Diy Bench power supply unstable with Inductive load - Page 1
Forum section: electronics / projects

author: diydidi
author_rank: Contributor
timestamp: April 10, 2016, 08:22:09 am
content: HiI have been woking on a bench power supply for the past two months or so.I decided on a discrete regulator design with adjustable voltage and and current limit.Yesterday I boxed most of it, and decided to test using a couple of parallel connected 12V car bulbs. The bulbs were flickering slightly. Scoped the output and revealed a 16- 25HZ almost square wave/ round edges type oscillation or motor boating.If i turn down the current limiting it gets even worse.With my DC load all works as it should.My electronics knowledge doesn't stretch as far as to fix this. The sense leads come together at the croc clips of the output leads. So there are four wires coming out of the box. The negative output terminal connects directly to neg of main filter caps.Connecting a 4700uF capacitor across the leads at the load doesn't help much.Any ideas? See schematic attached.

author: Lovely_Santa
author_rank: Contributor
timestamp: April 10, 2016, 11:03:31 am
content: Hi!First of all, could you post some pictures of your scope? With just resistor load and with your inductive load? Maybe that will clear my mind a bit...I was thinking about the inductive power isn't all flowing through the diode, but also through your 10K pot (OP V), D7 and R18 (100R) what will decrease the current on Q4 transistor...

author: diydidi
author_rank: Contributor
timestamp: April 10, 2016, 12:40:56 pm
content: Here are two waveforms.The one is at 12V with 3,5amps draw with the bulbs.The other is with a 8,2ohm resistor also at 12V.

author: mij59
author_rank: Frequent Contributor
timestamp: April 10, 2016, 01:22:34 pm
content: The bulbs.jpg reminds me of a chaotic signal, not that great for a power supplyNot sure what's going on, there are two different control loops, voltage and current.Disconnect the current limit control and check the output under load.In the schematic the value of C2 is way to high, try something like 1nF or leave it out for now.The output stage is quite slow, put a resistor between base and emitter, for Q8 use 100 Ohm , and for Q7 use 560 Ohm.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 10, 2016, 02:48:34 pm
content: Unusual to still find a power supply design without OPs. General setup looks rather good, but a few parts are odd and it looks like it is set up rather slow. Today I would not use separate transistors for an analog circuit like this - temperature drift and offsets are just not good - also gain is limited in such a simple circuit - so not a very good regulation.Today one would use 2 OPs for the regulation, but the general setup as a floating regulator is good.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 10, 2016, 03:43:24 pm
content: I didn't go through all the schematics but seems to me that C1 and C2 are mainly responsible for the oscillation.Parallel a capacitor to ground in the loop will create a pole, and two poles will make the system unstable for sure.

author: Lovely_Santa
author_rank: Contributor
timestamp: April 10, 2016, 04:17:04 pm
content: First of all, don't lay your priority on the inductive load... Becose the main reasons about the oscilations is the variation in the voltage...I would first make sure your output is a more stable signal...Like the people above said...(mij59)The output stage is quite slowThere are multiple ways to solve it...Like mij59 says:Lower C2... the output will be faster to react, so the voltage regulation will be more unstable , but this won't be a problem, becose you have the output capasitor (C7) to slow this process down.. with C2 so high, you are slowing it down even more... This could be why you have those peaks... becose your circuit can't handle the peaks fast enough...If it isn't beter, you could filter the noise out of this by setting a little capasitor in parallel with your output.. (I would take something about 100nF) If this is helping, but not enough, place somethink about 10nF for the little noise.. (so then 3 capasitors in parallel)If it isn't working, switch off your current circuit (1K pot fully open, or just replace with 1K resistor) and try it again.. but this shouldn't be the problem...Quote from: mij59 on April 10, 2016, 01:22:34 pmDon't.. It's just a Darlington transistor setup... nothing wrong with that... It acts like a transistor with a much higher current gain than a single transistor.If this works, post again 2 new pictures plz, becose I think the big peaks from the inductive load is just comming from your voltage peaks....

author: diydidi
author_rank: Contributor
timestamp: April 10, 2016, 04:59:57 pm
content: Thanks for all the replies thus far. Just as a matter of interest; I was playing around this afternoon and installed a 1uF capacitor between D8 cathode (+v sensing) and D15 Anode (0V), and the oscillation was gone, Why would this be? Is this a solution or should I rather decrease C2 value as mentioned above?

