Op amp applications cw nonlinear applications

OP amp Applications

1. LINEAR APPLICATIONS OF OP-AMPS

2. SYLLABUS Applications of Op-Amps: Addition, Subtraction, Multiplication, Current to Voltage and Voltage to current conversion operations , Integration, Differentiation, Active Filters, Comparators, Schmitt trigger, Instrumentation Amplifier, f to V converter.

3. LINEAR APPLICATIONS Adder Subtractor Voltage follower Current to voltage converter Voltage to current converter Integrator Differentiator Active filters

4. Non-linear applications Comparators Logarithmic amplifiers Exponential amplifiers Peak detectors Precision rectifiers Waveform generators Clippers & clampers

5. If R f = R, then V out = v 1 +v 2 +v 3

6. Non-inverting Summing Amplifier Therefore, using the superposition theorem , the voltage V 2 = V 1 V b & V c = 0. Net resistance = R+R/2 If R F =2R 1 , 1+R F /R 1 =3 V o = V a +V b +V c    

7. Determine the output voltage =12

8. Averaging Amplifier A summing amplifier can be made to produce the mathematical average of the input voltages. The amplifier has a gain of R f /R, where R is the value of each input resistor. The general expression for the output of an averaging amplifier is V OUT =-(R f /R)(V IN1 +V IN2 +…+ V INn ) Averaging is done by setting the ratio R f /R equal to the reciprocal of the number of inputs (n); that is , R f /R=1/n.

9. Problem Show that the amplifier in produces an output whose magnitude is the mathematical average of the input voltages.

10. ADDER-SUBTRACTOR It is possible to perform addition and subtraction simultaneously with a single op-amp using the circuit shown above. The output voltage V o can be obtained by using superposition theorem. To find output voltage V o 1 due to V 1 alone, make all other input voltages V 2 , V 3 and V 4 equal to zero. The simplified circuit is shown in Fig.b. This is the circuit of an inverting amplifier and its output voltage is,

11. 4.3 (b) Op-amp adder- subtractor, (c) Simplifier circuit for V 2 = V 3 = V 4 = 0, (d) Simplified circuit for V1 = V 2 = V 4 = Similarly V 02 = -V 2 Now, the output voltage V o3 due to the input voltage signal V 3 alone applied at the (+) input terminal can be found by setting V 1 =V 2 =V 4 =0 . The circuit now becomes a non-inverting amplifier as shown in Fig. 4.3 (d). The voltage V a at the non-inverting terminal is Applying Thevinons theorem at inverting terminal, we get =R/2 =1/3 R =V 3

12. So, the output voltage V o3 due to V 3 alone is Similarly, it can be shown that the output voltage V 04: due to V 4 alone is Vo 4 = V 4 (4.14) Thus, the output voltage V o due to all four input voltages is given by

13. Scaling Adder A different weight can be assigned to each input of a summing amplifier by simply adjusting the values of the input resistors. The output voltage can be expressed as V OUT =((R f /R 1 )V IN1 +(R f /R 2 )V IN2 +…+(R f /R n )V INn )

14. Problem Determine the weight of each input voltage for the scaling adder in Figure 7-16 and find the output voltage.

15. Voltage to Current Converter (Transconductance Amplifier) In many applications, one may have to convert a voltage signal to a proportional output current. For this, there are two types of circuits possible. V-I Converter with floating load V-I Converter with grounded load Figure 4.8 (a) shows a voltage to current converter in which load Z L is floating. Since voltage at node 'a' is v i therefore, Fig. 4.8 Voltage to current converter V i i L proportional to V i is controlled by R 1 (b) Grounded load (a) Floating load i L = v i /R V i = i L R 1    

16. That is the input voltage v i is converted into an output current of v i /R 1 , it may be seen that the same current flows through the signal source and load and, therefore, signal source should be capable of providing this load current. Since the op-amp is used in non-inverting mode, the gain of the circuit is 1 + R/R= 2. The output voltage is, v o = 2v 1 = v i + v o – i L R V i = i L R i L = v i /R A voltage-to-current converter with grounded load is shown in Fig. 4.8 (b). Let v i be the voltage at node 'a'. Writing KVL, we get

17. As the input impedance of a non-inverting amplifier is very high, this circuit has the advantage of drawing very little current from the source. A voltage to current converter is used for low voltage dc and ac voltmeter, LED and zenar diode tester.

