Analog Circuits Chapter 9: Op-Amp Applications Overview

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Explore the applications of operational amplifiers in electronic circuits and systems, delving into the characteristics, analysis, and design considerations of op-amp circuits. Understand the ideal parameters and limitations of op-amps for effective circuit design.

  • Analog Circuits
  • Op-Amp Applications
  • Electronic Circuits
  • Operational Amplifiers
  • Circuit Design
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  1. Teaching Materials of Analog Circuits Chap 9 Weiwei Xia College of Physics Science and Technology Yangzhou University Yangzhou, 225002 Email: wwxia@yzu.edu.cn

  2. Chapter 8 OP-AMP Applications Signal Processing and Operation Circuit 8.0 Preview Op-amps are used extensively in the design of electronic circuits and systems because of their relat- ively low cost and versatility, and because integrated circuit op-amp characteristics approach the ideal. We will consider, in this chapter, a few of the many app- lications of op-amps.

  3. 8.0 Preview We developed the ideal set of op-amp parameters and then consider the analysis and design of a wide variety of op-amp circuits, which will aid in our un- derstanding of the design process of electronic circuits. We general assume, in this chapter, that the op-amp is ideal. But in practical op-amp circuits, these ideal parameter values are not realized. Ideal op-amp parameters 1. Aod ; 2. KCMR ; 3. BW 4. Rid ; 5. Rod 0;

  4. 8.0 Preview 6. VIO 0 , IIO 0 , dVIO/dT 0, dIIO /dT 0. Ideal op-amp properties 1. vN vP (8-1) vo Aod vP vN= 0 vP vN vo=Aod(vP vN) 2. iP= iN 0 (8-2) Rid iP= iN 0 Op-amp Transfer Characteristics

  5. 8.0 Preview Since the op-amp is composed of tran- sistors biased in the active region by the dc input voltages VCCand VEE, The output voltage is limited. When voapproaches VCC, it will saturate, or be limited to a value nearly equal to VCC , since it cannot go above the positive bias voltage. Similarly, when the output voltage approaches VEE, it will saturate at a value nearly equal to VEE. vo VOmax Aod vP vN VOmin Figure 8.1 Figure 8.1 is a simplified voltage transfer characteristics for the op-amp, showing this saturation effect.

  6. 8.1 Signal operation Circuits Proportional Amplifier 1. Inverting op-amp circuit if iP= iN 0 i1= if Rf i1 vS vN vI i1= R1 R1 vN vO vO if = Rf Rf vO Rf Avf = = vI R1 iN vI vN R1 A iP vN vP 0 + vO vP + RL R Figure 8.2 (8-3) R= R1 Rf Rf R1 vO = vI Rif R1 Rof 0

  7. 8.1 Signal operation Circuits 2. Noninverting op-amp circuit iP= iN 0 vN vP vI Rf R1 vO Rf vI R1 R1 iN vN= Avf= =1+ vO vN R1+Rf A + vO R iP + RL Rf R1 vI vP (8-4) vO = (1+ )vI R= R1 Rf Figure 8.3 Rif Rof 0 A + vO vI + RL vO = vI Voltage Follower Avf 1 Figure 8.4

  8. 8.1 Signal operation Circuits Summing Amplifier 1. Inverting Summing Amplifier iP= iN 0 vN vP 0 i1 R1 if Rf vI1 i1+i2=if i2 R2 iN vI1 vN vI2 vN vN vO + = R1 R2 Rf vI2 vN A + vO iP vP + RL vI1 vI2 vO + = R1 R2 Rf R Figure 8.5 Rf Rf R1 R2 vO = ( vI1+ vI2 ) (8-5) R= R1 R2 Rf

  9. 8.1 Signal operation Circuits 2. Noninverting Summing Amplifier R3 Rf vN iP= iN 0 vN vP R3 vN= R3+Rf R2 R1 vP= R1+R2 R1+R2 Rf vO = (1+ )vN R3 Rf R2 R1 = (1+ )( vI1+ vI2 ) R3 R1+R2 R1+R2 iN vO A + vO iP R1 vI1 + vI1+ vI2 RL vP vI2 R2 (8-6) Figure 8.6 Rf =R3 , R1 = R2 vO= vI1+vI2

