Basic logarithmic amplifier circuit

A log amplifier gives an output voltage proportional to the logarithm of the input voltage. The basic circuit places a diode or a transistor (transdiode connection) in the feedback path of an inverting op-amp. It uses the exponential current–voltage law of the PN junction.

Circuit

              Q1 (npn)
        C +----|<|----+ E
          |   B=GND   |
 Vin-[R]--+--(-)\     |
             A   >----+--> Vo
     GND--(+)/

(A diode with its anode at the inverting input and its cathode at the output can replace Q1.)

Working and derivation

1. The inverting input is at virtual ground, so the input current is I = Vin/R. All of it flows into the collector of Q1.
2. The base and collector are both at 0 V, so VBE = −Vo. The transistor law gives:

IC = Is · exp(VBE/VT)

3. Equating the two currents:

Vin/R = Is · exp(−Vo/VT)
Vo = −VT ln( Vin / (R·Is) )
   = −2.303 VT log10( Vin / (R·Is) )

At 300 K, VT = kT/q ≈ 26 mV. So the output changes by about −60 mV for each tenfold (decade) increase in Vin.

Features and limitations

• It works only for one input polarity (positive Vin for an npn transistor). A protective diode is often added.
• The output is small (hundreds of mV). A following amplifier scales it up.
• Is and VT both depend on temperature, so the basic circuit drifts. Matched transistor pairs and temperature-compensating resistors are used in practical log amps.
• The transistor version works over a wider current range (about 7 decades) than the diode version.

Uses: compressing wide-range signals, dB meters, and analog multiplication and division.