Staff Site Universitas Negeri Yogyakarta
RECTIFIERS
and
VOLTAGE REGULATION
(2)
Analog Electronics
Pujianto
Department of Physics Edu.
State University of Yogyakarta
Full Wave Rectifier
While discussing half wave rectifier we have noted
that it suffers from many disadvantages:
1.There is excessive ripple
2.Low efficiency
3. D.C. saturation of transformer secondary coil
We will develop this rectifier into a full wave rectifier
by addition of another diode.
The secondary of the transformer is center-tapped
and as such it has twice the voltage from line to line
when AC voltage is applied across the primary coil of
the transformer.
During positive half of input voltage the upper terminal
A is positive the diode D1 conducts and current flows
through RL. The upper end P of load RL is positive.
Path of current is AD1PQC.
The D2 does not conduct since the lower terminal B is
negative.
During negative half of input voltage the lower end B
is positive so the diode D2 conducts and current flows
through RL. The upper end P of RL is positive. Path of
current is BD2PQC.
The DC or average current Idc is given by
1
I dc =
2p
id
t
(
)
w
ò
2p
0
Each diode operates independently and under
exactly the same conditions as in half wave circuit,
only the load current are combined.
p
2p
é
ù
Em
Em
1
sin wtd (wt ) + ò sin wtd (wt )ú
I dc =
êò
2p ë 0 RL
RL
p
û
The current of Im
Vm
Im =
( Rs + R f + RL )
By substituting for i
2 I m RL 2 Em
Em = I dc xRL =
=
p
p
So the output DC voltage
2 Em 2 I m
I dc =
=
pRL
p
We can conclude that for a full rectifier the DC
output voltage is
2I m
I dc =
p
and
Vdc = I dc RL
2Vm
- I dc (Rs + R f )
so Vdc =
p
with
I rms =
Im
2
Efficiency of Rectifier
Pac = (I rms )
æ Im ö
(r f + RL ) = ç ÷ (r f + RL )
è 2ø
2
2
æ 2I m ö
Pdc = I RL = ç
÷ RL
è p ø
2
2
dc
[
]
ù
Pdc
4 I m2 / p 2 RL
8 é
1
h=
= 2
= 2ê
ú
Pac
I m / 2 r f + RL p êë r f RL + 1úû
8
= 2 = 0,812 = 81,2%
(
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
I
2
rms
r
=
=
I
+
2
dc
æ I
çç
è I
m
dc
(I r )
ö
÷÷ - 1
ø
2
rms
Substituting for Irms and Idc we have
2
æ Im / 2 ö
p
÷ -1 =
1
0
,
48
r = çç
=
÷
2
I
/
8
p
ø
è m
2
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 - Vfullload ù
%Vr = ê
ú x100%
Vfullload
úû
êë
Ratio of Rectification
It is used as measure of merit to compare rectifiers
dc _ power _ delivered _ to _ the _ load
RoF =
ac _ input _ power _ from _ transformer _ sec ondary
Pdc
RoF =
Pac
Transformer Utilization Factor (TUF)
æ 2I m ö
ç
÷ RL
p ø
è
=
Vm I m
x
2
2
2
TUF =
Pdc
Pac _ rated
(
Vm = I m R f + RL
)
The Bridge Rectifier
For a Bridge rectifier the DC output voltage is
2I m
I dc =
p
and
Vdc = I dc RL
2Vm
- I dc (Rs + 2 R f )
so Vdc =
p
with
I rms =
Im
2
The current of Im
Vm
Im =
( Rs + 2 R f + RL )
By substituting for i
2 I m RL 2 Em
Em = I dc xRL =
=
p
p
So the output DC voltage
2 Em 2 I m
I dc =
=
pRL
p
The R-C Filter or Shunt Capacitor filter
Charging-discharging of the Condenser
Filter
The average value of load current Idc is the average
value of the capacitor discharge current over an
interval of T2 from q2 to q1.
The amount of charge lost by the capacitor
Qdisch arg e = I dc xT2
This charge is replanished during short interval T1
(from q1 to q2) during which voltage across capacitor
changes by an amount (peak to peak voltage of the
ripple)Vrpp
Q ch
arg e
Q ch
arg e
= V rpp xC
= Q disch
V rpp xC = I dc xT
V rpp
V rpp
arg e
I dc
T2
=
C
I dc
1
x
=
C
2 f
2
and
VOLTAGE REGULATION
(2)
Analog Electronics
Pujianto
Department of Physics Edu.
