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L03Operational Amplifiers
Applications 2Chapter 9
Ideal Operational Amplifiers and Op-Amp Circuits
Donald A. Neamen (2009). Microelectronics: Circuit Analysis and Design,
4th Edition, Mc-Graw-Hill
Prepared by: Dr. Hani Jamleh, Electrical Engineering Department, The University of Jordan
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Op-Amp Applications
1. Inverting Amplifier2. Amplifier with T-
Network
3. Non-InvertingAmplifier4. Voltage Follower
(Buffer)
5. Summing Amplifier
6. Current to VoltageConverter
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3
65
42
1
Op-Amp Applications
7. Difference Amplifier
8. InstrumentationAmplifier
9. Integrator
10. Differentiator
11. Reference VoltageSource Design
12. Precision Half-waveRectifier
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8
9
10
12
11
9.3 Summing Amplifier
β’ To analyze the op-amp circuit shown inFigure 9.14(a), we will use:
1. The superposition theorem and
2. The concept of virtual ground.
β’ Using the superposition theorem, wewill:
1. Determine the output voltage due toeach input acting alone, then
2. Algebraically sum these terms todetermine the total output.
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Figure 9.14(a)
9.3 Summing Amplifier
β’ If we set π£πΌ2 = π£πΌ3 = 0, the current π1is:
π1 =π£πΌ1π 1
β’ Since π£πΌ2 = π£πΌ3 = 0 and the invertingterminal is at virtual ground, thecurrents π2 and π3 must both be zero.β’ Current π1 does not flow through either π 2
or π 3, but the entire current must flowthrough the feedback resistor π πΉ , asindicated in Figure 9.14(b).
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Figure 9.14(a)
Figure 9.14(b)
9.3 Summing Amplifier
π1 =π£πΌ1π 1
β’ The output voltage due to π£πΌ1 acting alone is:
π£π(π£πΌ1) = βπ1π πΉ = βπ πΉπ 1
π£πΌ1
β’ Similarly, the output voltages due to π£πΌ2 andπ£πΌ3 acting individually are:
π£π(π£πΌ2) = βπ2π πΉ = βπ πΉπ 2
π£πΌ2
π£π(π£πΌ3) = βπ3π πΉ = βπ πΉπ 3
π£πΌ3
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Figure 9.14(a)
Figure 9.14(b)
9.3 Summing Amplifier
β’ Now we have:
π£π(π£πΌ1) = βπ1π πΉ = βπ πΉπ 1
π£πΌ1
π£π(π£πΌ2) = βπ2π πΉ = βπ πΉπ 2
π£πΌ2
π£π(π£πΌ3) = βπ3π πΉ = βπ πΉπ 3
π£πΌ3
β’ The total output voltage is the algebraic sum of theindividual output voltages:
π£π = π£π(π£πΌ1) + π£π(π£πΌ2) + π£π(π£πΌ3)
π£π = βπ πΉπ 1
π£πΌ1 +π πΉπ 2
π£πΌ2 +π πΉπ 3
π£πΌ3
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Figure 9.14(a)
9.3 Summing Amplifier
π£π = βπ πΉπ 1
π£πΌ1 +π πΉπ 2
π£πΌ2 +π πΉπ 3
π£πΌ3
β’ The output voltage is the sum of the three input voltages,with different weighting factors.
β’ This circuit is therefore called the inverting summingamplifier.
β’ The number of input terminals and input resistors can bechanged to add more or fewer voltages.
β’ A special case occurs when the three input resistancesare equal. When π 1 = π 2 = π 3 β‘ π , then
π£π = βπ πΉπ 1
(π£πΌ1 + π£πΌ2 + π£πΌ3)
β’ This means that the output voltage is the sum of the inputvoltages, with a single amplification factor.
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Figure 9.14(a)
Discussion
β’ Up to this point, we have seen that op-amps can be used to:
β’ Multiply a signal by a constant (e.g.π πΉ
π 1)
β’ Sum a number of signals with prescribed weights.These are mathematical operations.
β’ Later in the chapter, we will see that op-amps can also be used tointegrate and differentiate.β’ These circuits are the building blocks needed to perform analog
computationsβhence the original name of operational amplifier.
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Discussion
β’ Opamps, however, are versatile and can do much more than justperform mathematical operations, as we will continue to observethrough the remainder of this course.
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DESIGN EXAMPLE 9.4
β’ Objective: Design a summing amplifier to produce a specified outputsignal.
β’ Specifications: The output signal generated from an ideal amplifiercircuit is:
π£π1 = 1.2 β 0.5 sinππ‘ (π)
β’ Design a summing amplifier to be connected to the amplifier circuitsuch that the output signal is:
π£π = 2 β sinππ‘ (π)
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DESIGN EXAMPLE 9.4
π£π1 = 1.2 β 0.5 sinππ‘ (π) π£π = 2 β sinππ‘ (π)
β’ Choices:1. Standard precision resistors with tolerances of
Β± 1 πππππππ‘ are to be used in the final design.
