Chap 2 basic electronics gp 1

A DC Power Supply
A typical Power supply unit consists of the following.
•Transformer − An input transformer for the stepping down of the 230v AC power supply.
•Rectifier − A Rectifier circuit to convert the AC components present in the signal to DC
components.
•Smoothing − A filtering circuit to smoothen the variations present in the rectified output.
•Regulator − A voltage regulator circuit in order to control the voltage to a desired output
level.
•Load − The load which uses the pure dc output from the regulated output.

Rectification
• Rectification is the process of converting the alternating voltage or
current into corresponding direct (DC) voltage or current.
• Need of Rectifier :
• An alternating current has the property to change its state continuously.
• Every electronic circuit such as amplifiers needs a dc power for its operation.
• This dc voltage has been obtained from the ac supply.
• For this the ac supply voltage has to be reduced (stepped down) first using a
step down transformer and then converted to dc by using rectifier.
• Rectifier :
• It is an electronic device which is used for converting an alternating voltage or
current into corresponding direct (DC) voltage or current.
• The input to a rectifier is ac whereas its output is unidirectional or dc.

Half Wave Rectifier (HWR)
• In half wave rectifier the rectifier is ON only during one half cycle of the
ac supply.
• So output is produced only in that half cycle. The output is suppressed
in the other half cycle.
• The conduction takes place only in one half cycle of supply. Hence the
name is Half Wave Rectifier (HWR).

Operation of Half Wave Rectifier
• Operation in the positive half cycle of AC supply (0-∏)
• In the positive half cycle (0-∏) of the ac supply the secondary voltage VAB is positive
i.e. A is positive with respect to B. Hence, the diode is forward biased and starts
conducting.
• As the diode starts conducting the secondary voltage VAB appears almost as it is
across the load resistance.
• The load voltage is thus positive and almost equal to the instantaneous secondary
voltage VAB.
• The load current has the same shape as that of the load voltage since the load is
purley resistive.

• Operation in the Negative half cycle
• In the negative half cycle of ac supply (∏ to 2∏) secondary voltage VAB is
negative i.e. A is negative with respect to B.
• Hence, the diode is reversing biased and offers a very high resistance. Hence,
we can replace it by open circuited switch.
• The load is disconnected from the secondary. Hence, the load voltage and
load current both are zero and the voltage across the diode is equal to the
instantaneous secondary voltage VAB.
• The circuit is called as half ware rectifier because it deliver power to the load
during only one half cycle of the ac supply voltage.
Operation of Half Wave Rectifier

Waveform For the Half Wave Rectifier

Advantages of Half Wave Rectifier
• Simple Construction
• Less number of Components are required
• Small size

Disadvantages of Half Wave Rectifier
• More amount of ripple content
• Transformer utilization factor is very low
• Rectification efficiency is low
• Generates harmonics
• Low output voltage or current

Applications of Half wave rectifier
• It is used for applications in which a constant DC voltage is not very
essential, we can use power supplies with half wave rectifier.

Full Wave Bridge Rectifier
• The circuit consists of four
diodes connected to form a
bridge.
• Bridge rectifier offers full wave
rectification. The diode
conducts in pairs i.e. at any
given instant of time one pair of
diode either D1 D2 or D3 D4
will be conducting.

Operation in positive half cycle
• In the positive half cycle of AC
supply the secondary voltage VAB
is positive.
• Therefore diodes D1 and D2 are
forward biased where as D3 and
D4 are reverse biased.
• The reverse biased diodes D3
and D4 acts as open switches.
• The load current and load
voltage both are positive.

Operation in Negative half cycle
• In the negative half cycle of ac
supply the secondary voltage VAB
becomes negative diode D3 and
D4 are forward biased and starts
conducting.
• D1 and D2 are reverse biased
hence do not conduct.

Waveform of the Full Wave Bridge Rectifier

Advantages of Bridge Rectifier
• It requires a small size transformer. Centre top transformer is not
required. This makes the bridge rectifier cost effective.
• The input transformer is not a must. It is possible to operate the bridge
rectifier directly on the 230 V ac supply.
• High average output voltage.
• Rectifier efficiency η is high.
• Transformer utilization factor TUF is high.
• This circuit is most suitable for high voltage application.
• Core saturation does not take place. Therefore transformer losses are
reduced.
• The PIV is only Vm volt which is half the PIV of FWR with center tap.

