When using a hall effect sensor like this one, naturally, it will be pretty close to a rapidly changing magnetic field. This means that the analog signal line coming out of the sensor will also be close to a rapidly changing magnetic field. I'm concerned about shielding this analog signal from the very field that the hall effect sensor is trying to measure. According to my back-of-the-envelope calcs, I shouldn't have to worry about this... can someone confirm my calculations?

I know that Lenz's Law tells us that induced voltage due to a changing B-Field is given by

enter image description here

and of course

enter image description here

Given the strength of the magnets in my physical setup and their position relative to my loop, I expect a maximum dB/dt of about 10Tesla / second. My pcb is the standard 0.1524mm trace/space, so I could conservatively have the loop of "analog signal + return path" be on the order of 10^-5 square meters.

All of this suggests that even with a rapidly moving magnetic field, I will have less than 1 milliVolt of noise in my (very delicate) analog signal.

Is my analysis sane, or is EMI protection a problem when measuring a rapidly changing B-field with a hall effect sensor?

  • \$\begingroup\$ "Twisted pair" could be an approach to reduce loop area to almost 0. \$\endgroup\$ – JimmyB Aug 9 '16 at 21:16
  • \$\begingroup\$ @JimmyB Does that work with PCB traces on a PCB that is about 1.5mm thick? \$\endgroup\$ – johnny_boy Aug 9 '16 at 21:27
  • \$\begingroup\$ Probably not. You'd have to wire the pins. \$\endgroup\$ – JimmyB Aug 9 '16 at 21:32

Your Answer

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

Browse other questions tagged or ask your own question.