11 DC Bridges Rev 3 080425

Chapter 11: DC Bridges

0908341 Measurements & Instrumentation Chapter 11 DC Bridges (Revision 3.0, 25/4/2008)

1. Introduction As discussed in Chapter 1, SCE (signal conditioning element) or VCE (variable conversion element) is a typical component of a measurement system. It converts the signal received from the sensor from an unsuitable format to a suitable format. A bridge is one of the most widely used forms of VCE/SCE’s. Bridges can have either a.c. or d.c. excitation voltages. D.C. voltages are needed for resistance measurements while a.c. voltages are needed for capacitance and inductance measurements. The bridge can be used either in null type mode or deflection type mode. 2. Wheatstone Bridge The Wheatstone bridge is a type of d.c. bridge that is used for precision measurement of resistance from approximately 1 ohm to the low mega-ohm range. A typical Wheatstone null type bridge is shown in Figure 1 below. An excitation voltage source is used to operate the bridge (Vi). A galvanometer is used to connect the mid-points of the right hand side voltage divider (made up of R2 and R4) and the left hand side voltage divider (made up of R1 and R3). The galvanometer connection between the two mid-points form a bridge between the two sides, hence the name of the device. When used as a null type device, it can produce an accurate measurement of resistance. When used in the deflection type mode, it can produce a change in an output voltage that is proportional to a change in the sensor under question (a resistor). The change in the resistance of the sensor is 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 deflection type mode, as the error in the former case will be in the mV or μV compared to fraction of a volt in the latter case.

C RX Vi

R2

A

G

B

Vo R4

R3 D

Figure 1: Null type Wheatstone Bridge.

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When used as a null type device, the unknown resistor Rx is placed in one limb 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 other two limbs. The value of R2 is gradually changed until a zero deflection on the Galvanometer (G) is achieved (a Galvanometer is an extremely sensitive current detector). The null type bridge obviously needs human intervention to adjust the value of R2 until perfect balance conditions are achieved. Thus the null type bridge would not be suitable for measuring dynamic signals (a voltage that is changing quickly in value). The output voltage can be derived as follows: ⎛ Rx R2 ⎞ ⎟⎟ vo = VAC − VBC = Vi ⋅ ⎜⎜ − ⎝ Rx + R3 R2 + R4 ⎠

At balance conditions, the output voltage is zero: ⎛ Rx R2 ⎞ ⎟⎟ − 0 = ⎜⎜ ⎝ Rx + R3 R2 + R4 ⎠ Rx R2 = Rx + R3 R2 + R4 Rx ⋅ R2 + Rx ⋅ R4 = Rx ⋅ R2 + R3 ⋅ R2 Rx =

R3 ⋅ R2 R4

In cases where R3=R4, then Rx=R2. So to summarise, equation (1) below is used for the deflection type bridge and equation (2) below for the null type bridge. ⎛ Rx R2 ⎞ ⎟⎟ ……….(1) − vo = Vi ⋅ ⎜⎜ ⎝ Rx + R3 R2 + R4 ⎠ R Rx = 3 ⋅ R2 ……….(2) R4

When used as deflection type bridge, then the output voltage is a function in the value of Rx as shown in equation (1) above. Example 1 A Wheatstone bridge is used to produce a voltage in response to the strain in a 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 excitation voltage 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 © Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif

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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 the excitation 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 output voltage will be zero. At maximum strain of 3%, the resistance will increase by 6%, to 371 Ω. At this value, the output voltage will be: ⎛ Rx R2 ⎞ 371 350 ⎞ ⎟⎟ = 3 ⋅ ⎛⎜ vo = Vi ⋅ ⎜⎜ − − ⎟ = 43.7 mV ■ ⎝ 721 700 ⎠ ⎝ Rx + R3 R2 + R4 ⎠

3. Null Type Bridge using Decade Resistance Boxes The variable resistor R2 that was shown in the null type bridge is usually a calibrated 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 knob can only be set to one position within the available 9 or 10 positions (i.e., not in between values). The lowest stage (or the least significant digit) of the decade resistance box represents its resolution and is a main contributory factor in the accuracy of the null bridge.

