Introduction to Operational Amplifier Designvers 1.0 October 22, 2007Copyright © 2007 Reprinted by permission
Introduction to Operational Amplifier Design                                             Table of Contents1 Preface .........
Introduction to Operational Amplifier DesignTOP                                                     List of CircuitsCircui...
Introduction to Operational Amplifier DesignTOP                                                   List of FiguresFigure   ...
1 PrefaceTOP1 PrefaceThe scope of this course is the design of basic voltage feedback operational amplifier circuits. Usin...
2 IntroductionTOP2 IntroductionThe voltage feedback differential amplifier (“op amp” as it is called) is used in a wide va...
3 The Ideal Voltage Feedback Op AmpTOP3 The Ideal Voltage Feedback Op AmpThe voltage feedback op amp is a discrete device ...
3 The Ideal Voltage Feedback Op Amp                                               3.1 Ideal CharacteristicsTOP3.1 Ideal Ch...
3 The Ideal Voltage Feedback Op Amp                                                  3.2 Ideal Model with FeedbackTOP3.2 I...
3 The Ideal Voltage Feedback Op Amp                                    3.2 Ideal Model with FeedbackTOP3.2.2 The zero-refe...
3 The Ideal Voltage Feedback Op Amp                                              3.3 Inverting or Non-invertingTOP3.3 Inve...
3 The Ideal Voltage Feedback Op Amp                                                       3.4 About the Transfer FunctionT...
3 The Ideal Voltage Feedback Op Amp                                              3.5 About Input ImpedancesTOP3.5 About In...
4 Linear Amplifiers and Attenuators                                                         4.1 Voltage FollowerTOP4 Linea...
4 Linear Amplifiers and Attenuators                                           4.2 Inverting Amplifier or AttenuatorTOP4.2 ...
4 Linear Amplifiers and Attenuators                                                                  4.3 Non-Inverting Amp...
5 Summation Amplifier                                                               5.1 Inverting Summation AmplifierTOP5 ...
5 Summation Amplifier                                                           5.1 Inverting Summation AmplifierTOPTo get...
5 Summation Amplifier                                                    5.2 Non-inverting Summation AmplifierTOP5.2 Non-i...
5 Summation Amplifier                                                 5.2 Non-inverting Summation AmplifierTOP            ...
5 Summation Amplifier                                                  5.2 Non-inverting Summation AmplifierTOP           ...
6 The Integrator                                                                               6.1 Inverting IntegratorTOP...
6 The Integrator                                                               6.1 Inverting IntegratorTOP6.1.2 Bandwidth ...
6 The Integrator                                                                                  6.2 Non-inverting Low-pa...
6 The Integrator                                                                  6.2 Non-inverting Low-pass FilterTOP6.2....
7 The Differentiator                                                                 7.1 The Inverting DifferentiatorTOP7 ...
7 The Differentiator                                                          7.1 The Inverting DifferentiatorTOP7.1.2 Ban...
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
Introduction to operational amplifier designn
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Introduction to operational amplifier designn

  1. 1. Introduction to Operational Amplifier Designvers 1.0 October 22, 2007Copyright © 2007 Reprinted by permission
  2. 2. Introduction to Operational Amplifier Design Table of Contents1 Preface ..................................................................................................... 1-12 Introduction............................................................................................... 2-13 The Ideal Voltage Feedback Op Amp .............................................................. 3-1 3.1 Ideal Characteristics .............................................................................. 3-2 3.2 Ideal Model with Feedback ...................................................................... 3-3 3.3 Inverting or Non-inverting ...................................................................... 3-5 3.4 About the Transfer Function .................................................................... 3-6 3.5 About Input Impedances ........................................................................ 3-74 Linear Amplifiers and Attenuators.................................................................. 4-1 4.1 Voltage Follower.................................................................................... 