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AZFAR AZZAMUDIN BIN ZAINAL ABIDIN   B011910140 2 BEKG S2

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ARIF HAIKAL BIN ABDUL HALIM   B011910032 2 BEKG S2

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ADHWA’ BIN MOHD RADHI   B011910017 2 BEKG S2

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MOHD FADZELY BIN SAMAD     B011910045 2 BEKG S2

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An amplifier is an electronic device used to increase the magnitude of power or voltage or  current.Hence  the power amplifier is designed to increase the magnitude of power of a given input signal   It also has multiple classes of power amplifiers which is Class A, Class B, Class AB, Class C, Class D. Class A conducts through the full 360  of the input. Class B conducts through 180 of the input. Class AB conducts somewhere between 180 and 360. Class C conducts less than 180 while Class D is an amplifier that is biased especially for digital signals

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Class B operation is provided when the dc bias leaves the transistor biased just off, the transistor turning on when the ac signal is applied. This is essentially no bias, the transistor conducts current for only one half of the signal cycle (when it’s turned on). In order to obtain output for the full cycle of signal, it is necessary to use two transistors and have each conduct on opposite half-cycles. This circuit is referred to as a push-pull circuit because one part of the circuit pushes the signal high during one half cycle and the other part pulls the signal low during the other half-cycle

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Complementary Symmetry Circuit The most popular types of push-pull class B amplifiers: •using 2 transistors (npn and pnp) & both +ve and –ve power supplies. •A single input signal is applied to the base of both transistors •The npn transistor provides the +ve half cycle of the AC signal across the load •The other pnp transistor provides the -ve half cycle of the AC signal across the load •Disadvantages: •Need 2 separate dc voltage supplies Produces crossover distortion in the output signal

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Given ; Po(ac) = 60W               RL = 8 Ω Po(ac) = VL ²(p) ÷ 2RL  60W = VL² (p) ÷ 2(8) VL²(p) = 960V VL (p) = 30.984 V

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Pin (dc) = Vcc Idc Idc = (2 ÷ л) IL(p) IL(p) = VL(p) ÷ RL         = 30.984 ÷ 8 = 3.673 A Idc = 2.466 A assume Vcc = 30V  Hence Pin (dc) = 30V x 2.466A = 73.98 W

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% ת = ( Po(ac) ÷ Pin (dc) ) 100 %      = ( 60W ÷ 73.98W ) 100 %    = 81.10 ％ P2Q = Pi(dc) - Po(ac)          = 73.98W - 60W         = 13.98W PQ = P2Q ÷ 2 = 6.99 = 7

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max Po(ac) = VCC² ÷ 2RL                   = 30² ÷ 2(8)                   = 56.25W max Pi(dc) = 2VCC² ÷ лRL                    = 2(30²) ÷ л(8)                   = 71.62W max P2Q (dc) = 2VCC² ÷ л²RL                         = 2(30²) ÷ л²(8)                         = 22.79W

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max PQ = max P2Q  ÷ 2             = 22.79 ÷ 2 = 11.39 W % ת (max) = ( po (ac) ÷ pi (dc) ) x 100 %          = (  56.25W ÷ 71.62W ) x 100 %         =  78.53 %

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Students can understand the concept of amplifier circuit Students can solve the problem given Students can develop a teamworks to each other Students can learn to use a new software properly Students can develop a new soft skills for presentation

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