Figure 2: Decade resistance box.

Figure 3: One of the stages of the decide resistance box set to 7 times 10,000 ohms.

Example 2 A resistor is to be measured using a DC bridge in the null type mode. The true value of the resistor is 35.78 Ω resistor. We have the following components available: 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.

1

The 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 με.

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x1000Ω

6 7

5 4 3

2 1 8 9 0

0908341 Measurements & Instrumentation

x100Ω

6 7

x10Ω

5 4 3

2 1 8 9 0

6 7

5 4 3

2 1 8 9 0

x1Ω

6 7

x0.1Ω

5 4 3

2 1 8 9 0

6 7

5 4 3

2 1 8 9 0

Figure 4: Decade resistance box.

Based on the information above, answer the following questions: (a) (b) (c) (d) (e)

Showing all the components above (resistors, galvanometer, voltage source), draw the setup that you will need to use to measure the value of the resistor. What do we call this bridge? Describe in detail the practical steps that you will need to follow in order to get to the final value of the resistor. What is the value of the resistor that you will obtain using this method? Calculate the percentage error.

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

RX Vi

R2

G Vo R3

R4

(b) This bridge is called a Wheatstone bridge. (c) Assuming that the galvanometer is not showing a zero reading, we start by 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 from positive to negative on the galvanometer. Once that takes place we set it to © Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif

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the value that gives the smallest difference to zero on the galvanometer. We then move to the lower range on the decade resistance box and do the same process 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 the decade resistance box. As R2 is equal to R3, then we conclude that the unknown resistor is equal to the value of R1 (i.e., the final setting on the decade resistance box). (d) As the lowest resolution of the decade resistance box is 0.1, then the nearest value to the value of the resistor is 35.8 ohm (35.8 ohm is nearer to the correct value than 35.7 ohm). (e) The percentage error would be the difference between the two readings divided 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 of manufacturers. Model 275597 from Yokogawa is an example of a portable Wheatstone bridge that can measure resistance from 1 Ω to 10 MΩ by operation of dials and switches (Figure 5).

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

4. Source of Error in the Wheatstone Bridge The 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 known resistors (R2, R3 and R4). These errors are discussed in more detail in the 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 of the null type bridge. 4. The effect of the internal resistance of the voltage indicator (loading effect on the bridge). This is only relevant in the case of deflection type bridges. © Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif

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5. Changes in resistance of the bridge arms due to the heating effect of the current through the resistors. Heating effect (I2R) of the bridge arm currents may change the resistance of the resistor in question. The rise in temperature not only affects the resistance during the actual measurement, but excessive currents may cause a permanent change in resistance values. This may not be discovered in time and subsequent measurements could well be erroneous. The power dissipation in the bridge arms must therefore be computed in advance, particularly when low-resistance values are to be measured, and the current must be limited to a safe value. This is important when using strain gauges as the sensing elements in detecting strains and forces in structures. 6. Thermal emfs (i.e., thermo-electric voltages) in the bridge circuit or the galvanometer circuit can also cause problems when low-value resistors are being measured. To prevent thermal emfs, the more sensitive galvanometers sometimes have copper coils and copper suspension systems to avoid having dissimilar metals in contact with one another and generating thermal emfs. 7. Errors due to the resistance of leads and contacts exterior to the actual bridge circuit play a role in the measurement of very low-resistance values. These errors are discussed in more detail later in the section on lead wire compensation. 5. Initial balancing of the bridge When the bridge is first setup, it is important to ensure that it is in an initial balanced condition. This is referred to as initial balancing. This balancing will overcome any errors that exist from the difference in value between R3 and R4 for example. Initial balancing is important for deflection type bridges to ensure that at zero condition of the input variable the output deflection of the bridge is zero. Zero condition is defined as follows: 1. Zero condition for a deflection type bridge occurs when all sensing elements of the bridge are at their initial default values (i.e, Rxo). 2. Zero conditions for a null bridge occur are when Rx is replaced/switched with/to a calibrated resistor R1, where R1 is equal to R2. Cropico CF6 portable Wheatstone bridge has to balancing knobs as an example on this [5]. For null type bridges it ensures that the bridge will be balanced when Rx is equal to R2. Initial balancing can be done in three different methods [3]: 1. Series balancing: A variable resistor is placed in series with one or more arms of the bridge. It is adjusted until the output of the bridge is zero, under zero conditions. This is shown as Rib in Figure 6.