4-1 4.2 Inverting Amplifier or Attenuator ............................................................. 4-2 4.3 Non-Inverting Amplifier .......................................................................... 4-35 Summation Amplifier................................................................................... 5-1 5.1 Inverting Summation Amplifier ................................................................ 5-1 5.2 Non-inverting Summation Amplifier.......................................................... 5-36 The Integrator............................................................................................ 6-1 6.1 Inverting Integrator ............................................................................... 6-1 6.2 Non-inverting Low-pass Filter .................................................................. 6-37 The Differentiator ....................................................................................... 7-1 7.1 The Inverting Differentiator..................................................................... 7-1 7.2 Non-inverting High-pass Filter ................................................................. 7-28 Practical Considerations ............................................................................... 8-1 8.1 Device Parameters................................................................................. 8-1 8.2 Power Supplies ..................................................................................... 8-2 8.3 Offsetting and Stabilizing........................................................................ 8-3 8.4 Impedance Matching and Phase Compensation .......................................... 8-5Appendix A. ..Commonly Used Terms ................................................................ A-1Appendix B. ..Bibliography ............................................................................... B-1 ii
  3. 3. Introduction to Operational Amplifier DesignTOP List of CircuitsCircuit 4-1 Voltage Follower .........................................................................................4-1Circuit 4-2 Inverting Amplifier or Attenuator...................................................................4-2Circuit 4-3 Non-inverting Amplifier ...............................................................................4-3Circuit 5-1 Inverting Summation Amplifier .....................................................................5-1Circuit 5-2 Non-inverting Summation Amplifier ..............................................................5-3Circuit 6-1 Inverting Integrator ....................................................................................6-1Circuit 6-2 The Non-inverting Integrator ........................................................................6-3Circuit 7-1 Inverting Differentiator ................................................................................7-1Circuit 7-2 Non-inverting Differentiator..........................................................................7-2 iii
  4. 4. Introduction to Operational Amplifier DesignTOP List of FiguresFigure 3.1-1 Loop analysis example ..............................................................................2-1Figure 3.1-1 Ideal operational amplifier .........................................................................3-1Figure 3.1-2 Ideal op amp input impedances ..................................................................3-1Figure 3.2-1 Ideal op amp feedback model 1 ..................................................................3-3Figure 3.2-2 Ideal op amp feedback model 2 ..................................................................3-4Figure 3.3-1 Non-inverting examples ............................................................................3-5Figure 3.3-2 Loop equations for Fig 3.3-1 ......................................................................3-5Figure 3.4-1 Black box diagram ....................................................................................3-6Figure 3.4-2 Cascaded transfer functions .......................................................................3-6Figure 3.5-1 Zero impedance source .............................................................................3-7Figure 3.5-2 Source with internal impedance ..................................................................3-7 - +Figure 4.2-1 E and E loops for Circuit 4-2 ...................................................................4-2 - +Figure 4.3-1 E and E loops for Circuit 4-3 ...................................................................4-3 -Figure 5.1-1 E loop for Circuit 5-1 ...............................................................................5-1Figure 5.1-2 Thevenized input for Circuit 5-1..................................................................5-2 +Figure 5.2-1 