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.
 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.
 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 theseries
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.
 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.
SERIES NEGATIVE CLIPPER WITH NEGATIVE BIAS (Example Q: Explain the working ofa
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.
 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 ofa
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.
 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 ofa
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.
 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 V i
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.
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.
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
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

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.

Positive Clamper:
The positive clamper is made up of a voltage source V i, capacitor C, diode D, and load
resistor RL. In the below 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:

During the negative 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 Vm. The capacitor charged in inverse polarity (positive) with the input
voltage. As input current or voltage decreases after attaining its maximum value -Vm, the
capacitor holds the charge until the diode remains forward biased.

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 charge
stored in the capacitor is discharged (released). Therefore, the voltage appeared at the
output is equal to the sum of the voltage stored in the capacitor (V m) 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 2V m. Therefore, the total swing of the output is same
as the total swing of the input.

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 flow 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
charge stored in 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.

CLAMPING CIRCUIT THEOREM

Under steady-state conditions, for any input waveform, the shape of the output waveform of a
clamping circuit is fixed and also the area in the forward direction and the area in the reverse direction are
related. According to clamping circuit theorem, under steady-state conditions, the ratio of area in forward
direction Af to that of reverse direction Ar of output voltage is equal to the ratio of diode forward
resistance Rf to resistance R connected across 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
𝑞 = ∫ 2 𝑖 𝑑𝑡
1 𝑡1 𝑓

𝑡
However, 𝑖 =𝑉𝑓 𝑤ℎ𝑒𝑟𝑒 𝑉 𝑖s the diode forward voltage thus 𝑞 = 1 ∫ 2 𝑉f 𝑑𝑡
𝑓 𝑅𝑓 𝑓 1 𝑅𝑓 𝑡1
During the interval 𝑡2 to 𝑡3 diode id OFF. Hence the capacitor discharges and the charge lost is

𝑡3
𝑞2 = ∫ 𝑖𝑟𝑑𝑡
𝑡2

𝑉𝑟 1 𝑡
Put 𝑖 = where 𝑉 is the diode reverse voltage. Thus 𝑞 = ∫ 3 𝑉 𝑑𝑡
𝑟 𝑅 𝑟 2 𝑅 𝑡2 𝑟

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

1 𝑡2 1 𝑡
Therefore ∫ 𝑉 𝑑𝑡 = ∫ 3 𝑉 𝑑𝑡
𝑅𝑓 𝑡1 𝑓 𝑅 𝑡2 𝑟

𝑡 𝑡
However, 𝐴 = ∫ 2 𝑉 𝑑𝑡 and 𝐴 = ∫ 3 𝑉 𝑑𝑡 where 𝐴 is the area under the curve when diode is
𝑓 𝑡1 𝑓 𝑟 𝑡2 𝑟 𝑓
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