Please help me understand\fix this discrete linear power supply

[Agent24]

I'm trying to fix an old (1987!) DSA-524 Digital Storage Adaptor for Oscilloscopes\PC.

So far I've only succeeded in screwing up the +5v rail (or perhaps, finishing it off - see below...)


The PSU in this thing supplies +/-7 volts to the analog circuitry with an LM358, a TIP31A and a TIP32A, as well as a -5 volt rail via an upside-down 7805.
The digital section is powered by a +5v rail based on a TL072 and TIP32A.
Both op-amps work from a 2.45v ZN404 reference, D309.
(See Service Manual, Page 19 for PSU schematic)


On all my previous testing the +5v rail seemed without issue. However when trying to verify voltages to ICs on the analog board, I noticed a burnt resistor (R303) next to the PSU's power transistors. This is the base resistor for the +5v rail's TIP32A. (Q302)

I pulled it and it measures 33 Ohms. Since it starts with a Orange band, this seems reasonable. Perhaps not burnt out then, just overheated. The schematic says 68 Ohms so 33 Ohms is at least in the right area and most likely correct.

After replacing R303 with a new 33 Ohm resistor, I tested the +5v rail (but without the digital board, and hence without a load, plugged in). I got about 7 volts, which confused me and made me think the regulator circuit was bad. It was only when I removed LK3 and LK1 that I found the +/-7 rails also rose, to +/-11v without load, so it seems the design of this circuit is such that the unloaded voltage is higher than normal.


That was when I had the idea of using a 15 Ohm resistor as a dummy load on the +5v rail. 300mA should be nothing for a TIP32A capable of 3A, right? Maybe wrong...

I got +5v on the +5v rail with the resistor in place, no worries there. That seemed to prove the unloaded rail theory. I turned the unit off to do something else. Then when I turned it on again, the +5v rail was +0.5v.
I removed the resistor and the rail increased to +3.5v but did not return to the ~7v unloaded it was before.


I've verified the new R303 is all connected OK and hasn't blown. I haven't changed anything else.


What the hell went wrong with this supply? Was 300mA too much for it and something blew?
Was it on its last legs (overheated resistor a sign of that?) and I finished it off?

What should I look at first? It seems so simple and yet I don't really know where to start. Analog circuits are not exactly something I have good experience with.

Also, why does an apparently regulated supply seem so .. UNregulated without a load? I don't get it.

[stj]

Re: Please help me understand\fix this discrete linear power supply

33ohm base drive?
cant see the point - 330ohm or 3k3 maybe.

[stj]

Re: Please help me understand\fix this discrete linear power supply

soo i looked.
fuck - i think it was designed by the same mad bastard that did the psu in the sinclair spectrum!!

-5v is from a 7905, not a 7805

the circuits could have used darlington transistors driven by ua723's
the problem is sinclair have always put cost ahead of good or simple designs.

i have several hundred ua723's new and in the tubes btw, they used to be used in a lot of arcade machine psu's.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

Ah you're right. It is a 7905. The schematic is awful quality and the regulator was covered in dust.

The base drive to (or from, since it's a PNP?) Q302 has to be 33 Ohms - I've never heard of resistors going low value before. The base resistors for the other transistors are 100 Ohms.

[budm]

Re: Please help me understand\fix this discrete linear power supply

It should maintain 5V even without the load, the Vref feeding the - input of the TL072, and the output is fed into + input.
If the output (5V) tries to go higer than the set point (set by P301), the output on pin 7 of the TL072 will go high so the PNP connected to pin is turned on less which in turn less bias for the TIP32A which then the output Voltage will be reduced. For that Base resistor to burn up, it means the transistor that drive TIP32A may be damaged or the E-B junction of TIP32A is bad, since the output is not regulating until you out quite a load on the output, it indicates that it is not regulating properly. So you need to check the transistors real well, if the output of the Opamp pin 7 is stuck low then the transistors will be on more than what it should be.
The lower the Base resistor, the more Bias current to turn on TIP32A. The bias current has to be high enough to deliver current at 5V at current rated requirement.
What is the input DCV feeding the TIP32A?
BTW, it is also setup as tracking power supply so if the + goes high or low, it will track each other so the OPAMP (loads) will always get equal +/- for VCC.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

That's what I figured - that it should regulate even without a load - but if that's the case, something must be wrong with the +/-7v rails also, then? They also overvolt without a load.

