DC And AC Amplifiers - Summing Amplifier – Difference Amplifier – Voltage Follower – Differentiator – Integrator- Clamper - Clipper– Filters.

1. Presented by
Dr. R. RAJA, M.E., Ph.D.,
Assistant Professor, Department of EEE,
Muthayammal Engineering College, (Autonomous)
Namakkal (Dt), Rasipuram – 637408
19EEC03-Linear Integrated Circuits and Its Applications
Unit-II Applications of Operational Amplifier
MUTHAYAMMAL ENGINEERING COLLEGE
(An Autonomous Institution)
(Approved by AICTE, New Delhi, Accredited by NAAC, NBA & Affiliated to Anna University),
Rasipuram - 637 408, Namakkal Dist., Tamil Nadu.

2. Unit-II Applications of Operational Amplifier
8/21/2020 2
DC And AC Amplifiers - Summing Amplifier – Difference Amplifier – Voltage
Follower – Differentiator – Integrator- Clamper - Clipper– Filters.

3. Summing Amplifier
The Summing Amplifier is another type of operational amplifier circuit
configuration that is used to combine the voltages present on two or more inputs into
a single output voltage.
We saw previously in the inverting operational amplifier that the inverting amplifier
has a single input voltage, (Vin) applied to the inverting input terminal. If we add
more input resistors to the input, each equal in value to the original input resistor,
(Rin) we end up with another operational amplifier circuit called a Summing
Amplifier, “summing inverter” or even a “voltage adder”.
8/21/2020 3

4. Contd..
Summing Amplifier Circuit
In this simple summing amplifier circuit, the output voltage, ( Vout ) now becomes
proportional to the sum of the input voltages, V1, V2, V3, etc. Then we can modify
the original equation for the inverting amplifier to take account of these new inputs
thus:
8/21/2020 4

5. Contd..
8/21/2020 5
However, if all the input impedances, ( RIN ) are equal in value, we can simplify
the above equation to give an output voltage of:
Summing Amplifier Equation

6. Contd..
 We now have an operational amplifier circuit that will amplify each individual
input voltage and produce an output voltage signal that is proportional to the
algebraic “SUM” of the three individual input voltages V1, V2 and V3. We can
also add more inputs if required as each individual input “sees” their respective
resistance, Rin as the only input impedance.
 This is because the input signals are effectively isolated from each other by the
“virtual earth” node at the inverting input of the op-amp. A direct voltage
addition can also be obtained when all the resistances are of equal value and Rƒ
is equal to Rin.
 Note that when the summing point is connected to the inverting input of the op-
amp the circuit will produce the negative sum of any number of input voltages.
Likewise, when the summing point is connected to the non-inverting input of
the op-amp, it will produce the positive sum of the input voltages.
8/21/2020 6

7. Contd..
 A Scaling Summing Amplifier can be made if the individual input resistors are
“NOT” equal. Then the equation would have to be modified to:
 To make the math’s a little easier, we can rearrange the above formula to make
the feedback resistor Rƒ the subject of the equation giving the output voltage as:
 This allows the output voltage to be easily calculated if more input resistors are
connected to the amplifiers inverting input terminal. The input impedance of
each individual channel is the value of their respective input resistors, ie, R1, R2,
R3 … etc.
8/21/2020 7

8. Contd..
 Sometimes we need a summing circuit to just add together two or more voltage
signals without any amplification. By putting all of the resistances of the circuit
above to the same value R, the op-amp will have a voltage gain of unity and an
output voltage equal to the direct sum of all the input voltages as shown:
 The Summing Amplifier is a very flexible circuit indeed, enabling us to
effectively “Add” or “Sum” (hence its name) together several individual input
signals. If the inputs resistors, R1, R2, R3 etc, are all equal a “unity gain
inverting adder” will be made. However, if the input resistors are of different
values a “scaling summing amplifier” is produced which will output a weighted
sum of the input signals.
8/21/2020 8

9. Summing Amplifier Example No1
Find the output voltage of the following Summing Amplifier circuit.
Summing Amplifier
Using the previously found formula for the gain of the circuit:
8/21/2020 9

