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The signal-to-noise ratio of a Hall-effect magnetic sensor is proportional to $$ H(f,p)=\frac{I_1 (f,p)}{\sqrt{KK'(\frac{1-f}{1+f})} \sqrt{KK'(\frac{1-p}{1+p})}} $$ with $KK'(x)=K(x)K'(x)$ and $K'(x)=K(\sqrt{1-x^2}) $ and $$ I_1(f,p)=\int_{\alpha=0}^{\pi/2} \frac{F(\alpha,\frac{1-p}{1+p})d\alpha}{\sqrt{\sin^2\alpha+(\frac{1-f}{1+f})^2\cos^2\alpha}} $$ with the incomplete elliptic integral of first type $F(\alpha,k)=\int_{t=0}^{\alpha}\frac{dt}{\sqrt{1-k^2\sin^2t}}$ and $K(k)=F(\pi/2,k)$. The two parameters $0\le f,p\le 1$ are related to input and output resistances $0\le\lambda_f,\lambda_p\le\infty $ via $$\lambda_f=\frac{2K(f)}{K'(f)}=\frac{K'(\frac{1-f}{1+f})}{K(\frac{1-f}{1+f})} \quad \lambda_p=\frac{K'(p)}{2K(p)}=\frac{K(\frac{1-p}{1+p})}{K'(\frac{1-p}{1+p})}=\frac{2K((\frac{1-\sqrt p}{1+\sqrt p})^2)}{K'((\frac{1-\sqrt p}{1+\sqrt p})^2)}$$

Questions:

Q1: Is it possible to compute $I_1$ in closed form, at least for specific values of $f,p$ (except 0 or 1)?

Q2: Is it possible to simplify $I_1$ for the special case of a symmetric Hall-plate with equal input and output resistances $\lambda_f=\lambda_p$, which means $\sqrt p=\frac{1-\sqrt f}{1+\sqrt f}$?

Q3: Numerical inspection shows that $H(f,p)$ remains constant if one transforms $(\lambda_f,\lambda_p)\to(2/\lambda_f,2/\lambda_p)$, which means $(f,p)\to(\frac{1-f}{1+f},(\frac{\sqrt{1+p}-\sqrt{2\sqrt{p}}}{\sqrt{1+p}+\sqrt{2\sqrt{p}}})^2)$

I have wrecked my brain for months about these points: fruitlessly. Any properties of $H(f,p)$ or even better $H(\lambda_f,\lambda_p)$, which can be seen in the integral above would be interesting.

$H(\lambda_f,\lambda_p)$ in 3D-plot and contour-plot

The plot shows that $H(\lambda_f,\lambda_p)$ has a maximum at $\lambda_f=\lambda_p=\sqrt 2$ and you can also see the symmetry in the contour plot (red circles). It becomes also apparent that there are infinitely many points on the iso-lines, for which $H(\lambda_f,\lambda_p)$ is constant. It would be great to have a closed formula that relates $\lambda_f$ versus $\lambda_p$ along an iso-line. For the symmetric case $\lambda_f=\lambda_p$ I have found a fairly accurate and astonishingly simple approximation (error < 2%) $$ H(\lambda_f=\lambda_p) \approx \frac{\lambda_f}{\sqrt{\lambda_f^4+\lambda_f^2/2+4}} $$ The accurate expression for this case can be cast in the form $$ H(\lambda_f=\lambda_p) = \frac{I_2(f)}{K(f)K(\frac{1-f}{1+f})} \quad I_2(f)=\frac{1}{1+f}\int_{\alpha=0}^{\pi/2}\frac{F(\alpha,\frac{2\sqrt f}{1+f})d\alpha}{\sqrt{1-\frac{4 f}{(1+f)^2}\cos^2\alpha}}$$ where $I_2$ is only a weak function of $f$ and $I_2(f)=I_2(\frac{1-f}{1+f})$ (by numerical inspection). $I_2(0)=I_2(1)=\pi^2/8$ and the maximum is $I_2(\sqrt 2-1)=(\sqrt 2/3) K^2(\sqrt{2}-1)$. Using the integral definition of $F(\alpha,k)$ one can rewrite $$ I_2(f)=\int_{x=0}^{1}\int_{y=0}^{\sqrt{1-x^2}}\frac{(1+f)dy dx}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{(1+f)^2-4 f x^2}\sqrt{(1+f)^2-4 f y^2}} $$ Transforming into cylindrical coordinates and computing over the 90° angle gives $$ I_2(f)=\int_{x=0}^{1} \frac{K(\frac{(1-f)x\sqrt{1+6f+f^2-4fx}}{(1+f)(2-x)\sqrt{1+2f+f^2-4fx}})}{(2-x)\sqrt{1+2f+f^2-4fx}}dx = \int_{z=0}^{1} \frac{K(\frac{1-f}{1+f}\frac{1-z}{1+z}\sqrt{\frac{(1+f)^2+4fz}{(1-f)^2+4fz}})}{(1+z)\sqrt{(1-f)^2+4fz}}dz $$ where we have only a complete instead of an incomplete elliptic integral - however, I still do not see the symmetry:- The last integral has some similarity to (131) in "Elliptic integral Evaluations of Bessel moments", by Bailey,Borwein,Broadhurst,Glasser (free download https://arxiv.org/pdf/0801.0891.pdf).

If we set $\hat f = 2f/(1+f^2)$ then there is a one-to-one relation between $f$ and $\hat f$ in the interval [0,1]. Then the transformation $f\to (1-f)/(1+f)$ means $\hat f\to\sqrt{1-\hat f^2}$ and the conjecture is $I_2(\hat f)=I_2(\sqrt{1-\hat f^2})$.

In the general case $\lambda_f \ne \lambda_p$ we set $\hat p = (1-p)^2/(1+6p+p^2)$. Again there is a one-to-one relation between $p$ and $\hat p$ in the interval [0,1]. Then the transformation $p\to (\sqrt{1+p}-\sqrt{2\sqrt{p}})^{2} (\sqrt{1+p}+\sqrt{2\sqrt{p}})^{-2}$ means $\hat p\to\sqrt{1-\hat p^2}$ and the conjecture is $H(\hat f,\hat p)=H(\sqrt{1-\hat f^2}, \sqrt{1-\hat p^2})$.

By partial integration one can show that $H$ and $I_1$ remain the same, if $\lambda_f$ and $\lambda_p$ are swapped.

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  • $\begingroup$ I'm very curious about the origin of this formula, especially since I work with Hall sensors. As for the closed form, I have very little hope, but I'll try to figure something out when I have more time $\endgroup$ – Yuriy S Dec 16 '16 at 18:19
  • $\begingroup$ (+1) for the effort put in this well posed and nicely formatted question $\endgroup$ – tired Dec 16 '16 at 18:29
  • $\begingroup$ @YuriyS The derivation of the main integral is partly done in: AUSSERLECHNER, Udo. A method to compute the Hall-geometry factor at weak magnetic field in closed analytical form. Electrical Engineering, 2015, S. 1-18. (there the integral has a less symmetric shape, but it is straightforward to cast it in the symmetric way). The approximate formula is discussed in: Ausserlechner Udo. Two simple formulae for Hall-geometry factor of Hall-plates with 90° symmetry. $\endgroup$ – ausserle Dec 17 '16 at 8:54
  • $\begingroup$ @YuriyS : I forgot the journal: in U.P.B.Sci.Bull., Ser. A, Vol. 78, Iss. 1, 2016, pp. 275-282 (free download). $\endgroup$ – ausserle Dec 17 '16 at 9:03

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