Operational Amplifiers Luke Gibbons CSUS Fall 2006 ME 114.
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Transcript of Operational Amplifiers Luke Gibbons CSUS Fall 2006 ME 114.
Operational Amplifiers
Luke GibbonsCSUS Fall 2006
ME 114
IntroIntro
An Operational Amplifier, or Op-Amp, is An Operational Amplifier, or Op-Amp, is a component often used in circuits a component often used in circuits because of its wide range of abilitiesbecause of its wide range of abilities
Op-Amps can be used to amplify, invert, Op-Amps can be used to amplify, invert, add, subtract, integrate, differentiate, add, subtract, integrate, differentiate, filter and compare different input filter and compare different input signalssignals
Op-Amps were created to perform Op-Amps were created to perform specific mathematic functions, such as a specific mathematic functions, such as a function to invert, integrate, and function to invert, integrate, and amplify different input signalsamplify different input signals
Op-AmpsOp-Amps
Ideal Vs PracticalIdeal Vs Practical
We will concentrate our efforts to We will concentrate our efforts to understanding ideal op-amps then analyze understanding ideal op-amps then analyze both ideal and practical op-amps using both ideal and practical op-amps using Simulink and Camp-GSimulink and Camp-G
Characteristics of ideal and practical op-Characteristics of ideal and practical op-amps are very high voltage gain and input amps are very high voltage gain and input impedance, very low output impedance, and impedance, very low output impedance, and wide bandwidthwide bandwidth
Operational Amplifier Operational Amplifier DiagramsDiagrams
The The op-amp op-amp block block diagramdiagram
The The basic basic op-ampop-amp
Types of Op-AmpsTypes of Op-Amps
The type of op-amps we will analyze The type of op-amps we will analyze include:include: IntegratingIntegrating DifferentiatingDifferentiating Inverting Inverting Non-InvertingNon-Inverting SummingSumming SubtractingSubtracting FilteringFiltering ComparingComparing
Integrating and Integrating and Differentiating Op-AmpsDifferentiating Op-Amps
Bond Graphs of Integrating and Differentiating Op-Amps
SF
R
SE
CIntegrator:
0 0 0
1
12
3 45
6 7
8
Bond Graphs of Integrating and Differentiating Op-Amps
SF
R
SE
C
Differentiator:
0 0 0
1
12
3 45
6 7
8
Integrating and Integrating and Differentiating Op-Amp Differentiating Op-Amp
Transfer FunctionsTransfer Functions Integrating op-amp:Integrating op-amp:
TF(s) = Vo(s)/Vi(s) TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF(s) = -Z2(s)/Z1(s) TF = -(1/RC)/sTF = -(1/RC)/s
Differentiating op-ampDifferentiating op-ampTF(s) = Vo(s)/Vi(s) TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF(s) = -Z2(s)/Z1(s) TF = -RCsTF = -RCs
Transfer Function and Matrices from Camp-G[u] = [SF SE][A] = [-1/(C*R)][B] = [-1 1/R][C] = [1/C][D] = [0 0]
TF(e8/f1) = e8/f1 = [-R/[(C*R)*s + 1] 1/[(C*R)*s + 1]]
SF
R
SE
C
Integrator:
0 0 0
1
12
3 45
6 7
8
Transfer Function and Matrices from Camp-G[u] = [SF SE][A] = [-1/(C*R)][B] = [1 1/R][C] = [-1/C][D] = [0 1]
TF(e8/f1) = e8/f1 = [-R/[(C*R)*s + 1] (C*R)*s /[(C*R)*s + 1]]
SF
R
SE
C
Differentiator:
0 0 0
1
12
3 45
6 7
8
Op-Amp Transfer Op-Amp Transfer FunctionsFunctions
Notice how the transfer functions can Notice how the transfer functions can either relate the input current or either relate the input current or voltage to the output current or voltagevoltage to the output current or voltage
