Regulation – 2019 Academic Year 2023-2024

IFET COLLEGE OF ENGINEERING

(An Autonomous Institution)
Department of Electronics & Communication Engineering
SUBJECT CODE: 19UECPC401 SEM: IV
SUBJECT NAME: Linear Integrated Circuits YEAR: II

UNIT III – APPLICATIONS OF OPAMP

(100% Theoretical)
Sign Changer, Scale Changer, Phase Shift Circuits, Voltage Follower, V-to-I and I-to-V
converters, Adder, Subtractor, Instrumentation amplifier, Integrator, Differentiator, Schmitt
Trigger, Comparators and Waveform generators.

3.1. INTRODUCTION
 The operational amplifier is generally used in closed loop configuration using feedback. Countless
simple circuits for various applications can be constructed using the operational amplifier
combined with resistors and capacitors.
 The variety of applications of operational amplifier can be classified into linear applications
and non-linear applications. The linear applications, the output voltage varies linearly with the
change in input voltage.
 In such applications, a negative feedback is used from the amplifier output to the inverting
input terminal of operational amplifier.
 Some of the examples of linear applications of operational amplifier are: Voltage follower,
adder, subtractor, differential amplifier, integrators, differentiators etc.

 In non-linear applications, the feedback is from output terminal to the non-inverting terminal of the
operational amplifier. The feedback may also be given to inverting terminal using nonlinear
elements like diode and transistors.

 The non-linear applications provide nonlinear input output characteristics. Some of the
examples of nonlinear applications are: precision rectifier, log and antilog amplifier, Schmitt trigger
circuits etc.
3.2. IDEAL INVERTING AMPLIFIER
 The ideal inverting amplifier, in which the output of the amplifier is 180° phase shift when
compared with the input. Hence the amplifier which provides a phase shift of 180° between input
and output is called inverting amplifier shown in fig 3.1.

Fig.3.1. Ideal Inverting Amplifier

Vin = I.Rin and Vout = -I.Rf ……………. (3.1)

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Where,
Vin = Input voltage
Rin = Input resistance
I = Current
Vout = Output voltage
Rf = Output resistance
The closed loop voltage gain (ACL) of an inverting op amp is given as
= Vout / Vin = -I.Rf / I.Rin

ACL = Vout / Vin = - (Rf / Rin) …………….. (3.2)

 The negative sign of the closed-loop gain equation indicates that the output is inverted with
respect to the input applied. In a practical inverting amplifier, the non-inverting input is not
connected to ground directly.
 It should be grounded by a resistor with the same value as R in to keep the input currents equal.
This gives a better chance of the output voltage being zero (or close to 0) volts when the input is
zero volts.

 Note: In an inverting amplifier circuit, if both the resistors Rin and Rf are of equal
magnitude Rf = Rin , then the gain of the inverting amplifier will be -1, producing an output that
is a complement of the applied input, Vout = - Vin as shown in figure 3.2

 This type of an inverting amplifier configuration is generally called Unity Gain Inverter or simply
Inverting Buffer.

Fig.3.2. Input and Output Waveforms of Inverting Amplifier

3.2.1. Features
 An inverting amplifier circuit employs a negative feedback and produces an inverted output with respect to
the input. The gain of an inverting amplifier is thus, indicated as negative.
 The voltage gain of inverting amplifier is independent of the op-amp open-loop gain, which is very large.
 The voltage gain of the inverting amplifier depends upon the resistor values used and hence the
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gain can be accurately controlled by choosing the values of R1 and Rf appropriately.
 If Rf > R1, the gain will be greater than 1.
 If Rf < R1, the gain will be less than 1.
 If Rf = R1, the gain will be unity.
 Thus, the output voltage can be greater than, less than or equal to the input voltage in magnitude and 180o
out of phase.

3.3 SIGN CHANGER

 In the ideal inverting amplifier, if Rf = R1 , then the gain ACL= -1. Thus the magnitude of output is
same as that of the input but its sign is opposite to that of the input.
i.e., Vo=-Vin for Rf = R1 ………….. (3.3)
 This circuit is called as sign changer (if n=1) or phase inverter as shown in figure.3.3.

Fig.3.3. Sign Changer/Scale Changer

3.4. SCALE CHANGER

 In an ideal inverting amplifier, if Rf ≠ R1, then the gain ACL= -K, Where, K= Rf / R1. Thus, the circuit
is used to multiply input by a constant K called scaling factor,
i.e Vo= - KVin ………….. (3.4)
 The resistors must be precision resistors to adjust the scaling factor K precisely. This circuit is
called as Scale changer. (If n= any constant K, K ≠1)

3.5. IDEAL NON-INVERTING AMPLIFIER

 An amplifier which uses the input without producing any phase shift between input and output is called
non-inverting amplifier.
 The basic circuit diagram of a non-inverting amplifier using Op-amp is shown. The input is applied to
the non-inverting terminal of the op-amp. In other words a non-inverting amplifier behaves like a
voltage follower circuit.
 A non-inverting amplifier also uses negative feedback connection, but instead of feeding the entire
output signal to the input, only a part of the output signal voltage is fed back as input to the inverting input
terminal of the op-amp.
 The high input impedance and low output impedance of the non- inverting amplifier makes the circuit
ideal for impedance buffering applications

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Fig.3.4. Ideal Non Inverting Amplifier

 From the circuit shown in fig.3.4, it can be seen that the output voltage is potentially divided
across resistors R1 and Rf, before it is applied to the inverting input.
 When the non-inverting input is connected to ground, i.e. Vin = 0, the voltage at the inverting input
terminal must also be at ground level; if not, any voltage difference between the input terminals would
be amplified to move the inverting input terminal back to ground level, because of the concept of
virtual ground.
 Since the inverting input terminal is at ground level, the junction of the resistors R1 and Rf must also
be at ground level. This implies that the voltage drop across R2 will be zero. As a result, the current
flowing through R1 and Rf must be zero.
 Thus, there is zero voltage drops across R1 , and therefore the output voltage is equal to the input voltage,
which is 0V.
 When a positive-going input signal is applied to the non-inverting input terminal, the output
voltage will shift to keep the inverting input terminal equal to that of the input voltage applied. Hence,
there will be a feedback voltage developed across resistor Rf,
VRf = Vin = I fRf ………….. (3.5)
Where, If is the current flowing at the junction of resistors R1 and Rf
Vout = I f (R1 + R f) ………….. (3.6)
Voltage Gain:
 From the above equations of Vin and Vout, the closed-loop voltage gain of the non-inverting
amplifier can be calculated as
ACL = Vout / Vin
= If (R1 + Rf) / IfRf
= (R1 + R f) / Rf
ACL = 1 + (R1 / R f) ………….. (3.7)
 The above gain equation is positive, indicating that the output will be in-phase with the applied
input signal. The closed-loop voltage gain of a non-inverting amplifier is determined by the
ratio of the resistors R1 and Rf used in the circuit.

