Analog exponential convert for MFOS VCO
This is called a differential amplifier







simulate this circuit &ndash; 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







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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







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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.