Op-amp circuit to control gain of two analog channels with one single-section potentiometer

Question

I'm working on designing my own headphone amplifier with extremely high fidelity. One thing I want to incorporate is a circuit to control the gains of the two channel's using one single-channel potentiometer, and I want to use op-amps and discrete components for high precision channel to channel matching. A circuit to do exactly this is described in this article, but the schematic picture has 404ed. I'm trying to recreate the schematic from the body of the article. The key is that the wiper of the potentiometer is grounded, so we have two resistors varying inversely. There are some concepts like positive feedback which are difficult for me to intuitively understand, so I was only able to get partway. I'd really appreciate some help in recreating this circuit.

Definitions, from a lot of rereading the article:

• V1, V2: input voltages to the first stages. The signals which should be amplified by the same factor.
• R: the total resistance of the potentiometer, as well as a resistance value which is used elsewhere for other resistors
• Rg: another common resistor value
• K: the position of the wiper, from 0 to 1

Here's what I have so far, and my rationale.

the signal at A1's noninverting input is ... K x V1. This voltage is boosted by A1's noninverting gain of G = 1 + RG/R to produce overall gain as a function of K given by VOUT/VIN = KG.

A1 is in a non-inverting configuration with the gain network defined by passive resistors with values RG and R, and the voltage on the noninverting input is K*V1.

Positive feedback from A1's output to P's CCW terminal results in constant current drive to the pot, given by I = V1/R. Therefore, the signal at A1's noninverting input is equal to I x K x R = (V1/R) x K x R = K x V1

One of the potentiometer's end terminals is on the non-inverting input of A1, and is receiving a constant current from a positive feedback network. This current is setting the voltage on the op-amp's input terminal. As to the network that is giving it that constant current, I have no idea.

Surrounding A2 and P's CW terminal is a topology that produces a signal of V2(1 - K)(1 + R/RG) at A2's output

This sounds like the same topology as U1, since the other potentiometer resistor section would be changing inversely to K.

After attenuation by A3's feedback network to V2(1 - K), this voltage appears at A3's inverting input. So the differential voltage seen by A3 is V2 - V2(1 - K) = V2 x K. When amplified by A3's gain of (1 + RG/R) = G, this voltage becomes V2 x K x G. A3's gain, accordingly, is equal to KG, just like A1's

This is some form of differential amplifier circuit, with the output of A2 going into the inverting input network and unmodified V2 signal going into the non-inverting input. I'm not sure how the mentioned "attenuation in the feedback network" or the final gain value were attained. Presumably we have a bit more freedom here.

Answer 1

The article says:-

Positive feedback from A1's output to P's CCW terminal results in constant current drive to the pot... the differential voltage seen by A3...

From this I figure that each half of the pot is driven with a current proportional to the input signals. The voltage across each half is proportional to its resistance, so as the pot is turned the volume 'fades' from one channel to the other. A differential amplifier then subtracts the second channel's pot signal from the input signal, to produce a signal whose volume tracks the other channel.

We may never find out exactly what was in the circuit described in the article, but the schematic below reproduces its basic functionality.

^{simulate this circuit – Schematic created using CircuitLab}

In this circuit op amps OA1 and OA2 are configured as constant current generators controlled by the input signals. R2 and R3 set the current through each pot half to 0.1mA/V, which produces 1V/V (ie. 0dB attenuation) across the pot half when at maximum resistance. OA3 takes the signal from one half of the pot and subtracts it from the input signal to 'invert' the volume level on that channel.

Answer 2

Here's an LTSPICE simulation of Woodward's circuit done by someone on DIYAudio:

Obviously R5/R5A represents the potentiometer.

It's interesting (in theory anyway).