1. 1 ECE 222 Electric Circuit Analysis II Chapter 8 Operational Amplifiers Op Amps Herbert G. Mayer, PSU Status 5/11/2016 For use at CCUT Spring 2016

2. 2 Syllabus Definition, History, Features Terminal Currents & Voltages Ideal Op Amp Noninverting Op Amp Inverting Op Amp Difference Op Amp Summing Op Amp Integrating Op Amp References

3. 3 Definition High Level Operational Amplifiers, AKA Op Amps, amplify voltages! Are nonlinear electronic devices that amplify the difference of their 2 input signals v p and v n, shown at output terminal v o Sources +V CC and -V CC must satisfy: +V CC > -V CC, referred to as rails Rails are frequently omitted from circuit drawings to reduce clutter; assumed to be there Though Op Amps are non-linear devices, like transistors, they are almost always used in their narrow, linear subrange We can model them abstractly as a simple circuit element via dependent voltage source

4. 4 Note See separate presentation about cascaded Op Amps: l_11_op_amp_cascade

5. 5 Op Amp Definition High Level +V cc -V cc vpvp vnvn + - vovo =>

6. 6 Op Amp Definition Operational Amplifiers, AKA Op Amps, are solid state, low-cost, integrated circuits performing electric transformations of two (or one of two) input signals: one inverting, the other noninverting Transformations include Amplification Inversion Summation Difference Differentiation Integration Comparison A/D conversion Characteristic is the high-gain amplification of 10 3 to 10 5 within a narrow band of input voltages v p - v n Output is named v o

7. 7 Op Amp Definition Amplifies the difference of its two input voltage signals v p and v n The amplification factor A, known as gain, can be very high; the spectrum of voltages amplified is very narrow With feedback loop, A can become quite low, at times deliberately A = 1 Possible to hold one of the inputs down to 0 V, in which case solely the other input signal contributes to the amplified output signal v o

8. 8 Op Amp Definition Op Amps use external power supply, a positive +V CC and a second negative -V CC voltage pin Not necessarily of same voltage, but usually absolute values are equal: |+V CC | = |-V CC | Common power voltages are + - 5 V to + - 15 V, rarely over 20 V Op Amps generally are used in their linear range of amplification; part of the curve with inclination A Due to the high gain A, Op Amps operate only in a very narrow band of signal amplitudes, else saturate! Without feedback loop (below) they saturate even more rapidly Thus without feedback the voltage difference v p - v n must be very narrow for Op Amp to operate linearly

9. 9 Saturated Op Amp From Wikipedia: sin() input signal saturates Op Amp, so v o becomes rectangular; only brief vertical lines show gain A at very low input voltage (difference) Quick quiz: Is this Op Amp amplifying its sin() input signal v s at the inverting or noninverting input pin?

10. 10 Op Amp History Built during ww2 with vacuum tubes With 100-300 V power supplies then Solid state Op Amps since 1960s Early solid state Op Amps: μA702 Bob Widlar 1964, retired at age 30 after invention! 9 transistors only in first solid state Op Amps! μA741 Dave Fullagar 1968, Fairchild Most popular Op Amp of all time Packaged in 8-pin DIP OP-07 Precision Monolithics, 1975 Special-purpose, high gain

11. 11 Op Amp History, Early Contributors Harold S. Black, US Western Electric, 1930s Paul Voigt, UK, 1920s Alan Blumlein, UK, 1930s Hendrick Bode, US 1930s

12. 12 Op Amp Nomenclature +V in noninverting input,we’ll use: v p -V in inverting input, we’ll use: v n +V S positive source, we’ll use: +V CC -V S negative source, we’ll use: -V CC +V out output voltage, we’ll use: v o vovo +V cc -V cc vpvp vnvn + - V out +V s -V s +V in -V in + -

13. 13 More Likely You Just See: vovo vpvp vnvn + - V out +V in -V in + -

14. 14 Op Amp Circuit Symbol

15. 15 Op Amp Circuit Symbol Above circuit symbol shows enlarged shape of typical, contemporary Op Amp Packaged in a DIP with 2 * 4 = 8 pins; DIP stands for dual in-line package Two of the pins (offset null) are rarely used: they provide a method to scale the Op Amp function to overcome shifts caused by material aging One pin is unused; AKA no connection Leaves 5 pins for noninverting input +, inverting input -, output, and + - power supply

16. 16 Actual Op Amp 741 Circuit Detail

17. 17 Actual Op Amp 741 Circuit Detail Circuit detail of typical μA741 with 20 transistors is way more complex than its ancestor of 9 transistors Shown here purely for entertainment purposes We’ll treat an Op Amp as a black-box, with defined functions And that black box Op Amp will be idealized, e.g. assumed to be operating in linear region, i n = i p = 0, and v p = v n Those are the key equations for ideal Op Amp! Yet ideal Op Amp operates largely like a real Op Amp within a narrow linear input range!

