Wheatstone Bridge and Instrumentation Amplifier for Study Material

This article discusses the use of Wheatstone Bridge and Instrumentation Amplifier for measuring strain, temperature, and other physical quantities. It includes solutions to problems related to strain gauges, RTD, sensitivity, and more.

Question 1
One arm of the Wheatstone bridge is replaced with 1 kΩ strain gauge with gauge factor GF=3.
This strain gauge is bonded to a metal fixture as shown in Figure 1. The structure cross-sectional
area is 3 cm x 0.1 cm. The Young modulus E=2.5 x 109 Pa. The other three arms are fixed
resistors of 1 kΩ. The bridge is powered by 5V DC supply and voltage EO is the output voltage.
Gravity is g=9.81 m/s2.
a. Derive a relationship for output voltage, EO, as a function of mass m which is hanged on
the metal fixture.
Solution:
For a balanced wheatstone bridge, the ratio of the first two resistors is equal to the ratio
of the second pair of resistors and they are both equal to an output voltage of zero
R2
R1
= R4
Rg
=0
It is therefore right to say that the output voltage when the gauge is strained would be
given by
R4
Rg
=V Out
The strain in the gauge is caused by the mass hanging on the fixture.
Let E = young’s modulus;
GF = gain factor
Stress = E × Strain
But
Strain= change∈ Rg / Rg
GF
Rg is therefore given by;
Rg= strain ×GF
change∈ Rg
And
strain= mass × gravity
E
Hence;
Rg=
mg
E ×GF
change∈ Rg
The output voltage is therefore;
V Out= R4
Rg
= R4
mg
E ×GF
change ∈Rg

Simplifying the equation, we get;
EO =V Out= E R4 × change∈Rg
mgGF
b. Determine the output voltage EO when mass m=1 kg is hanged on the fixture?
Determine also EO when m=50 kg hanged on the fixture. How big is the nonlinearity
error?
Using the formula obtained;
EO =V Out= E R4 × change∈Rg
mgGF
We know that R4 = 1K ohm
EO = 2.5 x 109 × change∈ Rg
1 ×9.81 ×3
change ∈Rg= strain ×GF
Rg
And strain= mass × gravity
E
¿ 1000× 9.81
2.5 x 109
= 3.924 ×10−6
change ∈Rg= 3.924 ×10−6 ×3
1000
= 11.772 ×10−9
Therefore: for 1kg mass;
EO = 2.5 x 109 × 11.772× 10−9
1000 ×9.81 ×3
= 0.86612 V
For 50 kg mass;
EO = 2.5 x 109 × 11.772× 10−9
50,000 ×9.81 ×3
= 2.0 x 10−5 V

c. If the thickness of the fixture is doubled, how does it affect the output voltage?
Solution;
The thickness of the fixture is directly proportional to the stress subjected on the gauge.
Therefore, doubling the thickness would also double the output voltage.
d. If the output voltage of the Wheatstone bridge is to be amplified by a three op-amp
Instrumentation amplifier (as shown in Question 3a) and the maximum output voltage
of the IA is 5V, determine the measurement sensitivity when m=1 kg?
Solution;
At 1 kg mass the output is 0.86612 V
Output of the amplifier is;
V out =V 1 (1+2 R
Rg
)
= 0.86612 ×(1+ 2× 1
1+11.772× 10−9 )
= 2.59836 V
Sensitivity is therefore;
Max voltage – output voltage
max voltage
= 5−2.59836
5
= 0.480328
Question 2
A simple weight sensor consists of a steel cantilever clamped at one end with the free end
subject to a download force F. Four exact identical strain gauges (specified below) are bonded
onto the cantilever at halfway along its length as shown in Figure 2 below.
Following are the dimensions for the cantilever: length l=30cm, width w=7cm and thickness t=
0.3cm. The young modulus of steel E=200 G 𝑵⁄𝒎𝟐. It is known that the stress on this
cantilever is σ(𝒙) = 𝒕/ 𝟐 𝒙 𝑭 (𝒍−𝒙)/ 𝑰 and the moment of inertia I = 𝟏/𝟏𝟐 𝒘𝒕 𝟑. It is also
known: Gauge Factor GF=2.1 and the bridge supply voltage VS= 10 Volts and unstressed
resistance R0=120 Ω.

a. If force F=100 N, use the data given to calculate the strain and the output voltage of the
bridge halfway along the cantilever (x=0.5 l).
Solution;
The output of a wheatstone bridge is given by;
V out =V ¿ { R3
R3 +R4
− R2
R1 + R2 }
Also
strain= change ∈R ÷ R
GF = stress
youn g' s modulus , E
stress= t
2 × F(l−x)
l
= 0.3
2 × 100(30−15)
30
= 7.5N
strain= stress
youn g' s modulus , E
= 7.5
2× 109
= 3.75 ×10−9
Finding the values of resistances;
R1 = R3 = tensile
R2 = R4 = compressive
Hence
Change in R = strain × GF
R
= 3.75× 10−9 ×2.1
120
=± 6.563 ×10−11
R1 = R3 = 120 + 6.563 ×10−11
= 120.0 Ohms

