Analog exponential convert for MFOS VCO

I'm trying to understand the exponential converter in the MFOS VCO. Here is a link to the complete MFOS Noise Box synth schematic. http://musicfromouterspace.com/analogsynth_new/NOISETOASTER/NOISETOASTER.php However I tried to isolate just the exponential converter of the VCO here. https://www.circuitlab.com/circuit/8emjqv7bj65n/mfos-exponential-converter/ (warning, when simulated, this circuit does not work. I'm new to circuitlab so I'm not sure why)

Is this what going? The input (at node in) coming from the VCO mixer goes through an inverting opAmp (not shown), so represents a negative DC voltage at the base of transistor Q1. The - input of OpAmp OA1 is positive but near ground because + input is at ground. This drive the output of the OpAmp very negative. This turns on both PNP transistors. Because current is exponentially related to voltage across a transistor, the output is an exponentially related negative current. ?

(Bonus question, why doesn't the circuitlab simulation work?)

• which node is the linear-ramp Vin? Cannot be the opamp (-), because a forced voltage breaks the feedback behavior. – analogsystemsrf Aug 18 at 2:35
• I just added the node Vin at the opamp(-) so that in the simulation I could check to see that this value was close to the opamp (+) / ground. – honkskillet Aug 18 at 2:45

This is called a differential amplifier simulate this circuit – Schematic created using CircuitLab

A constant current source draws a constant combined current through the emitters of transistors Q1 and Q2.

If the voltage at V_in1 is equal to the reference voltage REF, and assuming that the transistors are identical, then the currents through Q1 and Q2 are the same, and hence the currents into I_out1+ and I_out1- are also the same.

Lets say we start increasing the base current of Q1 (by applying an input signal) then the emitter current of Q1 is going to increase and since the combined emitter currents from Q1 and Q2 must be constant so must the emitter current of Q2 decrease an equal amount. This causes an increase of current into I_out1+ and a decrease of current into I_out1-.

Now since the current through the base of Q1 is logarithmically related to the voltage across the base so if we let the voltage on V_in1 increase then the base current is going to increase logarithmically.

This circuit produces a constant current proportional to V_in2 simulate this circuit

Any voltage at V_in2 greater then 0v is going to cause a current to flow through R1 and out of I_out2-. If this causes any increase in voltage at the inverting input of the opamp then it is going to decrease the voltage on the output in an attempt to keep the voltage at the inverting input at 0v. This causes a constant current to flow between I_out2- and I_out2+.

Now lets combine them simulate this circuit

Does it start to look familiar?

The current at I_out1- is proportional to the voltage at V_in2 and log proportional to the voltage at V_in1.

• Awesome explanation. Unfortunately I haven't been able to get the simulation on circuit lab to actual produce exponential output current. I'm not quite sure what I'm doing wrong. I did find this circuit which is very similar but uses PNP transistors. circuitlab.com/circuit/pk72tb/… But, when I try to replace the transistors with NPN transistors and rewire I don't get exponential output. (Similar to my simulation linked in the question.) Do you have an example of you combined circuit with component values that produces the desired log output? – honkskillet Aug 18 at 21:07
• The circuits I have shown are just for explanation/ showing the principles. From what I have told you it should be possible for you to do your own research and learn how to use the circuit. In practice you need more components to make it work properly, and designing the final circuit and choosing the right component values is a whole weeks work on its own, and not something I think you are likely to get someone here to do for you. Regarding the simulation in circuitlab; don't wast your time, if you really want to simulate then use LTSpice instead, it is also free. – Vinzent Aug 19 at 6:04
• One question. I've been watching some EE circuits university lectures online. They tend to talk about OpAmps as voltage sources as opposed to current sources. Is the resistor R3 between the opAmp outpur and the transistor pair needed conceptually or is it just there practically to limit current through the transistors? – honkskillet Aug 22 at 2:27
• I am sorry to sound like a dick. But it is obvious from your question that you have a lot left to learn about op-amp circuits. I could try to give you a simple explanation, but it is not going to be of much use to you because it is not going to be enough for you to be able to understand the circuit in full. I would advise you to read a lot more about op-amps and op-amp circuits. For instance, google op-amp constant current source and op-amp feedback circuits etc. Again I'm sorry but what you're asking is kind of like; "How do I design a vaccine? recipe please!, only household ingredients!". – Vinzent Aug 25 at 6:20