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Thread: How do you read a Hall effect amperage sensor?

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    How do you read a Hall effect amperage sensor?

    Someone asked me how you read a Hall effect sensor that is of the 'donut' type.

    The sensor data sheet is here: http://www.tamuracorp.com/clientuplo...L01ZXXXS05.pdf

    Let's say you were measuring current draw of a 5 volt device. You would tap the sensor input power to the wire that supplies the 5 volt device, pass the wire through the donut on its way to the device and ground the sensor appropriately (I presume the same ground location would work).

    There's only one more wire coming out of the sensor and it has a 10k resistor on it. It appears to cut the voltage in half, so ZERO amperage to the 5v device would read 2.5 volts, correct?

    As the device uses more power, it would read a higher and higher voltage, correct?

    If the sensor is rated for, say, 60 amps then I presume a draw of 60 amps on the wire would cause the output of the sensor to read 5 volts, correct? At that point, interpolation would let you figure out what the various voltages would mean:

    2.5 volt range = 24 amps per volt or .24 per 1/100th volt

    Is this right?
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    woe! i have that exact same sensor and was wondering the same thing! You must have read my mind!!!
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    Quote Originally Posted by Bugbyte View Post
    Someone asked me how you read a Hall effect sensor that is of the 'donut' type.

    The sensor data sheet is here: http://www.tamuracorp.com/clientuplo...L01ZXXXS05.pdf

    Let's say you were measuring current draw of a 5 volt device. You would tap the sensor input power to the wire that supplies the 5 volt device, pass the wire through the donut on its way to the device and ground the sensor appropriately (I presume the same ground location would work).

    There's only one more wire coming out of the sensor and it has a 10k resistor on it. It appears to cut the voltage in half, so ZERO amperage to the 5v device would read 2.5 volts, correct?

    As the device uses more power, it would read a higher and higher voltage, correct?

    If the sensor is rated for, say, 60 amps then I presume a draw of 60 amps on the wire would cause the output of the sensor to read 5 volts, correct? At that point, interpolation would let you figure out what the various voltages would mean:

    2.5 volt range = 24 amps per volt or .24 per 1/100th volt

    Is this right?
    Vout is read before the 10K resistor-thats just for stabilization. But yea it should read max amperage at 5v and the negative of that at 0V.

    edit:
    looking at the spec it seems to indicate 1V-4V which is kind of strange...might want to double check

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    Quote Originally Posted by justchat_1 View Post
    looking at the spec it seems to indicate 1V-4V which is kind of strange...might want to double check
    I thought that was weird, too. Vcc + 1.5 = 4volts if Vcc is 2.5.
    Quote Originally Posted by ghettocruzer View Post
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    confirmed... vcc is 2.5 reading from the fusion brain. I'm still confused though....

    I need some sort of equation that solves for Amps and my head hurts already.
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    It should say +-1.5....not sure why it doesn't


    No equation needed though-its linear:

    Current=(TotalAmps/1.5)*(Vsense-2.5)
    Where TotalAmps is the current rating of the sensor.

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    what's Vsense? the output voltage?

    Edit: running the igep through it (3 loops through the whole) gives me an output of 2.52199. (600/1.5) * (2.52199 - 2.5) = 8.796, which doesn't seem right...
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    Quote Originally Posted by tripzero View Post
    what's Vsense? the output voltage?
    Yea

    Vcc=5v supply
    Vout or Vsense is the voltage before that 10k resistor

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    You have done well.
    I was considering the 5V series except for that +2.5V offset (yes - with +1V & +4V being FSD).

    It's fine with uPCs & PICs (FB etc), but a pain for ordinary cheap voltmeters.
    (Hence I'm using the +/-15V with cheap PSU.)

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    what's it going to look like after the 10k resistor? why would I read it before?
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