E loop for Circuit 5-2...............................................................................5-3 -Figure 5.2-2 E loop for Circuit 5-2 ...............................................................................5-3Figure 5.2-3 Thevenized input for Circuit 5-2..................................................................5-4 - +Figure 6.1-1 E and E loops for Circuit 6-1 ...................................................................6-1 +Figure 6.2-1 E loop for Circuit 6-2...............................................................................6-3 - +Figure 7.1-1 E and E loops for Circuit 7-1 ....................................................................7-1 +Figure 7.2-1 E loop for Circuit 7-2 ..............................................................................7-2Figure 8.2-1 Filtering the Power Supply .........................................................................8-2Figure 8.3-1 Avoiding common mode noise ....................................................................8-3Figure 8.3-2 DC baseline shift example..........................................................................8-3 - +Figure 8.3-3 E loop with Vs = 0 and E loop with bias VB ................................................8-4Figure 8.4-1 Impedance matching ................................................................................8-5Figure 8.4-2 Canceling reactance..................................................................................8-6 iv
  5. 5. 1 PrefaceTOP1 PrefaceThe scope of this course is the design of basic voltage feedback operational amplifier circuits. Using theideal op amp model and solving for the currents and voltages at each terminal we get the transferfunction as a Laplace Transform. This course provides a practical way of going from paper design toprototyping working circuits.This course is intended for professional electrical engineers. The course-taker should be familiar withthe Laplace and inverse Laplace Transforms and basic AC network analysis.After completing the course, there is a quiz consisting of 16 multiple choice questions. On completion,4 professional development hours will count towards satisfying PE licensure renewal requirements.Navigating the course is facilitated by hyperlinked table of contents on each page or the tags in thebookmark pane. 1-1
  6. 6. 2 IntroductionTOP2 IntroductionThe voltage feedback differential amplifier (“op amp” as it is called) is used in a wide variety ofelectronic applications such as: linear amplifier/attenuator, signal conditioner, signal synthesizer,computer, or simulator.A practical way to approach designing and implementing an op amp circuit is to start with the idealmodel and get an expression that relates the output to the input, regardless of the input. This is [1]accomplished by working with the loop equations in the frequency or s domain .In summary, the key to getting the transfer function is that the voltages at the input terminals of aclosed-loop op amp circuit mirror each other. Also, no current flows into or out of an input terminal. omWhen a signal is applied to either or both terminals, the output will adjust itself to meet theseconstraints. .c ng Using the figure to the left : ri + - if V2 and Z2 = 0 then E = 0 ∴E = 0 ee 02 ngin We know that no current flows into an input + - terminal so I = I = 0 ∴I1 = If and I2 = 0 -0 e - - V1 − E E − Vo = − Vo 04 ed V1 I1 = = and If = Z1 Z1 Zf Zf E .c V1 Vo so I1 = If gives us =− Z1 Zf se ww Vo Z ur /w Figure 3.1-1 Loop analysis example hence =− f V1 Z1 o :/C tpThe closed loop transfer function is in the frequency domain, A v (s) = Vo (s) V1(s) Z =− f . Z1 ht Vo ZFor the remainder of the course, we ll use the shorter notation as, A v = =− f . V1 Z1Also note : in AC network analysis, impedances are represented as phasors and do not vary withtime but with frequency. So there is no time domain representation or time variation of an impedance.In the following sections, the same method is used for application specific circuits where the voltagesand impedances are arbitrary.[1] To get the output in the time domain vo(t) we would have to multiply Vi(s) by Av(s) and then take the inverse Laplace -1Transform ; vo(t)=L [ Vi(s)·Av(s) ]. 2-1
