ECE2262 Electric Circuits. Chapter 4: Operational Amplifier (OPAMP) Circuits


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1 ECE2262 Electric Circuits Chapter 4: Operational Amplifier (OPAMP) Circuits 1
2 4.1 Operational Amplifiers 2
3 4. Voltages and currents in electrical circuits may represent signals and circuits can perform mathematical operations on these signals such as: scaling (amplification)! signal follower (buffer)! inverting! addition and subtraction! differentiation and integration! Op  Amps provide solutions to all these problems! 3 # " d dx
4 4
5 5
6 An operational amplifier ("opamp") is a DCcoupled highgain electronic voltage amplifier with a differential input and, usually, a singleended output. In this configuration, an opamp produces an output potential (relative to circuit ground) that is typically hundreds of thousands of times larger than the potential difference between its input terminals. 6
7 4.2 OpAmp Terminals! i + = V CC i " = "V CC Terminals of Op Amp inverting input (! ) noninverting input (+ ) output positive power supply (V + = V CC ) negative power supply (V! =!V CC ) 7
8 v + v! i + v 0 i! Terminal Voltage Variables Terminal Current Variables 8
9 i + i! v + v v +! v 0 v! The op amp with power supply terminals removed When the amplifier is operating within its linear region, the dc voltages ±V CC do not enter into the circuit equations 9
10 4.3 OpAmp Characteristic and Circuit Models The voltage transfer characteristic of an op amp v 0 v +! v! # % v 0 = $ % &% A v +! v!!v CC if v +! v! <!V CC / A ( ) if! V CC / A " v +! v! " V CC / A V CC if v +! v! > V CC / A 10
11 v 0 v +! v! typical values for V CC! 10, 15, 20 V the gain A is no less than 10, 000!!10 6 in the linear region: v +! v! " 20 /10 4 = 2mV 11
12 v 0 v +! v! in the linear region: v +! v! " 20 /10 4 = 2mV node voltages in the circuits we study are much larger than 2mV thus, if an op amp is constrained to its linear operation region we can assume that v + = v!! there is a virtual short between the input terminals! 12
13 An analysis of the op amp integrated circuit reveals that the equivalent resistance ( R i ) seen by the input terminals of the op amp is at least 1 M! This implies that i + = i! = 0 13
14 The model of ideal op amp i + = 0 i! = 0 v + v + = v! v! v 0 v + = v! i + = i! = 0 14
15 Example (a) Find v 0 if v a = 1 V, v b = 2 V. Assume ideal op amp i! i +!10 < 6 < 10 v + v 0 v! v + = v b = 2V, since v + = v!! v! = 2V i 25 = v a! v! 25k = 1! 2 25k =! 1 25 ma i 100 = v 0! v! 100k = v 0! ma KCL at : i 25 + i 100! i! = 0! i 100 =!i 25! v 0! = 1 25! v 0 = 6V 15
16 (b) For v a =1.5 V specify the range of v b that avoids op amp saturation i! i + v + v 0 v! Since i 100 =!i 25! v 0! v b 100k =! v! v a b 25k ( ) v b = 1 5 v and v a =1.5 The op amp is in the linear region if!10v " v 0 " 10V!!0.8V " v b " 3.2V 16
17 The linear (more realistic) circuit model of the op amp i + v + i! A( v +! v! ) v 0 = A( v +! v! )+ R o i 0 v 0 v! In the linear region of operation this model represents a more realistic approximation of an op amp. It includes: 1: A finite input resistance R i. 2: A finite openloop gain A. 3: A nonzero output resistance R 0 17
18 Example An invertingamplifier using the more realistic op amp model Inverting Amplifier i! v! v + v 0!! R f R s v s 18
19 Negative Feedback + v! v s A v +! v! v + + ( ) + v 0 0!!! v 0!! R f R s v s R o! 0, R i! ", A! " 19