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RX

R2

Rib Vi

G Vo R4

R3

Figure 6: Series initial balancing.

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

R5 RX Vi

R7

R2

G Vo

R6

R3

R4

Figure 7: Parallel (differential shunt) initial balancing.

3. Apex balancing (differential series balancing [8]): A potentiometer R5 is placed between resistors R3 and R4, and the variable terminal © Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif

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of the potentiometer forms the corner of the bridge and is connected to one end of the excitation power supply. The potentiometer setting is varied under zero condition to achieve the initial balance of the bridge.

RX

R2

Vi G Vo R4

R3 R5

Figure 8: Apex (differential series) initial balancing.

6. Lead wire compensation In many applications the sensing element of the bridge is placed remotely from bridge itself (e.g., in temperature measurements where the sensing element is a thermistor). In this case, an extra error is introduced represented by the resistance of the connecting leads, as shown in . The effective measured value of the sensing element is now RX+2·RW, where RW is the resistance of each wire.

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Rw

RX

R2 Rw

V

Vi

Vo R4

R3

Figure 9: Error in 2 wire connection.

Several methods have been developed to overcome this error. These are called lead wire compensation. There are three main methods of lead wire compensation [9]: 1. Siemens-three-wire-connection: In this connection three leads are connected to the sensing element as shown in Figure 10 below. This provides compensation in the null mode where the central wire carries zero current at balance conditions. Rw

RX

R2

Rw V Rw

Vo

Vi

Rw

R3

R4

Figure 10: Siemens 3 wire connection.

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

RX

R2 Rw V

Short

Vo

Rw Rw Rw R3

Vi

R4

Figure 11: Callender 3 wire connection.

3. Floating potential four wire connection: In this arrangement 4 wires are also 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 are swapped, and wire 2 and 3 are swapped. Another reading is then taken and the average of the two readings is used.

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RX

0908341 Measurements & Instrumentation

Rw

1

Rw

2 R2

Rw

3 V

4

Vo

Vi

Rw

R3

R4

Figure 12: Floating potential 4 wire connection.

7. Non-linearity in the Wheatstone Bridge When the Wheatstone bridge is used in the deflection type mode, a nonlinearity exists between the input and output. Example 3 Repeat Example 1, but assuming that strain is 1.5% rather than 3%. What do you expect the output voltage to be, considering that the output voltage at 3% strain was 43.7 mV? Solution Intuitively, 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: ⎛ Rx R2 ⎞ 360.5 350 ⎞ ⎟⎟ = 3 ⋅ ⎛⎜ vo = Vi ⋅ ⎜⎜ − − ⎟ = 22.17 mV ⎝ 710.5 700 ⎠ ⎝ Rx + R3 R2 + R4 ⎠

The value is in fact slightly more than half the value of the output voltage when the strain was 3%. This is due to the non-linear relationship between the change in RX and the output voltage.■ The relationship between the input and output is: ⎛ Rx R2 ⎞ ⎟⎟ − vo = Vi ⋅ ⎜⎜ + + R R R R 3 2 4 ⎠ ⎝ x

