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Chapter 11: DC Bridges 0908341 Measurements & Instrumentation

Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif Page 1 of 23

Chapter 11DC Bridges

(Revision 3.0, 25/4/2008)

1. Introduct ion

As discussed in Chapter 1, SCE (signal conditioning element) or VCE(variable conversion element) is a typical component of a measurementsystem. It converts the signal received from the sensor from an unsuitableformat to a suitable format. A bridge is one of the most widely used forms ofVCE/SCEs.

Bridges can have either a.c. or d.c. excitation voltages. D.C. voltages areneeded for resistance measurements while a.c. voltages are needed forcapacitance and inductance measurements. The bridge can be used either innull type mode or deflection type mode.

2. Wheatstone Bridge

The Wheatstone bridge is a type of d.c. bridge that is used for precisionmeasurement of resistance from approximately 1 ohm to the low mega-ohmrange. A typical Wheatstone null type bridge is shown in Figure 1 below.

An excitation voltage source is used to operate the bridge (Vi). Agalvanometer is used to connect the mid-points of the right hand side voltagedivider (made up ofR2and R4) and the left hand side voltage divider (madeup ofR1and R3). The galvanometer connection between the two mid-pointsform a bridge between the two sides, hence the name of the device.

When used as a null type device, it can produce an accuratemeasurement of resistance. When used in the deflection type mode, it canproduce a change in an output voltage that is proportional to a change in thesensor under question (a resistor). The change in the resistance of the sensoris representative of a change in the value of an external variable (e.g., stress,force, temperature). The null type mode is more accurate than the deflectiontype mode, as the error in the former case will be in the mV orV comparedto fraction of a volt in the latter case.

Figure 1: Null type Wheatstone Bridge.

R2

G

RX

R4R3

Vo

ViA B

C

D

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When used as a null type device, the unknown resistorRx is placed in onelimb of the bridge, a calibrated resistor R2 is placed in the adjacent limb.Another two fixed resistors of known value R3 and R4 are placed on the othertwo limbs. The value ofR2 is gradually changed until a zero deflection on the

Galvanometer (G) is achieved (a Galvanometer is an extremely sensitivecurrent detector). The null type bridge obviously needs human intervention toadjust the value ofR2 until perfect balance conditions are achieved. Thus thenull type bridge would not be suitable for measuring dynamic signals (avoltage that is changing quickly in value).

The output voltage can be derived as follows:

+

+==

42

2

3RR

R

RR

RVVVv

x

x

iBCACo

At balance conditions, the output voltage is zero:

2

4

3

23242

42

2

3

42

2

3

0

R

R

RR

RRRRRRRR

RR

R

RR

R

RR

R

RR

R

x

xxx

x

x

x

x

=

+=+

+=

+

+

+=

In cases where R3=R4, then Rx=R2.

So to summarise, equation (1) below is used for the deflection type bridge andequation (2) below for the null type bridge.

+

+=

42

2

3RR

R

RR

RVv

x

x

io.(1)

2

4

3

RR

R

Rx = .(2)

When used as deflection type bridge, then the output voltage is a function inthe value ofRxas shown in equation (1) above.

Example 1A Wheatstone bridge is used to produce a voltage in response to the strain ina steel column. A Constantan strain gauge (SG-2/350-LY47 from Omega)that has a nominal value of 350 ohms is used as RX. The maximum excitationvoltage on the strain gauge is not allowed to exceed 5 V rms. The other three

resistors are also selected as 350 ohms. When the strain attains its maximum

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value (3% strain1), the strain gauge resistance increases by 6% (i.e.,

sensitivity equals 2). Find the maximum output voltage of the bridge, if theexcitation voltage for the bridge circuit is set at 3 V.

