Question

4.70 A full-wave bridge-rectifier circuit with a 500-22 load operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 step-down transformer having a single Secondary winding. It uses four diodes, each of which can be modeled to have a 0.7-V drop for any current. What is the peak value of the rectified voltage across the load? For what fraction of a cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load?

Answer 1

Solution:

Given that

V1 =1200(sms) 3 € Ve Oy ܘܘܩܐ 3 R=5002

Kvi V2

Vi = 1 126 – 20 Vrms = 20 Vrms

V2(peak) = 20 X V2 = 28.2842 V

Here, positive half cycle D₁ and D₂ conducts during negative half cycle D₃ and D₄ conducts

- 28.280 7² -28.28V Vload 1 26.884 DI D2 D₂ D4 I QD2 D₂ DH

Here, load voltage peak value

= V2mar - 2 x voltage drop of diode during conduction

= V2mar - 2 x 0.7

= 28.2842 -1.4

= 26.884V

Peak value of the rectified voltage across load is

V mar(Load) = 26.8841

Conduction time of each diode $\infty$ = $\pi$ rad/sec

T = 60 Hz

T= ==0.01666 sec

Conduction time

T 0.01666 -= 8.333 msec

Average voltage:

V.= = | a Jo Vm sin wt

9 (7m soɔ - )"A7=

2Vm [:. Vm = max voltage across load]

Average voltage of load :

2 x 26.884 V.=

V. = 17.1148V

Average current through load:

V. R 17.1148 500 -= 0.0342A

1.= 0.0342A = 34.2m A