A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250, $$R_C = 1$$ k$$\Omega$$ and $$V_{CC} = 10$$ V. The minimum base current for $$V_{CE}$$ to reach saturation is:

In a common emitter amplifier using an NPN transistor, the saturation state and base current are calculated as follows:

At the saturation state, the collector-emitter voltage $$V_{CE}$$ reaches its minimum value (approximately zero).

The collector current $$I_C$$ is determined by:

$$I_C = \frac{V_{CC}}{R_C}$$

Substituting the given values ($$V_{CC} = 10\text{ V}$$ and $$R_C = 1\text{ k}\Omega$$):

$$I_C = \frac{10\text{ V}}{1000\ \Omega} = 10\text{ mA}$$

The relationship between collector current and base current is defined by the DC current gain ($$\beta$$):

$$\beta = \frac{I_C}{I_B}$$

Rearranging to solve for the minimum base current $$I_B$$:

$$I_B = \frac{I_C}{\beta}$$

Substituting $$I_C = 10\text{ mA}$ and $\beta = 250$$:

$$I_B = \frac{10\text{ mA}}{250} = 0.04\text{ mA}$$

Converting to microamperes:

$$\boxed{I_B = 40\ \mu\text{A}}$$