Solution

At zero strain, the value of the resistor will be 350 ohms, and hence the outputvoltage will be zero. At maximum strain of 3%, the resistance will increase by6%, to 371 . At this value, the output voltage will be:

mVRR

R

RR

RVv

x

x

io7.43

700

350

721

3713

42

2

3

=

=

+

+=

3. Null Type Bridge using Decade Resistance BoxesThe variable resistor R2 that was shown in the null type bridge is usually acalibrated decide resistance box, an example of which is shown in Figure 2.

A single stage of the box is shown Figure 3, and it can be seen how the knobcan only be set to one position within the available 9 or 10 positions (i.e., notin between values). The lowest stage (or the least significant digit) of thedecade resistance box represents its resolution and is a main contributoryfactor in the accuracy of the null bridge.

Figure 2: Decade resistance box. Figure 3: One of the stages of the decideresistance box set to 7 times 10,000

ohms.

Example 2A resistor is to be measured using a DC bridge in the null type mode. The truevalue of the resistor is 35.78 resistor. We have the following componentsavailable:

1. Two calibrated resistors of value 50 ohms (R3 and R4).2. A galvanometer.3. A voltage source (Vi).4. A decade resistance box (R2) as shown below.

1The strain is the ratio of the deflection to the original length. It is sometimes expressed in

micro-strain, which is equal to the strain multiplied by 1x106, and denoted as .

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Figure 4: Decade resistance box.

Based on the information above, answer the following questions:

(a) Showing all the components above (resistors, galvanometer, voltagesource), draw the setup that you will need to use to measure thevalue of the resistor.

(b) What do we call this bridge?(c) Describe in detail the practical steps that you will need to follow in

order to get to the final value of the resistor.(d) What is the value of the resistor that you will obtain using this

method?(e) Calculate the percentage error.

Solution(a) We use the bridge setup shown below.

(b) This bridge is called a Wheatstone bridge.

(c) Assuming that the galvanometer is not showing a zero reading, we startby adjusting the highest resistor stage in the decade box (largest 1000 ohm).

We keep on changing the setting of this range until a change takes place frompositive to negative on the galvanometer. Once that takes place we set it to

R2

G

RX

R4R3

Vo

Vi

0

12

3456

78 9 0

12

3456

78 9 0

12

3456

78 9 0

12

3456

78 9 0

12

3456

78 9

x1000 x100 x10 x1 x0.1

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the value that gives the smallest difference to zero on the galvanometer. Wethen move to the lower range on the decade resistance box and do the sameprocess until we get the smallest deviation from zero on the galvanometer.This is then repeated on all the five ranges on the decade resistance box (i.e.,1000 ohm stage; 100 ohm stage; 10 ohm stage; 1 ohm stage; 0.1 ohm stage).

Once all the stages have been adjusted, the value of the resistance on thedecade resistance box. As R2 is equal to R3, then we conclude that theunknown resistor is equal to the value of R1 (i.e., the final setting on thedecade resistance box).

(d) As the lowest resolution of the decade resistance box is 0.1, then thenearest value to the value of the resistor is 35.8 ohm (35.8 ohm is nearer tothe correct value than 35.7 ohm).

(e) The percentage error would be the difference between the two readingsdivided by the true value. So the percentage error will be:

error%=(35.8-35.78)/35.78 = 0.056%

The Wheatstone bridge is available as a portable device from a number ofmanufacturers. Model 275597 from Yokogawa is an example of a portableWheatstone bridge that can measure resistance from 1 to 10 M byoperation of dials and switches (Figure 5).

Figure 5: Model 275597 Portable Wheatstone Bridge (Courtesy of Yokogawa).

4. Source of Error in the Wheatstone BridgeThe sources of errors in the Wheatstone Bridge are (section 5.2.2 of [4]):

1. The main source of error is the limiting errors of the three knownresistors (R2, R3and R4). These errors are discussed in more detail inthe next section on initial balancing.

2. The resolution of the resistor R2 (usually a decade resistance box).The resolution is governed by the number of stages in the decade box.

