Question 6.SP.24: The cascaded amplifier of Fig. 6-24(a) uses a CC first stage......

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.

6.24
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(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)} =36.38

and 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.

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