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I found these schematics for an audio mixer.

I reproduce the mainpart of the schematics. A first stage of input buffers before each in line consisting of an opamp is not reproduced. I am only interested in unipolar mode.

I recognize a summing inverting op amp circuit. After the first op amp we have

$$ V = - \sum_i V_i \frac{1}{1 + R_i / 100k} $$

Where \$R_i\$ is the resistance value of the i-th potentiometer.

If the potentiometer is on the far right, we then have a coefficient of \$1\$ in front of \$V_i\$. It is ok.

When you turn the potentiometer on the far left, for a potentiometer of 25k, you then have a coefficient of \$ \frac{1}{1 + 25k / 100k} = 0.8\$.

The documentation says:

When in unipolar mode, all pots feeding that output behave in regular fashion, that is when the knob is fully counter-clockwise, no signal from the associated input passes into the mix. As the knob is advanced clockwise, a greater portion of the signal passes into the mix.

I don't understand how 80% of the signal can be considered no signal.

It seems that I am missing something.

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    \$\begingroup\$ Note that the schematics read "* multiple decoupling caps, as needed". You might get away with it as drawn, but I'd suggest adding (the pads at least) for 0.1µF caps near each TL072 + and - pin and ground. That way, if it is built and squeals badly or generates high-frequency oscillations, just add the caps. \$\endgroup\$ Commented Jun 14, 2022 at 12:09
  • \$\begingroup\$ Do you mean decoupling capacitors between power supply pins and ground or between inverting and noninverting inputs and ground ? \$\endgroup\$ Commented Jun 26, 2022 at 11:23
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    \$\begingroup\$ Power supply and ground. Adding caps to the inputs will essentially add a low-pass filter to the input and block all but the lowest audio signals. \$\endgroup\$ Commented Jun 27, 2022 at 12:07

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I am only interested in unipolar mode.

And

I don't understand how 80% of the signal can be considered as a no signal.

That's because you have misread the circuit. When a wiper is fully counter-clockwise (unipolar mode), its associated input becomes grounded and hence, does not contribute any signal to the output of the op-amp mixer.

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  • \$\begingroup\$ Indeed, i thought that it was connected to nothing and not to ground. Thanks Andy. \$\endgroup\$ Commented Jun 14, 2022 at 11:17

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