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Signal Conditioning Circuit of RTD

A platinum RT...

Question

A platinum RTD having nominal resistnace of $106 Ω$ at $20^{o} C$ and $d = 0.004 /^{o} C$ is used in wheatstone bridge circuit as shown in the figure. If $R_{1} = R_{2} = R_{3} = 100 Ω$ and the supply voltage is $10.0 V$ , then the amplifier output voltage that must be able to resovle in order to resolve a ${1.0}^{o} C$ change in temperature is (Gain of amplifier =1)

A

0.155 V

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B

10 m V

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C

1 m V

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D

0.145 V

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Solution

The correct option is B $10 m V$
Redraw the circuit,

$V_{0} = 10 [V_{1} - V_{2}]$

$Case - I: T = 20^{o} C, R_{T} = 106 Ω$

$V_{0} = 10 [\frac{106}{106 + 100} - \frac{100}{100 + 100}] = 0.145 V$

$Case - II: T = 21^{o} C, R_{T} = 106 [1 + 0.004 \times 1] = 106.4 Ω$

$V_{0} = 10 [\frac{106.4}{106.4 + 100} - \frac{100}{100 + 100}] = 0.155 V$

$Resolution for 1^{o} C change in temperatue,$

$= 0.155 V - 0.145 V$

$= 0.010 V$

$= 10 m V$

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