# Lecture 5: Using electronics to make measurements

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1 Lecture 5: Using electronics to make measurements As physicists, we re not really interested in electronics for its own sake We want to use it to measure something often, something too small to be directly sensed As an example, we ll assume we want to measure the strain on an object strain is the degree to which an object changes its shape due to an external force:! L strain = L We can convert this to an electrical signal by using a strain gauge, a device that changes its resistance under strain

2 The measuring power of a strain gauge is quantified by the gauge factor:! /! / GF = = strain! L / L A higher gauge factor makes a better gauge One example is a metallic wire bonded to a piece of material: wire become thinner and longer as material stretches resistance increases Typically these have resistance of 20Ω and GF 2

3 Let s say we want to be sensitive to strains of the order 0-5 or so using this gauge That means the change in resistance is:! 5 " = # S # GF = 20\$ # 0 # 2 = \$ How can we measure such a small change in resistance? One answer is the Wheatstone bridge:

4 Analysis of Wheatstone bridge Think of it as a set of two voltage dividers: 4 VA = Vo + So the difference in voltages is: Note that when =, V B = V # V! = # " V & ' 4 3 A B \$ o % V o V A! V = 0 B The bridge is balanced when this happens

5 For simplicity, we ll choose 2 = 3 = 4 = The bridge is then balanced when = Let s see what happens when is close to, but not equal to, : = +! " # \$ V = V % V = & % ' V ( +! + + ) A B o " # = & % ' Vo 2 (! /( 2) + ) "! #! * & % % ' Vo = % Vo 2 ( 2 ) 4 So we see that the voltage difference varies linearly with δ But since the change in resistance is small, so is the change in voltage! V -6 can be ~0 V 0

6 So we need to measure a very small voltage difference, without being sensitive to fluctuations in V o itself sounds like a job for a differential amplifier! We might try a variation on the inverting amplifier discussed in the last lecture This is called a follower with gain : V A V B V - V + V o

7 This has some nice features for example, the input impedance is very high (equal to the internal resistance of the op-amp) Here s how it works as an amplifier: The op-amp makes sure that V + = V - We also know that: V! I " 0k# = V = V A V! = VB = Vo + I " 0M# Current is the same through both resistors since no current flows into the op-amp So we have two expressions for I that must be equal: VA! VB VB! Vo I = = 0k" 0M" 0M" 0M" V = V! V + V # V! V 0k" 0k"! ( ) ( ) o B A B B A B

8 This has high gain (000), depending only on resistor values that s good! But the output is only approximately equal to the difference between the inputs there was still that V B term all by itself means there is some common-mode gain as well This circuit is good enough for many differentialamplification uses but not good enough for our strain gauge!

9 Instrumentation Amplifier To do the job we want, we need a circuit like this (called an instrumentation amplifier): V A v v o V B v 2 We ll break this up into pieces to see how it works

10 We ll start with the two left op-amps: Each op-amp has equal voltage at its inputs so, voltage at top of is V A, and at bottom it s V B Current through is: I V = A! V B This current can t go into op-amps must go across 2 and 3

11 Putting that information together, we have: I = V A! V ( ) 2 A A B ( ) 3 B A B " # v! v2 = ( VA! VB ) \$ + % & ' So this is also a differential amplifier gain determined by resistors, which is good! B v! I = V v + I = V v = V + V! V v = V! V! V A B

12 What about the common-mode gain of these two op-amps? Using the equations on the previous slide, we have: v = V + V! V 2 ( ) 2 A A B v = V! V! V ( ) 3 B A B "! # v + v2 = V + V + V! V & ' If we choose and 2 to be equal, the common-mode output is the same as the input in other words, common-mode gain is one CM is already pretty good for this circuit but not good enough for our strain gauge! ( ) 2 3 A B \$ A B %

13 That brings us to the final op-amp in our instrumentation amplifier: Applying the ideal op-amp rules here, we have: V = v! I = v + I 4 o 4 V = v! I = I V! + = V +! We can solve for the currents: v2 I2 = 2 4 I = v! v + I = v! v I v! v = / 2

14 With this, we can solve for v o :! v " vo = v # 2\$ v # % = v # v & 2 ' 2 2 This part of the circuit is a subtractor Gets rid of the common-mode signal that makes it through the first set of op-amps Note that exact subtraction requires all four resistors to have the same value Large CM will require the use of high-precision resistors!

15 Active low-pass filter We studied filters in the first lecture, and built them in the first lab Those were passive filters they could transmit or suppress a signal, but they couldn t amplify it With an op-amp, we can build an active filter an amplifier where the gain depends on the frequency of the signal Here s how it might look: Impedance Z F

16 This looks like an inverting amplifier, so we know the gain is: Vo Z F =! V The impedance is given by: i + i! 2C = + = + i! C = Z Z Z F F 2 c = + i! C So the gain is: Vo " 2 # = \$ Vi % + i! 2C & ' ( 2

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