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UNIT V
RECTIFIERS AND POWER SUPPLIES
PART –A (2 MARK QUESTIONS)
1. Draw the full wave bridge rectifier circuit (NOV/DEC 2009)
2. What are the advantages of SMPS over conventional regulators?
(NOV/DEC 2009) (APR/MAY 2010)
1.
Light weight since the transformer is too small and it it operates at high
frequency of 50Hz-1MHz.
2.
Output voltage is well regulated and controlled by duty cycle and there is
little resistive loss since the transistor fully on or off during switching.
3.
Greater efficiency since the switching transistor dissipates very little heat.
The SMPS can fail and can cause very high output voltage that destroys the
equipment.
3. Compare the half-wave and full-wave rectifiers. (APR/MAY 2010)
(NOV/DEC’12)



Efficiency is double for a full wave bridge rectifier.
The residual ac ripples (before filtering) is very low in the
output of a bridge rectifier. The same ripple percentage is very
high in half wave rectifier.
higher output voltage, Higher transformer utilization factor
(TUF) and higher output power.
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4. Write down the expression for ripple factor of LC filter in FWR.
(NOV/DEC 2011)
Ripple factor = 1.194/LC
5. Distinguish between line regulation and load regulation. (NOV/DEC
2011)
Load regulation is the capability to maintain a constant voltage (or current) level
on the output channel of a power supply despite changes in the supply's load.
Line regulation is the capability to maintain a constant output voltage level on
the output channel of a power supply despite changes to the input voltage level.
6. Briefly explain the working of Zener regulator. (NOV/DEC 2010)
As
the
input
voltage
increases
the
current
through
the Zener
diode increases but the voltage drop remains constant - a feature of zener
diodes. Therefore since the current in the circuit has increased the voltage
drop across the resistor increases by an amount equal to the difference
between the input voltage and the zener voltage of the diode.
7.
Give the ripple factor of inductance filter connected to FWR.
(NOV/DEC 2010)
Ripple factor =RL/3 2 ωL
8. What are the advantage of bridge rectifier over the centre tapped
part counter.
1. Bulky centre tapped transformer is not required.
2. Transformer utilization factor is high.
9. Draw half wave voltage doubler circuit.
10. Define transformer utilization factor.
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TUF=dc power delivered to the load/ac rating of the transformer
secondary.
11. What is voltage Multiplier?
Voltage multiplier is a modified capacitor filter circuit that delivers a dc
voltage twice or rnore times of the peak value (amplitude) of the input ac
voltage. Such power supplies are used for high-voltage and low-current
devices such as cathode-ray tubes (the picture tubes in TV receivers,
oscilloscopes and computer display).
12. What is meant by ripple factor?
It is a measure of percentage of ac component presented at the load.
13. What are the limitations of using zener diode regulator?
The zener diode regulator has limitations of range. The load current range
for which regulation is maintained, is the difference between maximum
allowable zener current and minimum current required for the zener to
operate in breakdown region.
14. Indicate two advantages of bleeder resistor in bridge rectifier?
A bleeder resistor is an electrical component that absorbs electrical power in
unregulatedpower supply outputs to improve voltage regulation.
15. Differentiate series and shunt voltage regulator.
16. What are the different types of rectifiers?
Half wave rectifiers
Full wave rectifiers
Bridge rectifiers
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17. A power supply has 4% voltage regulation and an open circuit
voltage of 48V DC. Calculate the full load voltage. (NOV/DEC’12)
Load regulation = (Vno load – Vfull load) / Vfull load
18. Define transformer utilization factor.
TUF = dc power delivered to the load/ac rating of transformer secondary
= Pdc /Pac . rated
PART- B (16 MARK QUESTIONS
1. Explain the circuit of voltage regulator and also discuss the short
circuit protection mechanism.(NOV/DEC 2009)
An ideal power supply maintains a constant voltage at its output terminals
under all operating conditions. The output voltage of a practical power
supply changes with load generally dropping as load current increases as
shown in fig. 1.
Fig. 1
The terminal voltage when full load current is drawn is called full load
voltage (VFL). The no load voltage is the terminal voltage when zero
current is drawn from the supply, that is, the open circuit terminal voltage.
