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Wheatsone bridge thermistor ADC

A thermistor has a resistance of 9.0 k ohms in ice-water and 0.5 k ohms in boiling water. The thermistor obeys the law... see attachment

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a.
R(T) = K*exp(B/T)

T1 = 0+273 = 273 K
T2 = 100+273 = 373 K
R(T1) = 9 ohm
R(T2) = 0.5 ohm

Hence,
9 = K*exp(B/273)
0.5 = K*exp(B/373)

9/0.5 = exp(B/273)/exp(B/373)
=> 18 = exp(B*(1/273 - 1/373))
=> ln(18) = B*(1/273 - 1/373)
=> B = ln(18)/(1/273 - 1/373) = 2943.24

Hence,
9 = K*exp(2943.24/273)
=> K = 9 /exp(2943.24/273) = 1.87*10^(-4)

b.
As, Vo =0 for T = 0 + 273 = 273 K,
This is a balanced wheat stone bridge at 0 degree C
Hence,
R2 = R3 = R4 = RT(0) = 9 ohm

Let us assume, when Vo=1V, i.e., T = 50+273 = 323 K, ...

Solution Summary

Variation of resistance of a thermistor with temperature is established. The thermistor utilized in a Wheatstone bridge as a transducer, hence other resistances are estimated and potential difference across the bridge is estimated.

This system is used as an input to a 12 bit ADC. Temperature change/bit is estimated. A successive approximation method is used for ADC.

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