E40M. Instrumentation Amps and Noise. M. Horowitz, J. Plummer, R. Howe 1

Transcription

1 E40M Instrumentation Amps and Noise M. Horowitz, J. Plummer, R. Howe 1

2 ECG Lab - 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 set of lecture notes M. Horowitz, J. Plummer, R. Howe 2

3 STRUMENTATION AMP M. Horowitz, J. Plummer, R. Howe 3

4 Starting Point: Differential Amplifier 1.0 If R 3 = R 1 and R 4 = R 2 v o = v 2 v 1 ( ) R 2 R 1 This amplifier requires that the input voltage sources provide input currents (i 1 and i 3 are not zero) not OK for the ECG project or a general-purpose instrumentation amplifier. M. Horowitz, J. Plummer, R. Howe 4

5 We Need A Differential Amplifier With No Input Current Want really want a differential amplifier with no input current Make sure the input isolation resistance isn t a problem This is a common situation for many types of instruments There is a special part for this situation Called instrumentation amplifier It can be thought of as 3 amplifiers Two non-inverting amplifiers (so there is no input current) One differential amplifiers These parts are built to match very well So it is better than building the circuit yourself M. Horowitz, J. Plummer, R. Howe 5

6 Instrumentation Amp (Used in ECG Lab) Kind of looks like two noninverting amplifiers But they are connected together in a funny way Fortunately the IA can be solved using the Golden Rules: Write KCL for - input of the op amp Find the output voltage that satisfies KCL when the voltage at the - input is equal to the voltage on the + input M. Horowitz, J. Plummer, R. Howe 6

7 Start with KCL at Inverting Input of Op Amp #1 At node v 1 and assuming no op amp input current, we have i 1 = i 2 + i 3 v 2 v 2 v 1 R G = v 1 v o1 10kΩ + v 1 v ref 40kΩ i 1 Since v = v 1 and v + = v 2 v 1 i 3 i 2 v o1 v + v R G = v v o1 10kΩ + v + v ref 40kΩ v ref M. Horowitz, J. Plummer, R. Howe 7

8 Now Find V o1 -- the Output Voltage of Op Amp #1 v + v R G = v v o1 10kΩ + v + v ref 40kΩ v 2 i 1 i 2 v o1 10kΩ = v 10kΩ + v + v ref 40kΩ v + v R G v 1 i 3 v o1 v ref ( ) v o1 = 5v 4 v 10kΩ v + ref 4 v R G M. Horowitz, J. Plummer, R. Howe 8

9 Next: KCL at Inverting Input of Op Amp #2 At node v 2 and assuming no op amp input i, we have i 4 = i 1 + i 5 v 2 v o v 2 40kΩ = v 2 v 1 R G + v 2 v o1 10kΩ i 1 i 5 i 4 Since v = v 1 and v + = v 2 v 1 i 2 v o v + 40kΩ = v + v R G + v + v o1 10kΩ i 3 v o1 v ref M. Horowitz, J. Plummer, R. Howe 9

10 Step n+1: Solve for V o v o v + 40kΩ = v + v R G + v + v o1 10kΩ v 2 i 1 i 5 i 4 v o 40kΩ = v + 40kΩ + v + v R G + v + v o1 10kΩ i 2 v 1 i 3 v o1 v o = 5v kΩ( v + v ) R G 4v o1 v ref M. Horowitz, J. Plummer, R. Howe 10

11 The Finale: Combining The Results ( ) v o1 = 5v 4 v 10kΩ v + ref 4 v R G i 1 v 2 i5 i 4 v o = 5v kΩ( v + v ) R G 4v o1 i 2 v 1 v o = 80kΩ + 5 R v + v G ( ) + v ref i 3 v o1 This confirms the gain expression v ref given in the 1NA126 data sheet! (using v ref = 0). M. Horowitz, J. Plummer, R. Howe 11

12 Another Instrumentation Amplifier (Bonus) (we are not using this architecture) Most instrumentation amplifiers are actually built with 3 op amps. The analysis is quite similar to the past few pages v ref M. Horowitz, J. Plummer, R. Howe 12