R2 = R4 = 120 - 6.563 ×10−11
= 119.9 Ohms
Therefore;
V out =V ¿ { R3
R3 +R4
− R2
R1 + R2 }
¿ 10 { 120
120+119.9 − 119.9
120+119.9 }
= 0.002V
b. Compare the output voltage and sensitivity if only Strain gauges 1 and 3 are used and
strain gauges 2 and 4 are replaced with fixed resistors.
Solution;
The output voltage would be twice the output when only 1 and # are used since they are
both tensile. This causes an unbalanced state in proportion to the difference in strain.
The sensitivity of the bridge and the strain gauges increase
c. Assuming that only strain gauges 1 and 3 are used (as stated in question b) and the
ambient temperature on the cantilever causes a measurement error. Suggest a solution
to resolve this problem! Draw the bridge set-up.
Solution;
Since both the strain gauges 1 and 3 are tensile, their output is positive and hence they
will both be affected by change in temperature. The solution would be to replace either
of resistors R2 and R4 with a variable resistor to regulate it and balance the bridge when
out of balance.
The bridge looks as shown below;

Question 3
The output signal VOUT from the Wheatstone-bridge in Question 1 is to be amplified such that
it can be read by further controller circuit. For this, this signal has to be amplified and filtered
out from noise and other interference picked up in the circuit and along the transmission-line
between sensor and signal conditioning circuit. One of the signal conditioning circuit that is
suitable for this is known as instrumentation amplifier.
a. The instrumentation amplifier’s input signal is the bridge’s output (V1-V2) = VOUT from
Question 1.
R1=R2=R3=R4=R5=R6=R=25 kΩ and RG=25 Ω. It is known for the instrumentation
amplifier: V out =(V ¿¿ 1−V 2)(1+2 R
Rg
) ¿
determine the output voltage, VOUT of the amplifier.
Solution;
The output of the bridge with a unit mass is 0.86612 V
Given that V1 – V2 =VOUT(B) then;
V out =(V ¿¿ 1−V 2)(1+2 R
Rg
) ¿
¿ 0.86612(1+2 25000
25 )
= 1733.11 V
b. An instrumentation amplifier is used to amplify the output signal of a thermocouple.
Assuming this circuit has an output voltage of 100 mV for a temperature difference of
100 0C and the resistance of R1 = R2 = R3 = R4 = R5 = R6 = 1 kΩ, it is known that the
thermocouple has a sensitivity of 43 μV/0C. Determine the gain resistance RG of the
instrumentation amplifier?
Solution;
The output of the thermocouple is the output of the bridge that is fed into the
instrument amplifier.
Sensitivity = 43 μV/ 0C

Temperature change = 100 0C
Output of the thermocouple is therefore;
V out (T )= 100 × 43
1
=4,300 μV
Output of the IA;
V out =V out(T )(1+2 R
Rg
)
Gain resistance RG;
RG = 2× V out(T )× R
V out−V out(T )
= 2× 0.0043× 1000
0.1−0.0043
= 89.864 kilo-Ohms
c. State the advantages of using Instrumentation amplifier in terms of;
- input impedance
The instrumentation amplifiers provide nearly infinite input impedance and gain.
- common mode rejection ratio/CMRR
Instrument amplifiers offer very high CMRR compared to other amplifier circuits.
- input bias current
The input bias current in an instrument amplifier is less due to the low value of resistors
that are mainly used.
Question 4
An RTD forms one arm of an equal-arm Wheatstone bridge, as shown in Figure 4 below. The
fixed resistances, R2 and R3 are equal to 25 Ω at temperature of 0 0C and is used to measure a
temperature that is steady in time.

The resistance of the RTD over a small temperature range maybe expressed in the following
equation:
RRTD= R0 [1 + α (T – T0)]
Where R0 is a reference resistance measured at temperature T0 and α is the temperature
coefficient of resistivity, here α = 0.003925 0𝐶 −1
A temperature measurement is done by placing the RTD in the ambient temperature to be
measured and balancing the bridge by adjusting R1. The value of R1 required to balance the
bridge (meaning output voltage G= 0 Volt) is 37 Ω
a. Determine the temperature of the RTD.
Solution;
For a balanced wheatstone bridge;
R2
R3
= R1
RRTD
=0 V
RRTD= R3 × R1
R2
¿ 25× 37
25
= 37
Therefore;
RRTD=R0 {1+ ∝ ( T −T 0 ) }
37=25 {1+0.003925 ( T −0 ) }
37 = 25 + 0.003925T(25)
T = 12
0.098125
= 122.293 0C
b. If the RTD is subjected to ambient temperature of 100 0C and the supply voltage of the
bridge is 5 V, determine the output voltage?
Solution;
RRTD=R0 {1+ ∝ ( T −T 0 ) }

¿ 25 {1+0.003925 ( 100−0 ) }
= 34. 8125 Ohms
In a wheatstone bridge, output voltage is given by;
V out =V ¿ { R3
R3 +R2
− RRTD
R1+ RRTD }
= 5 { 25
25+25 − 34.8125
25+34.8125 }
= - 0.41 V
c. Determine the sensitivity of the measurement?
Sensitivity is the measurement per degree of temperature
Hence
Sensitivity = 34.8125
100
= 0.34 V/ 0C