Disadvantages of Bridge Rectifier
• The number of diodes used is four instead of two for Full wave
rectifier.
• As two diodes conducts simultaneously, the voltage drop increases
and the output voltage reduces.

Application of Bridge Rectifier
• Laboratory dc power supplies.
• High current power supplies.
• Battery charger.
• Dc power supplies for various electronic circuits.

Full wave rectifier with center tapped transformer
• A rectifier circuit whose secondary
winding of transformer is tapped to get
the desired output voltage, using two
diodes alternatively, so as to rectify the
complete cycle is called as a Center-
tapped Full wave rectifier circuit.
• The full wave rectifier consists of a step
down center tapped transformer , two
diodes and a purely resistive load RL.
• In HWR the load current flows only for
one half cycles and in FWR the load
current flows for both the half cycles of
ac supply.
• The induced voltage in the two halves of
the secondary winding are always 180
out of phase with respect to each other.

Operation of Full wave rectifier with center tapped
transformer • Operation in the positive half cycle (0-∏)
• In the positive half cycle of ac supply the
polarities of the secondary induced voltages
are shows that VAO is positive and VBO is
negative.
• Due to the center tapped secondary, VAO and
VBO are always equal and opposite to each
other.
• Hence the diode is D1 forward biased and D2
is reverse biased. The load current starts
flowing from A through D1, load resistance RL
back to point O.
• The instantaneous load voltage is positive and
equal to VAO.
• AS the load is purely resistive , the load
current has the same shape as that of the
load voltage.

Operation of Full wave rectifier with center tapped
transformer • Operation in the negative half cycle (0-∏)
• In the negative half cycle of ac supply the
polarities of the secondary induced
voltages are shows that VAO negative is
and VBO is positive.
• Hence D1 is reverse biased and D2 is
forward biased so D1 acts as open
circuited and D2 carries the entire load
current.
• The direction of load current is same as
that in the positive half cycle. That means
even in the negative half cycle, the load
current continues to be positive.
• he instantaneous load voltage VL is positive
and almost equal to VBO .

• Load voltage and load current both are positive in both the half cycles of
the ac supply.
• Output voltage is available in both the half cycles of the ac supply.
• The full wave rectifier circuit consists of two half wave rectifier which
works independently and feed the common load.

Advantages of Full wave rectifier with center
tapped transformer
• Low ripple factor as compared to Half wave rectifier .
• Better rectification efficiency.
• Better transformer utilization factor.
• Higher values of average load and average load current.
• No possibility of transformer core saturation because transformer
current flows equally in both the half cycles.

Disadvantages of Full wave rectifier with
center tapped transformer
• Since PIV of the diodes is 2Vm, size of the diodes is larger and they
are most costly.
• Cost of the center tapped transformer is high.

Applications of Full wave rectifier with center
tapped transformer
• Laboratory power supply
• High current power supplies
• Battery charges.
• Power supplies for various electronic circuits.

Analysis of Half Wave Rectifier
• Average load current (IL dc)
• Average load voltage (VL dc)
• RMS load current (IL rms )
• RMS load voltage (VL rms)
• Ripple factor
• Rectification efficiency and Transformer Utilization Factor(TUF)

DC or Average load current (IL dc)
• The average value of periodic function is given by the area under one cycle
of the function divided by the period.
• From the diagram in the figure we have to consider one complete cycle of
this waveform from ωt = 0 to ωt = 2Π

• The average current IL dc is given by :
IL dc = Iavg =
1
2𝜋 0
2𝜋
Im 𝑠𝑖𝑛𝜔𝑡 𝑑𝜔𝑡
Where Im is given by ;
Im =
Vm
RS + RF + RL
It is Peak Load Current
RF = Diode forward resistance
RS = Transformer Secondary resistance
Vm = Maximum or Peak secondary voltage

DC or Average load voltage (VL dc)
• As the load is purely resistive the average load voltage is given by ;
VL dc = ILdc * RL
By substituting the value of ILdc
VL dc =
Im
𝜋
∗ RL
=
Vm
𝜋 (RS + RF + RL )
* RL
As RS and RF are very small as compare to RL
RS + RF + RL = RL thus
VL dc =
Vm
𝜋
= 0.318Vm
Where ;

AC or RMS load current (IL rms )
• From the load current waveform , consider one complete cycle of load
current waveform (0 - 2 𝜋 )
IL rms =
Im
2
This is the rms or effective value of the load current

