Question:

The current gain of a transistor in a common emitter configuration is 80. The resistances in collector and base sides of the circuit are 5 k$\omega$ and 1 k$\omega$ respectively. If the input voltage is 2 mV, the output voltage is:

Show Hint

For transistors, use the formula $ A_v = \beta \frac{R_C}{R_B} $ to determine voltage gain quickly.

Updated On: May 16, 2025

$ 4V $
$ 0.4V $
$ 0.8V $
$ 8V $

Hide Solution

Verified By Collegedunia

The Correct Option is C

Approach Solution - 1

Step 1: The voltage gain $ A_v $ of a transistor in a common emitter configuration is given by: \[ A_v = \beta \times \frac{R_C}{R_B} \]
Step 2: Given: \[ \beta = 80, \quad R_C = 5k\Omega, \quad R_B = 1k\Omega \]
Step 3: Compute voltage gain: \[ A_v = 80 \times \frac{5}{1} = 400 \]
Step 4: Compute output voltage: \[ V_{\text{out}} = A_v \times V_{\text{in}} \] \[ V_{\text{out}} = 400 \times 2mV = 0.8V \]

Was this answer helpful?

Hide Solution

Verified By Collegedunia

Approach Solution -2

To find the output voltage ($ V_{\text{out}} $) in a common emitter transistor configuration, we need to use the formula that relates transistor gain ($ \beta $), input voltage ($ V_{\text{in}} $), and resistances in the circuit.

The transistor gain ($ \beta $) is given as 80. The input voltage ($ V_{\text{in}} $) is 2 mV. The resistance on the collector side ($ R_C $) is 5 kΩ, and the resistance on the base side ($ R_B $) is 1 kΩ.

First, convert all values to compatible units:

$ V_{\text{in}} = 2 \text{ mV} = 0.002 \text{ V} $
$ R_B = 1 \text{ k}\Omega = 1000 \Omega $
$ R_C = 5 \text{ k}\Omega = 5000 \Omega $

The output voltage ($ V_{\text{out}} $) in terms of the current gain and input voltage is given by:

\[ V_{\text{out}} = \beta \times \frac{R_C}{R_B} \times V_{\text{in}} \]

Substitute in the given values:

\[ V_{\text{out}} = 80 \times \frac{5000}{1000} \times 0.002 \]

Calculate step by step:

Simplify the ratio of resistances: $\frac{5000}{1000} = 5$
Now substitute back: \[ V_{\text{out}} = 80 \times 5 \times 0.002 \]
Calculate: \[ V_{\text{out}} = 80 \times 0.01 = 0.8 \text{ V} \]

Therefore, the output voltage is $ 0.8V $.