18. Current to Voltage Converter (Trans-resistance Amplifier) Photocell, photodiode and photovoltaic cell give an output current that is proportional to an incident radiant energy or light. The current through these devices can be converted to voltage by using a current-to-voltage converter and thereby the amount of light or radiant energy incident on the photo-device can be measured. Figure 4.9. shows an op-amp used as I to V converter. Since the (-) input terminal is at virtual ground, current i i flows through the feedback resistor R f . Thus the output voltage v = - i i R f . It may be pointed out that the lowest current that this circuit can measure will depend upon the bias current of the op-amp. This means that for 741 (bias current is 3 nA) can be used to detect lower currents. The resistor R f is sometimes shunted with a capacitor C f to reduce high frequency noise and the possibility of oscillations.

19. Inverting Current to Voltage Converter

20. Non-inverting Current to Voltage Converter.

21. INTEGRATORS AND DIFFERENTIATORS Integrators produce output voltages that are proportional to the running time integral of the input voltages. In a running time integral, the upper limit of integration is t . Passive Integrator Gain is Less than Unity. Attenuation is more. For perfect integration, large values of RC are required. Gain decreases with increase of frequency.    

22. where C is the integration constant and is proportional to the value of the output voltage v o at time t = 0 seconds. For example , if the input is a sine wave, the output will be a cosine wave. If the input is a square wave, the output will be a triangular wave, as shown in Figure 7-23(c) and (b), respectively. Equation-7-23

23. When v in = 0, the integrator of Figure 7-23(a) works as an open-loop amplifier. This is because the capacitor C F acts as an open circuit (X CF = ∞ ) to the input offset voltage V io . In other words, the input offset voltage V io and the part of the input current charging capacitor C F produce the error voltage at the output of the integrator. Therefore, in the practical integrator shown in Figure 7-25, to reduce the error voltage at the output, a resistor R F is connected across the feedback capacitor C F . Thus, R F limits the low-frequency gain and hence minimizes the variations in the output voltage.

24. A practical Integrator

25. The frequency response of the basic integrator is shown in Figure 7-24. In this figure, f b , is the frequency at which the gain is 0 dB and is given by Both the stability and the low-frequency roll-off problems can be corrected by the addition of a resistor R F as shown in the practical integrator of Figure 7-25. The frequency response of the practical integrator is shown in Figure 7-24 by a dashed line. In this figure, f is some relative operating frequency, and for frequencies up to f a the gain R F /R i is constant. However, after f a , the gain decreases at a rate of 20 dB/decade. In other words, between f a , and f b , the circuit of Figure 7-25 acts as an integrator. The gain-limiting frequency f a is given by Frequency Response

26. If the op-amp was ideal, an integrator as shown in above Figure would require just one resistor, R, and one capacitor, C, and the relation between the output and input voltages would be given by f a < f b . If f a = f b /10, then R F = 10R 1 . The input signal will be integrated properly, if time period T of the input signal =f a =f b The Integrating range is in between f a & f b

27. Frequency Response of an Ideal & Lossy Integrator =f a =f b Integrating Range 20db/decade

28. Square waves are integrated to triangles, triangles to parabolas etc.

29. EXAMPLE 7-15 In the circuit of Figure 7-23, R 1 C F = 1 second, and the input is a step (dc) voltage, as shown in Figure 7-26(a). Determine the output voltage and sketch it. Assume that the op-amp is initially nulled.