3. Insufficient sensitivity of the null detector (Galvanometer) in the case ofthe null type bridge.

4. The effect of the internal resistance of the voltage indicator (loadingeffect on the bridge). This is only relevant in the case of deflectiontype bridges.

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5. Changes in resistance of the bridge arms due to the heating effect ofthe current through the resistors. Heating effect (I2R) of the bridge armcurrents may change the resistance of the resistor in question. Therise in temperature not only affects the resistance during the actual

measurement, but excessive currents may cause a permanent changein resistance values. This may not be discovered in time andsubsequent measurements could well be erroneous. The powerdissipation in the bridge arms must therefore be computed in advance,particularly when low-resistance values are to be measured, and thecurrent must be limited to a safe value. This is important when usingstrain gauges as the sensing elements in detecting strains and forcesin structures.

6. Thermal emfs (i.e., thermo-electric voltages) in the bridge circuit or thegalvanometer circuit can also cause problems when low-value

resistors are being measured. To prevent thermal emfs, the moresensitive galvanometers sometimes have copper coils and coppersuspension systems to avoid having dissimilar metals in contact withone another and generating thermal emfs.

7. Errors due to the resistance of leads and contacts exterior to the actualbridge circuit play a role in the measurement of very low-resistancevalues. These errors are discussed in more detail later in the sectionon lead wire compensation.

5. Initial balancing of the bridgeWhen the bridge is first setup, it is important to ensure that it is in an initialbalancedcondition. This is referred to as initial balancing. This balancing willovercome any errors that exist from the difference in value between R3 and R4for example. Initial balancing is important for deflection type bridges to ensurethat at zero condition of the input variable the output deflection of the bridge iszero. Zero condition is defined as follows:

1. Zero condition for a deflection type bridge occurs when all sensingelements of the bridge are at their initial default values (i.e, Rxo).

2. Zero conditions for a null bridge occur are when Rx isreplaced/switched with/to a calibrated resistorR1, where R1 is equalto R2. Cropico CF6 portable Wheatstone bridge has to balancingknobs as an example on this [5].

For null type bridges it ensures that the bridge will be balanced when Rx isequal to R2. Initial balancing can be done in three different methods [3]:

1. Series balancing: A variable resistor is placed in series with one ormore arms of the bridge. It is adjusted until the output of the bridgeis zero, under zero conditions. This is shown as Ribin Figure 6.

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Figure 6: Series init ial balancing.

2. Parallel balancing (or differential shunt balancing [8]): An exampleof this is shown in Figure 7. Two resistors (R5 and R6) and apotentiometer (R7) are placed in parallel with two arms of thebridge. The potentiometer is adjusted until the output of the bridgeis zero, under zero conditions. The main disadvantage with thismethod is that it reduces the sensitivity of the bridge.

Figure 7: Parallel (dif ferential shunt) initial balancing.

3. Apex balancing (differential series balancing [8]): A potentiometerR5 is placed between resistors R3 and R4, and the variable terminal

R2

G

RX

R4R3

Vo

Vi

R5

R6

R7

R2

G

RX

R4R3

Vo

Vi

Rib

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of the potentiometer forms the corner of the bridge and isconnected to one end of the excitation power supply. Thepotentiometer setting is varied under zero condition to achieve theinitial balance of the bridge.

Figure 8: Apex (diff erential series) ini tial balancing.

6. Lead wire compensation

In many applications the sensing element of the bridge is placed remotelyfrom bridge itself (e.g., in temperature measurements where the sensingelement is a thermistor). In this case, an extra error is introduced representedby the resistance of the connecting leads, as shown in . The effectivemeasured value of the sensing element is now RX+2RW, where RW is theresistance of each wire.

R2

G

RX

R4R3

Vo

Vi

R5

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Figure 9: Error in 2 wire connection.

