As part of the synthesizer project I have been working to design a simple voltage controlled amplifier. Due to the complexity of other parts of the circuit and the amount of time I've spent on this project overall I am seriously considering opting for a variable amplifier chip; an OTA perhaps. Still, working with what I have now, I have drawn up what I think might function as a voltage controlled amplifier from discrete components and an op amp.

As I understand it the circuit should behave as follows: T1 acts as a low impedance emitter follower, biasing the incoming signal at VCC/2 so as to allow for continuous operation of T2. This second transistor functions as a current source with the right end of R4 effectively at GND. The amount of current flowing through this resistor is determined by voltage across the base and emitter junctions of that transistor (Vbe/R4). Because of op amp rules the current flowing through T2's collector/emitter must be mirrored across R5. This produces a voltage at the output proportional to the voltage at the collector of Tr2.

During testing, however, the circuit didn't exactly behave as above. To account for possible errors in the values I chose for that test I have ignored assigning values to this particular circuit. Does this circuit make sense? Is my analysis correct.my circuit

P.S. Perhaps this is a different topic but please feel free to critique my EAGLE labelling conventions and circuit syntax as well. It's still new to me.

EDIT: Fixed an error in my drawing in which the non-inverting pin of the op amp was erroneously tied to GND and not 1/2 Vcc.

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    \$\begingroup\$ How fast do you need the amplifier to respond to the gain signal? \$\endgroup\$ Commented Apr 11, 2015 at 20:10
  • \$\begingroup\$ @OlinLathrop Hmm, I hadn't thought about that. Let's see: I intend to use a VCA in two places, once to set the cutoff frequency of a voltage controlled filter and again as a stand-alone amplifier. In the first instance I will need to accommodate frequencies up to 20KHz but for the later stand-alone section the requirements are more forgiving. There the gain will be routed to a low frequency oscillator with a maximum frequency of only 30 Hz or so. Does this help? \$\endgroup\$ Commented Apr 11, 2015 at 20:37
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    \$\begingroup\$ Um, OK, so how fast do you need the amplifier to respond to the gain signal? \$\endgroup\$ Commented Apr 11, 2015 at 21:41
  • \$\begingroup\$ Oh, was that not helpful? Sorry :P I suppose instantaneously would be best but with the 20KHz limit I believe 50 microseconds is the highest delay time. But I'm not sure I understand what you're asking. \$\endgroup\$ Commented Apr 11, 2015 at 22:09
  • \$\begingroup\$ I was asking about the response time to the gain signal, not the signal that the amplifier is amplifying. \$\endgroup\$ Commented Apr 12, 2015 at 12:37

1 Answer 1


Obviously you didn't build this, right? This is a simulation? Connecting the virtual ground of the opamp to GND is a no-no. Your op amp will have no swing. You either need to run both positive and negative supplies and define a ground in the middle or use voltage dividers all over the place to run the circuit off a single supply. If you try to build this the opamp output will be clipped all to hell since it can't get below ground.

The big problem with your circuit is that it is brutally non-linear in its gain programming characterisitic. The current through T2 is exponentially related to CV, not linear. If you want to connect your circuit to any kind of modular gear you're probably in a 1V/octave system or something. Your CV range for this circuit is probably less than 100mV.

  • \$\begingroup\$ Whoops, I didn't catch the GND issue. On my breadboard the non-inverting pin is in fact connected to 1/2*Vcc. I'll re-upload the proper the drawing to reflect that. Thanks for seeing such a silly error :P You're note on the non-linearity was an oversight of mine as well. \$\endgroup\$ Commented Apr 11, 2015 at 22:03
  • \$\begingroup\$ Is everything correct now? I harbored concerns that the op-amp might load down the emitter follower. \$\endgroup\$ Commented Apr 11, 2015 at 22:14
  • \$\begingroup\$ looks fine, bit it is still a weird circuit. The DC output level will also track the CV, which is undesired. Are you AC coupling the output to look at it? If so it should be doing what you want at this point. \$\endgroup\$
    – crgrace
    Commented Apr 11, 2015 at 22:37
  • \$\begingroup\$ btw, T2 will be the thing that loads down the emitter follower, not the opamp. Run a DC analysis of your circuit at a couple of different values for the CV and you'll see what I'm talking about. \$\endgroup\$
    – crgrace
    Commented Apr 11, 2015 at 22:38
  • \$\begingroup\$ I agree that it's a weird circuit and I hope I can think of something better later. In the meantime I'll toss a capacitor between the transistors, do a DC analysis, test it and then get back to you. \$\endgroup\$ Commented Apr 11, 2015 at 23:31

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