Applications of a Wheatstone Bridge (Research Paper Sample)

Application Activity
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Table of Contents
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Introduction  PAGEREF _Toc388354378 \h
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Calculations  PAGEREF _Toc388354379 \h
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Conclusion  PAGEREF _Toc388354380 \h
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References  PAGEREF _Toc388354381 \h
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Introduction
This paper mainly focuses on the applications of a Wheatstone bridge. It is mainly used with applications that use sensors like thermistors which change in resistance with the effect of temperature. The bridge has the capacity to provide null measurements which make it quite effective in giving great sensitivity. The main application depicted herein is the control circuit, which has a tank with a thermistor and a circuit control board. The whole applications work in coherent to heat water in the tank. Several calculations are discussed to show how the whole arrangement works at different temperatures and the amount of resistance the thermistors offers.
Calculations







Using the exponential equation RT = R0 QUOTE


Where:
RT = the resistance at a given temperature
R0 = the resistance at a reference temperature
T0 = the reference temperature in Kelvin (K), typically 298 K, which is 25oC
T = temperature in K
β = a constant (K) provided by the manufacturer
The resistance of the thermistor at a temperature of 40oC will be calculated as follows:
R0 = 25 kΩ
β = 4615K
T0 = 298K
T = 40oC + 273 = 313K
RT = 25
= 25e-0.7422
= 25 x 0.4761
= 11.90kΩ
The calculation is correct since the value 11.90kΩ found from the calculation corresponds to 40oC in figure 3.
At 25oC the voltage will be zero
Therefore, the balance will occur when = = 1
But RT = 25 kΩ
R1 = 11.90kΩ
Therefore; = 1
= 1 Thus R3 = 11.9 kΩ
In order to find the output voltage of the bridge at 40oC. We will have to find the current first. Thus;
I = But V = 15V and RTotal = 25kΩ
I = = 0.3A
The voltage drop across resistor VR2 will be given by:
VR2 = 0.3A x 25kΩ
= 7.5kV
A similar current flows through resistors R3 and R4. Therefore,
VR4 = 0.3A x 11.9kΩ
= 3.57 kΩ
Therefore, the output voltage will be the difference:
Vout = 7.5 – 3.57 = 3.93kΩ
A simple change that will be added to the circuit is a voltage follower.
First and foremost, we will find the resistance of the thermistor at the reference temperature 0 oC (273K).
Therefore;
RT = 25e4615 (298 – 273) / (298 x 273)
= 25e1.4182
= 25 x 4.13
= 103.24kΩ
Thus, if a voltage follower is used with all arms at 103.24kΩ then it will null at 0 oC.
The calculations that follow are:
 V = -
The voltage across the system is 15V and an off-null Voltage at 25 oC will be given as;
 V = 15= 4.576V
The revised Schematic diagram below will provide the result
In order to find the power that will be dissipated by the thermistor at 25 oC, we will use the formula:
P = IV
Where:
P = Power
I = Current
V = Voltage
But V has been given as 7.5V and I is known to be 0.3A
Therefore;
P = 0.3A x 7.5V
= 2.25W
As the temperature of the thermistor increases the resistance decreases. That can be depicted from the negative slope on the graph on figure 3. The analogy behind this principle is that as the temperature of the thermistor increases, the atoms of the thermistor start vibrating more and more. Such vibrations make the electrons flowing through the circuit to collide with the atoms thus increasing resistance (Herman, 2011).
However, the effect of temperature on the thermistor is contrary to this analogy. Therefore, the explanation here is that thermistors are made from semiconductors. Semiconductors, on the other hand, have an energy gap between the valence band and the conduction band. When the temperature is zero, there are no charges in the conduction band and so resistance is to its maximum, to some extent it behaves as an insulator (Whitaker, 1996).
When temperature starts to increase, electrons start to move in to the conduction band thus aiding in conduction thus lowering the resistance of the thermistor.
Resistors are devices meant to loose electrical energy in for...