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 Practically non-inverting amplifiers will have a resistor in series with the input voltage source, to
keep the input current same at both input terminals.The waveforms are given in fig 3.5.

Fig.3.5.Input and Output Waveforms of Non-Inverting Amplifier

3.5.1. Virtual Short in Non-Inverting amplifier
In a non-inverting amplifier, there exists a virtual short between the two input terminals. A virtual short
is a short circuit for voltage, but an open-circuit for current. The virtual short uses two properties of an ideal op-
amp:
 Since Rin is infinite, the input current at both the terminals is zero.
 Since AOL is infinite, the difference voltage (V1 -V2) is always zero.
 Although virtual short is an ideal approximation, it gives accurate values when used with heavy
negative feedback. As long as the op-amp is operating in the linear region (not saturated,
positively or negatively), the open loop voltage gain approaches infinity and a virtual short exists
between two input terminals.
 Because of the virtual short, the inverting input voltage follows the non-inverting input voltage. If
the non-inverting input voltage increases or decreases, the inverting input voltage immediately increases
or decreases to the same value. This action is often referred to as Bootstrapping.
3.5.2. Features
 A non-inverting amplifier uses a voltage-divider-bias negative feedback connection.
 The voltage gain is always greater than one.
 The voltage gain is positive, indicating that for AC input, the output is in-phase with the input signal
and for dc input the output polarity is same as the input polarity.
 The voltage gain of non-inverting op-amp depends only on the resistor values, and is independent of the
open-loop gain of the op-amp.
 The desired voltage gain can be obtained by choosing the appropriate values of the resistors.

3.5.3. Comparison
S.No Ideal Inverting Amplifier Ideal Non - Inverting Amplifier
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1 Voltage gain = - Rf / R1 Voltage gain = 1+Rf / R1
2 The output is inverted with respect to No phase shift between input and
input. output
3 The voltage gain can be adjusted as The voltage gain is always greater than
greater than, equal to or less than one. one.
4 The input impedance is R 1 . The input impedance is very large.

3.6. PHASE SHIFT CIRCUIT

 The phase shift circuits produce phase shifts that depend on the frequency and maintain a constant gain.
 These circuits are also called constant-delay filters or all-pass filters.
 That constant delay refers to the fact the time difference between input and output remains constant when
frequency is changed over arrange of operating frequencies.
 This is called all-pass because normally a constant gain is maintained for all the frequencies within the operating
range. The two types of circuits, for lagging phase angles and leading phase angles.

3.6.1. Phase Lag Circuit

 Phase log circuit is constructed using an op-amp, connected in both inverting and non-inverting
modes.
 To analyze the circuit operation, it is assumed that the input voltage V1 drives a simple inverting
amplifier with inverting input applied at (-) terminal of op-amp and a non-inverting amplifier with a
RC low-pass filter.
 It is also assumed that inverting gain is -1 and non-inverting gain after the low-pass circuit is

ACL = 1 + R f / R1
= 1 + 1 (Since R f = R1 )
ACL = 2
 In complex representation, the Phase lag circuit when the inputs are tied together, the ACL is given
as,
𝑉𝑜 1−𝑗𝜔𝑅𝐶
=( ) ………….. (3.8)
𝑉𝑖 1+𝑗𝜔𝑅𝐶
 The phase angle can be expressed as, θ

θ = - tan-1(𝜔𝑅𝐶) - tan-1(𝜔𝑅𝐶)
= - 2 tan-1(𝜔𝑅𝐶) ………….. (3.9)
Here when 𝜔 = 0, the phase angle approaches zero.
when 𝜔 = α, the phase angle approaches -1800.
= - 2 tan-1(2πf𝑅𝐶)
𝑓
θ = - 2 tan-1 ( ) ………….. (3.10)
𝑓0

 When f = f0, the phase angle θ = 90 0. The Phase Lag circuit and the bode plot for the phase lag

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circuit is as shown in the figure 3. 6 (a) & (b)

Fig. 3.6(a).Phase Lag Circuit

Fig. 3.6(b).Bode plot for the Phase lag Circuit

3.6.2. Phase Lead Circuit
 To analyze the circuit operation, it is assumed that the input voltage V 1 drives a simple inverting
amplifier with inverting input applied at (-) terminal of op-amp and a non-inverting amplifier with a RC
High -pass filter.

Fig. 3.7(a).Phase Lead Circuit

 In complex representation, the Phase lag circuit when the inputs are tied together, the A CL is given
as,
𝑉𝑜 −1+𝑗𝜔𝑅𝐶
=( ) ………….. (3.11)
𝑉𝑖 1+𝑗𝜔𝑅𝐶

𝑓
θ = 1800 - 2 tan-1 ( ) ………….. (3.12)
𝑓0

 When f = 0, the phase angle θ = 180 0. The Phase Lag circuit and the bode plot for the phase lead
circuit is as shown in the figure 3.7(a) & (b)

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Fig. 3.7(b).Bode plot for the Phase Lead Circuit

3.7. 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 in figure.3.8.

Fig.3.8.Voltage Follower
 `First, we can consider a circuit of low impedance load and a power source is feeding it shown
below in fig .3.9. 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.

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Fig.3.9. Impedance diagram

 Now consider that the same power is given to the voltage follower. Because of its very high
input impedance, a minimal amount of current is taken by this circuit shown in fig 3.10. The
output of the circuit will be same as that of the input due to the lack of feedback resistors.

