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RTD/Wheatstone Bridge Problem

  • Thread starter Illgresi
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  • #1
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Homework Statement



Currently writing a report about a Wheatstone Bridge circuit with RTD sensor. The sensor is to work between 0°C and 100°C and output 0 mV to 10 mV.

Rx has a nominal resistance, Rn of 100 Ω, Vs=10 V and α=0.00385 °C-1. Therefore RT = 138.5 Ω

Assuming RT is in the bottom right of a diamond circuit, R1 to the bottom left, and R2 and R3 are at the top, I have calculated that for a balanced equation R1=100 Ω and R2=R3=3608.1 Ω

The question asks us to vary the values of R2,3 and note the changes, such as if they are set
a low value i.e. R2,3=10 Ω, the output loses it's linearity.

I also noticed that the peak output voltage occurs around 118 Ω, and have a graph to show this, however I don't have the knowledge to calculate the Rmax and Vmax mathematically, only approximate it.

Homework Equations



wheat_16.gif


The Attempt at a Solution



My inclination was to reduce the formula to a polynomial of one variable and solve for 0 as you would for a quadratic equation. Inserting the values Vs=10, Rx=138.5, R1=100 and assuming R2,3=R I have reduced the equation to:

385R/(R2+238.5R+13850)

I have no idea where to go from here. Even mathcad sputters to a halt when trying to graph the equation.

Many thanks for any help!

Edit:

I've just noticed in the above equation R1 and R2 are reversed.
 
Last edited:

Answers and Replies

  • #2
Bystander
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Rn of 100 Ω,
Therefore RT = 138.5 Ω
I also noticed that the peak output voltage occurs around 118 Ω,
This is not a monotonic function of T --- you might want to look back through your calculations.
 
  • #3
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This is not a monotonic function of T --- you might want to look back through your calculations.
Hi, thanks for the reply. Not sure I follow what you mean. I understand what a non-monotonic function is, a quadratic is non-monotonic, but I can still solve a quadratic equation to find the stationary points.
 
  • #4
Bystander
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No, monotonic, meaning single-valued, over its range. For a Pt resistance thermometer, you expect R to increase with T over the entire range for which it is used; you state that you have a "peak" output voltage for a Pt resistance value of 118 Ω, yet your resistance ranges from100 to 138.5. Your "peak" output voltage in a Wheatstone bridge will occur at that peak resistance value. This means that you've got some arithmetic to check.
 
  • #5
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No, monotonic, meaning single-valued, over its range. For a Pt resistance thermometer, you expect R to increase with T over the entire range for which it is used; you state that you have a "peak" output voltage for a Pt resistance value of 118 Ω, yet your resistance ranges from100 to 138.5. Your "peak" output voltage in a Wheatstone bridge will occur at that peak resistance value. This means that you've got some arithmetic to check.
Ah I see the confusion. I'll use this image to try to explain more clearly.

bridge.jpg


Let's assume Rx is our RTD.

Rx = 100 Ω when T = 0 °C and Rx = 138.5 Ω when T = 100 °C. At 0 °C we want Vb = 0 V, therefore Rb must = 100 Ω i.e. in order for the circuit to be balanced. Using the bridge formula, Ra = Rc = 3608.1 Ω. This gives us a linear Vb output between T = 0 and T = 100 °C with Vb = 100 mV when T = 100 °C. This is the first part of the report question sheet.

The sheet then asks us to vary the values of Ra and Rc and comment on the results. I have chosen to do this for the single case where T = 100 °C. I am therefore assuming that Rx = 138.5 Ω, and Rb = 100 Ω as per above. I noticed that as Ra,b increase above 3608.1 Ω, Vb remains approximately linear, but simply decreases. However, as Ra,b decrease, the output loses its linearity - I'm guessing that's what he wants us to notice and comment on. I decided to plot this on a graph:

a9utf8.png


From observation, I noticed that Vb peaks at ≈ 118 °C. What I would like to know, is how to calculate the value precisely.

As I say, finding the stationary points of a quadratic etc, is easy. For the bridge equation, I have no clue.

Thanks again.
 
  • #6
gneill
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Your bridge really consists of two independent voltage dividers (diagram below). Ra and Rb form one, Rc and Rx the other. It should be clear that you can scale Ra and Rb up or down in value so long as they divide the voltage appropriately to accommodate the lowest value that can appear at A:

Fig1.gif

That means when Rx is 100 Ohms the voltage at B should equal the voltage at A if you want the bridge to null when the temperature is 0C. Changing the ratio of Ra and Rb will effectively alter the the temperature at which the output of the bridge shows zero volts. The voltage range of the sensor will still be determined by the variation of voltage at A, but will have an offset determined by the voltage at B. Again, you can scale Ra and Rb without changing the bridge output. It's their ratio that "sets the zero point".

Varying Rc will change the range of the output voltage as well as the offset compared to point B. So you might end up with more or less than a 10 mV range at the output.

You might consider fixing Ra and Rb and then plotting a family of curves for the output voltage versus Rx (over the complete range of Rx) for several values of Rc. That way you can discuss how the output shifts and range scales with Rc.
 

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