2. A diode is connected
to an ac source that
provides the input
voltage, V , and to a
inload resistor, RL,
forming a half-wave
rectifier
On the positive half-
cycle, the diode is
forward biased and
vice versa.
3. A rectifier circuit in which the half part of the
cycle either positive or negative of AC is
converted to DC is defined as the half-wave
rectifier.
If the positive half cycle is considered then
the negative half cycle of the source gets
blocked or if the negative cycle of the source
is considered in that case positive cycle gets
blocked.
4. The circuit consists of a single diode in series
with the AC supply and the load resistor. As
the sufficient supply is provided the diode
converts AC to DC the resultant will be
unidirectional by utilizing the half cycle of the
supply.
5. In positive half cycle sufficient amount of voltage
reaches the diode resulting it to function in
forwarding bias condition.
The process of rectification occurs at the load
resistance where the generated voltage in the
circuit is consumed by the load.
Load resistor is to block the excess current
produced in the circuit due to the diode.
Here the positive side of the supply is considered
so it is termed as a positive half-wave rectifier.
6. Here the diode direction is changed so it
begins to conduct during the applied negative
supply of the voltage and the positive cycle
gets blocked.
The current produced in the circuit gets
measured at the load resistor.
The output generated consists of pulsating Dc
but the intention of rectification is to produce
constant DC.
7.
8. • RMSValue of a Half-Wave Rectifier
• RMS is defined as the root mean square value. For
load current, the RMS value can be given as
• IRMS = Im/2
• The RMS value of the output voltage is given as
• VRMS = IRMS RL
• VRMS = Im/2 * RL
9. The efficiency of the rectifier is the ratio
between the generated output power to the
applied input power.
E = Pdc / Pac
The maximum efficiency produced is 40.6%.
Question: if input signal is being rectified half,
then why the efficiency is not 50%?
10. The output produced consists of pulsating DC.
Pulses in the output referred to as ripples.
The number of ripples present at the output
can be measured in terms of ripple factor.
Ripple Factor, γ =(Ripple voltage at the output)
( dc output voltage )
This ripple factor effects the practical
implementation of half wave rectifier.
11. Requirements list of the components is less.
The cost for the construction is low.
Less number of components presence results
in the construction of rectifier in the easiest
manner.
It is simple to analyze because the designed
circuit is straight forward.
12. The output generated in this rectifier is in the
form of pulses.
The ripple factor is high.
Results in the power loss of the circuit.
The transformer utilization factor (TUF) of the
half-wave rectifier is low.
Here the output generated requires the filter to
be connected across the load because of the
ripples generated at the output voltage.
13. The requirement of generating dc output
voltage paves the way for the application of
the half-wave rectifier circuit with the filter
attached across the load.
In the power supplies circuitry where the
constant DC at the output is not considered as
the major requirement at that case, the half-
wave rectifier can be used.
16. A rectifier that utilizes both the positive and
the negative half of the input cycle and
undergoes rectification is defined as a full-
wave rectifier.
17. The full-wave bridge rectifier uses four
diodes, as shown on the next slide
For +ve input cycle (a), diodes D1 and D4 are
forward-biased and conduct current, while diodes
D2 and D3 are reverse-biased
For –ve input cycle (b), diodes D2 and D3 are
forward-biased and conduct current, while diodes
D1 and D4 are reverse-biased
18.
19. The full-wave rectifier can be designed by
using with a minimum of two basic diodes or
it can use four diodes based on the topology
suggested.
Full-wave rectifier can be classified into two
types. They are
(1) Center – Tapped
A center-tapped full-wave rectifier circuit
consists of two diodes, a transformer, and a
resistive load.
20. A center-tapped transformer is a normal
transformer that has a slight modification in it.
Its secondary winding has a wire connected at
the center. Hence the input supply AC voltage
while passing through the secondary winding
its voltage is divided into two halves. The one
half is referred to positive half of the voltage.
Whereas the remaining half of the voltage is
for the negative part of the cycle.
21. The circuit consists
of two diodes that
are connected in
parallel to each
other along with the
resistive load. The
load is connected at
the center-tapped
wire of the
secondary winding.
22. The input is provided to the center-tapped
transformer as it reaches the secondary
winding the voltage is divided into two halves.
During the positive half of the input cycles,
the diode D1 is in forwarding bias condition
indicating the conducting mode and the diode
D2 is in the non-conducting mode because it is
in reverse bias condition. The flow of
current is observed at the terminal of diode
D1.
23. During the negative half of the cycle, the
diode D2 conducts because of the center-
tapped transformer property and the diode D1
is in reverse bias condition that is in non-
conduction mode. During this consequence the
terminal at D2 one can find the flow of current
through it and at D1 there is no evident flow
of current or the current is blocked at D1
terminal.
24. Hence both the cycles are utilized here for
rectification without any loss of input power.
The average output voltage of the half-wave
rectifier is doubled compared to that of the
half-wave rectifier.
25. A full-wave bridge rectifying circuit consists
of four diodes connected in a diagonal manner
that is nothing but in bridge topology.
No center-tapped transformer is used.
26.
27. The four diodes connected in a bridge form.
Diagonally opposite diodes conduct during
one half of the cycle.
Second half the remaining two diode
conducts.
28. The diode D1 and D4 are diagonally opposite to
each other whereas D2 and D3 are also
connected in a diagonally opposite manner.
During this the diodes D1 and D4 are in
forwarding bias condition that is in conduction
mode acts like closed switch then the path is
established for the flow of current. Whereas
diodes D2 and D3 are in reverse bias mode.
29. As the negative cycle of the supply reaches the
bridge rectifying circuit diodes D1 and D4
becomes reverse bias and the diodes D2 and D3
act in forwarding bias mode. Hence the path is
established for the flow of current between
and D3. Where D1 and D4 diodes act like an
open circuit and block the flow of current
30. The full-wave rectifier has more efficiency compared
to that of a half-wave rectifier.
There is the utilization of both the cycles. Hence there
is no loss in the output power.
As both the cycles used in rectification. There will be
no loss in the input voltage signal.
Ripple factor is less compared to that of the half-wave
rectifier.
Greater mean in DC value is achieved.
Compare to the center-tapped full-wave rectifier bridge
rectifier is cost-effective because the center-tapped is
more costly.
31. The only disadvantage of this circuit is that
the center-tapped transformer utilized here is
costly.
To overcome this disadvantage bridge
rectifier circuit is constructed where the four
diodes are connected in a bridge topology.
32. The amplitude for the modulating radio signal is
detected using the full-wave bridge rectifier circuit.
In electric wielding to supply steady DC voltage in a
polarized way, this circuit is preferred.
As the efficiency of rectification is high in this rectifier
circuit, it is used in various appliances as a part of the
power supply unit.
It has the capability of converting high AC voltage to
low DC value.
In case of powering up of the devices like motors and
LED devices these are used.