Question 4.12: Determine the minimum and maximum voltage gain of the amplif...
Determine the minimum and maximum voltage gain of the amplifier in Figure 4–46 (a) based on the data sheet information in Figure 4–46 (b).
On the data sheet, g_{m} is shown as y_{fs} . The range is 2000 μS to 8000 μS (shown as 2000 μmhos on the data sheet). The maximum value of r^{'}_{s} is
r^{'}_{s} = \frac{1}{g_{m} } = \frac{1}{2000 \mu S} = 500 \Omega
The ac source resistance , R_{s} ,is simply the load resistor, R_{L} , which is R_{S} . Substituting into Equation (4–9),
A_{v} = \frac{R_{s} }{r^{'}_{s} + R_{s} }
the minimum voltage gain is
A_{v(min)} = \frac{R_{s} }{r^{'}_{s} + R_{s} } = \frac{10 k\Omega }{500 \Omega + 10 k \Omega } = 0.95
The minimum value of r^{'}_{s} is
r^{'}_{s} = \frac{1}{g_{m} } = \frac{1}{8000 \mu S} = 125 \Omega
The maximum voltage gain is then
A_{v(max)} = \frac{R_{s} }{r^{'}_{s} + R_{s} } = \frac{10 k\Omega }{125 \Omega + 10 k \Omega } = 0.99
Notice that the gain is slightly less than 1. When r^{'}_{s} is small compared to the ac source resistance, then a good approximation is A_{v} = 1. Since the output voltage is at the source, it is in phase with the gate (input) voltage.