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part of scheme for alto-s-8-mixer-service-manual.pdf

I have come across a mixer that I am going to rebuild, and maybe try to replace the volume potentiometers with a digital variant.

The picture shows part of the schematics (after 3-band EQ, and before bus-summation).

The mixer is an Alto S-8, the op-amps on the board are all 4580s.

Signal in this part comes in on thr non-inverting input, also shared with CCW on the volume knob. The wiper is grounded, CW has feedback from the op-amp's output.

Scenario 1 - volume knob is turned "off", the signal is grounded and we wish silence. What then is the function of the components for negative input?

Scenario 2 - volume knob is turned on, or full on, components above is grounded - what then is the function of the components for negative input?

Why is the pot wired like this? I have never seen a similar op-amp configuration.

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  • \$\begingroup\$ It's the non-inverting version of an active gain control. Remember that in the non-inverting configuration, R262, C121, and VR47A are also part of the NFB network. \$\endgroup\$
    – user207421
    Commented Mar 17 at 22:55
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    \$\begingroup\$ It gives you a closer to exponential gain control, while using a linear pot. Smart IMO. \$\endgroup\$
    – tobalt
    Commented Mar 18 at 5:17
  • \$\begingroup\$ biggest issue with this is that when you change the gain you also change the LF -3dB point. If this is a mic amp (given the high gain it likely is) you want a LF point of around 10-20Hz without any extra filter. If you look at the gain graphs below you can see that you get a bit of LF cut at the highest gain, but it's actually flatter than you ideally want at lower gains. \$\endgroup\$
    – danmcb
    Commented Mar 18 at 15:25
  • \$\begingroup\$ DennisHolm - Hi, Where did those schematics originally come from? In order to comply with the site rule on referencing, details of the original source of copied / adapted material must be provided next to that material. If the original source is online (webpage, PDF, video etc.) then edit the question & add its name & link (URL). If the source was a printed book or other offline material, then edit the question & add a normal citation (see the linked rule for details). TY (Please see the tour & help center for the site rules.) \$\endgroup\$
    – SamGibson
    Commented Mar 18 at 15:51

4 Answers 4

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If 50k is a linear potentiometer, then we have "quasi" a linear (log) decrease of output.

Steps defined as 0.1 ... and not 0.2.

enter image description here

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The potentiometer accomplishes three tasks

  1. Adjusting the input level to the filter, effecting the output level of the associated frequency band.
  2. Adjusting the gain of the filter; at low input levels, the amplifier noise is still present. Reducing the gain at the same time as reducing the input level also reduces the output noise so that the band is quiet when its output level is cut.
  3. The “quasi-log” behavior in [Antonio51’s answer][1].

This could be done with two rheostats mechanically ganged together, but, this method requires only a single potentiometer. Much simpler and more cost effective.

Thanks @JosBergervoet for the third task inclusion. [1]: https://electronics.stackexchange.com/a/706412/319836

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    \$\begingroup\$ Three tasks, if you add the "quasi-log" behavior mentioned in the other answer. (And I think those three do sum it up.) \$\endgroup\$ Commented Mar 18 at 8:11
  • \$\begingroup\$ Thanks @JosBergervoet. I have included a reference. \$\endgroup\$
    – RussellH
    Commented Mar 18 at 15:16
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The potentiometer can be thought of as two variable resistors in series. Each is a fraction of the nominal one of the device, i.e. kR and (1-k)R (with 0≤k≤1 ) whose sum gives us the resistance R of the potentiometer itself. The circuit can therefore be designed and studied much more conveniently in the following way:

enter image description here

Circuit analysis and Bode plots:

enter image description here

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Because the wiper of the pot is grounded, the internal resistance elements above and below the wiper position can be treated and two independent resistors whose values are inversely proportional to each other - as one goes up, the other goes down.

The upper one is in series with the shunt leg of the opamp feedback loop, basically a resistance that is added to R262 as the pot rotates. As the pot rotates upwards, this resistance decreases, increasing the circuit gain.

The lower one is the shunt leg of a voltage divider formed with R246. As the pot rotates upwards, the resistance of this element increases, decreasing the attenuation and increasing the signal available to the opamp.

When the pot is rotated all the way down, the input signal is grounded and there is no circuit output. As the pot rotates upwards, both the input signal (at the opamp non-inverting input) and the circuit gain increase.

A fallout from the math is that a linear pot has a logarithmic effect on the output signal.

Note that while we don't know the input signal frequency range, this circuit almost certainly is not intended as a filter. With the pot rotated all the way up, the two corner frequencies of the gain stage are over 200 kHz (R283 and C192) and less than 1 Hz (R262 and C121).

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