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UNIT-II

DIODE APPLICATIONS

Rectifier- Half Wave Rectifier, Full wave Rectifier, Bridge Rectifier, Rectifiers with Filters
Capacitive and Inductive, Clippers-Clipping at two independent levels, Clamper- Clamping
Circuit Theorem, Clamping operation, Types of Clampers.

COURSE OUTCOME (CO2)

By the end of this Unit, students will be able to Construct Rectifiers and wave-shaping circuits
using PN junction diodes

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DEPARTMENT OF ECE GNITC

UNIT – II

DIODE APPLICATIONS

PN junction diode finds numerous applications including Switches, Rectifiers, Clippers,

Clampers, etc.

Rectifiers are electronic circuits that play a very important role in the conversion of AC power to
DC power. Rectifiers are part of Linear Power Supply systems.

Regulated DC power supply (Example Q. Explain about Regulated DC power supply)

• For the operation of most of the electronic devices and circuits, a D.C. source is required.
• So, it is advantageous to convert domestic a.c. supply into d.c. voltages.
• The process of converting a.c. voltage into D.C. voltage is called rectification.

This is achieved with

i) Step-down Transformer, ii) Rectifier, iii) Filter, and iv) Voltage regulator circuits.
These elements constitute D.C. regulated power supply shown in the figure below.

Fig. Block diagram of Regulated D.C. Power Supply

Transformer – steps down 230V AC mains to low voltage AC.

Rectifier – converts AC to pulsating DC, which is a combination of DC and ripple.
Smoothing(filter) – reduces the amount of ripple present in the output of the rectifier.
Regulator – eliminates ripple by setting DC output to a fixed voltage.

• The block diagram of a regulated D.C. power supply consists of a step-down transformer,
rectifier, filter, voltage regulator, and load.
• An ideal regulated power supply is an electronics circuit designed to provide a
predetermined D.C. voltage Vo which is independent of the load current and variations in
the input voltage ad temperature.
• If the output of a regulator circuit is an AC voltage then it is termed a voltage stabilizer,
whereas if the output is a DC voltage then it is termed a voltage regulator. The elements
of the regulated DC power supply are discussed as follows:

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TRANSFORMER:
• A transformer is a static device that transfers the energy from primary winding to
secondary winding through the mutual induction principle, without changing the
frequency.
• The transformer winding to which the supply source is connected is called the primary,
while the winding connected to the load is called secondary.
• If N1, N2 are the number of turns of the primary and secondary of the transformer then α
= N1/N2 is called the turn’s ratio of the transformer.

The different types of transformers are

1) Step-Up Transformer
2) Step-Down Transformer
3) Centre-tapped Transformer
The voltage, current, and impedance transformation ratios are related to the turn ratio of
the transformer by the following expressions

𝑉2 𝑁2
Voltage transformation ratio: =
𝑉1 𝑁1

𝐼2 𝑁2
Current transformation ratio: =
𝐼1 𝑁1

𝑍2 𝑁 2
impedance transformation ratio: = ( 2)
𝑍1 𝑁 1

RECTIFIER (Example Q. Explain the Characteristics of a Rectifier)

• Any electrical device which offers a low resistance to the current in one direction but a
high resistance to the current in the opposite direction is called a rectifier.
• Such a device is capable of converting a sinusoidal input waveform, whose average value
is zero, into a unidirectional waveform, with a non-zero average component.
• A rectifier is a device that converts a.c. voltage (bi-directional) to pulsating D.C. voltage
(Uni-directional)

Characteristics of a Rectifier Circuit:

1. Load currents: They are two types of output currents. They are average or D.C. current
and RMS current

i) Average or DC current: The average current of a periodic function is defined as the

area of one cycle of the curve divided by the base
It is expressed mathematically as

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DEPARTMENT OF ECE GNITC

ii) Effective (or) R.M.S. current: The effective (or) R.M.S. current squared of a periodic
function of time is given by the area of one cycle of the curve which represents the square of
the function divided by the base.
It is expressed mathematically as

2. Load Voltages: There are two types of output voltages. They are average or D.C.
voltage and R.M.S. voltage.

i) Average or DC Voltage: The average voltage of a periodic function is defined as the

areas of one cycle of the curve divided by the base.
It is expressed mathematically as

ii) Effective (or) R.M.S Voltage: The effective (or) R.M.S voltage squared of a periodic
function of time is given by the area of one cycle of the curve which represents the
square of the function divided by the base

1. Ripple Factor (Γ): It is defined as the ratio of R.M.S. value of a.c. component to the D.C.
component in the output is known as the “Ripple Factor”.

2. Efficiency (η): It is the ratio of d.c output power to the a.c. input power. It signifies, how
efficiently the rectifier circuit converts a.c. power into d.c. power
It is given by

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3. Peak Inverse Voltage (PIV): It is defined as the maximum reverse voltage that a diode
can withstand without destroying the junction.

4. Regulation: The variation of the d.c. output voltage as a function of d.c. load current is
called regulation. The percentage regulation is defined as

For an ideal power supply, % Regulation is zero.

Using one or more diodes in the circuit, the following rectifier circuits can be designed.

1. Half - Wave Rectifier

2. Full – Wave Rectifier
3. Bridge Rectifier

Half-Wave Rectifier (Example Q. Explain Half-wave Rectifier and derive expressions for
ripple factor, efficiency, form, and peak factors. )

A Half – wave rectifier is one which converts a.c. voltage into a pulsating voltage using only one-
half cycle of the applied a.c. voltage.

• The half-wave rectifier circuit shown above consists of a resistive load; a rectifying
element i.e., a p-n junction diode, and the source of a.c. voltage, all connected in series.
• The a.c. voltage is applied to the rectifier circuit using a step-down transformer

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INPUT & OUTPUT WAVEFORMS OF HALF-WAVE RECTIFIER

• The input to the rectifier circuit, V= Vm sinωt

Where Vm is the peak value of secondary a.c. voltage

Operation:

• For the positive half-cycle of input a.c. voltage, the diode D is forward biased and hence it
conducts.
• Now a current flow in the circuit and there is a voltage drop across RL.
• The waveform of the diode current (or) load current is shown in the figure.
• For the negative half-cycle of input, the diode D is reverse biased, and hence it does not
conduct.
• Now no current flows in the circuit i.e., i=0 and Vo=0. Thus for the negative half cycle, no
power is delivered to the load.

