1. Op-amp used as a summing amplifier or adder It is possible to apply more than one input signal to an inverting amplifier. This circuit will then add all these input signals to produce their addition at the output. Such a circuit will then be called as an adder or a summing amplifier. Depending on the polarity or a sign of the output voltage the adder circuits can be classified into two categories: 1.Inverting adder and 2.Non-inverting adder.

2. Op-amp used as a summing amplifier or adder The summing amplifier is an application of the inverting op- amp configuration. The summing amplifier has two or more inputs and its output voltage is proportional to the negative of the algebraic sum of its input voltage

3. Inverting adder or inverting summing amplifier OP-AMP - + V1V1 RFRF VoVo R1R1 + - V2V2 I B2 = 0 + - - + I B1 = 0 IFIF I1I1 +V CC -V EE A V1V1

4. Inverting adder or inverting summing amplifier OP-AMP - + V1V1 RFRF VoVo R1R1 + - V2V2 I B2 = 0 + - - + I B1 = 0 IFIF I1I1 +V CC -V EE A R2R2 + - - + I2I2 V2V2 V1V1

5. OP-AMP - + V1V1 RFRF VoVo R1R1 + - V2V2 I B2 = 0 + - - + I B1 = 0 IFIF I1I1 +V CC -V EE A R2R2 + - - + I2I2 R3R3 - + I3I3 + - V3V3 V2V2 V1V1 Inverting adder or inverting summing amplifier

6. OP-AMP - + V1V1 RFRF VoVo R1R1 + - V2V2 I B2 = 0 + - - + I B1 = 0 IFIF I1I1 +V CC -V EE A R2R2 + - - + I2I2 R3R3 - + I3I3 + - V3V3 V2V2 V1V1 VoVo = - (V 1 + V 2 + V 3 ) I F = I 1 + I 2 + I 3 Vo = - I F R F Vo = - (I 1 +I 2 + I 3 ) R F = - (V 1 /R 1 +V 2 /R 2 +V 3 /R 3 ) R F If all three resistors are equal R F =R 1 =R 2 =R 3 Inverting adder or inverting summing amplifier

7. FIGURE 13-20 Two-input inverting summing amplifier. Vout = - (V IN1 + V IN2 )

8. FIGURE 13-21 Summing amplifier with n inputs. Vout = - (V IN1 + V IN2 ----------- +V INn )

9. Determine the output voltage in given figure

10. Ans = - 12V

11. Determine the output voltage in given figure

12. Ans = - 7 v

13. Thomas L. Floyd Electronic Devices, 6e and Electronic Devices: Electron Flow Version, 4e Determine the output voltage in given figure

14. Thomas L. Floyd Electronic Devices, 6e and Electronic Devices: Electron Flow Version, 4e Determine the output voltage in given figure Ans = - 2.5 V

15. Thomas L. Floyd Electronic Devices, 6e and Electronic Devices: Electron Flow Version, 4e Determine the output voltage in given figure

16. Thomas L. Floyd Electronic Devices, 6e and Electronic Devices: Electron Flow Version, 4e Determine the output voltage in given figure Ans = - 8.84 V

17. Non-Inverting adder OP-AMP - + V1V1 RFRF VoVo R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb = V a + V b

18. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb Non-Inverting adder

19. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R Non-Inverting adder

20. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R V’ 1 = 2 VaVa Non-Inverting adder

21. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R V’ 1 = 2 VaVa = 1 + RFRF R1R1 V’ 1 Non-Inverting adder

22. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R V’ 1 = 2 VaVa = 1 + RFRF R1R1 V’ 1 RFRF = R 1 = R Non-Inverting adder

23. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R V’ 1 = 2 VaVa = 1 + RFRF R1R1 V’ 1 RFRF = R 1 = R V’ o = 2V’ 1 Non-Inverting adder

24. OP-AMP - + V’ 1 RFRF V’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’ 1 = V a RaRa + R b RbRb RaRa = R b = R V’ 1 = 2 VaVa = 1 + RFRF R1R1 V’ 1 RFRF = R 1 = R V’ o = 2V’ 1 V’ o =VaVa Non-Inverting adder

