# Log amplifier

A log amplifier is an amplifier for which the output voltage Vout is K times the natural log of the input voltage Vin. This can be expressed as,

${\displaystyle V_{\text{out}}=K\ln \left({\frac {V_{\text{in}}}{V_{\text{ref}}}}\right)}$

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

The logarithm amplifier gives an output voltage which is proportional to the logarithm of applied input voltage. To design a logarithm amplifier circuit, high performance op-amps like LM1458, LM771, LM714 are commonly used and a compensated logarithm amplifier may include more than one.

## Logarithmic amplifier applications

Logarithmic amplifiers are used in many ways, such as:

1. To perform mathematical operations like multiplication, division and exponentiation.
2. To calculate the dB value of a given quantity.
3. As a True RMS converter.

## Drawbacks of basic logarithmic amplifier configuration

The reverse saturation current for the diode doubles for every ten degree Celsius rise in temperature. Similarly the emitter saturation current varies significantly from one transistor to another and also with temperature. Hence, it is very difficult to set the reference voltage for the circuit.

## Basic op-amp diode circuit

Basic op-amp diode log converter

The relationship between the input voltage ${\displaystyle V_{\text{in}}}$ and the output voltage ${\displaystyle V_{\text{out}}}$ is given by:

${\displaystyle V_{\text{out}}=-V_{\text{T}}\ln \left({\frac {V_{\text{in}}}{I_{\text{S}}\,R}}\right)}$

where ${\displaystyle I_{\text{S}}}$ and ${\displaystyle V_{\text{T}}}$ are the saturation current and the thermal voltage of the diode respectively.

## Transdiode configuration

A transdiode configuration with a BJT connected in the negative feedback loop.

A necessary condition for successful operation of a log amplifier is that the input voltage, Vin, is always positive. This may be ensured by using a rectifier and filter to condition the input signal before applying it to the log amp input. As Vin is positive, Vout 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,

{\displaystyle {\begin{aligned}V_{\text{BE}}&=-V_{\text{out}}\\I_{\text{C}}&=I_{\text{SO}}\left(e^{\frac {V_{\text{BE}}}{V_{\text{T}}}}-1\right)\approx I_{\text{SO}}e^{\frac {V_{\text{BE}}}{V_{\text{T}}}}\\\Rightarrow V_{\text{BE}}&=V_{\text{T}}\ln \left({\frac {I_{\text{C}}}{I_{\text{SO}}}}\right)\end{aligned}}}

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

${\displaystyle I_{\text{C}}={\frac {V_{\text{in}}}{R_{1}}}}$, and
${\displaystyle V_{\text{out}}=-V_{\text{T}}\ln \left({\frac {V_{\text{in}}}{I_{\text{SO}}R_{1}}}\right)}$

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