30. SOLUTION The input function is constant beginning at t = 0 seconds. That is, V m = 2 V for 0 < / =s 4. Therefore, using Equation (7-23), The output voltage waveform is drawn in Figure 7-26(b); the waveform is called a ramp function. The slope of the ramp is -2 V/s. Thus, with a constant voltage applied at the input, the integrator gives a ramp at the output.

31. Determine the rate of change of the output voltage in response to the input square wave, as shown for the ideas integrator in Figure 7-22(a). The output voltage is initially zero. The pulse width is 100μs. Describe the output and draw the waveform.

32. Differentiator Circuit Figure 7-27 Basic differentiator, (a) Circuit, (b) Frequency re­sponse. f c = Unity Gain BW Gain increases with frequency Practical Ideal     f c = Unity Gain BW Figure 7-28 Figure 7-27b

33. Since I B = 0, i c = i F Since v 1 = v 2 = 0 V,

34. f a is the frequency at which the gain is 0db and is given by At f a gain is 0db f C is the unity gain-bandwidth of the op-amp, and f is some relative operating frequency. Both the stability and the high-frequency noise problems can be corrected by the addition of two components: R 1 and C F , as shown in Figure 7-28(a), This circuit is a practical differentiator , the frequency response of which is shown in Figure 7-27(b) by a dashed line. From frequency f to f b , the gain increases at 20 dB/decade. However, after f b the gain decreases at 20 dB/decade. This 40-dB/ decade change in gain is caused by the R 1 C 1 and R F C F combinations. The gain-limiting frequency f b is given by (7-29) where R 1 C 1 = R F C F , help to reduce significantly the effect of high-frequency input, amplifier noise, and offsets. Above all, it makes the circuit more stable by preventing the increase in gain with frequency . Generally, the value of f b and in turn R 1 C 1 and R F C F values should be selected such that f a < f b < f c (7-30)

35. But v 1 = v 2 = 0 V, because A is very large. Therefore, Thus the output v is equal to the R F C times the negative instantaneous rate of change of the input voltage v in with time. Since the differentiator performs the reverse of the integrator's function, a cosine wave input will produce a sine wave output, or a triangular input will produce a square wave output. However, the differentiator of Figure 7-27(a) will not do this because it has some practical problems. The gain of the circuit (R F /X C1 ) increases with increase in frequency at a rate of 20 dB/decade. This makes the circuit unstable. Also, the input impedance X C decreases with increase in frequency, which makes the circuit very susceptible to high-frequency noise. When amplified, this noise can completely override the differentiated output signal. The frequency response of the basic differentiator is shown in Figure 7-27(b). In this figure, f a is the frequency at which the gain is 0 dB and is given by

36. Figure 7-28 Practical differentiator, (a) Circuit, (b) Sine wave input and resulting cosine wave output, (c) Square wave input and resulting spike output.

37. Steps For the Design of Practical Differentiator: Select f a equal to the highest frequency of the input signal to be differentiated. Then, assuming a value of C 1 < 1 μ F, calculate the value of R F . 2. Choose f b = 20 f a and calculate the values of R 1 , and C F so that R 1 C 1 = R F C F . The differentiator is most commonly used in wave shaping circuits to detect high-frequency components in an input signal and also as a rate-of-change detector in FM modulators.

38. EXAMPLE 7-16 Design a differentiator to differentiate an input signal that varies in frequency from 10 Hz to about 1 kHz. If a sine wave of 1 V peak at 1000 Hz is applied to the differentiator of part (a), draw its output waveform. SOLUTION (a) To design a differentiator, we simply follow the steps outlined previ­ously:

39. Problem Determine the output voltage of the ideal op-amp differentiator in Figure 7-26 for the triangular-wave input shown.

40. Comparison Between Integrator & Differentiator. The process of integration involves the accumulation of signal over time and hence sudden changes in the signal are suppressed. Therefore an effective smoothing of signal is achieved and hence, integration can be viewed as low-pass filtering. The process of differentiation involves identification of sudden changes in the input signal . Constant and slowly changing signals are supressed by a differentiator. It can be viewed as high-pass filtering.

