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Constructing a hall effect transducer

  1. Nov 26, 2011 #1
    Is it possible to construct a hall effect transducer without the use of p-type semiconductors? I've been looking into the fundamental design of a hall effect sensor, and it generally comprises some amplification/thermal control bits and pieces that all are bound to the most important part, the hall element itself. I'm familiar with all the little bits from other work. But I need the element first.

    Commercial hall effect sensors employ elements made out of superthin wafers of p-type semiconductor, from what I've seen. Gallium arsenide and indium antimonide are the norm in the industry (again, from what I've seen). However, I definitely do not have access to these materials.

    From a theoretical standpoint, the Hall effect creates a potential difference across any conducting plate that has a current running across it. However, theory often fails to translate into reality. I am seriously in doubt of the idea that I could use any old metal plate, scrub it down, and pipe a current across it to observe the Hall effect (in at least a response of tens of microvolts, which could be amplified with an ultralow offset opamp like an OP07). Is this possible, or (as I suspect) am I spouting impossibilities?

    PS: My theoretical knowledge is running a bit short here since I'm not quite sure how to analyze the hall response theoretically (which would let me figure out if building my own hall element was possible), when I don't know how to analyze the charge carrier density theoretically. Wikipedia tells me that the carrier density is usually determined by analysis with the Hall effect >_> not very useful for this problem. Theory of charge carrier density is a question for another time though.
  2. jcsd
  3. Nov 26, 2011 #2
    The hall voltage is the result of an electric field caused by the deflection of electrons (the current) when a conductor is in a magnetic field.
    The force due to the electric field is equal to the force due to the magnetic field.
    Analysing this equality give the equation for the HALL VOLTAGE :

    Vh = (BxI)/(net)
    where B = Magnetic flux density (T)
    I = current (A)
    n = electron concentration (electrons per m^3)
    e = electron charge (1.6 x 10^-19C)
    t = thickness of sheet

    With metals the electron concentration is high ≈ 10^29 and this means that the Hall voltage is small and therefore difficult to detect.
    If you make I = 1A, B= 10T and t =1mm (1 x 10^-3 m) and substitute these you should see that you get about 1μV.
    Semiconductors are used in practical Hall effect probes because they can be made with a very small free charge carrier concentration and therefore give a larger Vh
  4. Nov 26, 2011 #3
    First off, thanks for the quoted number on charge carrier density. that helps a lot. I guess I would have to pass through a ton of current to make the effect more visible? I don't think I'm going to be dealing with magnetic fields that strong. (probably on the order of magnitude of 0.00002T or possibly even less)
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