  7. 7. 3 The Ideal Voltage Feedback Op AmpTOP3 The Ideal Voltage Feedback Op AmpThe voltage feedback op amp is a discrete device that has 2 input terminals and one output terminal.Without feedback, the output is the difference between the input voltages, multiplied by the open-loopgain (transfer function) of the op amp. + E the non - inverting terminal - E the inverting terminal Eo the output terminal om Avol is the open loop gain (transfer function) of .cFigure 3.1-1 Ideal operational amplifier the op amp itself, without feedback ng + - The key is in maintaining E = E otherwise ri The open l oop equati ons : the output will saturate: ee ⎛ + −⎞ ⎜ E − E ⎟ ⋅ A vol ≡ Eo where A vol = ∞ + - ⎝ ⎠ If we apply a voltage at E and ground E 02 ngin ⎛ + −⎞ Eo ≡ 0 + − E o will saturate to positive supply voltage. ⎜E − E ⎟ ≡ so E −E ≡ 0 ⎝ ⎠ A vol -0 e + − - + ∴ E ≡E If we apply a voltage at E and ground E 04 ed E o will saturate to negative supply voltage. E .c se wwIn the open loop model, each input terminal has infinite impedance so no current can flow into aninput terminal even with a voltage source or a ground applied. The output terminal has zero output ur /wimpedance. o :/C tp ht E + + has Zin = ∞ so I + =0 E − − has Zin = ∞ so I − =0 Eo has Z out ≡0 Figure 3.1-2 Ideal op amp input impedancesThe ideal characteristics are summarized in Table 3-1 below 3-1
  8. 8. 3 The Ideal Voltage Feedback Op Amp 3.1 Ideal CharacteristicsTOP3.1 Ideal Characteristics Summary Of Ideal Characteristics Zin = ∞ the input impedance at each terminal is infinite Zout = 0 the output impedance is zero ± I =0 no current flows into either of the input terminals Avol = ∞ the open loop gain is infinite om Bandwidth = ∞ the bandwidth is infinitely wide .c No temperature drift ng + - Eo = 0 the output voltage is zero when E = E ri Table 3-1 Summary of ideal characteristics ee 02 ngin -0 e 04 ed E .c se ww ur /w o :/C tp ht 3-2
  9. 9. 3 The Ideal Voltage Feedback Op Amp 3.2 Ideal Model with FeedbackTOP3.2 Ideal Model with FeedbackWith feedback, all or a portion of the output is tied to either or both input terminals. The differencebetween the voltages at the input terminals is still equal to zero and again no current flows into eitherof the input terminals.The voltage at each input terminal is treated as a node and is determined by the loop equations. Theterminal voltages are equated and we get a transfer function or closed loop gain which provides theoutput over the input as a ratio (Laplace Transform).In the following subsections, we’ll look at the feedback models : the non-zero reference and the zeroreference feedback models. As the names imply, the non-zero reference model is characterized by the ±voltage at either input terminal, E ≠ 0 , and the zero-reference model is where the voltage at either om ±input terminal is referenced to 0 volts or ground; E .c =0. ng3.2.1 The non-zero reference model ri ee + The figure to the left models a source Vi connected to the E 02 ngin terminal through a series impedance Z1 and feedback k Eo − Z2 -0 e tied to E ; with k being a voltage divider = Z2 + Zf 04 ed The voltage at each input terminal balances out to k Eo E .c + − + So E = E = k Eo and we see that I = I1 . se ww + + V −E Vi − k Eo ur /w But I = 0 , giving us I1 = i = =0 Z1 Z1 o :/ Eo 1 Z2 + Zf Z Hence = = = 1+ fC tp Vi k Z2 Z2Figure 3.2-1 Ideal op amp feedback model Eo ⎛ Z ⎞ ht1 The transfer function A v = = ⎜1 + f ⎟ ⎜ Vi ⎝ Z2 ⎟ ⎠3.2.1.1 The input impedance VThe input impedance seen by Vi is Zin = i Iin Vi − k Eo ⎛ 1 − kA v ⎞The input current Iin = I1 , so Iin = Since kA v = 1, Vi ⋅ ⎜ ⎜ Z ⎟=0 ⎟ Z1 ⎝ 1 ⎠With Iin = 0, Zin = ∞, so the input impedance seen by Vi is infinite or an open circuit. 3-3
  10. 10. 3 The Ideal Voltage Feedback Op Amp 3.2 Ideal Model with FeedbackTOP3.2.2 The zero-reference model The figure to the left models a source Vi connected − to the E terminal through a series impedance Z1 , + with E tied to ground. The voltage at each input must balance to 0 volts; ± thus E =0 om − − We can see that I1 + If = I ; with I = 0 , I1 = −If .c - - Vi − E Eo − E - ng Substituting I1 = , If = and E = 0 Z1 Zf ri Figure 3.2-2 Ideal op amp feedback model 2 Vi E =− o ee into I1 = −If gives us Z1 Zf 02 ngin Eo Z So the transfer function A v = =− f Vi Z1 -0 e 04 ed E .c3.2.2.1 The input impedance se ww ur /w V V VThe input impedance seen by Vi is Zin = i ; with Iin = I1 = i , Zin = i = Z1 Iin Z1 Vi o :/ Z1C tp ht 3-4
  11. 11. 3 The Ideal Voltage Feedback Op Amp 3.3 Inverting or Non-invertingTOP3.3 Inverting or Non-invertingThe determining factors for whether the output is inverted or not, are the circuit configuration and theloop equations for the terminal currents and voltages. With voltage feedback op amp circuits, whichterminal the signal is connected to, by itself, does not determine whether the output is inverted or not.For example, the 2 circuits below are both inverting amplifier/attenuators circuits. The only differenceis the op amp input terminals are reversed but both provide an output that is a linear amplifier orattenuator with 180° phase shift or inversion: .c om ring ee 02 ngin -0 e Figure 3.3-1 Non-inverting examples 04 ed E .c E - =0 ∴E + =0 E + =0 ∴E - =0 se ww + + - - I1 + I f = I I = 0 ∴ I1 = − I f I1 + I f = I I = 0 ∴ I1 = − If ur /w V V Vi Vo I1 = i If = o I1 = If = R1 Rf R1 Rf o :/C tp V R1 V ∴ i = − o and A v = − R f Rf R1 ∴ Vi R1 = − Vo Rf R and A v = − f R1 ht Figure 3.3-2 Loop equations for Fig 3.3-1 3-5