25. OP-AMP - + V’’ 1 RFRF V’’ o R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb V’’ 1 = V b RaRa + R b RaRa RaRa = R b = R V’’ 1 = 2 VbVb = 1 + RFRF R1R1 V’’ 1 RFRF = R 1 = R V’’ o = 2V’’ 1 V’’ o =VbVb Non-Inverting adder

26. OP-AMP - + V1V1 RFRF VoVo R1R1 V2V2 I B2 = 0 + - - + I B1 = 0 +V CC -V EE RaRa - + RbRb - + VaVa VbVb = V’ o + V’’ o VoVo = V a + V b Non-Inverting adder

27. Difference amplifier and subtractor OP-AMP - + A RFRF VoVo B I B2 = 0 + - I B1 = 0 +V CC -V EE R1R1 - + R1R1 - + V1V1 V2V2 VoVo = V 1 - V 2 RFRF

28. The Op-Amp Integrator An op-amp integrator simulates mathematical integration which is basically a summing process that determines the total area under the curve of a function

29. An ideal Op-Amp is shown in figure: Feedback element is a capacitor that forms an RC circuit with the input resistor. The Op-Amp Integrator

30. How a capacitor charges: To understand how the integrator works, it is important to review how a capacitor charges. Charge Q on a capacitor is proportional to the charging current (Ic) and the time (t). The Op-Amp Integrator

31. How a capacitor charges: To understand how the integrator works, it is important to review how a capacitor charges. Charge Q on a capacitor is proportional to the charging current (Ic) and the time (t). Q = Ic t The Op-Amp Integrator

32. How a capacitor charges: To understand how the integrator works, it is important to review how a capacitor charges. Charge Q on a capacitor is proportional to the charging current (Ic) and the time (t). Q = Ic t Also, in terms of the voltage, the charge on a capacitor is Q = C Vc The Op-Amp Integrator

33. How a capacitor charges: To understand how the integrator works, it is important to review how a capacitor charges. Charge Q on a capacitor is proportional to the charging current (Ic) and the time (t). Q = Ic t Also, in terms of the voltage, the charge on a capacitor is Q = C Vc From these two relationship, the capacitor voltage can be expressed as The Op-Amp Integrator Vc = ( Ic / C ) t

34. This expression has the form of an equation for a straight line that begins at zero with a constant slop of Ic/C. From algebra that the general formula for a straight line is Vc = ( Ic / C ) t Y = mx + b In this case y = Vc, m = Ic/C, x = t, and b = 0 The Op-Amp Integrator

35. Recall that the capacitor voltage in simple RC circuit is not linear but is exponential. This is because the charging current continuously decreases as the capacitor charges and causes the rate of change of the voltage to continuously decrease. The key thing using an op-amp with an RC circuit to form an integrator is that the capacitors charging current is made constant, thus producing a straight line (linear) voltage rather than a exponential voltage Now let’s see why this is true. The Op-Amp Integrator

36. In figure, the inverting input of the op-amp is at virtual ground (0V), so the voltage across R 1 equals V in. Therefore, the input current is I in = V in / R 1 If V in is a constant voltage, then I in is also a constant because the inverting input is always remains at 0 v, keeping a constant voltage across R 1. Because of the very high input impedance of the op-amp, there is negligible current at the inverting input. This makes all of the input current go through the capacitor, as indicated in figure, so I c = I in The Op-Amp Integrator

37. The capacitor voltage Since I in is a constant, so is I c. The constant I c charges the capacitor linearly and produces a linear voltage across C. The positive side of the capacitor is held at 0v by virtual ground of the op-amp. The voltage on the negative side of the capacitor, which is the op-amp output voltage, decreases linearly from zero as the capacitor charges, as shown in figure. This voltage is called a negative ramp and is the consequence of a constant positive input. A linear ramp voltage is produced across C by the constant charging current. The Op-Amp Integrator

38. The output voltage V out is the same as the voltage on the negative side of the capacitor. When a constant positive input voltage in the form of a step a pulse ( pulse has a constant amplitude when high) is applied, the output ramp decreases negatively until the op-amp saturates at its maximum negative level. This is indicated in figure. A constant input voltage produces a ramp on the output of the integrator. The Op-Amp Integrator