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…where Rx is a sensor that changes in accordance with the variable that is being measured. With R2, R3, R4 and the nominal value of Rx (say Rx0 which is the value of Rx when the input is zero) at equal values, this equation suffers from nonlinearity between Vo and Rx. In order to reduce this non-linearity, the values of R2 and R3 have to be increased in relation to R1 and Rx0. If we assume that the R3 and R4 are n times the value of R2 and Rx0, and we assign δ as the percentage change in the value of Rx, then we can rewrite the equation above as: ⎛ Rx R2 ⎞ ⎟⎟ Vo = Vi × ⎜⎜ − ⎝ Rx + R3 R4 + R2 ⎠ ⎛ ⎞ Rx 0 × (1 + δ ) R2 ⎟ = Vi × ⎜⎜ − ⎟ ⎝ Rx 0 × (1 + δ ) + n × Rx 0 R2 + n × R2 ⎠ ⎛ (1 + δ ) 1 ⎞ ⎟⎟ = Vi × ⎜⎜ − ⎝ (1 + δ ) + n 1 + n ⎠

This expression is independent of the values of the resistors and only dependent on δ and n. It can be shown that the non-linearity is reduced by increasing the value of n. To give an example on the above, assume that: Vi=10 V R1, R2, R3, Rx0=120 Ω n=1, 10 δ 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 view but has been set at such a large value in order to show the non-linearity in the relationship when the graphs are plotted. Using the values above results in the graphs shown in Figure 13 below. It can be clearly seen how the linearity has dramatically improved when the values of R3 and R4 are set at 10 times R2.

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Non-linearity in deflection Wheatstone Bridge 1.2

1

Output Voltage (V)

0.8

R2=R3=1*R1 (i.e., n=1) Ideal straight line

0.6

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

0.4

0.2

0. 00 0 0. 02 0 0. 04 0 0. 06 0 0. 08 0 0. 10 0 0. 12 0 0. 14 0 0. 16 0 0. 18 0 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. 40 0 0. 42 0 0. 44 0 0. 46 0 0. 48 0 0. 50 0 0. 52 0 0. 54 0 0. 56 0 0. 58 0

0

Percentage change in Rx

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

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Example 4 A Platinum resistance thermometer is used in a Wheatstone deflection bridge to measure the temperature between 0 ºC to 50 ºC. Use the following information: • The nominal value of Rx at 0 ºC is 500 ohm (i.e., Rxo=500 Ω). • R2=R3=R4=500 Ω. • The value of Rx changes (increases) by 4 Ω/ºC • Vi=10 V i) Find a formula for the output voltage of the bridge as a function of the input temperature, T. ii) Calculate the output voltage of the bridge when the temperature is 0 ºC. iii) Calculate the output voltage of the bridge when the temperature is 50 ºC. iv) The device (which is a conventional device that uses electronic circuitry but no software) will use a straight line approximation to calculate the temperature as a function of the voltage. This straight line is the connection of the two end points, EPL (i.e., the points that the curve passes through at T= 0 ºC and at T = 50 ºC). Find the equation of this straight line. [this method of linearization is called ‘End Point Linearity’ (EPL) as opposed to the other method that is called: ‘Best Fit Straight Line’ (BFSL)]. See figure below. BFSL Bridge

EPL

Output variable (qo)

Input variable (qi)

v) Hence find a formula for the error (in volts) between the curve and the straight line as function of the temperature T. vi) By differentiating the error formula and equating to zero, find the value of T at which the error is maximised. Hence find the maximum error in output voltage. vii) Using this maximum value of error in output voltage, calculate the maximum 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?

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RX Vi

R2

V Vo R3

R4

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

⎛ Rx 500 ⎞ ⎟⎟ vo = 10 × ⎜⎜ − ⎝ Rx + 500 1000 ⎠ Rx = Rx 0 + 4T = 500 + 4T 500 ⎞ ⎛ 500 + 4T ⎛ 125 + T ⎞ − − 0.5 ⎟ vo = 10 × ⎜ ⎟ = 10 × ⎜ ⎝ 500 + 4T + 500 1000 ⎠ ⎝ 250 + T ⎠ ii) The output voltage of the bridge when the temperature is 0 ºC is:

⎛ 125 + 0 ⎞ vo T =0 = 10 × ⎜ − 0.5 ⎟ = 0V ⎝ 250 + 0 ⎠ iii) The output voltage of the bridge when the temperature is 50 ºC is:

⎛ 125 + 50 ⎞ vo T =50 = 10 × ⎜ − 0.5 ⎟ = 0.8 3V ⎝ 250 + 50 ⎠ iv) The two end points that the straight line will pass through are: (0 ºC,0 V) and (50 ºC, 0.83333 V). So the equation of the straight line is:

vo =

0.83333 − 0 T + 0 = 0.016T 50 − 0

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

⎞ ⎛ 125 + T evo = 10 × ⎜ − 0.5 ⎟ − 0.016T ⎠ ⎝ 250 + T

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vi) Differentiating gives the following:

⎞ d d ⎛ ⎛ 125 + T ⎞ evo = − 0.5 ⎟ − 0.016T ⎟⎟ ⎜⎜10 × ⎜ dT dT ⎝ ⎝ 250 + T ⎠ ⎠ ⎛ ⎛ ⎛ 125 + T ⎞ ⎛ 1 ⎞ ⎞ ⎞ ⎟ − 0 ⎟ − 0.016 ⎟+ = 10 × ⎜ ⎜⎜ ⎜⎜ − ⎜ ⎝ (250 + T )2 ⎟⎠ ⎜⎝ 250 + T ⎟⎠ ⎟ ⎟ ⎠ ⎠ ⎝⎝ ⎛ 1 125 + T ⎞ ⎟ − 0.016 = 10⎜⎜ − 2 ⎟ ⎝ 250 + T (250 + T ) ⎠ At maximum evo, the derivative should be zero. Hence:

⎛ 1 125 + T ⎞ d ⎟ − 0.016 − evo = 0 = 10⎜⎜ 2 ⎟ dT ⎝ 250 + T (250 + T ) ⎠ ⎛ 1 125 + T ⎞ ⎟ 0.016 = 10⎜⎜ − 2 ⎟ 250 + T ( ) 250 + T ⎝ ⎠ 1 125 + T 0.0016 = − 250 + T (250 + T )2 0.0016 × (250 + T ) = 250 + T − 125 − T 2

(250 + T )2 = 75000 250 + T = 273.86 T = 23.86°C So the maximum error in vo occurs when T = 23.86 ºC. The value of the maximum error can be found as follows:

evo

max

⎛ 125 + 23.86 ⎞ = 10 × ⎜ − 0.5 ⎟ − 0.016 × 23.86 = 37.96 mV ⎠ ⎝ 250 + 23.86

vii) The full scale deflection in output voltage is 0.8333V. So the percentage non-linearity is 37.96/833.3=4.55% viii)With a conventional device, the ratio of the values of R2 and R3 relative to R1 and Rx0 can be increased. This ratio is denoted as n (i.e., n=R2/R1) and by increasing n (say a value of 10) the non-linearity can be reduced, but will not be eliminated. With an intelligent device the bridge equation can be inverted (i.e., expressing T as a function of vo) and this equation can be entered into the microprocessor. In this case there is no need to approximate the relationship by linearising and the non-linearity error is reduced to zero. The inverted equation is derived below.

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⎛ 125 + T ⎞ vo = 10 × ⎜ − 0.5 ⎟ ⎝ 250 + T ⎠ 125 + T 0.1× vo + 0.5 = 250 + T 25 × vo + 125 + 0.1Tvo + 0.5T = 125 + T ■