The common op-amp transfer function The common op-amp transfer function relates the input voltage to the output relates the input voltage to the output voltage, as shown a few slides backvoltage, as shown a few slides back
Because of the “geometry” of the bond Because of the “geometry” of the bond graphs, we will relate the input current graphs, we will relate the input current to the output voltage, as shown on the to the output voltage, as shown on the two previous slidestwo previous slides
Integrating and Integrating and Differentiating Op-AmpsDifferentiating Op-Amps
Say we have an input voltage of Say we have an input voltage of VVinputinput = R = Rinputinput*sin (t)*sin (t)
wherewhere R Rinputinput is some input resistance and is some input resistance and the current is represented as the current is represented as i = sin (t)i = sin (t)
SaySay C = 2, R = 2, R C = 2, R = 2, Rinputinput = 1, V = 1, Vinputinput = 1*sin (t) = 1*sin (t) The output voltages of the integrating The output voltages of the integrating
and differentiating op-amps in and differentiating op-amps in comparison with the input voltage is comparison with the input voltage is shown on the next slideshown on the next slide
Simulink Model of Simulink Model of Integrating Op-AmpsIntegrating Op-Amps
Simulink Model of Simulink Model of Differentiating Op-AmpsDifferentiating Op-Amps
Simulink Comparison of Simulink Comparison of Integrating and Integrating and Differentiating Op-AmpsDifferentiating Op-Amps
Integrator:Integrator:
Differentiator:Differentiator:
Sine Wave:Sine Wave:
Integrating and Integrating and Differentiating Op-AmpsDifferentiating Op-Amps
Notice the expected behavior of the Notice the expected behavior of the integrating and differentiating op-ampsintegrating and differentiating op-amps
Notice how we used the gain function to Notice how we used the gain function to reduce the output voltage for the reduce the output voltage for the reference casereference case
Notice how SIMULINK forces the initial Notice how SIMULINK forces the initial current/voltage to be zerocurrent/voltage to be zero
Inverting & Non-Inverting Inverting & Non-Inverting Op-AmpsOp-Amps
Bond Graph of Inverting & Non-Inverting Op-Amps
SF
Ri
SE
RfInverting:
0 0 0
1
12
3 45
6 7
8
0C
Need to add compliance element to make the system operate properly
10 9
Bond Graph of Inverting & Non-Inverting Op-Amps Need to add
compliance element to make the system operate properly
SF
Rf
SE
RiNon-Inverting:
0
0
1
1 23
4 5
6
0C8 7
9
Inverting & Non-Inverting Inverting & Non-Inverting Op-Amp Transfer FunctionsOp-Amp Transfer Functions
Inverting op-amp:Inverting op-amp:TF(s) = Vo(s)/Vi(s) TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF(s) = -Z2(s)/Z1(s) TF = -Rf/RiTF = -Rf/Ri
Non-Inverting op-amp:Non-Inverting op-amp:TF(s) = Vo(s)/Vi(s) TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF(s) = -Z2(s)/Z1(s) TF = (Ri+Rf)/RiTF = (Ri+Rf)/Ri
Non-Inverting Op-AmpsNon-Inverting Op-Amps
Notice how the inverting and non-inverting op-Notice how the inverting and non-inverting op-amps have different inputs into each terminalamps have different inputs into each terminal
In order to represent a non-inverting op-amp In order to represent a non-inverting op-amp (and all op-amps which have the input voltage (and all op-amps which have the input voltage going into the positive terminal) we have to use going into the positive terminal) we have to use either the a bond graph type similar to the bond either the a bond graph type similar to the bond graph from the inverting op-amp or the previous graph from the inverting op-amp or the previous non-inverting op-amp bond graph, that does not non-inverting op-amp bond graph, that does not fully work properly, but is suitable for our fully work properly, but is suitable for our purpose, and manually invert the output signalpurpose, and manually invert the output signal
