lecture 5: using electronics to make measurementsvarnes/teaching/405-505... · •the measuring...
TRANSCRIPT
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!" #
= $ % &+' (