Operational amplifier
Operational Amplifiers
An OPERATIONAL AMPLIFIER (abbreviated as “Opamp”) consists of a high gain direct coupled differential amplifier. It is available as an integrated circuit chip [IC]. Earlier, Opamps were primarily used to perform a number of mathematical operations such as addition, subtraction, integration, differentiation, etc. and hence the term ‘ Operational amplifier ’ is in practice.
Today’s Opamps are linear integrated circuits that use relatively low supply voltages (compared to the earlier versions of Opamps which were constructed using vacuum tubes) and are very reliable and inexpensive.
Differential amplifier: It is a very high gain direct coupled amplifier with the ability to reject common mode signals and amplify only the difference of its two input signals. It is also called as difference amplifier. Figure shows the block diagram of differential amplifier. It consists of two inputs ν1 and ν2 and two outputs ν01 and ν02 respectively. The expression for the output voltage given by ν0 = Ad (ν1  ν2 ), where Ad is the differential gain of the amplifier. ν1ν2 ν01 ν02 Differential amplifier
Schematic symbol and terminals of an opamp:
Inverting input
Non Inverting
input
Output
− VEE
+
–
+VCC
The standard symbol of Opamp is shown in figure. It has two input terminals and an output terminal. The input terminals are, INVERTING  the output inphase with the input signal.
A signal at the inverting terminal has the output inverted with respect to the input.
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The Ideal Opamp:
In order to get a clear concept of an Opamp, it is better to study its ideal behaviour. A practical Opamp, of course, falls short of these ideal standards, but it is much easier to understand and analyse the device from an ideal point of view.
Characteristics of an ideal opamp:
1.
Infinite Voltage Gain: Output of an ideal opamp is maximum, even for a zero input. [This means that the output of a typical opamp is very large even for an extremely small input].
2.
Infinite bandwidth: The gain is constant for input signals of frequency from zero to infinity. [For a practical opamp, the bandwidth is high but not infinity. This means that the gain is constant over a large band of input signal frequencies].
3.
Infinite input impedance: This means that an ideal opamp will not draw any current from the input signal source or preceding circuit and hence avoids the loading effect. ( In practice, the input impedance of a practical opamp is very high but not infinity. It is in the order of 10 6 Ω).
4.
Zero output impedance: The ideal opamp delivers full power at the output irrespective of the load resistance value. [A typical opamp has very low o/p impedance of the order of a few ohms and hence delivers maximum power at the output]
5.
High Stability : Characteristics of an ideal Opamp do not drift with temperature [This means that the typical opamp exhibits high stability against temperature variations].
6.
Perfect balance: In an ideal opamp, the output is zero when the inputs at the two terminals are equal. [In a typical opamp, there exists a small output voltage when the two inputs are equal. This is due to the slight imbalance in the characteristics of the two amplifier sections that form the differential amplifier in the opamp IC.]
7.
CMRR (Common mode rejection ratio): It is the ratio of differential gain (Ad) to the common mode gain (Ac). For an ideal Opamp, the CMRR is infinity. This means that the amplifier completely rejects the common mode signals and amplifies only the differential signals with infinite gain. [CMRR = Ad /AC]. [For a typical Opamp, the CMRR will be very high. It is expressed in dB. Higher the CMRR, the better is the performance of IC).
To study the applications of Opamp, the Opamp is treated as ideal. This is required to simplify the mathematical analysis.
Applications of Operational amplifiers:
1. Inverting Amplifier:
An inverting amplifier is one whose output is 180° out of phase with respect to the input. The figure shows an inverting amplifier using Opamp. A small input V1 is applied to the inverting terminal through R1 and the non inverting terminal is grounded.
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The point G is called the virtual ground. [Since G is not directly connected to the ground, no current from G flows to the ground. Further, due to negative feedback, the voltage drop across R1 makes the potential at G to be nearly zero. Hence G is not the true ground, but the virtual ground].
Input current through R1 is given by,
I1 = (V1– 0)/R1 (because, the potential at node G = 0). … (1)
Due to the infinite input impedance of the Opamp, current through the Opamp is zero and hence the whole of the input current I1 flows through the feedback resistor Rf. It is given by,
If = (0 – Vo)/Rf … (2)
Where, Vo is the output voltage.
