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I don't tend to post questions on online forums so I apologise in advance if I've messed up here.

I'm currently trying to work on an old transistor guitar amp and in taking measurements managed to short one of the output transistors with my multimeter, blowing the other transistor and a fuse (as far as i could test nothing else was damaged). After finding a suitable replacement the amp works exactly as it did before, with the same issues it had before.

However when measuring the voltage across the speaker out BEFORE the DC blocking capacitor it measures 20v! (approximately the half rail voltage) After the capacitor across the speaker there is no DC. I stupidly didn't perform the measurement before trying to work on the amp so I don't know if this is how it's supposed to be and if this is normal, and if not what I need to change to fix it.

For a bit of background the amp is (as far as I can tell) a Quasi complimentary push pull with a single rail design. No service manuals available, I couldn't even find any documentation on the amp at all.

Edit: Thanks for the help!

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  • \$\begingroup\$ This isn't a forum; it's a question and answer site. \$\endgroup\$ – Andy aka Jul 15 at 11:50
  • \$\begingroup\$ That's what the blocking capacitor is for! If there wasn't voltage there, they wouldn't need it. \$\endgroup\$ – user253751 Jul 15 at 14:26
  • \$\begingroup\$ Yes, the circuit is supposed to do this. Blocking that 20 V (hopefully half the single rail supply voltage) is what that DC blocking capacitor is for. \$\endgroup\$ – Graham Nye Jul 15 at 19:07
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Before the capacitor (transistor side), the average DC output voltage will be half the total supply rail of the amplifier hence, it your amplifier supply rails are 40 volts and 0 volts you will see a quiescent DC voltage of 20 volts.

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Single rail push-pull stage would typically have half of the supply voltage at output stage before the DC blocking capacitor. This way, the output can swing about half-supply up or down from the midpoint.

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