# Antilog Amplifier Circuit Applications

AntiLog amplifier or antilogarithmic amplifier is an electronic circuit that produces output that is proportional to the anti-logarithm of the applied input. Basically it performs mathematical operation of an anti-logarithm. In this article, we will see the different antilog amplifier circuits, its working and antilog amplifier applications. Basically two circuits are there to perform the anti-logarithmic function. First, using diode and op-amp and second, using BJT and op-amp.

Contents

## Antilog amplifier using diode and op-amp

Diode, resistor and op-amp used in the antilog amplifier as shown in figure 1. The input Vis applied through diode D at the inverting terminal. Vo is the output voltage. The non-inverting terminal of the op-amp is connected to the ground. This means that the voltage of the non-inverting terminal is zero volts.

### Analysis

The analysis of the antilog amplifier is shown in figure 2. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V) is equal to the voltage of the non-inverting terminal (V+ = 0V), according to the virtual short concept.

V= V+ = 0V

The currents entering both terminals of the op-amp are zero since the op-amp is ideal.

Let current I flows through the resistor R.

\label{eq:poly}

I=\frac{0-V_{o}}{R}=-\frac{V_{o}}{R}

and diode current $I_{d}$,

\label{eq:poly}

I_{d}=I_{o}e^{\frac{V_{d}}{\eta V_{T}}}

where $V_d$ = forward bias voltage across diode D

Apply KCL at node P

$I_{d} = 0 + I$

\label{eq:poly}

I = I_{d}

and

$V_{d} = V_{i}-0$

$V_{d} = V_{i}$

Therefore, equation (2) becomes

\label{eq:poly}

I_{d}=I_{o}e^{\frac{V_{i}}{\eta V_{T}}}

From equation (1), (3) and (4), we have

$-\frac{V_{o}}{R}=I_{o}e^{\frac{V_{i}}{\eta V_{T}}}$

Therefore, we have

$\quicklatex{color=”#000000″ size=20} \boxed{V_{o}=-I_{o}Re^{\frac{V_{i}}{\eta V_{T}}}}$

Note: The negative sign in the output signifies that there is a 180° phase difference between output and the applied input.

## Antilog amplifier using diode and transistor

Another circuit comprises of a BJT (PNP) T, resistor (R) and op-amp used as the antilog amplifier as shown in figure 3. The input Vis applied through transistor T at the inverting terminal. Vo is the output voltage. The non-inverting terminal of the op-amp is connected to the ground. This means that the voltage of the non-inverting terminal is zero volts.

### Analysis

The analysis of the antilog amplifier is shown in figure 4. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V) is equal to the voltage of the non-inverting terminal (V+ = 0V), according to the virtual short concept.

V= V+ = 0V

The currents entering both terminals of the op-amp are zero since the op-amp is ideal.

Let current I flows through the resistor R.

\label{eq:poly}

I=\frac{0-V_{o}}{R}=-\frac{V_{o}}{R}

and collector current $I_{C}$,

\label{eq:poly}

I_{C}=I_{s}e^{\frac{V_{EB}}{\eta V_{T}}}

where $V_{EB} = V_E – V_B$

and Is = Reverse saturation current of emitter-base junction

As we know that

$I_C + I_B = I_E$

since IB ≅ 0 A

Therefore

\label{eq:poly}

$I_C = I_E$

Apply KCL at node P

$I_E = 0 + I$

\label{eq:poly}

I = I_{E}

and

$V_{EB} = V_{i}-0$

$V_{EB} = V_{i}$

Therefore, equation (6) becomes

\label{eq:poly}

I_{C}=I_{s}e^{\frac{V_{i}}{\eta V_{T}}}

From equation (5), (7), (8) and (9), we have

$-\frac{V_{o}}{R}=I_{s}e^{\frac{V_{i}}{\eta V_{T}}}$

Therefore, we have

$\quicklatex{color=”#000000″ size=20} \boxed{V_{o}=-I_{s}Re^{\frac{V_{i}}{\eta V_{T}}}}$

Note: The negative sign in the output signifies that there is a 180° phase difference between output and the applied input.

## Antilog Amplifier Applications

1. It is used in analog multiplier circuits.

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