Analysis:
In the analysis of a HWR, the following parameters are to be analyzed.
i) DC output current ii) DC Output voltage
iii) R.M.S. Current iv) R.M.S. voltage
v) Rectifier Efficiency ( η) vi) Ripple factor ( Γ)
vii) Regulation viii) Transformer Utilization Factor (TUF)
ix) Peak Factor (P)

Let a sinusoidal voltage Vi be applied to the input of the rectifier

Vi= Vm sinω t
Where Vm is the maximum value of the secondary voltage
Let the diode be idealized to piece-wise linear approximation with resistance Rf in the forward
direction i.e., in the ON state and Rr (=∞) in the reverse direction i.e., in the OFF state
Now the current’ I’ in the diode (or) in the load resistance RL is given by

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Average (or) DC Output Current (Iav or Idc):

The average dc current Idc is given by

Substituting the value of Im, we get Idc

If RL>>Rf then

Average (or) DC Output Voltage (Vav or Vdc):

The average dc voltage is given by

If RL>>Rf then

R.M.S. Output Current (Irms):

The value of the R.M.S. current is given by

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`
R.M.S. Output Voltage (Vrms):

R.M.S. voltage across the load is given by

Rectifier efficiency (η):

The rectifier efficiency is defined as the ration of d.c. output power to the a.c. input
power i.e.

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Theoretically the maximum value of rectifier efficiency of a half-wave rectifier is 40.6%

when Rf/RL = 0
Ripple Factor (γ):

The ripple factor γ is given by

Regulation:
• The variation of d.c. output voltage as a function of d.c. load current is called regulation.
• The variation of Vdc with Idc for a half-wave rectifier is obtained as follows

• This result shows that Vdc equals Vm/π at no load and that the dc voltage decreases linearly
with an increase in dc output current.
• The larger the magnitude of the diode forward resistance, the greater is this decrease for a
given current change

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Transformer Utilization Factor (TUF):

• The d.c. power to be delivered to the load in a rectifier circuit decides the rating of the
transformer used in the circuit.
• So, the transformer utilization factor is defined as

• The factor which indicates how much is the utilization of the transformer in the circuit is
called Transformer Utilization Factor (TUF).
• The a.c. power rating of transformer = Vrms Irms
1
• The secondary voltage is purely sinusoidal hence its rms value is times maximum
√2
1
while the current is half sinusoidal hence its rms value is 2 of the maximum

• The d.c. power delivered to the load

• The value of TUF is low which shows that in the half-wave circuit, the transformer is not
fully utilized.
• If the transformer rating is 1 KVA (1000VA) then the half-wave rectifier can deliver
1000 X 0.287 = 287 watts to resistance load

Peak Inverse Voltage (PIV):

• It is defined as the maximum reverse voltage that a diode can withstand without destroying
the junction.

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DEPARTMENT OF ECE GNITC

• The peak inverse voltage across a diode is the peak of the negative half cycle.

For a half-wave rectifier, PIV is Vm

Form factor (F):

• The Form Factor F is defined as F = rms value / average value

𝐼𝑚 /2
F=
𝐼𝑚 /𝜋

Peak Factor (P):

The peak factor P is defined as P= Peak Value / rms value

Disadvantages of Half-Wave Rectifier:

1. High ripple factor.
2. The rectification efficiency is low.
3. Transformer Utilization factor is the lowest.

Because of all these disadvantages, the half-wave rectifier circuit is normally not used as a power
rectifier circuit.

Full–Wave Rectifier (Example Q. Explain Full Wave Rectifier and derive expressions
for ripple factor, efficiency, PIV, form, and peak factors)

• A full-wave rectifier converts an ac voltage into a pulsating dc voltage using both half
cycles of the applied ac voltage.
• In order to rectify both the half cycles of ac input, two diodes are used in this circuit. The
diodes feed a common load RL with the help of a center-tap transformer.
• A center-tap transformer is one that produces two sinusoidal waveforms of same
magnitude and frequency but out of phase with respect to the ground in the secondary
winding of the transformer.

The full wave rectifier is shown in the figure below

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Operation:

• During positive half of the input signal, anode of diode D1 becomes positive and at the
same time the anode of diode D2 becomes negative.
• Hence D1 conducts and D2 does not conduct. The load current flows through D1 and the
voltage drop across RL will be equal to the input voltage.
• During the negative half cycle of the input, the anode of D1 becomes negative and the
anode of D2 becomes positive.
• Hence, D1 does not conduct and D2 conducts. The load current flows through D2 and the
voltage drop across RL will be equal to the input voltage.

It is noted that the load current flows in both the half cycles of ac voltage and in the same direction
through the load resistance.

INPUT & OUTPUT WAVEFORMS OF FULL WAVE RECTIFIER

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Analysis:
Let a sinusoidal voltage Vi be applied to the input of a rectifier.
It is given by Vi=Vm sinωt
The current i1 through D1 and load resistor RL are given by

𝒊𝟏 = 𝑰𝒎 𝒔𝒊𝒏𝝎𝒕 𝒇𝒐𝒓 𝟎 ≤ 𝝎𝒕 ≤ 𝝅

𝒊𝟏 = 𝟎 𝒇𝒐𝒓 𝝅 ≤ 𝝎𝒕 ≤ 𝟐𝝅

Where
𝑽𝒎
𝑰𝒎 =
𝑹𝒇 + 𝑹𝑳
Similarly, the current i2 through diode D2 and load resistor RL is given by

𝒊𝟐 = 𝟎 𝒇𝒐𝒓 𝟎 ≤ 𝝎𝒕 ≤ 𝝅

𝒊𝟐 = 𝑰𝒎 𝒔𝒊𝒏𝝎𝒕 𝒇𝒐𝒓 𝝅 ≤ 𝝎𝒕 ≤ 𝟐𝝅

Therefore, the total current flowing through RL is the sum of the two currents i1 and i2.

i.e., iL = i1 + i2.

