Question 6.SP.1: For the CB amplifier of Fig. 3-23, find the voltage-gain rat......
For the CB amplifier of Fig. 3-23, find the voltage-gain ratio A_v = v_L/v_S using the tee-equivalent small-signal circuit of Fig. 6-3.
The small-signal circuit for the amplifier is given by Fig. 6-9. By Ohm’s law,
i_{c} = {\frac{v_{c b}}{R_{C}||R_{L}}} = {\frac{(R_{C} + R_{L})v_{L}}{R_{C}R_{L}}} (1)
Substituting (1) into (6.27) and (6.28) gives, respectively,
v_{e b} = (r_{e} + r_{b})i_{e} – r_{b}i_{c} (6.27)
v_{c b} = (\alpha^{′}r_{c} + r_{b})i_{e} – (r_{b} + r_{c})i_{c} (6.28)
v_{S} = v_{eb} = (r_{e} + r_{b})i_{e} – r_{b}{\frac{(R_{C} + R_{L})v_{L}}{R_{C}R_{L}}} (2)
v_{L} = v_{c b} = (\alpha r_{c} + r_{b})i_{e} – (r_{b} + r_{c}){\frac{(R_{C} + R_{L})v_{L}}{R_{C}R_{L}}} (3)
where we also made use of (6.29). Solving (2) for i_e and substituting the result into (3) yield
\alpha^{\prime} \approx -h_{fb} = \alpha (6.29)
v_{L} = (\alpha r_{c} + r_{b}){\frac{v_{S} + {\frac{r_{b}(R_{C} + R_{L})}{R_{C} + R_{L}}}v_{L}}{r_{c} + r_{b}}} – (r_{b} + r_{c})\,{\frac{R_{C} + R_{L}}{R_{C}R_{L}}}\,v_{L} (4)
The voltage-gain ratio follows directly from (4) as
A_{v} = {\frac{v_{L}}{v_{S}}} = {\frac{(\alpha r_{c} + r_{b})R_{C}R_{L}}{R_{C}R_{L}(r_{e} + r_{b}) + (R_{C} + R_{L})[(1 – \alpha)r_{e}r_{b} + r_{e}(r_{b} + r_{c})]}}