10. Contd..
We can now substitute the values of the resistors in the circuit as follows:
We know that the output voltage is the sum of the two amplified input signals and is
calculated as:
Then the output voltage of the Summing Amplifier circuit above is given as -45 mV
and is negative as its an inverting amplifier.
8/21/2020 10

11. Contd..
Non-inverting Summing Amplifier
 But as well as constructing inverting summing amplifiers, we can also use the
non-inverting input of the operational amplifier to produce a non-inverting
summing amplifier. We have seen above that an inverting summing amplifier
produces the negative sum of its input voltages then it follows that the non-
inverting summing amplifier configuration will produce the positive sum of its
input voltages.
 As its name implies, the non-inverting summing amplifier is based around the
configuration of a non-inverting operational amplifier circuit in that the input
(either ac or dc) is applied to the non-inverting (+) terminal, while the required
negative feedback and gain is achieved by feeding back some portion of the
output signal (VOUT) to the inverting (-) terminal as shown.
8/21/2020 11

12. Contd..
8/21/2020 12
So what’s the advantage of the non-inverting configuration compared to the
inverting summing amplifier configuration.

13. Contd..
 Besides the most obvious fact that the op-amps output voltage VOUT is in phase
with its input, and the output voltage is the weighted sum of all its inputs which
themselves are determined by their resistance ratios, the biggest advantage of
the non-inverting summing amplifier is that because there is no virtual earth
condition across the input terminals, its input impedance is much higher than
that of the standard inverting amplifier configuration.
 Also, the input summing part of the circuit is unaffected if the op-amps closed-
loop voltage gain is changed. However, there is more maths involed in selecting
the weighted gains for each individual input at the summing junction especially
if there are more than two inputs each with a different weighting factor.
Nevertheless, if all the inputs have the same resistive values, then the maths
involved will be a lot less.
8/21/2020 13

14. Contd..
 If the closed-loop gain of the non-inverting operational amplifier is made equal
the number of summing inputs, then the op-amps output voltage will be exactly
equal to the sum of all the input voltages. That is for a two input non-inverting
summing amplifier, the op-amps gain is equal to 2, for a three input summing
amplifier the op-amps gain is 3, and so on. This is because the currents which
flow in each input resistor is a function of the voltage at all its inputs. If the
input resistances made all equal, (R1 = R2) then the circulating currents cancel
out as they can not flow into the high impedance non-inverting input of the op-
amp and the voutput voltage becomes the sum of its inputs.
8/21/2020 14

15. Contd..
 So for a 2-input non-inverting summing amplifier the currents flowing into the
input terminals can be defined as:
 If we make the two input resistances equal in value, then R1 = R2 = R.
8/21/2020 15

16. Contd..
8/21/2020 16
The standard equation for the voltage gain of a non-inverting summing amplifier
circuit is given as:

17. Contd..
 The non-inverting amplifiers closed-loop voltage gain AV is given as: 1 +
RA/RB. If we make this closed-loop voltage gain equal to 2 by making RA = RB,
then the output voltage VO becomes equal to the sum of all the input voltages as
shown.
Non-inverting Summing Amplifier Output Voltage
8/21/2020 17

18. Contd..
 Thus for a 3-input non-inverting summing amplifier configuration, setting the
closed-loop voltage gain to 3 will make VOUT equal to the sum of the three input
voltages, V1, V2 and V3. Likewise, for a four input summer, the closed-loop
voltage gain would be 4, and 5 for a 5-input summer, and so on. Note also that
if the amplifier of the summing circuit is connected as a unity follower with RA
equal to zero and RB equal to infinity, then with no voltage gain the output
voltage VOUT will be exactly equal the average value of all the input voltages.
That is VOUT = (V1 + V2)/2.
8/21/2020 18

19. Contd..
Summing Amplifier Applications
 These amplifiers are used in an audio mixer to add different signals with equal
gains.
 There are various resistors are used at the input of the summing amplifier to
give a weighted sum.
 This amplifier is used to apply a DC offset voltage with an AC signal voltage.
8/21/2020 19