  10. 8.1 Signal operation Circuits Subtraction Circuit 1. Differential Amplifier iP= iN 0 vN vP R3 R3 R1 vI1+ vO R1+R3 R1+R3 iN R1 vN= vI1 vN A iP R2 + vO vP vI2 + R4 RL vP= vI2 R4 R2+R4 R1 = R2 R3 =R4 Figure 8.7 R3 R1 vO = (vI2 vI1) (8-7) Rif = 2R1

  11. 8.1 Signal operation Circuits 2. Subtraction Circuit Rf2 R2 vO = (vI2+vO1) R2 vI2 Rf1 R2 Rf2 Rf1 R1 vO1 = vI1 R1 vI1 vO Rf1 = R1 A1 A2 vO1 + + Rf2 R2 R vO = (vI2 vI1) R Rf2 = R2 Figure 8.8 vO = vI1 vI2 (8-8)

  12. 8.1 Signal operation Circuits Rf1 Rf2 Rf2 R2 R1 vO = (vI2 vO1) R2 vO A1 vO1 A2 vI1 Rf1 R1 + R2 vO1 =(1+ )vI1 + vI2 Rf2 Rf1 = R1 Figure 8.9 Rf2 R2 vO = (vI2 2vI1) (8-9) Rif

  13. 8.1 Signal operation Circuits Example 8.1: Yang P289-290 Problem 9-6 v1 R3 R4 + vO1 R4 R3 vO = (vO1 vO2) A1 v1 v2 R1 vO1 vO2= (R1+2R2) R2 A3 R1 vO + R2 R3 R4 2R2 R3 A2 vO= (1+ ) (v1 v2) vO2 v2 R1 + R4 (8-10) Figure 8.10

  14. 8.1 Signal operation Circuits Integrating and Differential Circuit 1. Integrating Circuit vI VS i2 C i1 iN O vO vN vO vI R A O + vP iP VS Figure 8.11 VOm iP= iN 0 vN vP 0 1 1 1 vO= vC = i2 dt = i1 dt = vI dt C C RC t 0 t 0 t 0 (8-11)

  15. 8.1 Signal operation Circuits 2. Differential Circuit vI VS i2 R i1 iN C O t vO vN vO vI A O + vP t iP Figure 8.12 iP=iN 0 vN vP 0 vO= i2 R = i1 R dvI i1=Cdt dvI dt vO= RC (8-12)

  16. 8.1 Signal operation Circuits Logarithmic and Antilogarithmic Converter 1. Logarithmic Converter iP=iN 0 vN vP 0 T iC vO= vBE VT ln(iC/IS) i1 iN iC = i1= vI /R vN R vI A iP vO vP + vI RIS vO= VT ln( ) (8-13) Figure 8.13 vO 0.7V

  17. 8.1 Signal operation Circuits iC2 T1 T2 vO=(1+R3/R4)vB2 vB2 = vBE2 vBE1 VT ln(iC2/iC1) = 2.3VT lg(iC1/iC2) VCC iC1 R2 R5 i1 vB2 + A2 vI R1 A1 + + vO R3 R4 RL iC1 vI /R1 iC2 VCC/R2 Figure 8.14 R3 R4 R2 R3 R1VCC R4 vO= 2.3VT (1+ )lg( )vI = 2.3VT (1+ )lgKvI (8-14) VT T vO R4

  18. 8.1 Signal operation Circuits 2. Antilogarithmic Converter iF R vN vP 0 iE T vI vI /VT iF iE iC=IS e vN A vP vO + vI/VT v O= iFR= RIS e (8-15) Figure 8.15 Multiplier and Divider Example 8.2: Yang P264 Figure 9-21

  19. 8.1 Signal operation Circuits T1 Figure 8.16 vI1 R A1 R1 R vO1 + Rf RP1 A3 T3 vO3 T2 vO A4 + + R/3 RP vI2 R A2 R2 vO2 + RP2 1 vO= vI1vI2 RIS (8-16)