Transfer Function and Matrices from Camp-G
[u] = [SF SE][A] = [-1/(C*Ri) -1/(C*Rf)][B] = [-1 1/Ri][C] = [1/C][D] = [0 0]
TF(e8/f1) = e8/f1 =
[-Ri*Rf/[(C*Ri*Rf)*s+Ri+Rf] Rf/[(C*Ri*Rf)*s+Ri+Rf]]
SF
Ri
SE
RfInverting:
0 0 0
1
12
3 45
6 7
8
0C10 9
Transfer Function and Matrices from Camp-G
[u] = [SF SE]
[A] = [-1/(C*Ri) -1/(C*Rf)]
[B] = [-1 0]
[C] = [1/C]
[D] = [0 0]
TF(e8/f1) = e8/f1 = [-Ri*Rf/[(C*Ri*Rf)*s+Ri+Rf] 0]
SF
Rf
SE
RiNon-Inverting:
0
0
1
1 23
4 5
6
0C8 7
9
Inverting and Non-Inverting and Non-Inverting Op-AmpsInverting Op-Amps
Say we have an input voltage of Say we have an input voltage of VVinputinput = R = Rinputinput*sin (t)*sin (t)
wherewhere R Rinputinput is some input resistance and the is some input resistance and the current is represented as current is represented as i = sin (t)i = sin (t)
SaySay C = 1, Ri = 2, Rf = 2, R C = 1, Ri = 2, Rf = 2, Rinputinput = 1, V = 1, Vinputinput = = 1*sin (t)1*sin (t)
The output voltages of the integrating and The output voltages of the integrating and differentiating op-amps in comparison with differentiating op-amps in comparison with the input voltage is shown on the next the input voltage is shown on the next slideslide
Simulink Model of Simulink Model of Inverting Op-AmpsInverting Op-Amps
Simulink Model of Non-Simulink Model of Non-Inverting Op -AmpsInverting Op -Amps
Simulink Comparison of Simulink Comparison of Inverting and Inverting and Non-Inverting Op- Non-Inverting Op-AmpsAmps
Inverting:Inverting:
Non-Inverting:Non-Inverting:
Sine Wave:Sine Wave:
Inverting and Non-Inverting and Non-Inverting Op-AmpsInverting Op-Amps
Notice the expected behavior of the Notice the expected behavior of the inverting and non-inverting op-ampsinverting and non-inverting op-amps
Notice how we used the gain function to Notice how we used the gain function to reduce the output voltage for the reduce the output voltage for the reference casereference case
Notice the lag involved with the Notice the lag involved with the inverting and non-inverting op-amps inverting and non-inverting op-amps when compared with the reference casewhen compared with the reference case
Investigation: Input Investigation: Input VoltageVoltage
We will investigate a series of issues We will investigate a series of issues encountered while using CAMPG and encountered while using CAMPG and MATLABMATLAB
First, we will look into determining the First, we will look into determining the transfer function from the output voltage transfer function from the output voltage to the input voltageto the input voltage
For most situations, it is more valuable For most situations, it is more valuable to know the relationship between the to know the relationship between the input and output voltages than the input input and output voltages than the input current to output voltage relationship current to output voltage relationship determined on the previous slidesdetermined on the previous slides
Transfer Function and Matrices from Camp-G[u] = [SF SE][A] = [-1/(C*R)][B] = [-1/R 1/R][C] = [1/C][D] = [0 0]
TF(e8/e1) = e8/e1 = [-1/[(C*R)*s + 1] 1/[(C*R)*s + 1]]
SE
R
SE
C
Integrator:
0 0 0
1
12
3 45
6 7
8
Transfer Function and Matrices from Camp-G[u] = [SF SE][A] = [-1/(C*R)][B] = [1/R -1/R][C] = [1/C][D] = [-1 1]
TF(e8/f1) = e8/e1 = [-CR*s/[CR*s + 1] CR*s /[CR*s + 1]]
SE
R
SE
C
Differentiator:
0 0 0
1
12
3 45
6 7
8
Transfer Function and Matrices from Camp-G
[u] = [SF SE][A] = [-1/(C*Ri) -1/(C*Rf)][B] = [-1/Ri 1/Ri][C] = [1/C][D] = [0 0]
TF(e8/e1) = e8/e1 =
[-Rf/[(C*Ri*Rf)*s+Ri+Rf] Rf/[(C*Ri*Rf)*s+Ri+Rf]]
SE
Ri
SE
RfInverting:
0 0 0
1
12
3 45
6 7
8
0C10 9
Investigation: Derivative Investigation: Derivative CausalityCausality
Now we will look the CAMPG’s Now we will look the CAMPG’s derivative causality error for the derivative causality error for the bond graphs on the following slidesbond graphs on the following slides
CAMPG cannot interface to another CAMPG cannot interface to another program when there are any program when there are any derivative causality errorsderivative causality errors
CAMPG errors show up in red and CAMPG errors show up in red and can be diagnosed by using the peek can be diagnosed by using the peek and analyze functionsand analyze functions
Derivative Causality
The derivative causality error is shown
in red because CAMPG needs the
integral form of the compliance
element
SF
Ci
SE
R
0 0
1
12
3 67
4 5
8
0Cf910
0
Derivative Causality
The derivative causality error is shown
in red because CAMPG and MATLAB
need the integral form of the compliance
element
SF SE
Cf
0 0
1
12
5 67
8 9
10
0 Ci3 4
1
Derivative Causality
The derivative causality error is shown
in red because CAMPG and MATLAB
need the integral form of the compliance
element
SF SE0 012
5 67
0 Ci3 4
1
R
18 9
10
0Cf1112
Investigation: Output Effort Investigation: Output Effort Location Location
Now we will look into which terminal Now we will look into which terminal should be used as the output should be used as the output terminalterminal
We will look at the relationship We will look at the relationship between the input voltage across the between the input voltage across the 11stst terminal and the output voltage terminal and the output voltage across 2 different terminalsacross 2 different terminals
Output Effort LocationAs before, look at the effort output across the 8th
terminal, e8
SE
R
SE
C
Differentiator:
0 0 0
1
12
3 67
4 5
8
[u] = [SF SE][A] = [-1/(C*R)][B] = [1/R -1/R][C] = [1/C][D] = [-1 1]
TF(e8/f1) = e8/e1 = [-CR*s/[CR*s + 1] CR*s /[CR*s + 1]]
Output Effort LocationNow, look at the effort output across the 7th
Terminal, e7
SE
R
SE
C
Differentiator:
0 0 0
1
12
3 45
6 7
8
[u] = [SF SE][A] = [-1/(C*R)][B] = [1/R -1/R][C] = [0][D] = [0 1]
TF(e8/f1) = e8/e1 = ERROR USING sym.maple at offset 28, ‘)’ expected’
ERROR IN sym.collect at 36 r=reshape(maple(‘map’,’collect’,S(:(X),size(s));ERROR IN campgsym at 135 H = collect H
Investigation: Non-Investigation: Non-Inverting Op-AmpsInverting Op-Amps
Now we will try to produce a non-Now we will try to produce a non-inverting op amp in CAMPGinverting op amp in CAMPG
SESE1 1 represents the voltage going represents the voltage going into the positive terminal of the into the positive terminal of the op-ampop-amp
SESE2 2 represents the voltage going represents the voltage going into the negative terminal of the into the negative terminal of the op-ampop-amp
Non-Integrating Amplifier
SE1
Rb
SE
Ra
Non-Integrator:
0
0
1
1
0
Rc
C
SE2 1
1
0
The following configuration is the only one found that
can be created without any derivative causality
errors
However, this bond graph does not get past the DOS Interface
Filtering Op-AmpsFiltering Op-Amps Single pole active low pass filterSingle pole active low pass filter
Single pole active high pass filterSingle pole active high pass filter
Summing & Subtracting Summing & Subtracting Op-AmpsOp-Amps
Comparing Op-AmpsComparing Op-Amps