25 × vo = 0.5T − 0.1Tvo = T × (0.5 − 0.1vo ) T=

25vo 0.5 − 0.1vo

8. One, Two and Four arm bridges The bridge discussed so far, only one limb in the bridge was implemented as a 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 is possible 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 bridge and 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 with strain gauges. In a two arm bridge, two of the arms vary with the variable. If the two sensors are placed in opposite arms of the bridge, then both change in the same direction with the change in variable (i.e., increase together and decrease together). But if they are placed in adjacent arms fo the bridge, then they must change in the opposite direction when the variable changes (i.e., as one increases the other decreases). In a four element bridge, two of the sensors in opposite arms change in the same direction (e.g, increase) while the other two in the other two arms change in the opposite direction to the that (e.g., decrease). We can refer to the 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 opposing arm are sensors that are installed in a position of tension while the other two opposing strain gauges are installed in a position of compression. The use of 2 arm and 4 arm bridges has the advantage of increasing the sensitivity of the bridge, as shown below. Let us suppose that the input variable changes by a percentage change of δ. The resistors R+ will increase in resistance by the percentage of δ, while the resistors R- will decrease by the same percentage. 8.1 Sensitivity of the 1 arm bridge The equation for the 1 arm bridge will be: ⎛ (1 + δ ) ⋅ Rx 0 ⎛ Rx+ Rx 0 ⎞ R2 ⎞ ⎟⎟ ⎟⎟ = Vi ⋅ ⎜⎜ vo = Vi ⋅ ⎜⎜ − − ⎝ (1 + δ ) ⋅ Rx 0 + Rx 0 Rx 0 + Rx 0 ⎠ ⎝ Rx + R3 R2 + R4 ⎠ ⎛ 2 +δ −δ 2 1 ⎞ ⎛ (1 + δ ) 1 ⎞ ⎛ (1 + δ ) ⋅ (2 − δ ) 1 ⎞ δ = Vi ⋅ ⎜⎜ − ⎟⎟ = Vi ⋅ ⎜⎜ − ⎟⎟ = Vi ⋅ ⎜⎜ − ⎟⎟ ≈ Vi ⋅ 2 2⎠ 4 ⎝ (2 + δ ) 2 ⎠ ⎝ (2 + δ ) ⋅ (2 − δ ) 2 ⎠ ⎝ 4 −δ

(

(

)

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)

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Where we have assumed that is small enough to be ignored. The sensitivity of the one arm bridge is thus: dvo Vi = 4 dδ

8.2 Sensitivity of the 2 arm bridge Taking first the two arm bridge with adjacent variable resistors (i.e, R+ and R-), the equation is: ⎛ R+ ⎛ (1 + δ ) ⋅ Rx 0 (1 − δ ) ⋅ Rx 0 ⎞⎟ R− ⎞ vo = Vi ⋅ ⎜⎜ + x − − x ⎟⎟ = Vi ⋅ ⎜⎜ − ⎟ ⎝ Rx + R3 Rx + R4 ⎠ ⎝ (1 + δ ) ⋅ Rx 0 + Rx 0 (1 − δ ) ⋅ Rx 0 + Rx 0 ⎠ ⎛ (1 + δ ) (1 − δ ) ⎞ ⎛ (1 + δ ) ⋅ (2 − δ ) (1 − δ ) ⋅ (2 + δ ) ⎞ ⎟⎟ = Vi ⋅ ⎜⎜ ⎟⎟ = Vi ⋅ ⎜⎜ − − ⎝ (2 + δ ) (2 − δ ) ⎠ ⎝ (2 + δ ) ⋅ (2 − δ ) (2 − δ ) ⋅ (2 + δ ) ⎠

(

) (

⎛ 2 +δ −δ 2 2 −δ −δ 2 = Vi ⋅ ⎜⎜ − 2 4 −δ 2 ⎝ 4 −δ

(

)

(

)

) ⎞⎟ ≈ V ⋅ δ ⎟ ⎠

i

2

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

The

dvo Vi = dδ 2

…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: ⎛ R+ ⎛ (1 + δ ) ⋅ Rx 0 ⎞ R2 ⎞ Rx 0 ⎟ = Vi ⋅ ⎜⎜ ⎟⎟ vo = Vi ⋅ ⎜⎜ + x − − + ⎟ ⎝ Rx + R3 R2 + Rx ⎠ ⎝ (1 + δ ) ⋅ Rx 0 + Rx 0 Rx 0 + (1 + δ ) ⋅ Rx 0 ⎠ ⎛ (1 + δ ) ⎛ (1 + δ ) ⋅ (2 − δ ) (2 − δ ) ⎞⎟ 1 ⎞ = Vi ⋅ ⎜⎜ − ⎟⎟ = Vi ⋅ ⎜⎜ − ⎟ ⎝ (2 + δ ) (2 + δ ) ⎠ ⎝ (2 + δ ) ⋅ (2 − δ ) (2 + δ ) ⋅ (2 − δ ) ⎠