  20. 8.2 Nonideal Effects in Op-Amp Circuits In this section, Some of the practical op-amp par- amerers will be considered in detail. We will discuss and analyze the effect of these nonideal parameters in op-amp circuits. Finite Aodand Rid iF iI=iN+iF Rf R1 vO vI vN R1 vN vN vO = + Rid+R vI + N Rf iN iI Rid vId AodvId vO P + vN vId= Aod Rid>>R R Figure 8.17

  21. 8.2 Nonideal Effects in Op-Amp Circuits vI vN R1 vN vN vO = + Rid+R RN=R1 Rf (Rid+R ) Rf Rf R1Rf+AodRN Rf vO = vI R1 AodRN vO = vI (8-17) vO vN vId= Aod vO vO vO vO vO = 100% = 100% Rf = 100% (8-18) Rf+AodRN Aod Rid

  22. 8.2 Nonideal Effects in Op-Amp Circuits Finite KCMRand Aod Rf R1 vN = vOR1+Rf vP=vI R1 vO + R1 vN vId=vP vN=vI vOR1+Rf vP+vN vIc= vI 2 AodvId + AocvIc R vI vP vO=AodvId+AocvIc Figure 8.18 1 1+ Rf R1 KCMR (R1+Rf )/R1 1+ Aod vO = (1+ ) (8-19) vI

  23. 8.2 Nonideal Effects in Op-Amp Circuits 1 1+ KCMR (8-20) = 1 100% (R1+Rf )/R1 Aod 1+ KCMR Aod IIO 0 and VIO 0

  24. 8.2 Nonideal Effects in Op-Amp Circuits (IIB+IIO /2) (R1 Rf ) Rf R1 VIO VN VIO VN IIB IIO /2 R1 Rf A + R1 VO VOR1+Rf VO VP IIO /2 VP IIB R2 (IIB IIO /2) R2 R2 Figure 8.19

  25. 8.2 Nonideal Effects in Op-Amp Circuits VP= (IIB IIO /2)R2 R1 IIO (IIB+ )(R1 Rf) VIO R1+Rf 2 VN=VO VP=VN Rf 1 VO= (1+ ) VIO+IIB(R1 Rf R2) + IIO(R1 Rf +R2) (8-21) R1 2 R2=R1 Rf Rf VO= (1+ ) (VIO+IIOR2) (8-22) R1

  26. 8.3 Precision Rectifier Precision Half-Wave Rectifier vI R2 D2 t 0 D1 vO vI vO R1 A vO t 0 + RL Figure 8.21 Figure 8.20 vI = 0 vO = 0 D1 D2 vO = 0 vI 0 vO 0 D1 D2 vO = 0 vI 0 vO 0 D1 D2 vO = (R2/R1) vI

  27. 8.3 Precision Rectifier R2 vI D2 t 0 D1 vO vI vO R1 t 0 A vO + RL Figure 8.23 Figure 8.22 vI = 0 vO = 0 D1 D2 vO = 0 vI 0 vO 0 D2 D1 vO = (R2/R1) vI vI 0 vO 0 D2 D1 vO = 0

  28. 8.3 Precision Rectifier Precision Full-Wave Rectifier 2R R 2R D2 R vI D1 R R A1 vO vO1 A2 + + R1 R2 Figure 8.24

  29. 8.3 Precision Rectifier A1 Precision Half-Wave Rectifier ; A2 Inverting Summing Amplifier vI 0 t vO1 0 vO Figure 8.25 t 0 t vI 0 vO1= vi vO = (2vO1+ vI) = ( 2vI+ vI) = vI vI 0 vO1=0 vO = (2vO1+ vI) = vI

  30. 8.4 Active Filters An important application of op-amp is the active filter. The word filter refers to the process of remov- ing undesired portions of frequency spectrum. The word active implies the use of one or more active devices, usually an operational amplifier, in the filter circuit. Two advantages of active filters over passive filters are: 1. The maximum gain or the maximum value of the transfer function may be greater than unity.