Applying KCL at node G,
I1 = If. Therefore, from eqns. 1 and 2, we get,
(V1–0) / R1 = (0 – Vo) / Rf
i.e. 1f1oRRVV−=
i.e., Voltage gain of the inverting amplifier is given by, 1RRAfIN−= … (3)
Negative sign indicates the phase reversal between the input and the output signals.
G
Rf
R1
− I
Operational amplifiers
Equation (3) shows that the gain of an Inverting Amplifier using Opamp depends only on the values of external resistors but not on the gain ( Open loop gain)of the opamp.
2. Scale Changer (Multiplier):
Inverting amplifier with Rf > Ri acts as a scale changer or multiplier since the input V1 gets multiplied (i.e., scaled ) by a factor –Rf/Ri and the output voltage Vo = V1 × (− Rf/Ri).
This circuit is also called as the negative scale changer or negative scalar. RfG R1 −⎟⎟⎠⎞⎜⎜⎝⎛−×=1RfR1VoVV1 +Scaling factorScale changer
3. Noninverting amplifier:
A noninverting amplifier is one whose output is in phase with the input voltage. The figure below shows an opamp circuit with a small input V1 applied to the noninverting terminal. The inverting terminal is connected to the ground through a resistor R1. Rf is the feedback resistor. The current through R1 is the same as the current through Rf because the current through opamp is zero. Therefore, voltage drop across the input terminals is also zero.
RfX R1 −V1+I1 If ⎟⎟⎠⎞⎜⎜⎝⎛+=1RRVV1f1O
Applying KCL at node X , I1 = If
i.e. ( V1 – 0) / R1 = ( Vo  V1) / Rf
∴ ( Vo  V1) / V1 = Rf / Ri
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i.e. ( Vo /V1 )–1 = Rf / Ri
Where Vo/V1 is the voltage gain of the NonInverting amplifier.
i.e., .1RR1INAf+=
There is no phase shift between the input and the output signals.
[Since the output voltage vo = vi (1+ Rf/R1), the input voltage gets multiplied by a factor 1+(Rf/R1). Therefore, the circuit is also called as a positive scalar].
4. Voltage follower: This is a special case of Noninverting amplifier. Non inverting amplifier with Rf = 0 has unity voltage gain and is called voltage follower because, the output voltage simply follows the input voltage. This circuit has very high input impedance and very low output impedance. Therefore it is used as a Buffer amplifier for impedance matching purposes.
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5. Summing amplifier (adder):
A Summing Amplifier is one whose output is the sum of the inputs. Figure shows the opamp used as an adder (summing amplifier). In this case, the output voltage is equal to the negative sum of the input voltages.
Input voltages V1, V2 and V3 are fed to the inverting terminal through R1, R2 and R3 respectively. Let I1, I2 and I3 be the currents through the respective branches as shown in the figure. The noninverting terminal is connected to the ground. Node G is the virtual ground. Due to the infinite input impedance, the current through the opamp is zero and hence the input current is equal to the current in the feedback resistor. V1 RfGR1 V2 V3 R2 R 3 _ +VO = − (V1 + V2 + V3) I1 I2 I3 I If
Operational amplifiers
∴ Applying KCL at node G, If = Ι
where Ι = Ι1 + Ι2 + Ι3 . i.e. Ιf = Ι1 + Ι2 + Ι3 .
We know that Ιf = Vo/Rf, Ι1 = V1/R1, Ι2 = V2/R2, Ι3= V3/R3.
i.e. Vo/Rf = V1/R1 + V2/R2 + V3/R3
∴ Vo = Rf (V1/R1 + V2/R2 + V3/R3).
If Rf = R1 = R2 = R3 = R then,
Vo = R/R (V1 + V2 + V3)
⇒ Vo = − (V1 + V2 + V3)
The output voltage is the negative sum of all the input voltages. If the values of the resistors are different, each input is scaled by a different factor.