(

)

⎛ 2 +δ −δ 2 (2 − δ ) ⎞⎟ ≈ V ⋅ δ = Vi ⋅ ⎜⎜ − i 2 4 − δ 2 ⎟⎠ 2 ⎝ 4 −δ

(

)

(

)

Where we have assumed that δ 2 is small enough to be ignored. The sensitivity of the two arm bridge in this case is the same as the previous two arm bridge: dvo Vi = dδ 2

So we get the same sensitivity of the two arm bridge whether we use two adjacent 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 Sensitivity of the 4 arm bridge By making all four resistors in the bridge to vary with the change with the variable measured, we have a four arm bridge (two opposite arms will have R+ and the other two arms will have R-). The equation becomes: ⎛ ⎞ ⎛ R+ (1 + δ ) ⋅ Rx 0 (1 − δ ) ⋅ Rx 0 R− ⎞ ⎟⎟ = δ ⋅ Vi vo = Vi ⋅ ⎜⎜ + x − − − x + ⎟⎟ = Vi ⋅ ⎜⎜ − ⎝ (1 + δ ) ⋅ Rx 0 + (1 − δ ) ⋅ Rx 0 (1 − δ ) ⋅ Rx 0 + (1 + δ ) ⋅ Rx 0 ⎠ ⎝ Rx + Rx Rx + Rx ⎠

The sensitivity is thus: dvo = Vi dδ

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 the relationship has now become completely linear, and this method is another way of linearising the relationship between the input and output. 9. Bridges as example of the use of the method of opposing inputs A bridge is an example of using the principle of opposing inputs to reduce systematic errors. For example in the case of the null type bridge, the result of balancing the bridge is not dependent on the value of the excitation voltage. Moreover, in the case of the four element deflection bridge, the change in 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 International Editions, 1990. [5] “Cropico-Test: Type CF6”, user’s manual, CFHbk.doc, Issue 3/2002. [6] “Bridge-Type Sensor Measurements are Enhanced by Autozeroed Instrumentation Amplifier with Digitally Programmable Gain and Output Offset”, Reza Moghimi, Analogue Devices, 38-05, May 2004. [7] “Theory and Design for Mechanical Measurements”, Richard S. Figliola & Donald Beasley, 2nd Edition, John Wiley & Sons, 1995. [8] “Distributed temperature sensing and non-contact torsion measurement with fibre bragg gratings”, Ludi Kruger, Master’s Thesis, University of Johannesburg, 1/1/2005.

© Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif

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“Experimental Methods for Engineers”, J.P. Holman, 7th Edition, McGraw Hill International Edition.

Problems 1. A Platinum resistance thermometer is used in a Wheatstone deflection bridge to measure the temperature between 0 ºC to 50 ºC. Use the following information: • The nominal value of Rx at 0 ºC is 500 Ohm (i.e., Rxo=500 Ω). • R2=R3=R4=500 Ω. • The value of Rx changes (increases) by 4 Ω/ºC • Vi=10 V i) Calculate the output voltage of the bridge when the temperature is 0 ºC. ii) Calculate the output voltage of the bridge when the temperature is 10 ºC. iii) Hence calculate the average sensitivity between the points of 0 ºC and 10 ºC in mV/Ω.

RX Vi

R2

V Vo R3

R4

Advanced Problem with Solution (based on problem 7.3 of [1]) 2.a) Suppose that the unknown resistance Rx in the figure below is a resistance thermometer whose resistance at 100 C is 500 Ω and whose resistance varies with temperature at the rate of 0.5 Ω/ºC for small temperature changes around 100 ºC. Calculate the sensitivity of the total measurement system for small changes in temperature around 100 ºC, given the following resistance and voltage values measured at 15 ºC by instruments calibrated at 15 ºC: R2 = 500; R3 = R4 = 5000 ; Vi=10 V.

