Determine the small-signal voltage gain, input resistance and output resistance of the following circuit, given that g_{m} = 2 mS and r_{d} = 100 kΩ .
The first step in this problem is to determine the small-signal equivalent circuit of the amplifier.
Clearly, from the equivalent circuit
\frac{v_{o}}{v_{i}} = – g_{m}(r_{d}//R_{D}) = -g_{m} \frac{r_{d}R_{D}}{r_{d} + R_{D}} = -2 \times 10^{-3} \frac{100 \times 10^{3} \times 2 \times 10^{3}}{100 \times 10^{3} + 2 \times 10^{3}} = -3.9The minus sign indicates that the amplifier is an inverting amplifier. The small-signal input resistance of the amplifier is simply R_{G} , thus
r_{i} = R_{G} = 1 MΩThe small-signal output resistance is given by
r_{o} = r_{d}//R_{D} = \frac{r_{d}R_{D}}{r_{d} + R_{D}} = \frac{100 \times 10^{3} \times 2 \times 10^{3}}{100 \times 10^{3} + 2 \times 10^{3}} \approx 2.0 kΩThis example considers a circuit containing an n-channel DE MOSFET; a similar calculation could be performed for a circuit incorporating another type of FET.