Question 10.13: Objective: Calculate the small-signal voltage gain of an NMO...
Objective: Calculate the small-signal voltage gain of an NMOS amplifier with an active load.
For the amplifier shown in Figure 10.40(a) the transistor parameters are: λ_{n} = λ_{p} = 0.01 V^{−1} , V_{T N} = 1 V, and K_{n} = 1 mA/V² . Assume M_{1} and M_{2} are matched and I_{REF} = 0.5 mA. Calculate the small-signal voltage gain for load resistances of R_{L} = ∞ and 100 kΩ.
Since M_{1} and M_{2} are matched, then I_{O} = I_{REF}, and the transconductance is
g_{m} = 2 \sqrt{K_{n} I_{REF}} = 2 \sqrt{(1)(0.5)} = 1.41 mA/V
The small-signal transistor conductances are
g_{o} = g_{o2} = λ I_{REF} = (0.01)(0.5) = 0.005 mA/V
For R_{L} = ∞, Equation (10.101) reduces to
A_{v} = \frac{V_{o}}{V_{i}} = −g_{m} (r_{o} || R_{L} || r_{o2}) = \frac{−g_{m}}{g_{o} + g_{L} + g_{o2}} (10.101)
A_{v} = \frac{−g_{m}}{g_{o} + g_{o2}} = \frac{−1.41}{0.005 + 0.005} = −141
For R_{L} = 100 k \Omega (g_{L} = 0.01 mA/V), the voltage gain is
A_{v} = \frac{−g_{m}}{g_{o} + g_{L} + g_{o2}} = \frac{−1.41}{0.005 + 0.01 + 0.005} = −70.5
Comment: The magnitude of the small-signal voltage gain of MOSFET amplifiers with active loads is substantially larger than for those with resistive loads, but it is still smaller than equivalent bipolar circuits, because of the smaller transconductance for the MOSFET.