author: mij59
author_rank: Frequent Contributor
timestamp: April 10, 2016, 05:26:40 pm
content: Quote from: diydidi on April 10, 2016, 04:59:57 pmCould be a layout problem, maybe the capacitor cancelled some of the parasitic inductance.C2 improves the response to load changes, make it too big and the circuit will oscillate.

author: Lovely_Santa
author_rank: Contributor
timestamp: April 10, 2016, 05:30:16 pm
content: Quote from: Lovely_Santa on April 10, 2016, 04:17:04 pmWhen you set a big capasitor on the output, the time to charge and discharge are long (RC>>>)When you set a little capasitor on the output, the time to charge and discharge are fast (RC<<<)So when you setted that little capasitor, it will charge on the peaks, at that moment it will filter the noise out... you could set it between D8 and D15... (so that's like setting it like I said on the output leads)....So all the noise and peak values are filtered out by your capasitor...So when you now set your inductive load on the output, you still have those bumpy peaks?? Could you post a picture? becose I would decrease that capasitor 1µF to 100nF.. and I would set it on the same connection as your C7... then they are less wires between the output and the cap, so it will work even beter...Quote from: diydidi on April 10, 2016, 04:59:57 pmThis is indeed a solution, even a solution I've mensioned... I would decrease C2 anyway, but if it works fine, it's not required... If it works fine for your inductive loads even, then I would let that capastor over there unchanged...

author: Kleinstein
author_rank: Super Contributor
timestamp: April 10, 2016, 05:33:03 pm
content: So the extra capacitor is effectively parallel to the output. This might be a good idea if the Output cap of 220 µF is not low ESr low inductance. The proper place would be close the the output. The output usually should have something like 1 µF with really low ESR/ESL and than an electrolytic cap with some ESR, e.g. the 220 µF cap in the plan.The capacitor C2 does not introduce an extra pole, its actually speeding things up. For further compensation an additional resistor in series to C6 might help.A base to emitter resistor at Q7 might help a little to speed up the output stage, a resistor at Q8 would be an alternative to R16.I think there is a mistake in the plan on where D4 goes - I would expect the anode at the base of Q6 instead of Q7. Otherwise the supply should not turn off.

