Lab Power Supply - The Lost Current

[radoczi94]

Hi.

I built the power supply shown on the schematic.It looks like working as a treat, but not really.The CV mode is ok, the CC is what causing the nightmare. I started the testing with a massive short circuit on the output, slowly turning the current pot up. The comparator doesn't wanted to switch over. I probed around for a while, found out, there isn't enough voltage drop across the shunt (R35-R25) resistors.I started to question the quality of those, modified the voltage divider on the comparator input, then came the idea: measuring the current thru the shunts.Okay, now, here comes the funny part.I measured 1A on the shunt, while on the output was 3A. I lost track of 2amps!!! I measured over and over again with 4 different DMMs and analog meters, the results were the same. It turned out, the lost 2A is not a fixed value, it depends on the load on the output, it's about 2/3 of the output current. I removed the shunts and measured resistance to see, if there is anything weird, but only measured something in the 200MOhms range, which is basically nothing. Anyone has any idea, what should I try, or measure? Because I'm out of ideas.

Thanks.

[C]

just a quick look

If you have a power supply that has a CV mode & a CC mode, you have a OR.

Output = the lesser of (set voltage or set current)

[Kilo Tango]

Hi,

most power circuits I have come across measure current in the  +ve rail, otherwise it means your -ve output is different to your circuit common terminal, the thick blue line, so which one is earth ?.

You can't loose current  (easily !) , it has to flow somewhere. if it is flowing out of X6-2 and back into X6-1, it has to make its way back to the input along the thick blue line, which means either thro your shunt resistors or by another path. By the way you don't need to remove the shunt resistors to check them, there isn't anything significant in the circuit to disturb their measurement easily, also if they now measure 200M then they are blown.

I'm still trying to get my head round the Q4 emitter follower pulling up the bases of the 2 2N3055's and both op amps are running without any feedback ?. is this a published circuit ?

Cheers Ken

[radoczi94]

I know, the schematic is a bit annoying, working on a version for better readability. You can find many versions of this schematic on the web, even an ebay kit is based on this PSU.

The IC2 gets feedback directly from the output, IC2 has no feedback, since it works as a comparator.

[Damianos]

How you measured 1A on 200MOhm resistance? Is there a 200MV source?
Also how you measured the output current?

On the circuit:
- What is the role of T1?
- What about C16?
- Missing capacitors on D4 and D5 (bases of respective transistors)...
- The reference voltage (VREF 8V) is relative to the other side of the shunt, so the current limit adjustment will vary with load current...
- The 100nF capacitors, here and there, are like the designer had a bag of them, without knowing where to use them! I'm wondering if any calculation of time-constants / filter-poles had been made.
- ...

[C]

I know, the schematic is a bit annoying, working on a version for better readability. You can find many versions of this schematic on the web, even an ebay kit is based on this PSU.

A lot of junk on web

The IC2 gets feedback directly from the output, IC2 has no feedback, since it works as a comparator.

If IC2 is a comparator then it would be a switcher supply.
IC2 feedback includes Q4,Q5 & Q6

If you look at T1, as current drop changes across current sense resistors it's resistance changes effecting output.
As R17 voltage changes due to output voltage change, IC2's output changes.
This is not an OR, it is a voltage modulated by current.
With this happing then remaining parts of circuit has to compensate for this error or you have a voltage modulated by current output.

[radoczi94]

How you measured 1A on 200MOhm resistance? Is there a 200MV source?
Also how you measured the output current?

On the circuit:
- What is the role of T1?
- What about C16?
- Missing capacitors on D4 and D5 (bases of respective transistors)...
- The reference voltage (VREF 8V) is relative to the other side of the shunt, so the current limit adjustment will vary with load current...
- The 100nF capacitors, here and there, are like the designer had a bag of them, without knowing where to use them! I'm wondering if any calculation of time-constants / filter-poles had been made.
- ...

I did not measured 1A on 200MOhm resistance, I removed the shunts and tried to measure resistance on the pcb to see, if there is some sort of short circuit or something.

T1 transistor shorts out the output of IC2, when the negative supply dies, which happens almost instantly, when the mains goes off.Whitout that, after switching off, the output would jump up to the maximum voltage.

Honestly, I have no idea, what C16 supposed to do.

I placed caps there on the PCB, just missing on the schematic.

I know, the schematic is a bit annoying, working on a version for better readability. You can find many versions of this schematic on the web, even an ebay kit is based on this PSU.

A lot of junk on web

The IC2 gets feedback directly from the output, IC2 has no feedback, since it works as a comparator.

If IC2 is a comparator then it would be a switcher supply.
IC2 feedback includes Q4,Q5 & Q6

If you look at T1, as current drop changes across current sense resistors it's resistance changes effecting output.
As R17 voltage changes due to output voltage change, IC2's output changes.
This is not an OR, it is a voltage modulated by current.
With this happing then remaining parts of circuit has to compensate for this error or you have a voltage modulated by current output.

Sorry, I meant IC1 has no feedback.Which is actually not true, because it has a cap, so its an integrator.My bad.

