A bipolar junction transistor having common emitter current gain of 100 is operating in active region with collector current of 2 mA. Assuming that the thermal voltage V_T is 25 mV, what is the input impedance of the transistor?

Concept:

The input impedance (r_π) of a BJT in the active region is given by: \( r_\pi = \frac{\beta V_T}{I_C} \)

Given:

Common emitter current gain, \( \beta = 100 \)

Thermal voltage, \( V_T = 25~mV = 25 \times 10^{-3}~V \)

Collector current, \( I_C = 2~mA = 2 \times 10^{-3}~A \)

Calculation:

\( r_\pi = \frac{100 \times 25 \times 10^{-3}}{2 \times 10^{-3}} \)
\( r_\pi = \frac{2.5}{2 \times 10^{-3}} = 1250~\Omega = 1.25~k\Omega \)

Answer:

Option 3) 1.25 kΩ