author: Lovely_Santa
author_rank: Contributor
timestamp: April 10, 2016, 05:47:55 pm
content: Quote from: mij59 on April 10, 2016, 05:26:40 pmLike he write it, it's like he setted a cap on the output... so low cap filter noise out... and it will also help discharging the inductance...Quote from: mij59 on April 10, 2016, 05:26:40 pm@diydidi:Like he says, the lower C2, the faster the circuit will react to changes on the output..BUT if it reacts too fast, it will give lots of noise... (see picture attached)AND if it reacts too slow, it will give peaks (like you had)Quote from: Kleinstein on April 10, 2016, 05:33:03 pmWhen the output need to change, the cap will help so it reaches faster the value, but then it will react otherwise, when the transistor should stop working, it will still work while the cap is discharging, this will give some delay, and it's that delay that will extend the noise.. but yes, it speeds it up... But if the cap on the output helps, I wouldn't change anymore on the circuit, why would you, if it works ^_^Quote from: Kleinstein on April 10, 2016, 05:33:03 pmit's a zener diode, it just gives a 5.6V voltage drop (+ an other 0.7V by the diode inside Q8) for the rest of the circuit.. (referenced on the lower transformer).... If you connected it to the base of Q6 you would have an other 0.7V voltage drop... so it would be 7V instead of 6.3V...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 10, 2016, 06:04:38 pm
content: Quote from: Lovely_Santa on April 10, 2016, 04:17:04 pmNo, don't use that excuse!The circuit should be stable, quiet and low impedance with a minimum of output capacitance.The worst part about an output capacitor is this: if you set a voltage and current, then go to connect a current-mode load (like you're testing LEDs), and suddenly the LED doesn't work anymore, why? Because you forgot to obey current-mode connection rules* and discharged a huge assed cap into a poor little LED!*Constant voltage supplies produce zero power into an open load. So you disconnect leads from one load, and connect them to another (break before make). Constant current supplies produce zero power into a shorted load, though. So you short the supply, disconnect leads from one load, connect them to another, and unshort the supply (make before break). CV supplies are traditional, so this is easy to forget...I also don't understand R14-C5-D8.As for the circuit, the floating follower topology is good for a wide range supply, though it takes a bit more work to create it. The problem is gain. By referencing the feedback circuit to the output, you've removed the primary source of feedback in the system: the output transistors aren't emitter followers anymore, they're high gain common-emitter amplifiers. This still isn't a bad thing, in and of itself, but you must be very careful!And judging by those nasty waveforms, it looks like you aren't -- likely, too much gain has resulted in a relaxation oscillator or something.If there should be any other bit of advice to walk away with today, it should be this:Always bound your inputs. Always bound your outputs.You have some examples of this already, but it should be applied comprehensively. Examples:- Q6 collector current is limited to about 10V/470 ohm = 20mA- Q3-Q4 emitter current is limited to about 10V/6.8k = 1.47mAThe worst offenders are:- Q3-Q4 "tail" current is way more than needed- Q6 base voltage is unlimited (D4 is relative to its emitter, so R12 only acts as a current source)- Q8 collector current is unlimited, especially at high voltage (R8 only drops ~2V at the ~10A it's capable of drawing, so emitter degeneration doesn't save you here).http://www.onsemi.com/pub_link/Collateral/TIP35A-D.PDFSOA says you need a DC limit of 0.5A at 60V (it's in 2nd breakdown) or else it's kablewie!- Both Q2 and Q5 are wide open, so their gain is very high, and uncontrolled (basically hFE limited). This is a nightmare to try and stabilize...The first step is to reduce, and stabilize, the transistor gains:- R8 should be much larger, for the circuit's sake. (If you don't want to waste the voltage drop and power dissipation by adding more resistance here, there are other solutions, but this is the simplest.- Q3-Q4 can have emitter resistors (but this reduces the voltage accuracy, too), on the order of 100 ohms- Q5 needs a large emitter resistor, and get rid of C1. This makes it a simple current mirror, with minimal delay; now Q2 can have direct control over Q6 without panicking and making a mess.You should also be able to get rid of R13, D5, R2, Q1 (adjust R1 as needed), R19, C5, R14, D8, D4, R6, and considerably adjust C2 and C7.A topological overhaul would be better, though. Q6-Q8 has very high gain, both in voltage and current. You need to be able to put bounds on both. Conceptually, you should design this as a transconductance stage: voltage in (at the base of Q6), current out (total through R8). The voltage drop across R8 is proportional to that output; and that voltage subtracts from the input, which is why I suggest making R8 as large as possible. Your input voltage has 3*Vbe worth of error on it, which means the feedback voltage should be about as strong (several volts). The Vbe's can be subtracted by using complementary transistors, but you can't subtract all the errors (errors due to manufacturing differences, temperature drifts, etc.).Once you have a nice clean transconductance output stage, you have trivial current limiting: you can use a differential amplifier across the sense resistor (as shown, D9-Q2, but a pair of matched transistors, or an op-amp, would be more accurate), or simply clamp the input voltage (at Q6)!A transconductance amp makes a pretty bad voltage source, though. So we need to wrap an error amplifier around it. Q3-Q4 are basically doing this right now, but the gain is so high, it's unstable. Once the output stage has known gain, this becomes much easier to compensate -- a capacitor (or R+C) across Q4 base-collector keeps its voltage gain under control.The output characteristic will be low impedance near DC (good DC voltage regulation), rising (inductively) at higher frequencies (where the error amplifier isn't able to keep up, due to compensation). In this range, you can keep the output impedance low by adding a capacitor. There will be some resonance effect between the equivalent inductance of the supply, and the capacitor, which simply means adjusting that compensation R+C to keep it happy.Test step response by connecting a function generator to the relevant nodes. For example, testing voltage response by adding a signal in series with D3 or R3.This process still doesn't address foldback current limiting, unfortunately, but thermal and breakdown issues really are another article. The simplest solution will be to use a massive heatsink and multiple transistors in parallel; preferably better transistors, too (MJE15xxx series parts are pretty cheap, and "free from 2nd breakdown"). You'll still need a truly massive heatsink to handle short circuit conditions!Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: April 10, 2016, 06:24:06 pm
content: The circuit can't work with D4 installed as shown. This would give about 0.4 mA off current through R6 and D4 to the base of Q7 all the time. This would be good for some 4 A. On the other side this same current is missing to drive Q6 and the give Q4,Q5 some current to work against. The 100 K resistor R12 is rather high for this. R15 calls for a current in the 0.5 mA range - not 40 µA from R12.There is a slight problem with this type circuit, as the loop gain for voltage control changes with the voltage setting. Without a resistor in series to C6 it might get hard to make this regulator unconditionally stable with high capacitive loads and at all voltage setting - the lower voltages are the critical ones, as C2 gets less effective and gain is highest.A larger value of C2 makes the regulation faster - to fast might cause trouble, but to slow (small C2) also. Though for this supply C2 should not be critical as C2 gets less effective at low voltage setting anyway. So it should be tuned to be stable even without C2 and a large value for C2.Just a few light bulbs is still a rather well behaved load - more like an ohmic one. The critical ones are more pure capacitive with a current sink in parallel. Inductive load are usually not that bad for voltage regulation.

author: mij59
author_rank: Frequent Contributor
timestamp: April 11, 2016, 04:17:13 am
content: In the schematic the primary side of the transformers are in series, is this really the case ?

author: mikerj
author_rank: Super Contributor
timestamp: April 11, 2016, 03:22:07 pm
content: Quote from: mij59 on April 11, 2016, 04:17:13 amI hope not, that was one of the first things that stood out when I looked at the schematic. Regulation of the raw DC voltages will be terrible if that's really two separate transformers with the primaries in series.

author: diydidi
author_rank: Contributor
timestamp: April 12, 2016, 06:00:21 am
content: Do please notice that there is a 10k resistor missing from my circuit. It is connected between Q6 emitter and D17 Anode.I have attached the original schematic from the 70's. Apparently it was a very good supply. "Paul" went and simplified the circuit a bit for his needs, but most of the values are identical. The circuit works great except for the problem I mentioned in my first post!Have managed to get rid of the oscillation by adding a cap between D8 cathode and D15 Anode. I'm not sure how big this cap should be?So I guess my next step would be to try and better this circuit slightly as I have already made a pcb for it.Further suggestions welcome.Definitely gonna try reducing C2 as mentioned by you guys!

author: Kleinstein
author_rank: Super Contributor
timestamp: April 12, 2016, 11:53:12 am
content: With that extra resistor, the circuit might kind of work, but not good, as R12 still delivers way to little current and D4 makes not much sense where it is now. It would be still better to mode D4 -Anode from the base of Q7 to the base of Q6.Anyway the performance will not be that good. Especially current limiting will not be rather temperature depended - Ok as an overload protection but not for use as an constant current source. Also the output resistance will not be that low, despite of using sense wires. I would expect something in the 10 mOhms range. That was acceptable in the 1970.Also keep the SOA of the output transistor in mind. With 60 V from the transformer, the output transistor can easily break down in case of a short at the output. The SOA curve only allows for about 0.5 A with perfect cooling - even less with a real heat sink.So if you really want the 60 V range, I would change to circuit to use 2 output transistors in series and automatic use of both transformer taps - it's not that complicated and is less critical to the SOA limit: at 30 V the TIP35 might stand up to about 3 A.

author: mikerj
author_rank: Super Contributor
timestamp: April 12, 2016, 12:38:14 pm
content: Quote from: diydidi on April 12, 2016, 06:00:21 amAs above, are the two transformer primaries really connected in series (as shown in your first schematic), or is it a single transformer as sown in the schematic?

author: diydidi
author_rank: Contributor
timestamp: April 12, 2016, 04:49:58 pm
content: OK. Just confirmed my cap idea actually doesn’t work!!Tried the 100uf across sensor wires but no effect.It’s just strange that if I short the + sense wire to the main + out on the pcb instead of at the croc clip, everything works great, This confirms that the current board is actually working.Tried making C2 smaller, this too does nothing.Something is up here....

author: Kleinstein
author_rank: Super Contributor
timestamp: April 12, 2016, 06:50:19 pm
content: A capacitor across the sensor wires is not expected to work - it's more like a bad idea.The first thing would be to fix D4 -> don't expect it to work if this is left as drawn. The very low current for Q4 also makes the circuit very slow and thus the compensation is not expected to fit any more.What might help is having a really low ESR capacitor (film type or ceramic right at the output) - at least it will help to keep HF signals out. As this cap is more for something for the upper frequency limit, it's likely not the cause for oscillations at 25 kHz.I would also try something like 1 K in series to C6, though I am not sure about the right value.To make the whole thing less guesswork one could also enter the circuit into Ltspice and check for trouble in the simulation.Another principle weakness of this circuit is that changing the voltage setting also changes the feedback loop. A more stable feedback would be possible if the adjusting pot is changing the reference voltage side instead. However this is little more difficult without an OP.

author: mikerj
author_rank: Super Contributor
timestamp: April 13, 2016, 02:38:48 pm
content: Why don't you want to answer the transformer question?

author: diydidi
author_rank: Contributor
timestamp: April 13, 2016, 07:32:14 pm
content: Quote from: mikerj on April 12, 2016, 12:38:14 pmThe primaries are not in series. Just the way he drew it. Actually only one prim winding.