Q.With the circuit diagram show how to obtain an output which is anti-logarithm of input.

An anti-logarithmic amplifier gives an output proportional to the natural anti-logarithm (exponential) of the input; it places the diode at the input and the resistor in feedback — the reverse of the log amplifier. …

[!TLDR]

Swapping the diode and resistor of the log amplifier gives an anti-log amplifier: $V_O = -I_s R_f\,e^{V_i/(\eta V_T)}$, so $V_O \propto \text{antilog}(V_i)$.

Circuit: The input voltage $V_i$ is applied to the inverting terminal (node A) through a semiconductor diode; the non-inverting terminal (node B) is grounded; a resistor $R_f$ is connected in the feedback path between the inverting terminal and the output. (This is the log amplifier with the diode and resistor interchanged.)

Derivation: For an ideal op-amp, $A=\infty$ and $Z_i=\infty$, so $V_B = V_A = 0$ (virtual ground) and $I_B = 0$.

Applying KCL at node A:

The feedback current through $R_f$ is

The voltage across the diode is

From Shockley's diode equation,

Substituting (1) and (2) in (3):

$\boxed{V_O = -I_s R_f\,e^{V_i/(\eta V_T)}}$ …