A common emitter amplifier is designed with n-p-n transistor (α=0.99). The input resistance is 1 k Ω and load is 10 k Ω. The voltage gain will be

Question

A common emitter amplifier is designed with n-p-n transistor (α=0.99). The input resistance is 1 k Ω and load is 10 k Ω. The voltage gain will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

β=(α/1-α)=99 V text gain =β (R text out /R text in ) =99 × (10/1)=990

Q. A common emitter amplifier is designed with $n - p - n$ transistor $(α = 0.99)$ . The input resistance is $1 k Ω$ and load is $10 k Ω$ . The voltage gain will be

$β = \frac{α}{1 - α} = 99$
$V_{gain} = β \frac{R _{out}}{R _{in}}$
$= 99 \times \frac{10}{1} = 990$

Q. A common emitter amplifier is designed with n−p−n transistor (α=0.99). The input resistance is 1kΩ and load is 10kΩ. The voltage gain will be

β=1−αα​=99 Vgain ​=βRin ​Rout ​​ =99×110​=990

Q. A common emitter amplifier is designed with $n - p - n$ transistor $(α = 0.99)$ . The input resistance is $1 k Ω$ and load is $10 k Ω$ . The voltage gain will be

$β = \frac{α}{1 - α} = 99$
$V_{gain} = β \frac{R _{out}}{R _{in}}$
$= 99 \times \frac{10}{1} = 990$