Question 10.12: Objective: Calculate the small-signal voltage gain of an amp...

For R_{L} = ∞, Equation (10.96) reduces to

A_{v} = \frac{− \left( \frac{I_{Co}}{V_{T}} \right)}{\left( \frac{I_{Co}}{V_{AN}}  +  \frac{1}{R_{L}}  +  \frac{I_{Co}}{V_{A P}} \right)}           (10.96)

A_{v} = \frac{−  \left( \frac{1}{V_{T}} \right)}{\left( \frac{1}{V_{AN}}  +  \frac{1}{V_{A P}}\right)} = \frac{−  \left( \frac{1}{0.026} \right)}{\left(\frac{1}{120}  +  \frac{1}{80}\right)} = −1846
which is the same as that determined for the open-circuit configuration in Example 10.11.

For R_{L} = 200  k \Omega, the small-signal voltage gain is
A_{v} = \frac{− \left( \frac{0.2}{0.026} \right)}{\left( \frac{0.2}{120}  +  \frac{1}{200}  +  \frac{0.2}{80} \right)} = \frac{−7.692}{0.001667  +  0.005  +  0.0025} = −839  

and for R_{L} = 200  k \Omega, the voltage gain is
A_{v} = \frac{− \left( \frac{0.2}{0.026} \right)}{\left( \frac{0.2}{120}  +  \frac{1}{20}  +  \frac{0.2}{80} \right)} = \frac{−7.692}{0.001667  +  0.05  +  0.0025} = −142 
Comment: The small-signal voltage gain is a strong function of the load resistance R_{L}. As the value of R_{L} decreases, the loading effect becomes more severe.
Design Pointer: If an amplifier with an active load is to drive another amplifier stage, the loading effect must be taken into account when the small-signal voltage gain is determined. Also, the input resistance of the next stage must be large in order to minimize the loading effect