Several methods have been developed to overcome this error. These arecalled lead wire compensation. There are three main methods of lead wirecompensation [9]:

1. Siemens-three-wire-connection: In this connection three leads areconnected to the sensing element as shown in Figure 10 below. Thisprovides compensation in the null mode where the central wire carries

zero current at balance conditions.

Figure 10: Siemens 3 wire connection.

V

RX

R4R3

Vo Vi

R2

Rw

Rw

Rw

Rw

RX

V

R4R3

VoVi

R2

Rw

Rw

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2. Callender-four-wire-connection: In this connection four wires areconnected to the location of the sensing element, two are connected tothe sensing element and two are just shorted near the sensingelement. This shown in Figure 11 below.

Figure 11: Callender 3 wire connection.

3. Floating potential four wire connection: In this arrangement 4 wires arealso used but only three of those are connected as shown in Figure 12.A reading is taken in this position, and then then wires 1 and 4 areswapped, and wire 2 and 3 are swapped. Another reading is thentaken and the average of the two readings is used.

V

RX

R4R3

Vo Vi

R2

Rw

Rw

Rw

Rw

Rw

Short

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Figure 12: Floating potential 4 wire connection.

7. Non-linearity in the Wheatstone BridgeWhen the Wheatstone bridge is used in the deflection type mode, a non-linearity exists between the input and output.

Example 3Repeat Example 1, but assuming that strain is 1.5% rather than 3%. What doyou expect the output voltage to be, considering that the output voltage at 3%strain was 43.7 mV?

SolutionIntuitively, you would expect the value to be half the output voltage at 3%strain (or 30 000 micro-strain) or 21.85 mV.

Substituting the values in the equation gives:

mVRR

R

RR

RVv

x

xio 17.22

700

350

5.710

5.3603

42

2

3

=

=

++=

The value is in fact slightly more than half the value of the output voltagewhen the strain was 3%. This is due to the non-linear relationship betweenthe change in RXand the output voltage.

The relationship between the input and output is:

+

+

=42

2

3 RR

R

RR

RVv

x

x

io

V

RX

R4R3

Vo Vi

R2

Rw

Rw

Rw

Rw

1

2

3

4

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where Rx is a sensor that changes in accordance with the variable that isbeing measured.

With R2, R3, R4 and the nominal value ofRx (say Rx0 which is the valueofRx when the input is zero) at equal values, this equation suffers from non-linearity between Vo and Rx.

In order to reduce this non-linearity, the values ofR2and R3 have to beincreased in relation to R1 and Rx0.

If we assume that the R3 and R4 are n times the value ofR2 and Rx0,and we assign as the percentage change in the value ofRx, then we can re-write the equation above as:

( )

( )( )

( )

+

++

+=

+

++

+=

+

+=

nnV

RnR

R

RnR

RV

RR

R

RR

RVV

i

xx

x

i

x

x

io

1

1

1

1

1

1

22

2

00

0

24

2

3

This expression is independent of the values of the resistors and onlydependent on and n. It can be shown that the non-linearity is reduced byincreasing the value ofn.

To give an example on the above, assume that:

Vi=10 VR1, R2, R3, Rx0=120n=1, 10 is a change of 2% over 29 steps for the whole range (58% in total)

N.B. The percentage change above is too large from a practical point of viewbut has been set at such a large value in order to show the non-linearity in therelationship when the graphs are plotted.

Using the values above results in the graphs shown in Figure 13 below. It canbe clearly seen how the linearity has dramatically improved when the values

ofR3and R4 are set at 10 times R2.

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Non-linearity in deflection Wheatstone Bridge

0

0.2

0.4

0.6

0.8

1

1.2

0.00

0

0.02

0

0.04

0

0.06

0

0.08

0

0.100

0.120

0.140

0.160

0.180

0.20

0

0.22

0

0.24

0

0.26

0

0.28

0

0.30

0

0.32

0

0.34

0

0.36

0

0.38

0

0.400

0.420

0.440

0.46

Percentage change in Rx

OutputVoltage(V)

R2=R3=1*R1

(i.e., n=1)

Ideal straight lin

R2=R3=10*R1

(i.e., n=10)

V i has been increased

in this case to account

for the increase in the

values of the resistors

and give the same

output

Figure 13: Non-linearity in relationship between Vo and Rx.