Power supply performance is measured in terms of percent voltage
regulation, which indicates its ability to maintain a constant voltage. It is
defined as
The Thevenin's equivalent of a power supply is shown in fig. 2. The
Thevenin voltage is the no-load voltage VNL and the Thevenin resistance
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is called the output resistance Ro. Let the full load current be IFL.
Therefore, the full load resistance RFL is given by
Fig. 2
From the equivalent circuit, we have
and the voltage regulation is given by
An unregulated power supply consists of a transformer (step down), a
rectifier and a filter. These power supplies are not good for some
applications where constant voltage is required irrespective of external
disturbances. The main disturbances are:
As the load current varies, the output voltage also varies because of its
poor regulation.
The dc output voltage varies directly with ac input supply. The input
voltage may vary over a wide range thus dc voltage also changes.
The dc output voltage varies with the temperature if semiconductor
devices are used.
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An electronic voltage regulator is essentially a controller used along with
unregulated power supply to stabilize the output dc voltage against three
major disturbances
Load current (IL)
Supply voltage (Vi)
Temperature (T)
Fig. 3, shows the basic block diagram of voltage regulator. where
Vi = unregulated dc voltage.
Vo = regulated dc voltage.
Fig. 3
Since the output dc voltage VLo depends on the input unregulated dc
voltage Vi, load current IL and the temperature t, then the change ΔVo in
output voltage of a power supply can be expressed as follows
VO = VO(Vi, IL, T)
Take partial derivative of VO, we get,
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SV gives variation in output voltage only due to unregulated dc voltage.
RO gives the output voltage variation only due to load current. ST gives the
variation in output voltage only due to temperature.
The smaller the value of the three coefficients, the better the regulations of
power supply. The input voltage variation is either due to input supply
fluctuations or presence of ripples due to inadequate filtering.
2. Explain the power control method using SCR. (NOV/DEC 2009)
(NOV/DEC’12) (NOV/DEC 2009) (NOV/DEC 2011)
Silicon-controlled rectifier A silicon-controlled rectifier (or semiconductorcontrolled rectifier) is a four-layer solid state device that controls current. The
name "silicon controlled rectifier" or SCR is General Electric's trade name for a
type of thyristor. The SCR was developed by a team of power engineers led by
Gordon Hall and commercialised by Frank W. "Bill" Gutzwiller in 1957. Theory of
operation An SCR is a type of rectifier, controlled by a logic gate signal. It is a
four-layer, three-terminal device. A p-type layer acts as an anode and an n-type
layer as a cathode; the p-type layer closer to the n-type(cathode) acts as a gate.
It is unidirectional in nature. Modes of operation In the normal "off" state, the
device restricts current to the leakage current. When the gate to cathode voltage
exceeds a certain threshold, the device turns "on" and conducts current. The
device will remain in the "on" state even after gate current is removed so long as
current through the device remains above the holding current. Once current falls
below the holding current for an appropriate period of time, the device will switch
"off". If the applied voltage increases rapidly enough, capacitive coupling may
induce enough charge into the gate to trigger the device into the "on" state; this
is referred to as "dv/dt triggering." This is usually prevented by limiting the rate of
voltage rise across the device, perhaps by using a snubber. "dv/dt triggering"
may not switch the SCR into full conduction rapidly and the partially-triggered
SCR may dissipate more power than is usual, possibly harming the device.
SCRs can also be triggered by increasing the forward voltage beyond their rated
breakdown voltage (also called as breakover voltage), but again, this does not
rapidly switch the entire device into conduction and so may be harmful so this
mode of operation is also usually avoided. Also, the actual breakdown voltage
may be substantially higher than the rated breakdown voltage, so the exact
trigger point will vary from device to device.
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The advantages of SCR Controls over other temperature control methods






Improved response time.
Closer process control.
Extended heater life.
Reduced maintenance costs.
Silent operation. No arcing and sparking.
Reduced peak power consumption.
3. Derive the expressions for the rectification efficiency, ripple factor,
transformer utilization factor, form factor and peak factor of
(i) half wave rectifier (ii) full wave rectifier. (APR/MAY 2010)
Ripple factor for Half-wave rectification
By definition the effective (ie rms) value of total load current is given by
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ripple factor r = 1.21
It is clear that a.c. component exceeds dc component in the output of a half-wave
rectifier.