18. 18 Op Amp Voltage Transfer Function

19. 19 Terminal Current Names

20. 20 Terminal Voltage Names

21. 21 Ideal vs. Real Op Amps

22. 22 Ideal Op Amp Without feedback, the straight Op Amp generally saturates quickly, except for a very narrow band of input voltages v n and v p specifically: v p – v n To reduce saturation, Op Amps generally feed some output signal from v o back to input signal v n This negative feedback voltage subtracts from the actual input signal v n Thus decreases gain and input voltage difference And also decreases output voltage v o Renders difference of v p and v n extremely small; we simplify by saying: v p = v n for ideal Op Amp!!! Input resistance idealized to ∞ Ω, practically >> 1 MΩ Idealized i p = i n = 0 in our modeling of Op Amps

23. 23 Ideal Op Amp Simplifications v p = v n -- known as virtual Op Amp short! i p = i n = 0 A-- ∞ input Ω causes no current A= ∞-- infinite gain, real A is ~10 5 max i p + i n + i o + i C+ + i C- = 0 -- Kirchhoff C. Law From i p = i n = 0 it follows: i o = - ( i C+ + i C- )

24. 24 Summary Ideal vs. Real Op Amp Electric Parameters of Real Op Amp Value Large gain A > 10 5 Small input currents i p, i n Close to 0 A Small difference of input voltages: v p - v n Close to 0 V Steep incline of function v o = f( v p - v n )High in linear range

25. 25 Summary Ideal vs. Real Op Amp Electric Parameters of ideal Op Amp Value Infinite gain A ∞ Zero Ampere input currents i p, i n 0 A Very small difference of input voltages: v p - v n 0 V Extremely steep incline of function v o = f( v p - v n )∞

26. 26 Type 1 Noninverting Op Amp

27. 27 Type 1 Noninverting Op Amp, Ideal The noninverting Op Amp has the positive input pin connected to the signal source v s via resistor R p The inverting pin is connected to the common node (ground) also via a resistor, named R n The input signal arriving at the noninverting input is often referred to as v s or v p Output voltage v o is fed back to the inverting input pin via feedback resistor R f generally decreasing gain A Keeping in mind: i p = i n = 0 for ideal Op Amp Goal is to compute v o as function of input signal v s and the resistors; i.e. find the characteristic Op Amp function!

28. 28 Type 1 Noninverting Op Amp, Generic

29. 29 Type 1 Noninverting Op Amp, Generic i a + i f – i n =0– i a runs through R n i p = i n =0– i n runs into - input v n = v P =v s – no current in R p v s / v o =R n / ( R n + R f )– voltage divider v o =v s * ( R f + R n ) / R n v o =v s * A Characteristic function for output voltage v o of Noninverting Op Amp, with signal voltage v s

30. 30 Type 1 Noninverting Op Amp The characteristic function v o = v s * ( R f + R n ) / R n states that output voltage v o is a scaled, direct replica of the input signal v S, scaled by gain A, with A = ( R f + R n ) / R n Note that the characteristic function v o is valid, since the Op Amp is (close to) ideal, operating in the linear region –i.e. NOT saturated! Gain A can be controlled purely by resistors R f and R n for noninverting Op Amp

31. 31 Type 1 Noninverting Op Amp The amplification of a noninverting Op Amp with R f and R n is A = ( R n + R f ) / R n Provided Op Amp is ideal and is not saturated That means, |v o | must be <= |+-V CC | v s * ( R f + R n ) / R n <= V CC A = ( R n + R f ) / R n <= |V CC / v s |

32. 32 Type 1 Noninverting Op Amp, Example 1 Design some noninverting Op Amp with gain A = 10 Both resistors at the input signal pins R n and R p are predefined to be 1 kΩ, and v s = 1 V Compute v o and R f Check, whether under those conditions the Op Amp works in linear mode, assuming external power supplies of + -15 V

