Question 4.17: Objective: Determine the small-signal voltage gain of a casc...
Objective: Determine the small-signal voltage gain of a cascode circuit.
Consider the cascode circuit shown in Figure 4.51. The transistor parameters are K_{n1} = K_{n2} = 0.8 mA/V^{2}, V_{T N 1} = V_{T N 2} = 1.2 V, and λ_{1} = λ_{2} = 0. The quiescent drain current is I_{D} = 0.4 mA in each transistor. The input signal to the circuit is assumed to be an ideal voltage source
Since the transistors are identical and since the current in the two transistors is the same, the small-signal transconductance parameters are
g_{m1} = g_{m2} = 2 \sqrt{K_{n} I_{D}} = 2 \sqrt{(0.8)(0.4)} = 1.13 mA/VThe small-signal equivalent circuit is shown in Figure 4.52. Transistor M_{1} supplies the source current of M_{2} with the signal current (g_{m1} V_{i}) Transistor M_{2} acts as a current follower and passes this current on to its drain terminal. The output voltage is therefore
V_{o} = – g_{m1} V_{gs1} R_{D}Since V_{gs1} = V_{i} the small-signal voltage gain is
A_{v} = \frac{V_{o}}{V_{i}} = – g_{m1} R_{D}or
A_{v} = – (1.13)(2.5) = – 2.83Comment: The small-signal voltage gain is essentially the same as that of a single common-source amplifier stage. The addition of a common-gate transistor will increase the frequency bandwidth, as we will see in a later chapter