Efficiency:
η = Pdc/Pac
Vac = √Vrms2 - Vdc2
FF = Vrms / Vdc
RF = Vac/Vdc
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Ripple Factor
The ripple factor for a Full Wave Rectifier is given by
The average voltage or the dc voltage available across the load resistance is
RMS value of the voltage at the load resistance is
Efficiency
Efficiency,  is the ratio of dc output power to ac input power
The maximum efficiency of a Full Wave Rectifier is 81.2%.
Transformer Utilization Factor
Transformer Utilization Factor, TUF can be used to determine the rating of a
transformer secondary. It is determined by considering
the primary and the secondary winding separately and it gives a value of 0.693.
Form Factor
Form factor is defined as the ratio of the rms value of the output voltage to the
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average value of the output voltage.
Peak Factor
Peak factor is defined as the ratio of the peak value of the output voltage to the
rms value of the output voltage.
Peak inverse voltage for Full Wave Rectifier is 2Vm because the entire
secondary voltage appears across the non-conducting diode.
This concludes the explanation of the various factors associated with Full Wave
Rectifier.
4.
Explain the operation of Voltage multiplier (APR/MAY 2010)
Voltage multiplier is a modified capacitor filter circuit that delivers a dc voltage
twice or more times of the peak value (amplitude) of the input ac voltage. Such
power supplies are used for high-voltage and low-current devices such as
cathode-ray tubes (the picture tubes in TV receivers, oscilloscopes and computer
display). Here we will consider half-wave voltage doubler, full-wave voltage
doubler and voltage tripler and quadrupler.
Half-Wave Voltage Doubler
The circuit of a half-wave voltage doubler is given in figure shown below. During
the positive half cycle of the ac input, voltage, diode D1 being forward biased
conducts (diode D2 does not conduct because it is reverse-biased) and charges
capacitor C1 upto peak values of secondary voltage Vsmax with the polarity, as
marked in figure shown below.
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During the negative half-cycle of the input voltage
diode D2 gets forward biased and conducts charging capacitor C2. For the
negative half cycle, the lower end of the transformer secondary is positive while
upper end is negative. The polarity of the capacitor C2 has also been marked in
the figure. Now starting from the bottom of the transformer secondary and moving
clockwise and applying Kirchhoffs voltage law to the outer loop we have
-Vsmax – Vc1 + Vc2 = 0
Or
Vc2 = Vsmax + Vc1= Vsmax + Vsmax = 2Vsmax = Twice the peak value of the
transformer secondary voltage. (Since Vc1 = Vsmax)
During the next positive half-.cycle diode D2 is reverse-biased and so acts as an
open and capacitor C2discharges through the load If there is no load across the
capacitor, C2 both capacitors stay charged – C1 to Vsmax and C2 to 2Vsmax. If, as
expected there is a load connected to the output terminals of the voltage doubler,
the capacitor C2 discharges a little bit and consequently the voltage across
capacitor C2 drops slightly. The capacitor C2 gets recharged again in the next
half-cycle. The ripple frequency in this case will be the signal frequency (that is,
50 Hz for supply mains.)
Full-Wave Voltage Doubler
The circuit diagram for a full-wave voltage doubler is given in the figure shown
below. During the positive cycle of the ac input voltage, diode D1 gets forward
biased and so conducts charging the capacitor C1 to a peak voltage Vsmax with
polarity indicated in the figure, while diode D2 is reverse-biased and does not
conduct.During the negative half-cycle, diode D2 being forward biased conducts
and charges the capacitor C2with polarity shown in the figure while diode D1 does
not conduct. With no load connected to the output terminals, the output voltage
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will be equal to sum of voltages across capacitors C1 and C2 that is, VC1 + VC2or
(Vs max + Vs max) or 2 Vs max. When the load is connected to the output terminals,
the output voltage VLwill be somewhat less than 2 Vs max. The input voltage and
output voltage waveforms are also shown in the figure below.