6. Op amp Subtractor: R+ −RV1 V2 RRV0 =V2V1 V2 ⁄ 2
Subtractor provides the output equal to the difference of the two input voltages.
When V1 alone is acting by grounding the non inverting terminal, the circuit works as an inverting amplifier and the output is given by,
V01 =  (R/R) ×V1
=  V1 … (1)
When V2 is acting alone by grounding the inverting terminal through R1, the circuit works as a non inverting amplifier with net input at the non inverting terminal equal to V2/2. Therefore, the output is given by,
V02 = (V2/2) × (1+R/R) = V2 … (2)
Therefore, when both the inputs are simultaneously applied, the output is given by applying the super position theorem.
i.e, VO = V2 – V1 . Thus the circuit works as a Subtractor.
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7. Integrator:
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R + −C dtVRC1Vio∫−= Vi G i
An integrator circuit is one whose output is proportional to the time integral of input voltage. The figure shows the use of an Opamp as an integrator. G is the virtual ground. The input voltage Vi charges the capacitor to a voltage VO. Let the charge on the capacitor be Q
The charge on the capacitor is given by,
Q = C (0 – Vo) [because the voltage at G = 0] … (1)
But, Q = ∫ idt [where i is the charging current] … (2)
i.e. − CVO = ∫ idt [where i= Vi/R]
∴ − CVo= ∫ (Vi/ R) dt.
Hence, the output voltage dtVRC1Vio∫−= . Thus the circuit works as an integrator.
If the input voltage is a square wave, the output waveform will be triangular. For rectangular or pulse waveform applied at the input, the output waveform will be a sawtooth wave (ramp). The output will be a cosine wave for sine wave input. The output is a paraboloid waveform if the input is a triangular wave. InputOutput InputOutputttvo vi
But, the output waveforms of OpAmp integrator will be inverted versions of the waveform shown above.
8. Differentiator:
A differentiator circuit is one whose output voltage is proportional to time derivative of the input voltage. Figure shows a differentiating circuit using Op
Operational amplifiers
amp. The circuit is an inverting amplifier configuration with a capacitor at the input. G is the virtual ground.
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If Q is the charge required by the capacitor then Q = Vi × C.
Differentiating the above equation we get,
dQ/dt = C × d/dt (Vi).
But dQ/dt = i
∴i = C × d/dt(Vi) … (1)
Due to the infinite input impedance of the Opamp, the current
through the opamp is zero.
Applying KCL to node G,
Input current i.e., charging current I = Current through the feedback resistor.
∴Charging current or current through the feedback resistor is given by,
i = (0 – VO)/R = −VO / R … (2)Equating equations (1) and (2)
C d/dt (Vi ) =  Vo / R
∴VO= RC d/dt (Vi)
i.e. output = RC time constant × time derivative of the input. Therefore the circuit works as a differentiator.
If the input voltage is a square wave, the output waveform consists of positive and the negative spikes. The output will be a cosine wave for sine wave input and the output is a square waveform if the input is a triangular wave.
But, the output waveforms of OpAmp differentiator will be inverted versions of the above output.
G
R
C
vi
i

+
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Questions carrying ONE mark each.
1.
What is a differential amplifier?
2.
What is an operational amplifier?
3.
What is meant by virtual ground?
4.
Define CMRR.
5.
What is meant by the Common mode rejection?
6.
Define the term ‘differential input’.
7.
Draw the circuit of an inverting amplifier.
8.
Draw the circuit of a non inverting amplifier.
9.
Draw the circuit of an Adder using OpAmp.
10.
Draw the circuit of a Subtractor.
11.
Draw the circuit of a Differentiator.
12.
Draw the circuit of an Integrator.
13.
What is the application of a Buffer amplifier?
14.
Give one application of Integrator circuit.
15.
Give one application of differentiator circuit.
16.
What is meant by single ended output?
17.
What is meant by unbalanced output?
18.
What type of output is obtained when the input of a differentiator is fed with a Square wave?
19.
What type of output is obtained when the input of a differentiator is fed with a Sine wave?
20.
What type of output is obtained when the input of a differentiator is fed with a triangular wave ?