Fig.3.10.Voltage Follower VA=VB = Vin

 The node A is directly connected to the output. Hence,

Vo= VA ………….. (3.13)
Equating the above equations, Vo =Vin

3.7.1. Advantages of Voltage follower

 Very large input resistance, of the order of MΩ.
 Low output impedance, almost zero. Hence it can be used to connect high impedance source to
low impedance load, as a buffer.
 It has large bandwidth.
 The output follows the input exactly without a phase shift.

3.8. VOLTAGE TO CURRENT CONVERTER

 The circuits in instrumentation for analog representation of certain physical quantities (weight,
pressure, motion etc), DC current is preferred. This is because DC current signals will be
constant throughout the circuit in series from the source to the load.
 The current sensing instruments also have the advantage of less noise. So, sometimes it is essential to
create current which is corresponding or proportional to a definite voltage. For this purpose Voltage to
Current Converters are used.
 It can simply change the carrier of electrical data from voltage to current. In a voltage to current
converter, the output load current is proportional to the input voltage. According to the connection of
load, there are two types of V-I Conversion:
 Floating type and grounded type. In the floating type, RL is not connected to the ground whereas in
grounded type, one end of RL is connected to the ground shown in fig 3.11.

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Fig.3.11. Voltage to Current Converter

3.8.1. Voltage to Current Converter with Floating Load

 As the name indicates, the load resistor is floating in this converter circuit shown in fig 3.12. That is,
the resistor RL is not linked to ground.
 The voltage, VIN which is the input voltage is given to the non-inverting input terminal. The
inverting input terminal is driven by the feedback voltage which is across the RL resistor.

Fig.3.12.V-I Converter with floating load

 This feedback voltage is determined by the load current and it is in series with the VD,
which is the input difference voltage. So, this circuit is also known as current series negative
feedback amplifier.
 For the input loop, the voltage equation is
VIN = VD + VF ………….. (3.14)
 Since A is very large, VD = 0.
VIN = VF ………….. (3.15)
 Since, the input to the Op-amp, (i.e.,) I B’=0
VIN = I L x R
I L = VIN/R .………….. (3.16)
 From the above equation, it is clear that the load current depends on the input voltage and the input
resistance that is, the load current, which is the input voltage.
 The load current is controlled by the resistor, R. Here, the proportionality constant is 1/R. So, this
converter circuit is also known as Trans-Conductance Amplifier. Other name of this circuit is Voltage
Controlled Current Source.
 The type of load may be resistive, capacitive or non-linear load. The type of load has no role in the
above equation. When the load connected is capacitor then it will get charge or discharge at a steady rate.
Due to this reason, the converter circuit is used for the production of saw tooth and triangular wave
forms.
Note: Thus the load current is always proportional to the input voltage and circuit works as Voltage

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to current converter.

3.8.2. Voltage to Current Converter with Grounded Load

 This converter is also known as Howland Current Converter. Here, one end of the load is always
grounded as shown in fig 3.13. For the circuit analysis, we have to first determine the voltage, VIN and
then the relationship or the connection between the input voltage and load current can be achieved.

Fig.3.13.V-I Converter with grounded load

 For that, we apply Kirchoff’s current law at the node V1,

A=1+ (Rf/R1)
I L=VIN/R .………….. (3.17)

 Thus, we can conclude from the above equation that the current I L is related to the voltage,
VIN and the resistor, R.

3.8.3. Application of V-I Converter

 Low voltage AC voltmeter
 Zener diode tester
 Testing LEDs

 Testing diodes

3.9. CURRENT TO VOLTAGE CONVERTER

 A current to voltage converter s h o w n i n f i g 3 . 1 4 will produce a voltage
proportional to the given current. This circuit is required if your measuring instrument is
capable only of measuring voltages and you need to measure the current output.
 If the instrument or data acquisition module (DAQ) has input impedance that is several orders
larger than the converting resistor, a simple resistor circuit can be used to do the conversion.
 However, if the input impedance of the instrument is low compared to the converting resistor then
the following op-amp circuit should be used. This circuit is also referred as Current Controlled
Voltage Source (CCVS).
 If the resistance in circuit is replaced by the impedance Z, The circuit is called as Trans-
impedance amplifier.
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Fig.3.14. Current to Voltage Converter

 To analyse the current to voltage converter by inspection,

 If we apply KCL to the node at V- (the inverting input) and let the input current to the inverting
input be I-, then

Vout −(V−)
= 𝐼𝑝 + (𝐼−) …………… (3.18)
𝑅𝑓

 Since the output is connected to V- through R f, the op-amp is in a negative

feedback configuration. Thus,

(V−) = V+ = 0 …………… (3.19)

 And assuming that I- is 0 and simplifying,

Vout =Ip Rf …………… (3.20)

 One example of such an application is using the photodiode sensor to measure light intensity
shown in fig 3.15. The output of the photodiode sensor is a current which changes proportional to
the light intensity. Another advantage of the op-amp circuit is that the voltage across the
photodiode (current source) is kept constant at 0V.

Fig.3.15. Current to Voltage Converter with photodiode

Note: Thus the output voltage is proportional to the input current and circuit works as a current to voltage
converter.

3.9.1. Applications of I/V Converter

 Some of the most frequent applications of I/V converters are:

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 Photodiode detector
 Photo-FET detector

3.10. SUMMER OR ADDER CIRCUIT

 As the input impedance of an op-amp is extremely large, more than one input signal can be applied to
the inverting amplifier. Such circuit gives the addition of the applied signals to the output. Hence it is called
summer or adder circuit.

 Depending upon the sign of the output, the summer circuits are classified as inverting summer and
non-inverting summer.

 Many applications in electronic circuits require two or more analog signals to be added or combined
into a single output. The summing amplifier does the exact same thing. For this reason, summing amplifier
is also called as Voltage adder since its output is the addition of voltages present at its input terminal.

3.10.1. Inverting Summing Amplifier

 The summing amplifier uses an inverting amplifier configuration, i.e. the input is applied to the inverting
input terminal of the op-amp, while the non-inverting input terminal is connected to ground. Due to this
configuration, the output of voltage adder is out of phase with respect to the input by 1800.