Voltage Tripler and Quadruples
The half-wave voltage doubler, shown in the earlier figure can be extended to
provide any multiple of the peak input voltage (that is, 3 Vs max, 4 Vs max or 5
Vs max), as illustrated in the figure shown below. It is obvious from the pattern of
the circuit connections how additional diodes and capacitors are to be connected
to provide output voltage, 5,6,7 or 8 times the peak input voltage from a supply
transformer of rating only Vsmax, and each diode in the circuit of PIV rating 2
Vs max. If load is small and the capacitors have little leakage, extremely high dc
voltages can be obtained from such a circuit using many sections to step-up the
dc voltage.
In operation capacitor C1 is charged through diode Dl to a peak value of
transformer secondary voltage, Vsmax during first positive half-cycle of the ac input
voltage. During the negative half cycle capacitor C2 is charged to twice the peak
voltage 2 Vs developed by the sum of voltages across capacitor C1 and the
transformer secondary. During the second positive half-cycle, diode D3 conducts
and the voltage across capacitor C2 charges the capacitor C3 to the same 2
Vg max peak voltage. During the negative half-cycle diodes D2 and D4 conduct
allowing capacitor C3 to charge capacitor C4 to peak voltage 2 VS max. From the
fogure shown below it is obvious that the voltage across capacitor C2 is 2 Vs max,
across capacitors C1 and C3it is 3 Vs max and across capacitors C2 and C4 it is 4
Vs max.
If additional diodes (each diode of PIV rating 2 Vs max) and capacitors (each
capacitor of voltage rating 2 Vsmax) are used, each capacitor will be charged to 2
Vs max. Measuring from the top of the transformer secondary winding (figure
below) will give odd multiples of Vg max at the output, while measuring from the
bottom of transformer secondary winding will give even multiples of the peak
voltage, Vs max.
Tripler and Quadruplar
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Some electronic devices, such as cathode ray tubes (in picture tubes in TV
receivers, oscilloscopes and computer display) need dc power supply at high
voltage with low current. This requirement can be met with either by employing a
step-up transformer with a rectifier circuit or by employing voltage multiplier.
Since transformers are very bulky and costly, voltage multipliers are preferred. By
using voltage multipliers, the voltage level is usually raised well into the hundreds
or thousands of volts.
Generally such circuits are employed when both the supply voltage and load are
maintained constant.
5. Describe how output voltage can be regulated with respect to line
variations and load variations using SMPS. (16) (NOV/DEC 2010)
(NOV/DEC 2011) (NOV/DEC’12)
It
uses
a switching
regulator to
convert
electric
power
efficiently. SMPS transfers electric power from a source ( AC mains) to the load
by converting the characteristics of current and voltage. SMPS always provide a
well regulated power to the load irrespective of the input variations. SMPS
incorporates a Pass transistor that switches very fast typically at 50Hz and 1
MHz between the on and off states to minimize the energy waste. SMPS
regulates the output power by varying the on to off time using minimum voltage
so that efficiency is very high compared to the linear power supply.
The SMPS essentially has
1.
Input rectifier
2.
Inverter
3.
Voltage converter
4.
Output regulator
Input rectifier
The AC input from mains is first rectified in the SMPS using a rectifier to convert it
into DC. The rectifier consisting of a full wave diode bridge or module that
produces an unregulated DC voltage to the Smoothing capacitor. The input AC
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passing into the rectifier has AC voltage pulses that may reduce the power factor.
So control techniques are used to force the average input current to follow the
sine wave.
Inverter
This stage converts the rectified DC into AC using a power oscillator. The
power oscillator has a small output transformer with a few windings at the
frequency 20-100 kHz. Switching is controlled by a MOSFET amplifier. The
output AC voltage is usually isolated optically from the input AC by using
an Optocoupler IC for safety reasons.
Input
Input rectifier
and filter
Inverter
Output transformer
Output rectifier
and filter
Chopper/controller
6.
(i) Explain the working of FWR with CLC filter and derive for its
ripple filter. (12) (ii) Compare HWR and FWR with respect to output
average voltage and ripple factor. (4) (NOV/DEC 2010)
LC Filter: - The ripple factor is directly proportional to the load resistance RL in
the inductor filter and inversely proportional to RL in the capacitor
filter. Therefore if these two filters are combined as LC filter or L section filter as
shown in figure the ripple factor will be independent of RL.
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If the value of inductance is increased it will increase the time of conduction. At
some critical value of inductance, one diode, either D1 or D2 will always
conducting.
