Question 4.11: (a) What is the dc drain voltage and the ac output voltage o...
(a) What is the dc drain voltage and the ac output voltage of the amplifi er in Figure 4–43 ? The g_{m} is 1500 μS, I_{D} is 2.0 mA, and V_{GS(off)} is 3 V.
(b) What is the input resistance seen by the signal source?
(a) First, find the dc drain voltage.
V_{D} = V_{DD} – I_{D} R_{D} = 15 V – (2 mA)(3.3 kΩ) = 8.4 V
Next, find the voltage gain
A_{v} = – g_{m} R_{d} = – (1500 μS)(3.3 kΩ) = -5.0
Alternatively, the voltage gain could be found by computing r^{'}_{s} and using the ratio of ac drain resistance to ac source resistance.
r^{'}_{s} = \frac{1}{g_{m} } = \frac{1}{1500 \mu S} = 667 \Omega
A_{v} = – \frac{R_{d} }{r^{'}_{s}} = – \frac{3.3 k\Omega }{667 \Omega } = -5.0
The ac output voltage is the gain times the input voltage.
V_{out} =A_{v} V_{in} = (-5.0)(100 mV) = -0.5 V rms
The negative sign indicates the output waveform is inverted.
(b) The input resistance is
R_{in} \cong R_{G} = 10 MΩ