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Example 4A Platinum resistance thermometer is used in a Wheatstone deflection bridgeto measure the temperature between 0 C to 50 C.

Use the following information: The nominal value ofRxat 0 C is 500 ohm (i.e., Rxo=500 ). R2=R3=R4=500 . The value ofRxchanges (increases) by 4 /C Vi=10 V

i) Find a formula for the output voltage of the bridge as a function of the inputtemperature, T.

ii) Calculate the output voltage of the bridge when the temperature is 0 C.iii) Calculate the output voltage of the bridge when the temperature is 50 C.iv) The device (which is a conventional device that uses electronic circuitry

but no software) will use a straight line approximation to calculate thetemperature as a function of the voltage. This straight line is theconnection of the two end points, EPL (i.e., the points that the curvepasses through at T= 0 C and at T = 50 C). Find the equation of thisstraight line.

[this method of linearization is called End Point Linearity (EPL) asopposed to the other method that is called: Best Fit Straight Line(BFSL)]. See figure below.

v) Hence find a formula for the error (in volts) between the curve and thestraight line as function of the temperature T.

vi) By differentiating the error formula and equating to zero, find the value of Tat which the error is maximised. Hence find the maximum error in outputvoltage.

vii) Using this maximum value of error in output voltage, calculate themaximum non-linearity as a percentage of full scale deflection.

viii)What can be done to reduce this non-linearity with a conventional device?

With an intelligent device?

Bridge

Outputvariable (qo)

Inputvariable (qi)

BFSL

EPL

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Solution

i) Note that this resistor has a positive temperature coefficient as it is ametal (PTC).

+

+=

++

+=

+=+=

+=

5.0250

12510

1000

500

5004500

450010

45004

1000

500

50010

0

T

T

T

Tv

TTRR

R

Rv

o

xx

x

x

o

ii) The output voltage of the bridge when the temperature is 0 C is:

VvTo

05.00250

012510

0=

+

+=

=

iii) The output voltage of the bridge when the temperature is 50 C is:

VvTo

38.05.050250

5012510

50=

+

+=

=

iv) The two end points that the straight line will pass through are: (0 C,0 V)and (50 C, 0.83333 V). So the equation of the straight line is:

TTvo

601.00050

083333.0=+

=

v) The following is the formula for the error (in volts) between the curve andthe straight line as function of the temperature T.

TT

Tevo

601.05.0250

12510

+

+=

R2

V

RX

R4R3

Vo

Vi

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vi) Differentiating gives the following:

( )

( )601.0

250

125

250

110

601.00250

1

250

12510

601.05.0250

12510

2

2

+

+

+=

++

++=

+

+=

T

T

T

TT

T

TT

T

dT

de

dT

dvo

At maximum evo, the derivative should be zero. Hence:

( )

( )

( )

( )

( )

CT

T

T

TTT

T

T

T

T

T

T

T

T

Te

dT

dvo

=

=+

=+

+=+

+

+

+=

+

+

+=

+

+

+==

86.23

86.273250

75000250

1252502506001.0

250

125

250

16001.0

250

125

250

110601.0

601.0250

125

250

1100

2

2

2

2

2

So the maximum error in vo occurs when T = 23.86 C. The value of themaximum error can be found as follows:

mVevo

96.3786.23601.05.086.23250

86.2312510

max=

+

+=

vii) The full scale deflection in output voltage is 0.8333V. So the percentagenon-linearity is 37.96/833.3=4.55%

viii)With a conventional device, the ratio of the values ofR2 and R3 relative toR1 and Rx0can be increased. This ratio is denoted as n (i.e., n=R2/R1) andby increasing n (say a value of 10) the non-linearity can be reduced, butwill not be eliminated.

