Question 12.10: For the circuit of Fig. 12.19 , calculate the input power, o...
For the circuit of Fig. 12.19 , calculate the input power, output power, and power handled by each output transistor and the circuit efficiency for an input of 12 V rms.
The peak input voltage is
V_{i}( p )=\sqrt{2} V_{i}( rms )=\sqrt{2}(12 V )=16.97 V \approx 17 VSince the resulting voltage across the load is ideally the same as the input signal (the amplifier has, ideally, a voltage gain of unity),
V_{L}( p )=17 Vand the output power developed across the load is
P_{o}( ac )=\frac{V_{L}^{2}( p )}{2 R_{L}}=\frac{(17 V )^{2}}{2(4 \Omega)}= 3 6 . 1 2 5 WThe peak load current is
I_{L}( p )=\frac{V_{L}( p )}{R_{L}}=\frac{17 V }{4 \Omega}=4.25 Afrom which the dc current from the supplies is calculated to be
I_{ dc }=\frac{2}{\pi} I_{L}( p )=\frac{2(4.25 A )}{\pi}=2.71 Aso that the power supplied to the circuit is
P_{i}( dc )=V_{C C} I_{ dc }=(25 V )(2.71 A )= 6 7 . 7 5 WThe power dissipated by each output transistor is
P_{Q}=\frac{P_{2 Q}}{2}=\frac{P_{i}-P_{o}}{2}=\frac{67.75 W -36.125 W }{2}= 1 5 . 8 WThe circuit efficiency (for the input of 12 V, rms) is then
\% \eta=\frac{P_{o}}{P_{i}} \times 100 \%=\frac{36.125 W }{67.75 W } \times 100 \%= 5 3 . 3 \%Related Answered Questions
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