AC or RMS load voltage (VL rms)
• As the load is purely resistive the rms value of load voltage is given by ;
VL rms = IL rms * RL
=
Im
2
* RL
=
Vm
2 (RS +RF +RL)
* RL
But (Rs + RF ) <<< RL
Thus VL rms =
Vm
2

Ripple Factor (RF)
• The rectifier outputs consist of Ac as well as Dc component the ripple
factor measures percentage of Ac component in the rectifier output.
• The ripple factor indicates how close the rectified output is to the
pure ideal dc voltage waveform. So it is a fig. of merit for the rectifier.
• Small values if ripple factor indicates that the rectifier output
waveform is close to being pure Dc.
• The ideal value of RF should be zero and practically it should be as
small as possible.

Ripple Factor (RF)
Therefore
ϒ=
(VL rms ) 2−(V L dc ) 2
VL dc
ϒ=
(Vm
2
)2−(Vm
𝜋
)2
(Vm
𝜋
)
ϒ= 1.21 OR 121%
• From the above equation we can say that the ripple content in the output voltage
is 1.21 times the dc components .
• This is very high value of ripple factor which indicates that the output of Half
wave rectifier is not close to the pure dc voltage .
• A filter is required to reduce the ripple factor.

Rectifier Efficiency
• Any circuit needs to be efficient in its working for a better output. To
calculate the efficiency of a half wave rectifier, the ratio of the output
power to the input power has to be considered.
• The rectifier efficiency is defined as;
η =
PL dc
Pac

η =
PL dc
Pac
By substituting the values ;
η =
(ILdc)2 ∗ RL
(Isrms)2 ∗ RS +RF+RL
η = 0.4 or 40%
• So the rectifier efficiency of a rectifier indicates the percentage of ac input
power actually converted into the average load power .
• It should be as high as possible
• Ideally it should be 100% .
• But in case of Half wave rectifier it is very small .

Transformer Utilization Factor (TUF)
• Transformer Utilization Factor indicates how well the input
transformer is being utilized.
• So, the transformer utilization factor is defined as :
TUF =0.287 OR 28.7%
• Ideal value of TUF should be 100% and practically it should be as high
as possible .
• In case of half wave rectifier the value of TUF is too low .

Analysis of Full Wave Rectifier with center tapped
transformer and Full wave Bridge rectifier
• Average load current (IL dc)
• Average load voltage (VL dc)
• RMS load current (IL rms )
• RMS load voltage (VL rms)
• Ripple factor
• Voltage Regulation
• Rectification efficiency and Transformer Utilization Factor(TUF)

• The average value of periodic function is given by the area under one cycle
of the function divided by the period.
• From the diagram in the figure we have to consider one complete cycle of
this waveform from ωt = 0 to ωt = Π

• The average current IL dc is given by :
IL dc = Iavg =
1
𝜋 0
𝜋
Im 𝑠𝑖𝑛𝜔𝑡 𝑑𝜔𝑡
IL dc =
2Im
𝜋
= 0.636 Im
This indicates that the average load current is twice than half wave rectifier
• Where Im for Full Wave Rectifier with center tapped transformer is given by ;
Im =
Vm
RS +RF+ RL
• And Im for Full wave Bridge rectifier
Im =
Vm
RS +2 RF+ RL
• It is Peak Load Current
• RF = Diode forward resistance
• RS = Transformer Secondary resistance
• Vm = Maximum or Peak secondary voltage

DC or Average load voltage (VL dc)
• As the load is purely resistive the average load voltage is given by ;
VL dc = ILdc * RL
By substituting the value of ILdc
VL dc =
2Im
𝜋
∗ RL
As RS and RF are very small as compare to RL
RS + RF + RL = RL thus
VL dc =
2 Vm
𝜋
= 0.636 Vm
Where ;

AC or RMS load current (IL rms )
• From the load current waveform , consider one complete cycle of load
current waveform (0 - 𝜋 )
IL rms =
Im
√2
This is the rms or effective value of the load current

AC or RMS load voltage (VL rms)
• As the load is purely resistive the rms value of load voltage is given by ;
VL rms = IL rms * RL
=
Im
√2
* RL
=
Vm
√2(RS +RF +RL)
* RL
But (Rs + RF ) <<< RL
Thus VL rms =
Vm
√2

Ripple Factor (RF)
Therefore
ϒ=
(VL rms ) 2−(VL dc ) 2
VL dc
ϒ=
(Vm
√2
)2−(2Vm
𝜋
) 2
(2Vm
𝜋
)
ϒ=0.48 OR 48%
• This value is less than the ripple factor of half wave rectifier which is 1.21
• Thus quality of dc voltage of full wave rectifier is much better than half
wave rectifier .

