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Wheatstone bridge is used for temperature compensation for strain gauges

enter image description here

If 2 strain gauges are connected like this figure. Due to temperature variation; the resistance Rg will change in the same way.

Lets say; the bridge is balanced at start. R1/R2 = (RG+DeltaR)/(RG-DeltaR)

If strain gauge resistance increases due to temperature: R1/R2 = (RG+DeltaR+RT)/(RG-DeltaR+RT)

How does this compensation work?

(RG+DeltaR+RT)/(RG-DeltaR+RT) is not the same as (RG+DeltaR)/(RG-DeltaR)

So gets the bridge unbalanced???

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  • \$\begingroup\$ You would need also R1 and R2 to be strain gauges. \$\endgroup\$ – Marko Buršič May 10 '18 at 21:38
  • \$\begingroup\$ It's said, that half or full bridge includes temperature compensation. This is half bridge. \$\endgroup\$ – Kono May 10 '18 at 21:43
  • \$\begingroup\$ Where is this said? \$\endgroup\$ – Marko Buršič May 10 '18 at 21:55
  • \$\begingroup\$ But first of all make your equations in the form like used in this forum. At least put the brackets, since the math, rules say multiplication and division have a priority over subtraction and addition. \$\endgroup\$ – Marko Buršič May 10 '18 at 22:01
  • \$\begingroup\$ elektron.pol.lublin.pl/elekp/ap_notes/… - Page 3-4 \$\endgroup\$ – Kono May 10 '18 at 22:03
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A real 4 element strain gauge is designed to have compensation like this: -

enter image description here

As you should be able to see, the temperature compensation "modifies" the gain of "both paths" in the bridge because it is in series with the excitation voltage/current. That's how temperature compensation works on a real device whether it is full-bridge or half-bridged active elements.

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