Question 4.16: Objective: Determine the small-signal voltage gain of a mult...
Objective: Determine the small-signal voltage gain of a multistage cascade circuit. Consider the circuit shown in Figure 4.49. The transistor parameters are K_{n1} = 0.5 mA/V^{2}, K_{n2} = 0.2 mA/V^{2}, V_{T N 1} = V_{T N 2} = 1.2 V, and λ_{1} = λ_{2} = 0. The quiescent drain currents are I_{D1} = 0.2 mA and I_{D2} = 0.5 mA
The small-signal equivalent circuit is shown in Figure 4.50. The small-signal transconductance parameters are
g_{m1} = 2 \sqrt{K_{n1} I_{D1}} = 2 \sqrt{(0.5)(0.2)} = 0.632 mA/Vand
g_{m2} = 2 \sqrt{K_{n2} I_{D2}} = 2 \sqrt{(0.2)(0.5)} = 0.632 mA/VThe output voltage is
V_{o} = g_{m2} V_{gs2} (R_{S2} ||R_{L})
Also,
V_{gs2} + V_{o} = – g_{m1} V_{gs1} R_{D1}
where
V_{gs1} = \left( \frac{R_{i}}{R_{i} + R_{Si}} \right) \cdot V_{i}
Then
V_{gs2} = – g_{m1} R_{D1} \left( \frac{R_{i}}{R_{i} + R_{Si}} \right) \cdot V_{i} – V_{o}
Therefore
V_{o} = g_{m2} \left[ – g_{m1} R_{D1} \left( \frac{R_{i}}{R_{i} + R_{Si}} \right) \cdot V_{i} – V_{o} \right] (R_{S2} ||R_{L})
The small-signal voltage gain is then
A_{v} = \frac{V_{o}}{V_{i}} = \frac{- g_{m1} g_{m2} R_{D1} (R_{S2} ||R_{L}) }{1 + g_{m2} (R_{S2} ||R_{L})} \cdot \left( \frac{R_{i}}{R_{i} + R_{Si}} \right)or
A_{v} = \frac{- (0.632) (0.632)(16.1)(8||4) }{1 + (0.632) (8||4)} \cdot \left( \frac{100}{100 + 4} \right) = – 6.14Comment: Since the small-signal voltage gain of the source follower is slightly less than 1, the overall gain is due essentially to the common-source input stage. Also, as shown previously, the output resistance of the source follower is small, which is desirable in many applications