20 + v! v s A v +! v! v + + ( ) + v 0 0!!! Noninverting terminal: v + = 0! v Node a:!! v s + v! 0! + v! v! o = 0 R s R i R f Node b: ( ) v o! v! + v! A 0! v o! R f R o + v o R L = 0 20
21 ! v o = R s R f!a + R o R f " 1+ A + R o + R % o # $ R i R L & ' + " 1+ R o # $ R L % " & ' 1+ R s # $ R i % & ' + R o R f v s If R o! 0, R i! ", A! " for any R L! v 0 =! R f " v s + O R s # $ 1 % A& ' 21
22 R i  very large R 0  very small A  very large 22
23 4.4 Negative Feedback How does a circuit maintain a virtual short (v + = v! ) at the input of an op amp, to ensure operation in the linear region? Negative feedback The portion of the output voltage is applied to the inverting input, causing v +! v! to decrease and hence V o to decrease, until the op amp operates in its linear region. Positive feedback: the output quickly becomes saturated, i.e., V o is stuck at one extreme or the other ( ±V CC ) and does not respond to the input. 23
24 24
25 4.5 Voltage Follower and Noninverting Amplifier A. Voltage Follower : Use the ideal op amp approximation i! i + v! v +!V S + i + R S + v + = 0! v + = V S! v! = V s! negative feedback! v 0 = v!! v = V 0 S 25
26 Ordinary circuits connection R L V 0 = R L R S + R L V S Such a connection changes the behavior of the circuits, e.g., the voltage V 0 = R L R S + R L V S is reduced compared to V S 26
27 Connection of circuits through a voltage follower V 0! V S R L R L >> R 0 / A The voltage follower prevents loading down the source voltage, i.e., the entire source voltage V S appears across the load resistance R L. Hence, R L does not draw current from the source network (the current is supplied by the opamp). Generally, the connection of one circuit to another through a voltage follower allows both circuits to continue to operate as designed. 27
28 Example 50! i = 20mA + 1V " V s = 5V V o = 4V 200! Load Function Generator A function generator has an output resistance of 50! and so its output will experience a significant internal voltage drop when the attached load draws a large current, resulting in a drop in the output terminal voltage. Therefore, one needs to buffer the function generator with an amplifier which presents a high input resistance to the source and which also provides a low output resistance to the load. 28
29 An ideal buffer amplifier with a gain 1, when placed in between the function generator and the load, delivers the full source voltage to the load. i = 0 i = 25mA + V i = 5V V o = 5V! 29
30 B. Noninverting Amplifier: Use the ideal op amp approximation i! i + R g v v 0! v + Since i + = 0! v + = v g! v! = v g KCL at v! : v! R s + v!! v 0 R f + i! = 0! v! = Note: R f! 0! voltage follower R s! v 0! v 0 = 1+ R f R s + R f " # R s $ % & v g 30
31 ! v 0 = 1 + R f " # R s $ % & v g Operation in the linear region requires that v 0! V CC! 1+ R f R s! V CC v g A noninverting amplifier multiplies the input voltage v g by a gain 1+ R f R s that is independent of the source resistance R g. Hence, the gain remains unchanged when the circuits are terminated by an external load. 31
32 Example Find the output voltage when R x = 60k! i! i + v! v + KCL at : v + + i + + v! V + S R x 15k v KCL at v! :! 4.5k + v! v! 0 63k! 1 = 0! v + 15k + 1 $ " # R x % & = V S 15k! v = 320mV + 4.5k + i! = 0! v! = 63k + 4.5k v 0 ( v! = v + )! v 0 = 67.5k 4.5k v = 4.8V +! v! 5V 0 32
33 Nonvertingamplifier using the more realistic op amp model v! A( v +! v! ) v + v 0 33
34 v! A( v +! v! ) v 0 v + Node a: Node b: Node v + : v! R s + v!! v + R i + v!! v o R f = 0 ( ) v o! v! + v o + v! A v! v o +! = 0 R f R L R o v +! v g + v! v +! = 0 R g R i 34