From Fourier series, the output voltage can be expressed as
The dc output voltage,
The ripple factor
CLC or p Filter
The above figure shows CLC or p type filter, which basically consists of a
capacitor filter, followed by LC section. This filter offers a fairly
smooth output and is characterized by highly peaked diode currents and poor
regulation. As in L section filter the analysis is obtained as follows.
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Comparison between HWR and FWR
Parameter
HWR
FWR
No. of Diodes
1
2
Need of Centre tapping
No
Yes
Average dc voltage Vdc
Vm/π
2Vm/π
Average d.c. current Idc
Im/π
2Im/π
Ripple Factor γ
1.21
0.48
Maximum Rectification
40.6 %
81.2 %
Form Factor
1.57
1.11
T.U.F
0.287
0.693
in secondary of the
transformer winding
Efficiency η
7. Explain how Zener diode acts as a regulator.
As long as the input voltage is a few volts more than the desired output voltage,
the voltage across the zener diode will be stable.As the input voltage increases
the current through the Zener diode increases but the voltage drop remains
constant - a feature of zener diodes. Therefore since the current in the circuit has
increased the voltage drop across the resistor increases by an amount equal to
the difference between the input voltage and the zener voltage of the diode. the
current flowing in the diode is determined using Ohm's law and the known voltage
drop across the resistor R;
IDiode = (UIN - UOUT) / RΩ
The value of R must satisfy two conditions :
1.
R must be small enough that the current through D keeps D in reverse
breakdown. The value of this current is given in the data sheet for D. For
example, the common BZX79C5V6[7] device, a 5.6 V 0.5 W zener diode, has a
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recommended reverse current of 5 mA. If insufficient current exists through D,
then UOUT is unregulated and less than the nominal breakdown voltage (this
differs to voltage regulator tubes where the output voltage will be higher than
nominal and could rise as high as UIN). When calculating R, allowance must be
made for any current through the external load, not shown in this diagram,
connected across UOUT.
2.
R must be large enough that the current through D does not destroy the
device. If the current through D is ID, its breakdown voltage VB and its maximum
power dissipationPMAX correlate as such:
.
A load may be placed across the diode in this reference circuit, and as long as
the zener stays in reverse breakdown, the diode provides a stable voltage
source to the load. Zener diodes in this configuration are often used as stable
references for more advanced voltage regulator circuits.
Shunt regulators are simple, but the requirements that the ballast resistor be
small enough to avoid excessive voltage drop during worst-case operation (low
input voltage concurrent with high load current) tends to leave a lot of current
flowing in the diode much of the time, making for a fairly wasteful regulator with
high quiescent power dissipation, only suitable for smaller loads.
These devices are also encountered, typically in series with a base-emitter
junction, in transistor stages where selective choice of a device centered around
the avalanche or zener point can be used to introduce compensating
temperature co-efficient balancing of the transistor PN junction. An example of
this kind of use would be a DC error amplifier used in aregulated power
supply circuit feedback loop system.
Zener diodes are also used in surge protectors to limit transient voltage spikes.
Another notable application of the zener diode is the use of noise caused by
its avalanche breakdown in a random number generator.
8. Explain in detail about Bridge rectifier circuit. (16)
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In the circuit diagram, 4 diodes are arranged in the form of a bridge. The
transformer secondary is connected to two diametrically opposite points of the
bridge at points A & C. The load resistance RL is connected to bridge through
points B and D.
During the first half cycle
During first half cycle of the input voltage, the upper end of the transformer
secondary winding is positive with respect to the lower end. Thus during the first
half cycle diodes D1 and D3 are forward biased and current flows through arm
AB, enters the load resistance RL, and returns back flowing through arm DC.
During this half of each input cycle, the diodes D2 and D4 are reverse biased and
current is not allowed to flow in arms AD and BC.
During the second half cycle
During second half cycle of the input voltage, the lower end of the transformer
secondary winding is positive with respect to the upper end. Thus diodes D2 and
D4 become forward biased and current flows through arm CB, enters the load
resistance RL, and returns back to the source flowing through arm DA. Flow of
current has been shown by dotted arrows in the figure. Thus the direction of flow
of current through the load resistance RL remains the same during both half
cycles of the input supply voltage.
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