  31. 8.4 Active Filters 2. The loading effect is minimal, which means that the output response of the filter is essentially independent of the load driven by the filter. Fundamental Concept Vo (s) Vi (s) A(s)= (8-23) vI(t) vO(t) Filter A(j )= A(j ) ej ( ) (8-24) Figure 8.26 d ( ) d ( )= (s) (8-25)

  32. 8.4 Active Filters Classes Low Pass Filter Figure 8.27(a) BW = H High Pass Filter Figure 8.27(b) BW Band Pass Filter Figure 8.27(c) BW = H L Band Reject Filter Figure 8.27(d) BW : > H ; < L All Pass Filter Figure 8.27(e) BW A A A0 A0 Stopband Passband Passband Stopband 0 0 H L (a) (b)

  33. 8.4 Active Filters A A Passband A0 A0 Stopband Stopband Stopband Passband Passband 0 0 L 0 H L 0 H (c) (d) A A0 Passband (e) Figure 8.27 0

  34. 8.4 .1 Low Pass Filter One-Pole Low Pass Filter Rf Rf R1 vo vo A A VP(s) VP(s) + + vi vi R R C C (b) (a) Figure 8.28 Vo(s) Rf VP(s) A (s) = = (1+ ) Vi(s) R1 Vi(s) Rf 1 R1 1+sRC = (1+ ) (8-26)

  35. 8.4 .1 Low Pass Filter 1+Rf/R1 A0 1+ j RC A(j ) = = (8-27) 1+ j / 0 A(j ) A0 20lg /dB Rf R1 1 Where A0=1+ , 0 =RC ideal 0 A0 3 A(j ) = 1+ ( / 0)2 20dB/decade actual ( )= arctan / 0 20 1 10 / 0 Figure 8.29 Two-Pole Low Pass Filter

  36. 8.4 .1 Low Pass Filter Rf Vo(s) = A0VP(s) R1 VA(s) VP(s) = 1+sRC A vi VP(s) vA vo + R R C C Vi(s) VA(s) VA(s) Vo(s) VA(s) VP(s) = + R 1/sC R Figure 8.30 Vo(s) A0 A (s) = = Vi(s) 1+(3 A0)sRC +(sRC)2 1 0 =RC (8-28) 1 Q=3 A0

  37. 8.4 .1 Low Pass Filter Vo(s) A0 02 A (s) = = (8-29) Vi(s) s2+ s + 02 Q 0 A(j ) 1 (8-30) 20lg = 20lg A0 1 ( )2 2+( ) 2 0 0 Q /( 0 Q) (8-31) ( )= arctg 1 ( )2 0 =0 A(j ) = A0 Low Pass Filter A(j ) 0

  38. 8.4 .1 Low Pass Filter A(j ) A0 20lg /dB Q = 0.707 Q =10 20 / 0=1 5 10 2 20lg A(j ) / A0 = 3dB 0 1 3 10 0.707 / 0=10 0.5 20 40dB/decade 20lg A(j ) / A0 = 40dB 30 40 1 10 / 0 Figure 8.31

  39. 8.4 .1 Low Pass Filter A(j ) A0 A Butterworth filter is a maximally flat magnitude filter. The transfer function is designed such that the magnitude of the transfer function is as fiat as possible 20lg /dB 0 3 10 20 n =6 n =4 n =8 30 n = 2 n =10 40 1 10 / 0 Figure 8.32 with the passband of the filter.

