Simple Log Amplifier Circuit

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4 Working Explanation of the log amplifier

Introduction

Don’t you think sometimes that where mathematics is used in electronics? Why do the circuit designers solve difficult maths question if there is no usage? May you be wrong. Or maybe you haven’t made any circuit that Performs mathematics in it. But don’t worry, today we are going to make a circuit that performs mathematics. Yes, we are discussing the logarithmic amplifier, which is also known as the log amplifier.

A log amplifier is a circuit that generates the output which is proportional to the logarithm of the applied input. In other words, the output voltage of the circuit is the k times of the natural logarithm of the applied input voltage.

V_out = K ln (V_in/V_ref)

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Hardware Required

S.no	Component	Value	Qty
1.	PCB board	–	1
2.	Opamp IC	LM1458/ LM741/ LM771	1
3.	Resistor	–	1
4.	Diode	–	1

Circuit Diagram

Working Explanation of the log amplifier

Opam is the central element of the circuit. It is connected as the closed-loop inverting component. It is connected with a diode in a closed-loop inverting configuration. The diode has a voltage that is proportional to the log of the current through the diode. The relation of current and voltage through a diode is:

I_d= I_s(e^{(V_d/V_t )}-1)……………………….. (a)

Where, I_d= Diode Current, Is= Saturation Current, V_d= Diode voltage, V_t= Thermal voltage

But, the voltage across the diode is the positive voltage. And, thermal voltage is very less. Hence, the equation becomes:

I_d= I_s×e^{(V_d/V_t )}……………………. (b)

The opam has input current Ir and it flows to the diode because there is only one path to flow. Therefore

I_r=I_d…………………………….. (c)

Since, I_r=V_in/R. Therefore, I_d= V_in/R.……………… (d)

By comparing b and d, it would become

V_in = I_s×e^{(V_d/V_t )}

The negative voltage of the diode is the output voltage of the circuit. After rearranging the final output equation will be

V_out = -V_t In (V_in/I_sR)