing this, I can produce an experimental graph of the dependence of threshold current on the output voltage for several values of the resistance of the load.With current technology and abundant cheap opamps, everything can be done by piling up well-behaved components with predictable outcomes. We all got used to that. However, I think it is very interesting to figure out how the seemingly simple but functional solutions work, as they are not that simple when you dig deeper.

author: magic
author_rank: Super Contributor
timestamp: September 23, 2024, 08:20:42 am
content: As far as I see, Q2 and Q3 are a Sziklai pair which acts as a constant current sink controlled by RV1 voltage. RV1 voltage is controlled by Q1 collector current and RV1 (current limit potentiometer) setting.It looks like Q1 and D6 are meant to be a poor man's differential pair, which regulates output voltage to approximately track RV2 voltage. The pair's bias current is generated by R4, so the bias and the achievable RV1 voltage increase with increasing output voltage. This means that current limit increases with output voltage, as long as the PSU is in regulation and D7 sees approximately zero volts across.When current limiting happens, the output falls out of regulation and D7 forward biases, stealing base current from Q1 and further reducing the current limit. This appears to be some sort of foldback, but it seems dodgy. How exactly is this behavior determined? How is this circuit able to start at all when the output is initially zero?IMO if you want current limit to be independent of output voltage, you need to replace R4 with a constant current source.

author: iMo
author_rank: Super Contributor
timestamp: September 23, 2024, 09:17:38 am
content: I think it will not work with the 500ohm pot (original schematics) there..PS: soviet constructions, often offered as "used to use in space", are often rather exotic, with a lot of improvised "tricks" which usually work under special conditions and with special types of components..

author: wasedadoc
author_rank: Super Contributor
timestamp: September 23, 2024, 11:16:19 am
content: Q2 and Q3 are NOT a "constant current sink". They pass whatever variable current is required by the load on the output up to some limit. Current limit occurs when current through R2 makes the emitter of Q2 sufficiently less negative than its base such that Q2 becomes starved of base current. RV1 controls how many volts positive the base of Q2 is with respect to the IN- end of R2.

author: RoGeorge
author_rank: Super Contributor
timestamp: September 23, 2024, 11:16:27 am
content: - Prisoner #439998, what was your crime?- I've attached LTspice pics without the .asc file.

author: Aramp
author_rank: Contributor
timestamp: September 23, 2024, 11:24:33 am
content: Thank you for the detailed analysis! It sheds light on many aspects for me. I have also initially modeled the thing in LTSpice but then decided to try it with real components, as it is more fun.I have modified and forgot to mention the value of R2. It was originally 1Ohm, but the PSU was intended for just 1A, and my intention was to use it for 2A limit. I have experimentally tried both versions and didn't see much difference.Not only were the schematics obscure in the USSR, but most of the people in engineering also had a hard time explaining them (if they even understood by themselves). So, the magazine's text didn't describe the actual inner workings as intended by the author. It mentioned, however, the fact that the max output current is 1A for the maximum voltage and that the output range is from 1 to 12V. It also suggests that the maximum current limit should be set for the maximum current limit.The reference schematic this one is based upon is attached. The attached one has a pretty good theory behind it, and as it does not regulate the output voltage, it just provides a means to cap the current. According to the authors, the current limit is close to (Uref - Ube)*hFE/R3, where Uref is the Zener voltage, and Ube is some 0.6 Volts for silicon transistors. According to the authors, as soon as the diode connected to the emitter closes, V2 transistor turns into the current source, thus limiting the current at the output. The base diode is then used for thermal stabilization. It looks like the author of the schematic that I have used introduced some trickery to allow for voltage regulation but, indeed, has turned it into a hard device to control properly.

author: PGPG
author_rank: Frequent Contributor
timestamp: September 23, 2024, 12:01:59 pm
content: Quote from: Aramp on September 23, 2024, 06:05:47 amIt is your problemQ2+Q3+R3 works as single npn transistor with very high hfe.Delete mentally Q3. Short R3. Imagine Q2 has big hfe.Now Q2 is the regulating transistor. To make it being conducting about 0.7V is needed between its base and emitter.Read once more my previous post and if you think my calculations are incorrect point exact place where I made a mistake.Quote from: magic on September 23, 2024, 08:20:42 amAssume following:- germanium transistors had orders of magnitude greater Ice0 than currents we are used to now,- typical, containing semiconductors, load consumes no current until voltage reaches some value so assuming PSU has resistor load is wrong,- may be this PSU was intended to be switched on without load and then connected to load and it intentionally had to not start if switched on with load being connected.So imagine what is going on with no load. You get output capacitor loaded through R1, RV2 and D7 and through Q3 Ice0. Assuming that Q1 Vbe is little lower than D6 voltage drop (what is probably true) R4 not loads this output as current into it is given by Q1. When output reach 5.3V (according to calculations from my first post) Q2 is switched on and normal regulation begins.If all semiconductors were germanium (as probably were in original design) than minimum output voltage to start regulation would be 0.25*(3.3/0.5)+0.25 = 1.9V. Many loads consumes no (or very little) current when powered by 1.9V.