  40. 8.4 .2 High Pass Filter One-Pole High Pass Filter Vo(s) Rf VP(s) A (s)= =(1+ ) Vi(s) R1 Vi(s) Rf R1 Rf sRC =(1+ ) R1 1+sRC s vo A VP(s) = A0s + 0 (8-32) + vi C R j j / 0 =A0 j + 0 1+j / 0 A(j )=A0 Figure 8.33 A(j ) / 0 20lg = High Pass Filter A0 1+ ( / 0)2 1 L= 0 = RC

  41. 8.4 .2 High Pass Filter Two-Pole High Pass Filter Rf A0 A (s) = R1 1 1 2 sRC 1+(3 A0) + sRC A vi C VP(s) vA vo + (8-33) C 1 1 R R 0 =RC Q=3 A0 Figure 8.34 A0 s2 0 Q A0 2 A (s) = (8-34) s2+ s + 02 A (j ) = 02 2+ j 0 Q

  42. 8.4 .2 High Pass Filter A(j ) A(j ) A0 20lg /dB 20lg A0 Q =10 1 5 = 20lg 2 0 ( )2 1 2+( )2 0 Q 0 3 1 0.707 (8-35) 0.5 1 f3dB =2 RC 40 (8-36) 0.1 1 5 / 0 Figure 8.35

  43. 8.4 .3 Band Pass Filter Vi(s) Vo1(s) Vo(s) (a) LPF HPF Rf A1 R1 (b) O A Vo(s) H + A2 R Vi(s) C 2R (c) C O L low-pass high-pass A R (d) (e) Figure 8.36 O L H

  44. 8.4 .3 Band Pass Filter AvfsRC Rf R1 (8-37) A(s)= 1+(3 Avf)sRC+(sRC)2 Avf=1+ Avf A0 =3 Avf s A0Q 0 A (s) = (8-34) 1 + + Q 0 s 1 1 s ( )2 0 0 =RC Q=3 Avf 1 j A0Q 0 A0 0 ( ) 0 A (j ) = = (8-38) 1+ 1+jQ jQ 0 0 ( )2

  45. 8.4 .3 Band Pass Filter = 0 A(j ) A0 Avf 20lg /dB A0 =3 Avf A(j 0) = 0 0.5 Q BW 1 2 20 0 ( ) 0 =1 Q 5 Q =10 40 A0 2 0.1 1 / 0 A(j ) = Figure 8.37 0 f0 2 Q Q BW= = (8-39)

  46. 8.4 .3 Band Pass Filter Example 8.2: For the circuit shown in Figure 8.36 (e) To assure f0=1kHz, BW=100Hz. Determine values of each resistance and capacitor. 1 Solution: C=0.01 F R= 15.9k 2 f0 f0 BW= Q=10 Q Avf=2.9 1 Q=3 Avf Rf=1.9R1 Rf R1 Avf=1+ Rf//R1=2R R1 48.6k Rf 92.3k

  47. 8.4 .4 Band Reject Filter Frequency Response of Double-T Network 2R(1+sRC) 1+(sRC)2 Z1= Z1 R R + + + + C C R/2 R/2 vi Z2 Z3 vf vi vf R/2 2C 2C 2C Figure 8.38 1 1 (R+ ) 2 Vf(s) Z3 sC F (s) = = = (8-40) Vi(s) Z1+Z3 2R(1+sRC) 1+(sRC)2 1 1 (R+ ) 2 + sC

  48. 8.4 .4 Band Reject Filter 1 ( RC)2 1 ( / 0)2 F (j ) = = (8-41) 1 ( RC)2 +4 j RC 1 ( / 0)2+4 j / 0 1 0= 90o RC F(j ) 1 0o 1 / 0 0 1 / 0 90o Figure 8.39

  49. 8.4 .4 Band Reject Filter Band Reject Filter of Double-T Network s Avf (1+ ) 2 0 (8-42) A (s) = s s 1+ +( )2 Q 0 0 Rf R1 A C 0 =1/RC C + Vo(s) Vi(s) 1 R R Q = 2(2 Avf) R/2 2C Rf R1 Avf=1+ Figure 8.40

  50. 8.4 .5 All Pass Filter R R Vi(s) Vi(s) R R VN(s) VP(s) A R A C Vo(s) + Vo(s) + C R (b) Figure 8.41 (a) For Figure 8.42(b): R R R R Vo(s)=(1+ )VP(s) Vi(s) Vo(s) Vi(s) 1 sRC A(s)= = 1+sRC 1/sC R+1/sC VP(s)= Vi(s) (8-43)

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