Basic Electronics (Rectifiers) by Er. Swapnil Kaware

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*FEL NOTES BY Er. SWAPNIL KAWARE (svkaware@yahoo.co.in)*
Chapter No. 2. Power Supply.
Q(1). Draw & explain the circuit diagram of Half Wave Rectifier/with resistive load?.
In half wave rectifier conduction takes place during one half cycle of the supply, the AC voltage
to be rectified is connected to primary coil of the transfer. The P region of diode is connected to
one end of the secondary coil of the transformer while N region of the diode is connected to
load resistance (RL). Also one end of load resistance (RL) is connected to other end of secondary
coil of transfer. The AC voltage which is applied to the secondary coil of the transformer is in
series with load resistance (RL).
Q(2). Explain working of Half Wave Rectifier & also draw its waveforms?.
(a).During the positive alternation of the input voltage, the positive alternation of the sine
wave causes the anode of the diode to become positive with respect to the cathode. The diode is
now forward-biased and hence it will conduct. Current will flow from the negative side of the
transformer secondary through the load resistor through the diode to the positive side of the
transformer secondary. This path for current flow will exist during the complete positive
alternation of the input waveform because the diode will remain forward biased. Hence When
Vin is greater than 0V, diode shorts, so Vout = Vin.

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(b). During the negative alternation of the input voltage, the anode is negative with respect
to the cathode and hence diode will become reverse-biased. Hence however, when Vin is less
than 0, diode opens i.e disconnected , so no current flows through RL, hence Vout = ILRL = 0.
(i.e. value of load resistance & load current is zero).
Q(3). Define Ripple Factor, Efficiency (% regulation) & TUF (Transformer Utilization
Factor)?.
(a). Ripple Factor: It is defined as the ratio of R.M.S value of all AC components to the Average
value of waveform.
(b). Efficiency (% Regulation): It is defined as the percentage change in no load voltage to the
full load voltage.
(c). TUF: It is defined as the ratio of DC output power to volt ampere (VA) ratings of the
transformer.
Q(4). Draw the circuit diagram of Centre-Tap-Full Wave Rectifier & also give its working
principle with required waveforms?.
As seen from the above diagram. The full wave rectifier consists of step down transformer T1,
Resistive load (RL), two diodes which are connected to A & B ends of secondary transformer
coil. The centre tap C of the secondary coil is connected to N region of the diode through load
resistance (RL). Full-wave rectification converts both polarities of the input waveform to DC
(direct current), and yields a higher mean output voltage. In this value of the load current in
both half cycles is positive. The output voltage is available in both half cycles.

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(a). During positive half cycle of AC input voltage, according to polarities connected, diode D1
becomes forward biased & Hence due to this load current IL flows in the circuit. But as diode
D2 is in reverse biased, so load current will not conducts hence load current becomes zero. i.e.
IL=0.
(b). During Negative half cycle of AC input voltage, according to polarities connected, diode
D2 becomes forward biased & Hence due to this load current IL flows in the circuit. But as
diode D1 is in reverse biased, so load current will not conducts hence load current becomes
zero. i.e. IL=0.
Q(5). Draw the circuit diagram of Full Wave Bridge Rectifier also state its working
principle & Draw its required waveforms?.

