Question

(a) Differentiate between three segments of a transistor on the bias 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.

Answer 1

a) Emitter(E): It is the left hand side thick layer of the transistor which is heavily doped.

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.

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

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

c) n-p-n transistor as an amplifier:

The operating point is fixed in the middle of its active region.An ac input signal $v_i$ is superimposed on bias $V_{BB}$(dc). The o/p is taken between collector and ground.Applying Kirchoff's Law to the outer loop:

$I_E=I_B+I_C$

If $V_{CE}$ is the collector voltage then: $V_{CE}=V_{CC}-I_C{R}_L$

When the input signal voltage is fed to the emitter base circuit, it will change the emitter base voltage and the emitter current, in turn changing the collector current. Due to this the collector voltage variation will appear as amplified output.

Then, $V_i=\Delta I_B\times R_i$

where $R_i$ is the input resistance and $V_i$ is the input voltage of the emitter and $\Delta I_B$ is the change in base current.

The ac gain of the transistor is: ${\beta }_{AC}=\frac{\Delta I_C}{\Delta I_B}$

where, $\Delta I_C$ is the change in collector current.