35 v o =! # " R s + R o A 1+ K r R f + R s + R s R o AR i $ & % ( ) + R f R s + R f + R s ( )( R i + R ) g AR i v g K r = R + R s g R i If R o! 0, R i! ", A! " + R f + R s R L + R f R s + R f R g + R g R s R i R L for any R L! v 0!! " # R s + R f R s $ % & v g 35
36 4.6 Inverting Amplifier: assume an ideal op amp i! v! v + KCL at : i s + i f! i! = 0 v + = 0! v! = v + = 0! i s = v s and i f = v! v 0! = v 0 R s R f R f Since i! = 0! i f =!i s! v 0 =! R f v s R s 36
37 InvertingAmplifier Equation v 0 =! R f v s R s The upper limit on the gain R f is determined by the power supply voltage V CC R s and the value of the signal voltage v s : v 0! V CC! R f R s v s! V CC! R f R s! V CC v s 37
38 If R f is removed, the negative feedback path is opened and the amplifier is operating in open loop. v! Opening the feedback path drastically changes the behavior of the circuit. It can be shown (use the realistic model of opamp) that the output voltage is v 0 =!Av! Hence, we can call A the openloop gain of the op amp. 38
39 4.7 Summing and DifferenceAmplifier Circuits A. SummingAmplifier Circuit (SAC) i! v! KCL at the inverting terminal: v!! v a R a + v!! v b R b + v!! v c R c + v!! v 0 R f + i! = 0 Since v + = 0! v! = 0 and i! = 0! solving the above w.r.t. v 0 we obtain 39
40 " v 0 =! # $ R f v R a + R f v a R b + R f v b R c c % & ' InvertingSumming Amplifier Equation i! v! 40
41 If R a = R b = R c = R s! v 0 =! R f R s ( v a + v b + v c ) The scaling factors in summingamplifier circuits are entirely determined by the external resistors: R f, R a, R b, i! v! 41
42 Example (a) Find v 0 if v a = 0.1 V, v b = 0.25 V. v! v + KCL at v! : v!! v a 5k + v!! v b 25k + v!! v 0 250k = 0 (i! = 0, v! = v + = 0 ) v 0 =!{ 50v a +10v b } = 7.5 V! v 0! 10 V 42
43 (b) If v b = 0.25 V, how large can v a be before the opamp saturates? v! v + From the previous solution in (a), we have v 0 =!{ 50v a +10v b } v 0 =!{ 50v a + 2.5}!!10 " v 0 " 15 is satisfied if!0.35v " v a " 0.15V 43
44 Example Design a summing amplifier whose output is v 0 =!( 4v a + v b + 5v c ), assume R f = 20k! i! v! " v 0 =! # $ R f v a + R f v b + R f % v c R a R b R c & ' 44
45 5k! 20k! 4k! 20k! 12 "12 v 0 = "( 4v a + v b + 5v c ) 45
46 SAC 46
47 47
48 B. Difference Amplifier Circuit i! v! v + KCL at v! : v!! v a R a + v!! v 0 R b + i! = 0! v!! v a R a + v!! v 0 R b = 0 KCL at v + : v +! v b R c + v + R d + i + = 0! v +! v b R c + v + R d = 0! v!! v b R c + v! R d = 0 48
49 " $ # $ % $ v!! v a R a + v!! v 0 R b = 0 v!! v b R c + v! R d = 0 v 0 = R b R a! " # 1+ R a R b $ % & R d R c + R d v b ' R b R a v a i! v! v + 49
50 If R d = R b and R c = R a! v 0 = R b R a! " # 1+ R a R b $ % & R d R c + R d v b ' R b R a v a = R b R a!##" =1 ## $! 1+ R $ a R b " # R b % & R a + R b v b ' R b R a v a R a v! i! v 0 = R b R a ( v b! v a ) R b v + 50
51 R a v! i! v 0 = R b R a ( v b! v a ) R b v + The scaling is controlled by the external resistors. The relationship between the output voltage and the input voltage is not affected by connecting a nonzero load resistance across the output of the amplifier. 51
52 Difference Amplifier  Another Perspective 1. differential mode input: v dm = v b! v a 2. common mode input: v cm = v a + v b 2! v a = v cm! v dm 2 and v b = v cm + v dm 2! v 0 = A cm v cm + A dm v dm A cm = R R! R R a d b c R a ( R c + R d ) common mode gain ( ) + R b ( R c + R d ) and A dm = R d R a + R b 2R a ( R c + R d ) differential mode gain 52
