Lecture 5: Using electronics to makemeasurements

• As physicists, we’re not really interested in electronics for itsown 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 onan object– strain is the degree to which an object changes its shape due to an

external force:

• We can convert this to an electrical signal by using a straingauge, a device that changes its resistance under strain

strainL

L

!=

• The measuring power of a strain gauge is quantified by thegauge factor:

• 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 120Ω and GF ≈ 2

/ /

strain /

R R R RGF

L L

! != =

!

• Let’s say we want to be sensitive to strains of the order10-5 or so using this gauge

• That means the change in resistance is:

• How can we measure such a small change in resistance?• One answer is the Wheatstone bridge:

5120 10 2 0.0024R R S GF

!" = # # = $# # = $

Analysis of Wheatstone bridge• Think of it as a set of two voltage dividers:

• So the difference in voltages is:

• Note that when ,

• The bridge is balanced when this happens

4

1 4

3

2 3

A o

B o

RV V

R R

RV V

R R

=+

=+

34

1 4 2 3

A B o

RRV V V

R R R R

! "# = #$ %

+ +& '

4 2 3 1R R R R=

0A BV V! =

• For simplicity, we’ll choose R2 = R3 = R4 = R• The bridge is then balanced when R1 = R• Let’s see what happens when R1 is close to, but not equal

to, R:

• 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

( )

1

1 11

2 / 2 1

11 1

2 2 4

A B o

o

o o

R R

R RV V V V

R R R R

VR

V VR R

!

!

!

! !

= +

" #$ = % = %& '+ + +( )

" #= %& '

+( )" #* % % = %& '( )

-6

0

can be ~10V

V

!

• So we need to measure a very small voltage difference,without being sensitive to fluctuations in Vo itself– sounds like a job for a differential amplifier!

• We might try a variation on the inverting amplifierdiscussed in the last lecture– This is called a “follower with gain”:

VA

VB

V-

V+

Vo

• 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:

– Current is the same through both resistors since no currentflows into the op-amp

– So we have two expressions for I that must be equal:

10k

10M

A B

B o

V I V V

V V V I

!

!

! " # = =

= = + " #

( ) ( )

10k 10M

10M 10M

10k 10k

B oA B

o B A B B A

V VV VI

V V V V V V

!!= =

" "

" "= ! + # !

" "

• This has high gain (1000), depending only on resistorvalues– that’s good!

• But the output is only approximately equal to thedifference between the inputs– there was still that VB term all by itself– means there is some common-mode gain as well

• This circuit is good enough for many differential-amplification uses– but not good enough for our strain gauge!

Instrumentation Amplifier• To do the job we want, we need a circuit like this (called

an instrumentation amplifier):

• We’ll break this up into pieces to see how it works

v1

v2

vo

VA

VB

• We’ll start with the two left op-amps:

• Each op-amp has equalvoltage at its inputs• so, voltage at top of R1

is VA, and at bottomit’s VB

• Current through R1 is:

• This current can’t go intoop-amps• must go across R2 and

R3

1

A BV V

IR

!=

• Putting that information together, we have:

• So this is also a differential amplifier– gain determined by resistors, which is good!

( )

( )

( )

1

1 2

2 3

2

1

1

3

2

1

2 3

1 2

1

1

A B

A

B

A A B

B A B

A B

V VI

R

v IR V

v IR V

Rv V V V

R

Rv V V V

R

R Rv v V V

R

!=

! =

+ =

= + !

= ! !

" #+! = ! +$ %

& '

• What about the common-mode gain of these two op-amps?• Using the equations on the previous slide, we have:

• If we choose R1 and R2 to be equal, the common-modeoutput is the same as the input– in other words, common-mode gain is one

• CMRR is already pretty good for this circuit– but not good enough for our strain gauge!

( )

( )

( )

2

1

1

3

2

1

2 3

1 2

1

A A B

B A B

A B A B

Rv V V V

R

Rv V V V

R

R Rv v V V V V

R

= + !

= ! !

" #!+ = + + !$ %

& '

• That brings us to the final op-amp in our instrumentationamplifier:

• We can solve for the currents:

• Applying the ideal op-amprules here, we have:

1 1 4 1 4

2 2 4 2 4

oV v I R v I R

V v I R I R

V V

!

+

+ !

= ! = +

= ! =

=

2

2

4

1 4 1 2 2 4 1 2

1 2

1

4

2

/ 2

2

2

vI

R

I R v v I R v v

v vI

R

=

= ! + = !

!=

• With this, we can solve for vo:

• 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 CMRR will require the use of high-precision resistors!

2

1 1 2 12

2o

vv v v v v

! "= # # = #$ %

& '

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’tamplify 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

ZF

• This looks like an inverting amplifier, so we know the gainis:

• The impedance is given by:

• So the gain is:

1

o F

i

V Z

V R= !

2

2 2 2

2

2

11 1 1 1

1

F c

F

i R Ci C

Z R Z R R

RZ

i R C

!!

!

+= + = + =

=+

2

1 2

1

1

o

i

V R

V R i R C!" #

= $ % &+' (