Q5Analog Electronics

Question

(A) For the circuit shown in fig 6, the transistor parameters are \beta = 100 and VA = \infty. Design the circuit such that ICQ = 0.25 mA and VCEQ = 3 V. Find the small-signal voltage gain Av = vo / vs. Find the input resistance seen by the signal source vs.

Answer 1

Given: $β = 100$ , $I_{C Q} = 0.25 mA$ , $V_{C E Q} = 3 V$ , $V_{C C} = 9 V$ (assumed). $I_{B Q} = I_{C Q} / β = 2.5 μ A$ . Using KVL: $V_{C C} = I_{C} R_{C} + V_{C E} + I_{E} R_{E}$ . For $R_{E} = 0$ (emitter bypassed):

R_{C} = \frac{V _{C C} - V _{C E Q}}{I _{C Q}} = \frac{9 - 3}{0.25 \times 1 0 ^{- 3}} = 24 k Ω

g_{m} = \frac{I _{C Q}}{V _{T}} = \frac{0.25 mA}{26 mV} \approx 9.62 mA/V

r_{π} = \frac{β}{g _{m}} = \frac{100}{9.62} \approx 10.4 k Ω

A_{v} = - g_{m} R_{C} = - 9.62 \times 24 \approx - 231

R_{in} = R_{1} ∥ R_{2} ∥ r_{π}

The negative sign indicates 180Â° phase inversion in the CE configuration. High voltage gain makes CE the most commonly used amplifier configuration.

Q5Analog Electronics

Question

Answer