RMIT University | School of Engineering | EEET2105/EEET2388 Industrial Automation


EEET2105/EEET2388 Industrial Automation

Model Answers for Tutorial #7

1) A cantilever beam is the load-sensing element of a load cell. The distortion of the beam under load
is detected using four strain gauges (SG1, SG2, SG3, and SG4) bonded on the surface of the beam and
connected in a Wheatstone bridge configuration. Strain gauges SG1 and SG2 have a nominal
resistance of 200 Ω and are connected together at one end to form the negative output terminal of the
bridge. Strain gauges SG3 and SG4 have a nominal resistance of 400 Ω and are connected together at
one end to form the positive output terminal of the bridge. The bridge supply voltage is 10 V. Under
load, the resistance of each strain gauge changes by ± δ%, depending on the physical location of the
strain gauge on the beam.


a) Draw the circuit diagram of the Wheatstone bridge, indicating the resistance values under load.

All we want is an increasing output voltage Vout for a change in the strain gauge resistance.

A resistive voltage divider has its divider voltage increasing for an increasing bottom resistor and for
a decreasing top resistor, and this is what we want for the resistive voltage divider connected to the
positive output terminal Vpos.

However, for the resistive voltage divider connected to the negative output terminal Vneg, we want
the opposite, because Vout = Vpos – Vneg, so to maximise Vout we need to make Vneg smaller as the
resistors change.

So, in summary, we always want:
Rpos,bot = Rnom + ∆R
Rpos,top = Rnom − ∆R
Rneg,bottom = Rnom − ∆R
Rneg,top = Rnom + ∆R




Hence we have
Rpos,bot = Rnom + ∆R = 400(1+δ) = SG3
Rpos,top = Rnom − ∆R = 400(1−δ)= SG4
Rneg,bottom = Rnom − ∆R = 200(1−δ)= SG2
Rneg,top = Rnom + ∆R = 200(1+δ)= SG1

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RMIT University | School of Engineering | EEET2105/EEET2388 Industrial Automation


b) Sketch the load-sensing beam, indicating the orientation and the resistance of each strain gauge on
the beam under load.

The application of a force to the cantilever beam may cause an increase or a decrease in the resistance
of strain gauge with respect to its nominal value (i.e. for the condition where we don’t have any force
applied to the cantilever beam) depending if the strain gauge is located on the top or at the bottom of
the cantilever beam.

If the strain gauge is at the top, the application of a force to the cantilever beam causes an elongation
of strain gauge in its direction of sensing, and this results on an increase of strain gauge resistance. If
the strain gauge is at the bottom, the application of a force to the cantilever beam causes a compression
of strain gauge in its direction of sensing, and this results on a decrease of strain gauge resistance.

The direction of sensing of strain gauge is shown in the figure below.




Hence one correct location of the four strain gauges is as shown in the figure below.




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c) Derive the relationship between the output voltage of the load cell and δ.

 R pos ,bot Rneg ,bot 
Vout =  −  ⋅ Vdc
 R pos ,bot + R pos ,top Rneg ,bot + Rneg ,top 

 400 (1 + δ ) 200 (1 − δ ) 
Vout =  −  ⋅10
 400 (1 + δ ) + 400 (1 − δ ) 200 (1 − δ ) + 200 (1 + δ ) 

 1+ δ 1−δ 
Vout =  − ⋅10
1 + δ + 1 − δ 1 − δ + 1 + δ 

1 + δ 1 − δ  1 + δ − 1 + δ   2δ 
Vout =  −  ⋅10 =   ⋅10 =   ⋅10 = 10 ⋅ δ
 2 2   2   2 

2) A transducer contains a column member with four strain gauges located as shown in Figure 1.
Strain gauges R1 and R2 sense the axial strain, while strain gauges R3 and R4 sense the transverse
strain on the surface of the column. The nominal resistance of all four strain gauges is R. Under a
compressive force F acting on the column, the resistance of strain gauges R1 and R2 decreases by
3∆R, while the resistance of strain gauges R3 and R4 increases by ∆R.

F




Figure 1




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RMIT University | School of Engineering | EEET2105/EEET2388 Industrial Automation


a) Draw the circuit diagram showing how the strain gauges should be connected to a 10 V power
supply in order to obtain an output voltage, Vo, when F ≠ 0. In your circuit diagram, include the
designations shown in Figure 1 and the resistance of each strain gauge when a compressive force is
applied to the column.




b) Calculate the output voltage of the circuit from item 2.a).

R4 R R +∆R R − 3∆R
Vout = Vpos −Vneg = Vdc − 1 Vdc = Vdc − Vdc
R4 + R2 R1 + R3 R +∆R + ( R − 3∆R) R − 3∆R + R +∆R

4∆R 2∆R 20∆R
Vout = Vdc = Vdc =
2( R −∆R) R −∆R R −∆R

3) An RTD has a sensitivity α ( 20 °C ) = 0.004 / °C . If the RTD resistance is 106 Ω at 20 °C, find the
RTD resistance at 25°C.