33. 33 Type 1 Noninverting Op Amp, Example 1

34. 34 Type 1 Noninverting Op Amp, Example 1 With v o / v s = A = ( R f + R n ) / R n -- require gain A=10: v o / v s = ( R f + R n ) / R n = 10 10 * R n = ( R f + R n ) R f = 9 R n R f = 9 kΩ To see whether the Op Amp is saturated, compute v o and see, whether or not it falls within + - V CC : v o = v s * ( R f + R n ) / R n v o = 1 * ( 9 + 1 ) / 1 = 10 V-- is within range + - 15 V

35. 35 Type 1 Noninverting Op Amp, Example 2 Assuming a signal voltage v s = 1.5 V, compute the lowest possible external power +-V CC that still enables the Op Amp to operate in linear mode We take all other design parameters from Example 1 Knowing that v o cannot be above |V CC |, we compute v o for v s = 1.5 V. This yields the lowest possible +V CC Then we compute v o for v s = -1.5 V. This yields the most negative legal -V CC Use formula v o = v s * ( R f + R n ) / R n With R f, R p, and R n inherited from Example 1

36. 36 Type 1 Noninverting Op Amp, Example 2

37. 37 Type 1 Noninverting Op Amp, Example 2 With v o = v s * ( R f + R n ) / R n Case 1 for largest v s = 1.5 V v o+ = 1.5 * ( 9 + 1 ) / 1= 15 V v o+ = 15 V Case 2 for smallest v s = -1.5 V v o- = -1.5 * ( 9 + 1 ) / 1= -15 V V o- = -15 V Matching exactly and just barely our previous power situation + -V CC from Example 1

38. 38 Type 2 Inverting Op Amp

39. 39 Type 2 Inverting Op Amp All Op Amps are Difference Op Amps! Even if one of the inputs is connected to the common node, AKA ground Inverting Op Amp has noninverting input pin grounded, which is at 0 V Inverting input pin is connected to v s the actual input signal Connection generally realized via resistor R s AKA R n in the literature Output voltage v o is fed back, as is typical, to inverting input pin via resistor R f

40. 40 Type 2 Inverting Op Amp, Generic

41. 41 Type 2 Inverting Op Amp i s + i f - i n =0 A v n = v P =0 V -- even with R p = 0 Ω at the +pin i s =( v S - v n ) / R S =v S / R S i f =( v o - v n ) / R f =v o / R f with i n =0 it follows: i s = -i f v o = - v S ( R f / R S ) Characteristic function for output voltage v o of Inverting Op Amp

42. 42 Type 2 Inverting Op Amp The characteristic function v o = - v S ( R f / R S ) states that the output voltage is a scaled inverted replica of the input signal v S Note that the characteristic function v o is valid only if the Op Amp is close to ideal The gain A can be controlled purely by the ratio of R f to R S Moreover, if R f = R S then the Op Amp reaches a gain A = 1, i.e. it is reduced to a pure voltage inverter We see it is easy to create a voltage inverter from a simplistic, inverting Op Amp

43. 43 Type 2 Inverting Op Amp, Example 3 In Example 3 we use an inverting Op Amp with: signal source of voltage v a = 1 V v n = v p, and v p being grounded = 0 V Feedback resistor R f = 100 kΩ Resistor R S at signal source v a is: R S = 25 kΩ Source voltages +V CC and -V CC of + -10 V Compute v o

44. 44 Type 2 Inverting Op Amp, Example 3

45. 45 Type 2 Inverting Op Amp, Example 3 i 25 + i 100 - i n =0 A with i n = 0 therefore: i 100 =-i 25 i 25 =(v a - v n ) / 25 k i 25 =(v a – 0 ) / 25 k i 25 =0.04 mA i 100 =(v o – v n ) / 100k=-0.04 v o / 100k=-0.04 v o =-4 Vwithin operating range

46. 46 Type 2 Inverting Op Amp, Example 3 Or less complicated than above, with characteristic function v o = - v S ( R f / R S ) v o = -1 * ( 100 k / 25 k ) v o = -1 * 4 v o =-4 V

47. 47 Type 2 Inverting Op Amp, Example 4 In Example 4 we use an inverting Op Amp with 1. 1.signal source of voltage v s = 0.4 V 2. 2.v n = v p, v p grounded = 0 V 3. 3.Feedback resistor R f = 80 kΩ 4. 4.Resistor R S at signal source v s of: R S = 16 kΩ 5. 5.Source voltages +V CC +10 V and -V CC of -15 V Compute v o

