Question 18.6: For the circuit in Example 18.5, compare the values obtained......
For the circuit in Example 18.5, compare the values obtained for the gain of the circuit when using the expressions of Equations 18.7, 18.8 and 18.9, assuming that g_{m} = 72 mS and r_{d} = 50 kΩ.
\frac{v_{o}}{v_{i}} \approx – \frac{R_{D}}{R_{S}} (18.7)
\frac{v_{o}}{v_{i}} \approx – \frac{g_{m}R_{D}}{1 + g_{m}R_{S} + \frac{R_{D} + R_{S}}{r_{d}} } (18.8)
\frac{v_{o}}{v_{i}} \approx – \frac{g_{m}R_{D}}{1 + g_{m}R_{S}} (18.9)
Step-by-Step
From Equation 18.7 we have
\frac{v_{o}}{v_{i}} \approx – \frac{R_{D}}{R_{S}} = -\frac{3.3 kΩ}{1 kΩ} = -3.3From Equation 18.8 we have
\frac{v_{o}}{v_{i}} \approx – \frac{g_{m}R_{D}}{1 + g_{m}R_{S} + \frac{R_{D} + R_{S}}{r_{d}} } = – \frac{72 × 10^{-3} × 3.3 kΩ}{1 + 72 × 10^{-3} × 1 kΩ + \frac{3.3 kΩ + 1 kΩ}{50 kΩ} } = -3.251From Equation 18.9 we have
\frac{v_{o}}{v_{i}} \approx – \frac{g_{m}R_{D}}{1 + g_{m}R_{S}} = – \frac{72 × 10^{-3} × 3.3 kΩ}{1 + 72 × 10^{-3} × 1 kΩ }= -3.255It can be seen that, in this case, the simple expression in Equation 18.7 produces a value that agrees quite closely with the values obtained using the more complete expressions.
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