I do not know if this thing oscillates or not, I do not have access to a scope to see what is happening.One thing came into my mind: if I tear off IC1 from the circuit, and put a reasonable load on the output, I would be able to disqualify a lot of theories. If the missing 2A comes out, then something crazy happens with IC1, if not, then it's an another kind of problem.

Anyway, made a new schematic, I hope, it's better.

[C]

Most people think of Ground as something that does not move.
In latest where you labeled GND changes based on current and needs to float.

missed Q7 acting like a switch.


If you see problems on power off you could also have problems at power up.
I would say best cure is to control power up and down of power supplies and circuit.


Think of it this way on power down. Circuit is losing power, output is dropping  and circuit reacts by turning on harder using even more power and kills output in process.

If you replace D3 with a switch controlled by -5v supply getting to say -4v then when power loss happens, V+ is no longer one diode drop from X1-1.
All V+ current demands then drops V+ faster. The harder IC2 output goes positive the faster output dies.

Controlling Vref is another way to control output on power up/down.

Many ways to do something.

I just do not like that circuit.

[radoczi94]

You may not like that solution, but it works perfectly since the first power up. If you look, D1-D2 and C1-C2 makes a Villard voltage duobler. At the power up, the negative supply needs some time to stabilise,it takes a few sine cycle. When powering off it goes away rapidly, because there is no large capacity like on the positive side. That part of the circuit works just fine.

[xavier60]

It's odd that the Emitter resistors are 0.2 ohms and the shunt resistors are 0.22 ohms. I would run the output into an ammeter or a shunt resistor to know the output current then measure the voltage drop across all of the low value resistors in the current path to make sense of it.
Maybe this is what you have been doing?

[radoczi94]

It's odd that the Emitter resistors are 0.2 ohms and the shunt resistors are 0.22 ohms. I would run the output into an ammeter or a shunt resistor to know the output current then measure the voltage drop across all of the low value resistors in the current path to make sense of it.
Maybe this is what you have been doing?

I used the same kind of resistors for the shunt and the emitter resistors. Measured the voltage drop across the emitter resistors too, they had around 1,5 A flowing thru each. Nothing odd there.

Tried the same kind of resistors as a load on the output, they had 0,65V voltage drop at 3A, so the resistors are correct value.

[xavier60]

It's odd that the Emitter resistors are 0.2 ohms and the shunt resistors are 0.22 ohms. I would run the output into an ammeter or a shunt resistor to know the output current then measure the voltage drop across all of the low value resistors in the current path to make sense of it.
Maybe this is what you have been doing?

I used the same kind of resistors for the shunt and the emitter resistors. Measured the voltage drop across the emitter resistors too, they had around 1,5 A flowing thru each. Nothing odd there.

Tried the same kind of resistors as a load on the output, they had 0,65V voltage drop at 3A, so the resistors are correct value.

And the voltage drop across R25 or R23 indicates only 1 amp? 

[radoczi94]

It's odd that the Emitter resistors are 0.2 ohms and the shunt resistors are 0.22 ohms. I would run the output into an ammeter or a shunt resistor to know the output current then measure the voltage drop across all of the low value resistors in the current path to make sense of it.
Maybe this is what you have been doing?

I used the same kind of resistors for the shunt and the emitter resistors. Measured the voltage drop across the emitter resistors too, they had around 1,5 A flowing thru each. Nothing odd there.

Tried the same kind of resistors as a load on the output, they had 0,65V voltage drop at 3A, so the resistors are correct value.

And the voltage drop across R25 or R23 indicates only 1 amp?

Yes. And the series connected current meter shows that too.

[xavier60]

Although you have checked, there must be a short somewhere bypassing current around the shunt resistors. What supplies input power?
Can you take some photos?

[radoczi94]

Although you have checked, there must be a short somewhere bypassing current around the shunt resistors. What supplies input power?
Can you take some photos?

The power is supplyed by a 250VA transformer, a cooled rectifier bridge and 10kuF 100V caps.

That board ind the front-middle is just only a reverse polarity protection board, 2 diodes and a fuse. It's not attached to the circuit right now. The small board on the left is a small supply for the relays and the fan control.
 And yes, the non marked wires have a reversed polarity on the rectifier board output,
 I corrected that before powering on.https://drive.google.com/drive/folders/1BI_EM4ImJGpWINuNeS3gs_Wgnjbidrzf?usp=sharing

[xavier60]

Wow! that is one serious power supply.
All I can suggest, if you have not done so already, is to remove one of the shunt resistors and solder in PCB pins so that you can experiment and take measurements while the PCB is mounted.

[radoczi94]

Yeah,  next time will make some wire loops for test points. Always learning.

[Damianos]

How you measured 1A on 200MOhm resistance? Is there a 200MV source?
Also how you measured the output current?

On the circuit:
- What is the role of T1?
- What about C16?
- Missing capacitors on D4 and D5 (bases of respective transistors)...
- The reference voltage (VREF 8V) is relative to the other side of the shunt, so the current limit adjustment will vary with load current...
- The 100nF capacitors, here and there, are like the designer had a bag of them, without knowing where to use them! I'm wondering if any calculation of time-constants / filter-poles had been made.
- ...

