Allachikatillo
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The base to emitter voltage of the 2SC1627 is 9V but it is supposed to be about 0.7V. I think you have its B and E pins swapped. Its datasheet shows the collector pin is in the middle so simply turn the transistor around.
EDIT: Replace the transistor because you might have overheated its fragile base-emitter junction and install the new one with the correct pins.
See reply in post #5. The 2SC1627 must be operating if it is only dropping Vbe and a little in the series resistor. The voltage at it's output is already confrmed to be correct by measuring across C132.
Brian.
I find it strange that you stll hear output when the controls are all turned to minimum.
Can you try this please:
Using headphones, confirm that a signal fed to the right input is only heard in the right earphone and a signal fed to the left input is only heard in the left earphone.
I'n trying to work out if there is a common point between the channels that is cross coupling instead of gounding.
Brian.
It seems odd to me that the LED current isn't returned to the same PCB that provides it but a few mA wouldn't upset its working. The current probably flows to the 'blue' PCB ground then by various leakage paths to a lower voltage. With 12V at it's anode it only need to connect a point of about 10V or lower for it to light up.
It does seem that the voltage across C132 is wrong. It comes from the emitter of the 2SC1627 which is why shorting C and E makes the voltage rise. All the 2SC1627 does is provide a slow rise in supply voltage so you don't hear a loud 'pop' at the output when you switch on, it isn't a voltage regulator and when C138 (the delay capacitor) has charged up it should appear almost as a short circuit. What seems wrong is that if you manually short it out, making +VM about 15V, the output is distorted.
Can you measure the voltage at both ends of R194 and report them back please.
Brian.
It is a "capacitance multiplier" (look in Google). It smooths the voltage the same as a huge capacitor value on the voltage that is the capacitor value at its base to ground multiplied by the hfe of the transistor.Q1 isn't a voltage regulator, it's purpose is to make the supply to U7 rise fairly slowly so you don't hear a pop in the headphones when the amplifier is switched on.
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