Op Amps M. Horowitz, J. Plummer, R. Howe 1 Reading A&L: - PowerPoint PPT Presentation

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

Reading A&L: Chapter 15, pp. 863-866. Reader, Chapter 8 • Noninverting Amp – http://www.electronics-tutorials.ws/opamp/opamp_3.html • Inverting Amp – http://www.electronics-tutorials.ws/opamp/opamp_2.html • Summing Amp – http://www.electronics-tutorials.ws/opamp/opamp_4.html M. Horowitz, J. Plummer, R. Howe 2

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

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

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

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

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

Moving the Reference V dd = 2.5 V V dd = 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

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

Op Amp Behavior • Relationship between output voltage and input voltage: ( ) ( ) = A v p − v n v o = A v + − v − 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

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

What Does This Do? v out V cc = 5 V 5 + 4 v in + + v out - 3 - 2 + 2.5 V - 1 - 1 2 3 5 4 M. Horowitz, J. Plummer, R. Howe 12

Same Circuit Different Reference v out V dd = 2.5 V + 2 v in + 1 + - - 0 v out + -1 2.5 V - -2 - -2 -1 0 2 1 M. Horowitz, J. Plummer, R. Howe 13

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

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

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

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

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

Ideal Op Amps The Two Golden Rules for circuits with ideal op-amps* No voltage difference between 1. op-amp input terminals 2. No current into op-amp inputs * when used in negative feedback amplifiers M. Horowitz, J. Plummer, R. Howe 19

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

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

Non-inverting Amplifier • i p = 0 so v p = v s • V + = V - so v n = v p = v s i 1 = i 2 so v o − v s = v s i 1 R 1 R 2 ⎛ ⎞ ∴ v o 1 + 1 = v s i 2 ⎜ ⎟ ⎜ ⎟ R 1 R 1 R 2 ⎝ ⎠ ⎛ ⎞ R 1 + R 2 ∴ v o = v s ⎜ ⎟ ⎜ ⎟ R 2 ⎝ ⎠ M. Horowitz, J. Plummer, R. Howe 22

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

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 so v o = − i s R f R f M. Horowitz, J. Plummer, R. Howe 24

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

Adding Capacitors C f • Suppose we add a capacitor in the feedback • We can treat this exactly as we did the earlier circuits by using impedances. • Our earlier analysis showed R f v o = − v s R s 1 Z f = Z s = R s 1 + j ∗ 2 π FC f 1 R f Sinusoidal voltage 1 + j ∗ 2 π FC f ⎛ ⎞ Z f R f R f 1 ∴ v o = − v s = − v s ⎜ ⎟ = − ⎜ ⎟ Z s R s R s 1 + j ∗ 2 π FR f C f ⎝ ⎠ M. Horowitz, J. Plummer, R. Howe 26

Sketching the Bode Plot 20 log 10 | V o /V s | ⎛ ⎞ V o v o − R f 1 = ⎯ ⎜ ⎟ ⎜ ⎟ v s R s 1 + j ∗ 2 π FR f C f 60 V s ⎝ ⎠ 40 20 0 0.1 1 10 F [Hz] 100 10 3 10 4 10 5 -20 F c = 1/(2 p R f C f ) = 10 Hz R s = 1 kΩ, R f = 100 kΩ, C f = 160 nF M. Horowitz, J. Plummer, R. Howe 27

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

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

Summing Amplifier i i 1 • i p = i n = 0 3 • v n = v p = 0 i 2 KCL at the summing point (or summing node): i 1 + i 2 = i 3 so v 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 1 + R f v 2 ⎜ ⎟ ⎜ ⎟ R 1 R 2 ⎝ ⎠ M. Horowitz, J. Plummer, R. Howe 30

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 + v n − v o = 0 R s R f v n = v 2 so v 2 − v 1 = v o − v 2 R s R f v 1 ∴ v o = v 2 − v 1 + v 2 R f R s R f v2 R f R f + R s ∴ v o = − v 1 + v 2 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

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