# The Operational Amplifier

The video in Reference (1) provides a detailed introduction to why operational amplifiers were designed. As mentioned in the discussion of the CE amplifier, transistor characteristics vary with voltage and temperature. The operational amplifier was designed to provide a feedback mechanism to compensate for those changes, so the amplifier gain remains stable over changes in the power supply and temperature. Consider the operational amplifier circuit below:

Assume the base of Q1(Vin) is at 2V, then the emitter of Q1(A) is at 1.4V ( 0.6v diode drop). Extending this, the base of Q2 (Vout) must be at 2V. We will see that Vout will stay at 2V even if the temperature or the supply voltage changes. Note that Q3 is a PNP transistor so applying Kirchoff's voltage law, the voltage across R1 is equal to the diode drop of 0.6V when Q3 is on.

1. If Vout goes up to say 2.1V, then the base emitter drop in Q2 is greater than 0.6V, so Q2 starts conducting heavily causing the emitter of Q1/Q2 to go up to 1.5V to maintain the diode drop. This causes Q1 to turn off because the base-emitter voltage is less than 0.6V. Lesser current across R1 leads to lesser current at the base of Q3, which in turn leads to a decrease in Vout, taking it back down to 2V.
2. Consider the other case in which Vout is less than 2V, say 1.9V, then the emitter of Q1/Q2 goes to 1.3V (0.6v Q2 base-emitter diode drop). Now the base-emitter drop across Q1 is greater than 0.6v, so Q1 starts conducting heavily. This means increased current across R1 leading to an increase in the base current of Q3. This in turn leads to a increase in Vout bringing it back up to 2V.

So, Vout remains stable at 2V.

This is essentially the non-inverting operational amplifier. The base of Q1 is the positive terminal and the base of Q2 is the -ve terminal. To change the gain of this circuit, we need to add two resistors, Rf and Ri as in fig.3 below.

The equivalent circuit shown in fig.4 below

We know that the circuit tries to make the voltages at the +ve and -ve inputs equal, so the -ve input terminal must also be at Vin. Using Kirchoff's current law at this input,

$$\ {V_{out} - V_{in}\over R_f} = {V_{in} \over R_i}$$

Hence, $$\ {gain, A_v} = {V_{out} \over V_{in}} = {R_f+R_i \over R_i}$$

## References:

1. Prof. Gunasekharan, Circuits for Analog Systems Design, Lecture 1 (Video)
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