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The Bridge rectifier circuit is shown in the above figure. The circuit has four diodes connected
to form a bridge. The AC input voltage is applied to the diagonally opposite ends of the bridge.
The load resistance is connected between the other two ends of the bridge.
(a). During positive half of the input ac voltage, terminal A is positive with respect to terminal
C hence diodes D1 and D2 conduct i.e. forward biased, hence load current flows in the direction
path as S1-A-B-RL-D-C-S2. During this interval diodes D3 and D4 remain in the OFF state i.e.
reverse biased. The conducting diodes will be in series with the load resistance RL and hence
the load current flows through RL.
(b). During Negative half of the input ac voltage, terminal C is positive with respect to
terminal A. Hence diodes D3 and D4 conduct i.e. forward biased, hence load current flows in the
direction path as S2-C-B-RL-D-A-S1. During these interval diodes D1 and D2 remain OFF i.e.
reverse biased. The conducting diodes D3 and D4 will be in series with the load resistance RL
and hence the current flows through RL in the same direction as in the previous half cycle. Thus
a bi-directional wave is converted into a unidirectional wave.
Q(6). Compare/write differences between Half Wave Rectifier & Full Wave Rectifier &
Bridge Rectifier?.
Half Wave Rectifier Full Wave Rectifier Bridge Rectifier
(1). Number of diodes
required is 1.
required is 2.
required is 4.
(2). Ripple factor is poor. (2). Ripple factor is better. (2). Ripple factor is better.
(3). Ripple frequency is 50Hz. (3). Ripple frequency is
100Hz.
(3). Ripple frequency is
100Hz.
(4). Center tap transformer is
not required.
(4). Center tap transformer is
required.
(4).Center tap transformer is
not required.
(5). Efficiency is good. (5). Efficiency is better. (5).Efficiency is better.
(6). Circuit diagram of HWR. (6). Circuit diagram of FWR. (6). Circuit diagram of BR.
(7). Waveform of HWR. (7). Waveform of FWR. (7). Waveform of BR.
(8). TUF is 28.7% (8). TUF is 69.2% (8). TUF is 81.2%
(9). Average output
voltage=Vm/ᴨ
(9). Average output
voltage=2Vm/ᴨ
(9). Average output
voltage=2Vm/ᴨ
(10). R.M.S. output
voltage=Vm/2
(10). R.M.S. output
voltage=Vm/√2
(10). R.M.S. output
voltage=Vm/√2
(11). Average load
current=Im/ ᴨ
(11). Average load
current=2Im/ ᴨ
(11). Average load
current=2Im/ ᴨ
(12). R.M.S. load
current=Im/2
(12). R.M.S. load
current=Im/√2
(12). R.M.S. load
current=Im/√2
Q(7). Why Filter circuits are necessary?
In circuit theory, a filter is an electrical network that alters the amplitude and/or phase
characteristics of a signal with respect to frequency. Filter circuit’s gives ripple free DC output
which also removes AC variations. It produces smoothened or filtered output. Hence so as to
receive such output we required filter circuits.

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Q(8).List/Give the types of Filters?.
(a). LC filter, (b).RC filter, (c). Inductive filter, (d). Capacitor filter, (e). Pi(ᴨ) type filter.
Q(9).Draw circuit diagram of Half Wave Rectifier with filter?.
Q(10). Explain the operation of Bridge Rectifier with filter circuit or Explain LC type
filter?.
It is found that the ripple factor increases with increase in load resistance (RL). In inductive
filter & in capacitive filter ripple factor decreases with increase in load resistance (RL). A
combination of these two filters makes ripple factor independent of load resistance (RL). The L
type filter is always connected in series with load resistance so as to avoid higher output signals
entering into load & also it reduces ripples present in load current (IL) & this L type filter gives
better results for Higher value of loads & functions well with larger current. Also we know that
inductor opposes the changes in current due to presence of back e.m.f. If we connects inductor
in parallel with load resistance then higher range output signal may damage the load resistance.
The C type filter is always connected in parallel with load resistance & also it reduces ripples
present in load voltage (VL) & this C type filter gives better results for lower value of loads.
Hence we have used LC filter for better ripple free outputs. When value L increases then
capacitor will be allowed to charge up to Vm & it turns off the diode allowing only short pulses
of current. But when this value of L is greater, then the pulses of current are smoothened &
made to flow for larger durations with reduced amplitude. In above circuit we have connected
bleeder resistance (Rb) so as to provide minimum value current for better voltage regulation &
thus output becomes sharper.

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Q(11). Write difference between ‘L’ type Filter & ‘C’ type Filter?.
‘L’ type filter ‘C’ type filter
(a). Preferred for lower value of load
resistance (RL).
(a). Preferred for higher value of load
resistance (RL).
(b). Connected in series with load. (b). Connected in parallel with load.
(c). Reduces ripples from load current. (c). Reduces ripples from load voltage
(d). Better for heavy loads. (d). Better for lighter loads.
(e). Size is small & compact. (e). Size is larger & extended.
Q(12). Explain the operation of Center Tapped Full Wave Rectifier with filter circuit?.
Please refer to question numbers 4 & 11 for operation/working details.
Q(13). Explain the operation of ᴨ (pi) type filter circuit?.
Q(14). Explain the operation of shunt capacitor circuit?.
Q(15). Explain the operation of series induction circuit?.