• Any circuit needs to be efficient in its working for a better output. To
calculate the efficiency of a full wave rectifier, the ratio of the output
power to the input power has to be considered.
• The rectifier efficiency is defined as;
η =
PL dc
Pac

η =
PL dc
Pac
By substituting the values ;
η =
(ILdc)2 ∗RL
(Isrms)2 ∗ RS +RF+RL
η = 0.812 OR 81.2%
• So the rectifier efficiency of a rectifier indicates the percentage of ac input
power actually converted into the average load power .
• It should be as high as possible
• Ideally it should be 100% .
• Rectifier efficiency of full wave rectifier is almost twice the rectifier
efficiency of half wave rectifier.

Transformer Utilization Factor (TUF)
• Transformer Utilization Factor indicates how well the input
transformer is being utilized.
• So, the transformer utilization factor is defined as :
• TUF for Center tapped full wave rectifier is 0.693 and for Full wave
bridge rectifier is 0.812

Filters
• Filters are the electronics circuits used along with rectifier in order to get a
pure ripple free dc voltage .
• The output of rectifier is pulsating dc signal , but we need pure dc voltage .
• In order to get this we use filter circuit at the output of rectifier .
• We can use filter at the output of any type of rectifier such as half wave or
full wave rectifier .
• Filter uses Resistor , Capacitor and Inductor , therefore they are called as
Passive Filters .

Types of Filters
• Filters are classified depending on the components used .
• Some important filters used in the field of electrons are :
• Inductor Filter
• Capacitor Filter
• LC Filter
• Π Filter
• An inductor allows dc and blocks ac.
• A capacitor allows ac and blocks dc.

Inductor Filter
•
• It is also called as “Choke Filter”
• It consist of inductor (L) which is
connected between rectifier output
and load resistance.
• The rectifier output contains ac as well
as dc components .
• When output of rectifier passes
through an inductor (L) it offers very
high resistance to the ac components
and no resistance to the dc
components .

• Thus ac components of the rectifier
output is get blocked and only dc
components reaches to the load .
• The ripple factor of an inductor filter
(for 50 Hz supply ) is given as :
ϒ =
RL
3 2 ω 𝐿
=
RL
1330 𝐿
• From the above equation the ripple
will decrease if the value of RL gets
decrease and L gets increase .
• The inductor filter is more effective
only for heavy load currents that is
when the load resistance is small .
Inductor Filter

• Ideally the output of the inductor
filter should consist of only dc
components but practically it contains
small amount of ac components too.
• These ac components can be reduce
by using large value of inductor .
• But large value of inductor have
higher dc resistance which results in
the lower dc output.
Inductor Filter

Capacitor Filter
• As we know that the inductor filter is suitable only for heavy load .
• Thus an inexpensive filter for light load is available in the form of capacitor
filter .
• It is also called as shunt capacitor filter as it is connected parallel (shunt )
with the load resistance .

• Operation in the interval 0 to A:
• The initial voltage of capacitor is assume
to be zero (0).
• In the first positive half cycle of the ac
input signal D1 is forward biased and
starts conducting .
• D2 is reverse biased and acts as a open
switch .
• Thus diode D1 supplies charging current to
the capacitor and load current .
• Capacitor starts charging through D1and
at the end of the interval that is at point A
it charges to the peak value of secondary
voltage that is Vm.
• Thus at point A , Vc = Vm
• After point A the secondary voltage starts
reducing shown by dotted waveform of
rectifier output .
• This will reverse bias D1 , hence at A D1 is
OFF.

• Operation in the interval A to B :
• During this interval voltage on the
capacitor is higher than the rectifier
output , hence D1 and D2 both
remains OFF.
• The capacitor discharge exponentially
through load resistance (RL ).
• As value of RL is very high than RF , the
capacitor discharge very slowly .
• The discharging time constant is “RL C"
• The value of C is very large so the
discharging time constant is as large as
possible .
• This will reduce ripple in the output.

