Introduction to Operational Amplifiers
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1 P. R. Nelson ECE 322 Fall 2012 p. 1/50 Introduction to Operational Amplifiers Phyllis R. Nelson Professor, Department of Electrical and Computer Engineering California State Polytechnic University, Pomona
2 P. R. Nelson ECE 322 Fall 2012 p. 2/50 Ideal Op Amp Op amp symbol: - input (V m ) input (V p ) output ( ) Ideal op amp model: 1. If there is a branch connecting the output to the - input, V p V m = 0 2. There are no currents flowing into or out of the input terminals.
3 P. R. Nelson ECE 322 Fall 2012 p. 3/50 How It Works If there is a branch connecting the output to the - input, the op amp adjusts so that V p = V m... unless it can t. (Thinking ahead, what could go wrong?)
4 P. R. Nelson ECE 322 Fall 2012 p. 4/50 Inverting Amp R 2 input ( ) output ( )
5 P. R. Nelson ECE 322 Fall 2012 p. 4/50 Inverting Amp R 2 input ( ) V m = V p = 0 output ( )
6 P. R. Nelson ECE 322 Fall 2012 p. 4/50 Inverting Amp R 2 input ( ) V m = V p = 0 V m V m R 2 = 0 output ( )
7 Inverting Amp R 2 input ( ) V m = V p = 0 V m V m R 2 = 0 output ( ) = R 2 P. R. Nelson ECE 322 Fall 2012 p. 4/50
8 P. R. Nelson ECE 322 Fall 2012 p. 5/50 Noninverting Amp R 2 output ( ) input ( )
9 P. R. Nelson ECE 322 Fall 2012 p. 5/50 Noninverting Amp R 2 output ( ) input ( ) V m = V p =
10 P. R. Nelson ECE 322 Fall 2012 p. 5/50 Noninverting Amp R 2 output ( ) input ( ) V m = V p = V m 0 V m R 2 = 0
11 Noninverting Amp R 2 output ( ) input ( ) V m = V p = V m 0 V m R 2 = 0 = 1 R 2 P. R. Nelson ECE 322 Fall 2012 p. 5/50
12 P. R. Nelson ECE 322 Fall 2012 p. 6/50 Questions????
13 P. R. Nelson ECE 322 Fall 2012 p. 7/50 Example 1 R 2 V ref
14 P. R. Nelson ECE 322 Fall 2012 p. 7/50 Example 1 R 2 V ref V m = V p = V ref V m V m R 2 = 0
15 P. R. Nelson ECE 322 Fall 2012 p. 8/50 Example 1 cont d R 2 V ref
16 P. R. Nelson ECE 322 Fall 2012 p. 8/50 Example 1 cont d R 2 V ref = ( R2 ) ( 1 R ) 2 V ref
17 P. R. Nelson ECE 322 Fall 2012 p. 8/50 Example 1 cont d R 2 V ref SUPERPOSITION! = ( R2 ) ( 1 R ) 2 V ref
18 P. R. Nelson ECE 322 Fall 2012 p. 9/50 Example 2 V S R S I i R L I L V O
19 P. R. Nelson ECE 322 Fall 2012 p. 9/50 Example 2 V S R S I i R L V O I L I L = V S V O R L I i = V S = I L V S R S
20 P. R. Nelson ECE 322 Fall 2012 p. 10/50 V S R S I i R L I L V O I L I i = ( 1 R S )
21 P. R. Nelson ECE 322 Fall 2012 p. 10/50 V S R S I i R L I L V O I L I i = ( 1 R S ) This is a current-controlled current source (current amplifier).
22 P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2
23 P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2 Rule 1 doesn t apply.
24 P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2 Rule 1 doesn t apply. Now what?
25 P. R. Nelson ECE 322 Fall 2012 p. 12/50 A Comparison R 2 R 2
26 P. R. Nelson ECE 322 Fall 2012 p. 12/50 A Comparison R 2 R 2 The same external circuit, but different opamp input connections.
27 P. R. Nelson ECE 322 Fall 2012 p. 12/50 A Comparison R 2 R 2 The same external circuit, but different opamp input connections. Look again at a familiar example...
