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Operational Amplifiers

Operational Amplifiers. Luke Gibbons CSUS Fall 2006 ME 114. Intro. An Operational Amplifier, or Op-Amp, is a component often used in circuits because of its wide range of abilities

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Operational Amplifiers

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  1. Operational Amplifiers Luke Gibbons CSUS Fall 2006 ME 114

  2. Intro • An Operational Amplifier, or Op-Amp, is a component often used in circuits because of its wide range of abilities • Op-Amps can be used to amplify, invert, add, subtract, integrate, differentiate, filter and compare different input signals • Op-Amps were created to perform specific mathematic functions, such as a function to invert, integrate, and amplify different input signals

  3. Op-Amps

  4. Ideal Vs Practical • We will concentrate our efforts to understanding ideal op-amps then analyze both ideal and practical op-amps using Simulink and Camp-G • Characteristics of ideal and practical op-amps are very high voltage gain and input impedance, very low output impedance, and wide bandwidth

  5. Operational Amplifier Diagrams • The op-amp block diagram • The basic op-amp

  6. Types of Op-Amps • The type of op-amps we will analyze include: • Integrating • Differentiating • Inverting • Non-Inverting • Summing • Subtracting • Filtering • Comparing

  7. Integrating and Differentiating Op-Amps

  8. Bond Graphs of Integrating and Differentiating Op-Amps SF C Integrator: 8 1 7 6 SE 0 0 0 1 3 5 4 2 R

  9. Bond Graphs of Integrating and Differentiating Op-Amps SF R Differentiator: 8 1 7 6 SE 0 0 0 1 3 5 4 2 C

  10. Integrating and Differentiating Op-Amp Transfer Functions • Integrating op-amp: TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF = -(1/RC)/s • Differentiating op-amp TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF = -RCs

  11. Transfer Function and Matrices from Camp-G C Integrator: 8 [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]] 1 7 6 0 SF SE 0 0 1 3 5 4 2 R

  12. Transfer Function and Matrices from Camp-G R Differentiator: 8 [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]] 1 7 6 SF SE 0 0 0 1 3 5 4 2 C

  13. Op-Amp Transfer Functions • Notice how the transfer functions can either relate the input current or voltage to the output current or voltage • The common op-amp transfer function relates the input voltage to the output voltage, as shown a few slides back • Because of the “geometry” of the bond graphs, we will relate the input current to the output voltage, as shown on the two previous slides

  14. Integrating and Differentiating Op-Amps • Say we have an input voltage of Vinput = Rinput*sin (t) where Rinputis some input resistance and the current is represented as i = sin (t) • Say C = 2, R = 2, Rinput = 1, Vinput = 1*sin (t) • The output voltages of the integrating and differentiating op-amps in comparison with the input voltage is shown on the next slide

  15. Simulink Model of Integrating Op-Amps

  16. Simulink Model of Differentiating Op-Amps

  17. Simulink Comparison of Integrating and Differentiating Op-Amps Sine Wave: Integrator: Differentiator:

  18. Integrating and Differentiating Op-Amps • Notice the expected behavior of the integrating and differentiating op-amps • Notice how we used the gain function to reduce the output voltage for the reference case • Notice how SIMULINK forces the initial current/voltage to be zero

  19. Inverting & Non-Inverting Op-Amps

  20. Bond Graph of Inverting & Non-Inverting Op-Amps 10 9 • Need to add compliance element to make the system operate properly C Rf 0 Inverting: 8 1 7 6 SF SE 0 0 0 1 3 5 4 2 Ri

  21. Bond Graph of Inverting & Non-Inverting Op-Amps • Need to add compliance element to make the system operate properly Rf 9 Non-Inverting: 8 7 C Ri 0 6 1 5 4 0 SE SF 0 1 3 2

  22. Inverting & Non-Inverting Op-Amp Transfer Functions • Inverting op-amp: TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF = -Rf/Ri Non-Inverting op-amp: TF(s) = Vo(s)/Vi(s) TF(s) = -Z2(s)/Z1(s) TF = (Ri+Rf)/Ri

  23. Non-Inverting Op-Amps • Notice how the inverting and non-inverting op-amps have different inputs into each terminal • In order to represent a non-inverting op-amp (and all op-amps which have the input voltage going into the positive terminal) we have to use either the a bond graph type similar to the bond graph from the inverting op-amp or the previous non-inverting op-amp bond graph, that does not fully work properly, but is suitable for our purpose, and manually invert the output signal

