# How does Log Amplifier using transistor work?

A log amplifier is an electronics circuit that calculates the logarithm of input signal. That is the output signal of a log amplifier is the logarithm of an input signal. The log amplifier or logarithm amplifier are thus electronics circuit that performs mathematical operation. Such amplifier can be designed using operational with diode or transistor in the feedback path. In the earlier tutorial we explained how does Log and Anti-Log Amplifiers work where diode was used. In this tutorial we explain how the log amplifier with operational amplifier with transistor works.

See the following other tutorials on log and antilog amplifier.

### Basic Log Amplifier with Transistor

The circuit diagram of a basic log amplifier using transistor is shown below.

As shown in the above circuit diagram, a transistor is used in the feedback path.

Now we can show that this circuit performs logarithm operation on the input signal. Let the input signal be Vin and the output signal is Vout.

Applying KVL at the input side we obtain,

$$V_{in} = V_{2} + I_{c} R$$

The current through the transistor I=Ic is given by,

$$I_{c} R = V_{in} - V_{2}$$

or,

$$I_{c} = \frac{V_{in} - V_{2}}{R}$$

Since the non-inverting terminal 3 is grounded, the voltage at node A is 0. Hence because of this the node 2 is virtual ground, so V2=0. With $$V_{2}=0$$ we have,

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

For transistor the volt-ampere equation is,

$$I_{c} = I_{s} (e^{\frac{V_{BE}}{ V_{T}}} - 1)$$

where,  $$I_{c}$$ = collector current, $$I_{s}$$ = emitter saturation current, $$V_{BE}$$ = base emitter voltage and $$V_{T}$$ = 25mV is the thermal voltage.

Now since $$e^{\frac{V_{BE}}{ V_{T}}} >> 1$$ we can write,

$$e^{\frac{V_{BE}}{ V_{T}}} = \frac{I_{c}}{ I_{s}}$$

taking natural logarithm on both sides,

$$ln[e^{\frac{V_{BE}}{ V_{T}}}] = ln[\frac{I_{c}}{ I_{s}}]$$

or,

$$\frac{V_{BE}}{ V_{T}} = ln[\frac{I_{c}}{ I_{s}}]$$

or,

$$V_{BE} = V_{T} ln[\frac{I_{c}}{ I_{s}}]$$

From the circuit diagram above, at the output side we have,

$$V_{out}+V_{BE} = 0$$

or,

$$V_{out}= - V_{BE}$$

or,

$$V_{out}= - V_{T} ln[\frac{I_{c}}{ I_{s}}]$$

Using equation(1) and letting  $$V_{ref} =I_{s} R$$ we get,

$$V_{out}= - V_{T} ln[\frac{V_{in}}{ V_{ref}}]$$

Thus from the above log amplifier equation we can see that the output voltage Vout is proportional to the logarithm of the input voltage Vin.

The following shows circuit diagram of log amplifier designed using LM358 op-amp and 2N3904 transistor.

The following shows the waveform graph of the input sine signal of 100mV with frequency of 1KHz and the logarithm output signal from the log amplifier.