21.
What type of output is obtained when the input of an integrator is fed with a Square wave?
22.
What type of output is obtained when the input of an integrator is fed with a Sine wave?
23.
What type of output is obtained when the input of an integrator is fed with a rectangular waveform?
Questions carrying TWO mark each
1.
Write the characteristics of an ideal Opamp?
2.
Distinguish between Common mode gain and Differential gain?
3.
With circuit write a note on Buffer amplifier.
4.
What is an integrator? What are its applications?
5.
What is a differentiator? What are its applications?
6.
Draw the circuit of a differential amplifier using transistors.
Operational amplifiers
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7.
Explain the meaning of Virtual ground in the inverting configuration of Opamp?
8.
What are inverting and non inverting amplifiers?
9.
Draw the output waveform of an integrator when the input is a square wave?
10.
Draw the output waveform of an integrator when the input is a triangular wave?
11.
Draw the output waveform of a differentiator when the input is a square wave?
12.
What is meant by voltage follower?
13.
Gain of a Buffer amplifier is Unity. Justify.
14.
Buffer is used for impedance matching. Justify.
Questions carrying Four/Five marks each.
1.
With a circuit, explain the working of a differential amplifier.
2.
With a circuit, explain the working of a inverting amplifier using Opamp. Obtain the expression for its gain.
3.
With a circuit, explain the working of a noninverting amplifier using Opamp. Obtain the expression for its gain.
4.
With a circuit, explain the working of an Adder circuit using Opamp. Obtain the expression for its output.
5.
With a circuit, explain the working of a Subtractor circuit using Opamp. Obtain the expression for its output.
6.
With a circuit, explain the working of an integrator circuit using Opamp. Obtain the expression for its output.
7.
With a circuit, explain the working of a differentiator circuit using Opamp. Obtain the expression for its output.
8.
What is a Scale changer? Draw the circuit of an opamp Multiplier and explain its working.
9.
Define CMRR. Explain. List out the advantages of a differential amplifier.
10.
Obtain an expression for the Averaging amplifier circuit using an Opamp.
11.
Explain the ideal characteristics of an Opamp. Compare these with the characteristics of typical operational amplifiers.
12.
What is a negative Scalar? Draw the circuit and obtain the expression for its output.
Problems
1.
Calculate the output voltage of an inverting amplifier using a resistor of value 1 MΩ at the input and feedback resistance of 4.7 kΩ when the input voltage is 0.8 V.
2.
Design an op−amp inverting amplifier for a gain of 500 and input resistance 1 kΩ.
3.
An opamp has a mid band gain of 105 and a bandwidth of 1 kHz. Design an inverting amplifier with a gain of 400 and an input resistance of 5 kΩ. What is the bandwidth of the amplifier?
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4.
The gain of a noninverting amplifier is 10. If R1 = 5.6 kΩ, find the value of R1.
5.
The input signals to an adder circuit are 3 mV, 5mV and 18 mV through resistors of 3 kΩ, 5 kΩ and 9 kΩ respectively. Find the value of feedback resistor to get an output of 1V.
6.
An opamp adder uses input resistances 4 kΩ and 10 kΩ, with a feedback resistance of 25 kΩ. The output voltage of the circuit is 100 mV. If the two input voltages are equal, calculate their values.
7.
In an adder circuit R1 = 10 kΩ, R2 = 100 kΩ, R3 = 1MΩ and Rf = 1MΩ. Find the output voltage if V1 = 0.04, V, V2 = −2V and V3 = 0.5 V.
8.
Design an adder circuit to obtain an output V0 = −(6V1 + 4V2 − 3V3). Assume Rf = 100 kilo ohm.
9.
An integrator gives an output of 60 μV after 2 mS when an input of the type vi = 3t is applied at its input. Find the time constant of the circuit.
10.
A differentiator circuit makes uses of 1 MΩ resistor and 1 μF capacitor. If the input voltage is t2, find the output after 2 second, 5sec and 10 second.
11.
Design a differentiator circuit which yields an output of −2V when an input vi = t is applied after a time 2 second. Assume C to be 10 μF.