Fig.3.16. Inverting Summing Amplifier

Output Voltage Calculation:
 It is already been said that a summing amplifier is an inverting amplifier with more than one voltage at
the input terminal. For an inverting amplifier, the output voltage is given as,
VOUT = – (Rf/RIN) VIN ………………. (3.21)
 So for the summing amplifier shown above, the output equation would be,

VOUT = – {(Rf/RIN1) VIN1 + (R f/RIN2) VIN2 + (R f/RIN3) VIN3} ….…………. (3.22)

 In a summing amplifier, if the input resistances are not equal, the circuit is called a Scaling Summing
Amplifier.
 If all the input resistances are chosen to be of equal magnitude (RIN), then the output equation of the
summing amplifier can be rewritten as,
VOUT = – {(Rf/RIN) {VIN1 + VIN2 + VIN3} ….…………. (3.23)
 In general, a summing amplifier output equation is given as,
VOUT = – {(Rf/RIN) {V1 + V2 + V3 + + VN}
V0 = - (V1+V2) ….…………. (3.24)
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 Sometimes, it is necessary to just add the input voltages, without amplifying them. In such situations, the value
of input resistance RIN1, RIN2, RIN3, etc. must be chosen equal to that of the feedback resistor Rf. Then, the
gain of the amplifier will be unity. Hence, the output voltage will be an addition of the input voltages.
Note: Theoretically, one can apply as many input voltages to the input of the summing amplifier as is
required. However, it must be noted that all of the input currents are added and then fed back through
the resistor Rf, so one should also be aware of the power rating of resistors.
3.10.2. Non-Inverting Summing Amplifier
A non-inverting summing amplifier can also be constructed, using the non-inverting amplifier configuration.
A non-inverting summing circuit is shown in the figure below:

Fig.3.17. Non - Inverting Summing Amplifier

 That is, the input voltages are applied to the non-inverting input terminal and a part of the output is
fed back to the inverting input terminal, through voltage-divider-bias feedback. If the input
resistances are equal, the output equation of the above circuit is given as,
V0 = (V1+V2) ….…………. (3.25)
3.10.3. Average Circuit
If in the inverting summer circuit, the values of resistance are selected as, R1=R2=R and Rf=R/2, then,
V0= - (V1+V2)/2 ….…………. (3.26)

3.11. SUBTRACTOR OR DIFFERENCE CIRCUIT

 The Subtractor also called a differential amplifier uses both the inverting and non- inverting
inputs to produce an output signal which is the difference between the two input voltages
V1 and V2 allowing one signal to be subtracted from another.
 More inputs can be added to be subtracted if required. If resistances are equal (R = R3 and RA
= R4) then the output voltage is as given and the voltage gain is +1. If the input resistance is
unequal the circuit becomes a differential amplifier producing a negative output when V1 is
higher than V2 and a positive output when V1 is lower than V2 .

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Fig.3.18.Subtractor Circuit
 To find the output V01 due to V1 alone, make V2 = 0. Then the circuit of figure as
shown in the above becomes a non-inverting amplifier having input voltage, V1/2 at
the non-inverting input terminal and the output becomes
V01 = V1/2(1+R/R) = V1 ….…………. (3.27)
 When all resistances are R in the circuit.Similarly the output V02 due to V2 alone (with V1
grounded) can be written simply for an inverting amplifier as
V02 = -V2 ….…………. (3.28)
 Thus the output voltage Vo due to both the inputs can be written as
V0 =V01 - V02 = V1 - V2
 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.

V0=V1 -V2 ….…………. (3.29)

3.12. INSTRUMENTATION AMPLIFIER

 Instrumentation amplifier is a kind of differential amplifier with additional input buffer stages. The
addition of input buffer stages makes it easy to match (impedance matching) the amplifier with the
preceding stage.
 Instrumentation is commonly used in industrial test and measurement application. The instrumentation
amplifier also has some useful features like low offset voltage, high CMRR (Common mode
rejection ratio), high input resistance, high gain etc.

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Fig.3.19.Instrumentation Amplifier

 The circuit diagram of a typical instrumentation amplifier using op-amp is shown below. A circuit
providing an output based on the difference between two inputs (times a scale factor) is given in
the below figure.
 In the circuit diagram, op-amps labeled A1 and A2 are the input buffers. Anyway the gain of these
buffer stages is not unity because of the presence of R1 and Rg. Op-amp labeled A3 is wired as a
standard differential amplifier.
 R3 connected from the output of A3 to its non-inverting input is the feedback resistor. R2 is the input
resistor. The voltage gain of the instrumentation amplifier can be expressed by using the equation
below.

Voltage gain (Av ) = Vo / (V2 -V1 ) = (1 + 2R1 /Rg )x R3/R2 …………..(3.30)

 If needed a setup for varying the gain, replace Rg with a suitable potentiometer. Instrumentation
amplifiers are generally used in situations where high sensitivity, accuracy and stability are
required. Instrumentation amplifiers can be also made using two op-amps, but they are rarely used and
the common practice is to make it using three op-amps like what is shown here.
 The only advantage of making an instrumentation amplifier using 2 op-amps is low cost and improved
CMRR.
 A high gain accuracy can be achieved by using precision metal film resistors for all the resistances.
Because of large negative feedback employed, the amplifier has good linearity, typically about 0.01%
for a gain less than 10.
 The output impedance is also low, being in the range of milli-ohms. The input bias current of the
instrumentation amplifier is determined by the op-amps A1 and A2.
 A simplified instrumentation amplifier design is shown below. Here the resistances labeled R1 are
shorted and Rg is removed. This results in a full series negative feedback path and the gain of A1 and A2
will be unity. The removal of R1 and Rg simplifies the equation to Av = R3/R2.