1. Average or DC Value:
2𝜋 2𝜋
1 1
𝐼𝑑𝑐 = ∫ 𝑖 𝑑𝜃 = ∫ 𝐼𝑚 𝑠𝑖𝑛𝜃 𝑑𝜃
2𝜋 2𝜋
0 0

2𝜋
𝜋
𝐼𝑚
= [∫ 𝑠𝑖𝑛𝜃𝑑𝜃 − ∫ 𝑠𝑖𝑛𝜃𝑑𝜃]
2𝜋 0
𝜋
𝐼𝑚 2𝐼𝑚
= [(−2)(−2)] = = 0.637𝐼𝑚
2𝜋 𝜋

𝐼𝑑𝑐 = 0.637𝐼𝑚

∴ 𝐈𝐝𝐜 𝐅𝐖𝐑 = 𝟐 𝐈𝐝𝐜 𝐇𝐖𝐑

2. R.M.S load Current Ir.m.s:

𝑰𝒎𝒂𝒙
𝑰𝒓𝒎𝒔 = = 𝟎. 𝟕𝟎𝟕 𝐈𝐦𝐚𝐱
√𝟐

3. DC output voltage Vdc:

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DEPARTMENT OF ECE GNITC

𝟐𝑰𝒎 𝑽𝒎
𝑽𝒅𝒄 = 𝑰𝒅𝒄 ∙ 𝑹𝑳 = ∙ 𝑹𝑳 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑰𝒎 =
𝝅 𝑹𝑺 + 𝑹𝒇 + 𝑹𝑳

𝟐𝑽𝒎 𝑹𝑳
𝑽𝒅𝒄 =
𝝅(𝑹𝑺 + 𝑹𝒇 + 𝑹𝑳 )

𝒊𝒇 (𝑹𝑺 + 𝑹𝒇 ) ≪ 𝑹𝑳

𝟐𝑽𝒎
𝑽𝒅𝒄 = = 𝟎. 𝟔𝟑𝟕𝑽𝒎
𝝅

4. Ripple Factor:

𝑰𝒓𝒎𝒔 𝟐
√
𝜸= ( ) −𝟏
𝑰𝒅𝒄

𝑰𝒎𝒂𝒙 𝟐𝑰𝒎
For FWR 𝑰𝒓𝒎𝒔 = & 𝑰𝒅𝒄 =
√𝟐 𝝅

𝑰𝒎 𝟐𝑰𝒎 𝟐
∴ 𝜸𝑭𝑾𝑹 = √( / ) −𝟏
√𝟐 𝝅

𝝅 𝟐
√
= ( ) −𝟏
𝟐√𝟐

𝟑. 𝟏𝟒𝟏𝟔 𝟐
= √ ( ) − 𝟏 = 𝟎. 𝟒𝟖𝟑
𝟐 × 𝟏. 𝟒𝟏𝟒

∴ 𝜸𝑭𝑾𝑹 = 𝟎. 𝟒𝟖𝟑

5. Regulation:

The variation of d.c. output voltage as a function of d.c. load current is called regulation

𝑰𝒅𝒄 (𝑹𝑺 + 𝑹𝒇 )
𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 =
𝟐𝑽
[ 𝝅𝒎 − 𝑰𝒅𝒄 (𝑹𝑺 + 𝑹𝒇 )]

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DEPARTMENT OF ECE GNITC

6. Rectification Efficiency:

5. TRANSFORMER UTILIZATION FACTOR (TUF):

8. PIV = 2 Vm

Advantages:
1) Ripple factor = 0.482 (against 1.21 for HWR)
2) Rectification efficiency is 0.812 (against 0.405 for HWR)
3) Better TUF (secondary) is 0.574 (0.287 for HWR)
4) No core saturation problem

Disadvantages:
1) Requires center tapped transformer

BRIDGE RECTIFIER (Example Q. Explain about Bridge rectifier using neat diagram).

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• A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave
rectification.
• This is a widely used configuration, as the expensive center tapped transformer is not
required in this configuration.

• The input AC signal is applied across two terminals A and B and the output DC signal is
obtained across the load resistor RL which is connected between the terminals C and D.
• The four diodes D1, D2, D3, D4 are arranged in series with only two diodes allowing electric
current during each half cycle.
• During the positive half cycle, the terminal A becomes positive while the terminal B
becomes negative.
• This causes the diodes D1 and D3 forward biased and at the same time, it causes the diodes
D2 and D4 reverse biased.
• The current flow direction during the positive half cycle is shown in the figure A (i.e., A
to D to C to B).

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• During the negative half cycle, the terminal B becomes positive while the terminal A
becomes negative.
• This causes the diodes D2 and D4 forward biased and at the same time, it causes the diodes
D1 and D3 reverse biased.
• The current flow direction during negative half cycle is shown in the figure B (i.e., B to D
to C to A).

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DEPARTMENT OF ECE GNITC

The following values of bridge rectifier are same as that of a full wave rectifier:

2𝐼𝑚
a) Average current 𝐼𝑑𝑐 = 𝜋
𝐼𝑚
b) RMS current 𝐼𝑟𝑚𝑠 =
√2
2𝑉𝑚
c) DC output voltage (no load) 𝑉𝑑𝑐 = 𝜋
d) Ripple factor (𝛾) = 0.482
e) Rectification efficiency (ƞ) = 0.812
2𝑉𝑚
f) DC output voltage full load 𝑉𝑑𝑐𝐹𝐿 = 𝜋
− 𝐼𝑑𝑐 (𝑅𝑆 + 2𝑅𝑓 ) i.e less by one diode loss

TUF of Bridge Rectifier is 0.812 (better that centre-tapped full wave rectifier; 0.693).