• Operation in the interval B to C :
• At point B, rectifier voltage is equal to
the voltage on the capacitor and after B
the rectifier output is higher than Vc .
• Thus D2 starts conducting and at the
end of the interval that is at point C ,
the voltage on capacitor is again equal
to Vm .
• After point C, again the rectifier output
get reduce than Vc, thus D2 become
reverse biased and stop conducting.
• The ripple factor of an capacitor filter
(for 50 Hz supply ) is given as :
ϒ =
1
4 3 𝑓 𝐶 RL
=
2890
𝐶 RL

Π Filter
• It is also called CLC or capacitor input
filter .
• It is used in the applications where low
output current and high dc output
voltage is required .
• It consist of two capacitors C1 , C2 and
inductor L1 connected in the form of
greek letter Pi .

• The function of each component is as follows :
• Capacitor C1 :
• It offers low reactance to the ac components of the rectifier output.
• But it offers infinite resistance to the dc components .
• Thus C1 bypass ac components to the ground while dc components move toward L1.
• Inductor L1:
• It offers high reactance to the ac components of the rectifier output.
• But it offers zero resistance to the dc components .
• Therefore it allows dc components to pass through it and blocks ac components .
• Capacitor C2 :
• It behaves similar to C1.
• It bypasses the ac component which could not be blocked by L1.
• Due to which only dc components will be available at the output.
• The ripple factor of Π filter (for 50 Hz supply ) is given as :
ϒ =
1
4 2C1C2L1RL
=
5700
C1C2L1RL
Π Filter

• The process of producing non-sinusoidal output wave forms from sinusoidal
input, using non-linear elements is called as nonlinear wave shaping.
• The Non Linear Wave shaping Circuits consist of linear components such as R,
L, C as well as some Non Linear devices such as Diode , transistor, etc.
• Examples of Non Linear Wave shaping Circuits are Clippers ,Clampers
,rectifiers , etc.
Non Linear Wave shaping Circuits

Clippers
• A Clipper circuit is a circuit that uses diode to “Clip off” or remove
portion of the input signal without distorting the remaining part of
the waveform ..
• Diode Clippers are of Two types :
• Series clippers : Diode is connected in series with load
• Parallel Clipper : Diode is connected in parallel to the load

Series Negative Clipper • The “Clips off” the negative half cycle of the
input .
• It is nothing but the half wave rectifier circuit.
• In the positive half cycle (0-∏) of the ac supply
, the diode is forward biased and starts
conducting.
• As the diode starts conducting it acts as a
close switch and connect load across the
input.
• The load voltage is thus positive and almost
equal to the input voltage.
• In the negative half cycle of ac supply (∏ to
2∏) , the diode is reverse biased and offers a
very high resistance. Hence, it acts as a open
circuited switch.
• The load gets disconnected from the input.
Hence, the load voltage becomes zero .
• That means the negative half cycle gets clips
off or remove at the output.

Series Positive Clipper • The “Clips off” the Positive half cycle of the input .
• In the positive half cycle (0-∏) of the ac supply ,
the diode is reverse biased and offers a very high
resistance. Hence, it acts as a open circuited
switch.
• The load gets disconnected from the input. Hence,
the load voltage becomes zero .
• The load voltage is thus positive and almost equal
to the input voltage.
• In the negative half cycle of ac supply (∏ to 2∏) ,
the diode is forward biased and starts conducting.
• As the diode starts conducting it acts as a close
switch and connect load across the input.
• Thus the voltage across the load resistor will be
equal to the applied input voltage as it completely
appears at the output
• That means the positive half cycle gets clips off or
remove at the output.

Shunt clipper OR Parallel Clipper
• In shunt clipper, the diode is connected in parallel with the output
load resistance. The operating principles of the shunt clipper are
nearly opposite to the series clipper.
• The series clipper passes the input signal to the output load when the
diode is forward biased and blocks the input signal when the diode is
reverse biased.
• The shunt clipper on the other hand passes the input signal to the
output load when the diode is reverse biased and blocks the input
signal when the diode is forward biased.

Shunt positive clipper
• In shunt positive clipper, during the
positive half cycle the diode is forward
biased and hence no output is generated.
• On the other hand, during the negative
half cycle the diode is reverse biased and
hence the entire negative half cycle
appears at the output.

Shunt negative clipper
• In shunt negative clipper, during the positive half cycle the diode is
reverse biased and hence the entire positive half cycle appears at the
output.
• On the other hand, during the negative half cycle the diode is forward
biased and hence no output signal is generated.