R (T ) = R (T0 ) ⋅ 1 + α (T0 ) ⋅ (T − T0 ) 

T0 = 20 °C , R (T0 ) = 106 Ω, α (T0 ) = 0.004 / °C

R (T ) = 106 Ω ⋅ 1 + 0.004 / °C ⋅ (T − 20°C ) 

R ( 25°C ) = 106 Ω ⋅ 1 + 0.004 / °C ⋅ ( 25°C − 20°C )  = 106 Ω ⋅ [1.02] = 108.12 Ω




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4) Suppose the RTD of question 3) has a dissipation constant of 25 mW / °C and is used in a circuit
that drives 8 mA through the sensor. If the RTD is placed in a bath at 100 °C , what resistance will
the RTD have?

If the RTD is placed in a bath at 100 °C its resistance should be
R (100°C ) = 106 Ω ⋅ 1 + 0.004 / °C ⋅ (100°C − 20°C )  = 106 Ω ⋅ [1.32] = 139.92 Ω

However, if there are 8 mA through the sensor then self-heating will cause a temperature rise from
the ohmic loss. The power dissipated is
P = R (100°C ) ⋅ I 2 = 139.92 Ω ⋅ ( 0.008 A ) = 8.95 mW
2




Thus the temperature rise will be
P 8.95 mW
∆T (100°C ) = = = 0.36°C
PD 25 mW / °C

So the resistance will be
R (100°C ) + ∆R (100°C ) = 106 Ω ⋅ 1 + 0.004 / °C ⋅ (100.36°C − 20°C )  = 106 Ω ⋅ [1.32144] = 140.07 Ω

If you didn’t know about the self-heating temperature rise you would think the temperature was
100.36 °C instead of 100 °C.

5) Redesign the voltage divider discussed on slide 23 of lecture notes for week 08 so that self-heating
temperature increase is reduced to 0.1 °C. What is the divider voltage for 20 °C? What is the divider
voltage for 19 °C and 21 °C?

If the self-heating is to be 0.1 °C the power which can be dissipated by the thermistor is given by
P = PD ∆T = ( 5 mW / °C ) ⋅ ( 0.1°C ) = 500 µW

This means a nominal thermistor current of
P 500 µW
I= = = 378 µ A
R 3.5 k Ω

The divider upper resistor must therefore be given by
Vdc V 10 V
I= ⇒ Rup = dc − Rth = − 3.5 k Ω ≈ 23 k Ω
Rup + Rth I 378 µ A

With a slope of −10%/°C, the thermistor resistances are
Rth ( 20°C ) = 3.5 k Ω
Rth ( 21°C ) = Rth ( 20°C ) + (1°C ) ⋅ ( −0.1/ °C ) ⋅ ( 3.5 k Ω ) = 3.15 k Ω
Rth (19°C ) = Rth ( 20°C ) + ( −1°C ) ⋅ ( −0.1/ °C ) ⋅ ( 3.5 k Ω ) = 3.85 k Ω




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RMIT University | School of Engineering | EEET2105/EEET2388 Industrial Automation


Hence the divider voltages are
R ( 20 °C ) ⋅ Vdc 3.5k Ω ⋅10 V
Vdiv ( 20 °C ) = th = = 1.32V
Rup + Rth ( 20 °C ) 23 k Ω + 3.5k Ω
Rth ( 21°C ) ⋅Vdc 3.15k Ω ⋅10V
Vdiv ( 21°C ) = = = 1.2 V
Rup + Rth ( 21°C ) 23 k Ω + 3.15k Ω
Rth (19 °C ) ⋅Vdc 3.85k Ω ⋅10V
Vdiv (19 °C ) = = = 1.43V
Rup + Rth (19 °C ) 23 k Ω + 3.85k Ω

6) A type S thermocouple with a 21 °C reference measures 12.12 mV. What is the junction
temperature TM?

Since the reference used for measurement (i.e. 21 °C) is different than the table reference (i.e. 0 °C),
a voltage correction must be applied before using the table to find the junction temperature which
corresponds to 12.12 mV.

From the table on slide 32 of the lecture notes for week 08, the thermocouple voltage for 21 °C is
between 0.113 mV at 20 °C and 0.143 mV at 25 °C. Interpolation gives
 V − VL 
VS 0 ( 21°C ) = VM = VL +  H  ⋅ (TM − TL )
T
 H − TL 

 0.143 mV − 0.113 mV 
VS 0 ( 21°C ) = 0.113 mV +   ⋅ ( 21°C − 20 °C ) = 0.119 mV
 25 °C − 20 °C 
Hence VS0 (21 °C) = 0.119 mV is the correction.

Assuming the junction temperature (unk...