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.

PotentiometerPotentiometers – a resistive device with a linear or rotary sliding

contacts.

Where L is the total length, Θ the total angular displacement, and Rp is the total resistance

xL

RxR p=)(

θθΘ

= pRR )(

Linear displacement:

Angular displacement:

turnsofnumber ntdisplaceme scale fullResolution =

Basic type of potentiometricdisplacement transducer

Contact position

Res

ista

nce

resolution

PotentiometerEx. 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

shaft

0.8 m

Rotating angle 250o

Required resolution better than 0.1 cm

Solution The shaft provides a conversion

cm/3.125or m/5.312m 8.0

250 ooo

=

A resolution of 0.1 cm at the panel translates intooo 0.3125cm/3.125cm .10 =×

Required resolution for the potentiometer. The given potentiometer actually has a resolution of

oo

0.300 turns1000

300=

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 factornoise (electrical contact) starting and remain torquemoment of inertia

• Self-heatingSelf-heating occurs because of the power dissipation in sensor, PD=I2RT

The increase in temperature from self-heating ∆T due to PD=I2RT is

DP Tδ= ∆

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.

θDPT =∆or

PotentiometerEx. A control potentiometer is rated as

150 Ω1 W (derate at 10 mW/oC above 65oC)30oC/W thermal resistanceCan it be used with 10 V supply at 80oC ambient temperature?

Solution The power dissipated by the potentiometer is

mW 667 150V) 10( 22

=Ω

==R

VP

The actual temperature of the potentiometer

C100)C/W30)(mW 667(C80 ooo =+=

+= θPTT ambientpot

The allowable power dissipation must be derated (decreased) by 10 mW for each degree above 65oC

mW 650

)C/mW10)(C65C100(mW 1000

)C/mW10)(C65(

=−−=

−−=ooo

ooTpotPP ratedallowed

Thus, the potentiometer can be used with the situation stated.

Potentiometer: Linearity

inp

inideal VRRV

RRRV 1

21

1 =+

=

inL

Lactual V

RRRRRV

21

1

////

+=

inLL

L VRRRRRR

RR

2211

1

++=

Instrument

Linearity: terminal base line

%100scale full

% idealactual ×−

=VVLinearity

Here, full scale output = Vin

%100Linearity% 1

2211

1 ×

−

++=

pLL

L

RR

RRRRRRRR

Vin

R2

R1 RL

Rp

V+

-

Potentiometer

R1 (Ω)

R2 (Ω)

Vdesired (V)

Vactual (V)

Linearity (%FSO)

0 1000 0.0 0.000 0.00 50 950 0.5 0.495 0.05

100 900 1.0 0.982 0.18 150 850 1.5 1.463 0.37 200 800 2.0 1.938 0.62 250 750 2.5 2.410 0.90 300 700 3.0 2.879 1.21 350 650 3.5 3.348 1.52 400 600 4.0 3.817 1.83 450 550 4.5 4.288 2.12 500 500 5.0 4.762 2.38 550 450 5.5 5.241 2.59 600 400 6.0 5.725 2.75 650 350 6.5 6.217 2.83 700 300 7.0 6.718 2.82 750 250 7.5 7.229 2.71 800 200 8.0 7.752 2.48 850 150 8.5 8.289 2.11 900 100 9.0 8.841 1.59 950 50 9.5 9.411 0.89

1000 0 10.0 10.000 0.00

Ex. Plot the transfer curve and determine endpoint linearity of a 1 kΩ potential driving a 5 kΩload, powered from a 10 V source.

Table: Nonlinearity of a potential caused by loading

V 10 1000

11 ×Ω

==RV

RRV in

pdesired

( ) ( ) V 10 5000 1000 1000 5000

5000

1111

1 ×Ω×−Ω+−Ω×+Ω×

Ω×=

RRRRRVactual

V 10scale full == inVVSolution here, we have

R1

0 200 400 600 800 1000

Out

put (

V)

0

2

4

6

8

10

Potentiometer

desired

actual

Nonlinearity error 2.83%

Linearity = 2.83 %FSO at 650 Ω

R1/Rp

0.0 .2 .4 .6 .8 1.0

Non

linea

rity

erro

r (%

FSO

)

0

1

2

3

4

5

6

7

Potentiometer

RL/Rp = 2

10

50100500

Plot of %linearity vs R1/Rp as a function of RL/Rp

Loading effects on the nonlinearity of a potentiometer

Max Error (%)

RpRL

10 1.263 5 2.742 1 14.590

0.5 29.410 0.1 147.900

0.05 296.100

Linear Variable Differential Transformer (LVDT)

LVDT – an electromechanical device that produce an electrical output proportional to the displacement of a separate movable core

M1

M2

L1

L2

L'2

e1

eo1

eo2

eo

|eo|

x

linear range

Circuit diagram for an LVDT. The secondary windings are normally connected in series opposition.