State University of Yogyakarta
Full Wave Rectifier
While discussing half wave rectifier we have noted
that it suffers from many disadvantages:
1.There is excessive ripple
2.Low efficiency
3. D.C. saturation of transformer secondary coil
We will develop this rectifier into a full wave rectifier
by addition of another diode.
The secondary of the transformer is center-tapped
and as such it has twice the voltage from line to line
when AC voltage is applied across the primary coil of
the transformer.
During positive half of input voltage the upper terminal
A is positive the diode D1 conducts and current flows
through RL. The upper end P of load RL is positive.
Path of current is AD1PQC.
The D2 does not conduct since the lower terminal B is
negative.
During negative half of input voltage the lower end B
is positive so the diode D2 conducts and current flows
through RL. The upper end P of RL is positive. Path of
current is BD2PQC.
The DC or average current Idc is given by
1
I dc =
2p
id
t
(
)
w
ò
2p
0
Each diode operates independently and under
exactly the same conditions as in half wave circuit,
only the load current are combined.
p
2p
é
ù
Em
Em
1
sin wtd (wt ) + ò sin wtd (wt )ú
I dc =
êò
2p ë 0 RL
RL
p
û
The current of Im
Vm
Im =
( Rs + R f + RL )
By substituting for i
2 I m RL 2 Em
Em = I dc xRL =
=
p
p
So the output DC voltage
2 Em 2 I m
I dc =
=
pRL
p
We can conclude that for a full rectifier the DC
output voltage is
2I m
I dc =
p
and
Vdc = I dc RL
2Vm
- I dc (Rs + R f )
so Vdc =
p
with
I rms =
Im
2
Efficiency of Rectifier
Pac = (I rms )
æ Im ö
(r f + RL ) = ç ÷ (r f + RL )
è 2ø
2
2
æ 2I m ö
Pdc = I RL = ç
÷ RL
è p ø
2
2
dc
[
]
ù
Pdc
4 I m2 / p 2 RL
8 é
1
h=
= 2
= 2ê
ú
Pac
I m / 2 r f + RL p êë r f RL + 1úû
8
= 2 = 0,812 = 81,2%
(
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
I
2
rms
r
=
=
I
+
2
dc
æ I
çç
è I
m
dc
(I r )
ö
÷÷ - 1
ø
2
rms
Substituting for Irms and Idc we have
2
æ Im / 2 ö
p
÷ -1 =
1
0
,
48
r = çç
=
÷
2
I
/
8
p
ø
è m
2
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 - Vfullload ù
%Vr = ê
ú x100%
Vfullload
úû
êë
Ratio of Rectification
It is used as measure of merit to compare rectifiers
dc _ power _ delivered _ to _ the _ load
RoF =
ac _ input _ power _ from _ transformer _ sec ondary
Pdc
RoF =
Pac
Transformer Utilization Factor (TUF)
æ 2I m ö
ç
÷ RL
p ø
è
=
Vm I m
x
2
2
2
TUF =
Pdc
Pac _ rated
(
Vm = I m R f + RL
)
The Bridge Rectifier
For a Bridge rectifier the DC output voltage is
2I m
I dc =
p
and
Vdc = I dc RL
2Vm
- I dc (Rs + 2 R f )
so Vdc =
p
with
I rms =
Im
2
The current of Im
Vm
Im =
( Rs + 2 R f + RL )
By substituting for i
2 I m RL 2 Em
Em = I dc xRL =
=
p
p
So the output DC voltage
2 Em 2 I m
I dc =
=
pRL
p
The R-C Filter or Shunt Capacitor filter
Charging-discharging of the Condenser
Filter
The average value of load current Idc is the average
value of the capacitor discharge current over an
interval of T2 from q2 to q1.
The amount of charge lost by the capacitor
Qdisch arg e = I dc xT2
This charge is replanished during short interval T1
(from q1 to q2) during which voltage across capacitor
changes by an amount (peak to peak voltage of the
ripple)Vrpp
Q ch
arg e
Q ch
arg e
= V rpp xC
= Q disch
V rpp xC = I dc xT
V rpp
V rpp
arg e
I dc
T2
=
C
I dc
1
x
=
C
2 f
2