0
\$\begingroup\$

I want to measure weight of 200kg using load cell. i have to use 4 load cell ( each capacity is 50kg).

My question is : is this possible to increase the capacity of load cell from 50kg to 200kg by connecting it in series.

if possible then please share me the any reference schematic that how to i connect it with series.

Thank you

\$\endgroup\$
  • \$\begingroup\$ Do you mean in series electrically, or in series mechanically? And can you share what load cell you're planning on using? There's more than one arrangement. \$\endgroup\$ – TimWescott Jan 10 at 4:37
  • 1
    \$\begingroup\$ You could measure 200kg with 1 50kg load cell by giving it 4:1 mechanical advantage. Do you have room for pulleys, gears or levers? \$\endgroup\$ – K H Jan 10 at 5:09
  • 1
    \$\begingroup\$ is this possible to increase the capacity of load cell from 50kg to 200kg by connecting it in series ... no, it is not .............. putting a load cell on each of the corners of a rectangular platform will increase the capacity 4X, because each load cell will be subject to 1/4 of the total weight \$\endgroup\$ – jsotola Jan 10 at 5:14
  • \$\begingroup\$ If you mean mechanically putting them in series this will do nothing they should be in parallel like jsotola said but it's hard to evenly distribute the load. Follow K H, one load cell and lever 4:1 is the best option. \$\endgroup\$ – Dorian Jan 10 at 8:44
  • \$\begingroup\$ Dear Timwescott, yes , four 50kg load cell are electrically connected not meachnically. \$\endgroup\$ – Shashikant Jan 10 at 11:05
0
\$\begingroup\$

You have two options: As K H said, you can use MA of 4:1 to allow you to weigh the weight, bare in mind that any higher than 200KG you will get bad results. This could be done with levers and a fulcrum. Also consider platform weight.

OR

You could use four load sensors at the corners, then take the inputs form each into a microcontroller, find the average, multiply by four, and use that. Again, account for platform weight.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.