Question 15.13: Objective: Design the supply voltage required in the PA12 po...

For an average of 20 W delivered to the load, the peak output voltage is
V_{p} = \sqrt{2R_{L} \bar{P}_{L}} = \sqrt{2(10)(20)} = 20  V

and the peak load current is
I_{p} = \frac{V_{p}}{R_{L}} = \frac{20}{10} = 2  A
Assuming an ideal class-B condition, for a 50 percent conversion efficiency, the average power supplied by each V_{S} source must be 20 W. If we neglect power dissipation in the bias circuit, the average power supplied by each source is
P_{S} = V_{S} \left(\frac{V_{p}}{π R_{L}} \right)

and the required supply voltage is then
V_{S} = \frac{π R_{L} P_{S}}{V_{p}} = \frac{π(10)(20)}{20} = 31.4  V
Trade-offs: The required power supply must also be able to deliver the required current. For a power of 20 W delivered to the 10 Ω load, the load current (rms value) by itself is 1.41 A.
Comment: The actual conversion efficiency for class-AB operation is less than 50 percent. This reduced conversion efficiency ensures that harmonic distortion in the output signal is not severe.
Computer Simulation Verification: A computer simulation analysis of the circuit in Figure 15.45 was performed. The supply voltages were set at ±31.4 V and the input sinusoidal signal was adjusted so that the peak sinusoidal output voltage was 19.7 V across a 10 Ω load resistor. For these settings,the bias supply currents were 1.971 A. The average power delivered by the supply voltage sources is 39.4 W, so that the conversion efficiency is 49.25 percent, which is just slightly below the design value of 50 percent.