28 P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω
29 P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω = 0.5 V
30 P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω = 0.5 V = R 2 = 50 V
31 P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω = 0.5 V = R 2 = 50 V ALWAYS?
32 P. R. Nelson ECE 322 Fall 2012 p. 14/50 Inverting Amp Power R 2 = R L = 1 kω R L R 2 = 100 kω = 0.5 V = 50 V
33 P. R. Nelson ECE 322 Fall 2012 p. 14/50 Inverting Amp Power R 2 = R L = 1 kω R L R 2 = 100 kω = 0.5 V = 50 V P in = 2 = 0.25 mw P out = [] 2 R L = 2.5 W = 10 4 P in
34 Inverting Amp Power R 2 = R L = 1 kω R L R 2 = 100 kω = 0.5 V = 50 V P in = 2 = 0.25 mw P out = [] 2 R L = 2.5 W = 10 4 P in What supplies this output power? P. R. Nelson ECE 322 Fall 2012 p. 14/50
35 Ideal op amp model rule 1 (V p V m = 0 if the output is connected to the - input) means that the op amp adjusts. P. R. Nelson ECE 322 Fall 2012 p. 15/50
36 P. R. Nelson ECE 322 Fall 2012 p. 15/50 Ideal op amp model rule 1 (V p V m = 0 if the output is connected to the - input) means that the op amp adjusts. The ideal op amp model thus assumes a dependent voltage source inside the opamp.
37 P. R. Nelson ECE 322 Fall 2012 p. 15/50 Ideal op amp model rule 1 (V p V m = 0 if the output is connected to the - input) means that the op amp adjusts. The ideal op amp model thus assumes a dependent voltage source inside the opamp. This source must be able to supply power.
38 P. R. Nelson ECE 322 Fall 2012 p. 15/50 Ideal op amp model rule 1 (V p V m = 0 if the output is connected to the - input) means that the op amp adjusts. The ideal op amp model thus assumes a dependent voltage source inside the opamp. This source must be able to supply power. The op amp requires an external power supply.
39 P. R. Nelson ECE 322 Fall 2012 p. 15/50 Ideal op amp model rule 1 (V p V m = 0 if the output is connected to the - input) means that the op amp adjusts. The ideal op amp model thus assumes a dependent voltage source inside the opamp. This source must be able to supply power. The op amp requires an external power supply. The output voltage is limited by the power supply voltage(s)!
40 P. R. Nelson ECE 322 Fall 2012 p. 16/50 Output Voltage Limits V sp V m V p V sm Vo = V p lim V p > V m 0 V p = V m V p < V m V m lim
41 P. R. Nelson ECE 322 Fall 2012 p. 16/50 Output Voltage Limits V sp V m V p V sm Vo = V p lim V sp V m lim V sm V p lim V p > V m 0 V p = V m V p < V m V m lim
42 P. R. Nelson ECE 322 Fall 2012 p. 17/50 Ideal Op Amp VTC V p lim V p V m V m lim
43 P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued R 2
44 P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued V p = ( R1 R 2 ) R 2
45 P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued V p = ( R1 R 2 ) R 2 = V p = 0
46 P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued V p = ( R1 R 2 ) R 2 = V p = 0 If V p, equals either V p lim or V m lim.
47 P. R. Nelson ECE 322 Fall 2012 p. 19/50 ut = V p lim < V p 0 = V p > V p V m lim V p = ( R1 R 2 )
48 P. R. Nelson ECE 322 Fall 2012 p. 19/50 ut = V p lim < V p 0 = V p > V p V m lim V p = ( R1 R 2 ) These equations can t be solved for given,
49 P. R. Nelson ECE 322 Fall 2012 p. 19/50 ut = V p lim < V p 0 = V p > V p V m lim V p = ( R1 R 2 ) These equations can t be solved for given, but there is a range of for a given value of.