  12. 12. 3 The Ideal Voltage Feedback Op Amp 3.4 About the Transfer FunctionTOP3.4 About the Transfer FunctionThe transfer function gives you the output over the input expressed as a ratio. To get the output for aspecific input, multiply the Laplace Transform of the input by the transfer function. To convert to the [2]time domain, take the inverse Laplace Transform . Using Laplace Transform tables available on theinternet or in printed textbooks, is a very useful tool in op amp circuit design. Some online referencesare listed in Appendix B.Working in the s domain (using Laplace Transforms) is advantageous and readily gives any transientsthat might exist Vo (s) so Vo (s) = Vi (s) ⋅ A v (s) om A v (s) ≡ Vi (s) .c to get the time domain representation −1 [ Vi (s) ⋅ A v (s) ] ng v o (t) = L or ri ee by using convolut ion : Figure 3.4-1 Black box diagram t ∫ v o (t) = v i (τ)⋅av (t − τ)dτ 02 ngin 0 -0 e 04 ed When cascading circuits, the overall transfer function is the product of all the transfer functions in the cascade and the overall E .c transfer function can be viewed as a single se ww ratio or Laplace transform ur /w V (s) A v (s) o2 = A v1(s) ⋅ A v 2 (s) o :/ Vi (s)C tp Figure 3.4-2 Cascaded transfer functions htIn the time domain, you could not express the transfer function as a ratio. You would have to solvedifferential equations for the terminal voltages and currents and use the convolution integral to get theoutput as a function of time, because superposition does not apply. -1[2] The time domain representation of Av (that is av(t) = L [Av(s)] ) is actually the unit impulse response of the circuit. Theoutput vo(t) after applying an arbitrary input vi(t) is found by convolving vi(t) with av(t) { vo(t) ≡ vi(t)*av(t) } . 3-6
  13. 13. 3 The Ideal Voltage Feedback Op Amp 3.5 About Input ImpedancesTOP3.5 About Input ImpedancesAn input source can be either a directly connected zero impedance source, or a thevenin equivalentsource with an internal impedance. We need to be aware if a source does have internal impedance.This needs to be considered in determining both the input impedance seen by the voltage source andthe transfer function of the circuit.3.5.1 Directly connected sourceConsider a directly-connected source, with zero output impedance, connecting to an op amp inputterminal, through a series impedance Z1 : The input to the circuit is Vi = Vg . om Vo In general, the transfer function A v gives us .c Vg ng ± When E = 0 , the input impedance seen by Vg is ri ± Vg Vg − E Vg − 0 Vg ee Zin = ; with Iin = = , Zin = = Z1 . Iin Z1 Z1 Vg 02 ngin Figure 3.5-1 Zero impedance source Z1 -0 e ⎛ ± ⎞ ± Vi Vi − E ⎜ Vi ⎟ 04 edWhen E ≠ 0 , the input impedance seen by Vi is Zin = ; with Iin = , Zin = Z1 ⎜ ⎟ Iin Z1 ⎜ V − E± ⎟ ⎝ i ⎠ E .c* It is interesting to note that in a case like this, we have a capability of synthesizing a voltage se ww ±dependent input impedance that varies with Vi and E . ur /w o :/3.5.2 Source with built-in internal impedanceC tpConsider the case where the input is the end of a cable span or an input from a previous circuit stagein the cascade: ht Now we have a source Vg with internal impedance Z g . ⎛ Zin ⎞ We have to modify our calculatio ns because V = Vg ⎜ ⎟. i ⎜ Zin + Z g ⎝ ⎟ ⎠ Substituti ng into the transfer function A v , we get Vo V ⎛ Zin + Z g ⎞ Vo ⎛ Zg ⎞ = o ⎜ ⎜ ⎟ = ⎜ ⎟ V ⎜1 + Z ⎟ ⎟ Vi Vg ⎝ Zin ⎠ g ⎝ in ⎠ the total impedance seen by Vi is Zin the total impedance seen by Vg is Z g + Zin Figure 3.5-2 Source with internal impedance 3-7
  14. 14. 4 Linear Amplifiers and Attenuators 4.1 Voltage FollowerTOP4 Linear Amplifiers and AttenuatorsThe most basic circuit configurations are linear amplifiers and attenuators. Ideally, they provide flatgain or loss across the device-rated bandwidth.4.1 Voltage FollowerThe voltage follower circuit provides unity gain with no inversion. This circuit is used to isolate a highimpedance source input and provide a buffered output. For Circuit 4 - 1, the terminal voltages are : Vo = Eo om E+ = Vi .c - E = Eo ng - V E+ E , Vi = Vo giving us o = 1 . ri Since = Vi ee Circuit 4-1 Voltage Follower So A V = 1 is the transfer function for Circuit 4.1. 02 ngin -0 e4.1.1 The Input impedance 04 ed E .cThe input impedance seen by source Vi is Zin = Vi where Iin = Vi − E + se ww Iin R1 ur /w + ViSince E = Vi , Iin = 0 , making Zin = =∞ 0 o :/So the input impedance seen by Vi for Circuit 4 - 1 is an open circuit.C tp ht 4-1