[budm]

Re: Please help me understand\fix this discrete linear power supply

Vef from the Zener Diode is used for all 3 supplies for ref voltage, I cannot read the 1N44XX? It is next to the C312 in the drawing.
The pot 300 is used for adjusting the +/- 7v output, P301 is for adjusting the output to +5V.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

The Vref comes from a ZN404 (2.45v). It's the metal can device next to the overheated resistor in the photo. It seems to be OK and is supplying 2.463v to both Opamps.

I can adjust the 7v rails from 6.3-7.3v with P300.

[budm]

Re: Please help me understand\fix this discrete linear power supply

So what happen when you adjust P300? The loads right now are all the circuits that are connected to the output of the regulators. For 7V to go up as high as 11V is way off.
If you look at the internal working of the LM78XX series, you will see that it works the same as the discrete regulator like you have here, they add thermal protection, SOA protection, etc. but otherwise you still have Ref voltage and an error amplifier for regulation. and can maintain output of 5% from 5mA range up to 1A.

https://cdn.badcaps-static.com/pdfs/...49e01b5d66.pdf

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

When I adjust P300, the 7v rails can be changed between 6.3v and 7.3v which seems sensible.

Yes, with the analog circuits currently connected the +/-7v rails seem to be regulating OK, but when I had them disconnected before, they went to +/-11v or so. I don't know what P300 does when they (analog circuits\load) are not connected, I did not try that yet.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

With the analog circuitry disconnected, the +7v rail is +11.13v, the -7v rail is -11.16v, and adjusting P300 does nothing to change them.

In this situation (no load on any rail) the D309\VRef actually changed to 1.235v

And I just realised why. The datasheet of the ZN404 shows its forward voltage will drop if the forward current through it drops below 1mA, which is because it's getting no voltage from the main +7v rail, because I removed LK1 without realising that while this indeed cuts power to the analog circuitry, it also cuts power to the ZN404 so all it has is a tiny current from R309, obviously not enough to make it work properly.

OK, so the thing probably *DOES* regulate just fine without a load. It just doesn't regulate without a load and an incorrect Vref. My fault!


Unfortunately this does not explain the stuffed +5v rail though.....

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

Back to troubleshooting the +5v rail...

For that Base resistor to burn up, it means the transistor that drive TIP32A may be damaged or the E-B junction of TIP32A is bad, since the output is not regulating until you out quite a load on the output, it indicates that it is not regulating properly. So you need to check the transistors real well, if the output of the Opamp pin 7 is stuck low then the transistors will be on more than what it should be.
The lower the Base resistor, the more Bias current to turn on TIP32A. The bias current has to be high enough to deliver current at 5V at current rated requirement.
What is the input DCV feeding the TIP32A?

The resistor didn't actually burn up. The paint did but the resistor itself seems intact. I guess it was still passing too much current, but fault or bad design, I don't know.

The Input DCV feeding the TIP32A is 8.78 volts.