author: Aramp
author_rank: Contributor
timestamp: September 23, 2024, 12:41:05 pm
content: Quote from: PGPG on September 23, 2024, 12:01:59 pmI think I have pointed out where the mistake is. And it's still your problem because I have made the thing and it produces significant current at 2 volts with all silicon transistors. It's when I limit the current at 15 Volts to be around 2A, I get a low current at 1.5 Volts. If I let the current at 15 Volts be 3A, I have no problem drawing at least 0.5A at 2Volts. I'd suggest you explain how your explanation explains that.

author: PGPG
author_rank: Frequent Contributor
timestamp: September 23, 2024, 04:36:07 pm
content: Quote from: Aramp on September 23, 2024, 12:41:05 pmIf the schematic is as you shown then I don't believe that this circuit can regulate 2V giving significant current.What is significant current?Regulate 2V means that you change load current from some small minimum load current to some (significant) current and you all the time have (around) 2V until after reaching this current voltage drops down.If you see it being regulating 2V then set load at 1/2 of that significant current and give us voltages at all transistor pins.You can get some current having 2V at output (forced by load and not regulated) but it can only came from as I have written here:Quote from: PGPG on September 23, 2024, 12:01:59 pmFor silicon transistors you can ignore Q3 Ice0 part.I don't know what is the input voltage to this circuit, but as R1 is only 910 than current flowing through R1, RV1 and D7 can be 'significant' when output is 2V, but this output can't be regulated by this circuit - transistors Q2 and Q3 can't conduct yet.

author: coromonadalix
author_rank: Super Contributor
timestamp: September 23, 2024, 05:09:25 pm
content: this circuit is simple and crude, sure it can works BUT dont offer state of the art regulation

author: magic
author_rank: Super Contributor
timestamp: September 23, 2024, 05:57:38 pm
content: Quote from: RoGeorge on September 23, 2024, 11:16:27 amThere is no .asc file, I just redrew the schematic to follow civilized normsQuote from: Aramp on September 23, 2024, 11:24:33 amThis circuit has voltage regulation and it works exactly the same way as the 3-transistor one. When output voltage rises above V3 (reference) voltage, V5 forward biases and diverts R3 current away from V2 and therefore away from V1 base, so output voltage can't rise any further. When output voltage falls below V3 voltage, V5 is biased off and V2 current increases up to a limit determined by R3.The difference is that V1 simply multiplies V2 current (minus 3mA for R2) by its hFE, while the 3-transistor circuit attempts to control current amplification more precisely and provides an adjustable limit.In both circuits, the base diode pulls V2 base towards ground when output voltage is low and reduces R3 and V2 current. This is the exact reason why maximum load current is lower when output voltage is lower. It has nothing to do with any thermal compensation.Quote from: PGPG on September 23, 2024, 12:01:59 pmYes, it has to be something like that - D7 allows the output capacitor to charge even when nothing else works and the PSU can only start under light or no load.

author: Aramp
author_rank: Contributor
timestamp: September 24, 2024, 06:00:49 pm
content: Quote from: magic on September 23, 2024, 05:57:38 pmYes. I meant to say it does not allow to adjust the voltage.Quote from: PGPG on September 23, 2024, 04:36:07 pmI'll double-check tomorrow and will get back with the measurement results.

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 04:26:05 pm
content: Quote from: PGPG on September 23, 2024, 04:36:07 pmI came back with numbers that I have just measured.The input voltage for the circuit is 22V, provided by a regulated linear power supply. I am using the 13V Zener, actual voltage at Zener is 13.1V. The max output voltage without a load is 13.1V, almost exactly at the level set by the Zener. When loaded at max voltage the output level drops somewhat, depending on the load. At 2A load the output drops to 12.9V,. I have adjusted the pot so that the max output current before the cutoff is about 4A.Then, I adjusted the output voltage to 2.2V without a load. I loaded the output and measured the voltage at the output (Vout) and the current through the load (Iout)Iout = 0.39A, Vout = 2.14 V, the voltages on bases of transistors are as follows: Q1: -2.24V, Q2: -14.77V, Q3: 2.8VWhen Iout = 0.22A, Vout= 2.18, voltages at bases are: Q1: 2.25V, Q2: 15.08V, Q3: 2.8VI am able to increase Iout to 0.5A, then Vout=2.1V; above that, the protection kicks in, and the voltage drops to almost 0. I can decrease the output current to any level I want.All transistors are as on the schematic; for the power transistor, I have used an old USSR-made KT818BM (silicon), on the sink, as it is very hard to destroy.Any thoughts?

author: magic
author_rank: Super Contributor
timestamp: September 26, 2024, 04:50:38 pm
content: So these are voltages with respect to the positive rail (common for input and output)?And the negative input is -22V but you only have -15V on Q2 base? Suspicious...What's the voltage and current (Ohm's law) across RV1? And across R4? And if RV1 current is greater than R4 current, where is it coming from?

author: PGPG
author_rank: Frequent Contributor
timestamp: September 26, 2024, 05:10:24 pm
content: Quote from: magic on September 26, 2024, 04:50:38 pmEverything is fine except the comma. Vbe = 7V instead of 0.7V

