I managed to fry a piece of moderately rare equipment due to connecting it to the wrong voltage power supply, and I'm now trying to diagnose what's wrong. Luckily, reverse engineered schematics are available (http://electrickery.xs4all.nl/comp/px8/doc/f10pa.pdf).

I've managed to identify one possible problem; there's a -12V power supply, which is actually generating -7.8V. Circuit below. (I didn't draw this. It's using US-style zigzag resistors; note that L7 is actually an inductor.)

enter image description here

So I'm rather puzzled as to how this works. I can identify the general principle: the two Schmitt triggers form an oscillator; this combined with the L7 inductor produces voltage spikes; C9 decouples this; D17 grounds the positive spikes but lets the negative ones through to charge C10; C31 is a smoothing capacitor.

However, the closest thing I can find to a voltage reference is the potential divider / transistor arrangement at Q13 / R86 / R83. I would expect the divider to turn the transistor on if the output voltage is too great, causing C10 to be discharged and so lower the voltage (lower in this case meaning 'towards zero', because the output voltage is of course negative).

Except the potential divider appears to generate about -1V when the output voltage is -12V, which is higher than the turn-on voltage of about 0.6V. But it produces -0.6V when given the -7.8V I'm seeing, which is... suspicious. Also, why is the collector connected to the oscillator circuit and not to ground?

So my questions are: how does this all actually work, and how could this circuit ever have generated -12V?

  • \$\begingroup\$ Try a free simulation tool. \$\endgroup\$ – Andy aka Apr 9 '19 at 11:37
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    \$\begingroup\$ Quick calculation: Q13 starts to conduct when there's 0.6 V across R83 so 0.6 / 820 = 730 uA. Assuming 730 uA through R86: 730 uA * 10 k = 7.3 V. Add 0.6 V Vbe = 7.9 V. At -7.9 V Q13 starts to conduct. It is unclear to me how "hard" Q13 needs to be on to significantly influence the oscillator so that the output voltage is lowered. I do think that maybe the feedback needs to be dimensioned slightly differently for -12 V output (actually -12.6 V as there is a diode drop as well). \$\endgroup\$ – Bimpelrekkie Apr 9 '19 at 11:44
  • \$\begingroup\$ After browsing over the schematics, I cannot find where this -12 V is used. Maybe it is only used for the negative supply of (some) opamps? Then my guess would be that the -7.8 V you measured could be "normal", the opamps could work just as well with that voltage. My point: it might be that this circuit is undamaged and that the fault is somewhere else. \$\endgroup\$ – Bimpelrekkie Apr 9 '19 at 12:07
  • \$\begingroup\$ I believe (haven't followed the tracks) that the -12V line is fed into the RS232 line driver chip; the connection's not shown on the schematic. These are SN75188 and SN75189s. The datasheet for that says that the nominal negative input voltage should be -9V, so that kinda matches. The schematic has been spot on so far, though. I'll need to check that. (It'd be awesome if the line driver's fried, they're easy to replace!) \$\endgroup\$ – David Given Apr 9 '19 at 12:14

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