  15. 15. 4 Linear Amplifiers and Attenuators 4.2 Inverting Amplifier or AttenuatorTOP4.2 Inverting Amplifier or AttenuatorThe inverting configuration provides flat gain or attenuation with constant 180° phase shift. For Circuit 4 - 2, the terminal voltages are : Vo = Eo E+ =0 - E = E+ =0 * the terminal voltages balance out to zero .c omCircuit 4-2 Inverting Amplifier or Attenuator ng - ri From Figure 4.2 - 1 , for the E loop we have : ee - - Vi − E V −E I1 = and I f = o 02 ngin R1 Rf − − With I1 + I f = I and I = 0 , we have I1 = - I f -0 e - E 04 ed Since the terminal voltage at = 0, Vi V V R I1 = and I f = o and with I1 = - I f , o = − f E .c R1 Rf Vi R1 se ww R So A v = − f is the transfer function for Circuit 4.2 ur /w - + R1Figure 4.2-1 E and E loops for Circuit 4-2 o :/C tp ht4.2.1 The Input impedance Vi ViThe input impedance seen by source Vi is , Zin = ; where Iin = I1 = Iin R1 V Substituting Iin into the expression for Zin gives us Zin = i = R1 Vi R1So the input impedance seen by Vi for Circuit 4 - 2 is R 1 4-2
  16. 16. 4 Linear Amplifiers and Attenuators 4.3 Non-Inverting AmplifierTOP4.3 Non-Inverting AmplifierThe non-inverting configuration provides flat gain with no phase shift. For Circuit 4 - 3, the terminal voltages are : Vo = Eo E+ = Vi - R1 E = kVo ; where k is a voltage divider = R1 + R f - V Since E+ E , kVo = Vi ; readily giving us o = = 1 Vi k om 1 ⎛ R ⎞ = ⎜1 + f ⎟ . .c So by inspection, we see that A v = ⎜ k ⎝ R1 ⎟ ⎠ ng Circuit 4-3 Non-inverting Amplifier ri − Analysis of the E loop gives us the same result : ee - - E Vo − E From Fig 4.3 - 1 we see that I1 = and If = 02 ngin R1 Rf − − -0 e With If = I1 + I and I = 0 , If = I1 - + - + 04 ed Since E = kVo , E = Vi and E = E , - we substitute E = Vi , into the expressions for I1 and If E .c se ww V Vo − Vi - + giving us I1 = i and If = E E R1 Rf ur /wFigure 4.3-1 and loops for Circuit 4-3 o :/C tpWith If = I1 we get Vi R1 = Vo − Vi Rf ∴ ⎛ 1 Vi ⋅ ⎜ ⎜R + R ⎝ 1 1 f ⎞ ⎟ ⋅ R f = Vo ; giving us ⎟ ⎠ ht Vo ⎛ R ⎞Av = = ⎜1 + f ⎟ as the transfer function for Circuit 4 - 3. Vi ⎜ ⎝ R1 ⎟ ⎠4.3.1 The Input impedance + Vi V −EThe input impedance seen by source Vi is Zin = From Figure 4.3 - 1, Iin = I g where I g = i Iin Rg + Vi − kVo VSubstituting E = kVo into Iin gives us Iin = With kVo = Vi , Iin = 0 making Zin = i = ∞ Rg 0So the input impedance seen by Vi for Circuit 4.3 is an open circuit. 4-3
  17. 17. 5 Summation Amplifier 5.1 Inverting Summation AmplifierTOP5 Summation AmplifierThe summation amplifier provides flat gain or loss and can be configured in an inverting or non-inverting configuration. Typical uses are signal combiner, voltage comparator or summing junction.5.1 Inverting Summation AmplifierThe inverting summation amplifier provides flat gain or loss across the rated bandwidth with constant180° phase shift. For Circuit 5 - 1, the terminal voltages are : om Vo = Eo .c - E+ = 0 so E = 0 ng * the terminal voltages ri balance out to zero ee 02 ngin -0 e 04 ed E .c Circuit 5-1 Inverting Summation Amplifier se ww ur /w - From Figure 5.1 - 1, for the E loop we have : IS = I1 + I 2 + I 3 where o :/C tp I1 = V1 − E R − , I2 = V2 − E R − , I3 = V3 − E R − ht − V −E and I f = o Rf − Since the terminal voltage at E = 0 V1 V V V I1 = , I 2 = 2 , I 3 = 3 and I f = o R R R Rf − − - With I = IS + I f and I = 0 , IS = −I f Figure 5.1-1 E loop for Circuit 5-1 V1 V2 V3 VWith IS = −If , we have + + =− o R R R Rf RfSimplifyin g the expression for Vo , we get Vo = − ⋅ (V1 + V2 + V3 ) R 5-1
  18. 18. 5 Summation Amplifier 5.1 Inverting Summation AmplifierTOPTo get the transfer function as a single ratio, we have to thevenize the 3 sources into one V1 + V2 + V3 Requivalent source Vi with thevenin impedance Z eq ; giving us Vi = and Z eq = 3 3as illustrated in the figure below. - Analyzing the E loop in Figure 5.1 - 2 we have : - - V −E V −E Is = i and I f = o R Rf 3 om - Substitutin g E = 0 into I s and I f we have, .c 3 Vi V Is = and I f = o ng R Rf ri − − With I s + I f = I and I = 0 , Is = − If ee 3Vi V Vo 3R f ∴ 02 ngin = − o = − R Rf Vi R -0 e 3R f 04 ed So A v = − is the transfer function for Circuit 5 - 1. Figure 5.1-2 Thevenized input for Circuit 5-1 R E .c se ww5.1.1 The Input impedance ur /wBy inspection, the input impedance seen by our thenevized source Vi is Zin = 0 because, as Figure 5.1 - 2 o :/ R -illustrates, we are connecting a source with an internal impedance of directly to the E terminal.C tp 3 ht VnHowever, the input impedance seen by each of the sources V1 , V2 and V3 is Zn = ; where In − Vn − E −In = With E = 0 , Zn is going to be whatever impedance is in series with each source. ZnFor Circuit 5 - 1, each of the sources Vn sees Zn = R as the input impedance. 5-2
  19. 19. 5 Summation Amplifier 5.2 Non-inverting Summation AmplifierTOP5.2 Non-inverting Summation AmplifierThe non-inverting summation amplifier provides flat gain across the rated bandwidth with no phaseshift. For Circuit 5 - 2, the terminal voltages are : Vo = Eo E+ = kE o R1 om k = ; a voltage divider R1 + R f .c - E = E+ = kE o ring Circuit 5-2 Non-inverting Summation Amplifier ee 02 ngin From Figure 5.2 - 1, we see that the voltage at the -0 e E+ terminal is Eo through a voltage divider. 04 ed E .c So E+ = k Eo + + se ww Also , I f = I1 + I and with I = 0 , I1 = I f ur /w +Figure 5.2-1 E loop for Circuit 5-2 o :/ − From Figure 5.2 - 2, for the E loop we have :C tp IS = I1 + I 2 + I3 where ht − − − V1 − E V2 − E V3 − E I1 = , I2 = , I3 = R R R − V V V 3⋅E so IS = 1 + 2 + 3 − R R R R - - With I = IS and I = 0, IS = 0 − V1 V2 V3 3 ⋅ E ∴ R + R + R = R -Figure 5.2-2 E loop for Circuit 5-2 5-3
