Page 1034 - Basic Electrical Engineering

or,

V = 48 + 0.7 = 48.7 V
m

PIV = V = 48.7 V
m

Example 15.3 A half-wave rectifier circuit has been made using a step-
down transformer of turn ratio 10:1. The input voltage is v = 325 sin ωt the
diode forward resistance is 25 Ω. A load resistance of 1.2 kΩ has been

connected in the circuit. Assuming a secondary winding resistance of the
transformer as 1Ω, calculate the following: (a) RMS value of load current (b)

rectification efficiency, and (c) ripple factor.

Solution:

Input voltage, v = 325 sin ωt

Input, V = 325
m
Transformer has a turn ratio of 10:1

The output

where R is the secondary winding resistance, R is the forward resistance of
2
F
the diode and R is the load resistance.
L