With an intelligent device the bridge equation can be inverted (i.e., expressingT as a function of vo) and this equation can be entered into themicroprocessor. In this case there is no need to approximate the relationshipby linearising and the non-linearity error is reduced to zero. The invertedequation is derived below.

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( )

o

o

ooo

oo

o

o

v

vT

vTTvTv

TTTvv

T

Tv

T

Tv

1.05.0

25

1.05.01.05.025

1255.01.012525

250

1255.01.0

5.0250

12510

=

==+=+++

+

+=+

+

+=

8. One, Two and Four arm bridgesThe bridge discussed so far, only one limb in the bridge was implemented asa sensor that varies in response to the change in the measured variable (e.g.,a strain gauge that changes its resistance as the stress on it changes). It ispossible however to use more than one limb with a sensor in it (2 and up to 4

sensors in the bridge). A bridge that has two sensors is called a 2 arm bridgeand one that has four sensors in all its four limbs is called a four arm bridge.An example of this application is used in the measurement of stress withstrain gauges.

In a two arm bridge, two of the arms vary with the variable. If the twosensors are placed in opposite arms of the bridge, then both change in thesame direction with the change in variable (i.e., increase together anddecrease together). But if they are placed in adjacent arms fo the bridge, thenthey must change in the opposite direction when the variable changes (i.e., asone increases the other decreases).

In a four element bridge, two of the sensors in opposite arms change inthe same direction (e.g, increase) while the other two in the other two armschange in the opposite direction to the that (e.g., decrease). We can refer tothe sensors that increase with the increase in the measured variable as R

+

and the sensor that decrease with the increase in the measured variable as R-

. As an example of a four arm bridge used with strain gauges, two opposingarm are sensors that are installed in a position of tension while the other twoopposing strain gauges are installed in a position of compression.

The use of 2 arm and 4 arm bridges has the advantage of increasingthe sensitivity of the bridge, as shown below.

Let us suppose that the input variable changes by a percentage change of

.The resistors R+

will increase in resistance by the percentage of, while theresistors R- will decrease by the same percentage.

8.1 Sensit ivi ty of the 1 arm bridgeThe equation for the 1 arm bridge will be:

( )( )

( )

( )

( ) ( )

( ) ( )

( )

( )42

1

4

2

2

1

22

21

2

1

2

1

1

1

2

2

00

0

00

0

42

2

3

+=

+

+=

+

+=

+

++

+=

+

+=

+

iiii

xx

x

xx

x

i

x

x

io

VVVV

RR

R

RR

RV

RR

R

RR

RVv

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Where we have assumed that is small enough to be ignored. The sensitivityof the one arm bridge is thus:

4

ioV

d

dv=

8.2 Sensit ivi ty of the 2 arm bridgeTaking first the two arm bridge with adjacent variable resistors (i.e, R

+and R

-),

the equation is:

( )( )

( )( )

( )( )

( )( )

( ) ( )( ) ( )

( ) ( )( ) ( )

( )( ) ( )( ) 242

4

2

22

21

22

21

2

1

2

1

1

1

1

1

2

2

2

2

00

0

00

0

43

+=

+

+

+

+=

+

+=

+

++

+=

+

+=

+

+

ii

ii

xx

x

xx

x

i

x

x

x

x

io

VV

VV

RR

R

RR

RV

RR

R

RR

RVv

Where we have assumed that 2 is small enough to be ignored. Thesensitivity of the two arm bridge is thus:

2

ioV

d

dv=

which is twice the one arm bridge sensitivity.If we now take the 2 arm bridge with opposite variable resistors (i.e, R

+

and R+), the equation becomes:

( )( ) ( )

( )( ) ( )

( ) ( )( ) ( )

( )( ) ( )

( )( )

