Electrical – hx711 values non-linear

The strain gauge elements come with a positively stress-sensitive portion and a negatively stress-sensitive portion. If you wire them up carefully by flipping them around so the stress sensitive portions unbalance the bridge constructively, you can use all four sensors without any extra resistors. jonk's link to the blog post at http://www.nerdkits.com/forum/thread/900/ has a good hint with Mongo's diagram (copied below), and the jonk - user37977 comments on jonk's answer also help.

Basically, two diagonally-opposite sides of a wheatstone bridge are formed by the positive-strain elements of two gauges wired in series, while the other two sides of the bridge are formed from the two negative-strain elements.

With compression on all the positive-strain sensors, the active resistances are reduced, and it pulls the bridge out of balance one way, and under tension, the positive-strain resistances increase, pulling the bridge out of balance the other way.

Wire all four sensors in a big ring with maximum resistance, matching colors and initially ignoring the center tap wires. Choose two opposite center taps as E+ and E-, and the remaining two center taps as S+,S-. Put the excitation voltage on the E+/E- from the diagram above and read a force-sensitive voltage difference across S+/S-.

See https://electronics.stackexchange.com/a/75717/30711 for a good schematic and Arduino Leonardo + 3 wire Load Cells + INA125P – Analog Signal Bounce / Noise for a wiring diagram of the colored wires combining into a wheatstone bridge.

Edit: Actually, I am uncertain if OP's three wire load cells have only one active strain gauge as in Mongo's diagram. If they are like the 50kg load cell from SparkFun's https://www.sparkfun.com/products/10245 or Ebay's http://www.ebay.com/itm/4pcs-Body-Load-Scale-Weighing-Sensor-Resistance-Strain-Half-bridge-Sensors-50kg-/251873576571 they mught have a compression and tension gage both on the top surface. The Ebay site has a diagram like:

... which indicates a positive strain gauge on the red-white, and a negative strain on the red-black. (note that the coloring order in this diagram does not match the coloring order in this picture. I have a similar gauge with blue-red-black colors, and the positive strain gauge is the right pair, negative on the left.) The gauged surface on the center bar between the face-to-face coupled 'E's in the sensor should act like a parallel bar and has portions under compression and under tension, rather than purely under tension. In cross-section, the gauged bar in the center is sort of the cross-piece in a Z-shaped spring. In this case, the strains oppose each other, and, if manufactured well, the reduction of resistance in the negative strain portion will offset the increase in resistance in the positive strain portion and the total white-black resistance should be constant. One still needs to set up the bridge so that the voltage dividers move in opposite directions with added load, and 4 devices wired in a white-to-white and black-to-black loop should work as above.

Here's a schematic with gauges 1-4 as G1 G2, G3, G4 per the above specs, applying an excitation on the G1 and G3 reds, and reading the signals off the G2 and G4 reds. The G4 gauge is loaded a bit with some positive strain increasing the G4+ resistance, and some negative strain reducing the G4- resistance. Ideally, loading G4 with 25kg would produce 0.5mV/V times its 2.5V excitation voltage, producing 1.250mV across Sig+/Sig-, and stretching R8 to be 1001 ohms and compressing R7 to 999 ohms as shown. One could increase the sensitivity by a factor of 4 by increasing V1 up to the 20V (=2*10V) specification (The schematic/simulator thing is pretty cool.)

^{simulate this circuit – Schematic created using CircuitLab}

With only two devices, one should hook white-to-black and black-to white, imposing an excitation voltage from between these two junctions, and reading the differences across the reds, as increased load pulls one side high and the other side low.

I am quite a noob (I'm actually a programmer, not an electrical engineer!) - but I'm doing something similar and maybe my discoveries will help you out.

Firstly, I suggest you read this: http://www.instructables.com/id/Arduino-Load-Cell-Scale/

Yes - it's for a 4-wire load cell, but it's very similar.

Also, read this: http://airtripper.com/1626/arduino-load-cell-circuit-sketch-for-calibration-test/

FIRSTLY: The big difference between these two articles, is the latter shows exciting the load cell from the INA125 voltage reference... NOT the arduino supply. I would strongly suggest doing this - as my readings significantly stabilised (improved from 50g fluctuation to only 5g!).

SECONDLY: In your particular circuit, you cannot use pin 15 for your voltage reference (5v) - Page 11 (section "Precision Voltage Reference") of the specification says "Positive supply voltage must be 1.25V above the desired reference voltage."

http://www.ti.com/lit/ds/symlink/ina125.pdf

This means that because your circuit supply is 5v, you can only use a voltage reference pin that is less than 5v-1.25v=3.75v. (Why? It appears that the IC uses 1.25v to generate those reference voltages, meaning that the 5v and 10v pins will not actually be producing 5v and 10v for you!). That leaves only the 2.5v reference pin as a candidate. Unfortunately, that also means that if you use the same voltage reference as E+, you will be running your load sensor at 2.5v - which may not be enough excitation - you will need to read your load cell spec - but they usually want around 10v to really work well.

I originally made the same mistake, and used the 5v reference pin, with a circuit supply of 5v, but then I saw this on my scope:

That spike is a 100mV pulse every 200ms. With my calibrations, it resulted in 200g worth of error!! When I switched to the 2.5Vref, that spike went away.

SECONDLY: Why is your VrefOUT (pin 4) connected to your 5v supply? This pin should ONLY be connected to your VrefIN (pin 14 for 2.5v, pin 15 for 5v, pin 16 for 10v) AND your load cell E+. Here is my understanding of what it's for... The amplifier needs to have a consistent voltage reference, as the circuit supply may fluctuate throughout its life (i.e. depleting battery etc), so you need to give the INA125 a known voltage reference - luckily the INA125 produces 3 of them! (2.5, 5, and 10).

THIRDLY: your amplifier gain... I don't use Arduinos, but my analog inputs are referenced against 3.3v. My load cell produces about 4.1mv when loaded with 5kg - I needed to amplify that to near 3.3v, so my required gain was around 800!! If your cell output and Arduino requirements are anywhere near mine - then your gain resistor is FAR too big. Mine was 75 ohms. With such a huge resistor, I would expect you to see no change on your analog input.

So, to summarise:

1. Feed your load-cell E+ from your INA125P pin 4 - not your circuit supply. Pin 4 will be much smoother and more consistent.
2. Don't connect your pin4 to your circuit supply (marked as 5v in your diagram). I don't know why you did this.
3. You amplifier gain is probably too small, as a result of your gain resistor being far too large. If you can't be bothered calculating what resistor you need, grab a potentiometer in the range of 200R and play with it.