Exercise says:"An amplifier operating from a single 15 V supply provides a 12 V peak to peak sine wave signal to a 1 kohm load and draws negligible input current from signal source.. The DC current drawn from the 15V supply is 8 mA. What is the power dissipated in the amplifier and what is the ampifier efficiency?
I said:
Pdc= Vdc * Idc = 12 * 0.08 = 120 mW
Pload = (Vo /sqrt(2))^2/Rl = 72 mW
Pdiss + Pl = Pdc + Pi
Since no current is drawn from source, Pi=0 =>
Pdiss = Pdc-Pl = 48 mW
efficiency=Pl/Pdc = 60%.
Answer says: 102mw, 15%.
What am I doing wrong?
I said:
Pdc= Vdc * Idc = 12 * 0.08 = 120 mW
Pload = (Vo /sqrt(2))^2/Rl = 72 mW
Pdiss + Pl = Pdc + Pi
Since no current is drawn from source, Pi=0 =>
Pdiss = Pdc-Pl = 48 mW
efficiency=Pl/Pdc = 60%.
Answer says: 102mw, 15%.
What am I doing wrong?