In a transistor, connected in a common-emitter mode R_c = 4 kΩ; R_i = 1 kΩ; I_c = 1mA and l_b = 20 μA. Find the voltage gain.

Solution

Given, a common emitter configuration.

Collector resistance, R_c = 4 k $Ω$
Internal resistance, R_i = 1 k $Ω$
Current across collector, I_c = 1 mA
Current across base, I_b = 20 $μA$

Using the formula for current gain,

$β = \frac{I_{c}}{I_{b}} = \frac{10^{- 3}}{20 \times 10^{- 6}} = 50$

Therefore,
Voltage gain is,
$A_{V} = β \frac{R_{c}}{R_{i}} = 50 \times \frac{4000}{1000} = 200 .$

For Daily Free Study Material Join wiredfaculty

Wave Optics

In a transistor, connected in a common-emitter mode R_c = 4 kΩ; R_i = 1 kΩ; I_c = 1mA and l_b = 20 μA. Find the voltage gain.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Wave Optics

In a transistor, connected in a common-emitter mode Rc = 4 kΩ; Ri = 1 kΩ; Ic = 1mA and lb = 20 μA. Find the voltage gain.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

In a transistor, connected in a common-emitter mode R_c = 4 kΩ; R_i = 1 kΩ; I_c = 1mA and l_b = 20 μA. Find the voltage gain.