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 06:22:51 pm
content: Quote from: magic on September 26, 2024, 04:50:38 pmYes, of course. All voltages with respect to the positive rail (lower leg of RV2).QuoteOh, my bad. Here is where I created a bit of confusion. Firstly, the input is fed into the the diode bridge, so on the negative rail it's 20.3V (we lose voltage on two diodes).Also, I have originally set the input to 17.9V, as that was fine for the measurement in the case of 2V output. Then I checked to see how the circuit behaves at max voltage and found out that the output drops significantly at around 3A. But of course that was because of the voltage drop on diodes and on R2 becomes large enough to drop below the voltage of Zener. That's why I increased the voltage to 22. So ignore 22V and edited my message for the max current and drop, but forgot to keep the original voltage for measurement in 2V region.The voltage drop on VR1 is 1V when the load current is 0.4A.I can re-measure everything for 22V, but I'ts going to be almost proportional.

author: magic
author_rank: Super Contributor
timestamp: September 26, 2024, 06:35:08 pm
content: Quote from: Aramp on September 26, 2024, 06:22:51 pmOK, so since VR1 can't be more than 500Ω, this corresponds to at least 2mA current through VR1. This current should be coming from Q1 and from R4, so how is it possible if the output is set to 2.2V and there is less than 2V across R4?There is something paranormal happening here

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 06:36:51 pm
content: So, for easier reference. When the output current is 0.4A, and the voltage on input (after diodes) is 16.1V, then voltages on transistors are as follows:Q1:E=1.6B=2.2C=15.1Q2: E=15.8B=15.1C=2.8Q3: E=2.14B=2.75C=15.75If you want me to measure anything else, I'll be happy.

author: magic
author_rank: Super Contributor
timestamp: September 26, 2024, 06:45:50 pm
content: Well, this shouldn't be possible. Are you sure that R4 is 3k3 and not lower and that RV1 is 500Ω and not higher?If Q1:E is -1.6V and R4 is 3k3, then R4 current is 0.5mA and so is Q1 emitter current and so is Q1 collector current and 0.5mA should be not enough to produce 1V across RV1.

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 06:47:27 pm
content: Quote from: magic on September 26, 2024, 06:35:08 pmVR1 is 1K in my circuit (that's yet another thing I genuinely forgot to mention). It is set roughly in the middle. And we have 1.6V across R4.

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 06:55:03 pm
content: Well, I feel really stupid now.The problem is that I have published the schematic that I started with, but then I introduced some changes as I did some experiments and, of course, didn't document them.Of course, you are right that R4 can't be 3.3K. It is 680 Ohm.By the way, you are right that the circuit does not start under the load.

author: PGPG
author_rank: Frequent Contributor
timestamp: September 26, 2024, 07:43:53 pm
content: Quote from: Aramp on September 26, 2024, 06:55:03 pmYou should notice it after seeing in my first answer:Quote from: PGPG on September 23, 2024, 12:19:40 amAnd you were right - my calculations were not correct as input data were not correct.

author: magic
author_rank: Super Contributor
timestamp: September 26, 2024, 07:49:16 pm
content: Now it's starting to make sense. So it's working quite as expected, and as I said, if you want more current at low voltage, replace R4 with a current source of a few mA. Perhaps it could be as simple as a PNP with its base connected to the negative rail with 1MΩ or so. At this point foldback limiting will no longer work, so D7 could be removed and this should also help with starting under load.Beware that high load current and low output voltage means lots of power dissipated in Q3 - you will need good heatsinking (or reduced input voltage). It appears that the purpose of D7 was to idiot-proof the design

author: Aramp
author_rank: Contributor
timestamp: September 26, 2024, 08:21:34 pm
content: Quote from: magic on September 26, 2024, 07:49:16 pmThanks! I will experiment with a current source to see how it goes. The circuit designers were clearly concerned about the cost of transistors at the time (they were relatively expensive and in short supply in the USSR). D6 will still do the job by limiting current if the output is shorted. D7 was introduced as a safety measure because when transistors warm up, their hfe increases, then the output increases, and that will further heat the transistors up, which may create a runaway situation. I believe that a proper heatsink and a good headroom for the output power capacity of transistors used, along with reasonable limitation of the cutoff current will solve that problem. Again, in those times, powerful transistors were particularly expensive.BTW, for the school power supply I ended up using a trivial datasheet design using LM317 with a current-sensing shunt. The behavior was much more predictableIt has a pretty high output impedance and, thus, the regulation is not ideal, but nobody expects first-class performance from a crude power supply for simple lab experiments.@PGPG, sorry for the confusion. I perceived your argument as "this is only possible with germanium transistors" and having had a working circuit on my desk, was not seeing why it can't work.

author: enut11
author_rank: Super Contributor
timestamp: September 27, 2024, 01:05:50 am
content: Hi ArampCongrats on your persistence. I ask myself 'what lesson did I learn?' No need to answer that here, that is for you only.I do know that some of my best learnings have come from failures!enut11