Comments: Getting Started with Load Cells
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Hello! Nice article! But in the wheatstone bridge of a load cell how the resistance of R1,R2,R3,R4 varries with applied force? I mean which one increases and which one decreases? and how we can calculate the force(weight) using V out? That I think is not written in the article.
Hello! Thanks for the clarification re: load cell vs. load sensor. In a somewhat related question to the previous comment, is it possible to connect 4 load cells to the amplifier board? For example, could I connect them in parallel and then calibrate w/ the resulting effective circuit resistance? Any info you could provide about using multiple load cells w/ one amplifier would be greatly appreciated. Thank you!
Update (8/30/16): Figured it out -- can connect multiple load cells in parallel!
For better sensitivity, you can connect four of the three-terminal half-bridge cells in a ring by the excitation terminals, matching the colors, making a full wheatstone bridge consisting of two strain gauges per leg. You then excite with two of the sense terminals from opposite sides of the bridge, and sense on the two intervening sense terminals. See http://electronics.stackexchange.com/a/199470/30711 for a schematic.
Any way this could be used with just one of the single strain gauge load sensors (https://www.sparkfun.com/products/10245)? I'd like to put two sets of four of these in a rectangle and independently measure the forces to detect weight shifts. Getting the $56 full bridge load cells gets real expensive at qty. 8 though. Perhaps completing the bridge somehow with static resistors. Or do these need to be real fancy precision resistors?
Yes, those are the ones you would want to use with the load sensor compinator board, and the HX711 amplifier board. The load cells have 4 load sensors on them, and act as their own load combinator. So if you are looking for one dimensional measurement of weight, then I would definitely suggest the load sensors you mentioned. Hope that helps!
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 The listed equation: 2 * (500 x 10-6) = 0.1 got totally messed up somewhere. It should be: 2 * (500 * 10^-6) = 0.1% (but the parentheses aren't needed).  While the Wheatstone Voltage equation ( Vout = [(R3/(R3+R4) - R2/(R1+R2))]*Vin ) would need Ohm's Law to convert between resistance, current, and Voltage units, the equation is really derived using Kirchhoff's Circuit Law that allows one to calculate the resultant output Voltage.  Note that the convention for honoring all the great scientists whose names are used for various technical terms is to capitalize the term or reference: e.g. Ohm, Kirchhoff, Wheatstone, Ampere, Volt/Voltage (for Voltaire), Seimen, Henry, Newton, Pascal, etc.