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This is about Hall Effect from this youtube video. The voltage across a metal will change based on the strength of the magnetic field. I am wondering whether this will be accurate in reading small magnetic field such as the earth's magnetic field to point to the NORTH. From my knowledge, since the earth magnetic field is small and weak, it is very difficult to read on a multimeter, so perhaps we can amplify that small voltage to a big voltage by using op-amp.

Is this a possible solution because I am trying to make my own DIY magnetometer?

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The largest problem you will face when measuring small Hall signals is the geometric error of the Hall element itself. A typical Hall coefficient of a commercial Si Hall element is on the order of 200 ohm/tesla. This is plenty for measuring fields on the order of 10 µT! However, the zero-field offset can be several Ohm typically.

Let's say you get one with 1 ohm offset. If you read this element at zero field, you read 1 ohm Hall resistance and could infer a (erroneous) magnetic field of 1/200 Tesla = 5 mT -- or 100x the Earth magnetic field. So you see how tremendous the typical offset is compared to the Earth field. Worse, this offset drifts with temperature.

Using AC-current as suggested by Neil does not remove this offset, as this is not a voltage offset, but an offset of the Hall resistance itself. Instead, you need a quadrature excitation which means changing the current axis through the Hall cross and averaging both measured Hall resistances. This is fairly complicated to do manually without digital processing.

Therefore, many companies offer Hall ICs that integrate a Si Hall cross and the switching circuitry to implement the quadrature excitation and a gain stage to output an easily readily voltage. One example is the DRV5053.

Finally, using metals as the Hall element itself is a rather poor choice. Their normal Hall coefficients are far smaller than Si. If you insist on it because of reasons (metals scale better to very small dimensions, metals are more ductile and exhibit less strain sensitivity) some metals like Bi indeed have appreciable Hall effects when made very thin. Another option for large Hall coefficients from metals is exploiting the anomalous Hall effect. In any case, using metals is probably off the chart for hobbyists because it requires special thin film deposition equipment to fabricate the Hall element.

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  • \$\begingroup\$ There are innumerable papers using the idea a compass to measure the horizontal component of the earth’s magnetic field using compass needle deflections. These may use bar magnets with known magnetic moment (\$\text{A}\cdot \text{m}^2\$ or \$\text{N}\cdot\text{m}\cdot T^{-1}\$, which can be grossly estimated within a factor of 2 or 3 from basic physics, knowing the material used, and the mass, but would need to be improved upon through calibration.) Or they may use a well-designed electromagnet designed for the purpose, I suppose. I wonder if the OP should start there. Or maybe already has? +1 \$\endgroup\$
    – jonk
    Commented Apr 3, 2022 at 18:41
  • \$\begingroup\$ @jonk I agree if only a qualitative assessment of "north" is needed there may be much simpler - even non-electronic - solutions. But as the OP asked specifically about the Hall effect I gave my honest opinion. Difficulties notwithstanding, I believe Hall ICs are still the bread and butter for generic integrated magnetic field sensors such as in smart phones. \$\endgroup\$
    – tobalt
    Commented Apr 3, 2022 at 18:48
  • \$\begingroup\$ I've no complaints about your answer. It was on-target to the OP's question. I was just thinking out loud and hoping the OP might read the comment and explore what can be done other ways, first. One can actually use the deflection angle (or, perhaps better, the separated distance found by watching for a well-chosen deflection angle that minimizes the resulting error) to get useful results. And there's a lot of learning that takes place, given time into the project. All of which would help in better understanding what you wrote, which is kind of a "next level up" in my mind. \$\endgroup\$
    – jonk
    Commented Apr 3, 2022 at 18:55
  • \$\begingroup\$ HI sir, what is Bi metal? Are there any metals which I can buy easily from online store that can be used as Hall Effect plate :) ? Thank you \$\endgroup\$
    – Jasmine Su
    Commented Apr 4, 2022 at 2:16
  • \$\begingroup\$ @JasmineSu Bi is Bismuth. And no you probably can't buy a "plate" of metal that would be useful for a sensitive measurement of the Earth field. When using metals, the Hall element has to be very thin ( 1..100 nm), so it must be deposited , e.g. via sputtering, on an insulating carrier. I am not aware of such products. \$\endgroup\$
    – tobalt
    Commented Apr 4, 2022 at 4:37
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Amplifying a DC-excited Hall element with an opamp will be fraught with the amplifier's DC offsets, and any thermoelectric DC you get from junctions between the Hall and the amplifier.

Far better would be an AC-excited Hall element, as the output voltage is current * field, and so can be AC coupled, avoiding the DC offsets. It would of course have to be synchronously rectified after the amplifier.

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    \$\begingroup\$ The Hall effect itself gives a sizable offset for imperfect cross geometries (all crosses are imperfect). So while AC coupling could remove the DC offset of the opamp, it can't remove the much much stronger error of the Hall element itself. A quadrature excited Hall element is needed to remove the geometric error of the Hall element. \$\endgroup\$
    – tobalt
    Commented Apr 3, 2022 at 17:28

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