20. Difference Amplifier or Op Amp Subtractor
 A difference amplifier or op amp subtractor is a specially designed op amp
based amplifier circuit, which amplifies the difference between two input
signals and rejects any signals common to both inputs.
8/21/2020 20

21. Contd..
Let us consider the above op amp circuit. Now, by applying Kirchhoff Current Law
at node 1, we get,
We have written this equation by assuming that there is no current entering in the
inverting terminal of the op amp.
Now, by simplifying the above equation, we get,
8/21/2020 21

22. Contd..
Now, by applying Kirchhoff Current Law, at node 2, we get,
We know that, in ideal op amp, voltage at inverting input is same as the voltage at
non inverting input. Hence,
8/21/2020 22

23. Contd..
So, now from equation (i) and (ii), we get,
The difference amplifier must reject any signal common to both inputs. That means,
if polarity and magnitude of both input signals are same, the output must be zero.
This condition must be satisfied only when,
8/21/2020 23

24. Contd..
In that case, equation (iii) becomes
Again, if we make, R1 = R2, then equation (iv) becomes,
So, if R1 = R2 and also R3 = R4 then the difference amplifier becomes a perfect
subtractor, which subtracts directly the input signals.
Finally, the circuit of op amp substractor becomes,
8/21/2020 24

25. Voltage Follower
 Voltage follower is an Op-amp circuit whose output voltage straight away
follows the input voltage. That is output voltage is equivalent to the input
voltage. Op-amp circuit does not provide any amplification. Thus, voltage gain
is equal to 1. They are similar to discrete emitter follower. The other names of
voltage follower are Isolation Amplifier, Buffer Amplifier, and Unity-Gain
Amplifier. The voltage follower provides no attenuation or no amplification but
only buffering. This circuit has an advantageous characteristic of very high
input impedance.
 This high input impedance of voltage follower is the reason of it being used in
several circuits. The voltage follower gives an efficient isolation of output from
the input signal. The circuit of voltage follower is shown below.
8/21/2020 25

26. Contd..
8/21/2020 26

27. Contd..
 For understanding this concept and the use of voltage follower, we can go
through the following examples.
 First, we can consider a circuit of low impedance load and a power source is
feeding it shown below. Here, a large amount of current is drawn by the load
due to the low resistance load as explained by Ohm’s law. Thus, the circuit
takes a large amount of power from the power source, resulting in high
disturbances in the source.
8/21/2020 27

28. Contd..
Next, we can consider that we are giving the same power to the voltage follower.
Because of its very high input impedance, a minimal amount of current is taken by
this circuit. The output of the circuit will be same as that of the input due to the lack
of feedback resistors.
8/21/2020 28

29. Contd..
Voltage Follower in Voltage Divider Circuits
 In every circuit, voltage is shared or distributed to the impedance or resistance
of the connected components. When Op-amp is connected, the major part of
voltage will drop across it due to high impedance. So, if we use voltage
follower in voltage divider circuits, it will let adequate voltage to be supplied
across the load.
 Let us go through a voltage divider circuit with voltage follower as shown in
the figure below.
8/21/2020 29

30. Contd..
Here, the voltage divider is in the middle of two 10 KΩ resistors and the Op-amp.
This Op-amp will offer input resistance of some hundreds of megaohm. Now, we can
assume it to be 100 MΩ. So the equivalent parallel resistance will be 10 KΩ || 100
KΩ.
So, we get 10KΩ || 10KΩ. We know that the voltage divider which comprises of two
similar resistances will offer exactly the half of the voltage in the power source. We
can prove it using voltage divider formula as follows:
Thus, this 5V will drop across the 10KΩ resistance in the top and 5V drop across the
resistance 10KΩ in the bottom and the load resistance 100Ω (since 10 KΩ||100 Ω,
same voltage will drop in resistors which are in parallel).
8/21/2020 30