1 2o o oe e e= − Output voltages for a series-opposition connected LVDT.

movable core

Core at 0(Null Position)Core at A Core at B

Linear Variable Differential Transformer (LVDT)

AdvantageFrictionless (no physical contact between the movable core and

coil structure) Theoretical infinite resolution, resolution limited by the external

electronicsIsolation of exciting input and output (transformer action)

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)

Linear Variable Differential Transformer (LVDT)

Powersupply

Frequencygenerator LVDT Demodulator DC

amplifer

Referencesignal

Vout

Block diagram of the readout circuit for an LVDT

V

Phase sensitive demodulator based on (a) a half-wave rectifier and (b) full-wave rectifier

V

)()( 21 DCVDCVVout −=

fc

2fc

f

f

fc f

Linear Variable Differential Transformer (LVDT)

Assume the displacement varies with in the limited band.

A

B

AxB LPF Amp

A

B

AxB

Low pass filter

f

x

)(txx =

tfKtxS cπ2sin)(A =

tfaS cπ2sinB =

( )( )

24cos1)(

2sin)( 2AxB

tftaKxtftaKxS

c

c

ππ

−=

=

Phase sensitive demodulator based on carrier multiplication

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)

Incremental optical encoder Absolute optical encoder

01101001

photodetector LED

Optical Encoder

Output waveform

Code track on disk

Channel A wave form

Channel B wave form

Zero index

Code tracks on disk

Tachometer encoderoutput and code track

Quadrature encoderoutputs 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 D

L ∆L

Force

Definition of strain

Strain can be positive: tensile or negative strain: compressive. The practical unit is microstrain (µε), which is ε x 10-6.

Force

D

LL+∆L

Force

D-∆D When a bar is strained with a uniaxial force, a phenomenon known as Poisson strain causes the bar diameter, D to contract in transverse direction. The magnitude of this contraction is a material property indicated by its Poisson’s Ratio, ν/

/t D D

L Lενε

∆= − = −

∆ For most metal: ν ~ 0.3, polymer: ν > 0.3

σ - ε Curve, Brittle and Ductile materials

Elastic

Plastic

Stre

ss

Strain

σy

σTSM

P

O’O

necking

fracture

σf

σy –Yield strengthThe maximum stress

before the plastic deformation occurs.

σTS –tensile strengthThe maximum stress

atσ - ε curve

σf –fracture strengthThe 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)

The change of R can be affected by three quantities, and is given by

dR d dL dAR L A

ρρ

= + −

The electric resistance of a wire having length l, cross section A, and resistivity ρ is

lRA

ρ=

Consider a wire under an applied force F within the elastic limit, for a wire with a diameter D

2

4DA π

= 2 2dA dD dLvA D L

= = −

(1 2 )dR dL dR L

ρνρ

= + +

L

D

L+∆L

D-∆D

z

x

F

F

Therefore the geometry change results in the R change as follows

/ / /1 2/

dR R dR R dK vdL L

ρ ρε ε

= = = + +

Geometrical change

Strain dependence of resistivity

Derivation of Gage factor: continue

Sensitivity of a strain gage: Gage factor (K) is defined as

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

( )0 0 00

1 1dRR R dR R R KR

ε

= + = + = +

where R0 is the resistance where there is no applied stress

Strain GageEx 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

m 1046.2 402 013.0

20.2m .10

//

6−×=ΩΩ

×=

∆

=∆

∆∆

=

RR

KLL

LLRRK

εσ E=FromLL∆

=εwhere andAF

=σ

LLE

AF ∆

=Therefore ALLEF ∆

=or

24-24

22 m104

cm 10m 1cm 4 ×=×=A

N 10037.2

m104m 1.0

m1046.2mN107.20

3

4--6

210

×=

×××

××=F

Strain GageIdeally, 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)1( 00 TRR TT ∆+= α

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.

Ω=Ω=∆ 0.942) (240)C1)(C/003925.0( ooRSOLUTION

But the stress applied by the load in the previous example caused only a 0.0013 Ω change in the strain gage’s resistance.

5.72 013.0 942.0

=ΩΩ

=∆

∆

stress

Temp

RR

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

by ∆RR

R

R+∆R

R

E Vout+E

RRRRE

RRRVout )( ∆++

−+

=

ERR

R∆+

∆=

24

While ∆R is typically 0.01 Ω, so in practice R >> ∆R

ERRVout 4

∆≈

ε4

KEVout ≈

Temperature Compensation

R R+∆Rε+∆RT

R

E Vout+

R+∆RT

active

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) mV 13.0)240(4

)V 10)(013.0(4

=Ω

Ω=

∆= E

RRVout

The stress produces a rather small signal.

Temperature Compensation

(b) Using the voltage-divider give us

V 0 V 5 - V 5 ) 4.249 4.249(

)V 10)( 4.249() 240 240(

)V 10)( 240(

==Ω+Ω

Ω−

Ω+ΩΩ

=outV

The use of a dummy gage eliminates the effect of temperature.

(c) With both a temperature and a stress-induced resistance change,

mV 0.13 V 4.99987 - V 5 ) 013.0 4.249( 4.249(

)V 10)( 4.249() 240 240(

)V 10)( 240(

==Ω+Ω+Ω

Ω−

Ω+ΩΩ

=outV

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+∆R

E Vout+

R-∆R

R-∆R

R+∆R

tensioncompression

compressiontension

R R+∆R

R

E Vout+

R-∆R

tension

compression

ERRVout 4

∆≈

ERRVout 2

∆≈

Gage in tension(R +∆R)

Gage in compression

(R -∆R)

F

ERRVout

∆≈

Single active strain gage Four active strain gage (Full bridge)

Two active strain gage (Half bridge)

R R+∆Rε+∆RT

R

E Vout+

R+∆RT

active

dummy

Effect of Lead Wire Resistance

Effect of Lead Wire Resistance: Defection method

Vr

R1

R4

R2

R3=R0(1+x)

Rw1

Rw2

Rw3

Two-wire connections of Quarter-Bridge Circuit

Siemens or Three-wire connections of Quarter-Bridge Circuit

Vr

R1

R4

R2

R3=R0(1+x)

Rw1

Rw2

Vo +-

0 W 0 W 0

12 1 2 /

R R RR R R R R R

∆ ∆ ∆= = + +

4r

oV RV

R∆

≈

In strain gage application, we can write

But, with lead wire, R3 = R0+2RW

desensitivity

0 W 0 W 0

11 /

R R RR R R R R R

∆ ∆ ∆= = + +

desensitivityTherefore, with three-wire connection the reduction in sensitivity is smaller.

sensitivity

Effect of Lead Wire Resistance: Defection method

Vr

R1

R4

R2

R3=R0(1+x)

Rw1

Rw2

Rw3

Two-wire connections of Quarter-Bridge Circuit

Siemens or Three-wire connections of Quarter-Bridge Circuit

Vr

R1

R4

R2

R3=R0(1+x)

Rw1

Rw2

Vo +-

3 3 W

3 w 3 w 3 w

2

2

24 4 2 2 2

r ro

T T

T

R R RV R VVR R R R R R R

RR

ε

∆ ∆ ∆∆≈ = + + + + +

∆−

Temperature effect

Dummy gage

Dummy gage

The detrimental effect resulting from lead wires is loss of temperature compensation

The detrimental effect of long lead wires can be reduced by employing the three-wire system

+∆

−+

∆−

+∆

+

+∆

++

∆=

∆≈

Tw

w

Tw

Tw

w

Tww

rro

RRR

RRR

RRR

RRR

RRRV

RRVV

22

2

33

3

3

3

44ε

Load CellThe 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 CellEx 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

mV 20 V) mV/V)(10 2()capacity)( ratedat (output excitationout(max) === VV(b)

This is for 500 lb: V/lb 40 lb mV/500 20load scale-/fullout/lbout/lb µ=== VV

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) lb 25.0 lb) 500%)(05.0(tyNonlineari ±=±=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

lb 1 lb) (500 F)100)(F/%02.0(shift Zero oo +=+=So if the transducer experience significant changes in temperature, you must readjust zero.