41. INSTRUMENTATION AMPLIFIER In a number of industrial and consumer applications, one is required to measure and control physical quantities. Some typical examples are measurement and control of temperature, humidity, light intensity, water flow etc. These physical quantities are usually measured with the help of transducers. The output of transducer has to be amplified so that it can drive the indicator or display system. This function is performed by an instrumentation amplifier. The important features of an instrumentation amplifier are: (i) high gain accuracy (ii) high CMRR (iii) high gain stability with low temperature coefficient (iv) low dc offset (v) low output impedance There are specially designed op-amps such as (µA725 to meet the above stated requirements of a good instrumentation amplifier. Monolithic (single chip) instrumentation amplifier are also available commercially such as AD521, AD524, AD620, AD624 by Analog Devices, LM-363.XX (XX -»10,100,500) by National Semiconductor and INA1O1,1O4, 3626, 3629 by Burr-Brown.

42. Non Linear Applications : Precision rectifiers. The major limitation of ordinary diode is that it cannot rectify voltages below v γ (~ 0.6 V), the cut-in voltage of the diode. A circuit that acts like an ideal diode can be designed by placing a diode in the feedback loop of an op-amp as in Fig. 4.10 (a). Here the cut-in voltage is divided by the open loop gain A 0L (~ 10 4 ) of the op- amp so that v γ is virtually eliminated. When the input V i > V γ /A OL then v oA , the output of the op-amp exceeds V γ and the diode D conducts. Thus the circuit acts like a voltage follower for input v i > V γ /A 0L (i.e., 0.6/10 4 = 60 μ v) and the output v follows the input voltage v i during the positive half cycle as shown in Fig. 4.10 (b).

43. When v i is negative or less than V γ /A 0L , the diode D is off and no current is delivered to the load R L except for small bias current of the op-amp and the reverse saturation current of the diode. This circuit is called the precision diode and is capable of rectifying input signals of the order of mill volt. Some typical applications of a precision diode discussed are: Half-wave Rectifier Full-Wave Rectifier Peak-Value Detector. Clipper. Clamper.

44. Fig. 4.10 (a) Precision diode, (b) Input and output waveforms

45. When V i is +ve, V oa is –ve. D 2 is reverse biased and output is zero. When V i is -ve, R f =R 1 , V oa is +ve, and D 2 conducts even when the input is < 0.7V. The op-amp should be a high speed version as it alternates between open loop & closed loop operations. High Slew rate is required as the input passes through zero, V oa must change from 0.6V to -0.6V. If the diodes are reversed, -ve output occurs.

46. An inverting amplifier can be converted into an ideal half-wave rectifier by adding two diodes as shown in Fig. 4.11 (a). When v i is positive , diode D 1 conducts causing v 0A to go to negative by one diode drop (~ 0.6 V). Hence diode D 2 is reverse biased. The output voltage v is zero , because, for all practical purposes, no current flows through R f and the input current flows through D 1 . For negative input, i.e., v i < 0, diode D 2 conducts and D 1 is off. The negative input v i forces the op-amp output v 0A positive and causes D 2 to conduct. The circuit then acts like an inverter for R f =R 1 and output v becomes positive. The input, output waveforms are shown in Fig. 4.11 (b). The op-amp in the circuit of Fig. 4.11 (a) must be a high speed op-amp since it alternates between open loop and closed loop operations. The principal limitation of this circuit is the slew rate of the op-amp. As the input passes through zero, the op-amp output v oA must change from 0.6 V to - 0.6 V or vice-versa as quickly as possible in order to switch over the conduction from one diode to the other. The circuit of Fig. 4.11(a) provides a positive output. However, if both the diodes are reversed, then only positive input signal is transmitted and gets inverted. The circuit, then provides a negative output.

