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

Question

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

Tardigrade Admin · Accepted Answer

Voltage gain = Current gain × Resistance gain Current gain (β)=(α/1-α)=(0.99/1-0.99)=99 ⇒ Voltage gain =99 × (10 k Ω/1 k Ω)=990

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

9.9

99

990

9900

Voltage gain = Current gain $\times$ Resistance gain
Current gain $(β) = \frac{α}{1 - α} = \frac{0.99}{1 - 0.99} = 99$
$\Rightarrow$ Voltage gain $= 99 \times \frac{10 k Ω}{1 k Ω} = 990$

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

9.9

99

990

9900

Voltage gain = Current gain × Resistance gain Current gain (β)=1−αα​=1−0.990.99​=99 ⇒ Voltage gain =99×1kΩ10kΩ​=990

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

Voltage gain = Current gain $\times$ Resistance gain
Current gain $(β) = \frac{α}{1 - α} = \frac{0.99}{1 - 0.99} = 99$
$\Rightarrow$ Voltage gain $= 99 \times \frac{10 k Ω}{1 k Ω} = 990$