Question 15.12: Objective: Determine the output voltage and conversion effic...
Objective: Determine the output voltage and conversion efficiency for an LM380 power amplifier.
The required power for an 8 Ω is to be 4 W, with minimum distortion in the output signal.
From the curves in Figure 15.44, for an output of 4 W, minimum distortion occurs when the supply voltage is a maximum, or V^{+} = 22 V. For 4 W to be delivered to the 8 Ω load, the peak output signal voltage is determined by
\bar{P}_{L} = 4 = \frac{V^{2}_{P}}{2R_{L}} = \frac{V^{2}_{P}}{2(8)}
which yields V_{P} = 8 V.
The power dissipated in the device is 3 W, which means that the conversion efficiency is 4/(3 + 4) → 57 percent.
Comment: A reduction in the harmonic distortion means that the conversion efficiency is less than the theoretical value of 78.5 percent for the class-B output stages. However, a conversion efficiency of 57 percent is still substantially larger than would be obtained in any class-A amplifier
