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I found these schematics for an audio mixer.

The main part of the schematic is shown below. A first stage of input buffers before each in line consisting of an opamp is not reproduced. I am only interested in unipolar mode.

enter image description here

I recognize a summing inverting op amp circuit, but I don't get precisely the point of adding a 220 \$\Omega\$ resistor after the output of the second opamp. My hypothesis was that it limited the current going through the opamp from its output.

What is the purpose of this resistor? Is there a name for such a resistor? How do you choose the value of 220?

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  • \$\begingroup\$ Something really "strange" in this mixer? \$\endgroup\$
    – Antonio51
    Jul 5, 2022 at 11:48
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    \$\begingroup\$ What do you mean ? \$\endgroup\$ Jul 5, 2022 at 13:40
  • \$\begingroup\$ Will add simulation. I must check again. \$\endgroup\$
    – Antonio51
    Jul 5, 2022 at 14:09
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    \$\begingroup\$ use of linear pots is a bit odd. Would expect either log or a shunt resistor from wiper to ground to slug the law into "fake log". \$\endgroup\$
    – danmcb
    Jul 6, 2022 at 7:22

3 Answers 3

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The configuration shown is tolerant of output capacitance.

Without the resistor and (very important) the 47pF capacitor, the circuit would be prone to high frequency oscillation when loaded with even relatively small capacitances (such as a couple meters of shielded cable would have). That's because the load capacitance in conjunction with the AC output impedance of the op-amp (typically in the 100 ohm range) will create a pole. The 220 ohm resistor is a bit bigger than Ro and will isolate the load capacitance so the 47pF capacitor can provide AC feedback without the lag.

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  • \$\begingroup\$ Thanks Spehro, I placed myself in DC so erased the capacitor and indeed it was quite useless. In an AC setting it makes sense with your answer. I'll check on output impedence of opamp on some book and try to figure out where the value 220 \$\Omega\$ comes from. \$\endgroup\$ Jul 5, 2022 at 13:45
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enter image description here

The transfer function of this configuration is (assuming the load resistance is high enough): $$ \mathrm{\frac{V_o (s)}{V_i(s)}=-\frac{R_B}{R_A}\cdot \Bigg(\frac{1}{1+sC \ (R_B+R_C)}\Bigg)} $$

From the equation above, we can say:

  • The capacitor \$\mathrm{C}\$ interacts with RB and RC (100k, and 220R, respectively) to prevent HF oscillation.
  • Without the capacitor \$\mathrm{C}\$ (47p in your schematic) the configuration would be simple inverting amplifier with the well-known gain formula. RC (220R) would have no effect on gain.
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  • \$\begingroup\$ In your transfert function, you then have a factor \$R_B + R_C\$ that approximates \$R_B\$. I don't see the point of the B resistor then. \$\endgroup\$ Jul 5, 2022 at 13:17
  • \$\begingroup\$ @InfiniteLooper Yes, in your case RC seems to have no effect but DC current limiting. But if you place one that could not be neglected by RB then it'll have both mid-band gain and the frequency response. \$\endgroup\$ Jul 5, 2022 at 13:28
  • \$\begingroup\$ Your TF seems to have something wrong, at very high frequency gain approaches infinity. BTW if you don't place a load (capacive) and or study the output impedance that circuit is just a lowpass inverting amplifier. \$\endgroup\$
    – carloc
    Jul 5, 2022 at 16:43
  • \$\begingroup\$ @carloc I re-calculated the gain equation and yeah, I made a math mistake. Thanks for pointing out. Fixed. \$\endgroup\$ Jul 6, 2022 at 6:51
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EE&O ...

Here is what I found with the simulation of this circuit.
This configuration seems ok. Vo1 and Vo2 are attenuated.

enter image description here

I discover something "strange" about Vo2 ... Someone could confirm?
Pot cursor at the left or at the right ... "attenuated" input pass at Vo1.
Pot cursor at the middle, Vo2 ???

enter image description here

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