• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Wheatstone bridge offset voltage

Status
Not open for further replies.

bigheadpirate

New Member
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!
 
Last edited:

j.p.bill

New Member
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.
 

Leftyretro

New Member
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!
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
 

bigheadpirate

New Member
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!
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top