Theory of the MG-1 VCA

I will be referencing the schematic in the previous post.

Q: Why is Q28 needed?
A: We want an envelope to be based at 0V, go to some positive voltage, then back to 0V. This is the control signal we expect for a filter, oscillator, and VCA as well. The problem is the 3080's control current pin (5) is internally connected to a diode which is connected to the negative supply voltage. If we wanted to connect a control voltage (with series resistor) directly to this pin, our CV would have to based at the negative supply instead of ground. This single PNP transistor circuit is the simplest solution. Here's how it works. When the envelope is 0V, the transistor doesn't conduct and no current is passed to the OTA. Note the -15V on Q28's collector at the start and finish of the envelope. During the peak of the envelope the transistor conducts, so the base-emitter voltage will be roughly 0.6V. The voltage across R104 is (4.5-0.6), giving a current of (4.5-0.6)/15k = 0.26mA. The collector voltage is then roughly 0.26m*10k+(V_ABC). V_ABC is a diode drop above the negative supply, so the collector voltage is 2.6+(-15+0.6)=-11.8. This corresponds to the -12 shown at the envelope peak on the MG-1 schematic. Actually, if we neglected the diode drop voltages of the transistor and OTA instead of using 0.6V, we get -12V exactly.

Q: What are R96 and R97 for?
A: According to an article on OTAs from Nuts and Volts magazine posted here (follow the link at the bottom to UsingOTAs2.pdf), the input resistors "help equalize the source impedances of the two signals and thus maintain the DC balance of the OTA."

Q: Why are they 100 ohm?
A: If they were larger, they would do a better job of equalizing the source impedance. However, there are two concerns. The OTA has a finite input resistance. The data sheet gives a typical value of 27k. If the resistors were larger, the differential input voltage would be reduced. The output of the ladder filter is already quite low. The second concern is noise. The larger the resistor, the noisier. These resistors are at the input of an amplifier, so any noise will be amplified as well.

Q: What is the gain of the OTA?
A: We found I_ABC is 0.26 mA at the peak of the envelope. We also know the transconductance (g_m) of the OTA is 19.2*I_ABC. This gives a g_m of 5mS. We can see our input voltages are two reverse polarity 2mV_pp signals, so a total differential input of 4mV_pp. The output current swing is then 4mV*5mS=0.02mA.

Q: What is the output voltage?
A: The OTA output is fed to the master volume pot in series with a 1k resistor. This is a clever arrangement that allows a standard pot to be used while the resistance, dominated by the smaller resistor, can be designed. The max output voltage peak-to-peak is 0.02mA*1k//10k = .02*10//11=18mV_pp. This corresponds to the 17mV_pp shown on the schematic.