Question 10.11: Objective: Calculate the open-circuit voltage gain of a simp...

From Equation (10.86), the small-signal, open-circuit voltage gain is

A_{v} = \frac{d V_O}{d V_I} = −  \left( \frac{V_{AN}  V_{A P}}{V_{AN } +  V_{A P}} \right) \left(\frac{1}{V_{T}} \right)  =  \frac{− \left( \frac{1}{V_{T}} \right)}{\frac{1}{V_{AN}}  +  \frac{1}{V_{A P}}}          (10.86)
A_{v} = \frac{−  \left( \frac{1}{V_{T}} \right)}{\frac{1}{V_{AN}}  +  \frac{1}{V_{A P}}} = \frac{−  \left( \frac{1}{0.026} \right)}{\frac{1}{120}  +  \frac{1}{80}} = \frac{−38.46}{0.00833  +  0.0125} = −1846
Comment: For a circuit with an active load, the magnitude of the small-signal,
open-circuit voltage gain is substantially larger than the resulting gain when a
discrete resistor load is used.
Computer Verification: The voltage transfer characteristics of the active load circuit in Figure 10.29 were determined for a standard 2N3904 transistor as the npn device and standard 2N3906 transistors as the pnp devices. The circuit was biased at 5 V and the resistor was set at R = 1 kΩ. The transfer curve is shown in Figure 10.32

The input transition region, during which both Q_{0} and Q_{2} remain biased in the forward-active mode, is indeed very narrow. The slope of the curve, which is the voltage gain, is found to be −572. The reason for the smaller value compared to the hand calculation is that the Early voltages of these standard transistors are smaller than assumed in the previous calculation. The Early voltage of the npn device is 74 V and that of the pnp devices is only 18.7 V.
Design Pointer: From the transfer characteristics in Figure 10.32, we can see that, for this circuit, it would be very difficult to apply the required input voltage to bias both Q_{0} and Q_{2} in the active region. This particular circuit, therefore, is not practical as an amplifier. However, the circuit does demonstrate the basic properties of an active load.
In Chapters 11 and 13, we will see how an active load is applied to actual circuits