On Apr 16, 2012, at 7:57 AM, Holm Tiffe wrote: Ouch. Yeah, that looks like the wirewound resistor (at least) got WAY too hot. You can see the FR-4 fibers poking through, which means the board temperature was probably way above design (assuming a reasonable thermal design). Still, I'd be surprised if it were fatal damage, especially since it's probably a 2-layer board with no plated through holes. I imagine it's not too structurally sound anymore, though (I guess that's why you put in the perf? My German isn't so good, though I think I can guess what "brutalen Aktion" means). - Dave

I don;t know this supply at all, so I am totally guessing here... Is it possible that this is an inrush limiter circuit to prevent the PSU taking too high a surge current when it's first conencted to the maisn (and htus the capacitors are discharged)? At power-on, the wirewound resistor is in series with charging circuti to the capactors (either in series with the mains supply, or the output of the rectifier), thuis limitign the current. Once the supply has got going, the resistor is shorted out by that semiconductor device (are we sure it's an FET, I ahve seen triacs and SCRs used in such circuits too). If the cirucit fails, the PSU tries to run with the resistor in series. The resisotr gets hot and bothered, and the chopper is passing more current than it should be, due to the lower input voltage to that stage (some of the mains votlage is dropepd across the resistor). I seem to rememebr one of the DEC PSUs (H777?) has a relay used to short out the inrush limiter resistor, and there's a fuse in series with said resistor. If the relay or its drive fails, the fuse will blow, hopefully before too much damage is done. -tony

Ok, lets think for a second.. This wirewound resisotr is, perhaps, 4700 ohms. And it burnt up spectacualrly, so perhaps it disipated at least 5W. Now, P=V^2/R, so V= sqrt(P*R). That implies a voltage across that resistor of about 153V by my calculations. And a voltage that won't drop even when supplying such a load. Are you sure? What is it conencted to? Which side of the isulation barrier is it on (mains or output side?) -tony

Dorry, I was using the term in a 'logical' rather than 'physical' sense. Computer PSUs always (I think that's true) have the outputs electrically isiolated fro mthe mains input. There is no direct electrical connection between them. Pwoer is transfered from the mains to the outputs by a double-would (separate priamry and secondary windings) transformer, contorl feedback is passed fro mthe outputs to the chopper transistor either by double-wound pulse transoformers or by optoisolators. I regard that electrical 'gap', bridged by the transformers and optoisolators as the 'isolation barrier'. It's often convenient to regard an SMPUS ans being in 2 halves, one on each side of this. The maisn side cotnains the mains rectifier, smoothing capacitors and chopper cirucit. The otuptu side cotnains the putput rectifiers, smoothign capaictors, crowbar, ay aditional regualtors, and the voltage feedback circuit. The chopper drive cirucit could be on either side of the barrier, depeneding on the design of the PSU. My guess is that this is the centre-tapped primary of the chopper transformer and that you have a push-pull chopper circuit (not unusual). What is this resisotr, etc, connected to? If it really is a 4700 ohm one, it must have quite a voltage acorss it to burn it out. Is the FET associated with this resistor one of the 2 you have just mentioned as being connected to the chopper transformer (I susepct these are the choppers)? -tony

My apologies... It is very hard to judge somebody's level of experience the basics. In whcih case the PCB wit hthe trasnformer on it is probably the otupt uside, the PCB with the FETs o nit is the mains side. There will be a litle of the 'wrong' side on one of the PCBs, for example the optoisolator will eb on one PCB amd will bridge the 2 sides, but apart between the 2 PCBs. Wellm there is, it;s just not a single chip... There is something to drive the chpper transitors and vry the drive based on the output voltage, but is;ts discrete transsitors and Op-amps. And that makes life difficult. You cna use a SMPSU control IC( let's say an SG3524) in essentiually one way. But you can use trnasistors and op-amps in many ways, so figuring out what is going on is harder. But you know this. I would still try to work out just how the resitor and its FET are used. If necessary, trace out the circuit from the mains input through the rectifier to the capacitros and choppers. Then trace out the supply to the control circuitry (e.g. the power pins on that 358). Hopefully something will make sense. -tony