Question

SOLVE PROBLEM #2

A class B complementary symmetry amplifier using ONE power supply is shown. Let Vcc = 12 V, R1 = R3 = 1 k22, R2 = 1702, Rs =

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Answer #1

PROBLEM #2 SOLUTION

Consider the diagra below with two VCC = ± 12 V

1596189896206_image.png

To achieve maximum power operation the output voltage must be

  V_L(p)= V_{cc}= 12 \: V

Part a)

Here the maximum input (supply) power is given by,

maximum,\: P_i(dc)= \frac{2V_{cc}^2}{\pi R_L}= \frac{2(12)^2}{\pi \times 8}= 11.46\: W

maximum output (load) power is given by,

maximum,\: P_o(ac)= \frac{V_{cc}^2}{ 2R_L}= \frac{12^2}{2 \times 8}= 9\: W

Part b)

]At maximum input and output power points, maximum efficiency is achieved

\because\: \: \: \: \: \: \: \%\eta =\frac{P_o}{P_i} \times 100 \: \%= \frac{9}{11.46} \times 100

= 78.539 \approx 78.54 \%

.....................................................................................................................................................................................

NB: For reference use Electronic Devices and Circuit Theory by Robert L Boylestad and Louis Nashelsky

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