# The common emitter amplifier

## The transistor as a switch

In the circuit below, when Vin is switched from 0V to 2V, the current through R3 is 2mA (Ib) which results in 150mA of collector current (Ic) flowing through R1.

The Vcc power supply and the multimeter show 150mA collector current flowing through R1 below.

## Biasing the common emitter amplifier

Let us consider the NPN transistor amplifier shown in fig.1. This is the common-emitter configuration and a small AC signal at the base appears as a amplified AC signal at the collector. The input AC signal Vi is amplified by Q1 and it appears at VC

We'd like to find the ideal voltage at the base of Q1 (find values for resistors R1 and R2). In other words, we'd like to bias this circuit.

The base-emitter junction is forward-biased and the base-collector junction is reverse-biased.

Let's say $$V_{cc}=15V$$

We set the collector of transistor Q1 to half of the supply voltage, so we can have equal positive and negative AC swing.

$$\ \ Set \ V_c=7.5V$$

The voltage at the base, VB has a AC component from the input Vi and a DC component from the resistor divider R1 and R2

Because the base-emitter junction is basically a diode with a voltage drop of 0.6V, any change in the AC input signal Vi will result in a similar change in the emitter voltage, VE

Because the base current in a transistor is a order of magnitude less than the collector current, we can approximate current in emitter, $$\ I_E \approx I_C$$ A change in the AC input signal Vi will result in a varying current across RE, which will result in a varying current across RL Hence the ratio of RL/RE sets the gain. Assuming a gain of 10, we can set $$\ R_L=10K ,\ \ R_E=1K$$ Voltage drop across RL is $$\ V_{R_L} = 15V-7.5V = 7.5V$$ Current through RL, $$\ I_C = {7.5V \over 10K} = 0.75mA$$ So, $$\ I_E \approx I_C = 0.75mA$$ Since RE=1K, the emitter voltage is $$\ V_E = R_E*I_e = 1K*0.75mA = 0.75V$$ Since the base-emitter junction is basically a diode, which has a voltage drop of 0.6V, the base of Q1 is at $$\ V_B = 0.75V + 0.6V = 1.35V$$ Base current, $$\ I_B = {I_E \over h_{FE}}$$ Consider a transistor with $$\ h_{FE}=100, \ then \ I_B = 7.5uA$$ To minimize the effect of IB, set the current through R1 and R2 at ten times IB $$\ 10*I_B = 75uA$$ Now, we calculate R1 & R2 as $$\ R1 = {(15V - 1.35V) \over 75uA} = 182K$$ $$\ R2 = {1.35V \over 75uA} = 18K$$

As the input AC voltage Vi increases, the current through RL increases, so the drop across RL increases and voltage the collector, VC decreases. Hence VC is 180deg reversed in phase compared to Vi.

However, there are a couple of issues with the CE amplifier circuit in fig.1. One is that the base emitter voltage is not fixed at 0.6v but changes with temperature at about 2.2mV/°C. The other is that the supply voltage may vary from 15V and cause VC to change. So the amplified output may not be a true reflection of the input signal. To overcome these, the operational amplifier was designed.

## References:

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