31. Contd..
 From this, we have seen how the Op-amp works as a buffer for getting the
desired voltage to the connected load. In the same circuit with the absence of
voltage follower, it will not work due to the lack of supply of sufficient voltage
across the load.
 Mainly, voltage follower is implemented in circuits for two reasons. One is
isolating purpose, and the other is for buffering the output voltage from an
electrical or electronic circuit to get the desired voltage to the connected load.
8/21/2020 31

32. Contd..
Advantages of Voltage Follower
 Provides power gain and current gain.
 Low output impedance to the circuit which uses the output of the voltage follower.
 The Op-amp takes zero current from the input.
 Loading effects can be avoided.
Applications of Voltage Follower
 Buffers for logic circuits.
 In Sample and hold circuits.
 In Active filters.
 In Bridge circuits via transducer.
8/21/2020 32

33. Differentiator Op Amp
 The basic operational amplifier differentiator circuit produces an output signal
which is the first derivative of the input signal.
 Here, the position of the capacitor and resistor have been reversed and now the
reactance, XC is connected to the input terminal of the inverting amplifier while
the resistor, Rƒ forms the negative feedback element across the operational
amplifier as normal.
8/21/2020 33

34. Contd..
 This operational amplifier circuit performs the mathematical operation of
Differentiation, that is it “produces a voltage output which is directly
proportional to the input voltage’s rate-of-change with respect to time“. In other
words the faster or larger the change to the input voltage signal, the greater the
input current, the greater will be the output voltage change in response,
becoming more of a “spike” in shape.
 As with the integrator circuit, we have a resistor and capacitor forming an RC
Network across the operational amplifier and the reactance ( Xc ) of the
capacitor plays a major role in the performance of a Op-amp Differentiator.
8/21/2020 34

35. Contd..
 The input signal to the differentiator is applied to the capacitor. The capacitor
blocks any DC content so there is no current flow to the amplifier summing
point, X resulting in zero output voltage. The capacitor only allows AC type
input voltage changes to pass through and whose frequency is dependant on the
rate of change of the input signal.
 At low frequencies the reactance of the capacitor is “High” resulting in a low
gain ( Rƒ/Xc ) and low output voltage from the op-amp. At higher frequencies
the reactance of the capacitor is much lower resulting in a higher gain and
higher output voltage from the differentiator amplifier.
 However, at high frequencies an op-amp differentiator circuit becomes unstable
and will start to oscillate. This is due mainly to the first-order effect, which
determines the frequency response of the op-amp circuit causing a second-order
response which, at high frequencies gives an output voltage far higher than
what would be expected. To avoid this the high frequency gain of the circuit
needs to be reduced by adding an additional small value capacitor across the
feedback resistor Rƒ.
8/21/2020 35

36. Contd..
 Ok, some math’s to explain what’s going on!. Since the node voltage of the
operational amplifier at its inverting input terminal is zero, the current, i
flowing through the capacitor will be given as:
 The charge on the capacitor equals Capacitance times Voltage across the
capacitor
 Thus the rate of change of this charge is:
8/21/2020 36

37. Contd..
but dQ/dt is the capacitor current, I
from which we have an ideal voltage output for the op-amp differentiator is given as:
Therefore, the output voltage Vout is a constant –Rƒ*C times the derivative of the
input voltage Vin with respect to time. The minus sign (–) indicates a 180o phase
shift because the input signal is connected to the inverting input terminal of the
operational amplifier.
8/21/2020 37

38. Contd..
 One final point to mention, the Op-amp Differentiator circuit in its basic form
has two main disadvantages compared to the previous operational amplifier
integrator circuit. One is that it suffers from instability at high frequencies as
mentioned above, and the other is that the capacitive input makes it very
susceptible to random noise signals and any noise or harmonics present in the
source circuit will be amplified more than the input signal itself. This is because
the output is proportional to the slope of the input voltage so some means of
limiting the bandwidth in order to achieve closed-loop stability is required.
Op-amp Differentiator Waveforms
 If we apply a constantly changing signal such as a Square-wave, Triangular or
Sine-wave type signal to the input of a differentiator amplifier circuit the
resultant output signal will be changed and whose final shape is dependant upon
the RC time constant of the Resistor/Capacitor combination.
8/21/2020 38

