Question

A load resistor of 2kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain β = 50. The input resistance of the transistor is 0.50 kΩ. If the input current is changed by 50µA. (a) by what amount does the output voltage change, (b) by what amount does the input voltage change and (c) what is the power gain?

Answer 1

Given:
Base current gain,  \[\beta = 50\]

Change in base current, \[\delta I_b  = 50  \mu A\]

Load resistance, \[R_L\] = 2 kΩ
Input resistance, \[R_i\]  = 0.50 kΩ

(a) The change in output voltage is given by

\[V_0  =  I_c  \times  R_L \] 

\[ \because  I_c  = \beta \times  I_b \] 

\[ \therefore  V_0  = \beta \times  I_b  \times  R_L \] 

\[ \Rightarrow  V_0  = 50 \times 50  \mu A \times 2  k\Omega\] 

\[ \Rightarrow  V_0  = 5  V\]

(b) The change in input voltage is given by

\[\delta V_i  = \delta l_b  \times  R_i \] 

\[ \Rightarrow \delta V_i  = 50 \times  {10}^{- 6}  \times 5 \times  {10}^2 \] 

\[ \Rightarrow \delta V_i  = 25 \times  {10}^{- 3} \] 

\[ \Rightarrow \delta V_i  = 25 \text{  mV}\] 

(c) Power gain is given by 

\[\beta^2  \times \frac{R_L}{R_i}\] 

\[ \Rightarrow 2500 \times \frac{2}{0 . 5}\] 

\[ \Rightarrow 2500 \times \frac{20}{5} =  {10}^4\]