A common emitter amplifier is designed with NPN transistor ( = 0.… - Vidyadip

MCQ

A common emitter amplifier is designed with $NPN$ transistor ($\alpha = 0.99$). The input impedance is $1 K \Omega$ and load is $10 K \Omega$The voltage gain will be

A
$9.9$
B
$99$
✓
$990$
D
$9900$

✓

Answer

Correct option: C.

$990$

c
(c)Voltage gain $= \beta × $ Resistance gain
$\beta = \frac{\alpha }{{1 - \alpha }} = \frac{{0.99}}{{(1 - 0.99)}} = 99$
Resistance gain $ = \frac{{10 \times {{10}^3}}}{{{{10}^3}}} = 10$
==> Voltage gain $= 99 × 10 = 990.$

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