53 If R a = R c! A cm = 0! v 0 = A dm v dm! v 0 = A dm ( v b! v a ) R b R d i! v! v + An ideal difference amplifier ( perfect matching: R a R b = R c R d ) amplifies only the differential mode portion of the input signal voltage, and eliminates the common mode portion! provides immunity to noise common to both inputs 53
54 Example VR 200! R Since v! = v + and i! = i + = 0 then VD yields v 1 = R 6k + R v 0! Gain = v 0 = 6k + R v 1 R = 31 if R = 200! Note:!v 1 + ( v +! v! ) + V R = 0! V R = v 1 54
55 Example V in v! = v + = V in, VD: V in = R 1 R 1 + R 2 V 0! V 0 = R 1 + R 2 R 1 V in! Gain = V 0 V in = 1+ R 2 R 1 55 = 1+ 20k 3.3k = 7.06 V 0 = Gain!V in = V! I 0 = V k = k = 606 µa! i! = 0
56 Example V V KCL at v + : V! v 2 R 2 + V R I + i + = 0! v 2! V R 2 = V R I! V = R I R I + R 2 v 2 VD 56
57 KCL at v! : V! v 1 R 1 + V! v 0 R F + i! = 0! v 1! V R 1 = V! v 0 R F!! v 0 =! R " F v R % F R 1 # $ & ' V V = R 1 R I R I + R 2 v 2 v 0 =! R " F v R F R 1 # $ R 1 % & ' R I R I + R 2 v 2 Note: If R F R 1 = R I =!! v 0 =! ( v 2 " v 1 )! ideal difference amplifier R 2 57
58 4.8 Other OpAmp Circuits A. Inverting/Noninverting Amplifier V!! + i 0 i 2 V 0 R 4 V S V + i 4 R 3 58
59 V!! + i 0 i 2 V 0 R 4 V S V + i 4 R 3 V + : V + R 3 + V +! V 0 R 4 = 0! V + = R 3 R 3 + R 4 V 0 V! : V! V! S + V! V!! 0 = 0! V 0 = 1+ R 2 R 1 R 2 " #!! V 0 = 1+ R $ 2 " # % & R 1 ' R 3 ( V 0 ) R 3 + R 4 * +  R 2, R 1 $ % & V ' R 2 ' R 1 V S ** V! = V + ** ( ) V S! V 0 =! R 2 R 3 + R 4 V S R 1 R 1 R 4! R 2 R 3 59
60 V!! + i 0 i 2 V 0 R 4 V S V + i 4 R 3 V 0 =! R 2 ( R 3 + R 4 ) V S R 1 R 4! R 2 R 3 If R 1 R 4! R 2 R 3 > 0! Inverting Amplifier If R 1 R 4! R 2 R 3 < 0! Noninverting Amplifier 60
61 Currents V!! + i 0 i 2 V 0 R 4 V S V + i 4 R 3 i 0 = i 2! i 4! i 2 = V!! V 0 R 2! V! = R 3 ( ) V 0, V 0 =! R 2 R 3 + R 4 V S R 3 + R 4 R 1 R 4! R 2 R 3 i 4 = V 0! V + R 4! V + = R 3 ( ) V 0, V 0 =! R 2 R 3 + R 4 V S R 3 + R 4 R 1 R 4! R 2 R 3 61
62 B. Summing Amplifier:Noninverting Configuration R 5 R 4 V +! + V! V 0 V 1 V 2 R 3 62
63 R 5 R 4 V +! + V! V 0 V 1 V 2 R 3 V! : V! R 4 + V!! V 0 R 5! V! = R 4 R 4 + R 5 V >! V + : V +! V 1 R 1 + V +! V 2 R 2 + V + R 3 = 0!! " # $ R 1 R 2 R 3 % & V = 1 + V V 2 R 1 R 2 ** V! = V + ** 63
64 1 V 0 = R A V ! $ V 2 " # R 1 R 2 % &! R = R 4 A R 1 R 2 R 3 R 5 Let R 1 = R 2 = R 3 = R! V 0 = R 5! " # R 4 $ % & V + V 1 2 ( ) If R 5 R 4 = 2! V 0 = V 1 + V 2 64
65 C. VoltagetoCurrent Converter : V S! i L R 52 R 4 1 V!! + V 0 R 4 R 3 V L V L i L V S R 3 Load 65
66 R 52 R 4 1 V!! + V 0 R 4 R 3 V L V L i L V S R 3 Load V L = V + = V! V! : V! + V! V!! 0 = 0! V 0 = 1+ R 2 R 1 R 2 " # R 1 $ % & V! '! V 0 = 1+ R $ 2 " # % & V L R 1 66
67 R 52 R 4 1 V!! + V 0 R 4 R 3 V L V L i L V S R 3 Load Current thr. the load: i L = V S! V L R 3 + V! V! 0 L! V 0 = 1+ R 2 R 4 " # R 1 $ % & V L i L = V S R 3! V L R a! R a = R 4 R 4 R 3! R 2 R 1 If R 4 R 3 = R 2 R 1! R a =!! i L = V S R 3 67
68 R 52 R 4 1 V!! + V 0 R 4 R 3 V L V L i L V S R 3 Load R 4 R 3 = R 2 R 1! i L = V S R 3 The current of the load does not depend on the voltage across the load We have a perfect mapping V S! i L 68
69 D. Current Amplifier i S i 0 i 0 =! R 2 R 1 i S 69
70 Summary: Signal Processing Circuits 70
71 71
72 72
73 73
74 If R 1 R f = R 2 R g then V out = R f ( V 2! V 1 ) R 1 74
75 OpAmp Integrator v s OpAmp Differentiator v s 75
76 v 0 = A( v +! v! ) v + = v! i + = i! = 0 76
77 4.8 Problems! Set 4 77
78 LAB 2 78
79 79
The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A =
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