  20. 20. 5 Summation Amplifier 5.2 Non-inverting Summation AmplifierTOP - V1 V2 V 3 ⋅ kVo R1Substituting E = kVo , we get + + 3 = Substituting k = we get, R R R R R1 + R f ⎛ V1 V2 V3 ⎞ 3Vo ⎛ R 1 ⎞ ⎛ R ⎞ ⎜ ⎜ R + R + R ⎟ = R ⋅ ⎜R + R ⎜ ⎟ ⎟ ⎟ Simplifying we get, Vo = 1 (V1 + V2 + V3 ) ⋅ ⎜1 + f ⎟ ⎜ ⎝ ⎠ ⎝ 1 f ⎠ 3 ⎝ R1 ⎟ ⎠To get the transfer function in the form of a ratio, we have to thevenize the 3 sources into one V1 + V2 + V3 Requivalent source Vi with thevenin impedance Z eq ; so Vi = and Z eq = 3 3 -From Figure 5.2 - 3 below, we look at the loop equations for the E terminal. .c om We have : − ng Is = I =0 ri − V −E − with Is = i , substitute E = kVo ee R 02 ngin 3 Vi − kVo ⎛ Vi − kVo ⎞ Expanding Is we have ; Is = = 3⋅⎜ ⎟ =0 -0 e R ⎜ R ⎟ ⎝ ⎠ 04 ed 3 giving us Vi = kVo E .c ⎛ R1 ⎞ Vo ⎛ Rf ⎞ Substituti ng k = ⎜ ⎜ R + R ⎟ , we get V = ⎜1 + R ⎟ , so ⎟ ⎜ ⎟ se ww ⎝ 1 f ⎠ i ⎝ 1⎠ ⎛ R ⎞ ur /w A v = ⎜1 + f ⎟ is the transfer function for Circuit 5 - 2. ⎜ ⎝ R1 ⎟ ⎠ o :/Figure 5.2-3 Thevenized input for Circuit 5-2C tp ht5.2.1 The Input impedance Vi ⎛ Vi − kVo ⎞The input impedance seen by Vi is Zin = where Iin = Is = 3 ⋅ ⎜ ⎜ ⎟ ⎟ Iin ⎝ R ⎠ ViSubstituti ng kVo = Vi into Iin gives us Iin = 0 making Zin = =∞ 0The input impedance for Circuit 5 - 2 is an open circuit.The input impedance seen by each source is going to be voltage - dependent. ⎛ R1 ⎞We know that Vo = 1 (V1 + V2 + V3 ) , where k = ⎜ ⎜R + R ⎟ ⎟ 3k ⎝ 1 f ⎠ 5-4
  21. 21. 5 Summation Amplifier 5.2 Non-inverting Summation AmplifierTOP - VnWe also know that the voltage at E = kVo . The input impedance seen by each source is Zin n = In - Vn − Ewhere In = and Zn is the series impedance of each source. ZnFor example; - V1 − EIn Circuit 5 - 2, source V1 has a series impedance of R so the input current I1 = R - V1 − 1 (V1 + V2 + V3 ) 2V1 − (V2 + V3 ) om = (V1 + V2 + V3 ) , I1 = 1Substituting E 3 = 3 R 3R .c ng V V1so the input impedance seen by source V1 is Zin 1 = 1 = which simplifies to I1 2V1 − (V2 + V3 ) ri 3R ee ⎛ V1 ⎞Zin 1 = 3R ⋅ ⎜ ⎟ ⎜ 2V − (V + V ) ⎟ 02 ngin ⎝ 1 2 3 ⎠ + V3 ) (V2 (V + V3 )So Zin 1 is an open circuit when V1 − , and is a negative impedance when V1 < 2 -0 e 2 2A negative impedance means V1 is drawing current from the other sources and the output. 04 ed E .cTo get a nominal value for Zin 1 , we have to know something about the harmonic content and magnitude se wwcoefficients of the other sources so we can calculate an average. ur /wThe same calculations apply to getting the input impedances seen by the other sources at the input. o :/C tp ⎛Zin n = 3R ⋅ ⎜ Vn ⎜ 2V − ∑ V ⎝ n m ⎞ ⎟ ⎟ ⎠ ht 5-5
  22. 22. 6 The Integrator 6.1 Inverting IntegratorTOP6 The IntegratorTypical uses of the integrator circuit are: noise reduction, simulation of a first order RC low-passcircuit, low pass filter, calculating an integral or just phase compensation.6.1 Inverting IntegratorThe inverting integrator provides low-pass filtering with constant +90° phase shift. The roll-off startsat DC. Typical use is direct integration of a time domain function, noise reduction, or low pass filter. For Circuit 6 - 1, the terminal voltages are : om Vo = Eo .c E+ =0 ng - + E = E =0 ri * the terminal voltages balance out to zero ee 02 ngin − From Figure 6.1 - 1, for the E -0 e loop we have : − − V −E V −E 04 ed Circuit 6-1 Inverting Integrator ⎛ −⎞ I1 = i , and I f = o = sC ⎜ Vo − E ⎟ R1 1 ⎝ ⎠ E .c sC − se ww Substituting E = 0 into I1 and I f gives us ur /w Vi I1 = , and I f = sCVo R1 o :/ − − With I f + I1 = I and I = 0 , I f = −I1 soC tp sC Vo = − Vi ∴ Vo =− 1 ht R1 Vi sCR 1 1 - + Av = − is the transfer function for Circuit 6 - 1.Figure 6.1-1 E and E loops for Circuit 6-1 sCR 16.1.1 The Input impedance V VThe input impedance Zin seen by Vi is Zin = i in this case Iin = I1 = i Iin R1 Vi ViWith Iin = Zin = = R1 R1 Vi R1The input impedance seen by source Vi for Circuit 6 − 1 is R 1. 6-1
  23. 23. 6 The Integrator 6.1 Inverting IntegratorTOP6.1.2 Bandwidth Considerations 1For Circuit 6 − 1, A v = − ; substituting s = jω gives us the value of A v(s) at s = jω. sCR 1We see at s = 0 (DC), A v = − ∞ which means the output of the circuit will be saturated at the negativesupply voltage, or 0 volts if we are using a single supply. We may want to offset this if we are workingwith input signals that have DC components.See Section 8.3.2 for offsetting the baseline. π j 1 2 1We can rewrite A v =− as A v = A v ⋅ e where A v = ; noting that the constant 90° jωCR 1 ωCR 1 omphase shift is independent of ω. If we start the design at a fundamental frequency ω 0 , then we would .c 1select CR 1 to give us the desired gain magnitude at ω 0 ; i.e CR 1 = ω0 ⋅ A v ng 1 riOur - 3dB bandwidth is the ω at which A v (ω) = A v (ω 0 ) 2 ee A v (ω) ω 1This gives us the direct relationship = 0 = A v (ω 0 ) ω 02 ngin 2So our - 3dB bandwidth occurs at ω = 2ω 0 -0 e 04 ed E .c se ww ur /w o :/C tp ht 6-2
