- Fri Dec 28, 2018 2:17 pm
#201575
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
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
