Answers to Sample Problems

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1. Pt100 breaks down as Pt for Platinum, a pure metal, and 100 is the resistance in W at 0 ºC. The resistance increases almost linearly with temperature. Have a look at http://www.buerkert.com/COM/316.html for an example of the resistance v temp characteristic for a Pt100. You should notice that it is not perfectly linear. The equation in the next question only applies over a small temperature range. Copper and Nickel can also be used instead of platinum. They are cheaper but less linear.

Thermistors are highly non linear, made from semiconductor materials, resistive devices.

The thermocouple is made from two dissimilar metals and is based on the thermoelectric effect – an e.m.f. is generated at the junction and it’s size is proportional to temperature. Linear, mV output.

The LM35 is an ntegrated circuit, it provides a linear voltage output (10 mV/ºC) with no offset and the range is small (-55 to +15 ºC). Advantage: linear, mV out, cheap. Disadvantage: only for air, small range.

2. (i) Plot the data correctly

(ii) Expanding the equation above we get that

R(T) = RoaT + Ro

The intercept = Ro = 100W

Slope or Sensitivity = Roa = 0.38W/°C. Therefore, a = 0.38/Ro = 0.0038°C^-1.

The textbook value of temperature coefficient of resistance for Platinum is 0.00385 ºC^-1. An alternative to the Pt100 is the Pt1000 which is an industry standard and has a resistance of 1000W at 0 ºC. At 100 ºC, it has a resistance of 1385 W, i.e. a much higher sensitivity. It finds use where higher sensitivity is required.

3. A balanced bridge means that the output (volts) is zero. This happens when the potentials on either side of the bridge are the same and this happens if the top two resistors are equal to each other and likewise for the bottom. Both bridges are balanced when the RTDs are equal to 100 W. This happens at a temperature of 0 ºC.

The resistance at 75 ºC is given by the equation in question 2:

R = 100(1 + 0.00385x75) = 128.9 W

The equation for the bridge is:

Option 1 If RTD = 128.9 W, we have: We have 345 mV at 75 ºC

Option 2 If RTD = 128.9 W, we have: We have 345 mV at 75 ºC also

This makes sense since we are simply measuring the difference between two voltage dividers. It doesn’t matter that we have swapped them around.

What is different are the voltages at points D and B. For option 1 we see from above calculation that these are 0.091V and 0.114 V respectively. For option 2, they are much higher, D is 0.886 V and B is 0.909 V.

Both options are the same as regards linearity – the voltage differences are always the same. The bridge itself is not linear; the equation y = mx + c does not apply to it, it has a different equation, so it is inherently non linear. We can reduce the non-linearity (i.e. make it more linear which is better) if we make the upper resistors in option 1 very big. This means that changes in the RTD have less of an impact in the denominator (the part below the line, i.e. RTD + 1000) so the bridge becomes more linear. But the output becomes smaller so the sensitivity and resolution are not as good. Conclusion: there is a trade off to be made here between linearity and sensitivity and that is the goal of bridge design.

4. Only the differential amp and the instrumentation amp should be used because the output from the bridge is not a single ended voltage but a difference in two voltages.

5. The instrumentation amp has a gain that can be adjusted by changing only one resistor. It also has a high input impedance (no loading effects), a high common mode rejection ratio, low input offset voltage and low temperature coefficient of voltage. (Read Principles of Measurement Systems, Bentley for more information).

6. (i) Location. Doesn’t really matter (see above) but location does affect which side of the bridge will have a higher voltage as temperature increases. Choice is between top left hand side (R1) or bottom right hand side (R4). Either will work. Let’s choose the first one, R1.

(ii) Now we can pick the opposite resistor to give 0 V at -20 ºC. The resistance from the equation at this temp is R = 100(1 + 0.00385(-20)) = 92.3 Ω. This is R2.

(iii) Now calculate the other two. In fact, we only need to calculate one as we make the same so the bridge works, i.e. set R3 = R4. We need the resistance of the Pt100 at the upper end of the range. Use the equation again: R = 100(1 + 0.00385(30)) = 111.55 Ω. The desired Vout at this temperature is 300 mV. KEEP UNITS CONSISTENT!! Stick to Volts so Vout = 0.3 V. We also need the Wheatstone bridge equation (in solution 3 above) which in this case is:

In this case, Wheatstone bridge eqn is: If RTD = 111.55 and R2 = 92.3 W, and R3 = R4we have: (I haven’t made a mistake here in putting R3 where R4 is because R3 = R4) You need to cross multiply the two fractions in the brackets to get: Cross multiply:

Multiply all this out to get: You can find the general solution to a quadratic in you exam formulae booklet. Apply here to give: This has two solutions: R3 = 412.867 Ω or 0.748 Ω. Both will give 300 mV at 30 ºC but the smaller will result in a more non linear output from the bridge. Therefore choose the larger value, i.e. R3 = 413 Ω.

Here is the Wheatstone bridge to match the requirements in the question: