Your Mentor: Log & Anti-log Amplifier

A log amplifier (logarithmic converter) is one for which the output voltage V_out is K times the natural log of the input voltage V_in. This can be expressed as,

$V_\mathrm{out} = K \ln\frac{V_\mathrm{in}}{V_\mathrm{ref}}$

where V_ref is the normalization constant in volts and K is the scale factor.

A necessary condition for successful operation of a log amplifier is that the input voltage, V_in is always positive. This may be ensured by using a rectifier and filter to condition the input signal before applying to the log amp input. As V_in is positive, V_out is obliged to be negative (since the op amp is in the inverting configuration) and is large enough to forward bias the emitter-base junction of the BJT keeping it in the active mode of operation. Now,

$V_\mathrm{BE} = -V_\mathrm{out}\,\!$

$I_\mathrm C = I_\mathrm{SO}(e^{V_\mathrm{BE} / V_\mathrm T} - 1) \approx I_\mathrm {SO} e^{V_\mathrm{BE} /V_\mathrm T}$

$\Rightarrow V_\mathrm{BE} = V_\mathrm T \ln \frac{I_\mathrm C}{I_\mathrm{SO}}$

where $I_\mathrm{SO}\,$ is the saturation current of the emitter-base diode and $V_\mathrm T\,$ is the thermal voltage. Due to the virtual ground at the op amp differential input,

$I_\mathrm C = \frac{V_\mathrm{in}}{R_1}$ , and

$V_\mathrm{out} = -V_\mathrm T \ln \frac{V_\mathrm{in}}{I_\mathrm{SO} R_1}$

The output voltage is expressed as the natural log of the input voltage. Both the saturation current $I_\mathrm{SO}\,$ and the thermal voltage $V_\mathrm T\,$ are temperature dependent, hence, temperature compensating circuits may be required.