Question 6.SP.7: Use a small-signal h-parameter equivalent circuit to analyze......
Use a small-signal h-parameter equivalent circuit to analyze the amplifier of Fig. 3-10(a), given R_C = R_L = 800 Ω, R_i = 0, R_1 = 1.2 kΩ, R_2 = 2.7 kΩ, h_{re} ≈ 0, h_{oe} = 100 μS, h_{fe} = 90, and h_{ie} = 200 Ω. Calculate (a) the voltage gain A_v and (b) the current gain A_i.
(a) The small-signal circuit is shown in Fig. 6-13, where R_{B} = R_{1}R_{2}/(R_{1} + R_{2}) =831 \Omega. By current division in the collector circuit,
-i_{L} = {\frac{R_{C}(1/h_{oe})}{R_{C}(1/h_{oe}) + R_{L}(1/h_{oe}) + R_{L}R_{C}}}h_{fe}i_{b}
The voltage gain is then
A_{v} \equiv {\frac{v_{L}}{v_{i}}} = {\frac{R_{L}i_{L}}{h_{i e}i_{b}}} = -{\frac{h_{f e}R_{L}R_{C}}{h_{i e}(R_{C} + R_{L} + h_{o e}R_{L}R_{C})}} = -{\frac{(90)(800)^{2}}{200[1600 + (100 \times 10^{-6})(800)^{2}]}} = -173.08 (1)
(b) By current division,
i_{b} = {\frac{R_{B}}{R_{B} + h_{i e}}}\,i_{i}
so A_{i} \equiv {\frac{i_{L}}{i_{i}}} = {\frac{R_{B}}{R_{B} + h_{i e}}}{\frac{i_{L}}{i_{b}}} = \frac{R_{B}h_{i e}}{R_{L}(R_{B} + h_{i e})}\,A_{v} = \frac{(831)(200)(-173.08)}{(800)(1031)} = -34.87
