If an amplifier working with ±5V DC power supplies, requires a
minimum output voltage of 3V across a load resistance of 100 ohms,
calculate the minimum current to be drawn from the power supply, if
the efficiency of the amplifier is 35%.
η=UmImUmIm+U2R, ⟹ \eta=\frac{U_mI_m}{U_mI_m+\frac{U^2}{R}},\impliesη=UmIm+RU2UmIm,⟹ Im=ηU2(1−η)UmR,I_m=\frac{\eta U^2}{(1-\eta)U_mR},Im=(1−η)UmRηU2,
Im=0.35⋅520.65⋅3⋅100=44.9 mA.I_m=\frac{0.35\cdot 5^2}{0.65\cdot 3\cdot 100}=44.9~mA.Im=0.65⋅3⋅1000.35⋅52=44.9 mA.
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