2. Assume an ideal op-amp is available.
β’ Solution: In this case, we need only two inputs tothe summing amplifier, as shown in Figure 9.14.β’ One input to the summing amplifier is the output of the
ideal amplifier circuit and
β’ The second input should be a DC voltage to cancel the+ 1.2π signal from the amplifier circuit.
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Figure 9.14
ππ1
β1.2π
DESIGN EXAMPLE 9.4
π£π1 = 1.2 β 0.5 sinππ‘ (π) π£π = 2 β sinππ‘ (π)
β’ If the voltage gains of each input to the summingamplifier are equal, then an input of β1.2π at the secondinput will cancel the +1.2π from the amplifier circuit.
β’ For a β0.5π sinusoidal input signal and a desired 2πsinusoidal output signal, the summing amplifier gainmust be:
π΄π£ = βπ πΉπ 1
=2
β0.5= β4
β’ If we choose the input resistances to be:π 1 = π 2 = 30πΞ©
β’ Then the feedback resistance must be:π πΉ = 4 Γ 30πΞ© = 120πΞ©
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Figure 9.14
ππ1
β1.2π
DESIGN EXAMPLE 9.4
π£π1 = 1.2 β 0.5 sinππ‘ (π) π£π = 2 β sinππ‘ (π)π 1 = π 2 = 30πΞ©
π πΉ = 4 Γ 30πΞ© = 120πΞ©
β’ Trade-offs: From Appendix C, we can choose precisionresistor values of π πΉ = 124πΞ© and π 1 = π 2 = 30 πΞ©.The ratio of the ideal resistors is 4.13.
β’ Considering the Β±1 πππππππ‘ tolerance values, the outputof the summing amplifier is:
π£π = βπ πΉ 1 Β± 0.01
π 1 1 Β± 0.01β 1.2 β 0.5 sinππ‘ β
π πΉ 1 Β± 0.01
π 2 1 Β± 0.01β (β1.2)
β’ The DC output voltage is in the range:β0.1926 β€ π£π(π·πΆ) β€ 0.1926 π
β’ The peak ac output voltage is in the range:1.967 β€ π£π(ππ) β€ 2.047 π
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Figure 9.14
ππ1
β1.2π
EXERCISE PROBLEM Ex. 9.4
β’ Design an inverting summing amplifier that willproduce an output voltage of:
π£π = β3(π£πΌ1 + 2π£πΌ2 + 0.3π£πΌ3 + 4π£πΌ4)
β’ The maximum resistance is to be limited to:π πππ₯ = 400 πΞ©
β’ Answer: Recall the summing amplifier output for threeinputs:
π£π = βπ πΉπ 1
π£πΌ1 +π πΉπ 2
π£πΌ2 +π πΉπ 3
π£πΌ3
β’ We expand it for Four inputs as:
π£π = βπ πΉπ 1
π£πΌ1 +π πΉπ 2
π£πΌ2 +π πΉπ 3
π£πΌ3 +π πΉπ 4
π£πΌ4
β’ Let π 3 = π πππ₯ = 400πΞ©. Why π 3?
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π£πΌ4π4
π 4
EXERCISE PROBLEM Ex. 9.4
π£π = β3(π£πΌ1 + 2π£πΌ2 + 0.3π£πΌ3 + 4π£πΌ4)
π£π = βπ πΉπ 1
π£πΌ1 +π πΉπ 2
π£πΌ2 +π πΉπ 3
π£πΌ3 +π πΉπ 4
π£πΌ4
β’ Let π 3 = π πππ₯ = 400πΞ©. 390πΞ© + 10kΞ©
β’ π πΉ = 3 Γ 0.3 β 400πΞ© = 360πΞ©
β’ π 1 =360k
3= 120kΞ©
β’ π 2 =360k
3Γ2= 60kΞ© 47πΞ© + 13πΞ©
β’ π 4 =360k
3Γ4= 30kΞ©
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EXERCISE PROBLEM Ex. 9.4
β’ Using the results of previous part for π βπ values. Determine π£π for:
a) π£πΌ1 = 0.1π, π£πΌ2 = β0.2π, π£πΌ3 = β1π, π£πΌ4 = 0.05π;β’ Answer: π£π = +1.2π
b) π£πΌ1 = β0.2π, π£πΌ2 = 0.3π, π£πΌ3 = 1.5π, π£πΌ4 = β0.1π;β’ Answer: π£π = β1.35π
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9.4 Noninverting Amplifier
β’ In our previous discussions, the feedbackelement π 2 or π πΉ was connected betweenthe output and the inverting terminalcreating a negative feedback loop.
β’ However, a signal can be applied to thenoninverting terminal while stillmaintaining negative feedback.
2018-09-27 Medical Electronics - Dr. Hani Jamleh - JU 18Figure 9.14(a)
Figure 9.10