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I want to use a load cell (Capable of measuring up to 1 ton) connected to a arduino board. I am a beginner, my questions are as follows : 1. How to calculate the current generated by the load cell ? 2. If current is too high, how will We make the arduino use the current to display weight ?

p.s If you know some helpful articles that might help me learn, please suggest.

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    What load cell do you use exactly?
    – chrisl
    Dec 28, 2018 at 18:01

4 Answers 4

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My suggestion would be to make a spreadsheet with some items of known weights. Generally while using scales you’ll have a 5lb calibration weight that has a very tight tolerance. You can find the output voltage changes and map it from 0-1023.

This should mean for every 2.2lbs you will get another increment. Not extremely accurate but fine when measuring at the ton scale

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    Load cells are extremely linear. There is little reason to do such small incremental changes. I designed some scale electronics that we manufacture at my work. The automated test checks four points in order to check linearity and offsets. I have never found any reason to have more that that tested. It will show anything worth looking at.
    – Rudy
    May 29, 2019 at 0:33
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You need to use a load cell amplifier between the Arduino and the load cell. Load cells are around 350 Ohms. Four wire. And the signal that needs to be measured are very small.

The current will not be too high. The question indicates that you don't understand the application. (and therefore you are looking for help)

Here are some useful links.

https://www.hackster.io/MOHAN_CHANDALURU/hx711-load-cell-amplifier-interface-with-arduino-fa47f3

https://circuits4you.com/2016/11/25/hx711-arduino-load-cell/

A typical load cell has an output of 2 millivolts per volt of excitation voltage at the full rated weight. For example, if 5 volts were used for the excitation voltage, and the output is 2 mV/volt, then the voltage out would be 10mV at full scale. The voltage difference (compared to the unloaded load cell) is what is measure.

If the excitation voltage (just a technical name for applied voltage) were 5 volts, the voltage at the output terminals, measured from the negative terminal, would be 2.5 volts, half of the applied voltage. But this 2.5 volts is not the measured output. What is measured is the difference between the +signal and the -signal wires.

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That makes sense, most of my experience is with Mettler Toledo but I have never used the voltage for reference. A lot of the time hacking something together on Arduino requires mapping to values it likes. I had a few ultrasonic distance sensors that were just slightly different in the way they responded so I spent a while getting them just right.

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Get a multimeter and check the max output. That would be the best way to find it. If you actually have items around that weigh a literal ton that’d be the best way to find out the max current. Once that’s done you can split the current with a few resistors. If I recall correctly twice as much current would flow through a resistor with twice the value of the other. Make a y and pick the end that outputs the value you like. If you need 1/4 you should have a resistor for example with 47k and 3 together would be 141k. Connect to the 3 to get 1/4 the output current

Wikipedia has a pretty good article on Current Dividers if the output current is too high.

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