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


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:


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


[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


[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



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



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

operational amplifiers luke gibbons csus fall 2006 me 114

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