  24. Transfer Function and Matrices from Camp-G 10 9 C Rf 0 Inverting: [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]] 8 1 7 6 SF SE 0 0 0 1 3 5 4 2 Ri

  25. Transfer Function and Matrices from Camp-G Rf 9 8 Non-Inverting: 7 C Ri 0 [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] 6 1 5 4 0 SE SF 0 1 3 2

  26. Inverting and Non-Inverting Op-Amps • Say we have an input voltage of Vinput = Rinput*sin (t) where Rinputis some input resistance and the current is represented as i = sin (t) • Say C = 1, Ri = 2, Rf = 2, Rinput = 1, Vinput = 1*sin (t) • The output voltages of the integrating and differentiating op-amps in comparison with the input voltage is shown on the next slide

  27. Simulink Model of Inverting Op-Amps

  28. Simulink Model of Non-Inverting Op -Amps

  29. Simulink Comparison of Inverting and Non-Inverting Op-Amps Sine Wave: Inverting: Non-Inverting:

  30. Inverting and Non-Inverting Op-Amps • Notice the expected behavior of the inverting and non-inverting op-amps • Notice how we used the gain function to reduce the output voltage for the reference case • Notice the lag involved with the inverting and non-inverting op-amps when compared with the reference case

  31. Investigation: Input Voltage • We will investigate a series of issues encountered while using CAMPG and MATLAB • First, we will look into determining the transfer function from the output voltage to the input voltage • For most situations, it is more valuable to know the relationship between the input and output voltages than the input current to output voltage relationship determined on the previous slides

  32. Transfer Function and Matrices from Camp-G C Integrator: 8 [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]] 1 7 6 0 SE SE 0 0 1 3 5 4 2 R

  33. Transfer Function and Matrices from Camp-G R Differentiator: 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]] 1 7 6 SE SE 0 0 0 1 3 5 4 2 C

  34. Transfer Function and Matrices from Camp-G 10 9 C Rf 0 Inverting: [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]] 8 1 7 6 SE SE 0 0 0 1 3 5 4 2 Ri

  35. Investigation: Derivative Causality • Now we will look the CAMPG’s derivative causality error for the bond graphs on the following slides • CAMPG cannot interface to another program when there are any derivative causality errors • CAMPG errors show up in red and can be diagnosed by using the peek and analyze functions

  36. Derivative Causality 10 9 Cf R 0 The derivative causality error is shown in red because CAMPG needs the integral form of the compliance element 8 1 5 4 SF 0 SE 0 0 1 3 7 6 2 Ci

  37. Derivative Causality Cf The derivative causality error is shown in red because CAMPG and MATLAB need the integral form of the compliance element 10 1 9 8 SF SE 0 1 0 1 5 7 6 2 3 4 0 Ci

  38. Derivative Causality 12 11 Cf R 0 The derivative causality error is shown in red because CAMPG and MATLAB need the integral form of the compliance element 10 1 9 8 SF SE 0 1 0 1 5 7 6 2 3 4 0 Ci

  39. Investigation: Output Effort Location • Now we will look into which terminal should be used as the output terminal • We will look at the relationship between the input voltage across the 1st terminal and the output voltage across 2 different terminals

  40. Differentiator: Output Effort Location R 8 As before, look at the effort output across the 8th terminal, e8 [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]] 1 5 4 SE SE 0 0 0 1 3 7 6 2 C

  41. Differentiator: Output Effort Location R 8 Now, look at the effort output across the 7th Terminal, e7 [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 1 7 6 SE SE 0 0 0 1 3 5 4 2 C

  42. Investigation: Non-Inverting Op-Amps • Now we will try to produce a non-inverting op amp in CAMPG • SE1 represents the voltage going into the positive terminal of the op-amp • SE2 represents the voltage going into the negative terminal of the op-amp

  43. Non-Integrating Amplifier Non-Integrator: The following configuration is the only one found that can be created without any derivative causality errors C Rb 0 1 Rc SE1 1 0 1 SE 0 SE2 1 0 However, this bond graph does not get past the DOS Interface Ra

  44. Filtering Op-Amps • Single pole active low pass filter • Single pole active high pass filter

  45. Summing & Subtracting Op-Amps

  46. Comparing Op-Amps

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