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Q(16). Define Voltage Regulator?.
It is the circuit which maintains output voltage constant under all operating conditions. In other
words it is the change in voltage from no load to full load.
Q(17).Draw & explain block diagram of Regulated Power Supply?.
In regulated power it produces DC output voltage which is regulated i.e. good result oriented
signal. As observe from the above block diagram, Whenever we gives AC input supply to the
circuit comprising of transformer & rectifier then it converts that AC waveform (AC input) into
rectified output (waveform), then this rectified output is given to the filter circuit, then filter
circuit converts that output into filtered output, this filtered output is then given to the voltage
regulator circuit for maintaining proper or desirable values of DC output voltages. Hence by this
way in regulated power supply we get regulated output voltage.
Q(18). Define Load Regulation& also give its formula?.
It is defined as the change in output voltage when value of load current changes from no load
(small) to full load (maximum).
Load Regulation= VNL-VFL, & % change in Load Regulation is given as,
%L.R.= ((VNL-VFL)*100))/VFL.
Q(19). Define Line Regulation& also give its formula?.
It is defined as change in load voltage due to change in supply voltage. for e.g. a line regulation
of 3% means that the output voltage with change 3V for every change in input voltage.
Q(20). Draw & explain circuit diagram of Shunt Voltage Regulator?.
As seen from the above block diagram, it is observed that the control element is connected in
parallel with load resistance, so hence we called it as shunt voltage regulator. The output from
the rectifier & filter circuit known as Vin. Then finally the total current (I) which is taken by
shunt control element & load resistance is given as, I=Ish+IL.
Then the sampling network samples the output signal & it gives that sampled output (feedback
signal) to the comparator circuit. Then the main function of the comparator circuit is to
compare the feedback signal with constant reference voltage & then after comparing, it

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produces control signal which is then given to the shunt control element. Then this shunt
control element produces constant output voltage by changing its shunt current (Ish). Hence by
this way value of load current (IL) also decreases & output voltage is maintained in regulated
range. Hence if load voltage decreases then it will take exactly opposite action to produce
regulated output voltage.
Q(21). Give applications of Shunt Voltage Regulator?.
(a). In emitter follower regulator,
(a). In low load voltage regulators,
Q(22). Draw &explain circuit diagram of Series Voltage Regulator?.
As seen from the above block diagram, it is observed that the control element is connected in
series with load resistance, so hence we called it as series voltage regulator. The required
output voltage is given by, Vo=Vin-Vs. (Vs is the voltage drop across control element). Then the
sampling network samples the output signal & it gives that sampled output (feedback signal) to
the comparator circuit. Then the main function of the comparator circuit is to compare the
feedback signal with constant reference voltage & then after comparing it produces control
signal. Which is then given to the series control element. Then this series control element
produces constant output voltage by changing its voltage drop (Vs) across element. As control
element is connected in series with load resistance so hence control element should have high
current rating as well as high power rating. Hence Vs has to be higher as compared to Vo.
Q(23). Give applications of Series Voltage Regulator?.
(a). In 3-pin IC regulator,
(b). In IC-723 regulator,
(a). In high quality voltage regulators,
Q(24). Compare/write differences between Shunt Voltage Regulator & Series Voltage
Regulator.
Shunt Voltage Regulator Series Voltage Regulator
(1). The control element is in parallel with
load.
(1). The control element is in series with
load.
(2). Produces constant output voltage by
changing its shunt current (Ish).
(2). Produces constant output voltage by
changing its voltage drop (Vs).

