Exponential VCA with temperature correction

As promised in the previous post, here is a new take on the exponential VCA. This time the CV input has been modified and a tempco resistor added. This is the same scheme used in many VCOs for the pitch CV.

As the 1k 3300ppm tempco resistor is the most used tempco resistor in DIY designs, possibly making it easier to get hold of, I have decided to use it for this design as well. This, however, means that the input CV has to be amplified before it is attenuated by the resistor voltage divider formed by R21 and TR1.

There are two ways to build this circuit - trimmable and not trimmable. For the not-trimmable version, use the exact values found in the diagram - 57k (47k + 10k) on the CVTRIM input and 23k (22k + 1k) in the feedback loop. All resistors should be 1% or better metal film resistors.

For a trimmable version you may replace the CVTRIM input resistors with a 10k resistor. The input voltage should be around -2.6V, and you may connect a 20k or 25k potmeter between 0 and -15V to achieve this. This potmeter will subtract a voltage from the input CV, which affects the maximum and minimum amplification possible. Both move in the same direction, if the maximum drops, the minimum will also get lower.

You may also change R11 to a 18k resistor and R19 to a 10k potentiometer. This pot will affect the "distance" between the maximum and minimum amplification. Turning the potmeter to the right and increasing the resistance of R19 will make the maximum amplification drop, but at the same time the minimum gets higher (i.e. you will not be able to attenuate the signal as much).

CV (linear) and response (Exponential) without trimming. No tempco used but temperature is 24 degrees celcius. Trim voltage is -15V, not 15V as it says in the picture.

The maths behind this

The maximum input to the transistor base when no correction CV is added should be -0.2072V. Using the R21/TR1 resistor voltage divider means that the output from the IC1A opamp must be -11.8V.

This voltage should be reached when the CV is 5V, which means that the CV has to be amplified -11.8V/5V = -2.362 times.

We will almost achieve this if we select a 23k feedback resistor and a 10k input resistor.

To reach -70dB, the correction CV must be 0.106V at the transistor input. This means it has to be 6.042V at the opamp output due to the resistor voltage divider. If we  choose to input -15V at the CVTRIM input, we need an input resistor (R1+R3) = 23k * -15V/6,042V = 57.1k

Or

If we choose a 10k CVTRIM resistor, the input correction CV must be 6.042V * -10k/23k = -2.63V

UPDATE: As pointed out by MB in the comments, there is an error in the schematics. The output from IC1A/B should go to the top of the 56k resistor and the Q1/Q2 transistor bases should be connected to where the 56k meets the 1k tempco, as shown in my original sketch here: