Position Transducer
• Potentiometer
•Linear and angular displacement
•Inexpensive, easy to use
• Linear variable differential Transformer (LVDT)
•Linear and angular position
•Capable to measurement small displacement
• Encoder
•Angular position
•Rug and easily to interface with computer (digital
output in nature)
• etc.
Potentiometer
Potentiometers – a resistive device with a linear or rotary sliding
contacts.
Linear displacement:
R( x) =
Angular displacement: R(θ ) =
Rp
L
Rp
Θ
x
θ
Where L is the total length, Θ the
total angular displacement, and Rp is
the total resistance
Basic type of potentiometric
displacement transducer
full scale displacement
number of turns
Resistance
Resolution =
resolution
Contact position
Potentiometer
Ex. It is necessary to measure the position of a panel. It moves 0.8 m. Its position
must be know within 0.1 cm. Part of the mechanism which moves the panel is shaft
that rotate 250o when Panel is moved from one extreme to the other. A control
potential has been found which is rated at 300o full scale movement. It has been
1000 turns of wire. Can this be used?
panel
Rotating
angle 250o
Solution The shaft provides a conversion
250o
= 312.5o / m or 3.125o / cm
0.8 m
A resolution of 0.1 cm at the panel translates into
shaft
0.1 cm × 3.125o / cm = 0.3125o
0.8 m
Required resolution for the potentiometer. The
given potentiometer actually has a resolution of
Required resolution better than 0.1 cm
300o
= 0.300o
1000 turns
The potentiometer can detect a change of 0.3o, which is finer than the required
0.3125o. So the available potentiometer will work.
Potentiometer
Important Characteristics:
resolution, linearity, accuracy, output span,
power rating and derating factor
noise (electrical contact)
starting and remain torque
moment of inertia
• Self-heating
Self-heating occurs because of the power dissipation in sensor, PD=I2RT
The increase in temperature from self-heating ∆T due to PD=I2RT is
PD = δ∆T
or
∆T = PDθ
Where δ is heat dissipation factor (mW/K)
θ is thermal resistance (K/mW)
To minimize self-heating effect, the power dissipation must be limited.
Potentiometer
Ex. A control potentiometer is rated as
150 Ω
1 W (derate at 10 mW/oC above 65oC)
30oC/W thermal resistance
Can it be used with 10 V supply at 80oC ambient temperature?
Solution The power dissipated by the potentiometer is
V 2 (10 V) 2
=
= 667 mW
P=
R
150 Ω
The actual temperature of the potentiometer
Tpot = Tambient + Pθ
= 80o C + (667 mW)(30o C/W ) = 100o C
The allowable power dissipation must be derated (decreased) by 10 mW for each degree
above 65oC
o
o
Pallowed = Prated − (Tpot − 65 C)(10mW / C)
= 1000 mW − (100o C − 65o C)(10mW / o C)
= 650 mW
Thus, the potentiometer can be used with the situation stated.
Potentiometer: Linearity
Videal =
R2
Vin
Rp
R1
+
V
-
RL Instrument
R1
R
Vin = 1 Vin
R1 + R2
Rp
R1 // RL
Vin
R1 // RL + R2
R1 RL
Vin
=
R1 RL + R1 R2 + R2 RL
Vactual =
Linearity: terminal base line
% Linearity =
Vactual − Videal
×100%
full scale
Here, full scale output = Vin


R
R
R
L
1
1
 × 100%
%Linearity = 
−
RR +RR +R R
R p 
1 2
2 L
 1 L
Potentiometer
Ex. Plot the transfer curve and determine endpoint linearity of a 1 kΩ potential driving a 5 kΩ
load, powered from a 10 V source.
R1
R
Vfull scale = Vin = 10 V
× 10 V
Solution here, we have Vdesired = 1 Vin =
Rp
1000 Ω
Vactual =
R1 × 5000 Ω
× 10 V
R1 × 5000 Ω + R1 × (1000 Ω − R1 ) + (1000 Ω − R1 )× 5000 Ω
Table: Nonlinearity of a potential caused by loading
R1
(Ω)
0
50
100
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
R2
(Ω)
1000
950
900
850
800
750
700
650
600
550
500
450
400
350
300
250
200
150
100
50
0
Vdesired
(V)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
Vactual
(V)
0.000
0.495
0.982
1.463
1.938
2.410
2.879
3.348
3.817
4.288
4.762
5.241
5.725
6.217
6.718
7.229
7.752
8.289
8.841
9.411
10.000
Linearity
(%FSO)
0.00
0.05
0.18
0.37
0.62
0.90
1.21
1.52
1.83
2.12
2.38
2.59
2.75
2.83
2.82
2.71
2.48
2.11
1.59
0.89
0.00
Potentiometer
10
Output (V)
8
6
desired
Nonlinearity
error 2.83%
4
actual
2
0
0
200
400
600
800
R1
Linearity = 2.83 %FSO at 650 Ω
1000
Potentiometer
Plot of %linearity vs R1/Rp as a function of RL/Rp
Nonlinearity error (%FSO)
7
Loading effects on the
nonlinearity of a potentiometer
RL/Rp = 2
6
5
4
10
3
2
50
100
500
1
0
0.0
.2
.4
.6
R1/Rp
.8
1.0
Max
Error (%)
RL
Rp
10
5
1
0.5
0.1
0.05
1.263
2.742
14.590
29.410
147.900
296.100
Linear Variable Differential Transformer (LVDT)
LVDT – an electromechanical device that produce an electrical
output proportional to the displacement of a separate movable
core
M1
e1
|eo|
L2
eo1
L'2
eo2
eo
L1
x
M2
linear range
movable core
Circuit diagram for an LVDT. The secondary
windings are normally connected in series
opposition.
e = e −e
o
o1
o2
Core at A
Core at 0
(Null Position)
Core at B
Output voltages for a series-opposition
connected LVDT.
Linear Variable Differential Transformer (LVDT)
The output voltage is linear function of core position as long as
the motion of the core is within the operating of the LVDT. The
direction of motion can be determined from the phase of the
output voltage relative to the input voltage (primary coil)
Advantage
™ Frictionless (no physical contact between the movable core and
coil structure)
™ Theoretical infinite resolution, resolution limited by the external
electronics
™ Isolation of exciting input and output (transformer action)
Linear Variable Differential Transformer (LVDT)
Power
supply
Frequency
generator
LVDT
Demodulator
DC
amplifer
Vout
Reference
signal
Block diagram of the readout circuit for an LVDT
V
V
Vout = V1 ( DC ) − V2 ( DC )
Phase sensitive demodulator based on (a) a half-wave rectifier and (b) full-wave rectifier
Linear Variable Differential Transformer (LVDT)
B
AxB
LPF
Amp
Assume the displacement
varies with in the limited
band.
x
A
f
Phase sensitive demodulator based on
carrier multiplication
x = x(t )
A
S A = x(t ) K sin 2πf c t
S B = a sin 2πf c t
S AxB = aKx (t )(sin 2πf c t )
B
2
=
aKx(t )(1 − cos 4πf c t )
2
fc
f
fc
f
Low pass filter
AxB
2fc
f
Optical Encoder
• Incremental optical encoder provides a pulse each time the shaft
has rotated a defined distance (fast application: velocity)
• Absolute optical encoder provides a output code (Gray, binary or
BCD) corresponding to the angular position. These codes are
derived from independent tracks on the encoder disk
corresponding to photodetectors. (slow application)
0
1
1
0
1
0
0
1
photodetector
Incremental optical encoder
LED
Absolute optical encoder
Optical Encoder
Output waveform
Code track on disk
Tachometer encoder
output and code track
Channel A wave form
Channel B wave form
Zero index
Code tracks on disk
Quadrature encoder
outputs and code tracks
Incremental optical encoder
Force Transducer: Strain Gages
Strain (ε) is defined as a fractional change in length of a body due to an applied
force.
Force
Force
D
L
∆L
Definition of strain
Strain can be positive: tensile or negative strain: compressive. The practical unit
is microstrain (µε), which is ε x 10-6.
When a bar is strained with a
D-∆D
D
uniaxial force, a phenomenon
known as Poisson strain causes
Force
Force
the bar diameter, D to contract in
transverse direction.
The
L
magnitude of this contraction is a
L+∆L
material property indicated by its
Poisson’s Ratio, ν
εt
∆D / D
ν =− =−
For most metal: ν ~ 0.3, polymer: ν > 0.3
ε
∆L / L
σ - ε Curve, Brittle and Ductile materials
M
σTS
P
Elasti
c
Plastic
necking
Stress
σy
σf
fracture
O
O’
Strain
σy –Yield strength
The maximum stress
before the plastic
deformation occurs.
σTS –tensile strength
The maximum stress
at
σ - ε curve
σf –fracture strength
The stress at instance
of fracture
Typical σ - ε curve from a tensile test on a ductile
polycrystalline metal (e.g. aluminum alloys, brasses,
bronzes, nickel etc.)
Strain Gages: Derivation of Gage
Factor
Strain gage is a device whose electrical resistance varies in proportion to the
amount to of strain the device. (discovery by Lord Kelvin in 1856)
F
The electric resistance of a wire having length l, cross
section A, and resistivity ρ is
l
R=ρ
A
The change of R can be affected by three quantities, and
is given by
dR d ρ dL dA
=
+
−
ρ
R
L
A
D-∆D
Consider a wire under an applied force F within the elastic
limit, for a wire with a diameter D
L
L+∆L
D
dA
dD
dL
=2
= −2v
A
D
L
4
Therefore the geometry change results in the R change as
follows
dR
dL d ρ
= (1 + 2ν ) +
ρ
R
L
A=
z
x
F
π D2
Derivation of Gage factor: continue
Sensitivity of a strain gage: Gage factor (K) is defined
as
dR / R dR / R
dρ / ρ
=
= 1 + 2v +
K=
ε
ε
dL / L
Geometrical
change
Strain dependence
of resistivity
Note:The change in resistivity as a result of a mechanical strain is
called piezoresistive effect.
Therefore, the resistance of the strain gage under an applied force is
 dR 
R = R0 + dR = R0 1 +
 = R0 (1 + K ε )
R0 

where R0 is the resistance where there is no applied stress
Strain Gage
Ex A strain gage is bonded to a steel beam which is 10.00 cm long and has a cross-sectional
area of 4.00 cm2. Young’s modulus of elasticity for steel is 20.7x1010 N/m2. The strain gage has
a nominal (unstrained) resistance of 240 Ω and a gage factor 2.20. When a load is applied, the
gage’s resistance changes by 0.013 Ω. Calculate the change in length of steel beam and the
amount of force applied to the beam.
SOLUTION
From
Therefore
∆R / R
∆L / L
L  ∆R  0.1 m 0.013 Ω
∆L = 
×
= 2.46 ×10 −6 m
=
K  R  2.20 240 Ω
F
∆L
and σ =
where ε =
σ = Eε
A
L
K=
F
∆L
=E
A
L
or
F=E
∆L
A
L
1 m2
A = 4 cm × 4
= 4 × 10-4 m 2
2
10 cm
2
N 2.46 × 10-6 m
F = 20.7 ×10
×
× 4 × 10-4 m
2
m
0 .1 m
= 2.037 × 103 N
10
Strain Gage
Ideally, we would prefer the strain gage to change resistance only in respect
to the stress-induced strain in the test specimen, but the resistivity and the
strain sensitivity of all known strain sensitive materials vary with
temperature.
The resistance of a conductor at a given temperature, T is
RT = RT 0 (1 + α 0 ∆T )
Where RT0 is the resistance at the reference temperature T0 , α0 is the
temperature coefficient and ∆T is the change in temperature from T0
Ex Calculate the change in resistance caused by a 1oC change in temperature for the strain
gage in the previous example. The temperature coefficient α0 for most metals is 0.003925/oC.
SOLUTION
∆R = (0.003925 / o C)(1o C)(240 Ω) = 0.942 Ω
But the stress applied by the load in the previous example caused only a 0.0013 Ω change
in the strain gage’s resistance.
∆RTemp 0.942 Ω
=
= 72.5
∆Rstress 0.013 Ω
Therefore, it is necessary to compensate for temperature effect on the strain gage
Wheatstone Bridge: Deflection Method
Wheatstone bridges are often used in the deflection mode: This
method measures the voltage difference between both dividers or the
current through a detector bridging them.
With no stress ∆R = 0, all four resistors are equal, so Vout = 0.
When stress is applied, the strain gage changes its resistance
R
R+∆R
E
+
R
R
by ∆R
Vout
R
R
E−
E
R+R
R + ( R + ∆R )
∆R
=
E
4 R + 2∆R
Vout =
While ∆R is typically 0.01 Ω, so in practice R >> ∆R
Vout ≈
∆R
E
4R
Vout ≈
KE
ε
4
Temperature Compensation
R
E
R+∆Rε+∆RT
active
+
R
Vout
R+∆RT
dummy
Placement of active and dummy gages for
temperature compensation
Ex Determine Vout given that R0 = 240 Ω, E = 10 V and
(a) Stress cause the upper resistor (the active gage) on the right to increase by 0.013 Ω.
(b) Temperature causes both resistors (active and dummy) on the right to increase by 9.4 Ω.
(c) Stress cause the active gage to increase by 0.013 Ω and temperature both resistors to
increase by 9.4 Ω.
SOLUTION (a)
Vout =
∆R
(0.013Ω)(10 V )
E=
= 0.13 mV
4R
4(240Ω)
The stress produces a rather small signal.
Temperature Compensation
(b) Using the voltage-divider give us
(240 Ω)(10 V)
(249.4 Ω)(10 V)
−
(240 Ω + 240 Ω) (249.4 Ω + 249.4 Ω)
=5V -5V =0V
Vout =
The use of a dummy gage eliminates the effect of temperature.
(c) With both a temperature and a stress-induced resistance change,
(240 Ω)(10 V )
(249.4 Ω)(10 V)
−
(240 Ω + 240 Ω) (249.4 Ω + (249.4 Ω + 0.013 Ω)
= 5 V - 4.99987 V = 0.13 mV
Vout =
So even in the presence of both stress and temperature resistance changes, the use of
a dummy gage eliminates the effect of the temperature change
Strain gage arrangements
R+∆Rε+∆RT
R
active
E
+
Vout
Vout
∆R
≈
E
4R
+
R+∆R
Single active strain gage
Vout
Vout ≈
R-∆R
compression
Four active strain gage (Full bridge)
R
F
R+∆R
tension
E
Gage in
compression
(R -∆R)
tension
tension
dummy
Gage in
tension
(R +∆R)
R+∆R
E
R+∆RT
R
R-∆R
compression
+
R
Vout
R-∆R
compression
Two active strain gage (Half bridge)
Vout ≈
∆R
E
2R
∆R
E
R
Effect of Lead Wire Resistance
Effect of Lead Wire Resistance: Defection method
sensitivity
In strain gage application, we can write
R2
R1
- Vo +
Vr
Vo ≈
Rw1
R4
Rw2
But, with lead wire, R3 = R0+2RW
R3=R0(1+x)
Two-wire connections of Quarter-Bridge
Circuit
R1
Vr
R2
Vr ∆R
4 R

