# E40M. Op Amps. M. Horowitz, J. Plummer, R. Howe 1

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1 E40M Op Amps M. Horowitz, J. Plummer, R. Howe 1

2 Reading A&L: Chapter 15, pp Reader, Chapter 8 Noninverting Amp Inverting Amp Summing Amp M. Horowitz, J. Plummer, R. Howe 2

3 How to Measure Small Voltages? Arduino input has full-scale around 5V It produces a 10 bit answer (1024) This means a LSB (least significant bit) is 5mV Need to make the signal bigger before input to Arduino So, we will use an amplifier Many ways to build amplifiers One often uses a standard building block for amplifiers Called an Operational Amplifier, or Op-Amp A circuit with very high gain at low frequencies (< 10 khz) M. Horowitz, J. Plummer, R. Howe 3

4 Electrical Picture Signal amplitude 1 mv Noise level will be significant will need to amplify and filter We ll use filtering ideas from the last two lectures M. Horowitz, J. Plummer, R. Howe 4

5 OP AMPS M. Horowitz, J. Plummer, R. Howe 5

6 Op Amp Is a common building block It is a high-gain amplifier Output voltage is A (V+ V-) Gain, A, is 10 K to 1 M LM741 Output voltage can be + or Often can swing between +Vdd and -Vdd supplies Huh? M. Horowitz, J. Plummer, R. Howe 6

7 Op-Amp Power Supply Up to now we had one supply voltage, Vdd All voltages were between Vdd and Gnd Generally measured relative to Gnd So all voltages were positive. A sinewave goes positive and negative And most input signals do that too It is convenient to have a reference where The output can be positive and negative Can do that by changing what we call the reference M. Horowitz, J. Plummer, R. Howe 7

8 Moving the Reference V dd = 5 V V dd = 2.5 V v out v out 2.5 V - The voltages are all the same, only the reference voltage has moved M. Horowitz, J. Plummer, R. Howe 8

9 What You Will Actually Do Use the USB supply Just change the reference voltage 5 V + - R R + - M. Horowitz, J. Plummer, R. Howe 9

10 Op Amp Behavior Relationship between output voltage and input voltage: v o = A( v + v ) = A( v p v ) n A is the op-amp gain (or open-loop gain), and is huge 10K-1M The input currents are very, very small so i p 0 and i n 0. M. Horowitz, J. Plummer, R. Howe 10

11 Since the Output Swing is Limited The high gain only exists for a small range of input voltages If the input difference is too large, the output saturates Goes to the max positive or negative value possible Close to supply voltages M. Horowitz, J. Plummer, R. Howe 11

12 What Does This Do? V cc = 5 V v out v in v out V M. Horowitz, J. Plummer, R. Howe 12

13 Same Circuit Different Reference V dd = 2.5 V v out + 2 v in v out V M. Horowitz, J. Plummer, R. Howe 13

14 How To Get A Useful Amplifier The gain of the op amp is too high to make a useful amplifier We need to do something to make it useful We will use analog feedback to fix this problem Feedback makes the input the error between the value of the output, and the value you want the output to have. Let s see how to do this M. Horowitz, J. Plummer, R. Howe 14

15 Connect V out to V in- v out = A(V + V ) = A(v in v out ) + V cc = 5 V v in V cc = -5 V v out ( A +1)v out = Av in v out = A ( A +1) v in v in - M. Horowitz, J. Plummer, R. Howe 15

16 What Is Going On We solved the equation to find the answer But how does the op-amp get this answer? Think about what happens when the input increases in voltage From 0 V to 0.1 V Initially the output can t change There is capacitance at every node The op-amp thinks it needs to create a huge output voltage So it drives current into the output Which charges the capacitor Causing the output to increase This then decreases the input difference M. Horowitz, J. Plummer, R. Howe 16

17 Feedback in an Op-amp Circuit As the output rises The input difference decreases So A* DV in also decreases The system is stable when A* DV in is exactly equal to V out If A is large (10 6 ) for any V out Say in the range of ± 10V Dv in will be very, very small Can approximate that by saying Dv in will be driven to 0 Output will be set so v in+ v in M. Horowitz, J. Plummer, R. Howe 17

18 BUT This is only true if you connect the output feedback To the negative terminal of the amplifier What happens if you connect it to the positive terminal? M. Horowitz, J. Plummer, R. Howe 18

