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, $\delta I_{\mathrm{b}}=50\mathrm{\mu A}$
Load resistance,$R_{\mathrm{L}}$ = 2 kΩ
Input resistance, $R_{\mathrm{i}}$ = 0.50 kΩ

(a) The change in output voltage is given by
$$\begin{gathered} V_0=I_{\mathrm{c}}\times R_{\mathrm{L}} \\ \because I_{\mathrm{c}}=\beta \times I_{\mathrm{b}} \\ \therefore V_0=\beta \times I_{\mathrm{b}}\times R_{\mathrm{L}} \\ \Rightarrow V_0=50\times 50\mathrm{\mu A}\times 2\mathrm{k\Omega } \\ \Rightarrow V_0=5\mathrm{V} \end{gathered}$$

(b) The change in input voltage is given by
$$\begin{gathered} \delta V_{\mathrm{i}}=\delta l_{\mathrm{b}}\times R_{\mathrm{i}} \\ \Rightarrow \delta V_{\mathrm{i}}=50\times {10}^{-6}\times 5\times {10}^2 \\ \Rightarrow \delta V_{\mathrm{i}}=25\times {10}^{-3} \\ \Rightarrow \delta V_{\mathrm{i}}=25\mathrm{mV} \end{gathered}$$

(c) Power gain is given by
$$\begin{gathered} {\mathrm{\beta }}^2\times \frac{R_{\mathrm{L}}}{R_{\mathrm{i}}} \\ \Rightarrow 2500\times \frac{2}{0.5} \\ \Rightarrow 2500\times \frac{20}{5}={10}^4 \end{gathered}$$