47. Full-wave Rectifier Fig. 4.12 (a) Precision full wave rectifier, (b) Equivalent circuit for v i > 0; D 1 is on and D 2 is OFF; op-amp A 1 and A 2 operate as inverting amplifier A full wave rectifier or absolute value circuit is shown in Fig. 4.12 (a). For positive input, i.e. v i > 0, diode D 1 is on and D 2 is off. Both the op-amps A 1 and A 2 act as inverter as shown in equivalent circuit in Fig. 4.12 (b). It can be seen that v = v i

48. For negative input, i.e. v i < 0, diode D 1 is off and D 2 is on. The equivalent circuit is shown in Fig. 4.12 (c). Let the output voltage of op-amp A 1 be v. Since the differential input to A 2 is zero, the inverting input terminal is also at voltage v. KCL at node 'a' gives The equivalent circuit of Fig. 4.12 (c) is a non-inverting amplifier as shown in Fig. 4.12 (d). The output v is, (4.31) (4.32) Hence for v i < 0, the output is positive. The input and output waveforms are shown in Fig. 4.12 (e). The circuit is also called an absolute value circuit as output is positive even when input is negative. For example, the absolute value of | +2 | and | -2 | is +2 only. It is possible to obtain negative outputs for either polarity of input simply by reversing the diodes.

49. Fig. 4.12 (c) Equivalent circuit for v, < 0, (d) Equivalent circuit of (c) Fig. 4.12 (e) Input and output waveforms

50. LOG -AMPLIFIER

51. Log Amplifiers The basic log amplifier produces an output voltage as a function of the logarithm of the input voltage; i.e., V out = -K ln(V in ), where K is a constant. Recall that the a diode has an exponential characteristic up to a forward voltage of approximately 0.7 V. Hence, the semiconductor pn junction in the form of a diode or the base emitter junction of a BJT can be used to provide a logarithm characteristic.

52. There are several applications of log and antilog amplifiers. Antilog computation may require functions such as In x, log x or sinh x. These can be performed continuously with log-amps. One would like to have direct dB display on digital voltmeter and spectrum analyzer. Log-amp can easily perform this function. Log-amp can also be used to compress the dynamic range of a signal. Log Amplifier The fundamental log-amp circuit is shown in Fig. 4.34 (a) where a grounded base transistor is placed in the feedback path. Since the collector is held at virtual ground and the base is also grounded, the transistor's voltage-current relationship becomes that of a diode and is given by,

53. Since, I C = I E for a grounded base transistor I S = emitter saturation current = 10 -13 A k = Boltzmann's Constant T = absolute temperature (in K) Since I E = I C Taking natural Log on both sides, we get From Fig.4.18(a)

54. The circuit, however, has one problem. The emitter saturation current I S varies from transistor to transistor and with temperature. Thus a stable reference voltage V ref cannot be obtained. This is eliminated by the circuit given in Fig. 4.18 (b). The input is applied to one log-amp, while a reference voltage is applied to another log-amp. The two transistors are integrated close together in the same silicon wafer. This provides a close match of saturation currents and ensures good thermal tracking.

55. V 1 V 2 Thermistor

56. Assume, Is 1 = Is 2 = I s (4.39) and then, V 1 = 4.41 4.42 4.43 4.44 Thus reference level is now set with a single external voltage source. Its dependence on device and temperature has been removed. The voltage V o is still dependent upon temperature and is directly proportional to T. This is compensated by the last op-amp stage A 4 which provides a non-inverting gain of (1 + R 2 /R TC ). Now, the output voltage is, Where R TC is a PTC thermistor.