39. Contd..
8/21/2020 39

40. Contd..
Improved Op-amp Differentiator Amplifier
 The basic single resistor and single capacitor op-amp differentiator circuit is not
widely used to reform the mathematical function of Differentiation because of
the two inherent faults mentioned above, “Instability” and “Noise”. So in order
to reduce the overall closed-loop gain of the circuit at high frequencies, an extra
resistor, Rin is added to the input as shown below.
8/21/2020 40

41. Contd..
 Adding the input resistor RIN limits the differentiators increase in gain at a ratio
of Rƒ/RIN The circuit now acts like a differentiator amplifier at low frequencies
and an amplifier with resistive feedback at high frequencies giving much better
noise rejection.
 Additional attenuation of higher frequencies is accomplished by connecting a
capacitor Cƒ in parallel with the differentiator feedback resistor, Rƒ. This then
forms the basis of a Active High Pass Filter as we have seen before in the filters
section.
8/21/2020 41

42. Integrator Op Amp
 The integrator Op-amp produces an output voltage that is both proportional to
the amplitude and duration of the input signal.
 Operational amplifiers can be used as part of a positive or negative feedback
amplifier or as an adder or subtractor type circuit using just pure resistances in
both the input and the feedback loop.
 But what if we where to change the purely resistive ( Rƒ ) feedback element of
an inverting amplifier with a frequency dependant complex element that has a
reactance, ( X ), such as a Capacitor, C. What would be the effect on the op-
amps voltage gain transfer function over its frequency range as a result of this
complex impedance.
8/21/2020 42

43. Contd..
 By replacing this feedback resistance with a capacitor we now have an RC
Network connected across the operational amplifiers feedback path producing
another type of operational amplifier circuit commonly called an Op-amp
Integrator circuit as shown below.
 As its name implies, the Op-amp Integrator is an operational amplifier circuit
that performs the mathematical operation of Integration, that is we can cause
the output to respond to changes in the input voltage over time as the op-amp
integrator produces an output voltage which is proportional to the integral of
the input voltage.
8/21/2020 43

44. Contd..
 In other words the magnitude of the output signal is determined by the length of
time a voltage is present at its input as the current through the feedback loop
charges or discharges the capacitor as the required negative feedback occurs
through the capacitor.
 When a step voltage, Vin is firstly applied to the input of an integrating
amplifier, the uncharged capacitor C has very little resistance and acts a bit like
a short circuit allowing maximum current to flow via the input resistor, Rin as
potential difference exists between the two plates. No current flows into the
amplifiers input and point X is a virtual earth resulting in zero output. As the
impedance of the capacitor at this point is very low, the gain ratio of XC/RIN is
also very small giving an overall voltage gain of less than one, ( voltage
follower circuit ).
 As the feedback capacitor, C begins to charge up due to the influence of the
input voltage, its impedance Xc slowly increase in proportion to its rate of
charge. The capacitor charges up at a rate determined by the RC time constant, (
τ ) of the series RC network. Negative feedback forces the op-amp to produce
an output voltage that maintains a virtual earth at the op-amp’s inverting input.
8/21/2020 44

45. Contd..
 Since the capacitor is connected between the op-amp’s inverting input (which is
at virtual ground potential) and the op-amp’s output (which is now negative),
the potential voltage, Vc developed across the capacitor slowly increases
causing the charging current to decrease as the impedance of the capacitor
increases. This results in the ratio of Xc/Rin increasing producing a linearly
increasing ramp output voltage that continues to increase until the capacitor is
fully charged.
 At this point the capacitor acts as an open circuit, blocking any more flow of
DC current. The ratio of feedback capacitor to input resistor ( XC/RIN ) is now
infinite resulting in infinite gain. The result of this high gain (similar to the op-
amps open-loop gain), is that the output of the amplifier goes into saturation as
shown below. (Saturation occurs when the output voltage of the amplifier
swings heavily to one voltage supply rail or the other with little or no control in
between).
8/21/2020 45

46. Contd..
8/21/2020 46
The rate at which the output voltage increases (the rate of change) is determined by
the value of the resistor and the capacitor, “RC time constant“. By changing this RC
time constant value, either by changing the value of the Capacitor, C or the Resistor,
R, the time in which it takes the output voltage to reach saturation can also be changed
for example.

