Question 4.15: Objective: Determine the small-signal voltage gain and outpu...
Objective: Determine the small-signal voltage gain and output resistance of the common-gate circuit shown in Figure 4.48(a).
Assume the reference bias current is I_{Bias} = 0.20 mA and the bias voltage is V_{D D} = 3.3 V. Assume that the transistor parameters are V_{T N} = +0.4 V, V_{T P} =−0.4 V, K_{n} = 0.20 mA/V^{2}, K_{p} = 0.20 mA/V^{2}, and λ_{n} = λ_{p} = 0.01 V^{−1} .
We may note that, since M_{2} and M_{3} are matched transistors and have the same source-to-gate voltage, the bias current in M_{1} is I_{D1} = I_{Bias} = 0.20 mA
(voltage gain): From Figure 4.48(b), we can sum currents at the output node and obtain
\frac{V_{o}}{r_{o2}} + g_{m1} V_{gs} + \frac{V_{o} – (- V_{gs} )}{r_{o1}} = 0or
V_{o} \left( \frac{1}{r_{o2}} + \frac{1}{r_{o1}} \right) + V_{gs} \left( g_{m1} + \frac{1}{r_{o1}} \right) = 0From the circuit, we see that V_{gs} = – V_{i} We then find the small-signal voltage gain to be
A_{v} = \frac{\left( g_{m1} + \frac{1}{r_{o1}} \right) }{\left( \frac{1}{r_{o2}} + \frac{1}{r_{o1}} \right) }We find the small-signal equivalent circuit parameters to be
g_{m1} = 2 \sqrt{K_{n} I_{D1}} = 2 \sqrt{(0.20)(0.20)} = 0.40 mA/Vand
r_{o1} = r_{o2} = \frac{1}{\lambda I_{D1}} = \frac{1}{(0.01)(0.20)} = 500 k \OmegaWe then find
A_{v} = \frac{\left(0.40 + \frac{1}{500} \right) }{\left( \frac{1}{500} + \frac{1}{500} \right) }or
A_{v} = 101(output resistance): The output resistance can be found from Figure 4.48(c).
Summing currents at the output node, we find
However, V_{gs} = 0 so that g_{m1} V_{gs} = 0 We then find
R_{o} = \frac{V_{x}}{I_{x}} = r_{o1} || r_{o2} = 500 || 500or
R_{o} = 250 k \OmegaComment: A voltage gain of A_{v} = + 101 is typical of a common-gate amplifier. The output signal is in phase with respect to the input signal and the gain is relatively large. Also, a large output resistance of R_{o} = 250 k \Omega is typical of a common-gate amplifier in that the circuit acts like a current source.