(a) Differentiate between three segments of a transistor on the basis of their size and level of doping.

(b) How is a transistor biased to be in active state?

(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain.  

On the basis of size and level of doping,

(a)
Emitter (E) - Emitter is heavily doped and is the left hand side thick layer of the transistor. 

Base (B) - It is the central thin layer of the transistor, which is lightly doped.

Collector (C) - It is the right hand side thick layer of the transistor, which is moderately doped.
(b)
There are two conditions for a transistor to be into an active region.

1. The input circuit should be forward biased by using a low voltage battery.

2. The output circuit should be reverse biased by using a high voltage battery.

(c)
n-p-n Transistor as an amplifier:

The input circuit is forward biased by using a low voltage battery. Hence, the resistance of the input circuit is small. The output circuit is reverses biased using a high voltage battery. Hence, the resistance of the output circuit is large.

The operating point is fixed in the middle of its active region.
 

The output is taken between the collector and the ground.

Applying Kirchhoff’s law to the output loop:
V_CC = V_CE + I_CR_{C
If Vc is the collector voltage then,
Vc = VCE − ICRC                                                      ... (A)}

When the input signal voltage is fed to the emitter base circuit, it will change the emitter voltage and hence to the emitter current, which in turn will change the collector current. Due to this the collector voltage V_C will vary in accordance with relation (A). This variation in collector voltage appears as the amplified output.

V_BB = V_BE + I_BR_B

Here, v_i ≠ O

Then, V_BB + v_i = V_BE + I_BR_B + ΔI_B (R_B + r_i)

 
 ; is the required expression for current gain.