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I am involved with a 4H project looking to take some weight measurements with bee hives and am trying to figure out some 3-wire load sensors to do just that.

I have four 3-wire sensors (load sensors/strain gauges) from one bathroom scale (each sensor was at a corner). Each sensor has a red, black and white wire. The resistance between the red wire and either the black or white wire is 1k ohms. The resistance between the white and black wires is 2k ohms (the resistance between the leads on my load cells and each came away with R->B=1K, R->W=1K, B->W=2k).

Because of this, I was told each 3-wire load sensor represents 1/2 of a Wheatstone bridge (each sensor containing two 1k resistance legs).

I can get my head around the single wheatstone application but I'm confused how a scale would work when made from two wheatstone bridges. My question is, if this is so, why would a scale require two Wheatstone bridges (remember, all four 3-wire sensors came from one bathroom scale)?

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  • \$\begingroup\$ The three-wire bathroom scale load sensors have two resistive elements, one of them is force sensitive and the other is not. With yours the center tap is the red wire, but it is unclear whether the force sensitive one is the R->B or the R->W. \$\endgroup\$ – Dave X Nov 6 '15 at 18:49
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    \$\begingroup\$ The person told you something misleading. Each sensor could be well be half of a bridge, but you might want to think of each sensor as two quarters, or two eighths of a bridge and you want to distribute the variable-resistance portions of the sensors to create a balanced, sensitive bridge that measures what you want. \$\endgroup\$ – Dave X Nov 6 '15 at 20:49
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    \$\begingroup\$ Remember that these bathroom type scale sensors have unspecified time and temperature drift so not ideal for in-place applications. They are intended to re-zero before each reading so do some work characterising the devices before making expensive choices. \$\endgroup\$ – KalleMP Nov 25 '16 at 7:48
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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.

Mondo's pic from the blog post

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:

three-wire 50kg load cell ... 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.)

schematic

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.

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If all the half-bridge sensors changed their resistance exactly the same when a load was applied, can you see that they could be mounted in parallel - the effective end-to-end resistance would drop from 2 kohm to 1 kohm but that is of no consequence to a bridge measurement circuit. Even if there are disparities in the resistance between two paralleled devices then I'd bet on the error introduced being insignificant.

Maybe they used two wheatstone bridges and two differential amplifiers and summed the signals internally to get an average but I doubt it because cost would be an issue to them.

Why couldn't they use two half-bridges and two dummy load cells? It's probably cheaper and more accurate to use four half-bridges.

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    \$\begingroup\$ Is this an answer or a guess? The total weight on a bathroom scale is the sum of the weights on each support (which are not necessarily equal), and four supports are used for maximum stability when stepping on or off the scale. Therefore, four load cells are used, and their outputs are summed appropriately. \$\endgroup\$ – Dave Tweed Mar 7 '14 at 16:45
  • \$\begingroup\$ @DaveTweed No it's not a guess. I've used gauges in parallel before and they work fine. \$\endgroup\$ – Andy aka Mar 8 '14 at 12:30
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    \$\begingroup\$ Then how about rewording it to eliminate all the "I bet", "Maybe", "I doubt", "probably", etc.? \$\endgroup\$ – Dave Tweed Mar 8 '14 at 12:36
  • \$\begingroup\$ @DaveTweed It was a while ago (probably about 1990) and I can't remember how the error stuff panned out and whether it needed non-linear compensation or not. I don't think there was but, given my doubts in this area and because I'm not sitting in front of a simulator to check (the lazy way) I'm going to leave it as it is until I am more sure. \$\endgroup\$ – Andy aka Mar 8 '14 at 12:40
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I think Omega may discuss a possible explanation and schematic.

Four strain gages are used to obtain maximum sensitivity and temperature compensation. Two of the gauges are usually in tension, and two in compression, and are wired with compensation adjustments as shown in Figure 7-2 (ed: see below.) When weight is applied, the strain changes the electrical resistance of the gauges in proportion to the load.

Omega strain gauge diagram

Source from Omega for the above description

I also found a blog from 2010 that may help, too.
Blog on hacking three-wire scale gauges

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    \$\begingroup\$ the circuit from the bottom of the blog looks like a possibility. He's suggesting that the four 3-wire load cells actually get wired up into a big wheatstone bridge, with whites wired together, blacks wired together and the reds being used as E+,E-,S+,S-. Never contemplated that. I'll see that works (hopefully over the weekend). Thanks for the info. \$\endgroup\$ – user37997 Mar 7 '14 at 19:00
  • \$\begingroup\$ Yes, that one caught my attention, too! \$\endgroup\$ – jonk Mar 7 '14 at 19:06
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    \$\begingroup\$ I wonder if it matters which red wires are used for S+ & S- (the toe or heel ones) and which ones are used for E+ & E-? \$\endgroup\$ – user37997 Mar 7 '14 at 19:27
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    \$\begingroup\$ Shouldn't matter. It is symmetrical and there are no PN junctions to worry about. You will probably need a circuit to get the most out of it, though. Not just an ADC input and code. \$\endgroup\$ – jonk Mar 7 '14 at 20:03
  • \$\begingroup\$ If all the gauges are all in tension, or all the gauges are in compression, then you have to be careful about how you connect them so they don't cancel each other out. The blog post diagram is good about this--There are two stress-sensitive Rs between E+ and S-, and two stress-sensitive Rs between E- and S+, with the paired insensitive ones on the other legs. Then, if the gauges all compress at the same time, R drops, and S- is pulled towards E+ and S+ is pulled towards E-. Or vice-versa with tension. Cool. \$\endgroup\$ – Dave X Nov 6 '15 at 19:05
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How you make these work is make each strain guage into a full wheatstone bridge by using fixed 1k resistors to make up the other half. This way the scales take the weight on each of the four corners and adds them up to give the total weight on the plate.

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