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I have a kitchen weight scale containing four half-bridge load cells, each cell associated to one foot of the scale.

I want to hack it to use it as a pressure sensor and link it to an Arduino microcontroller.

With resistance measurement, I have deducted the sensors were linked using this configuration:

mia scale load cell circuit_schéma

I have removed the original PCB and connected the sensors together the same way and also to an opamp (INA122) using the E+, E-, S+ and S- connections.

However I do not have consistent values according to the place where the pressure is applied.

I am now thinking about building four parallel bridges, each containing one load cell and three regular resistors. Each bridge would be connected to its own opamp. This would allow to better calibrate the output of each bridge and improve the balance between all of them.

  1. Is it a good way to combine the sensors and get a position-independant measurement ?
  2. Is there a better combination ?

Thank you in advance.

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  • \$\begingroup\$ What is the weight range you are trying to cover? For kitchen scale I believe the range should be 0-10 kg, and each load cell usually ranges from 0-50 kg. So 4 load cells = 200 kg, which I suppose is too large for a kitchen scale? \$\endgroup\$
    – user72213
    Commented Apr 13, 2015 at 19:29

4 Answers 4

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In order for the bridge to produce an output between S- and S+ the bridge must become "unbalanced" when weight is applied. So if all 4 load cells produce equally increasing or decreasing resistances when weight is applied the bridge will remain balanced and the output will not change.

If the load cells are arranged such that the left and right halves of the bridge move in opposite directions, then the difference between S- and S+ will increase when weight is applied.

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  • \$\begingroup\$ So if I understand correctly it is a normal consequence to have a different measurement when the pressure is applied in the center or when it is applied closer to a specific load cell. \$\endgroup\$
    – nesdnuma
    Commented Dec 8, 2013 at 19:30
  • \$\begingroup\$ @nesdnuma : No. If you flip the active/passive parts of the cells around properly when you wire them up into the bridge, then when the weights are physically translated into tensions and electrically into increases in resistance, you can make them add constructively and increase the S+/S- voltage imbalance. If you have (for example) R7 & R6 swapped backwards, so that tension/resistance in R7 increases with weight/tension, then it would end up reducing the S+ value. \$\endgroup\$
    – Dave X
    Commented Nov 6, 2015 at 21:01
  • \$\begingroup\$ ... It also depends on the gauge. Rather an one active element and a temperature compensating element, some gauges have two active elements but opposing in sense. Still, nedsnuma needs to set up the polarities so that the E-/S-/E+ voltage divider moves opposite to the E-/S+/E+ voltage divider. See electronics.stackexchange.com/a/199470/30711 for \$\endgroup\$
    – Dave X
    Commented Nov 9, 2015 at 19:10
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For maximum sensitivity and linearity, you want to have one Wheatstone bridge with four active elements, and not four separate bridges.

See Kester's AD chapter on bridge circuits for an exhaustive treatment: http://www.analog.com/static/imported-files/seminars_webcasts/49470200sscsect2.PDF

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  • \$\begingroup\$ The 4 strain gauges are all in tension on the bathroom scale sensors, and should grouped 2 each on opposite diagonals, with their integral temperature-compensating resistors on the other two diagonals. \$\endgroup\$
    – Dave X
    Commented Nov 6, 2015 at 21:08
  • \$\begingroup\$ Oops -- Some load sensors have two strain gauges-one in tension when the other is in tension. You can still build a wheatstone with 8 active elements, if you are careful about the signs. See electronics.stackexchange.com/a/199470/30711 \$\endgroup\$
    – Dave X
    Commented Nov 9, 2015 at 19:04
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I think your wiring arrangement is ok. However, I would bet that one of the half bridges is wired backwards.

To determine which one might be wired backwards, I would connect a digital volt meter to the points labelled S+ and S-. I would make a note of the reading with no pressure applied.

I would then apply the same known weight to each corner (half bridge) one at a time.

If I guessed correctly three voltage readings will increase (or decrease) in one direction and the fourth will decrease (or increase). The odd one is wired incorrectly. If two increase and two decrease, I would reverse the wiring on decreasing ones.

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  • \$\begingroup\$ I have made that test but whatever the load cell tested the output voltage always increases. \$\endgroup\$
    – nesdnuma
    Commented Dec 8, 2013 at 4:06
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    \$\begingroup\$ When you apply the same weight to each load cell the output should change the same amount. Half bridge load cells are more sensitive to off center loading and side pressures. So you need to make sure the four cells have continuous contact on a flat surface. \$\endgroup\$
    – Jeff
    Commented Dec 8, 2013 at 15:08
  • \$\begingroup\$ So how is it possible to have a measurement without being dependant on the position ? \$\endgroup\$
    – nesdnuma
    Commented Dec 8, 2013 at 19:35
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    \$\begingroup\$ The outputs from the load cells need to match and all of the mechanical mounting and forces need to match. If the load is applied in the middle between two load cells they each receive half of the load. As the load is shifted towards one load the weight divides 40/60, 25/75, etc. \$\endgroup\$
    – Jeff
    Commented Dec 8, 2013 at 20:29
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Think of each load cell as a cantilevered beam with two Strain gages (SG). One is in the direction of the strain, the other in the direction of no strain. This one acts as a temperature compensator. Because the SG is a long multi-folded fine wire element occupying a small area epoxied to the beam. So, temperature strain can change its resistance by many %, therefore the compensating no strain SG. This visualization will help to find why you have a problem. The cantilever beams may be not equalized or the balance bridge is incorrect.

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    \$\begingroup\$ Technically both strain gauges are strained, one in tension, the other in compression. And this really doesn't answer the question \$\endgroup\$
    – Matt Young
    Commented Jan 22, 2014 at 15:43

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