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Fig.3.20.Simplified Instrumentation Amplifier

3.12.1. Requirements of a Good Instrumentation amplifier

The Requirements of a Good Instrumentation Amplifier are used to amplify the low level differential
signals very precisely, in presence of the large common mode noise and interference signals. Hence a good
instrumentation amplifier has to meet the following specifications:

Easier gain adjustment ,High input impedance ,Low output impedance ,High CMRR,Low power
consumption ,Low thermal and Time drifts,High slew rate .
3.12.2. Working of an Instrumentation Amplifier
 The most commonly used Instrumentation amplifiers consist of three op-amps. In this circuit, a non-
inverting amplifier is connected to each input of the differential amplifier.
 This instrumentation amplifier provides high input impedance for exact measurement of input data from
transducers. The op-amps 1 & 2 are non-inverting amplifiers and together form an input stage of the
instrumentation amplifier. The op-amp 3 is a difference amplifier that forms the output stage of the
instrumentation amplifier.

 The output stage of the instrumentation amplifier is a difference amplifier, whose output Vo is the
amplified difference of the input signals applied to its input terminals. If the outputs of op- amp 1
and op-amp 2 are Vo1 and Vo2 respectively, then the output of the difference amplifier is given by,

Vo = (R3/R2) (Vo1 -Vo2) …………..(3.31)

 The expressions for Vo1 and Vo2 can be found in terms of the input voltages and resistances. Consider the
input stage of the instrumentation amplifier as shown in the figure. 3.22 .

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Fig.3.21. Input stage of instrumentation amplifier

 The potential at node A is the input voltage V1 . Hence the potential at node B is also V1, from the virtual
short concept. Thus, the potential at node G is also V1 .
 The potential at node D is the input voltage V2 . Hence the potential at node C is also V2, from the virtual
short. Thus, the potential at node H is also V2. Ideally the current to the input stage op-amps is zero.
Therefore the current I through the resistors R1, Rg and R1 remains the same.
 Applying Ohm’s law between the nodes E and F,
I = (Vo1 -Vo2)/(R1+R g+R1)

I = (Vo1 -Vo2)/(2R1+Rg) …………..(3.32)

 Since no current is flowing to the input of the op-amps 1 & 2, the current I between the nodes G and H
can be given as,

I = (VG -VH)/R g = (V1 -V2)/R g ..………… (3.33)

 Equating equations 3.32 and 3.33,

(Vo1 -Vo2)/(2R1+Rg) = (V1-V2)/Rg

(Vo1 -Vo2) = (2R1+Rg)(V1 -V2)/R g ………………(3.34)

 The output of the difference amplifier is given as,

Vout = (R3 /R2) (Vo1 -Vo2) …………… (3.35)

 Therefore, (Vo1 – Vo2) = (R2/R3)Vout Substituting (Vo1 – Vo2) value in the equation 3.35, we get
(R2/R3)Vout = (2R1+Rg)(V1-V2)/R g ……………... (3.36)

i.e., Vout = (R3/R2){(2R1+Rg)/R g}(V1-V2) ..……………. (3.37)

 The above equation gives the output voltage of an instrumentation amplifier. The overall gain of the
amplifier is given by the term (R3/R2){(2R1+Rg)/Rg}.
Note:
 The overall voltage gain of an instrumentation amplifier can be controlled by adjusting the value of resistor

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( R g).
 The common mode signal attenuation for the instrumentation amplifier is provided by the difference
amplifier.

3.12.3. Advantages
 The gain of a three op-amp instrumentation amplifier circuit can be easily varied and controlled by adjusting
the value of R g without changing the circuit structure.
 The gain of the amplifier depends only on the external resistors used. Hence, it is easy to set the gain
accurately by choosing the resistor values carefully.
 The input impedance of the instrumentation amplifier is dependent on the non-inverting amplifier circuits in the
input stage. The input impedance of a non-inverting amplifier is very high.
 The output impedance of the instrumentation amplifier is the output impedance of the difference amplifier,
which is very low.
 The CMRR of the op-amp 3 is very high and almost all of the common mode signal will be rejected.
3.12.4. Applications
The instrumentation amplifier, along with a transducer bridge can be used in a wide variety of applications.
These applications are generally known as data acquisition systems. At the input stage, there is a transducer
device that converts the change in the physical quantity to an electrical signal. The electrical signal is fed to
an instrumentation amplifier. The amplified signal is then fed to a display device, which is calibrated to detect the
change in the quantity being measured.
(i).Light Intensity Meter
(ii). Temperature Indicator
(iii). Temperature Controller
3.13. INTEGRATOR
 Major application of Op-amp is its use in mathematical applications. An Operational Amplifier can be
configured to perform the mathematical operations of Integration and Differentiation.
 In this section, the working of an Operational Amplifier as Integrator will be analyzed in detail.
 Operational amplifier can be configured to perform calculus operations such as differentiation and
integration. In an integrating circuit, the output is the integration of the input voltage with respect to time.
A passive integrator is a circuit which does not use any active devices like op-amps or
transistors.
 An integrator circuit which consists of active devices is called an Active integrator. An active integrator
provides a much lower output resistance and higher output voltage than is possible with a simple RC
circuit. Op-amp differentiating and integrating circuits are inverting amplifiers, with appropriately
placed capacitors.
 Integrator circuits are usually designed to produce a triangular wave output from a square wave input.
Integrating circuits have frequency limitations while operating on sine wave input signals.

3.13.1. Ideal Active Op-Amp Integrator

An op-amp integrating circuit produces an output voltage which is proportional to the area
(amplitude multiplied by time) contained under the waveform. An ideal op-amp integrator uses a capacitor C1,
connected between the output and the op-amp inverting input terminal, as shown in the figure below.

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Fig.3.22. Ideal Active Op-amp Integrator

o The negative feedback to the inverting input terminal ensures that the node X is held at ground potential
(virtual ground). If the input voltage is 0 V, there will be no current through the input resistor R1, and
the capacitor is uncharged. Hence, the output voltage is ideally zero.
o If a constant positive voltage (DC) is applied to the input of the integrating amplifier, the output
voltage will fall negative at a linear rate, in an attempt to keep the inverting input terminal at
ground potential.
o Conversely, a constant negative voltage at the input results in a linearly rising (positive) voltage
at the output. The rate of change of the output voltage is proportional to the value of the applied
input voltage.