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Chapter 11: DC Bridges

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RX Vi

R2

V Vo R3

R4

Figure 14: Wheatstone Bridge.

Solution: In order to calculate the sensitivity, we can find the output voltage at a temperature of 100 ºC and the output voltage at a temperature of 101 ºC, and then find the sensitivity between the two points. ⎛ Rx R2 ⎞ 500 500 ⎞ ⎟⎟ = 10 ⋅ ⎛⎜ − − Vo T =100C = Vi ⋅ ⎜⎜ ⎟ = 0 mV ⎝ 500 + 5000 500 + 5000 ⎠ ⎝ Rx + R3 R4 + R2 ⎠ ⎛ Rx R2 ⎞ 500.4 500 ⎞ ⎟⎟ = 10 ⋅ ⎛⎜ − − Vo T =101C = Vi ⋅ ⎜⎜ ⎟ = 0.826mV + + + + R R R R 500 . 4 5000 500 5000 ⎝ ⎠ 3 4 2 ⎠ ⎝ x

So the sensitivity around 100 ºC is 0.826 mV/ºC.■ 2.b) If the resistance thermometer is measuring a fluid whose true temperature is 104 ºC, calculate the error in the indicated temperature if the ambient temperature around the bridge circuit is 20 ºC instead of the calibration temperature of 15 ºC, give the following additional information:

• • • •

Voltage-measuring instrument zero drift coefficient = +1.3 mV/ºC Voltage-measuring instrument sensitivity drift coefficient=0 mV/V/ºC Resistance R2, R3 and R4 have a positive temperature coefficient of +0.2% of nominal value/ ºC Voltage source Vi is unaffected by temperature changes.

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

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Chapter 11: DC Bridges

0908341 Measurements & Instrumentation

104 ºC RX Vi

R2

V Vo R3

R4

20 ºC

Figure 15: Temperature changes on the various components.

Resistor Rx will be at a temperature of 104 ºC. So the value of the resistor Rx will be: Rx =500 + (104 – 100)x 0.5=502 Ω The other resistors in the bridge will be at a temperature of 20 ºC which is 5 degrees more than the calibration temperature of 15 ºC. So the value of R3 and R4 at the temperature of 20 ºC will be: 5000 + (5000x0.2/100)x(20-15)= 5050 Ω The value of R2 at the temperature of 20 ºC will be: 500+ (500x0.2/100)x(20-15)= 505 Ω The voltmeter will be affected by the temperature change only in the zero drift. The zero drift will be: +1.3 mV/ ºC x (20-15)= 6.5 mV So the voltmeter will indicate 6.5 mV when the actual output voltage of the bridge is 0 V. The voltage source is unaffected by the temperature change, so will remain at 10 V.

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Chapter 11: DC Bridges

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

⎛ Rx R2 ⎞ ⎟⎟ = Vo T =104C = 6.5mV + Vi × ⎜⎜ − R + R R + R 3 4 2 ⎠ ⎝ x 502 505 ⎛ ⎞ − 6.5mV + 10 × ⎜ ⎟ = 1.588mV ⎝ 502 + 5050 505 + 5050 ⎠ We need to find that this indicated voltage would correspond to in terms of temperature. When the indicated voltage is 0 mV, the corresponding temperature is 100 C. The sensitivity around 100 C has been found from the last part as 0.826 mV/ºC. So the relationship between the indicated voltage and the corresponding temperature will be:

T = 100°C +

vo mV 0.826 mV / °C

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

T = 100°C +

1.588 mV = 101.922°C 0.826 mV / °C

But we know that the actual temperature of the fluid is 104 ºC. So the error is: Error = 104 ºC – 101.922 ºC = 2.077 ºC■

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