50 P. R. Nelson ECE 322 Fall 2012 p. 20/50 Choose some values to make the calculations concrete: = 1 kω R 2 = 5 kω V ± lim = ±6.0 V
51 P. R. Nelson ECE 322 Fall 2012 p. 20/50 Choose some values to make the calculations concrete: = 1 kω R 2 = 5 kω V ± lim = ±6.0 V If = V p lim = 6.0 V
52 P. R. Nelson ECE 322 Fall 2012 p. 20/50 Choose some values to make the calculations concrete: = 1 kω R 2 = 5 kω V ± lim = ±6.0 V If = V p lim = 6.0 V ( ) 1 kω V p = 1 kω 5 kω 6 V = 1 V
53 P. R. Nelson ECE 322 Fall 2012 p. 20/50 Choose some values to make the calculations concrete: = 1 kω R 2 = 5 kω V ± lim = ±6.0 V If = V p lim = 6.0 V ( ) 1 kω V p = 1 kω 5 kω 6 V = 1 V If = V m lim = 6 V
54 Choose some values to make the calculations concrete: = 1 kω R 2 = 5 kω V ± lim = ±6.0 V If = V p lim = 6.0 V ( ) 1 kω V p = 1 kω 5 kω 6 V = 1 V If = V m lim = 6 V ( ) 1 kω V p = 1 kω 5 kω (6 V) = 1 V P. R. Nelson ECE 322 Fall 2012 p. 20/50
55 P. R. Nelson ECE 322 Fall 2012 p. 21/50 1 V < < 1 V: Two values of > 1 V: = 6 V < 1 V: = 6 V
56 P. R. Nelson ECE 322 Fall 2012 p. 22/50 Schmitt Trigger V p lim Stable point Switch point For between the switch points, the circuit has memory. Switch point Stable point V m lim depends on history as well as the current value of.
57 P. R. Nelson ECE 322 Fall 2012 p. 23/50 Example 4 R 2 Now it s your turn. Sketch vs.
58 P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2
59 P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2 WHY?
60 P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2 WHY? superposition and the voltage divider formula, or KCL at the op amp terminal plus algebra
61 P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2 WHY? superposition and the voltage divider formula, or KCL at the op amp terminal plus algebra Voltage dividers, superposition, current dividers, and the Thevénin and Norton equivalent circuits can save a lot of algebra, and potentially errors!
62 P. R. Nelson ECE 322 Fall 2012 p. 25/50 Hints 2 & 3 Hint 2: The switch points are at V p = 0 V.
63 P. R. Nelson ECE 322 Fall 2012 p. 25/50 Hints 2 & 3 Hint 2: The switch points are at V p = 0 V. Hint 3: switch = ( R1 R 2 )
64 P. R. Nelson ECE 322 Fall 2012 p. 26/50 Another Schmitt Trigger V lim R 2 V p lim R 2 V m lim V m lim
65 P. R. Nelson ECE 322 Fall 2012 p. 27/50 Questions????
66 P. R. Nelson ECE 322 Fall 2012 p. 28/50 Finite Open Loop Gain - input (V m ) input (V p ) Rule 1 becomes = A vo (V p V m ) output ( )
67 P. R. Nelson ECE 322 Fall 2012 p. 28/50 Finite Open Loop Gain - input (V m ) input (V p ) Rule 1 becomes = A vo (V p V m ) output ( ) A vo = A vo finite V p V m
68 P. R. Nelson ECE 322 Fall 2012 p. 28/50 Finite Open Loop Gain - input (V m ) input (V p ) Rule 1 becomes = A vo (V p V m ) output ( ) A vo = A vo finite V p V m V p V m 0
69 P. R. Nelson ECE 322 Fall 2012 p. 29/50 Finite Gain Circuit Model If V m lim < < V p lim the op amp can be modeled as a voltage-controlled voltage source: V p A(V p V m ) V m
70 P. R. Nelson ECE 322 Fall 2012 p. 30/50 Inverting Amp, Finite Gain Rule 1: R 2 = A vo (0 V V m ) Rule 2: V m V m R 2 = 0 = R 2 ( ) 1 1 A vo 1 R 2
71 P. R. Nelson ECE 322 Fall 2012 p. 31/50 Differential Input Offset Voltage The internal components that connect to the and - terminals of an op amp are not perfectly identical due to process variations. A small voltage difference V IO must be used at the input of a real op amp to make the output voltage zero. V IO = 0 V
72 P. R. Nelson ECE 322 Fall 2012 p. 32/50 The differential input offset voltage is modeled by a voltage source in series with the terminal of an ideal op amp. V IO V p V m The value of V IO can be either positive or negative.
73 P. R. Nelson ECE 322 Fall 2012 p. 33/50 Example 5: Integrator Although I used resistors to derive the closed-loop gain of the inverting amplifier, the same reasoning works with complex impedances. C 2
74 P. R. Nelson ECE 322 Fall 2012 p. 33/50 Example 5: Integrator Although I used resistors to derive the closed-loop gain of the inverting amplifier, the same reasoning works with complex impedances. C 2 = = 1 jω C 2 C 2 dt
75 P. R. Nelson ECE 322 Fall 2012 p. 34/50 C 2 V IO
76 P. R. Nelson ECE 322 Fall 2012 p. 34/50 C 2 V IO = V IO 1 C 2 ( V IO ) dt
77 P. R. Nelson ECE 322 Fall 2012 p. 34/50 C 2 V IO = V IO 1 C 2 ( V IO ) dt If = 0, there is still a signal (V IO ) to integrate. The steady-state output with = 0 V will be at V p lim or V m lim.
