Question

For the circuit shown trace Vo Assume ideal diode model. Initial conditions@ t-0 Vin versus time 12 10 RC time constant-lus 4 Vc O -2 -4 -8 -10 -12 0.000 0.250 0.500 0.750 1.000 .250 1.500 1.750 2.000 time (seconds) Vin

Answer 1

Given input period T = 1 sec Then w = 2 pi/T = 2 pi rad/sec Then v_in = -5 sin 2 pi t circuit in - Negative cycle: - diode is forward bias to v_d is +ve and capacitor charged to v_in + v_c + 1 = 0 v_c = 6 v v_c = -6 v \r\n Then in +ve cycle Let - v_in - v_c + v_a = 0 v_a = v_in + v_c v_a = v_in + 6 v diode is reverse biased due v_d is -ve v_0 = v_a = v_in + 6 v if v_in - 5 to 5 v rightarrow v_0 = 1 to 11 v rightarrow It is a which clamps the -ve part of the input to total +ve
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