Information about a circuit

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Here is the Pioneer M22 amp schematic. As Q4 and Q6 have push pull signal going to the output stage what is the function of C12-C13 in this circuit. For me they mix opposite signal together (Cancellation ? Why ?). Thanks


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Output transistors Q1 and Q2 have to remain turned on slightly at all times, to prevent a nasty sound of them turning on and off, called "crossover distortion". The current with no AC signal going through R25 and R26 should be constant, as evidence by the voltage written next to the emitter resistors R25 R26. Potentiometer VR100 series the spreader diodes D8 D9 is adjusted to keep the voltage across R25 & R26 as shown in the schematic, .4 v each. As this is a DC condition, the capacitors C12 & C13 keep the voltage difference between the driver transistor bases Q7 Q8 constant. Two different technology capacitors are used because C12 as an electrolytic, has inductance and will not follow high frequency signals perfectly. C13 is smaller and hence faster.
If the output transistors Q1 and Q2 were left to switch on and off with each waveform, the amps would be "class B". With the slight leakage promoted by the bias spreader diodes and potentiometer, the amp is "class AB". These have been found to sound better than Class B, which was used in very early transistor amps.
In more advanced amps D8 & D9 are mounted on the heat sink with the output transistors, to collapse the idle current a little as the output transistors heat up. This would be to prevent "thermal runaway". Even more modern amps put a small transistor in that position instead of the diodes, called a "vbe multiplier".
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They appear to be bias stabilization components. The idea is that the charge across them keeps the driver bias voltages on the bases of Q7 and Q8 at a more or less fixed voltage with respect to each other.

They are not mixing opposing signals since the signal coming into Q7 and Q8 are in phase but separated by about 3 volts.

VR2 is your idle bias adjustment.
The bias circuit needs constant current through it to maintain a stable bias voltage. These capacitors carry the signal current across the bias circuit allowing it to see steady state conditions. The signal currents can be quite high and are in particular required to drain stored charge out of the top driver transistor as it switches off so this isn't delayed (cross conduction and switching distortion would result otherwise.)

The circuit would also benefit from a speed-up cap between the emitters of the drivers to help with switching of the main output transistors.

The drivers and output transistors in this configuration do turn off fully on each half cycle (*), it takes a more complex scheme to prevent it, and frankly doesn't gain you anything useful.

(*) the large emitter resistors on the outputs will pull several volts at full power, so that the "off" transistors actually have negative base drive.
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As this amp is sounding a little bright in the treble, would it be a good idea to take out C13 ?
Or is there a kind of compensation that could be done in the negative feedback path ? Thanks

C13 is there because C12 is an electrolytic capacitor and electrolytics do not have a great frequency response. Removing it could cause instability at higher frequencies and may open a door for Radio Frequency Interference.

I generally advise against manipulating the frequency response of an amplifier. There is a strong risk you will destabilise it leading either to low frequency motor-boating or high frequency oscillations.

Ask if it's too hot in the treble or too weak in the bass...
Better, I think, to simply use your tone controls to trim it.
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