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COMPARISON OF RECTIFIERS (Example Q. Compare Half Wave Rectifier, Full Wave

Rectifier, Bridge rectifier)

Full Wave
S. No. Parameter Half Wave Rectifier Bridge Rectifier
Rectifier
1 No. of diodes 1 2 4
2 Peak inverse voltage Vm 2Vm Vm
Secondary Voltage
3 (rms) V V-0-V V

DC output voltage at no 𝑉𝑚 2𝑉𝑚 2𝑉𝑚

4 load = 0.318𝑉𝑚 = 0.636 𝑉𝑚 = 0.636 𝑉𝑚
𝜋 𝜋 𝜋
5 Ripple factor (γ) 1.21 0.482 0.482
6 Ripple frequency f 2f 2f
Rectification efficiency
7 (ƞ) 0.406 0.812 0.812

Transformer Utilization
8 Factor (TUF) 0.287 0.693 0.812

FILTERS (Example Q. what are the different types of Filters?)

• Since the output produced by the rectifier consists of a large value of ripple, it is very much
required to reduce the ripple present in the output of the rectifier to achieve a better dc
output.
• Filters are electrical circuits that are made of capacitors and/or inductors, the purpose of
which is to minimize the ripple.
Based on the element being used filters can be classified as
1. Capacitor filters (C-filters)
2. Inductor filters (L-filters)
3. LC filter
4. CLC or π filter

FULL WAVE RECTIFIER WITH CAPACITOR FILTER (Example Q. Derive the

expression for ripple factor of a full wave rectifier with capacitor filter.)

• The main duty of the capacitor filter is to short the ripples to the ground and blocks the
pure DC (DC components), so that it flows through the alternate path and reaches output
load resistor RL.

FWR WITH CAPACITOR FILTER

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DEPARTMENT OF ECE GNITC

• During the positive half cycle, the diode (D1) current reaches the filter and charges the
capacitor. However, the charging of the capacitor happens only when the applied AC
voltage is greater than the capacitor voltage.
• Initially, the capacitor is uncharged. That means no voltage exists between the plates of
the capacitor. So when the voltage is turned on, the charging of the capacitor happens
immediately.
• During this conduction period, the capacitor charges to the maximum value of the input
supply voltage. The capacitor stores a maximum charge exactly at the quarter positive half
cycle in the waveform. At this point, the supply voltage is equal to the capacitor voltage.

• When the AC voltage starts decreasing and becomes less than the capacitor voltage, then
the capacitor starts slowly discharging.
• The discharging of the capacitor is very slow as compared to the charging of the capacitor.
So, the capacitor does not get enough time to completely discharged.
• Before the complete discharge of the capacitor happens, the charging again takes place. So
only half or more than half of the capacitor charge get discharged.

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DEPARTMENT OF ECE GNITC

• When the input AC supply voltage reaches the negative half cycle, the diode D1 is reverse
biased (blocks electric current) whereas the diode D2 is forward biased (allows electric
current).
• During the negative half cycle, the diode (D2) current reaches the filter and charges the
capacitor.
• However, the charging of the capacitor happens only when the applied AC voltage is
greater than the capacitor voltage.
• The capacitor is not completely uncharged, so the charging of the capacitor does not
happen immediately.
• When the supply voltage becomes greater than the capacitor voltage, the capacitor again
starts charging.
• In both positive and negative half cycles, the current flows in the same direction across the
load resistor RL.
• So, we get either complete positive half cycles or negative half cycles. In our case, they
are complete positive half cycles.

Expression for Ripple factor of FWR with Capacitor filter:

During the rising quarter wave (output voltage of the rectifier) of the applied input voltage to the
capacitor
Charge acquired by the capacitor = Vr, p-p x C

During the falling quarter cycle of the input to the capacitor (when applied voltage is less than
capacitor voltage)
Charge lost by the capacitor = Idc x t2
Charge lost by the capacitor = charge gained by the capacitor

i.e. Vr, p-p x C = Idc x t2 from the figure t2 = T/2

Thus Vr, p-p x C = Idc x T/2 = Idc/2f as T = 1/f

𝐼
𝑑𝑐
𝑉𝑟,𝑝−𝑝 = 2𝑓𝐶
𝑉𝑟,𝑝−𝑝
𝐹𝑜𝑟 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑤𝑎𝑣𝑒𝑓𝑜𝑟𝑚 𝑉𝑟,𝑟𝑚𝑠 = ⁄
2√3

𝑉𝑟,𝑝−𝑝 𝐼
Thus 𝑉𝑟,𝑟𝑚𝑠 = ⁄ = 4 𝑑𝑐
2√3 √3𝑓𝐶

𝑉𝑑𝑐
As 𝐼𝑑𝑐 = 𝑉𝑑𝑐 /𝑅𝐿 => 𝑉𝑟,𝑟𝑚𝑠 = 4
√3𝑓𝐶𝑅𝐿

𝑉𝑟,𝑟𝑚𝑠 1
∴ 𝑅𝑖𝑝𝑝𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 (𝛾) = =
𝑉𝑑𝑐 4√3𝑓𝐶𝑅𝐿

Ripple frequency of FWR = 2 x ripple frequency of HWR

Thus the ripple factor may be decreased by increasing either RL or C or both

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DEPARTMENT OF ECE GNITC

INDUCTOR FILTER WITH FULL WAVE RECTIFIER (Example Q: Explain the

working of an inductor filter ?)