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(3). It is high voltage low current device. (3). It is high current low voltage device.
(4). Voltage regulation is poor. (4). Voltage regulation is better.
(5). Current equation as, I=Ish+IL. (5). Voltage equation as, Vo=Vin-Vs
(6). Used in fixed output voltage
applications.
(6). Used in variable output voltage
applications.
Q(25). Draw & explain circuit diagram of Zener Diode as a Voltage Regulator?.
It is observed from above figure that, positive terminal of battery is connected to negative
terminal of diode whereas negative terminal of battery is connected to positive terminal of diode,
so hence we can say that zener diode is in reverse bias mode. As long as the Zener diode is
reverse biased, the diode is selected at Vz (zener voltage) equal to the desired voltage across the
load. Here the zener diode acts as a bypass valve through which more current can pass when an
increase in input voltage or decreases in load current occurs by maintaining the voltage at the
output nearly constant at Vz. The input voltage Vin is the unregulated output of regulator. It can
also be observed from the above diagram that, it satisfies the following equations.
i.e. I=Iz+IL,
it will be seen under all conditions that,
Vout=Vz,
Hence Vin=IRs+Vout =IRs+Vz,
Vin=(Iz+IL)Rs+Vz,
The value of Rs can be calculated using the above relation as below,
Vout(or Vz)=Vin-(IL+IZ)Rs,
Rs=Vin-Vout/(IL+Iz).
Q(26). Explain IC Voltage Regulators & state its advantages with examples?.
It is a three terminal device in which we have to set a predetermined output. This IC is having
terminals as, (1). Input, (2). Output, (3). Ground.
Advantages:-
(a). Easily available, (b). Simple connections to external circuit, (c). External components
required are less, (d). Cheaper in cost, (e). Consumes low power, (f). Output current range
required is 100mA to 3A.
IC
number
7805 7806 7808 7812 7815 7818 7824
Output
voltages
+5V +6V +8V +12V +15V +18V +24V

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The maximum output voltage is 40V for +24V output voltage & it is 35V for remaining output
voltages.
IC
number
7905 7906 7908 7912 7915 7918 7924
Output
voltages
-5V -6V -8V -12V -15V -18V -24V
e.g. IC 78XX is a series of fixed positive voltage regulators & IC 79XX is a series of fixed negative
voltage regulators.
Q(27). Define Drop Out Voltage?.
It is defined as the difference between the unregulated voltage (Vin) & regulated voltage (Vout).
It is the voltage which is required for proper operation of regulator. It should be minimum 2V.
It can be calculated as, Drop Out Voltage=Vin-Vout.
Q(28). Explain 78XX series IC?.
In above circuit diagram, we have used IC 7812 which generates +12V output voltage.
From the circuit diagram we can observe that the input from the bridge rectifier & capacitive
filter is connected to the IC 7812. As we know the minimum drop out voltage should be 2V,
hence because of IC 7812 the input voltage should be 14V. In above circuit C1 is the filter
capacitor which is connected between input terminal & ground. Here C1 capacitor is used to
cancel out further inductive effect due to long distribution loads. While C2 is connected at the
output terminal to improve the response of regulator at changing load conditions. This
capacitor also helpful for removing noise level (ripples) from the output.
Q(29). Explain 79XX series IC?.
In above circuit diagram we have used IC 7912 which generates -12V output voltage.
From the circuit diagram we can observe that the input from the bridge rectifier & capacitive
filter is connected to the IC 7912. As we know the minimum drop out voltage should be 2V,

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hence because of IC 7912 the input voltage should be -10V. In above circuit C1 is the filter
capacitor which is connected between input terminal & ground. Here C1 capacitor is used to
cancel out further inductive effect due to long distribution loads. While C2 is connected at the
output terminal to improve the response of regulator at changing load conditions. This
capacitor also helpful for removing noise level (ripples) from the output.
Q(30). Draw pin diagram for IC 723?.
Q(31). Explain working of IC 723 as a low voltage & low current regulator?.
The required block diagram for the low voltage regulator using IC 723 is as shown in above
figure. The above circuit gives the value of load voltage between 2V to 7V & gives the value of
load current up to 150mA. R1 & R2 forms a voltage divider between Vref. & ground. The voltage
across is connected to non-inverting amplifier. Hence voltage across non inverting terminal is
given by,
V(N.I.)=(R2/(R1+R2))*Vref.
As we know gain of the internal error amplifier is large,
Hence, VN.I.=VI.
Hence the output voltage (Vo) which is connected to the inverting terminal through current
sensing resistor(Rsc) & R3 Must be equal to V(N.I.)
So, Vo=V(N.I.)=(R2/(R1+R2))*Vref.
Also, R3=R1||R2=(R1*R2)/(R1+R2).
The current sensing resistor is connected between current sense(CS) pin & current limit(CL)
pins which prevents IC from further damaging.
Q(32). Draw & explain block diagram of Switched Mode Power Supply (SMPS)?.
+VCC (Unregulated DC supply)

