A resistance temperature detector (RTD) is placed in a Wheatstone bridge as acco

Question

A resistance temperature detector (RTD) is placed in a Wheatstone bridge as according to Figure 1.
The RTD can be assumed to have 1st order dynamics where the temperature coefficient is given as:
= 0.001K ^-1. It is given that when R1 = 500, R2 = 600, R3 = 300, the Wheatstone bridge is
balanced when T = 280K. The input voltage is also given as: V = 8 Volts.
1. Please find the RTD resistance under the balanced condition. Subsequently, determine the
output voltage V_0 of the Wheatstone bridge when T = 340K.
2. Suppose the temperature coefficient has an uncertainty tolerance of 3% and the voltage
source V has an uncertainty of +/- 0.1 Volts. The other components of the circuit do not
contain any uncertainty. Please determine the resulting uncertainty of V_0 when T = 340K.

Explanation / Answer

[1]

For a PT-100 RTD, Res. at 0 deg. C (or -273 deg. K) = R0 = 100 Ohms

And res. at temp. T = Rt = R0 * (1 + Alpha * t), where Alpha is temp. cofficient of resistance.

The bridge is balanced at T = 280 deg. K = 280 -273 = 7 dec. C

Therefore, RTD res. at T=280 deg. K = Rt = R0 x (1 + Alpha * t) = 100 x (1 + 0.001 * 7) = 100.7 Ohms.

At balance, RTD res. = 100.7 Ohms

Furyter, at T=340 deg K = 340-273 = 67 deg. C, Rt = 100 * (1 + 0.001 * 67) = 106.7 Ohms

Vout = Vsx(R1 / ()R1 + RTD)) - (Vs/2) = 8 x (500/(500+106.7)) - (8/2)

= 8 x (500/606.7) - 4 = 6.59 - 4 = 2.59 V

[2] If Vs has uncertainity of +/- 0.1 V, then Vs = 7.9 V to 8.1 V

Therfore, Vout at 7.9 V = 7.9 x (500/606.7) - (7.9/2) = 2.53 V

Vout at 8.1V = 8.1 x (500/606.7)-(8.1/2) = 2.62 V

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