• Inductor offers very low impedance R for the DC component and offers a very high
impedance Z which is √𝑅 2 + (𝜔𝐿)2 for the AC component of the rectified output.
• As such AC components are blocked to a large extent and DC component is easily allowed
to the load, external output contains fewer ripples, which means ripple content is attenuated
much by the filters.
• For a full wave rectifier circuit, the ripple frequency is 2fs and the pulsating output load
current can be considered to contain

4𝑉𝑚 ∙𝑐𝑜𝑠(2𝜔𝑡−∅)
AC component =
3𝜋√𝑅𝐿 2 +4𝜔2 𝐿2

2𝑉𝑚 0.638𝑉𝑚
DC Component = =
𝜋𝑅𝐿 𝑅𝐿

If (4𝜔2 𝐿2 /𝑅𝐿 2 ) is much greater than unity, then

4𝑉𝑚 1
[ ∙ ]
3𝜋√2
√𝑅𝐿 2 +4𝜔2 𝐿2
Ripple factor 𝛾 = 2𝑉
( 𝜋𝑅𝑚 )
𝐿

[√2𝑅𝐿 ] 𝑅𝐿
= =
3√2∙𝜔𝐿
3√𝑅𝐿 2 +4𝜔2 𝐿2

Therefore Ripple factor will be small for low values of load or for large values of load currents
and high values of inductances (L).

L-SECTION FILTER OR LC FILTER (Example Q: What is an L-section filter? How does

it work?)

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DEPARTMENT OF ECE GNITC

• This filter is also known as choke input filter as the filter element looking from the rectifier
is an inductance L1.
• Merits of both inductor and capacitance filters can be combined in L-section filter.
• Series inductor L1 drops the AC, which is further by-passed by the capacitor C1, which
results in reduced ripple and improved filtering action.

The final expression for ripple factor of L-C filter is

1
𝑉𝑅𝑀𝑆 ⌈ ⌉ √2 1 √2 1
𝜔𝐶1
𝛾=( ) ∙[ 1 ]= ∙ ≅ ∙
𝑉𝐷𝐶 𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟 𝜔𝐿1 − 3 (𝜔2 𝐿1 𝐶1 −1) 3 (𝜔2 𝐿1 𝐶1 )
𝜔𝐶1

√2 1 1 √2 𝑋𝐶 𝑋𝐶
= ∙ ∙ = ∙ = 0.471
3 𝜔𝐿1 𝜔𝐶1 3 𝑋𝐿 𝑋𝐿

Here the ripple factor is independent of the load currents. Hence the L-section filter is used in
applications having wide variations of load currents.

Π-SECTION FILTER OR CLC FILTER: (Example Q: What is a CLC filter? Give

expression for its ripple factor.)

• Π section filter is a combination of Capacitor input and L-section filters.

• The ripple is very much reduced by the double filtering action.

𝟐
√𝟐𝑿𝑪
• The expression for ripple factor is 𝜸 =
𝑹𝑳 𝑿𝑳 𝟐

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DEPARTMENT OF ECE GNITC

Analysis of Π section filter:

• The Π section filter provides higher output voltage (that approaches the peak value of the
AC potential of the source) at light loads and a very smooth output.
• Capacitor C1 offers a low impedance path for the harmonics of rectifier output. At the same
time, it offers high impedance to DC and second harmonic content.
• Output across C1 is a triangular wave with vertical sides. The remaining ripple content is
reduced by the LC filter.

COMPARISON OF FILTERS: (Example Q: Compare different filters.)

1. A capacitor filter provides Vm volts at fewer load currents. But regulation is poor.
2. An Inductor filter gives high ripple voltage for low load currents. It is used for high load
currents.
3. L – Section filter gives a ripple factor independent of load current. Voltage regulation can
be improved by use of bleeder resistance.
4. Multiple L – Section filter or filters give much less ripple than the single L –Section Filter.

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DEPARTMENT OF ECE GNITC

DIODE APPLICATIONS: CLIPPERS AND CLAMPERS

CLIPPERS: (Example Q: What are Clippers? Classify the types.)

• A clipper is a device that removes either the entire or a portion of the positive half (top
half) or negative half (bottom half), or both positive and negative half cycles of the input
AC signal.
• Clippers are also known as amplitude limiters, and slicers.

Clippers are broadly classified into

1. Series Clippers
2. Shunt Clippers
3. Two level Clippers

Series clippers

Series clippers are again classified under

1. Series Positive Clippers

2. Series Positive Clippers with bias voltage
3. Series Negative Clippers
4. Series Negative Clippers with bias voltage

SERIES POSITIVE CLIPPERS: (Example Q: What are series positive Clippers?)

• In series positive clipper, the positive half cycles of the input AC signal are removed.
• If the diode is arranged in such a way that the arrowhead of the diode points towards the
input and the diode is in series with the output load resistance, then the clipper is said to
be a series positive clipper.

• During the positive half cycle, terminal A is positive and terminal B is negative.
• That means the positive terminal A is connected to n-side and the negative terminal B is
connected to p-side of the diode.
• As we already know that if the positive terminal is connected to n-side and the negative
terminal is connected to p-side then the diode is said to be reverse biased.

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DEPARTMENT OF ECE GNITC

• Therefore, the diode D is reverse biased during the positive half cycle and it acts as an
open circuit.
• As current does not flow through the load, no output voltage will be developed across the
load. Positive cycle will be blocked.
• During negative half cycle of input signal, terminal A will be negative and diode D gets
forward biased.
• Diode acts as a closed switch allowing current to flow through the load.
• Output voltage will be generated across the load, the magnitude of which is equal to the
negative half cycle of input signal.
• So negative half cycle will be produced at the output.

SERIES POSITIVE CLIPPERS WITH POSITIVE BIAS VOLTAGE (Example Q: Explain the
working of a positively biased series positive Clipper)

• Sometimes it is desired to remove a small portion of positive or negative half cycles. In

such cases, the biased clippers are used.
• The construction of the series positive clipper with bias is almost similar to the series
positive clipper.
• The only difference is an extra element called battery is used in series positive clipper with
bias.
• During the positive half cycle, terminal A is positive and terminal B is negative. That
means the positive terminal is connected to n-side and the negative terminal is connected
to p-side.
• The diode becomes reverse biased. Therefore, the diode is reverse biased by the input
supply voltage Vi.

• Whenever the input supply voltage Vi is less than the battery voltage VB (Vi < VB), diode
remains forward biased as a result the signal appears at the output.
• When the input signal voltage Vi exceeds the battery voltage diode becomes reverse biased
and acts as an open circuit.
• No current flows through the load, no output voltage will be developed, thus input signal
(Vi > VB) will not appear at the output.

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DEPARTMENT OF ECE GNITC

• During the negative half cycle, terminal A is negative and terminal B is positive. That
means the diode D is forward biased due to the input supply voltage.
• Battery voltage VB keeps diode forward biased during the entire negative half cycle of the
input signal. Thus, output signal which is similar to the input signal will be developed
across the load during the negative cycle.

SERIES POSITIVE CLIPPERS WITH NEGATIVE BIAS VOLTAGE (Example Q: Explain the
working of a series positive clipper with negative bias?)

• During the positive half cycle, the diode D is reverse biased by both input supply voltage
Vi and battery voltage VB.
• So, no signal appears at the output during the positive half cycle. Therefore, the complete
positive half cycle is removed.

• During the negative half cycle, the diode is forward biased by the input supply voltage Vi
and reverse biased by the battery voltage VB.
• However, initially, the battery voltage VB dominates the input supply voltage Vi. So, the
diode remains to be reverse biased until the Vi becomes greater than VB.
• When the input supply voltage Vi becomes greater than the battery voltage VB, the diode
is forward biased by the input supply voltage Vi. So, the signal appears at the output.

SERIES NEGATIVE CLIPPER (Example Q: What are negative Clippers? Explain about the
series negative clipper )

• In series negative clipper, the negative half cycles of the input AC signal is removed at the
output.
• If the diode is arranged in such a way that the arrowhead of the diode points towards the
output and the diode is in series with the output load resistance, then the clipper is said to
be a series negative clipper.
• During the positive half cycle, terminal A is positive and terminal B is negative.
• That means the positive terminal A is connected to p-side and the negative terminal B is
connected to n-side of the diode and the diode gets forward biased.

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DEPARTMENT OF ECE GNITC

• During forward biased condition, electric current flows through the diode. So, the positive
half cycle is allowed at the output.
• Therefore, a series of positive half cycles appears at the output.
• During the negative half cycle, the terminal A is negative and the terminal B is positive .
Diode gets reverse biased and no current flows through it.
• So negative cycles of the input signal do not appear at the output.

SERIES NEGATIVE CLIPPER WITH POSTIVE BIAS VOLTAGE (Example Q: Explain how a
series negative clipper work when positive bias is applied?)

• Sometimes it is desired to remove a small portion of positive or negative half cycles of the
input AC signal. In such cases, the biased clippers are used.
• In this clipper circuit, P-side of the diode is connected towards the input. A battery is
connected to the to the N-side of the diode, such that positive terminal of the battery is
connected to the N-side of the diode.

• Diode remains reverse biased till the time the input signal voltage (Vi) is less than the
battery voltage (VB). So, it behaves like an open circuit and output signal will not be
produced.
• When the input signal voltage (Vi) is more than the battery voltage (VB), diode becomes
forward biased and behaves like a closed switch.
• Output signal (Vi > VB) will now be produced which is similar to the input signal.

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DEPARTMENT OF ECE GNITC

SERIES NEGATIVE CLIPPER WITH NEGATIVE BIAS (Example Q: Explain the working of
a series negative Clipper with negative bias?)

• During the positive half cycle, the diode D is forward biased by both input supply voltage
Vi and the battery voltage VB.
• So it doesn’t matter whether the input supply voltage is greater or less than battery voltage
VB, the diode always remains forward biased.
• Therefore, during the positive half cycle, the signal appears at the output.

• During negative half cycle of the input signal when the voltage Vi is less than the battery
voltage VB, the diode will be forward biased by the battery voltage VB. As a result, the
signal appears at the output.
• When the input supply voltage Vi becomes greater than the battery voltage VB, the diode
will become reverse biased. As a result, no signal appears at the output.

SHUNT POSITIVE CLIPPERS: (Example Q: What are shunt positive clippers? Explain.)

• 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 shunt clipper 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.

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DEPARTMENT OF ECE GNITC

• 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 CLIPPER WITH POSITIVE BIAS VOLTAGE (Example Q: What is a shunt clipper?
How does it work with a positive bias?)

• Initially, when the input supply voltage Vi is less than the battery voltage VB, the diode
gets reverse biased. Therefore, the signal appears at the output.
• When the input supply voltage Vi becomes greater than the battery voltage VB, the diode
D is forward biased by the input supply voltage Vi. As a result, no signal appears at the
output.

• During the negative half cycle, the diode is reverse biased by both the input supply
voltage and battery voltage, always. As a result, the complete negative half cycle
appears at the output.

SHUNT POSITIVE CLIPPERS WITH NEGATIVE BIAS (Example Q: Explain the working of
a shunt negatively biased positive Clipper)

• During the positive half cycle, the diode is forward biased by both input supply voltage
Vi and battery voltage VB.
• Therefore, no signal appears at the output during the positive half cycle.

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DEPARTMENT OF ECE GNITC

• During negative half cycle of the input signal initially, the input supply voltage Vi is less
than the battery voltage VB. So the battery voltage makes the diode to be forward biased.
• As a result, no signal appears at the output.
• However, when the input supply voltage Vi becomes greater than the battery voltage VB,
the diode is reverse biased by the input supply voltage Vi.
• As a result, the signal appears at the output.

SHUNT NEGATIVE CLIPPER (Example Q: How will a shunt negative Clipper work?)

• 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.
• During the negative half cycle the diode is forward biased and hence no output signal is
generated.

SHUNT NEGATIVE CLIPPER WITH POSITIVE BIAS (Example Q: Explain the working of
a shunt positively biased negative Clipper?)

• During the positive half cycle initially, the input supply voltage is less than the battery
voltage (VB).
• So the diode is forward biased by the battery voltage. As a result, no signal appears at the
output.

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DEPARTMENT OF ECE GNITC

• When the input supply voltage (Vi) becomes greater than the battery voltage (VB) then the
diode is reverse biased by the input supply voltage. As a result, the signal appears at the
output.

