Half Wave Rectifier with Efficiency,Ripple factor Pdf

Half Wave
Rectifier
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Table Of Contents
Rectifier
Half Wave Rectifier
Mathematical Analysis
DC value of current
RMS value of current
Efficiency
Ripple factor
Peak Inverse Voltage

Rectifier:
Rectification is the process of converting AC to DC. The circuit which is
used for this purpose is known as rectifier. Rectifiers are of two types:
1. Half wave rectifier
2. Full wave rectifier
In a half wave rectifier, DC is available at its output terminals during one
half cycle of the AC input, whereas in a full wave rectifier DC is obtained
during both half cycles of the AC input.
Half wave rectifier using diode:
Assemble the half wave rectifier circuit using P-N junction diode as shown
in image(1). Terminal A in the secondary is connected to the P section of
the diode and the other end N is connected to B terminal through load RL .
Image(1)
Image(2)

• The AC is supplied across the primary of a transformer.
• During one half of the cycle, A is positive with respect to B. This
makes the P-N junction diode to conduct as it is forward biased and
the current flows through the load RL as shown in the image(2).
• During the next half cycle the point A is negative with respect to B.
In this state, the diode does not conducts because it is reverse biased
and hence no current passes through RL.
• Thus current passes through the RL only during positive cycles.
Hence this circuit is known as half wave rectifier.
Image(3)

Mathematical Analysis:
Let Vi=Vpsinω t−−−(1) be the input voltage to the rectifier, where Vp
is the peak input voltage. While the diode is conducting, let id be the
current flowing through the circuit and Vd be the voltage across the diode.
Let us consider the ohmic resistance of the secondary of transformer as
negligible and applying Kirchoff's voltage law to the closed circuit, we
have:

Vi = Vd + id RL
= id Rf + id RL
where Rf is the forward resistance of the diode.
Vi=id(Rf +RL)−−−(2)
Comparing (1) and (2), we get:
Vp sin ωt=id(Rf +RL)
id=Im sin(ωt)
where Im=
Vp
Rf +RL
isthe peak current.
If the resistance of the diode is negligible compared to RL , then
Im=
V
RL
For a half wave rectifier, we have
Id=
{Im sin ω t for 0 < ωt < π
0 for π < ω t < 2π}
DC value of current(Idc):
From the image(3), it is seen that the output current is not steady but
contains fluctuations despite of being a DC current. The average value of
this fluctuating DC current can be calculated as follows:

Idc=
1
2π
∫
0
2π
Id d(ω t)
=
1
2π [∫
0
π
id d (ω t)+∫
π
2π
id d(ω t)
]
=
1
2π
∫
0
π
id d(ωt)
=( 1
2π )∫
0
π
Im sinω t d(ω t)
=
Im
2π
[−cos(ω t)]0
π
∴Idc=
Im
π
[∵ id=0 in the range  to 2]

RMS value of output current(Irms):
Irms=
√ 1
2π
∫
0
2π
id
2
d(ω t)
=
√ 1
2π
[∫
0
π
id
2
d(ω t)+∫
π
2π
id
2
d(ω t)]
=
√ 1
2π
∫
0
π
id
2
d(ω t)
=
√ 1
2π
∫
0
π
Im
2
sin
2
ω t d(ω t)
=
√Im
2
2π
[
ω t
2
−
sin2ω t
4
]
0
π
=
√Im
2
2π
π
2
=
√Im
2
4
∴Irms=
Im
2
∴Irms>Idc
[∵ id=0 in the range  to 2]

Efficiency of a half wave rectifier():
η=
Output DC power
Input AC power
x 100%
Input AC power Piac =
1
2 π
∫
0
2π
(instantaneous power)d(ω t)
Piac=
1
2π
∫
0
2π
e x id d(ω t)
=
1
2π
∫
0
2π
id(Rf+RL) x id d(ω t)
=
Rf +RL
2π
∫
0
2π
id
2
d(ωt)
=(Rf+RL)Irms
2
∴Piac=
(Rf +RL)Im
2
4
Output DC power Podc=Idc
2
RL
Podc=
Im
2
RL
π
2
Rectifier efficiency =
Podc
Piac
x 100%
=
Im
2
RL
π
2 x
4
(Rf+RL)Im
2 x 100%

=
40.6 RL
Rf+RL
%
If Rf << RL, = 40.6%
If Rf = RL, = 20.3%
Ripple factor(r):
At the output of half wave rectifier, periodically varying components are
still present even though we have achieved a unidirectional current. Filters
are used in the rectifier to reduce the varying components. A measure of
the varying component is given by the ripple factor as follows:
r=
Irms
'
Idc
=
Erms
'
Edc
where I'
rms and E'
rms represent the RMS value of ripple current and voltage
respectively.
Instantaneous value of ripple current I'
= id – Idc
Irms
'
=
√ 1
2π
∫
0
2π
id
'2
(ω t)
=
√ 1
2π
∫
0
2π
(id−Idc)
2
d(ω t)
=
√ 1
2π
∫
0
2π
(id
2
−2id Idc+Idc
2
)d (ω t)
Using equations
Idc=
1
2π
∫
0
2π
Id d (ω t) & Irms=
√ 1
2 π
∫
0
2π
id
2
d(ω t)

we have,
Irms
'
=√(Irms
2
−2Idc
2
+Idc
2
)
=√(Irms
2
−Idc
2
)
Ripple factor is given by
r=
Irms
'
Idc
=
√(Irms
2
−Idc
2
)
Idc
2
=
√(Irms
Idc
)
2
−1
=
√(Im /2
Im /π )
2
−1
∴r=
√(π
2 )
2
−1
∴r=1.21
The above calculation shows that the RMS value of the ripple exceeds that
of the DC potential of the output. This shows that the half wave rectifier
without a filter is relatively a poor device for converting AC into DC.
Peak Inverse Voltage:
It is defined as the maximum voltage applied across the diode when the
diode is reverse biased. In the case of half wave rectifier, the maximum
voltage across the diode when it is not conducting is equal to Vp, the peak

input voltage.
Full Wave & Half Wave Rectifier
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