( )( ) 242

4

2

22

2

22

21

2

1

2

1

11

1

22

2

00

0

00

0

2

2

3

+=

+

+

+=

+

+

+=

++

++

+=

+

+=

++

+

ii

ii

xx

x

xx

x

i

xx

x

io

VV

VV

RR

R

RR

RV

RR

R

RR

RVv

Where we have assumed that 2 is small enough to be ignored. Thesensitivity of the two arm bridge in this case is the same as the previous twoarm bridge:

2

ioV

d

dv=

So we get the same sensitivity of the two arm bridge whether we use twoadjacent variable resistors (R

+and R

-) or two opposite variable resistors (R

+

and R+). However, this bridge arrangement has the advantage that there is

no nonlinearity error in this arrangement [6].

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8.3 Sensit ivi ty of the 4 arm bridgeBy making all four resistors in the bridge to vary with the change with thevariable measured, we have a four arm bridge (two opposite arms will haveR

+and the other two arms will have R

-). The equation becomes:

( )( ) ( )

( )( ) ( ) i

xx

x

xx

x

i

xx

x

xx

x

ioV

RR

R

RR

RV

RR

R

RR

RVv =

++

++

+=

+

+

=+

+

+

00

0

00

0

11

1

11

1

The sensitivity is thus:

i

oV

d

dv=

Which is twice that of the 2 arm bridge and 4 times that of the 1 arm bridge.Note that there is no approximation in this result as in the case of the one arm

and two arm bridges. So the other important result from this equation is therelationship has now become completely linear, and this method is anotherway of linearising the relationship between the input and output.

9. Bridges as example of the use of the method of opposing inputsA bridge is an example of using the principle of opposing inputs to reducesystematic errors.

For example in the case of the null type bridge, the result of balancingthe bridge is not dependent on the value of the excitation voltage.

Moreover, in the case of the four element deflection bridge, the changein temperature affects all four resistors in the bridge equally, cancelling the net

effect of this temperature change on the output.

References & Bibliography[1] Measurement & Instrumentation Principles, Alan S. Morris, Elsevier,

2001.[2] Principles of Measurement Systems, John P. Bentley, Fourth Edition,

Pearson Prentice Hall.[3] An introduction to Electrical Instrumentation and Measurement

Systems, B.A.Gregory, Macmillan & English Language Book Society,Second Edition, 1981.

[4] Modern Electronic Instrumentation and Measurement Techniques,Albert D. Helfrick and William D. Cooper, Prentice Hall InternationalEditions, 1990.

[5] Cropico-Test: Type CF6, users manual, CFHbk.doc, Issue 3/2002.[6] Bridge-Type Sensor Measurements are Enhanced by Autozeroed

Instrumentation Amplifier with Digitally Programmable Gain and OutputOffset, Reza Moghimi, Analogue Devices, 38-05, May 2004.

[7] Theory and Design for Mechanical Measurements, Richard S. Figliola& Donald Beasley, 2

ndEdition, John Wiley & Sons, 1995.

[8] Distributed temperature sensing and non-contact torsion measurementwith fibre bragg gratings, Ludi Kruger, Masters Thesis, University of

Johannesburg, 1/1/2005.

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[9] Experimental Methods for Engineers, J.P. Holman, 7th

Edition, McGrawHill International Edition.

Problems1. A Platinum resistance thermometer is used in a Wheatstone deflection

bridge to measure the temperature between 0 C to 50 C.

Use the following information:

The nominal value ofRxat 0 C is 500 Ohm (i.e., Rxo=500 ). R2=R3=R4=500 . The value ofRxchanges (increases) by 4 /C Vi=10 V

i) Calculate the output voltage of the bridge when the temperature is 0 C.ii) Calculate the output voltage of the bridge when the temperature is 10 C.iii) Hence calculate the average sensitivity between the points of 0 C and 10

C in mV/.