48. 48 Type 2 Inverting Op Amp, Example 4

49. 49 Type 2 Inverting Op Amp, Example 4, Hard i 16 + i 80 + i n =0 A with i n = 0 therefore: i 80 =-i 16 i 16 =(v s - v n ) / 16 k i 16 =(v s - 0) / 16 k i 16 =0.025 mA i 80 =(v o – v n ) / 80k=-0.025 mA v o / 80k=-0.025 v o =-2 V within operating range

50. 50 Type 2 Inverting Op Amp, Example 4, Easy Students use easier way of computing v o : Use characteristic equation! v o = -2 V

51. 51 Type 3 Difference Op Amp

52. 52 Type 3 Difference Op Amp, Terminology Easy to get confused by terminology: Difference Amplifier vs. Differential Amplifier When the circuit containing Op Amps is purely resistive, some literature uses the two terms interchangeably  Literally, all Op Amps are Difference Amplifiers, as the output voltage v o is a function of the difference of the input voltages v p and v n Some authors use the term Differential Amplifier as long as the gain is > 1; then when A = 1, switch to Difference Amplifier as Op Amp literally generates the difference v o = v p - v n with no gain! We use Difference Amplifier to reduce confusion with Differentiating Amplifier

53. 53 Type 3 Difference Op Amp For Difference Op Amp output voltage v o is proportional to the difference of voltages of input signals v a and v b Again we assume an ideal Op Amp... And with feedback loop from v o to inverting input v n via feedback resistor R f Both input signals v a and v b are scaled by resistors; one is R a connected to inverting input, the other R b Noninverting input often has a separate resistor R p to common node (ground) as well Goal to express v o as a function of v a and v b, specifically, as a function of the difference v b - v a

54. 54 Type 3 Difference Op Amp, Generic

55. 55 Type 3 Difference Op Amp (1) KCL, summing currents flowing to inverting pin: ( v n – v a ) / R a + ( v n – v o ) / R f - i n =0 (2) Voltage Division: v p = v n = v b * R p / ( R b + R p ) Substitute v n (2) into (1), and with i n = 0: v b *R p /(R a *(R b +R p ))-v a /R a + v b *R p /(R f *(R b +R p )) - v o /R f = 0 v o = v b * (R p *R f / (R a *(R b +R p ) ) + R p / (R b +R p )) - v a * R f / R a Characteristic function for output voltage v o of a general Difference Op Amp

56. 56 Type 3 Difference Op Amp Special case: R a / R f = R b / R p Or specifically: R b  R a, and R p  R f then v o = v b * (R f *R f + R a *R f ) / (R a *R a +R a *R f ) - v a * R f /R a v o = v b * R f / R a - v a * R f / R a v o = R f / R a * ( v b - v a ) Characteristic function for output voltage v o of Difference Op Amp with R b and R p sized after R a and R f

57. 57 Type 3 Difference Op Amp Special case: R f = R a v o = v b * R f / R a - v a * R f / R a v o = v b * R f / R f - v a * R f / R f v o = v b * 1 - v a * 1 v o = ( v b - v a ) Characteristic function for output voltage v o of Difference Op Amp with all R adjusted

58. 58 Type 3 Difference Op Amp, Example 5

59. 59 Type 3 Difference Op Amp, Example 5 i 100 =-i 20 as before, since i n = 0 i 20 =(v a - v n )/ 20k i 20 =( 1 - 2 )/ 20k = -0.05 mA i 100 =(v o – v n )/ 100k i 100 =v o /100k - 2/100k= +0.05 mA V o /100k=0.05 mA + 2/100k A=0.07 mA V o =7 Voutside operating range

60. 60 Type 4 Summing Op Amp

61. 61 Type 4 Summing Op Amp, Generic Summing Op Amp has multiple input signals all joining at the inverting or noninverting input pin Shown here: 3 inputs v a, v b, v c, at inverting input Scaled by resistors R a, R b, R c Ideal Op Amp still requires currents i n and i p = 0 A Also, with noninverting input pin connected to common node (or ground), we know that v p = 0, and therefore: v n = v p = 0 Goal to compute v o