3.13.2. Output Voltage Calculation

From the circuit, it is seen that node Y is grounded through a compensating resistor R1. Node X will also
be at ground potential, due to the virtual ground.

VX = VY = 0 ……………..(3.38)
Since the input current to an op-amp is ideally zero, the current flowing through the input resistor,
due to Vin, also flows through the capacitor Cf.
From the input side, the current I is given as,
I = (Vin – VX) / R1 = Vin / R1 ……………..(3.39)
From the output side, the current I is given as,
I = Cf [d(VX – Vout)/dt] = -Cf [d(Vout)/dt] ..…………….(3.40)
Equating the above two equations of I, we get,
[Vin / R1] = – Cf [d(Vout)/dt] ……………….(3.41)
Integrating both the sides of the above equation,
Vout = -(1/R1Cf)ʃVin.dt ….…………..(3.42)
In the above equation, the output is -{1/(R1.Cf)} times the integral of the input voltage, where the term
(R1.Cf) is known as the time constant of the integrator. The negative sign indicates that there is a phase shift of

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180o between input and output, because the input is provided to the inverting input terminal of the op-amp. The
main advantage of an active integrator is the large time constant, which results in the accurate integration of
the input signal.

3.13.3. Integrator as Ramp Generator

 If the step input of the integrating amplifier is replaced by a continuous time square wave, the change in the
input signal amplitude charges and discharges the feedback capacitor.
 This results in a triangular wave output with a frequency that is dependent on the value of (R1.Cf),
which is referred to as the time constant of the circuit. Such a circuit is commonly called a Ramp
Generator.
 During the positive half-cycle of the square wave input, a constant current I flows through the input
resistor R1. Since the current flowing into the op-amp internal circuitry is zero, effectively all of the
current flows through the feedback capacitor Cf.
 This current charges the capacitor. Since the capacitor connected to the virtual ground, the voltage across the
capacitor is the output voltage of the op-amp.
 During the negative half-cycle of the square wave input, the current I is reversed. The capacitor is now linearly
charged and produces a positive - going ramp output.

Fig.3.23. Ramp Waveform

3.13.4. Applications
 Op-amp integrating amplifiers are used to perform calculus operations in analogue computers.
 Integrating circuits are most commonly used in analogue-to-digital converters, ramp generators and also in wave
shaping applications.
 Another application would be to integrate a signal representing water flow, producing a signal
representing the total quantity of water that has passed by the flow meter. This application of an
integrator is sometimes called a totalizer in the industrial instrumentation trade.

3.14. DIFFERENTIATOR
 An op-amp differentiator or a differentiating amplifier is a circuit configuration which produces output
voltage amplitude that is proportional to the rate of change of the applied input voltage.
 A differentiator with only RC network is called a passive differentiator, whereas a differentiator
with active circuit components like transistors and operational amplifiers is called an active
differentiator.
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 Active differentiators have higher output voltage and much lower output resistance than simple RC
differentiators. An op-amp differentiator is an inverting amplifier, which uses a capacitor in series
with the input voltage.
 Differentiating circuits are usually designed to respond for triangular and rectangular input
waveforms. For a sine wave input, the output of a differentiator is also a sine wave, which is out
of phase by 180o with respect to the input (cosine wave).
 Differentiators have frequency limitations while operating on sine wave inputs; the circuit attenuates all
low frequency signal components and allows only high frequency components at the output. In
other words, the circuit behaves like a high-pass filter.

3.14.1. Ideal Differentiator Circuit

 An op-amp differentiating amplifier uses a capacitor in series with the input voltage source, as
shown in the figure below.

Fig.3.24. Ideal Differentiator Circuit

 For DC input, the input capacitor C 1 remains uncharged and behaves like an open-circuit.
 The non-inverting input terminal of the op-amp is connected to ground through a resistor Rcomp ,
which provides input bias compensation, and the inverting input terminal is connected to the
output through the feedback resistor Rf. Thus, the circuit behaves like a voltage follower.
 When the input is a positive - going voltage, a current I flows into the capacitor C1 , as shown in the
figure.
 Since the current flowing into the op-amp internal circuit is zero, effectively all of the current I flows
through the resistor Rf. The output voltage is,
Vout = – (I x Rf) ……………. (3.43)
 Here this output voltage is directly proportional to the rate of change of the input voltage. From the
figure, node X is virtually grounded and node Y is also at ground potential. i.e.,
Vx = Vy = 0 ……………. (3.44)
 From the input side, the current I can be given as,
I = C1 [d(Vin-Vx)/dt] = C1 [d(Vin)/dt] ……………….(3.45)
 From the output side, the current I is given as,
I = -{(Vout -Vx)/R f} = - {Vout/Rf} ……………….(3.46)
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 Equating the above two equations of current we get
C1{d(Vin)/dt} = -Vout/Rf

Vout = -C1 .Rf {d(Vin)/dt} ………………. (3.47)

 Above equation indicates that the output is C1.R f times the differentiation of the input voltage.
The product C1 .Rf is called as the RC time constant of the differentiator circuit.

 The negative sign indicates the output is out of phase by 180o with respect to the input. The main
advantage of such an active differentiating amplifier circuit is the small time constant required
for differentiation.

3.14.2. Input and Output Waveforms

 When a step input with amplitude Vm is applied to an op-amp differentiator, the output can be
mathematically expressed as, Vout = – C1.Rf {d(Vm)/dt}. For simplicity, assume the product
(C1.Rf) is unity. Therefore, Vout = 0 because the amplitude V is constant.
 But practically, the output is not zero since the input step wave takes a finite amount of time
to rise from 0 volts to Vm volts. Hence the output appears like a spike at time t = 0, as shown in
the figure below.

Fig.3.25. Input (Unity) and Output (Spike) Waveform

 If the input to the differentiator is changed to a square wave, the output will be a waveform consisting of
positive and negative spikes, corresponding to the charging and discharging of the capacitor, as
shown in the figure below.