47. Contd..
 If we apply a constantly changing input signal such as a square wave to the
input of an Integrator Amplifier then the capacitor will charge and discharge
in response to changes in the input signal. This results in the output signal being
that of a sawtooth waveform whose output is affected by the RC time constant
of the resistor/capacitor combination because at higher frequencies, the
capacitor has less time to fully charge. This type of circuit is also known as a
Ramp Generator and the transfer function is given below.
8/21/2020 47

48. Contd..
We know from first principals that the voltage on the plates of a capacitor is equal to
the charge on the capacitor divided by its capacitance giving Q/C. Then the voltage
across the capacitor is output Vout therefore: -Vout = Q/C. If the capacitor is
charging and discharging, the rate of charge of voltage across the capacitor is given
as:
But dQ/dt is electric current and since the node voltage of the integrating op-amp at
its inverting input terminal is zero, X = 0, the input current I(in) flowing through the
input resistor, Rin is given as:
8/21/2020 48

49. Contd..
The current flowing through the feedback capacitor C is given as:
Assuming that the input impedance of the op-amp is infinite (ideal op-amp), no
current flows into the op-amp terminal. Therefore, the nodal equation at the inverting
input terminal is given as:
To simplify the math’s a little, this can also be re-written as:
8/21/2020 49

50. Contd..
 Where: ω = 2πƒ and the output voltage Vout is a constant 1/RC times the
integral of the input voltage VIN with respect to time.
 Thus the circuit has the transfer function of an inverting integrator with the gain
constant of -1/RC. The minus sign ( – ) indicates a 180o phase shift because the
input signal is connected directly to the inverting input terminal of the
operational amplifier.
The AC or Continuous Op-amp Integrator
 If we changed the above square wave input signal to that of a sine wave of
varying frequency the Op-amp Integrator performs less like an integrator and
begins to behave more like an active “Low Pass Filter”, passing low frequency
signals while attenuating the high frequencies.
 At zero frequency (0Hz) or DC, the capacitor acts like an open circuit due to its
reactance thus blocking any output voltage feedback. As a result very little
negative feedback is provided from the output back to the input of the amplifier.
8/21/2020 50

51. Contd..
 Therefore with just a single capacitor, C in the feedback path, at zero frequency
the op-amp is effectively connected as a normal open-loop amplifier with very
high open-loop gain. This results in the op-amp becoming unstable cause
undesirable output voltage conditions and possible voltage rail saturation.
 This circuit connects a high value resistance in parallel with a continuously
charging and discharging capacitor. The addition of this feedback resistor, R2
across the capacitor, C gives the circuit the characteristics of an inverting
amplifier with finite closed-loop voltage gain given by: R2/R1.
 The result is at high frequencies the capacitor shorts out this feedback resistor,
R2 due to the effects of capacitive reactance reducing the amplifiers gain. At
normal operating frequencies the circuit acts as an standard integrator, while at
very low frequencies approaching 0Hz, when C becomes open-circuited due to
its reactance, the magnitude of the voltage gain is limited and controlled by the
ratio of: R2/R1.
8/21/2020 51

52. Contd..
The AC Op-amp Integrator with DC Gain Control
 Unlike the DC integrator amplifier above whose output voltage at any instant
will be the integral of a waveform so that when the input is a square wave, the
output waveform will be triangular. For an AC integrator, a sinusoidal input
waveform will produce another sine wave as its output which will be 90o out-
of-phase with the input producing a cosine wave.
 Further more, when the input is triangular, the output waveform is also
sinusoidal. This then forms the basis of a Active Low Pass Filter as seen before
in the filters section tutorials with a corner frequency given as.
8/21/2020 52

53. Contd..
8/21/2020 53
n the next tutorial about Operational Amplifiers, we will look at another type of
operational amplifier circuit which is the opposite or complement of the Op-amp
Integrator circuit above called the Differentiator Amplifier.
As its name implies, the differentiator amplifier produces an output signal which is
the mathematical operation of differentiation, that is it produces a voltage output
which is proportional to the input voltage’s rate-of-change and the current flowing
through the input capacitor.