62. 62 Type 4 Summing Op Amp, Generic

63. 63 Type 4 Summing Op Amp, Generic (v n -v a ) / R a + (v n -v b ) / R b + (v n -v c ) / R c + (v n -v o ) / R f - i n = 0 with v n = 0, i n = 0  v o / R f = -v a / R a - v b / R b - v c / R c v o = -R f *( v a / R a + v b / R b + v c / R c ) Characteristic function for output voltage v o of a Summing Op Amp with 3 inputs, scaled by resistors R a, b, c

64. 64 Type 4 Summing Op Amp, i Reversed Direction of 4 currents unclear? Then try reversed direction, with: (v a - v n ) / R a + (v b - v n ) / R b + (v c - v n ) / R c + (v o - v n ) / R f - i n = 0 And v n = 0 [V], i n = 0 [A] still yields the same equation: v o / R f = -v a / R a - v b / R b - v c / R c v o = -R f *( v a / R a + v b / R b + v c / R c ) Characteristic function for output voltage v o of Summing Op Amp with 3 inputs, scaled by resistors R a, b, c

65. 65 Type 4 Summing Op Amp, Generic with R s = R a = R b = R c  v o = -R f * ( v a + v b + v c ) / R s Characteristic function for output voltage v o of a Summing Op Amp with 3 inputs, identical resistors R S And with R s = R f  v o = -( v a + v b + v c ) Characteristic function for output voltage v o of a Summing Op Amp with 3 inputs at inverting input, R S = R f

66. 66 Type 4 Summing Op Amp, Example 6

67. 67 Type 4 Summing Op Amp, Example 6 (v n -v a )/5k + (v n -v b )/25k + (v n -v o )/250k - i n = 0 with v a = 0.1 V, v b = 0.25 V -0.1 / 5k - 0.25 / 25k=v o / 250k v o / 250k = -0.02 mA - 0.01 mA = -0.03 mA v o = -7.5 V

68. 68 Type 5 Integrating Op Amp

69. 69 Type 5 Integrating Op Amp Goal: devise an Op Amp circuit that integrates Output signal v o is integral of input function v S Here v S is linear voltage function, integral is easy: Function integration accomplished by capacitor C f in feedback loop, from signal v o to one of the inputs Recall: all Op Amps are Difference Amplifiers! Even with one input connected to ground Here: integrating Op Amp has noninverting input grounded at 0 [V] Inverting input connected to input signal v s via R S

70. 70 Type 5 Integrating Op Amp

71. 71 Type 5 Integrating Op Amp Recall KCL, valid of course also for Op Amp: Add all currents Σi = 0 A at inverting input pin Summary of attributes of ideal Op Amp: Current i n into (or out of) inverting input pin = 0 A Ditto, current i p into noninverting input pin = 0 A And always v p = v n If v p = 0 which it is due to grounding, then also v n = 0 Adding all non-zero currents at inverting input pin: i f + i S =0 And we know: i f =C f dv o (t) / dt v n =v p

72. 72 Type 5 Integrating Op Amp i f + i S =0 i S =v S / R S i f =C f dv o (t) / dt v S / R S + C f dv o (t) / dt =0 d v o (t) / dt=-1 / C f * v S / R S d v o (t) / dt=-1 / ( C f R S ) * v S d v o (t)=-1 / ( C f R S ) * v S dt – integrate from t 0 to t v o (t)=-1 / ( C f R S ) v S dt + v o (t 0 ) -- over the time period from t 0 to t

73. 73 Type 5 Integrating Op Amp Starting at time t = t 0 = 0, with v 0 (t 0 ) = 0 it follows: v o (t)=-1 / ( C f R S ) v S dt v o (t)=-1 / ( C f R S ) v S t Which is exactly our goal: show that output voltage v o (t) is the integral over time of the input signal v S For general case, with v S constant: v o (t) = -1 / ( C f R S ) * v S t + v o (t 0 ) With v o (t 0 ) = 0 [V] OpAmp integrates linearly over time: v o (t) = -1 / ( C f R S ) * v S t

74. 74 Example 7 For Integrating Op Amp

75. 75 Integrating Op Amp, Example 7 Example 7 uses an input signal v S that is a step function, i.e. constant values over time segments Assuming capacitor C f in feedback loop, resistor R S at input pin, and step function for v S : At time t < 0 assumev S = 0 From time 0 to t 1 v S = V m From time t 1 to t 2 = 2 t 1 v S = -V m V m being a constant voltage, i.e. horizontal v(t) line Compute output signal v o (t) as a function of –i.e. the integral of– v S v o (t)=-1 / ( C f R S ) * v S t