19 Ideal Op Amps The Two Golden Rules for circuits with ideal op-amps* No voltage difference between op-amp input terminals No current into op-amp inputs * when used in negative feedback amplifiers M. Horowitz, J. Plummer, R. Howe 19

20 USEFUL OP AMPS CIRCUITS M. Horowitz, J. Plummer, R. Howe 20

21 Approach To Solve All Op-amp Circuits First check to make sure the feedback is negative If not, STOP! Find the output voltage that makes the input difference 0 Assume V + = V - Find V out such that KCL holds We ll do some examples M. Horowitz, J. Plummer, R. Howe 21 E40M Lecture 19

22 Non-inverting Amplifier i p = 0 so v p = v s V + = V - so v n = v p = v s i 1 i 1 = i 2 so v o v s R 1 = v s R 2 i 2 v o 1 = v R s R 1 R 2 R v o = v 1 +R 2 s R 2 M. Horowitz, J. Plummer, R. Howe 22

23 Inverting Amplifier At node v n v n v s R s + v n v o R f + i n = 0 But v n =v p = 0 and i n = 0, so v s R s v o R f = 0 or v o = v s R f R s M. Horowitz, J. Plummer, R. Howe 23

24 Current-to-Voltage Converter i 2 i p = i n = 0 i 1 v n = v p = 0 i R So i R = 0 as well KCL at the v n node: i 1 = i s = i 2 = v o R f so v o = i s R f M. Horowitz, J. Plummer, R. Howe 24

25 OP AMP FILTERS M. Horowitz, J. Plummer, R. Howe 25

26 Adding Capacitors Sinusoidal voltage v o = v s Z f Z s = C f Suppose we add a capacitor in the feedback j 2πFC R f f R s We can treat this exactly as we did the earlier circuits by using impedances. Our earlier analysis showed vo = v s R f R s 1 Z s = R Z s f = 1 + j 2πFC R f f R = v f 1 s R s 1+ j 2πFR f C f M. Horowitz, J. Plummer, R. Howe 26

27 Sketching the Bode Plot 20 log 10 V o /V s 60 v o R f 1 = v s R V s 1+ j 2πFR f C s f V o F [Hz] -20 R s = 1 kω, R f = 100 kω, C f = 160 nf F c = 1/(2pR f C f ) = 10 Hz M. Horowitz, J. Plummer, R. Howe 27

28 Learning Objectives Understand how living things use electricity Understand what an op amp is: The inputs take no current The output is 10 6 times larger than the difference in input voltages The two Golden Rules of op amps in negative feedback Input currents are 0; V in- = V in+ Be able to use feedback to control the gain of the op amp For inverting and non-inverting amplifiers Understand op amp filters and differential amplifiers M. Horowitz, J. Plummer, R. Howe 28

29 More Examples M. Horowitz, J. Plummer, R. Howe 29

30 Summing Amplifier i 1 i 2 i 3 i p = i n = 0 v n = v p = 0 KCL at the summing point (or summing node): v i 1 + i 2 = i 3 so 1 + v 2 = v o R 1 R 2 R Output voltage is a scaled sum of the input voltages: v o = R f v R 1 + R f v 1 R 2 2 M. Horowitz, J. Plummer, R. Howe 30

31 A Subtracting (Difference) Amplifier? Take an inverting amplifier and put a 2 nd voltage on the other input? i 1 + i 2 = 0 so v n v 1 R s + v n v o R f = 0 v 1 v2 v n = v 2 so v 2 v 1 R s = v o v 2 R f v o R f = v 2 v 1 R s + v 2 R f v o = v 1 R f R s + v 2 R f +R s R s Not quite what we wanted. We d like v o a (v 1 v 2 ). M. Horowitz, J. Plummer, R. Howe 31

32 Differential Amplifier 1.0 v 1 v n R 1 = v n v o R 2 v 1 v 2 R 4 R 3 +R 4 R 1 = v 2 R 4 R 3 +R 4 v o R 2 v o = v 1 + v 2 R 4 + R 1 R 4 R 2 R 1 R 1 R 3 +R 4 R 2 R 3 +R 4 But if R 3 = R 1 and R 4 = R 2 v o = v 2 v 1 ( ) R 2 R 1 M. Horowitz, J. Plummer, R. Howe 32

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