Page 1032 - Basic Electrical Engineering

value of reverse voltage that appears across the diode when it gets reverse
biased. Here

PIV = V m

Rectifier diodes are specified for their average forward current-carrying

capacity and their reverse voltage capacity, i.e., their PIV capacity. For
example, low-power rectifier diode series, IN 4000 to IN 4007 are rated for

forward current of 1000 mA and maximum reverse voltage of value varying
from 50 V to 1000 V.

Voltage Regulation of the rectifier is calculated using the relation,

The difference between no-load voltage and full-load voltage is the voltage
drop in the transformer winding and across the diode. The value of voltage
regulation, which is generally expressed in percentage should be low.

Example 15.1 A half-wave diode rectifier has a forward voltage drop, i.e.,
voltage drop across the diode when conducting is 0.7 V. The load resistance

is 600 Ω. The RMS value of the ac input is 28.87 V. Calculate I , I rms , PIV,
dc
and form factor.

Solution:

Figure 15.3

V (RMS) = 28.27 V
i
V (max) = V (RMS)
i
i