Advanced Problem with Solut ion (based on problem 7.3 of [1] )2.a) Suppose that the unknown resistance Rx in the figure below is aresistance thermometer whose resistance at 100 C is 500 and whoseresistance varies with temperature at the rate of 0.5 /C for smalltemperature changes around 100 C. Calculate the sensitivity of the total

measurement system for small changes in temperature around 100 C, giventhe following resistance and voltage values measured at 15 C by instrumentscalibrated at 15 C: R2= 500; R3 = R4 = 5000 ; Vi=10 V.

R2

V

RX

R4R3

Vo

Vi

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Figure 14: Wheatstone Bridge.

Solution: In order to calculate the sensitivity, we can find the output voltageat a temperature of 100 C and the output voltage at a temperature of 101 C,and then find the sensitivity between the two points.

mVRR

R

RR

RVV

x

x

iCTo0

5000500

500

5000500

50010

24

2

3

100=

+

+=

+

+=

=

mVRR

R

RR

RVV

x

x

iCTo826.0

5000500

500

50004.500

4.50010

24

2

3

101=

+

+=

+

+=

=

So the sensitivity around 100 C is 0.826 mV/C.

2.b) If the resistance thermometer is measuring a fluid whose truetemperature is 104 C, calculate the error in the indicated temperature if theambient temperature around the bridge circuit is 20 C instead of thecalibration temperature of 15 C, give the following additional information:

Voltage-measuring instrument zero drift coefficient = +1.3 mV/C Voltage-measuring instrument sensitivity drift coefficient=0

mV/V/C

Resistance R2, R3 and R4 have a positive temperaturecoefficient of +0.2% of nominal value/ C

Voltage source Vi is unaffected by temperature changes.

Solution to part b): The various parts of the circuit will be affected by thedifferent temperatures as shown in the diagram below. Rx will be at 104 C,while all the other bridge components, the voltmeter and the voltage sourcewill be at 20 C.

R2

V

RX

R4R3

Vo

Vi

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Figure 15: Temperature changes on the various components.

ResistorRx will be at a temperature of 104 C. So the value of the resistorRxwill be:

Rx=500 + (104 100)x 0.5=502

The other resistors in the bridge will be at a temperature of 20 C which is 5degrees more than the calibration temperature of 15 C.

So the value ofR3 and R4 at the temperature of 20 C will be:

5000 + (5000x0.2/100)x(20-15)= 5050

The value ofR2at the temperature of 20 C will be:

500+ (500x0.2/100)x(20-15)= 505

The voltmeter will be affected by the temperature change only in the zero drift.The zero drift will be:

+1.3 mV/ C x (20-15)= 6.5 mV

So the voltmeter will indicate 6.5 mV when the actual output voltage of thebridge is 0 V.

The voltage source is unaffected by the temperature change, so will remain at10 V.

R2

V

RX

R4R3

Vo

Vi

104 C

20 C

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Copyright held by the author 2008: Dr Lutfi R Al Sharif Page 23 of 23

Using the values above, we can find the indicated voltage when thetemperature of the fluid is 104 C and the ambient temperature around thebridge is 20 C.

mVmV

RR

R

RR

R

VmVVx

x

iCTo

588.15050505

505

5050502

502105.6

5.624

2

3

104

=

+

++

=

+++==

We need to find that this indicated voltage would correspond to in terms oftemperature. When the indicated voltage is 0 mV, the correspondingtemperature is 100 C. The sensitivity around 100 C has been found from thelast part as 0.826 mV/C.

So the relationship between the indicated voltage and thecorresponding temperature will be:

CmV

mVvCT

o

+=

/826.0100

So at an indicated voltage of 1.588 mV, the temperature will be shown as:

CCmV

mVCT =

+= 922.101

/826.0

588.1100

But we know that the actual temperature of the fluid is 104 C. So the error is:

Error = 104 C 101.922 C = 2.077 C

11 dc bridges rev 3 080425

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