• During the negative half cycle, the diode is forward biased by both input supply voltage Vi
and battery voltage VB. So the complete negative half cycle is removed at the output.

SHUNT NEGATIVE CLIPPER WITH NEGATIVE BIAS (Example Q: How will a negatively
biased parallel negative Clipper work?)

• During the positive half cycle, the diode is reverse biased by both input supply voltage Vi
and battery voltage VB. As a result, the complete positive half cycle appears at the output.

• During the negative half cycle, initially, when the input supply voltage is less than the
battery voltage the diode will be reverse biased by the battery voltage.
• As a result, the signal appears at the output.
• When the input supply voltage becomes greater than the battery voltage, the diode will be
forward biased by the input supply voltage.
• As a result, the signal does not appear at the output.

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DEPARTMENT OF ECE GNITC

TWO LEVEL CLIPPER (Example Q: Describe the working of a two-level clipper?)

• When portions of positive and negative half cycles of the input wave form have to be
removed dual (two level) clippers are used.
• Dual clippers are made by combining the biased shunt positive clipper and biased shunt
negative clipper.

• During the positive half cycle, the diode D1 is forward biased by the input supply voltage
Vi and reverse biased by the battery voltage VB1.
• On the other hand, the diode D2 is reverse biased by both input supply voltage Vi and
battery voltage VB2.
• Initially, the input supply voltage is less than the battery voltage. So the diode D1 is reverse
biased by the battery voltage VB1.
• Similarly, the diode D2 is reverse biased by the battery voltage VB2. As a result, the signal
appears at the output.
• However, when the input supply voltage Vi becomes greater than the battery voltage VB1,
the diode D1 is forward biased by the input supply voltage. As a result, no signal appears
at the output.
• During the negative half cycle, the diode D1 is reverse biased by both input supply voltage
Vi and battery voltage VB1.
• On the other hand, the diode D2 is forward biased by the input supply voltage Vi and
reverse biased by the battery voltage VB2.
• Initially, the battery voltage is greater than the input supply voltage. Therefore, the diode
D1 and diode D2 are reverse biased by the battery voltage.
• As a result, the signal appears at the output.
• When the input supply voltage becomes greater than the battery voltage VB2, the diode D2
is forward biased. As a result, no signal appears at the output.

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DEPARTMENT OF ECE GNITC

APPLICATIONS OF CLIPPERS (Example Q: What are the applications of Clipper

circuits?)

• 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.

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DEPARTMENT OF ECE GNITC

CLAMPERS (Example Q: What is a Clamper? Explain in detail)

• 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.
• A clamper circuit adds the positive dc component to the input signal to push it to the
positive side.
• 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.
• Similarly, a clamper circuit adds the negative dc component to the input signal to push it
to the negative side.

Types of clampers (Example Q: Discuss the different types of clampers)

Clamper circuits are of three types:

• Positive clampers
• Negative clampers
• Biased clampers

• 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.

• 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.

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POSITIVE CLAMPER WITH POSITIVE BIAS (Example Q: Explain positive clamper

with a positive bias voltage)

• During the positive half cycle, the battery voltage forward biases the diode when the input
supply voltage is less than the battery voltage.
• This current or voltage will flows to the capacitor and charges it.
• When the input supply voltage becomes greater than the battery voltage then the diode
stops allowing electric current through it because the diode becomes reverse biased.

• During the negative half cycle, the diode is forward biased by both input supply voltage
and battery voltage. So, the diode allows electric current. This current will flow to the
capacitor and charges it.

POSITIVE CLAMPER WITH NEGATIVE BIAS (Example Q: Explain positive clamper

with a negative bias voltage)

• During the negative half cycle, the battery voltage reverse biases the diode when the input
supply voltage is less than the battery voltage.
• As a result, the signal appears at the output.
• When the input supply voltage becomes greater than the battery voltage, the diode is
forward biased by the input supply voltage and hence allows electric current through it.
• This current will flow to the capacitor and charges it.

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• During the positive half cycle, the diode is reverse biased by both input supply voltage and
the battery voltage. As a result, the signal appears at the output.
• The signal appeared at the output is equal to the sum of the input voltage and capacitor
voltage.

NEGATIVE CLAMPER WITH POSITIVE BIAS

• During the positive half cycle, the battery voltage reverse biases the diode when the input
supply voltage is less than the battery voltage.
• When the input supply voltage becomes greater than the battery voltage, the diode is
forward biased by the input supply voltage and hence allows electric current through it.
• This current will flow to the capacitor and charges it.

• During the negative half cycle, the diode is reverse biased by both input supply voltage
and battery voltage. As a result, the signal appears at the output.

NEGATIVE CLAMPER WITH NEGATIVE BIAS

• During the positive half cycle, the diode is forward biased by both input supply voltage
and battery voltage. As a result, current flows through the capacitor and charges it.

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• During the negative half cycle, the battery voltage forward biases the diode when the input
supply voltage is less than the battery voltage.
• When the input supply voltage becomes greater than the battery voltage, the diode is
reverse biased by the input supply voltage and hence signal appears at the output.

CLAMPING CIRCUIT THEOREM

Statement: Under steady state conditions, for any input waveform the ratio of area in forward
direction 𝐴𝑓 to that of the reverse direction 𝐴𝑟 of the output voltage is equal to the ratio of forward
resistance 𝑅𝑓 of the diode to the resistance R connected across the diode.

