Question

4.2 For the circuit in Fig. 4.3(a), sketch the waveform of v,. Ans. vp = v, - vo, resulting in the waveform in Fig. E4.2 Up - V, Figure E4.2

Answer 1

4.2) The given circuit is. T + V 3R Vo oy Also, given the input waveform as v o I have divided the input waveform V into & intervals as t,, tz, tz, ty as marked in figure. I

Now, we can analyse the output waveform in lach interval Seperately case 1; oftet, During this period, V, ir negative So the diode get reverse biased, so the circuit wil 3 + - + SR. AV.

so to ir V, - vo (by KVL) Also; Vo will be zero (since input is not connected othur Vo=V to output) case 2; t, statz Input is Positive in this interval, so Diode get forward biased, so the circuit will be. io vo OZ Revo In this case Vo= V, - Vo = 0 (since the En output Vo=vi) so vo will be zero here.

case 3 ; ta I t I tz Here the case will be same as that of caset, since input is negative, Diode get reverse biased. So Vo=V-Vo = V (since vois 0 ) so Vo=y here, circuit will be same as that of case !

case & talt ette case & will be same as that of case 2, since V, is Positive, Diode gets forward biased, so. Vo = V-Vo = 0 (since Vo=",) so vo=0 in this case. (circuit will be raran rendah dir an Same as that of case 2) so we have; Vo = 4,; Oftat, -0, t, at eta =vi; teststa = 0; tzat eta

Now; We can Plot the Vo waveform. v prostor p ti to tz Vo vp Vet

So the Vo waveform will be. 1 so voltage appears across the diode only when V, is negative at which diode is reverse biased.
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