the operational amplifierxuanqi-net.com/circuit/chapter5.pdfthe operational amplifier qi xuan...
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1 Electric Circuits
The Operational Amplifier
Qi Xuan Zhejiang University of Technology
October 2015
Structure
• Opera&onal Amplifier Terminals • Terminal Voltages and Currents • The Inver&ng-‐Amplifier Circuit • The Summing-‐Amplifier Circuit • The Noninver&ng-‐Amplifier Circuit • The Difference-‐Amplifier Circuit • A More Realis&c Model for the Opera&onal Amplifier
2 Electric Circuits
Strain Gages
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A strain gage is a grid of thin wires whose resistance changes when the wires are lengthened or shortened.
R: the resistance of the gage at rest; ΔL/L: is the frac&onal lengthening of the gage (which is the defini&on of "strain”); 2: is typical of the manufacturer's gage factor; ΔR: the change in resistance due to the bending of the bar.
The change in resistance experienced by the strain gage is typically much smaller than could be accurately measured by an ohmmeter.
In order to make an accurate measurement of the voltage difference, we use an opera&onal amplifier circuit to amplify, or increase, the voltage difference.
Opera&onal Amplifier Terminals
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A simplified circuit symbol for an op amp
Terminal Voltages
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Saturate
Saturate
Linear region
Ideal Op Amp
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A = +∞
VCC ≤ 20 V vp = vn Virtual short condi&on
Linear region
Maintain the virtual short condi&on to ensure linear opera&on:
Negative feedback
Puzzle
• Even if the circuit provides a nega&ve feedback path for the op amp, linear opera&on is not ensured. So how do we know whether the op amp is opera:ng in its linear region?
• The answer is: we don’t know!
• We first assuming linear opera&on, performing the circuit analysis, and then checking our results for contradic&ons, if we have -VCC≤vo≤VCC, it’s in the linear region, if not, it’s saturate!
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Terminal Currents
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Ideally, the equivalent input resistance is infinite, resul&ng in the current constraint:
Note that the current constraint is not based on assuming the op amp is confined to its linear opera&ng region as was the voltage constraint.
Kirchhoff's current law:
Notes
• The posi&ve and nega&ve power supply voltages do not have to be equal in magnitude. In the linear opera&ng region, vo must lie between the two supply voltages.
• Be aware also that the value of A is not constant under all opera&ng condi&ons. For now, however, we assume that it is.
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Example #1
The op amp in the circuit is ideal. a) Calculate vo, if va = 1 V and vb= 0 V. b) If va = 1.5 V, specify the range of vb that avoids
amplifier satura&on.
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Solu:on for Example #1
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vn = vp = vb = 0
i25 + i100 = in = 0
vo = -4 V
a) vn
vp
The Inver&ng-‐Amplifier Circuit
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vn = vp = 0 is + if = in = 0
Open-‐loop Gain
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vo = -Avn = -Avs
A: open-loop gain
|vs| ≤ VCC / A
Example #2
a) Design an inver&ng amplifier with a gain of 12. Use ±15 V power supplies and an ideal op amp.
b) What range of input voltages, vs, allows the op amp in this design to remain in its linear opera&ng region?
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Solu:on for Example #2 a) We need to find two resistors whose ra'o is 12 from the
realis&c resistor values listed in Appendix H.
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Rs = 1 kΩ Rf = 12 kΩ verify
b) Solve two different versions of the inverting-amplifier equation for vs: first using vo = +15 V and then using vo = -15 V:
The Summing-‐Amplifier Circuit
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The Noninver&ng-‐Amplifier Circuit
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Opera&on in the linear region requires that
Example #3 a) Design a noninver&ng amplifier with a gain of 6. Assume the
op amp is ideal. b) Suppose we wish to amplify a voltage vg, such that -1.5 V ≤ vg ≤ 1.5 V. What are the smallest power supply voltages that could be used with the resistors selected in part (a) and s&ll have the op amp in this design remain in its linear opera&ng region?
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Solu:on for Example #3 a) Using the noninver&ng amplifier equa&on
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Therefore, we have
We want two resistors whose ra&o is 5. Look at the realis&c resistor values listed in Appendix H. Let's choose Rf = 10 kΩ, so Rs = 2 kΩ. But there is not a 2 kΩ resistor in Appendix H. We can create an equivalent 2 kΩ resistor by combining two 1 kΩ resistors in series. We can use a third 1 kΩ resistor as the value of the resistor Rg.
Electric Circuits 20
b) Solve two different versions of the noninver&ng amplifier equa&on for vo—first using vg = +1.5 V and then using vg = -1.5V:
Thus, if we use ±9 V power supplies for the noninver&ng amplifier designed in part (a) and -1.5 V ≤ vg ≤ +1.5 V, the op amp will remain in its linear opera&ng region.
✗
✗
The Difference-‐Amplifier Circuit
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The Difference Amplifier—Another Perspec:ve
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Insights
• In many applica&ons it is the differen:al mode signal that contains the informa:on of interest, whereas the common mode signal is the noise found in all electric signals.
• Thus, we hope RaRd = RbRc
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Common Mode Rejec:on Ra:o • An ideal difference amplifier has zero common mode gain and
nonzero (and usually large) differen&al mode gain. • Two factors have an influence on the ideal common mode
gain—resistance mismatches or a nonideal op amp.
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or
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The common mode rejec'on ra'o (CMRR) can be used to measure how nearly ideal a difference amplifier is. It is defined as the ra&o of the differen&al mode gain to the common mode gain
A More Realis&c Model for the Opera&onal Amplifier
A realis&c model includes three modifica&ons to the ideal op amp: (1) a finite input resistance, Ri; (2) a finite open-‐loop gain, A; and (3) a nonzero output resistance, Ro.
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For the juA741 op amp, the typical values of Ri, A, and Ro are 2 MΩ, 105, and 75 Ω, respec&vely.
vn = vp
in = ip = 0
Analysis of the More Realis&c Op Amp Model
• Take the Inver&ng-‐Amplifier Circuit for example:
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RL
Strain Gages
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The pair of strain gages that are lengthened once the bar is bent have the values R + ΔR in the bridge feeding the difference amplifier, whereas the pair of strain gages that are shortened have the values R - ΔR.
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Summary
• Ideal op amp, linear region, saturate • Voltage constraint, current constraint • Inver&ng amplifier, summing amplifier, noninver&ng amplifier, difference amplifier
• Common mode and difference mode • Difference-‐amplifier, common mode rejec&on ra&o (CMRR)
• More realis&c Op Amp model
Electric Circuits 30