Objective: Calculate the small-signal voltage gain of an amplifier with an active load and a load resistance $R_{L}$ .

For the circuit in Figure 10.37(a), the transistor parameters are $V_{AN} = 120 V$ and $V_{A P} = 80 V$ . Let $V_{T} = 0.026 V$ and $I_{Co} = 0.2 mA$ . Determine the small-signal voltage gain for load resistances of $R_{L} = ∞, 200 kΩ$ , and 20 kΩ.

Question

Objective: Calculate the small-signal voltage gain of an amplifier with an active load and a load resistance [latex]R_{L}[/latex].

For the circuit in Figure 10.37(a), the transistor parameters are [latex]V_{AN} = 120 V[/latex] and [latex]V_{A P} = 80 V[/latex]. Let [latex]V_{T} = 0.026 V[/latex] and [latex]I_{Co} = 0.2 mA[/latex]. Determine the small-signal voltage gain for load resistances of [latex]R_{L} = ∞, 200 kΩ[/latex], and 20 kΩ.

Accepted Answer

For [latex]R_{L} = ∞[/latex], Equation (10.96) reduces to

[latex]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)} [/latex] (10.96)

[latex]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[/latex]
which is the same as that determined for the open-circuit configuration in Example 10.11.

For [latex]R_{L} = 200 k \Omega[/latex], the small-signal voltage gain is
[latex]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 [/latex]

and for [latex]R_{L} = 200 k \Omega[/latex], the voltage gain is
[latex]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 [/latex]
Comment: The small-signal voltage gain is a strong function of the load resistance [latex]R_{L}[/latex]. As the value of [latex]R_{L}[/latex] 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

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

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