39. Rate of change of the output voltage The rate at which the capacitor charges, and therefore the slope of the output ramp, is set by the ratio I C /C, as you have seen Ic = V in /R 1, the rate of change or slope of the integrator’s output voltage is ΔV out /Δt. A constant input voltage produces a ramp on the output of the integrator. ΔV out ΔtΔt V in R1R1 = The Op-Amp Integrator

40. The Op-Amp Differentiator An ideal differentiator is shown in figure. The capacitor is now input element. A differentiator produces an output that is proportional to the rate of change of input voltage

41. Apply a positive-going ramp voltage to the as indicated in figure In this case I C =I in and voltage across the capacitor is equal to V in at all times (V C =V in ) because of virtual ground on the inverting input. The Op-Amp Differentiator

42. From the basic formula, Vc = ( Ic / C ) t The Op-Amp Differentiator

43. From the basic formula, Vc = ( Ic / C ) t The capacitor current is The Op-Amp Differentiator Ic = ( Vc / t ) C

44. The capacitor current is Ic = ( Vc / t ) C Since the current at the inverting input is negligible, I R = I C Both currents are constant because the slop of the capacitor voltage (V C /t) is constant. The output voltage is also constant and equal to the voltage across R f because one side of the feedback resistor is always 0V (virtual ground) The Op-Amp Differentiator

45. V out = I R R f Since the current at the inverting input is negligible, I R = I C Both currents are constant because the slop of the capacitor voltage (V C /t) is constant. The output voltage is also constant and equal to the voltage across R f because one side of the feedback resistor is always 0V (virtual ground) The Op-Amp Differentiator

46. V out = I R R f = I C R f Since the current at the inverting input is negligible, I R = I C Both currents are constant because the slop of the capacitor voltage (V C /t) is constant. The output voltage is also constant and equal to the voltage across R f because one side of the feedback resistor is always 0V (virtual ground) The Op-Amp Differentiator

47. V out = I R R f = I C R f Since the current at the inverting input is negligible, I R = I C Both currents are constant because the slop of the capacitor voltage (V C /t) is constant. The output voltage is also constant and equal to the voltage across R f because one side of the feedback resistor is always 0V (virtual ground) V out = ( ) CR f VcVc t The Op-Amp Differentiator

48. The output is negative when input is a positive going ramp. And positive when input is a negative going ramp as shown in figure. V out = ( ) CR f VcVc t The Op-Amp Differentiator

49. During the positive slope of the input, the capacitor is charging from the input source and the constant current through the feedback resistor is the direction shown. V out = ( ) CR f VcVc t The Op-Amp Differentiator

50. During the positive slope of the input, the capacitor is charging from the input source and the constant current through the feedback resistor is the direction shown. During the negative slope of the input, the current is in the opposite direction because the capacitor is discharging. V out = ( ) CR f VcVc t The Op-Amp Differentiator

51. V out = ( ) CR f VcVc t Frthe slope increases, V out increases. If the slope decreases, V out decreases. So the output voltage is proportional to the slope ( rate of change) of the input. The constant of proportionality is the time constant R f Com the above equation the term V C / t is the slope of input. The Op-Amp Differentiator

52. Voltage to current converters: ( V to I) The voltage to current converters can be classified in to two categories, depending on the position of the load. They are: 1. V to I converter with floating load. 2. V to I converter with grounding load

53. Figure shows a voltage to current converter with floating load. This load ( RL) is called as floating load because it is not connected to ground. + - OP-AMP - + - R1R1 Io +V CC -V EE RLRL Io V in VdVd VFVF V to I converter with floating load with inverting amplifier

54. The input voltage is applied to the non-inverting terminal of the op-amp. Load resistance RL is connected in place of feedback resistor RF. + - OP-AMP - + - V in R1R1 Io +V CC -V EE RLRL Io VdVd VFVF V to I converter with floating load with inverting amplifier

55. Apply KVL to the input loop V in = V d + V F + - OP-AMP - + - V in R1R1 Io +V CC -V EE RLRL Io VdVd VFVF V to I converter with floating load with inverting amplifier