Here's some more voltage readings:

Opamp U212 \ TL072:
Pin 4: -7.03 volts
Pin 5: Adjustable from 1.874 - 1.65v via P301
Pin 6: 2.463v (VRef)
Pin 7: Seems different everytime I turn it on. Seen from -0.300v to -0.210 volts, but does not change as Pin 5 changes. (Does that qualify for 'stuck low'?)
Pin 8: 8.78 volts

Transistor Q303, ZTX214:
Base: Follows U212 Pin 7.
Emitter: Seen to jump around anywhere from -0.125v to +0.100v. Does small jumps then will occasionally do a big jump and change polarity.
Collector: -0.001 volts
(Also tried a BC327 for Q303 and got the same thing)

Transistor Q302, TIP32A:
Base: Follows Q303's Emitter, small mV voltage which jumps about from negative to positive occasionally
Emitter: 8.78v
Collector: 3.53v

[stj]

Re: Please help me understand\fix this discrete linear power supply

pull and check q303

it may be pulling the opamp down,
and if the resistor got hot then it's been passing a LOT of current for an E-line package!

and on that note, where did the current go?
best to pull and check q302 and d308 for leaks.

[budm]

Re: Please help me understand\fix this discrete linear power supply

Voltages jumping around it may may be due to bad filter caps and the circuit is going into oscillation. The output of the opamp is low so the transistor is on, since the Vref is 2.45V on the - pin for it to have + output, the + input has to be >2.45V which is adj by P301.
The input VDC is 8.78V, so the E-B of TIP32 is around 0.6V and Vsat Q303 is may be 0.6V or if it is leaky (Q303 should never be fully on if the circuit is working), so then you can calulate Vr33 Ohms, or just measure the voltage drop on the 33 Ohms to get the current draw that will be Base current, so time that by the Beta of TIP32A will give you approx Iec current when Q303 is fully on (which it should not be doing right now if the circuit is working).

[stj]

Re: Please help me understand\fix this discrete linear power supply

i assumed this thing had been re-capped.

[budm]

Re: Please help me understand\fix this discrete linear power supply

One simple way to understand how this regulator functions is to think of TIP32 as a variable resistor for now for simplification.
Let's say the input Voltage is 10V (instead of 8.78V) for simplification, and the output V on the load is to be maintained at 5V withe load resistance between 10 Ohm to 1000 Ohms range. So the Vce of TIP32 must always be at 5V.
At 1K Ohms load, the current flowing through the circuit will be 5mA so the TIP32 will be bias on to provide 5mA of current to the load and maintain 5V on Vce and on the load.
AT 100 Ohms load, the current will be 50mA so the TIP32 will be bias to provide 50mA to the load and maintain 5V on Vce and on the load.
At 10 Ohms load, the current will be 500mA so the TIP32 iwll be bias to provide 500mA to the load and maintain 5V on Vce and on the load.
As you see in the circuit, the error detection circuit drive the Base circuit to provide the current needed by the load and maintain the 5V output.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

Voltages jumping around it may may be due to bad filter caps and the circuit is going into oscillation.

I had that idea too, checked the output of the Opamp on my oscilloscope, but couldn't see anything, just a flat line. Maybe the voltmeter was affecting it somehow?

i assumed this thing had been re-capped.

I have replaced all the electrolytic capacitors on the digital board already, as I had assumed my original problem with the unit was in that section, they were originals but actually all tested OK out of circuit.

The analog board has been fully recapped by the previous owner (though I don't know when) except the 3 big filter capacitors. Everything seems to be Panasonic\Rubycon.

The output of the opamp is low so the transistor is on, since the Vref is 2.45V on the - pin for it to have + output, the + input has to be >2.45V which is adj by P301.
The input VDC is 8.78V, so the E-B of TIP32 is around 0.6V and Vsat Q303 is may be 0.6V or if it is leaky (Q303 should never be fully on if the circuit is working), so then you can calulate Vr33 Ohms, or just measure the voltage drop on the 33 Ohms to get the current draw that will be Base current, so time that by the Beta of TIP32A will give you approx Iec current when Q303 is fully on (which it should not be doing right now if the circuit is working).

I assume you mean those 0.6v drops across transistor junctions are what should be there if it's working? Because right now E-B of the TIP32 is about 8.78 volts, right? (since Emitter is 8.78v and Base is about 0v).

I noticed the datasheet for TIP32A says V-EBO max -5v. Does this mean the current situation is running the transistor out of spec?

One simple way to understand how this regulator functions is to think of TIP32 as a variable resistor for now for simplification.
Let's say the input Voltage is 10V (instead of 8.78V) for simplification, and the output V on the load is to be maintained at 5V withe load resistance between 10 Ohm to 1000 Ohms range. So the Vce of TIP32 must always be at 5V.
At 1K Ohms load, the current flowing through the circuit will be 5mA so the TIP32 will be bias on to provide 5mA of current to the load and maintain 5V on Vce and on the load.
AT 100 Ohms load, the current will be 50mA so the TIP32 will be bias to provide 50mA to the load and maintain 5V on Vce and on the load.
At 10 Ohms load, the current will be 500mA so the TIP32 iwll be bias to provide 500mA to the load and maintain 5V on Vce and on the load.
As you see in the circuit, the error detection circuit drive the Base circuit to provide the current needed by the load and maintain the 5V output.

And in real operation with the 8.78v input, the TIP32 should drop 3.78v across it and there will be 5v across the load?
This is real close to the 3.53v I am currently getting out of the TIP32 transistor with no load. Coincidence?


Thanks, I think I'm getting it now. The whole thing wasn't helped by my murky understanding of opamps, but I found this http://www.talkingelectronics.com/pr.../OP-AMP-1.html last night and it starts to make sense.

[budm]

Re: Please help me understand\fix this discrete linear power supply

"Because right now E-B of the TIP32 is about 8.78 volts, right? (since Emitter is 8.78v and Base is about 0v)." Emitter Base junction should be only around 0.6~.07V (same as Diode junction), if it shows 8.78V (measured measured between the E leg and the B leg) then it has open junction. Red probe on E, black on the B, the meter then should show 0.6V or so for the P-N junction (Anode is at E, Cathode is at B)

"And in real operation with the 8.78v input, the TIP32 should drop 3.78v across it and there will be 5v across the load?" yes.
You can replace the 5V section with LM7805, if the max current requirement is 0.50A and about 3.5V Vdrop, it will dissipate about 1.75W (same as the TIP32 has to dissipate right now) so you will need heatsink to keep Tj below 125c. It already uses LM7905 for -5V already.

[Agent24]

Re: Please help me understand\fix this discrete linear power supply

I measured across the E-B junction and got 8.78 volts. Seems like the TIP32 is blown then?


That couldn't have been from my 15 Ohm test resistor on the +5v rail, surely?
Because, with a 15 Ohm load drawing 333mA, that would only be about 33mA of base current at worst case, since the datasheet says Hfe is 10 minimum.
The datasheet also says TIP32 max base current is 1 Amp, and the Q303 would be surely burnt out before that happened.

And now that I think about that... the ZTX214 can handle up to 200mA. At that current, the 33 Ohm resistor would have to dissipate 1.32 Watts. I'm guessing it's a quarter-watt, maybe half a watt rated. I'm thinking the resistor would burn before Q303 would anyway. Perhaps.


Now, since I never noticed smoke at any point when the unit was in my possession, I assume the resistor overheated before I received it.

I tried to think some way the circuit could have drawn too much base current from the TIP32 and blown it but I don't see it being possible. There isn't enough voltage and the 33 Ohm resistor is too high. The resistor or Q303 would go up in flames long before the TIP32 (in theory)

So if the TIP32 is bad, it must have just died on its own accord, nothing to do with the rest of the circuit. Does that make sense or have I got it horribly wrong?

[budm]

Re: Please help me understand\fix this discrete linear power supply

"I measured across the E-B junction and got 8.78 volts. Seems like the TIP32 is blown then?" Remove it and check the B-E junction.
If the small transistor shorted V-C, then worst case will be about 250mA flowing through 33 Ohms and the E-B of TIP32A, so more likely the transistor just fail like you suggested.