1
\$\begingroup\$

I am currently reading data from my 30kg load cell with my Arduino UNO board and passing this data to my PC. I am using INA125P to amplify the voltage of my load cell and it is established on a bread board. Here is a picture of my circuit:

enter image description here

Everything works fine, however when move/touch/shake my circuit a little bit the voltage output differs a lot. When this happens, I have to re-calibrate my load cell again. So here comes my question;

Why is my system messed up every time I touch my board ?

Why does my load cell needs to be re-calibrated, aren't they suppose to create a certain amount of millivolts for certain amount of strain ?

Thanks.

\$\endgroup\$
  • \$\begingroup\$ What's changing? The gain, or the offset? \$\endgroup\$ – Nick Johnson Jul 23 '15 at 13:38
  • \$\begingroup\$ Breadboards are not meant for sensitive circuits. \$\endgroup\$ – brhans Jul 23 '15 at 13:50
  • 1
    \$\begingroup\$ Because contact resistance and parasitic capacitance is a thing \$\endgroup\$ – PlasmaHH Jul 23 '15 at 13:58
  • \$\begingroup\$ @NickJohnson I am not sure if the gain or offset is changing. What I know is the output voltage amount for a certain kg changes, the voltage does not shift in other words the interpolation system gets messed up therefore I need to re-calibrate system. \$\endgroup\$ – 0014 Jul 23 '15 at 14:16
  • \$\begingroup\$ @brhans I`ll try establishing my circuit on a PCB I hope the problem is because the bread board itself. \$\endgroup\$ – 0014 Jul 23 '15 at 14:18
2
\$\begingroup\$

Thanks for all the suggestions it helped me a lot.

Yes it seems like the problem was the bread board. After I set my PCB circuit the shacking or touching did not impair my calibration.

Working on PCB brought another problem to occur. When my hand got close to the circuit, the output value from load cell was increasing in large amounts. This problem is solved by placing two parallel parasitic capacitors, 1uF and 100uF, between Vcc and ground.

\$\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.