Question

A. Explain the following statements with regards to PN junction Diode with proper diagrams to illustrate your explanation "Njunction diode is a unidirectional bipolar device" [1 marks] B. With the given input voltage for the Full-wave rectifier circuit below, cakulate the peak output voltage and the de voltage. Assume ideal diodes [2 marks] Page 3 Final Assignment diades Ideal diodes Figures C. From the circuit given with load resistor Re-870. (Assume Si diode). [2 marks D Figure 6 a Find peak input voltage b. Find peak output voltage c. Find average output voltage 4. Find Peak Inverse Voltage

Answer 1

2 A. → Diode is a unidirectional device since it allows current to flow from 'p' to 'N and blocks the current from W to 'P' I P N 4 Forward bias -IS VON reverse bias so, diode conducts current only when forward biased. 40 reverse basobias only a minimal current Hows which is of the order 1015 A

A Diode is a bipolar device since the current flows due to the conduction of both electrons and holes N P (MOI) I 사 VE Ipt In I-

Given circuit is a full wave rectifier, Vp 160V st - Vp M epona -1607 Output peak voltage VR = 80V QVpo Dc Value g FWR= 2 x 80 50.92V IT IT

DI 5:1 +60V Vs - Vo + M 8 Tor coms) -1 * De Vscoms) 160V Vrms VP Vs,peak i Vs coms) trz Vs, peak = 160x82 = 226.27 v Alu Vi slå slut 160 Vi - Vi = 32 V (oms) Vicpeak) = 32 x 12 = 45.25V Positive halt cycle: Diode di conducts Vo + M From KVL 2x Vi -Vo:0 Via Vo

Negative half cycle: Diode de conducts et ogn -ve halt cycle, current flows from positive terminal to negative terminal of scarce u Voltage Foom KVL, - Vi-Vo:0 Vo:- Vi Vi N A t 7 -45-25 V vi vo Vi 45.25V Vo =- Vi means -ve cycle of input becomes positive cycle in output as Peak output voltage : 45.25V - average output Voltage : 2 Vp । 2X45-25 TI 28.8V

peak inverse Vottages -t is the maximum voltage across the reverse biased diode - Vot Vi Vpart By KVL vitvi - VPIV VPIV = 2 Vi VeIV = 2 Vp (maximum Voltage) VPIV = 90.5 V