Question

A common emitter amplifier is designed with an n-p-n transistor (α=0.99). The input impedance is 1kΩ and load is 10kΩ. The voltage will be :
A. 9900
B. 99
C. 9.9
D. 990

Answer 1

Hint: If we combine two extrinsic semiconductors we get diodes. Extrinsic semiconductor in which electrons are excess are called n-type semiconductor whereas extrinsic semiconductor in which holes are excess are called p-type semiconductor. Now if we combine two diodes we get a transistor. It can be a p-n-p transistor or n-p-n transistor. That transistor has emitter base and collector and can be used as switches, amplifiers and as many other devices
Formula used:
$\eqalign{
  & {\text{alpha = }}\dfrac{{{{\text{i}}_{\text{c}}}}}{{{{\text{i}}_{\text{e}}}}} \cr
  & {\text{beta = }}\dfrac{{{{\text{i}}_{\text{c}}}}}{{{{\text{i}}_{\text{b}}}}} \cr
  & {{\text{i}}_{\text{e}}}{\text{ = }}{{\text{i}}_{\text{c}}}{\text{ + }}{{\text{i}}_{\text{b}}} \cr} $

Complete answer:
The characteristics of emitter, base and collector will be different. Emitter is highly doped while the collector and base are lightly doped. The width of the base is very less. Moreover emitter-base junction is always forward biased whereas base-collector junction is always reverse biased. Forward biased in the sense diode provides very low resistance and reverse biased in the sense diode provides huge resistance
We have
${\text{voltage gain = }}\beta \times {\text{resistance gain}}$
Where
${\text{beta = }}\dfrac{{{{\text{i}}_{\text{c}}}}}{{{{\text{i}}_{\text{b}}}}}$
Where ${{\text{i}}_{\text{c}}}$ is the collector current and ${{\text{i}}_{\text{b}}}$ is base current and ${{\text{i}}_{\text{e}}}$ is emitter current
${\text{alpha = }}\dfrac{{{{\text{i}}_{\text{c}}}}}{{{{\text{i}}_{\text{e}}}}}$
And $\beta {\text{ = }}\dfrac{\alpha }{{1 - \alpha }}$
${\text{alpha }}$is given as 0.99 hence $\beta {\text{ = }}\dfrac{alpha }{{1 - alpha }} = \dfrac{{0.99}}{{1 - 0.99}} = 99$
Resistance gain will be
$\dfrac{{{\text{load resistance}}}}{{{\text{input impedance}}}}{\text{ = }}\dfrac{{{\text{10KOmega }}}}{{{\text{1KOmega }}}} = 10$
As load resistance and input impedance are given in the problem.
Now voltage gain will be ${\text{voltage gain = }}\beta \times {\text{resistance gain}}$= 99 x 10 = 990

So, the correct answer is “Option D”.

Note:
Transistor can be used as an amplifier. An ideal amplifier should have high input impedance and very low output impedance i.e it should not draw any current from the input. Impedance is the combination of the resistance and the reactance. Since the answer which we got has no units it’s voltage gain rather than voltage as voltage will have units. Impedance is the hindrance to the flow of current.