57. Vin is converted in to current I c = I EBO e Vin/K

58. Antilog Amplifier The circuit is shown in Fig. 4.19. The input V; for the antilog-amp is fed into the temperature compensating voltage divider R 2 and R TC and then to the base of Q 2 . The output V o of the antilog-amp is fed back to the inverting input of A 1 through the resistor R 1 . The base to emitter voltage of transistors Q l and Q 2 can be written as

59. OP-AMP COMPARATORS

60. Non-Inverting Comparator. (b) Input and Output wave- forms when V ref is +ve (c) Input and Output wave-forms when V ref is -ve

61. Inverting Comparator. b) Input and Output Wave Forms when V ref is +ve and c) Input and Output Wave Forms when V ref is –ve

62. WINDOW COMPERATOR Used in A.C Voltage Stabilizers .

63. Input(Volts) LED3 LED2 LED1 Less than 2V ON OFF OFF Less than 4V & More than 2V OFF ON OFF More than 4V OFF OFF ON If V cc = 6V

64. Fig. 5.5 (a) Zero crossing detector and (b) Input and output waveforms (a) (b)

65. (a) Inverting Schmitt Trigger circuit (b) } (c) and (d) Transfer Characteristics of Schmitt Trigger INVERTING SCHMITT TRIGGER The input voltage v i triggers the output v o every time it exceeds certain voltage levels, V LT & V UT If V ref = 0, then the voltage at the junction of R1 & R2 will form will determine V UT & V LT . If V i < V LT , V o = +V sat V i > V LT , V o = -V sat ZERO if V ref is Zero.

66. (a) Input and Output waveforms of Schmitt Trigger and (b) Output v versus V i plot of the hysteresis voltage. If a sine wave frequency f=1/T is applied, a symmetrical square wave is obtained at the output. The vertical edge is shifted in phase by  from zero crossover Where sin  = V UT /V m and V m is the peak sinusoidal voltage.

67. NON-INVERTING SCHMITT TRIGGER The input is applied to the non-inverting input terminal of the op-amp. To understand the working of the circuit, let us assume that the output is positively saturated i.e. at +V sat . This is fedback to the non-inverting input through R 1 . This is a positive feedback. Now though V in is decreased, the output Continues its positive saturation level unless and until the input becomes more negative than V LT . At lower threshold, the output changes its state from positive saturation + V sat to negative saturation - V sat . It remains in negative saturation till V in increases beyond its upper threshold level V UT . Now V A = voltage at point A =I in R2 = V UT As op-amp input current is zero, I in entirely passes through R 1 .

68. Chattering can be defined as production of multiple output transitions the input signal swings through the threshold region of a comparator. This is because of the noise. Eliminates Comparator Chatter.

69. S.No . Schmitt Trigger. Comparator. 1. The feedback is used. No feedback is used. 2. Op-amp is used in closed loop mode. Used in open loop mode. 3. No false triggering. False Triggering. 4. Two different threshold voltages exists as V UT & V LT Single reference voltage V ref or –V ref . 5. Hysteresis exists. No Hysteresis exists. Comparison.

70. Square & Triangular waveform generation

71. Square Wave Generator

72. V ref = β V sat Let V initially be + V sat. The capacitor charges through R to + β V sat . Then V goes to – V sat . The cycle repeats and output will be a Square Wave. Where β = R 2 /(R 1 +R 2 )

73. V Sat + _ Triangular/rectangular wave generator.

74. Operation of the Circuit Let the output of the Schmitt trigger is + Vsat. This forces current + V sat /R 1 through C 1 , charging C 1 with polarity positive to left and negative to right. This produces negative going ramp at its output, for the time interval t 1 to t 2 . At t 2 when ramp voltage attains a value equal to LTP of Schmitt trigger , the output of Schmitt trigger changes its stage from + V sat to -V sat , Now direction of current through C reverses. It discharges and recharges in opposite direction with polarity positive to right and negative to left. This produces positive going ramp at its output, for the time interval t 2 to t 3 . At t 3 when ramp voltage attains a value equal to UTP of Schmitt trigger , the output of Schmitt trigger changes its state from - V sat to + V sat and cycle continues. The circuit acts as free running waveform generator producing triangular and rectangular output waveforms.

75. V o ’ =+V sat = V in -V ramp + - 0V

76. V o (PP) = 2V ramp V in = V Sat Substitute V (pp) from Eqn. (4)