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A switched-mode power supply (SMPS) is an electronic power supply that incorporates a
switching regulator to convert electrical power efficiently. An SMPS is usually employed to
efficiently provide a regulated output voltage, typically at a level different from the input
voltage.
As seen from the above block diagram, the main function of pulse generator is to provide
rectangular pulses to the control terminal of electronic switch. It can be either transistor or
MOSFET. This switch can be turned ON & OFF by using the rectangular pulses. When this
electronics switch is ‘ON’ then this switch connects unregulated DC input (Vin) to the filter &
When this electronics switch is ‘OFF’ then this switch disconnects unregulated DC input (Vin) to
the filter. Hence we get rectangular waveform at the input of filter. The average value of the
waveform can be adjusted by changing duty cycle or frequency of the rectangular pulses from
the pulse generator. The filter circuit is helpful for removing ripples from DC output voltage &
thus makes DC output as smoother.
Duty Cycle(D)= tON/(tON+tOFF) (since T=tON+tOFF).
Where tON= ON time pulse of waveform,
tOFF= OFF time pulse of waveform,
So, T=1/F.
Hence it satisfies the following equation,
Vo=(tOn/T)*Vin. Or Vo=D*Vin.
Hence we can say that the value of the average output voltage is depends on duty cycle (D).
Q(33). State advantages/applications of SMPS?.
(a). Higher power handling applications.(b).Personal computers, (c). Printers, (d). Cable
networks, (e). TV receivers. (f). Battery chargers.
Q(34). What is UPS (Uninterruptable Power Supply)?.
The main function of UPS system is to handle all problems in mains supply, including complete
power failure for time intervals greater than half hour. It uses a battery for energy storage. It
also supplies error free signal to the load. It makes normal user independent of mains supply.
Mainly there are two types of UPS as follows,
(a). ON line UPS (Inverter preferred/Continuous UPS system).
(a). OFF line UPS (Line preferred/Transfer UPS system).
Q(35). Explain ON line UPS (Inverter preferred/Continuous UPS system)?.
Online UPS on the other hand uses an Inverter which always gives sine wave. The incoming AC
signal is first converted into DC signal by a transformer to charge the battery as well as to give
power to the inverter transformer. The inverter transformer converts the DC to AC

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continuously for giving power to the load. If power fails, the battery backup circuit switches on
and takes the load. The advantage of the Online UPS is that, it cleans up the AC waveform by
converting it into DC then reconverting this DC to fresh AC.
In this the load is always connected to the inverter through the UPS static switch. This static
switch contains anti parallel SCR’s so hence conduction through static switch is bidirectional.
The UPS static switch is normally maintained in ‘ON’ mode it only turned off only when the UPS
system fails. In above circuit, we have used the mains static switch to connect AC mains directly
to the load. This mains switch is usually keep in ‘OFF’ mode for bypassing the UPS.
The main operation/working of online UPS is as explained below in two modes.
(a). Mains switch ‘ON’:- when this switch is ON then rectifier provides power to the inverter as
well as battery. So, hence this circuit acts as a rectifier cum charger. The output from the
inverter is connected to the load via UPS static switch. Hence battery will be charged in this
mode. The required circuit diagram is shown below.
(b). Mains switch ‘OFF’:- In this situation the rectifier output becomes zero. Hence we use the
battery bank for supplying power to the inverter. The inverter then delivers the load through
UPS static switch. This transfer from rectifier to battery bank is instantaneous hence, during
this transfer period there is no power interruption. The required circuit diagram is shown
below.
Q(36). Explain OFF line UPS (Line preferred/Transfer UPS system)?.
As seen from the below diagram, The mains static switch is normally ‘ON’ mode switch. It
connects AC mains directly to the load. The main function of battery is to charge the charger. In
this the UPS static switch is normally ‘OFF’. It is closed only when the mains fails. Thus in offline
UPS the concept of inverter comes into action only when mains fails. Here we have used the