Applications of clippers
• Clippers are commonly used in power supplies.
• Used in TV transmitters and Receivers
• They are employed for different wave generation such as square,
rectangular, or trapezoidal waves.
• Series clippers are used as noise limiters in FM transmitters.

Clamper circuits
• A clamper is an electronic circuit that changes the DC level of a signal to
the desired level without changing the shape of the applied signal.
• In other words, the clamper circuit moves the whole signal up or down to
set either the positive peak or negative peak of the signal at the desired
level.
• The dc component is simply added to the input signal or subtracted from
the input signal. A clamper circuit adds the positive dc component to the
input signal to push it to the positive side.
• Similarly, a clamper circuit adds the negative dc component to the input
signal to push it to the negative side.

• If the circuit pushes the signal
upwards then the circuit is said to be
a positive clamper.
• When the signal is pushed upwards,
the negative peak of the signal meets
the zero level.
• if the circuit pushes the signal
downwards then the circuit is said to
be a negative clamper.
• When the signal is pushed
downwards, the positive peak of the
signal meets the zero level.
Clamper circuits

• The construction of the clamper circuit is almost similar to the clipper
circuit.
• The only difference is the clamper circuit contains an extra element
called capacitor.
• A capacitor is used to provide a dc offset (dc level) from the stored
charge.
• A typical clamper is made up of a capacitor, diode, and resistor.
• Some clampers contain an extra element called DC battery.
• The resistors and capacitors are used in the clamper circuit to maintain
an altered DC level at the clamper output.
• The clamper is also referred to as a DC restorer, clamped capacitors, or
AC signal level shifter.
Clamper circuits

Types of clampers
• Clamper circuits are of three types:
• Positive clampers
• Negative clampers
• Biased clampers

Positive clamper
• The positive clamper is made up of
a voltage source Vi, capacitor C,
diode D, and load resistor RL.
• In the circuit diagram, the diode is
connected in parallel with the
output load.
• So the positive clamper passes the
input signal to the output load
when the diode is reverse
biased and blocks the input signal
when the diode is forward biased.

• During negative half cycle :
• In this the diode is forward biased and
acts as a close switch .
• Therefore it allows the current to flow
through it .
• This current will also charge the capacitor
to its maximum value Vm.
• Polarity of charging capacitor is opposite
to the applied voltge
• That is Vc = +Vm
• The capacitor will hold the charge until
the diode become forward biased .
Positive clamper

• During positive half cycle:
• During the positive half cycle of the input AC
signal, the diode is reverse biased and hence the
signal appears at the output.
• In reverse biased condition, the diode does not
allow electric current through it. So the input
current directly flows towards the output.
• When the positive half cycle begins, the diode is
in the non-conducting state and the capacitor is
discharged (released).
• Therefore, the voltage appeared at the output is
equal to the sum of the voltage stored in the
capacitor (Vm) and the input voltage (Vm) .
• I.e. Vo = Vm+ Vm = 2Vm which have the same
polarity with each other.
• As a result, the signal shifted upwards.
• The peak to peak amplitude of the input signal is
2Vm, similarly the peak to peak amplitude of the
output signal is also 2Vm. Therefore, the total
swing of the output is same as the total swing of
the input.
Positive clamper

Negative clamper
• During positive half cycle:
• During the positive half cycle of the input AC
signal, the diode is forward biased and hence
no signal appears at the output.
• In forward biased condition, the diode allows
electric current through it.
• This current will flows to the capacitor and
charges it to the peak value of input voltage in
inverse polarity -Vm.
• As input current or voltage decreases after
attaining its maximum value Vm, the capacitor
holds the charge until the diode remains
forward biased.

• During negative half cycle:
• During the negative half cycle of the input AC
signal, the diode is reverse biased and hence the
signal appears at the output.
• In reverse biased condition, the diode does not
allow electric current through it.
• So the input current directly flows towards the
output.
• When the negative half cycle begins, the diode is
in the non-conducting state and the capacitor is
discharged (released).
• Therefore, the voltage appeared at the output is
equal to the sum of the voltage stored in the
capacitor (-Vm) and the input voltage (-Vm)
• I.e. Vo = -Vm- Vm = -2Vm} which have the same
polarity with each other.
• As a result, the signal shifted downwards.
Negative clamper

Applications of Clamper
• In the voltage multiplier
• In order to provide DC shift to the input waveform.

Chap 2 basic electronics gp 1

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Chap 2 basic electronics gp 1