76. 76 Input Signal v S for Example 7 vSvS VmVm t t1t1 t2t2 0 -V m

77. 77 Integrating Op Amp, Example 7 The applicable function: v o (t) = -1 / ( C f R S ) * v S t For time t < 0: v o (t) = 0 For v S = V m and time t from 0 to t 1 : v o (t) = -1 / ( C f R S ) * V m t For time t from t 1 to t 2 with t 2 = 2 * t 1 : v o (t) = -1 / ( C f R S ) -V m dt - 1 / ( C f R S ) V m t 1 v o (t) = V m / ( C f R S ) t - 2 V m / ( C f R S ) t 1 v o (t) = V m / ( C f R S ) * ( t - 2 t 1 ) Curve would intersect with the v(t) axis at -2 V m t 1 / ( C f R S ); but is not valid in range 0.. t 1

78. 78 Output Signal v o for Example 7 t2 t2 t1t1 vovo 0 -t 1 *V m R S C f -2t 1 *V m R S C f t

79. 79 Integrating Op Amp, Example 7 Recalling that input signal v S is fed into the inverting input pin, we observe: The output signal v o is a sign-inverted integral (ascending-descending straight line) of the input signal (horizontal line, above-below 0) But v o is valid only within the defined ranges, 0..t 1 and t 1..t 2

80. 80 Example 8 For Integrating Op Amp

81. 81 Example 8 Integrating Op Amp

82. 82 Integrating Op Amp, Example 8 Example 8 uses step input function v S = ±V m from 0.. t 1, and v S = -V m in range t 1.. t 2 Capacitor C f = 0.1 μF is in feedback loop, and resistor R S = 100 kΩ at inverting input pin Initial capacitor voltage v o (t 0 ) = 0 [V] At time t < 0 assumev S = 0[V] From time 0 to t 1 = 1 [s]v S = 50[mV] From time t 1 to t 2, t 2 = 2 [s]v S = -50[mV] v S = ±V m is constant voltage ± 50[mV] Compute output signal v o (t) as a function of v S v o (t) = -1 / ( C f R S ) v S dt + v o (t 0 )

83. 83 Input Signal v S for Example 8 VmVm t1t1 t2t2 0 -V m v S [mV] t

84. 84 Integrating Op Amp, Example 8 For time t < 0:v o = 0 For time t from 0 to 1 [s]: v o = -1 / ( C f R S ) * V m t + 0 v o = -1 * 50 * 10 -3 t / ( 100 * 10 3 * 0.1 * 10 -6 ) v o = -5 t For time t from 1 [s] - 2 [s]: v o (t) = V m t / ( C f R S ) - 2 V m t 1 / ( C f R S ) v o = 50t*10 -3 /(100*10 3 *0.1*10 -6 ) - 2*50*10 -3 /(100*10 3 *0.1*10 -6 ) v o (t) = 5 t - 10[V] In range 1.. 2 [s]. Curve would intersect at v o = -10 [V]

85. 85 Output Signal v o for Example 8 t2 t2 t1t1 v o (t) 0 -5 t

86. 86 References 1. 1.Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13: 9780-13-376003-3 2. 2.Wiki page: http://en.wikipedia.org/wiki/Operational_amplifier 3. 3.A. D. Blumlein, Improvements in and relating to Thermionic Valve Amplifiers, UK Patent 425,553, issued March 18, 1935 4. 4.Hendrick Bode, Relations Between Attenuation and Phase In Feedback Amplifier Design, Bell System Technical Journal, Vol. 19, No. 3, July, 1940 5. 5.Hendrick Bode: Amplifier, US Patent 2,123,178, issued July 12, 1938

87. 87 References 6. 6.Dave Fullagar: http://www.edn.com/electronics- news/4326905/Voices-Dave-Fullagar-analog-IC- designer-and-entrepreneur 7. 7.Widlar mirror: https://en.wikipedia.org/wiki/Widlar_current_source 8.Differentiation rules: http://www.codeproject.com/KB/recipes/Differentiatio n.aspx 9.Euler’s Identities: https://en.wikipedia.org/wiki/Euler%27s_identity 10.Table of integrals: http://integral- table.com/downloads/single-page-integral-table.pdf