1
∆R
∆R
∆R 
=
=


R R0 + 2RW R0  1 + 2RW / R0 
desensitivity

∆R
∆R
∆R 
1
=
=


R R0 + RW R0  1 + RW / R0 
Rw1
desensitivity
Rw2
Therefore, with three-wire
Rw3
connection the reduction in
sensitivity is smaller.
Siemens or Three-wire connections of Quarter-Bridge
Circuit
R4
R3=R0(1+x)
Effect of Lead Wire Resistance: Defection method
Temperature effect
R2 Dummy gage
R1
- Vo +
Vr
R4
Rw1
Rw2
Two-wire connections of Quarter-Bridge
Circuit
R1
Vr
Vo ≈
R2 Dummy gage
Rw1
Rw2
∆R3
2∆RW
Vr ∆R Vr  ∆R3
= 
+
+
4 R 4  R3 + 2Rw ε R3 + 2Rw T R3 + 2Rw
∆R 
− 2 
R2 T 
T
R3=R0(1+x)
The detrimental effect resulting from
lead wires is loss of temperature
compensation
∆R3
Vr ∆R Vr  ∆R3
=
+
Vo ≈
4 R
4  R3 + Rw ε R3 + Rw
−
∆R2
R2 + Rw
−
T
+
T
∆Rw
R2 + Rw
∆Rw
R3 + Rw
T



T 
The detrimental effect of long lead
R3=R0(1+x) wires can be reduced by employing
R4
Rw3
the three-wire system
Siemens or Three-wire connections of Quarter-Bridge
Circuit
Load Cell
The performance of strain gage depends heavily on the installation, the
proper alignment with an applied force, gage adhesion with the specimen etc.
Load cell – a force transducer consists of a beam with properly
mounted strain gages (usually four)
Load cell specifications
Load Cell
Ex A GSE5352 load cell has a full-scale rating of 500 lb.
(a) What is the recommended excitation voltage?
(b) Using that excitation, what is the output voltage per pound?
(c) What is the nonlinearity in pounds?
(d) What is the zero shift (in pounds) if the temperature varies across its rated range?
SOLUTION
(a) The manufacturer recommends a 10-V dc excitation
(b)
Vout(max) = (output at rated capacity)(Vexcitation ) = (2 mV/V)(10 V) = 20 mV
This is for 500 lb: Vout/lb = Vout/lb /full - scale load = 20 mV/500 lb = 40 µV/lb
This means that your electronics must be able to clearly and accurately amplify
differential signals of 40 µV or less if you expect to resolve and display 1-lb increments.
(c)
Nonlinearity = (±0.05%)(500 lb) = ± 0.25 lb
So no matter how good your electronics, there will be a ±0.25-lb uncertainty.
(d) The temperature range is +25 to +125oF. This represents a 100oF shift. This zero shift
with temperature is +0.002% FS/oF
Zero shift = (+0.02% / o F)(100o F) (500 lb) = + 1 lb
So if the transducer experience significant changes in temperature, you must readjust zero.