# How does Log and Anti-Log Amplifiers work?

Log and Antilog amplifier are electronics circuit that outputs log or antilog values for a given input. If x is the input then the output from log amplifier is $$ln(x)$$ and the output from antilog amplifier is $$e^x$$. Log and Anti-log amplifier are build using operational amplifier and diodes or transistors. In case of log or logarithmic amplifiers a diode or transistor is placed in the feedback loop. In case of anti-log(anti-logarithmic) or exponential amplifier the diode is placed at the input. Here we will discuss how log and anti-log amplifiers works.

### Basic Log or Logarithmic Amplifiers using Diode

The circuit diagram of a basic Log amplifier using diode is shown below.

In the circuit diagram above, the diode is used in the negative feedback. The analysis of this basic log amplifier is as follows.

In the above circuit, the node A is grounded and therefore the node B is at virtual ground. Therefore the voltage V2 at node B is grounded. Hence $$V_{2}=0$$ and we have,

$$V_{in} = V_{2} + I R$$

or,

$$I R = V_{in} - V_{2}$$

or,

$$I = \frac{V_{in} - V_{2}}{R}$$

With $$V_{2}=0$$ we have,

$$I = \frac{V_{in}}{R} ------------------------->(1)$$

As the op-amp input current is zero, we have, $$I = I_{f}$$, which is the current that flows through the diode path.

Now the basic volt-ampere relation for a diode is given by,

$$I = I_{0} (e^{\frac{V}{\mu V_{T}}} - 1)$$

where, I = diode current

$$I_{0}$$ = reverse current

V= diode voltage

$$\mu$$ = 1 for Ge diode and 2 for Si diode

$$V_{T}$$ = kT = voltage equivalent of temperature

k = Boltzmann's constant =  $$8.62 \times 10^5 eV/^o K$$

T = temperature in Kelvin

From solid state physics it can be derived that at room temperature of $$T=27^oC$$, $$V_{T}= 26mV$$

When the diode is forward biased, the current through the diode $$I$$ is $$I_{f}$$ and in this case the term $$e^{\frac{V}{\mu V_{T}}}$$ is much greater than 1, that is $$e^{\frac{V}{\mu V_{T}}} >> 1$$ and we can write,

$$I_{f} = I_{0} e^{\frac{V}{\mu V_{T}}}$$

Taking natural logrithm on both sides and solving for voltage V across the diode,

$$ln[I_{f}] = ln[I_{0} e^{\frac{V}{\mu V_{T}}}]$$

or,   $$ln[I_{f}] = ln[I_{0}] + ln[ e^{\frac{V}{\mu V_{T}}}]$$

or,   $$ln[I_{f}] = ln[I_{0}] + \frac{V}{\mu V_{T}}$$

or,   $$V = \mu V_{T} [ln(I_{f}) - ln(I_{0})]$$

therefore, $$V = \mu V_{T} ln[ \frac{I_{f}} {I_{0}} ]$$

which is the basic equation for log and anti-log amplifiers

The voltage across the diode V is,

$$V=V_{B}-V_{out}$$

and since $$V_{B}=0$$ because of virtual ground

$$V= -V_{out}$$

And we can write ,

$$-V_{out} = \mu V_{T} ln[ \frac{I_{f}} {I_{0}} ]$$

using eqn(1),

$$V_{out} = - \mu V_{T} ln[ \frac{V_{in}} { I_{0} R} ]$$

Let $$V_{ref} =I_{0} R$$ which is just a constant voltage then we get,

$$V_{out} = - \mu V_{T} ln[ \frac{V_{in}} { V_{ref}} ] -------------->(2)$$

From the above log amplifier equation we can see that the output voltage $$V_{out}$$ is function of logarithm of the input voltage $$V_{in}$$. The above equation gives natural logarithm with base e. To obtain logarithm to the base 10 using the following relation,

$$log_{10}X = 0.4343 ln(X)$$

Consider the following log amplifier designed with LM358 operational amplifier and 1N4148 switching diode.

If the input signal is a sine wave of amplitude 100mV and frequency of 1KHz, then the waveform of the input signal and the output signal from the log amplifier is shown below.

In the above waveform graph, the green colored signal is the input sine wave signal and the red signal is the output from the logarithm amplifier.