54. Clamping Circuit
The circuit which can set the positive or negative peak of the input AC signal at a
required level is what we call as a clamping circuit. This circuit quickly adds or
subtracts a DC element to the input AC signal. Other than the name clamping
circuit, Baseline Stabilizer or DC Reinserted or Level Shifter or D.C Restorer are
also frequently used.
 The requirement of clamper circuit is similar to in TV receivers; where the
signal moves through the capacitive coupling network, then the dc component
of these signal will get lost. This DC component (not exactly same as that of the
lost one) will get restored using this clamping circuit. This circuit will add the
DC element which will be positive or negative to the AC input signal. It pushes
the signal towards the positive or the negative side (shown below). Here; when
it shifts the signal to upward side or positive side, both negative peak and the
zero level will meet which is called positive clamper circuit.
8/21/2020 54

55. Contd..
 When it shifts the signal to the downside or negative side, both positive peak
and the zero level will meet which we refer as the negative clamper circuit.

8/21/2020 55

56. Contd..
 The minimum number of components of a clamping circuit is three – a capacitor,
a resistor and a diode. In some cases, DC supply is also needed to give an
additional shift. The nature of the waveform remains alike, but the difference is in
the shifted level. The peak to peak value of the waveform will never change.
The peak value and average value of the input waveform and the clamped output
will be different. The time constant of the circuit (RC) must have to be ten times
the time-period of the entering (input) AC voltage for better clamping action.
 Now, we can assume a negative clamper shown in figure 2. Throughout the
positive half cycle of input, the D diode will conduct, and the output voltage will
be same as barrier potential of the diode (V0). At that time, the capacitor will get
charged to (V – V0). Throughout the negative half cycle of input, the diode will
become negative biased, and it has no role on capacitor voltage. The capacitor
cannot discharge a lot because of the high value of R. Therefore output voltage
will be – (2V- V0). The peak to peak voltage will be 2V. The output waveform that
we get will be the original signal shifted in the downward direction.
8/21/2020 56

57. Contd..
8/21/2020 57
Next, we can assume a positive clamper shown in figure 3. The one and only
distinction from the previous circuit is that diode is in reverse polarity. So the output
will be the shifted original signal in the upward direction.

58. Contd..
 The explanation and working are same as above clamper circuit. Thus we can
conclude that we obtain the positive clamper if the diode in the circuit points
the upward direction and we get negative clamperwhen the diode points in the
downward direction.
8/21/2020 58

59. Contd..
Principle of Operation of Clamping Circuit
In general, the clamper circuit depends on a variation in the capacitor time constant
The time constant should be sufficient that the capacitor voltage does not discharge
considerably throughout the non-conducting diode period. One should select the
values of capacitance and resistance in such a way that the circuit keeps the time
constant high. For preventing quick the discharge of capacitor, the resistance value
should be high. All through the diode conducting period, the capacitor charging
should be very fast. For this, we select a small value of capacitance.
The C in the positive clamper charges quickly throughout the first negative phase of
AC input voltage. When Vin becomes +ve, the C serves as a voltage doubler and
when Vin is –ve, the C operates as a battery with voltage Vin. Thus, we can conclude
that the capacitor and input voltage act against each other. This results in zero net
voltage as seen by the load.
8/21/2020 59

60. Contd..
8/21/2020 60
Biased positive clamping circuit

61. Contd..
Biased negative clamping circuit
8/21/2020 61

62. Classification of Clipping Circuit
According to non linear devices used
- Diode Clippers.
- Transistor Clippers.
According to biasing
- Unbiased Clippers.
- Biased Clippers.
According to configuration used
- Series diode clippers.
- Shunt or Parallel diode clippers.
- A series combination of reference supply, resistor and diode.
- Multi-diode clippers comprises of a number of diodes, resistors and
reference voltage.
- Two emitter-coupled transistors functioning as an over driven difference
amplifier.
8/21/2020 62

