Question

Briefly explain how the diodes(D, and D2) conduct the current in each half-cycle of the input Vin. Sketch the output voltage Vout (on the same scale as Vin) if voi = VD2 = 0.7 V(on conducting). D + Vin +V01- + VO out +V02 D2 Vin (V) 6 t(ms) 10 20 30 40 -6

Answer 1

Page 1 Follows: The giver diagram can be simplified as wer t Viola DI + Vin Vout Voz + As the input is altern ternating in nature We can consider the operation of the circuit in two steps: Step 1: During Negative half cyde During negative half cycle, the polarity of the electronical components of the circuit will be D, OFF + li Vout A in COFFDD, is in Current is in Here. Di reverse biosed condition COFFEDA Forward biased condition (on).so the Path will be: Secondary of Transformer >> D2 → Load. Тbe. in pula voltage gelo halved due to centre tapped transformer, hence available VO voltage the Secondary sida is side across Load Available voltage across secondary voltage cloop actass diocle De Vine - 0.7 (cut in voltage, giveo in questions) at Vin The voltage il

Page-2 Step 2: During positive half cycle In this half cycle, the polarity of the electrical components of the circuit will be AL vo D + + + + Vout D2 OFF in Here Di in Forward biased condition Con). D2 orverse biased condition . so the Current path will be Secondary of Pransformer >0, → Load. The input voltage gel halvad at Secondary Side of transformed due centoe tapping.ie available voltage at the secondary side 6 Vin The voltage across Lood = Available voltage across Secondary side - Vokage drop acto55 diode Dj to 2 = Vin -0.7 2 current througth Note that the the load, dusing both positive & negative half cycle is same Hence the voltage across load has one Polarity (Rectified volt oltage) with Vout Vin -0.7 Value 2. Vinman 0:7 17 Maximum Vour Vmuscous 2 -0.7. = 3-0.7 = 2-3V

Page 3 Output voltage Vow (v) 2-3 t (ms) lo 20 30 40 with a maximum 19 rectified voltage 3 a Output voltage value of 2.3V