RECTIFIERS
and
VOLTAGE REGULATION
(1)
Analog Electronics
Pujianto
Department of Physics Edu.
State University of Yogyakarta
We have studied that a PN Junction conducts easily
when forward biased and practically does not
conduct when reverse biased (very small current
flow).
This characteristic of PN Junction is very similar to
that of vacuum diode and we may also say that it
acts like a switch which conducts current in one
direction in the ON position and does not conduct in
the OFF position.
RECTIFIERS
A rectifier may be defined as a device which
converts AC voltage/current into DC voltage/current.
Half Wave Rectifier
Whenever an AC signal e = Em sin wt is applied
across a circuit consisting of a PN diode junction with
some resistance RL called load, during the positive
half cycle of input AC signal the diode conducts and
there is current through load, but during negative half
cycle we find no current through it.
The output voltage across load appears during
positive half of input cycle.
We call that the half wave is rectified in the output
The current flowing through diode during positive
half cycle
Applied _ voltage
ib =
Total _ resis tan ce _ of _ the _ circuit
Em sin wt
i = ib =
Rs + R f + RL
When 0 < wt <p
ib = 0 When p < wt < 2p
Here, Rs = the Resistance of secondary coil
Rf = the Forward resistance of diode
RL = the Resistance of the load
If Rs and Rf are very small in comparison with the
load RL than
Em sin wt
i = ib =
When 0 < wt <p
RL
or
where
Im
i = ib = I m sin wt
Em
=
RL
is called the peak value current
A Fourier analysis of the half sinusoidal voltage
pulse appearing across the load yield
Em Em
0,2 Em
cos 2wt
e=
+
sin wt p
2
3p
A diode is a frequency converter in which an input
frequency “w” is changed to a large number of
output frequencies.
In rectification we desire the DC or zero frequency
component and it is E
m
p
The DC or average current Idc is given by
p
2p
é
ù
1
I dc =
ê ò id (wt ) + ò id (wt )ú
2p ë 0
p
û
The second term in the bracket is zero since there
is no conduction during p to 2p interval of input.
So
p
1
I dc =
id (wt )
ò
2p 0
By substituting for i
1
I dc =
2p
(
Em
- )Em
p
ò0 RL sin wtd (wt ) = 2pRL [cos wt ]o
p
Or I = Em = I m
dc
pRL
p
This current has been shown as a dashed
straight line in the following figure
We can conclude that for a half rectifier the DC
output voltage is
Im
I dc =
p
so
and
Vdc = I dc RL
Vm
Vdc =
- I dc (Rs + R f )
p
with
I rms
Im
=
2
Efficiency of Rectifier
Pac = (I rms )
2
2
æ Im ö
(rf + RL ) = ç ÷ (rf + RL )
è 2 ø
2
æ Im ö
Pdc = I RL = ç ÷ RL
èp ø
2
dc
[
]
é
ù
Pdc
I m2 / p 2 R L
4
1
h =
= 2
= 2 = ê
ú
Pac
I m / 4 (r f + R L ) p
êë (r f R L ) + 1 úû
4
= 2 = 0 , 406 = 40 , 6 %
p
(
)
Ripple Factor
Ripple factor r is defined as the ratio of two current
(or voltage) components.
(
I r )rms (Vr )rms
r=
=
I dc
Vdc
Voltage Regulation
The degree to which a power supply varies in output
voltage under conditions of load variations is
measured by the voltage regulation which is usually
expressed as percentage
éVnoload - V fullload ù
%Vr = ê
ú x100%
V fullload
úû
êë
Ratio of Rectification
It is used as measure of merit to compare rectifiers
RoF =
dc _ power _ delivered _ to _ the _ load
ac _ input _ power _ from _ transformer _ sec ondary
Pdc
RoF =
Pac
Transformer Utilization Factor (TUF)
TUF
=
P dc
P ac
_ rated
æ Im
ç
p
è
=
Vm
2
2
ö
÷ RL
ø
Im
x
2
Vm = I m (R f + RL )
Problem
A half-wave rectifier consists of a diode having a
dynamic resistance of 20 ohm and an output voltage
is 12 volt.
It has a secondary resistance of 5 ohm and a load
resistance of 100 ohm. Determine:
a. An open-circuit secondary transformer voltage
b. DC output current
c. % Voltage regulation
d. Decreasing voltage because of transformer and
diode