Hello everyone. A while a go, I got an old Allen Bradley PLC from an acquaintance. I played around with it a bit until it's PSU failed. There was no output from it whatsoever. I decided to repair it, as I can still use this PLC for some home project. At first, I just quickly analysed it and found out that electrolytic cap (C11) failed and PWM IC couldn't start (it was also making squeeking noises). I replaced it and found out that IC itself is also shorted. In the process of making measurements, I also made a short with probe and burned power switching mosfet. I replaced both of them and it started working - well...somewhat. This is a dual output power supply (5V and 24V) and I'm measuring around 35-40V and 4.1V respectively. So, obviously something is not functioning ok. So, I decided to analyze it more deeply and I reverse engineered it. I took quality high res photos of pcb and I drew PCB from which I could generate schematic - really fun process and now I want to do more of this.

However, I've come up with following schematics. On the first image you can see transformer and optocoupler, connecting output section with control section (switching). Then we have input section, which is fine, control section (switching), which is also fine I suspect and lastly output section, which is the most interesting part of this design. Based on transformer and connections in input and control section, I think this is a forward converter type of SMPS - please correct me if I'm wrong. Secondary winding, that's used for +5V is also used for feedback, although I don't fully understand how the part which goes into U4 reference pin works.
The other part of this section is really interesting to me. There are 3 transistors. Q2 and Q4 are some kind of emmiter followers. Q2 is driven by U5 voltage reference, then Q4 is driven by Q2. And lastly, Q5 is driven by Q4. Q5 controls voltage reference U5 and serves as some kind of feedback. I don't know if pin 6 of J2 is input from PLC circuit or is actually output. The section that I tried to explain is switching control for Q3 which is used as buck or boost converter? How?
I'm having some hard time to fully understand this design so I'm looking for some advice/tips/explanations to help me better understand it and possibly fix the problem. Please correct me if I misunderstood something.
I'm also attaching image of the PCB if by any chance someone recognizes it or if it helps in some other way.

 

I solved the problem with help of a member on another forum. Apparently, C26 which is secondary reservoir capacitor on rail that's used for feedback was faulty. I replaced it and this solved the problem. I also replaced C22 and C23, just in case.

 

EDIT: I'd still like to better understand section with Q2, Q3, Q4, Q5 and U5 though.

 

Are you sure about this part of the circuit? Is there a negative voltage on C17?
This part of the circuit looks more like a soft start + power-good signal.

 

Are you sure about this part of the circuit? Is there a negative voltage on C17?
This part of the circuit looks more like a soft start + power-good signal.

I'm not 100% sure about it, but I think it's quite accurate. Also, I can't tell if there is negative voltage on C17. I already assembled whole PLC together. Can you please elaborate a bit on this assumption - soft start + power-good signal. Could you "walk me through" functioning of this circuit a bit please?

 

I see it this way:
At power up Q3 if OFF (no current flow ). Q2 is saturated, thus, C30 is discharged. The Q4 and Q5 (complementary Schmitt trigger) are cut off too. U5 is also cut off.
As the voltage 5V rail rises. The negative voltage at C17 will "rise" to (more negative). This will turn ON U5 and Q3 and cut off Q2.
Therefore after some delay needed to charge C30 the Q4 and Q5 will turn ON and latch the circuit in the ON state.

Maybe someone else has better ideas. Because I'm not convinced that the schematic is correct.