13 Another Instrumentation Amplifier (Bonus) Consider a simplified case in which all resistors are the same (except R gain ) and v ref = 0. The analysis is quite similar to the past few pages. We won t cover this in class try it yourself, you should be able to analyze this! Try it to test your understanding. v + v o v M. Horowitz, J. Plummer, R. Howe 13

14 Front End of Instrumentation Amplifier (Bonus) v + v 1 v o1 i 1 G.R. #1: v + = v 1 and v = v 2 KCL: i 1 = i 2 = i 3 + v o1 v = v + v = v v o2 R R gain R i 2 v + o v o1 R = v R + v v R gain + R gain v 2 i 3 v v o2 v o1 = R R gain ( v + v ) + v + Similarly, v o2 = R ( v + R v ) + v gain M. Horowitz, J. Plummer, R. Howe 14

15 Back End of Instrumentation Amplifier (Bonus) v + G.R. #2: i 4 = i 5 and i 6 = i 7 v o1 v 3 R = v 3 v o R v 1 so that v o = 2v 3 v o1 v o1 i 4 i 5 v 3 v o2 v 4 R = v 4 R so that v 4 = v o2 2 = v 3 v 4 v o Combining, v o = v o2 v o1 Using the results from the previous page, v o = v o2 v o1 = v o = v v + v v 2 R ( v + R v ) + v gain ( ) 2 R +1 R gain M. Horowitz, J. Plummer, R. Howe 15 R R gain v o2 i 6 i ( 7 v + v ) v +

16 NOISE M. Horowitz, J. Plummer, R. Howe 16

17 ECG Measurement Need to measure the difference between L1 and L2 We think the circuit looks like M. Horowitz, J. Plummer, R. Howe 17

18 The Circuit Really Looks Like This: There are many unwanted signals coupling into our circuit Both capacitive (stray electric fields) and inductive (magnetic fields) These signals can be larger than what we want to measure! How to prevent them from obscuring our signal? M. Horowitz, J. Plummer, R. Howe 18

19 Noise Protection For Wires Shield the signal (literally cover it with metal) Try to make the noise common mode Twist wires to each other M. Horowitz, J. Plummer, R. Howe 19

20 Model of the Capacitive Noise (if it is common to both wires) V L1 V L2 The voltage at the two outputs will depend on ECG and Noise But if the capacitors and resistors are the same (V L1 - V L2 ) will not depend on noise This is only true if the capacitance on both wires is identical Which means we need a balanced differential amplifier M. Horowitz, J. Plummer, R. Howe 20

21 Balanced Amplifier This is a completely differential system Good for reducing noise coupling M. Horowitz, J. Plummer, R. Howe 21

22 New Problem in Our Balanced Amplifier v 1 v 2 What sets the voltage at v 1, v 2? V ECG only sets v 1 - v 2 They are not referenced to our chip s reference (Gnd)! Chip won t work unless inputs are between +/- supply voltage. M. Horowitz, J. Plummer, R. Howe 22

23 The Reason for the Third Wire Need to measure the difference between L1 and L2 L3 is used to set the common-mode of the person M. Horowitz, J. Plummer, R. Howe 23

24 Why Does the ECG Circuit Look Like This? M. Horowitz, J. Plummer, R. Howe 24

25 Noise: Skin Voltage A voltage forms when metal contacts skin The size of the voltage depends on the skin condition This means if the conditions at the two electrodes differ You can generate a voltage This voltage will change very slowly with time Log f M. Horowitz, J. Plummer, R. Howe 25

26 Why Does the ECG Circuit Look Like This? M. Horowitz, J. Plummer, R. Howe 26

27 Noise: 60Hz Wall Voltage The main capacitive noise comes from AC power 120 to 240V, 60 Hz This signal can be quite large (Volts!) 1000x your signal Differential circuit cancels most of it out But some will still get through due to imperfect symmetry Log f M. Horowitz, J. Plummer, R. Howe 27

28 Why Does the ECG Circuit Look Like This? M. Horowitz, J. Plummer, R. Howe 28

29 Learning Objectives Understand how an instrumentation amplifier works And how to set its gain through resistor selection Understand what noise is Other electrical signals that you don t want on your wires And how to minimize their effects on your circuit through differential amplifiers and filtering Understand the design philosophy behind our E40M ECG circuit M. Horowitz, J. Plummer, R. Howe 29