Introduction to Operational Amplifiers

P. R. Nelson ECE 322 Fall 2012 p. 1/50 Introduction to Operational Amplifiers Phyllis R. Nelson prnelson@csupomona.edu Professor, Department of Electrical and Computer Engineering California State Polytechnic University, Pomona

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.

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?)

P. R. Nelson ECE 322 Fall 2012 p. 4/50 Inverting Amp R 2 input ( ) output ( )

P. R. Nelson ECE 322 Fall 2012 p. 4/50 Inverting Amp R 2 input ( ) V m = V p = 0 output ( )

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 ( )

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

P. R. Nelson ECE 322 Fall 2012 p. 5/50 Noninverting Amp R 2 output ( ) input ( )

P. R. Nelson ECE 322 Fall 2012 p. 5/50 Noninverting Amp R 2 output ( ) input ( ) V m = V p =

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

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

P. R. Nelson ECE 322 Fall 2012 p. 6/50 Questions????

P. R. Nelson ECE 322 Fall 2012 p. 7/50 Example 1 R 2 V ref

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

P. R. Nelson ECE 322 Fall 2012 p. 8/50 Example 1 cont d R 2 V ref

P. R. Nelson ECE 322 Fall 2012 p. 8/50 Example 1 cont d R 2 V ref = ( R2 ) ( 1 R ) 2 V ref

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

P. R. Nelson ECE 322 Fall 2012 p. 9/50 Example 2 V S R S I i R L I L V O

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

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 )

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).

P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2

P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2 Rule 1 doesn t apply.

P. R. Nelson ECE 322 Fall 2012 p. 11/50 Example 3 R 2 Rule 1 doesn t apply. Now what?

P. R. Nelson ECE 322 Fall 2012 p. 12/50 A Comparison R 2 R 2

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.

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...

P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω

P. R. Nelson ECE 322 Fall 2012 p. 13/50 Inverting Amp Voltage R 2 = 1 kω R 2 = 100 kω = 0.5 V

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

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?

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. 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

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

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

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.

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

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

P. R. Nelson ECE 322 Fall 2012 p. 17/50 Ideal Op Amp VTC V p lim V p V m V m lim

P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued R 2

P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued V p = ( R1 R 2 ) R 2

P. R. Nelson ECE 322 Fall 2012 p. 18/50 Example 3 Continued V p = ( R1 R 2 ) R 2 = V p = 0

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.

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 )

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,

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.

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

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

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

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

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

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

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.

P. R. Nelson ECE 322 Fall 2012 p. 23/50 Example 4 R 2 Now it s your turn. Sketch vs.

P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2

P. R. Nelson ECE 322 Fall 2012 p. 24/50 Hint 1 ( R1 ) ( R2 ) V p = R 2 R 2 WHY?

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

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!

P. R. Nelson ECE 322 Fall 2012 p. 25/50 Hints 2 & 3 Hint 2: The switch points are at V p = 0 V.

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 )

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

P. R. Nelson ECE 322 Fall 2012 p. 27/50 Questions????

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 ( )

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

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

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

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

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

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.

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

P. R. Nelson ECE 322 Fall 2012 p. 34/50 C 2 V IO

P. R. Nelson ECE 322 Fall 2012 p. 34/50 C 2 V IO = V IO 1 C 2 ( V IO ) dt

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.

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.

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.

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.

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

P. R. Nelson ECE 322 Fall 2012 p. 38/50 Example 6: Inverting Amp With I B and I OS R 2 I m

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

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

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

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

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?

P. R. Nelson ECE 322 Fall 2012 p. 40/50 Redesign Ideas Idea 1: Smaller R 2, constant R 2 /

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.

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...

P. R. Nelson ECE 322 Fall 2012 p. 41/50 New Design R 2 I m R 3 I p

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

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

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 )

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

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!

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.

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

P. R. Nelson ECE 322 Fall 2012 p. 44/50 Example 7: Audio Amplifier 24 kω 1 kω 1 kω 8 Ω

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

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

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 = 0.894 V RMS = 1.26 V peak = 50.6 mv peak

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

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

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

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)!

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

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

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

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

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

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.