78 P. R. Nelson ECE 322 Fall 2012 p. 35/50 Stability of V IO If the value of V IO could be measured, the error it introduces could be eliminated by adding a voltage to cancel it. However, V IO changes with temperature and drifts over time as the op amp ages. Approaches to minimizing errors due to V IO will be discussed later.
79 P. R. Nelson ECE 322 Fall 2012 p. 36/50 Input Bias and Input Offset Currents Some current flows into the input terminals of a real op amp.
80 P. R. Nelson ECE 322 Fall 2012 p. 36/50 Input Bias and Input Offset Currents Some current flows into the input terminals of a real op amp. V p V m I p I m These currents can be modeled as current sources I p and I m external to an ideal op amp.
81 Designer s Model I OS /2 V p V m I B I B The input bias current I B and input offset current I OS are in data sheets. I B = I m I p 2 I OS = I m I p I m = I B I OS 2 I p = I B I OS 2 P. R. Nelson ECE 322 Fall 2012 p. 37/50
82 P. R. Nelson ECE 322 Fall 2012 p. 38/50 Example 6: Inverting Amp With I B and I OS R 2 I m
83 P. R. Nelson ECE 322 Fall 2012 p. 38/50 Example 6: Inverting Amp With I B and I OS R 2 I m R 2 I m = 0
84 P. R. Nelson ECE 322 Fall 2012 p. 38/50 Example 6: Inverting Amp With I B and I OS R 2 I m R 2 I m = 0 = R 2 R 2 I m I m = I B ± I OS 2
85 P. R. Nelson ECE 322 Fall 2012 p. 39/50 Numerical Example I m R 2 = 10 kω R 2 = 1 MΩ n = 0 V I B = 100 na I OS = 0 na
86 P. R. Nelson ECE 322 Fall 2012 p. 39/50 Numerical Example I m R 2 = 10 kω R 2 = 1 MΩ n = 0 V I B = 100 na I OS = 0 na = 0.1 V
87 P. R. Nelson ECE 322 Fall 2012 p. 39/50 Numerical Example I m R 2 = 10 kω R 2 = 1 MΩ n = 0 V I B = 100 na I OS = 0 na = 0.1 V How can the circuit be redesigned to minimize this error?
88 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /
89 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /... but is the input resistance of this circuit, which is often specified.
90 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /... but is the input resistance of this circuit, which is often specified. Idea 2: Add a voltage at the terminal.
91 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /... but is the input resistance of this circuit, which is often specified. Idea 2: Add a voltage at the terminal.... but we don t know what value to use because we don t know I m.
92 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /... but is the input resistance of this circuit, which is often specified. Idea 2: Add a voltage at the terminal.... but we don t know what value to use because we don t know I m. Idea 3: I B flows in both op amp terminals...
93 P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /... but is the input resistance of this circuit, which is often specified. Idea 2: Add a voltage at the terminal.... but we don t know what value to use because we don t know I m. Idea 3: I B flows in both op amp terminals so we could use a resistor at the terminal to generate a voltage proportional to I B.
94 P. R. Nelson ECE 322 Fall 2012 p. 41/50 New Design R 2 I m R 3 I p
95 P. R. Nelson ECE 322 Fall 2012 p. 41/50 New Design R 2 I m R 3 I p = ( R2 ) R 2 I m ( 1 R ) 2 R 3 I p
96 P. R. Nelson ECE 322 Fall 2012 p. 41/50 New Design R 2 I m R 3 I p = ( R2 ) R 2 I m R 3 = R 2 = ( 1 R 2 ( R2 ) ) R 3 I p R 2 I OS
97 P. R. Nelson ECE 322 Fall 2012 p. 42/50 Improvement = 10 kω R 2 = 1 MΩ n = 0 V I B = 90 na I OS = 20 na I m = 100 na = 20 mv (100 mv without R 3 )
98 P. R. Nelson ECE 322 Fall 2012 p. 42/50 Improvement = 10 kω R 2 = 1 MΩ n = 0 V I B = 90 na I OS = 20 na I m = 100 na = 20 mv (100 mv without R 3 ) A more typical case with the same gain: = 1 kω R 2 = 100 kω = 2 mv
99 P. R. Nelson ECE 322 Fall 2012 p. 42/50 Improvement = 10 kω R 2 = 1 MΩ n = 0 V I B = 90 na I OS = 20 na I m = 100 na = 20 mv (100 mv without R 3 ) A more typical case with the same gain: = 1 kω R 2 = 100 kω = 2 mv Small R 2 minimizes I OS errors!