56. Apply KVL to the input loop V in = V d + V F But the open loop gain Av of this op-amp is very large Vd=0 Therefore V in = V F + - OP-AMP - + - V in R1R1 Io +V CC -V EE RLRL Io VdVd VFVF V to I converter with floating load with inverting amplifier

57. But V in = R 1 X I o Therefore I o = V in / R 1 Equation indicates that the input voltage Vin is converted into a proportional output current I o = V in /R 1 + - OP-AMP - + - V in R1R1 Io +V CC -V EE RLRL Io VdVd VFVF V to I converter with floating load with inverting amplifier

58. V to I converter with floating load with Non-inverting amplifier The V-I converter with input applied to the inverting terminal is shown in figure. The type of load is still floating type. Point B in figure is at virtual ground and the current flowing into the inverting terminal is zero OP-AMP - + - V in R1R1 Vo +V CC -V EE B VFVF I in - RLRL IoIo + A

59. Therefore OP-AMP - + - V in R1R1 Vo +V CC -V EE B VFVF I in - RFRF IoIo + A = I o V to I converter with floating load with Non-inverting amplifier

60. Therefore But OP-AMP - + - V in R1R1 Vo +V CC -V EE B VFVF I in - RFRF IoIo + A = I o I in = V in / R 1 V to I converter with floating load with Non-inverting amplifier

61. Therefore But Therefore OP-AMP - + - V in R1R1 Vo +V CC -V EE B VFVF I in - RFRF IoIo + A = I o I in = V in / R 1 IoIo = V in / R 1 V to I converter with floating load with Non-inverting amplifier

62. Therefore Thus the current flowing through the load is proportional to the input voltage Vin. OP-AMP - + - V in R1R1 Vo +V CC -V EE B VFVF I in - RFRF IoIo + A IoIo = V in / R 1 V to I converter with floating load with Non-inverting amplifier

63. V to I converter with grounded load OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R1R1 V1V1 I1I1 + RFRF I2I2 - RLRL I B = 0 -+ ILIL

64. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 V to I converter with grounded load

65. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 But I 1 V in - V 1 = R V to I converter with grounded load

66. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 But I 1 V in - V 1 = R and I 2 VoVo - V 1 = R V to I converter with grounded load

67. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 But I 1 V in - V 1 = R and I 2 VoVo - V 1 = R Substitute in above equation. ILIL V in - V 1 = R VoVo R + V to I converter with grounded load

68. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 But I 1 V in - V 1 = R and I 2 VoVo - V 1 = R Substitute in above equation. ILIL V in - V 1 = R VoVo R + ILIL V in + = VoVo - 2V 1 R V to I converter with grounded load

69. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Applying KCL at V1 ILIL = I 1 + I 2 But I 1 V in - V 1 = R and I 2 VoVo - V 1 = R Substitute in above equation. ILIL V in - V 1 = R VoVo R + ILIL V in + = VoVo - 2V 1 R V1V1 = V in +VoVo - I L R 2 V to I converter with grounded load

70. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Op-amp in figure is connected In non-inverting mode A VF 1 + R = R = 2 A VF = VoVo V1V1 VoVo = 2V 1 V1V1 = V in +VoVo - I L R 2 V1V1 = V in +VoVo - I L R 2 VoVo = V in +VoVo - I L R V in = I L R V to I converter with grounded load

71. OP-AMP - + - R Vo = 2V 1 +V CC -V EE B V2V2 R A V in R V1V1 I1I1 + R I2I2 - RLRL I B = 0 -+ ILIL Op-amp in figure is connected In non-inverting mode V in = I L R I L = V in / R V to I converter with grounded load

72. I to V converter Non-inverting amplifier OP-AMP - + - Vo +V CC -V EE V2V2 - RFRF I in + V1 VoVo I in = -R F I in I B = 0

73. OP-AMP - + - Vo +V CC -V EE V2V2 - RFRF I in + V1 VoVo I in = -R F I in I B = 0 Above equation indicates that the output voltage is proportional to the input current Iin. Thus the circuit converts the input current proportional to voltage I to V converter Non-inverting amplifier