𝑨𝒇 𝑹𝒇
𝑖. 𝑒. =
𝑨𝒓 𝑹

Proof: Let for the first cycle of the input signal the diode be ON i.e during the time interval 𝑡1 to
𝑡2 . Hence during this period charge builds up on the capacitor C

If 𝑖𝑓 is the diode current, then charge gained by the capacitor during 𝑡1 to 𝑡2 is

𝑡
𝑞1 = ∫𝑡 2 𝑖𝑓 𝑑𝑡
1

𝑉 1 𝑡
However, 𝑖𝑓 = 𝑅𝑓 𝑤ℎ𝑒𝑟𝑒 𝑉𝑓 𝑖s the diode forward voltage thus 𝑞1 = 𝑅 ∫𝑡 2 𝑉𝑓 𝑑𝑡
𝑓 1 𝑓

During the interval 𝑡2 to 𝑡3 diode id OFF. Hence the capacitor discharges and the charge lost is

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DEPARTMENT OF ECE GNITC

𝑡3
𝑞2 = ∫ 𝑖𝑟 𝑑𝑡
𝑡2

𝑉𝑟 1 𝑡
Put 𝑖𝑟 = where 𝑉𝑟 is the diode reverse voltage. Thus 𝑞2 = 𝑅 ∫𝑡 3 𝑉𝑟 𝑑𝑡
𝑅 2

At steady state charge lost is equal to the charge gained i.e. 𝑞1 = 𝑞2

1 𝑡 2 13 𝑡
Therefore 𝑅𝑓
∫𝑡1 𝑉𝑓 𝑑𝑡 = 𝑅 ∫𝑡2 𝑉𝑟 𝑑𝑡

𝑡 𝑡
However, 𝐴𝑓 = ∫𝑡 2 𝑉𝑓 𝑑𝑡 and 𝐴𝑟 = ∫𝑡 3 𝑉𝑟 𝑑𝑡 where 𝐴𝑓 is the area under the curve when diode is
1 2
ON and 𝐴𝑟 is the area of the curve when the diode is OFF. From both the equations

𝑨𝒇 𝑨𝒓 𝑨𝒇 𝑹𝒇
= 𝑜𝑟 =
𝑹𝒇 𝑹 𝑨𝒓 𝑹

TOPIC BEYOND SYLLABUS

VOLTAGE REGULATORS:

• A voltage regulator is a device that regulates the voltage level.

• It essentially steps down the input voltage to the desired level and keeps it at that same
level during the supply.
• This ensures that even when a load is applied the voltage doesn’t drop. The voltage
regulator is used for two main reasons, and they are:
➢ To vary or regulate the output voltage
➢ To keep the output voltage constant at the desired value in spite of variations in the
supply voltage.
• Voltage regulators are used in computers, power generators, alternators to control the
output of the plant.

ZENER VOLTAGE REGULATOR

• There is a series resistor connected to the circuit in order to limit the current into the diode.
It is connected to the positive terminal of the d.c.
• It works in such a way the reverse-biased can also work in breakdown conditions.
• We do not use ordinary junction diode because the low power rating diode can get damaged
when we apply reverse bias above its breakdown voltage

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DEPARTMENT OF ECE GNITC

• When the minimum input voltage and the maximum load current is applied, the Zener
diode current should always be minimum.
• Since the input voltage and the required output voltage is known, it is easier to choose a
Zener diode with a voltage approximately equal to the load voltage, i.e., VZ = VL.
• Current through the diode increases when the voltage across the diode tends to increase
which results in the voltage drop across the resistor.
• Similarly, the current through the diode decreases when the voltage across the diode tends
to decrease.
• Here, the voltage drop across the resistor is very less, and the output voltage results
normally.

IMPORTANT QUESTIONS

1. What is a rectifier? Mention the different types of rectifiers.

2. Derive the expression for efficiency, ripple factor, form and peak factors of full-wave
rectifier with centre-tap transformer.
3. Explain the construction and working of Bridge rectifier. Derive expressions for ripple
factor, efficiency, form and peak factors.
4. Compare Halfwave, Full wave and Bridge rectifiers.
5. State and explain clamping circuit theorem.
6. With the help of a circuit diagram and necessary waveforms explain the working of a
positive clipper with negative bias voltage.
7. Draw the output waveform of negative clipper with positive bias voltage. How does this
circuit work?
8. Differentiate positive and negative clippers.
9. With the help of necessary waveforms explain the working of a negative clamper with
positive bias.
10. How does a two-level clipper circuit work?

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OBJECTIVE TYPE QUESTIONS

1. Pulsating DC is a combination of ___ & ____ [ A ]

A. DC, AC ripple B. DC, DC
C. AC, AC D. Ripple, Ripple

2. ____ circuit converts AC to pulsating DC [ C ]

B. Filter B. Regulator
C. Rectifier D. Clipper

3. Value of ripple factor of a half wave rectifier is ___ [ B ]

A. 0.48 B. 1.21
C. 0.72 D. 0.32

4. Number of diodes needed to construct a half wave rectifier is [ D ]

A. 4 B. 3
C. 2 D. 1

5. Value of efficiency of a centre tapped full wave rectifier is [ B ]

A. 0.48 B. 0.81
B. 1.21 D. 0.40

6. Value of Peak Inverse Voltage of bridge rectifier is ____ [ C ]

A. 2Vm B. Vm/2
C. Vm D. Vm/4

7. CLC filter is also known as___ filter [ A ]

A. π B. Inductor
C. Capacitor D. α

8. The purpose of filter is to minimize___ present in the output of rectifier [ C ]

A. DC B. DC & Ripple
C. Ripple D. resistance

9. Bridge rectifier requires __ number of diodes [ B ]

A. 5 B. 4
C. 3 D. 2

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10. Value of ripple factor of capacitor filter is [ D ]

3√2𝑅𝐿
A. ωRL B. 𝜔𝐿

𝜔𝐿𝑅𝐿 𝑹
𝑳
C. D. 𝟑√𝟐𝝎𝑳
3√2

FILL IN THE BLANKS:

1. Clippers are also known as slicers or amplitude limiters.

2. Clampers circuit consists of diode, resistor and capacitor

3. DC restorers is the other name of Clampers.

4. Value of efficiency of a Half wave rectifier is 0.408

5. Transformer utilization factor of Bridge rectifier is 0.82

6. Ratio of rms value of ripple to the dc voltage is known as ripple factor

7. Clamping is the process of adding DC level to an AC signal

8. Circuit diagram of a positive clamper is

9. Ripple factor of centre-tap full wave rectifier is 0.48

10. Advantage of bridge rectifier over Centre tap full wave rectifier is no centre tap transformer

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