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rectifier/charger to charge the battery bank. Hence size & power ratting of OFF line UPS is
smaller as compared to ON line UPS.
The main operation/working of online UPS is as explained below in two modes.
(A). Mains switch ‘ON’:- When this switch is ‘ON’ the power is supplied to the load through
mains static switch. Hence the battery gets charged through charger.
(A). Mains switch ‘OFF’:- In this situation the mains static switch gets disconnected i.e. open
circuited. The UPS static switch gets connected/closed automatically. Hence battery charging
will gets stop & thus battery starts supplying power through inverter circuit. So hence in this
mode the total time taken to identify the power failure is about 5msec.
Hence by this way the OFF line UPS works.
Q(37). Compare ON line & OFF line UPS system?.
ON line UPS system OFF line UPS system
(A). Efficiency is poor. (A). Efficiency is better.
(b). Transfer time is 0sec. (b). Transfer time is 5msec.
(c). It is costlier. (C). It is cheaper.
(d). Bigger in size & weight. (d). Smaller in size & weight.
(e). Operates continuously. (e). Operates only when the mains fails.
(f). Rating of rectifier/charger is higher. (f). Rating of rectifier/charger is lower.
(e). Handles heavy loads. (e). Handles smaller loads.
Q(38). State advantages/applications of UPS system?.
(a). As a Power supply device.
(b). In computer systems.
(c). In communications systems.
(d). In process control systems.
Q(39). Explain the concept of constant current limiting?.
Current limiting is the practice in electrical or electronic circuits of imposing an upper limit on
the current that may be delivered to a load with the purpose of protecting the circuit generating
or transmitting the current from harmful effects due to a short-circuit or similar problem in
the load. As seen from the below figure the Rsc also known short circuit protection resistance
is connected in series with the load. Therefore the through Rsc is equal to the load current (IL).

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In given circuit there are two transistors are present as Q1 as a series pass transistor & Q2 as a
additional transistor for diverting base current.
(a). When the load current through Rsc is less than the predetermined value. Then voltage drop
across Rsc will be less than 0.6V. i.e VBE terminal of Q2 transistor. Hence due to this Q2
transistor becomes ‘OFF’ & there will be no further operation/working of circuit is possible.
(b). When the load current through Rsc is greater than the predetermined value. Then voltage
drop across Rsc will be sufficient i.e. higher than 0.6V. Hence it forward biases the VBE terminal
of Q2 transistor. Hence due to this Q2 transistor becomes ‘ON’. As Q2 transistor is ‘ON’ then
collector current starts flowing, this provides a bypass path for the base current of Q1
transistor. As the base current of Q1 reduces, the load current also reduces. Hence as long as
transistor Q2 conducting VBE remains almost constant. This will keep the voltage Rsc constant.
The final value of the load current is given by,
Isc =VBE2/Rsc=0.6/Rsc,
As value of IL remains constant hence the above concept is known as constant current limiting.
Q(40). Explain concept of foldback current limiting in regulators?.
The following circuit actually decreases the output current while working in overloading
conditions. The main advantage of foldback current circuit is that it reduces the power
dissipation happening in series pass transistor Q1 under short circuit condition. Foldback is
a current limiting feature (a type of overload protection) of power supplies and power
amplifiers. When overcurrent is drawn by the load, foldback reduces both the
output voltage and current to well below the normal operating limits. Under a short circuit,
where the output voltage has reduced to zero, the current is typically limited to a small fraction
of the maximum current.
For operation/ working principle, please refer the question no. 39.
*************************************** ALL THE BEST **********************************

Basic Electronics (Rectifiers) by Er. Swapnil Kaware

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