OpenScale - Load-cell vs strain-gauge

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fmr300
Posts: 7
Joined: Sun Dec 20, 2009 8:56 am

OpenScale - Load-cell vs strain-gauge

Post by fmr300 » Fri Dec 28, 2018 2:17 pm

I'm confused on how a 4-wire load-cell, having 4 strain-gauges (like a bathroom scale) works, and am hoping someone can help me. Just for reference, I'm an EE so you shouldn't have to start with Adam&Eve.

When the Combinator 'combines' 4 strain-gauges, it appears to put the 4 resistances into a Wheatstone bridge config (ie, 2 separate resistor-dividers in parallel, input (Vin) is the top/bottom of the two dividers, output (Vout) is the center of the two dividers). But I don't see how this works - at rest, assuming equal distribution on the scale's platform and similar strain gauge resistances ('R') both divider voltages would be Vin/2 so Vout=0. Now, if you add weight to the platform, and again assume equal distribution to the 4 strain gauges, each gauge would be R x X.... but again, the dividers voltage would be Vin/2, and Vout=0.

Of course equal distribution will never happen, but it's the easy case for me to describe....

I'll try to answer my question with some web research, but thought someone here can point out the error in my thinking.....

Thanks in advance,

-mark

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DanV
Posts: 396
Joined: Sat Apr 25, 2015 9:29 am

Re: OpenScale - Load-cell vs strain-gauge

Post by DanV » Sat Dec 29, 2018 6:17 am

Research provided many tutorials on this subject.
Pay attention to the terms load sensor (needs multiple units and a combinator) and load cell (1 unit - no combinator).
Ultimately, both configurations need an amplifier.

https://www.allaboutcircuits.com/textbo ... in-gauges/

fmr300
Posts: 7
Joined: Sun Dec 20, 2009 8:56 am

Re: OpenScale - Load-cell vs strain-gauge

Post by fmr300 » Sat Dec 29, 2018 11:18 am

Thanks DanV - that's a far better article than what I had found so far, and does a GREAT job in explaining.... appreciated !

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