Fig.3.26. Input and Output Waveform

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 For sine wave input, which is mathematically represented as V (t) = V m sin ωt, where Vm is the
amplitude of the input signal and t is the period, the output of the differentiator is given as,
Vout = -C1.Rf {d(Vm sin ωt)/dt} ………………(3.48)
 For simplicity, let us assume the product (C1 .Rf) is unity.
Vout = – Vm. ω. cos ωt ………………(3.49)
 Thus the output of a differentiator for a sine wave input is a cosine wave and the input-
output waveforms are shown in the figure below.

Fig.3.27. Input (Sine) and Output (Cosine)Waveforms

3.14.3. Frequency Response of Ideal Differentiator

 The gain of an op-amp differentiator is directly dependent on the frequency of the input signal. Hence,
for DC inputs where f = 0, the output is also zero. As the frequency of the input signal increases, the
output also increases. The frequency response of an ideal differentiator is as shown in the figure below.

Fig.3.28. Frequency Response of Ideal Differentiator

 The frequency f1 is the frequency for which the gain of the differentiator becomes unity. It can be seen
from the figure that for frequency less than f1 , the gain is less than unity. For f1, the gain becomes the
unity (0 dB) and beyond f1, the gain increases at 20dB per decade.

3.14.4. Disadvantages
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 The gain of the differentiator increases as frequency increases. Thus at some high frequency, the
differentiator may become unstable and break into oscillations. There is a possibility that Op-amp may
go into saturation.
 Also, the input impedance decreases as frequency increases. This makes circuit very much sensitive to
the noise.
3.14.5. Applications
 Differentiating amplifiers are most commonly designed to operate on triangular and rectangular signals.
 Differentiators also find application as wave shaping circuits, to detect high frequency components in the
input signal.

3.15. SCHMITT TRIGGER

 A Schmitt trigger circuit is also called a regenerative comparator circuit. The circuit is designed with a
positive feedback and hence will have a regenerative action which will make the output switch
levels.
 Also, the use of positive voltage feedback instead of a negative feedback, aids the feedback voltage to
the input voltage, instead of opposing it. The use of a regenerative circuit is to remove the
difficulties in a zero-crossing detector circuit due to low frequency signals and input noise voltages.
 Shown below is the circuit diagram of a Schmitt trigger. It is basically an inverting comparator circuit
with a positive feedback.
 The purpose of the Schmitt trigger is to convert any regular or irregular shaped input waveform
into a square wave output voltage or pulse. Thus, it can also be called a Squaring circuit.

Fig.3.29. Schmitt trigger and its waveform

 As shown in the circuit diagram, a voltage divider with resistors R1 and R2 is set in the positive
feedback of the 741 IC op-amp. The same values of Rdiv1 and Rdiv2 are used to get the resistance value
Rpar = R1 ||R2 which is connected in series with the input voltage. Rpar is used to minimize the offset
problems.
 The voltage across R1 is fedback to the non-inverting input. The input voltage Vi triggers or changes
the state of output Vout every time it exceeds its voltage levels above a certain threshold value called
Upper Threshold Voltage (Vupt) and Lower Threshold Voltage (Vlpt).

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 Let us assume that the inverting input voltage has a slight positive value. This will cause a negative
value in the output. This negative voltage is fedback to the non-inverting terminal (+) of the op-amp
through the voltage divider.
 Thus, the value of the negative voltage that is fedback to the positive terminal becomes higher.
The value of the negative voltage becomes again higher until the circuit is driven into negative
saturation (-Vsat).
 Now, let us assume that the inverting input voltage has a slight negative value. This will cause a
positive value in the output.
 This positive voltage is fedback to the non-inverting terminal (+) of the op-amp through the voltage
divider. Thus, the value of the positive voltage that is fedback to the positive terminal becomes higher.
The value of the positive voltage becomes again higher until the circuit is driven into positive
saturation (+Vsat). This is why the circuit is also named a regenerative comparator circuit.
 When Vout = +Vsat, the voltage across Rdiv1 is called Upper Threshold Voltage (Vupt). The input
voltage, Vin must be slightly more positive than Vupt inorder to cause the output Vo to switch from
+Vsat to -Vsat. When the input voltage is less than Vupt, the output voltage Vout is at +Vsat.

Upper Threshold Voltage, Vupt = +Vsat (R 1/[R1+R 2])

 When Vout = -Vsat, the voltage across Rdiv1 is called Lower Threshold Voltage (Vlpt). The input
voltage, Vin must be slightly more negaitive than Vlpt inorder to cause the output Vo to switch from
-Vsat to +Vsat. When the input voltage is less than Vlpt, the output voltage Vout is at -Vsat.

Lower Threshold Voltage, Vlpt = -Vsat (R 1/[R1+R2])

 If the value of Vupt and Vlpt are higher than the input noise voltage, the positive feedback will eliminate
the false output transitions. With the help of positive feedback and its regenerative behaviour, the
output voltage will switch fast between the positive and negative saturation voltages.

Applications of Schmitt trigger:

 Schmitt trigger is mostly used to convert a very slowly varying input voltage into an output having
abruptly varying waveform occurring precisely at certain predetermined value of input voltage.
Schmitt trigger may be used for all applications for which a general comparator is used.
 Any type of input voltage can be converted into its corresponding square signal wave. The only
condition is that the input signal must have large enough excursion to carry the input voltage beyond the
limits of the hysteresis range.
 The amplitude of the square wave is independent of the peak-to-peak value of the input waveform.

3.15 Comparators:
 The Op-amp comparator compares one analogue voltage level with another analogue voltage level,
or some preset reference voltage, VREF and produces an output signal based on this voltage comparison.
In other words, the op-amp voltage comparator compares the magnitudes of two voltage inputs and
determines which is the largest of the two.
 We have seen in previous tutorials that the operational amplifier can be used with negative feedback to
control the magnitude of its output signal in the linear region performing a variety of different
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functions. We have also seen that the standard operational amplifier is characterised by its open-
loop gain AO and that its output voltage is given by the expression: VOUT = AO(V+ – V-) where V+
and V- correspond to the voltages at the non-inverting and the inverting terminals respectively.
 Voltage comparators on the other hand, either use positive feedback or no feedback at all (open- loop
mode) to switch its output between two saturated states, because in the open-loop mode the amplifiers
voltage gain is basically equal to AVO.
 Then due to this high open loop gain, the output from the comparator swings either fully to its
positive supply rail, +Vcc or fully to its negative supply rail, -Vcc on the application of varying input
signal which passes some preset threshold value.

 The open-loop op-amp comparator is an analogue circuit that operates in its non-linear region as changes
in the two analogue inputs, V+ and V- causes it to behave like a digital bistable device as triggering
causes it to have two possible output states, +Vcc or -Vcc.

 Then we can say that the voltage comparator is essentially a 1-bit analogue to digital converter, as the
input signal is analogue but the output behaves digitally.

 Consider the basic op-amp voltage comparator circuit below.

Fig:30: Comparators
 With reference to the op-amp comparator circuit above, lets first assume that VIN is less than the DC
voltage level at VREF, ( VIN < VREF ).
 As the non-inverting (positive) input of the comparator is less than the inverting (negative) input, the
output will be LOW and at the negative supply voltage, -Vcc resulting in a negative saturation of the
output.

 If we now increase the input voltage, VIN so that its value is greater than the reference voltage
VREF on the inverting input, the output voltage rapidly switches HIGH towards the positive supply
voltage, +Vcc resulting in a positive saturation of the output.

 If we reduce again the input voltage VIN , so that it is slightly less than the reference voltage, the op-
amp’s output switches back to its negative saturation voltage acting as a threshold detector.

 Then we can see that the op-amp voltage comparator is a device whose output is dependant on the
value of the input voltage, VIN with respect to some DC voltage level as the output is HIGH when the
voltage on the non-inverting input is greater than the voltage on the inverting input, and LOW when the

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non-inverting input is less than the inverting input voltage.

 This condition is true regardless of whether the input signal is connected to the inverting or the non-
inverting input of the comparator.
 We can also see that the value of the output voltage is completely dependent on the op-amps power
supply voltage. In theory due to the op-amps high open-loop gain the magnitude of its output
voltage could be infinite in both directions.
 However practically, and for obvious reasons it is limited by the op-amps supply rails giving VOUT =
+Vcc or VOUT = -Vcc.

 We said before that the basic op-amp comparator produces a positive or negative voltage output by
comparing its input voltage against some preset DC reference voltage.

 Generally, a resistive voltage divider is used to set the input reference voltage of a comparator, but a
battery source, zener diode or potentiometer for a variable reference voltage can all be used as shown.

3.16 WAVEFORM GENERATORS:

There are different types of wave form generators which are given below. 1 Square wave generator
2 Triangular wave generator 3 Saw tooth wave generator
1 . Square wave generator:
 It is also called a Free running oscillator.The principle of generation of square wave output is to force an
op-amp to operate in saturation region.
 In Figure 3.31(a), fraction β = R2/(R1+R2) of the output is fed back to the +ve input terminal. Thus the
reference voltage Vref is βVo and may take values as +βVsat or – βVsat.
 The output is also fed back to the –ve input terminal after integrating by means of a low pass RC
combination.Whenever input at the –ve input terminal just exceeds Vref, switching takes place resulting in
a square wave output. In Astable multivibrator, both the states are quasi stable.

Fig 3.31 (a)Circuit diagram (b) I/P ,O/P waveforms

Frequency Derivation:
 The frequency is determined by the time it takes the capacitor to charge from –βVsat to +βVsat and
vice versa
 The voltage across the capacitor as a function of time is given by
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Vc(t) = Vf + (Vi – Vf) eʌ -t/RC
Where, the final value, Vf = +Vsat
And the initial value, Vi = -βVsat
Therefore Vc(t) = +Vsat + (-βVsat -Vsat) eʌ -t/Rc
Vc(t) = Vsat – Vsat(1+β) e ʌ -t/RC
 At t = T1, voltage across the capacitor reaches βVsat and switching takes place.
Therefore
Vc(T1) = βVsat = Vsat – Vsat(1+β) e ʌ -T1/RC
After algebraic manipulation, we get
T1 = Rc ln (1 + β)/(1 - β)
This gives only one half of the period.
Therefore the total time period, T = 2*T1 = 2RC ln(1+β)/(1-β) and the output waveform is
symmetrical.
If R1 = R2, Then β = 0.5 and T = 2RC ln 3 and for R1 = 1.16 R2, it can be seen that
T = 2RC or fo = 1/RC
2.TRIANGULAR WAVEFORM GENERATOR

Fig 3.32 Triangular waveform generator (a) Circuit b)Waveform

 Triangular waveform generator Circuit and waveform are shown in fig 3.32.Basically,
triangular wave is generated by alternatively charging and discharging a capacitor with a
constant current.

 This is achieved by connecting integrator circuit at the output of square wave generator. Assume
that V’ is high at +Vsat. This forces a constant current (+Vsat/R3) through C (left to right) to drive
Vo negative linearly.

 When V’ is low at -Vsat, it forces a constant current (-Vsat/R3) through C (right to left) to drive

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Vo positive, linearly.

 The frequency of the triangular wave is same as that of square wave. This is illustrated in Fig.
2.86.

 Although the amplitude of the square wave is constant (± Vsat), the amplitude of the triangular
wave decreases with an increase in its frequency, and vice versa.

 This is because the reactance of capacitor decreases at high frequencies and increases at low
frequencies.

 The time period for oscillation

T = 4R1 C1 R2/R3
 Hence the frequency of oscillation fo is fo = 1/T = R3/4 R1 C1 R2
3. Sawtooth Waveform Generator:
 The difference between the triangular and sawtooth waveform is that the rise time of the
triangular wave is always equal to its fall time while in sawtooth wave generator, rise time
may be much higher than its fall time or vice versa.
 The triangular wave generator can be converted to a sawtooth wave generator by injecting a
variable dc voltage into the noninverting terminal of the integrator.
 This can be done by using a potentiometer as shown in figure 3.33. When the wiper of the
potentiometer is at the centre, the output will be a triangular wave since the duty cycle is 50%.
 If the wiper moves towards –V, the rise time of the sawtooth becomes longer than the fall time.
If the wiper moves towards +V, the fall time becomes more than the rise time.

Fig.3.33 Saw tooth waveform generator (a) Circuit b)Waveform

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