Wheatstone bridge offset voltage

Hi guys,

I'm trying to figure out why using a Wheatstone bridge eliminates the baseline offset voltage compared to using a simple voltage divider. But I can't figure it out for the life of me. I really appreciate any comments/help!

Did you Google "wheatstone bridge"? There is a lot of info out there.

You take the signal off it differentially, so the offset becomes a common-mode term. Use the right amp and it doesn't show up in the output.

bigheadpirate said:

Hi guys,

I'm trying to figure out why using a Wheatstone bridge eliminates the baseline offset voltage compared to using a simple voltage divider. But I can't figure it out for the life of me. I really appreciate any comments/help!

Click to expand...

A W bridge can be thought of as two parallel voltage dividers of two elements each. By making all three fixed elements the same value as the variable element static value, then any offset is cancelled out when measuring across the junction of the two voltage divider legs, one would only measure the effect of the variable element. These elements can made of resistors, capacitors or inductors depending on the variable elements characteristics.

Make sense?

Lefty

thanks Lefty! that answers my question.

I did google it, and that was the impression that I got, but I couldn't find the answer explicitly and didn't want to take any chances. thanks again!