The typical transfer characteristic of a transistor in CE configuration is shown in figure. A load resistor of $$2 \text{ k}\Omega$$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $$0.50 \text{ k}\Omega$$. The voltage gain of the transistor is ______.

The given load resistance $$R_L = 2 \text{ k}\Omega$$ and input resistance $$R_{in} = 0.50 \text{ k}\Omega$$ require determining the voltage gain from the CE transfer characteristic.

Since the transfer characteristic of a CE transistor plots the collector current ($$I_C$$) versus the base current ($$I_B$$), we extract two points from the typical curve provided.

For example, at $$I_B = 10 \, \mu A$$, $$I_C = 0.5 \text{ mA}$$ and at $$I_B = 20 \, \mu A$$, $$I_C = 1.0 \text{ mA}$$.

From these values, the current gain $$\beta$$ is given by

$$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{(1.0 - 0.5) \text{ mA}}{(20 - 10) \, \mu A} = \frac{0.5 \times 10^{-3}}{10 \times 10^{-6}} = \frac{0.5 \times 10^{-3}}{0.01 \times 10^{-3}} = 50$$

The voltage gain of a CE amplifier is given by $$A_v = \beta \times \frac{R_L}{R_{in}}$$, and substituting the values yields

$$A_v = 50 \times \frac{2 \text{ k}\Omega}{0.50 \text{ k}\Omega} = 50 \times 4 = 200$$

Therefore, the voltage gain of the transistor is $$\textbf{200}$$.