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I would like to test the hall effect sensor of a motor (DC brushless Papst dual coil) I have for a small project of mine. I have found a bit of information online, but I cannot seem to figure out what I am doing wrong.

I tried this kind of setup to test it : garagelab.com/profiles/blogs/tutorial-how-to-use-the-hall-effect-sensor-with-arduino

Here is a picture of the motor : http://s11.postimg.org/sqfr6jmar/IMG_20150301_161751.jpg and here is the info that I have with it : http://s18.postimg.org/ra5e6gz61/IMG_20150301_161803.jpg

I am trying to read the output with an an oscilloscope. Here is what I tried : positive battery side to the violet cable, negative battery side to the green cable, positive battery side to 10ohm resistor to my oscilloscope probe and finally the blue (or yellow) cable to the probe.

The only thing I see is a constant 5V from the battery when I manually rotate the motor.

Is something wrong with the hall effect sensor ? With my setup ? With my very basic and limited knowledge of electronics ?

Thanks a lot for you help.

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You should measure the difference between the blue and yellow cable. If you use a battery (not ground-referenced) to supply the sensor, you can connect the oscilloscope ground to the blue cable and the signal to the yellow cable. You should get some signal in the range of tens of millivolts at least.

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  • \$\begingroup\$ Hi, thanks for your answer ! I ran the test with scope signal on Yellow ground on Blue. 3 V battery on Violet + and Green – Scope set up 1 V, 50ms I read the DC offset ( No AC coupling the PC scope!) and some 60Hz noise. Nothing in relation with the hand force shaft motor turning. (s17.postimg.org/ilx7yazlb/image001.png) Any tips/ideas ? Thanks a lot again ! :) \$\endgroup\$
    – Charles
    Mar 2, 2015 at 16:48
  • \$\begingroup\$ What current does the hall sensor take when you apply 3 Volts? Typical operating currents for semiconductor hall sensors are around 10 to 20 mA. If the setting (10x attenuating probe) is correct, the DC offset sound implausible, as it is around -5V which exceeds the supply voltage. I'm sorry, I'm out of ideas what you could try. \$\endgroup\$
    – Michael Karcher
    Mar 3, 2015 at 1:27

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