63. Classification of Clipping Circuit
According to non linear devices used
- Diode Clippers.
- Transistor Clippers.
According to biasing
- Unbiased Clippers.
- Biased Clippers.
According to configuration used
- Series diode clippers.
- Shunt or Parallel diode clippers.
- A series combination of reference supply, resistor and diode.
- Multi-diode clippers comprises of a number of diodes, resistors and
reference voltage.
- Two emitter-coupled transistors functioning as an over driven difference
amplifier.
.
8/21/2020 63

64. Clipping Circuit
 It is in fact a wave shaping circuit, which can control the shape of the output
waveform by eliminating or clipping a part of applied wave. This is done
without distorting the other (remaining) part of waveform.
 The clipping circuit does not have energy storage elements (capacitors) but it
includes both linear (Resistors) and nonlinear elements (transistors or junction
diodes). This circuit is normally used for the selection in the transmission
purpose.
 In transmission, a portion of a signal wave form occupied below or above a
particular reference voltage level is selected. Other than the name – Clipping
circuits; Slicers, Clippers, Limiters or Amplitude selectors are also often used.
8/21/2020 64

65. Contd..
Diode Clippers
 At least two components – an ideal diode and resistor are employed for the
formation of these clippers. In some cases; for fixing the clipping level, a DC
battery is also used (Figure 1). When the circuit is forward biased, the ideal
diode used operates as a closed switch.
 When the circuit is reversed biased, the ideal diode used operates as an open
switch. Here; by altering the voltage of the battery and by exchanging the
position of the various elements, the input waveform can be clipped.
8/21/2020 65

66. Contd..
Positive clippers
It actually removes the positive half cycles of the input voltage. Here in positive
series clipper, when the input is positive then the diode is in reverse biased condition
(output is zero) and when the input is negative, then the diode is in forward biased
condition (figure 2)
8/21/2020 66

67. Contd..
In positive shunt clipper, when the input is negative, the whole input voltage appears
across the resistor RL or diode (if R<L). When R<L, this circuit will operates as
voltage divider and its output voltage is
8/21/2020 67

68. Contd..
When the diode is connected in reverse polarity in the circuits of positive series
clipper and positive shunt clipper, it becomes negative series clipper and negative
shunt clipper respectively (figures 4 and 5). These clippers can eliminate the entire
negative half cycle of input voltage.
8/21/2020 68

69. Contd..
8/21/2020 69
The clippers discussed above are considered as the circuits with ideal diode. But if
the barrier potential (V0) is considered (Si = 0.7 V and Ge = 0.3 V), the output
voltage of positive and negative clippers are shown below.

70. Contd..
8/21/2020 70
Biased Clippers
In some cases, we need to eliminate a tiny portion of positive or negative half cycles of
the input signal voltage. In that case we use biased clippers.
While during the negative half cycle of input signal in biased negative clipper circuit, it
removes the input signal voltage when it is greater than the battery voltage. When the
battery and diode is reversed, the clipping can be changed to biased positive clipper
(Figure 7).

71. Contd..
8/21/2020 71

72. Contd..
Combination Clipper
 For removing a portion of both positive and negative half cycle of input signal,
we use this combination clipper (figure below).
8/21/2020 72

73. Contd..
Zener Diode as a Peak Clipper
Zener diodes can be employed for removing (clipping off) the portion of input
waveform in the wave shaping circuit (see figure below).
8/21/2020 73

74. Contd..
Application of Clipper
 TV Transmitters and Receivers.
 Noise Limiters.
 Protection of different circuits.
 Different wave generation.
8/21/2020 74

75. Filters
 Passive Low Pass Filter
 Passive High Pass Filter
 Passive Band Pass Filter
 Active Low Pass Filter
 Active High Pass Filter
 Active Band Pass Filter
 Butterworth Filter Design
 Second Order Filters
 State Variable Filter
 Band Stop Filter
 Sallen and Key Filter
8/21/2020 75

76. Thank You
8/21/2020 76