In the circuit below, when V_{in} is switched from 0V to 2V, the current through R3 is 2mA (I_{b}) which results in 150mA of collector current (I_{c}) flowing through R1.

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

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 V_{i} is amplified by Q_{1} and it appears at V_{C}

We'd like to find the ideal voltage at the base of Q_{1} (find values for resistors R_{1} and R_{2}). 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, V_{B} has a AC component from the input V_{i} and a DC component from the resistor divider R_{1} and R_{2}

Because the base-emitter junction is basically a diode with a voltage drop of 0.6V, any change in the AC input signal V_{i} will result in a similar change in the emitter voltage, V_{E}

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 V_{i} will result in a varying current across R_{E}, which will result in a varying current across R_{L}
Hence the ratio of R_{L}/R_{E} sets the gain. Assuming a gain of 10, we can set
$$\ R_L=10K ,\ \ R_E=1K $$
Voltage drop across R_{L} is $$\ V_{R_L} = 15V-7.5V = 7.5V $$
Current through R_{L}, $$\ I_C = {7.5V \over 10K} = 0.75mA $$
So, $$\ I_E \approx I_C = 0.75mA $$
Since R_{E}=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 I_{B}, set the current through R_{1} and R_{2} at ten times I_{B}
$$\ 10*I_B = 75uA $$
Now, we calculate R1 & R2 as
$$\ R1 = {(15V - 1.35V) \over 75uA} = 182K $$
$$\ R2 = {1.35V \over 75uA} = 18K $$

As the input AC voltage V_{i} increases, the current through R_{L} increases, so the drop across R_{L} increases and voltage the collector, V_{C} decreases. Hence V_{C} is 180deg reversed in phase compared to V_{i}.

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 V_{C} to change. So the amplified output may not be a true reflection of the input signal. To overcome these, the operational amplifier was designed.

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