The diagram shows part of the schematic for the late-model Minimoog VCO. In the circuit, Q7 and Q8 form a current sink, drawing a current through the collector of Q7 that is exponentially related to the voltage present at the base of Q7. This current is integrated by C7 until the voltage at the drain of Q4 (which is at present switched off) reaches 0V as detected via the voltage follower IC7 by the comparator IC8b, whose + input is held close to ground by a resistor not shown. At this point, the capacitor is discharged by Q4, and the drain of Q4 rises to 5V again. Thus the oscillator generates a falling sawtooth waveform with a frequency proportional to the current. Compensating resistor R31 can be neglected at low frequencies, and the purpose of Q5 is just to lift the sawtooth at the base of Q6 (which converts the sawtooth to a triangle wave) by 1 V_BE drop. IC8b has a bit of positive feedback to ensure a clean reset.

Further analysis is given in the Minimoog Service Manual, Section 2.3, which can be found online: http://www.cyborgstudio.com/synthmp3s/moog/minimoog/manual/minimoogservicemanual.pdf

Note that

(a) The collector of Q8 is held at ground by the negative feedback around op-amp IC6.

(b) So 5 microamps flow in R30 as claimed on the schematic [though in the opposite direction to that marked].

(c) The emitters of Q7 and Q8 are held at a V_BE drop (appropriate to I_C = 5 μA) below ground by action of the op-amp.

(d) When the base of Q7 is at ground potential, 5 μA will flow in Q7 also.

(e) This will charge C7 to 5V in 1ms, and give the oscillator a frequency of 1kHz, assuming reset time is negligible. Indeed the legend 200Hz/μA appears above R30.

(f) But the legend 0.00mV = 2' lo F = 350Hz appears at the base of Q7, implying that setting the oscillator's range switch to the highest (2 foot) setting and pressing the bottom note on the keyboard will produce a voltage of 0.00 mV here and a frequency of 350Hz. And indeed, taking A4 = 440Hz gives F4 = 349.23Hz according to [an online source].

These things can't all be true at once! Is it just (f) that's wrong, or is there some reason why the current sink doesn't give 5 μA but something more like 5/3 μA when Q7's base is at ground?

Bonus question: what is the purpose of CR2 in this circuit?

Minimoog schematic

  • \$\begingroup\$ Schematic from fantasyjackpalance.com/fjp/sound/synth/synthdata/… \$\endgroup\$ Commented May 16, 2016 at 15:49
  • \$\begingroup\$ Is the oscillation a sin wave, square wave, or sawtooth at this point? \$\endgroup\$ Commented May 16, 2016 at 17:38
  • \$\begingroup\$ I added a brief summary of the circuit operation. \$\endgroup\$ Commented May 16, 2016 at 18:27
  • \$\begingroup\$ It's possible that the op amp acts as a comparator, in which case pin 2 is not necessarily at ground \$\endgroup\$ Commented May 16, 2016 at 20:46
  • \$\begingroup\$ Interesting idea, but (sorry to pull a trump card) that's not part of the circuit analyses I've read -- and not consistent with the 5 μA current labelling R30. I'll give a link in the question to an analysis. \$\endgroup\$ Commented May 16, 2016 at 21:26

3 Answers 3


Actually, more information can be gleaned by looking at annotations surrounding the two op-amps that compute the control voltage, which tell us things like low F = 0V on the keyboard, and 2' = -5V on the octave switch. Many of the scaling resistors involved are trimmable, but it's possible to deduce their nominal values from the fact that they are supposed to give 1V/octave overall. It turns out that "0.00 mV" is not in the possible range of values at the node in question, but values around -30mV are possible, at least according to my calculations, and would give 350Hz from an oscillator with a natural frequency of 1kHz. So I think the annotation (f) is just wrong, and the other voltages given are consistent with each other. I won't give the details here.

Thanks for the suggestions; perhaps more detail could be given only by those who have themselves built and tested the circuit, a club I hope to join very soon.


I think you're assuming that the text "0.00 mV = 2' LO F = 350 Hz" directly relates to this circuit.

I think it does not. I think the text means that there should be no DC voltage (0.00mV) at this point and that ther should be an AC signal of nominally 350 Hz (coming from oscillator number 2 perhaps ?).

CR2 prevents the output of opamp IC6 from having a voltage above 0.7 V. My guess is that this helps the opamp reach the desired state more quickly (after power up for example) and it might prevent a "lock up" situation if the output voltage of the opamp goes too high. It's a bit of crude solution but the TL081 probably cannot source much current anyway so it would work in this case.

  • \$\begingroup\$ That's not right. That node is the output of an op amp that sums the control voltages for the oscillator, and is normally a DC voltage that determines the frequency, increasing by 18-21 mV per octave. (18mV would be normal for an exponential pair at room temperature, but Q7 and Q8 are in an oven at about 80C, and 21 mV/oct is the right figure here.) \$\endgroup\$ Commented May 16, 2016 at 18:34
  • \$\begingroup\$ If you know it so well, then why do you ask ? \$\endgroup\$ Commented May 16, 2016 at 19:12
  • \$\begingroup\$ I'm sorry, FakeM, but by "that's not right," I meant, "that's a helpful suggestion, but it's inconsistent with what I understand about the rest of the circuit." And the problem is precisely that the information given on the circuit diagram seems inconsistent. I don't want to build it and then find it can't be tuned! \$\endgroup\$ Commented May 16, 2016 at 19:49

IC6 output will be -ve as its inverting input is +ve but will be clamped at around -0.7V by CR2. This means that the exponential function starts at 0v rather than 0.7v

  • \$\begingroup\$ Isn't CR2 the wrong way round for that? \$\endgroup\$ Commented May 12, 2021 at 8:14

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