I did not measured 1A on 200MOhm resistance, I removed the shunts and tried to measure resistance on the pcb to see, if there is some sort of short circuit or something.

T1 transistor shorts out the output of IC2, when the negative supply dies, which happens almost instantly, when the mains goes off.Whitout that, after switching off, the output would jump up to the maximum voltage.

Honestly, I have no idea, what C16 supposed to do.

I placed caps there on the PCB, just missing on the schematic.

So, the 1A is imaginary or real current?  How you lost the 2A? How you know that the total was 3A? 

T1(Q7) and IC2: See the notes of the maximum ratings section in the Op-Amp datasheet.

C16: see what does C14 and you will see that it, at least, cancels the compensation.

Consider the current reference, noted above...

I know, the schematic is a bit annoying, working on a version for better readability. You can find many versions of this schematic on the web, even an ebay kit is based on this PSU.

A lot of junk on web

The IC2 gets feedback directly from the output, IC2 has no feedback, since it works as a comparator.

If IC2 is a comparator then it would be a switcher supply.
IC2 feedback includes Q4,Q5 & Q6

If you look at T1, as current drop changes across current sense resistors it's resistance changes effecting output.
As R17 voltage changes due to output voltage change, IC2's output changes.
This is not an OR, it is a voltage modulated by current.
With this happing then remaining parts of circuit has to compensate for this error or you have a voltage modulated by current output.

Sorry, I meant IC1 has no feedback.Which is actually not true, because it has a cap, so its an integrator.My bad.

I do not know if this thing oscillates or not, I do not have access to a scope to see what is happening.One thing came into my mind: if I tear off IC1 from the circuit, and put a reasonable load on the output, I would be able to disqualify a lot of theories. If the missing 2A comes out, then something crazy happens with IC1, if not, then it's an another kind of problem.

Anyway, made a new schematic, I hope, it's better.

The IC1 is neither a comparator nor an integrator, it is intended to operate linearly by reducing the voltage setting when the current tends to be greater than the set limit...

[radoczi94]

No, I measured that 1A series with the shunt resistors, 3A was present AT THE SAME TIME on the output, parallel(Intentionally).
If IC1 oscillates, it may have tricked the dmm.

I do not know which data should I look in the table, but it has unlimited short circuit protection in the temperature range, if you think abot that.

Yes, it moves a little bit, but this was not meant to be a high precision thing. It can be solved by whacking in an another reference chip, tied to the fixed GND. I was aware of that, but probably it will be good enough for the girls I go out with.

[xavier60]

It wouldn't be too surprising if it is oscillating in CC mode because both op-amps are in the loop making the response rather complex.
I see in other designs that either the CV op-amp or the CC op-amp can take sole control of the power transistors.  The response of both loops can be individually optimized.

[Damianos]

I lost the communication!
If someone gives me an algorithm, to decode the messages, it may be reestablished!

[TimNJ]

Try a "diode-OR" to switch between the CV and CC control loops. This is a common and effective way to implement constant current limiting on a "normally CV" power supply.

Check out the attached schematic of an HP6214A lab power supply.

Also check out this Keysight article: http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.html

[xavier60]

There is a 5W 0.68 ohm resistor on the regulator PCB. I can't see the marking of the resistor next to it? Are these the shunt resistors?

[radoczi94]

Try a "diode-OR" to switch between the CV and CC control loops. This is a common and effective way to implement constant current limiting on a "normally CV" power supply.

Check out the attached schematic of an HP6214A lab power supply.

Also check out this Keysight article: http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.html

Thanks, will try it.

Edit: One thing came into my mind. I am not shure, that I can put a diode in series with the CV opamp. What I could do is to move the diode on the CV opamp output (D7) and tie it's anode to the driver transistors' (Q4) base.

There is a 5W 0.68 ohm resistor on the regulator PCB. I can't see the marking of the resistor next to it? Are these the shunt resistors?

Yes, it was in there, when I took the picture. When I put those 0,68R resistors,then I realised, that the "missing" 2 amps is more or less linear with the output load current.

[xavier60]

Don't modify it until you have found a definite reason for the problems. Have you checked for oscillations?
When you checked the regulator PCB for a possible short bypassing the shunt, did you have all of the wires connected?
 

Try a "diode-OR" to switch between the CV and CC control loops. This is a common and effective way to implement constant current limiting on a "normally CV" power supply.

Check out the attached schematic of an HP6214A lab power supply.

Also check out this Keysight article: http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.html

Thanks, will try it.

Edit: One thing came into my mind. I am not shure, that I can put a diode in series with the CV opamp. What I could do is to move the diode on the CV opamp output (D7) and tie it's anode to the driver transistors' (Q4) base.

There is a 5W 0.68 ohm resistor on the regulator PCB. I can't see the marking of the resistor next to it? Are these the shunt resistors?

Yes, it was in there, when I took the picture. When I put those 0,68R resistors,then I realised, that the "missing" 2 amps is more or less linear with the output load current.