100 P. R. Nelson ECE 322 Fall 2012 p. 43/50 Output Resistance So far we have neglected the internal impedance of the voltage-controlled voltage source.
101 P. R. Nelson ECE 322 Fall 2012 p. 43/50 Output Resistance So far we have neglected the internal impedance of the voltage-controlled voltage source. We need to use a Thevénin equivalent circuit instead of an ideal source.
102 P. R. Nelson ECE 322 Fall 2012 p. 43/50 Output Resistance So far we have neglected the internal impedance of the voltage-controlled voltage source. We need to use a Thevénin equivalent circuit instead of an ideal source. A vo (V p V m ) A vo (V p V m ) R o
103 P. R. Nelson ECE 322 Fall 2012 p. 43/50 Output Resistance So far we have neglected the internal impedance of the voltage-controlled voltage source. We need to use a Thevénin equivalent circuit instead of an ideal source. A vo (V p V m ) A vo (V p V m ) R o Output resistance is important when the op amp supplies significant current.
104 P. R. Nelson ECE 322 Fall 2012 p. 44/50 Example 7: Audio Amplifier 24 kω 1 kω 1 kω 8 Ω
105 P. R. Nelson ECE 322 Fall 2012 p. 44/50 Example 7: Audio Amplifier 24 kω 1 kω 1 kω 8 Ω Want 100 mw RMS... to 8 Ω speaker
106 P. R. Nelson ECE 322 Fall 2012 p. 44/50 Example 7: Audio Amplifier 24 kω 1 kω 1 kω 8 Ω Want 100 mw RMS... to 8 Ω speaker / = 25
107 P. R. Nelson ECE 322 Fall 2012 p. 44/50 Example 7: Audio Amplifier 24 kω 1 kω 1 kω 8 Ω Want 100 mw RMS... to 8 Ω speaker / = 25 = V RMS = 1.26 V peak = 50.6 mv peak
108 P. R. Nelson ECE 322 Fall 2012 p. 45/50 With 15 Ω R o 24 kω 1 kω X 1 kω 15 Ω 8 Ω V x R o = R L R 2
109 P. R. Nelson ECE 322 Fall 2012 p. 46/50 V x = 1 R o ( 1 R L 1 R 2 [1 1Av ]) = 2.88
110 P. R. Nelson ECE 322 Fall 2012 p. 46/50 V x = 1 R o ( 1 R L 1 R 2 [1 1Av ]) = 2.88 V x = 3.64 V peak
111 P. R. Nelson ECE 322 Fall 2012 p. 46/50 V x = 1 R o ( 1 R L 1 R 2 [1 1Av ]) = 2.88 V x = 3.64 V peak You would not like the sound if you run this circuit from a 3 V supply (two AA batteries in series)!
112 P. R. Nelson ECE 322 Fall 2012 p. 47/50 Finite Input Impedance Model: V p R p C P R d C d V m R m C m
113 P. R. Nelson ECE 322 Fall 2012 p. 48/50 Example 8: Amp for High-R Sensor 24 kω 1 kω 0.1µA sensor 1 MΩ
114 P. R. Nelson ECE 322 Fall 2012 p. 48/50 Example 8: Amp for High-R Sensor 24 kω 1 kω 0.1µA sensor 1 MΩ Ideal op amp model = 0.1µA 1 MΩ ( 1 ) 24 kω 1 kω = 2.5 V
115 P. R. Nelson ECE 322 Fall 2012 p. 49/50 With Finite R p 24 kω 1 kω 1 MΩ R p = 2 MΩ 0.1µA sensor
116 P. R. Nelson ECE 322 Fall 2012 p. 49/50 With Finite R p 24 kω 1 kω 1 MΩ R p = 2 MΩ 0.1µA sensor = 0.1µA (1 MΩ 2 MΩ) ( 1 ) 24 kω 1 kω = 1.67 V
117 P. R. Nelson ECE 322 Fall 2012 p. 50/50 Other Effects This discussion has covered only a few of the most important op amp parameters. For an extensive listing of op amp parameters, see section 11-2 of Op Amps For Everyone.
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