Question 6.SP.2: Assume that rc is large enough so that ic ≈ αie for the CB a......
Assume that r_c is large enough so that i_c ≈ αi_e for the CB amplifier of Fig. 3-23, whose small-signal circuit is given by Fig. 6-9. Find an expression for the current-gain ratio A_i = i_L/i_s and evaluate it if r_e = 30 Ω, r_b = 300 Ω, r_c = 1 MΩ, R_E = 5 kΩ, R_C = R_L = 4 kΩ, and α = 0.99.
Letting i_c ≈ αi_e in (6.27) allows us to determine the input resistance R_{in}:
v_{e b} = (r_{e} + r_{b})i_{e} – r_{b}i_{c} (6.27)
v_{e b} = (r_{e} + r_{b})i_{e} – r_{b}(αi_e)
from which R_{\mathrm{in}} = {\frac{v_{e b}}{i_{e}}} = r_{e} + (1 – \alpha)r_{b}
By current division at node E,
i_{e} = {\frac{R_{E}}{R_{E} + R_{\mathrm{in}}}}~i_{s}
Solving for i_{s} gives
i_{s} = {\frac{R_{E} + R_{\mathrm{in}}}{R_{E}}}\,\,i_{e} = {\frac{R_{E} + r_{e} + (1 – \alpha)r_{b}}{R_{E}}}\,\,i_{e} (1)
Current division at node C, again with i_c ≈ αi_e, yields
i_{L} = {\frac{R_{C}}{R_{C} + R_{L}}}\ i_{c} = {\frac{R_{C}\alpha i_{e}}{R_{C} + R_{L}}} (2)
The current gain is now the ratio of (2) to (1):
A_{i} = {\frac{i_{L}}{i_{s}}} = {\frac{\alpha R_{C}/(R_{C} + R_{L})}{[R_{E} + r_{e} + (1 – \alpha)r_{b}]/R_{E}}} = {\frac{\alpha R_{C}R_{E}}{(R_{C} + R_{L})[R_{E} + r_{e} + (1 – \alpha)r_{b}]}}
Substituting the given values results in
A_{i} = \frac{(0.99)(4 \times 10^{3})(5 \times 10^{3})}{(4 \times 10^{3} + 4 \times 10^{3})[5 \times 10^{3} + 30 + (1 – 0.99)(300)]} = 0.492