The cascaded amplifier of Fig. 6-24(a) uses a CC first stage followed by a CE second stage. Let R_S = 0, R_{11} = 100 kΩ, R_{12} = 90 kΩ, R_{21} = 10 kΩ, R_{22} = 90 kΩ, R_L = R_C = 5 kΩ, and R_E = 9 kΩ. For transistor Q_1, h_{oc} ≈ 0, h_{ic} = 1 kΩ, h_{rc} ≈ 1, and h_{fc} = -100. For Q_2, h_{re} = h_{oe} ≈ 0, h_{fe} = 100, and h_{ie} = 1 kΩ. Find (a) the overall voltage-gain ratio A_v = v_L/v_s and (b) the overall current-gain ratio A_i = i_L/i_s.
(a) The small-signal equivalent circuit is drawn in Fig. 6-24(b), where
R_{B1} = R_{11}\Vert R_{12} = \frac{(90 \times 10^{3})(100 \times 10^{3})}{90 \times 10^{3} + 100 \times 10^{3}} = 47.37\,\mathrm{k}\Omega
and R_{B2} = R_{22}\Vert R_{21} = \frac{(90 \times 10^{3})(10 \times 10^{3})}{90 \times 10^{3} + 10 \times 10^{3}} = 4.5\,\mathrm{k} \Omega
From the results of Problem 6.44,
A_{v1} = -{\frac{h_{f c}(R_{E}\Vert R_{B2}\Vert h_{ie})}{h_{i c} – h_{r c}h_{fc}(R_{E}\Vert R_{B2}\Vert h_{ie})}} = -{\frac{(-100)(818.2)}{1 \times 10^{3} – (1)(-100)(818.2)}} = 0.9879
and from the results of Problem 6.7,
A_{v2} = -{\frac{h_{f e}R_{L}R_{C}}{h_{ie}(R_{L} + R_{C})}} = -{\frac{(100)(5 \times 10^{3})(5 \times 10^{3})}{(1 \times 10^{3})(5 \times 10^{3} + 5 \times 10^{3})}} = -100
Then A_{v} = A_{v1}A_{v2} = (0.9879)(-100) = -98.79
(b) From the results of Problem 6.21,
A_{i1} = {\frac{-i_{e1}}{i_{s}}} = -{\frac{ h_{fc}R_{B1}}{R_{B1} + h_{i e} + h_{r e}h_{fc}(R_{E}\Vert R_{B2}\Vert h_{i e})}} = -\frac{(-100)(47.37 \times 10^{3})}{47.37 \times 10^{3} + 1 \times 10^{3} + (1)(-100)(818.2)} =36.38and again from Problem 6.7,
A_{i2} = {\frac{(R_{E}\|R_{B2})h_{i e}}{R_{L}(R_{E}\|R_{B2} + h_{i e})}}~A_{v2} = {\frac{(4.5 \times 10^{3})(1 \times 10^{3})}{(5 \times 10^{3})(4.5 \times 10^{3} + 1 \times 10^{3})}}(-100) = -16.36
Then A_{i} = A_{i1}A_{i2} = (36.38)(-16.36) = -595.2
Note that, in this problem, we made use of the labor-saving technique of applying results determined for single-stage amplifiers to the individual stages of a cascaded (multistage) amplifier.