  24. 24. 6 The Integrator 6.2 Non-inverting Low-pass FilterTOP6.2 Non-inverting Low-pass FilterThe non-inverting low-pass filter provides filtering with phase shift that varies with the complextransfer function. For Circuit 6 - 2, the terminal voltages are : Vo = E o ⎛ 1 ⎞ ⎜ ⎟ ⎛ ⎞ E = Vi ⎜ sC 1 ⎟ = Vi ⎜ 1 ⎟ + ⎜ sCR + 1 ⎟ ⎜R + ⎟ ⎝ 1 ⎠ ⎜ 1 ⎟ ⎝ sC ⎠ om ⎛ 1 ⎞ ⎜ ⎜ sCR + 1 ⎟ being a voltage divider ⎟ ⎝ 1 ⎠ .c - R2 E = kVo where k = ng R2 + R 3 ri k being a voltage divider Circuit 6-2 The Non-inverting Integrator ee - ⎛ 1 ⎞ Vo 1⎛ 1 ⎞ ⎛ R3 ⎞ ⎛ 1 ⎞ 02 nginWith E+ = E , Vi ⎜ ⎜ sCR + 1 ⎟ = kVo ⎟ so = ⎜ ⎜ sCR + 1 ⎟ = ⎜1 + R ⎟ ⋅ ⎜ sCR + 1 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ Vi k⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 1 ⎠ -0 e Vo ⎛ R ⎞ ⎛ 1 ⎞Av = = ⎜1 + 3 ⎟ ⋅ ⎜ ⎜ ⎟ is the transfer function for Circuit 6 - 2. R 2 ⎟ ⎜ sCR 1 + 1 ⎟ 04 ed Vi ⎝ ⎠ ⎝ ⎠ E .c6.2.1 The input impedance se ww + E ur /wTo calculate the input impedance seen by Vi we analyze the loop. o :/ Vi The input impedance seen by source Vi is Zin =C tp Iin For Circuit 6 - 2 , Iin = I1 ht + Vi − E From Figure 6.2 - 1, we have I1 = R1 Vi Vi − + Vi sCR 1 + 1 + Recall E = so I1 =Figure 6.2-1 E loop for Circuit 6-2 sCR 1 + 1 R1 ⎛ sC ⎞ Vi sCR 1 + 1 1I1 simplifies to Vi ⋅ ⎜ ⎜ sCR + 1 ⎟ so Zin = ⎟ = = R1 + ⎝ 1 ⎠ ⎛ sC ⎞ sC sC ⎜ sCR + 1 ⎟ Vi ⋅ ⎜ ⎟ ⎝ 1 ⎠ 1The input impedance seen by source Vi for Circuit 6 - 2 is R 1 + . sC 6-3
  25. 25. 6 The Integrator 6.2 Non-inverting Low-pass FilterTOP6.2.2 Bandwidth considerationsCircuit 6 - 2 is a first order low - pass filter with a fundamental frequency of ω 0 = 0. Vo ⎛ R ⎞ ⎛ 1 ⎞ ⎛ R ⎞Previously, we found the transfer function to be A v = = ⎜1 + 3 ⎟ ⋅ ⎜ ⎟ ; let s replace ⎜1 + 3 ⎟ Vi ⎜ ⎝ R 2 ⎟ ⎜ sCR 1 + 1 ⎟ ⎠ ⎝ ⎠ ⎜ ⎝ R2 ⎟ ⎠ 1 1 ⎛ 1 ⎞ - jθwith so A v = ⎜ ⋅⎜ ⎟ Rewriting A v in phasor form, and substituing s = jω gives us A v = A v ⋅ e ⎟ k k ⎝ sCR 1 + 1 ⎠ ⎛ ⎞ 1 ⎜ 1 ⎟ -1where A v = ⋅⎜ ⎟ and θ = tan ωCR 1 k ⎜(ωCR 1 ) + 1 ⎟ 2 ⎝ ⎠ om 1We see that at ω 0 , A v (ω 0 ) = and θ = 0 radians .c k A v (ω) ng 1 1The - 3dB bandwidth is where A v (ω) = ⋅ A v (ω 0 ) or = A v (ω 0 ) ri 2 2 ⎛ ⎞ ee 1 1 ⎜ 1 ⎟ A v (ω) 1 1Since A v (ω 0 ) = and A v (ω) = ⋅ ⎜ ⎟ , = = k k ⎜ (ωCR 1 ) 2 ⎟ + 1⎠ A v (ω 0 ) (ωCR 1 ) 2 +1 2 02 ngin ⎝ 1 1 3 -0 eThe - 3dB point is where = or where ω = (ωCR 1 )2 +1 2 CR 1 04 ed E .c se ww ur /w o :/C tp ht 6-4
  26. 26. 7 The Differentiator 7.1 The Inverting DifferentiatorTOP7 The DifferentiatorTypical uses of the differentiator circuit are: simulation of a first order RC high-pass circuit, high passfilter, calculating a derivative or just phase compensation.7.1 The Inverting DifferentiatorThe inverting differentiator provides high-pass filtering with constant - 90° phase shift. The roll-offstarts at DC. Typical use is direct differentiation of a time domain function, or high pass filter. For Circuit 7 - 1, the terminal voltages are : Vo = Eo om E+ =0 .c - E = E+ = 0 ring From Figure 7.1 - 1, for the E − loop we have : ee − − IC + If = I ; with I = 0 , IC = − If where 02 ngin Circuit 7-1 Inverting Differentiator − − Vi − E Vo − E -0 e IC = and If = 1 R1 04 ed sC − With the voltage at E = 0 , E .c V V se ww IC = i and If = o 1 R1 ur /w sC Substituting into IC = − If , o :/ Vo VoC tp we get sCVi = − ∴ = −sCR1 R1 Vi ht Vo Av = = −sCR1 is the transfer function - + ViFigure 7.1-1 E and E loops for Circuit 7-1 for Circuit 7 − 1 .7.1.1 The input impedance VThe input impedance seen by Vi is Zin = i where Iin = IC Iin Viwith Iin = sCVi Zin = sCVi 1The input impedance seen by source Vi for Circuit 7 - 1 is sC 7-1
  27. 27. 7 The Differentiator 7.1 The Inverting DifferentiatorTOP7.1.2 Bandwidth considerationsFor Circuit 7 − 1, A v = −sCR1 ; substituting s = jω gives us the value of A v(s) at s = jω.We see at s = 0 (DC), A v = 0 which means the output of the circuit will be 0 volts π −j 2We can rewrite A v = − jωCR 1 as A v = A v ⋅ e where A v = ωCR 1 ; noting that the constant - 90°phase shift is independent of ω. If we start the design at a fundamental frequency ω 0 , we would Avselect CR 1 to give us the desired gain magnitude at ω 0 ; i.e CR 1 = ω0 omOur - 3dB bandwidth is the ω at which A v (ω) = 2 ⋅ A v (ω 0 ) .c A v (ω) ωThis gives us the direct relationship = =2 A v (ω 0 ) ωo ngSo our - 3dB bandwidth occurs at ω = 2ω 0 ri ee 02 ngin -0 e 04 ed E .c se ww ur /w o :/C tp ht 7-1

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