An Analytical Theory of Hall-Effect Devices with Three Contacts

Methods: In contrast to devices with four contacts, it is not possible to determine the sheet resistance of devices with three contacts by electrical measurements like the one of van der Pauw. However, for both types of devices, the output voltage over input current depends only on resistances of the ERC, the sheet resistance, and the Hall angle, irrespective of the exact shape of the devices and the size of the contacts.


INTRODUCTION
Traditional Hall plates have four contacts and two perpendicular mirror symmetries [1 -4].Current I in is sent through two opposite contacts and the output voltage V out is tapped between the other two opposite contacts (Fig. 1a).
(1) with the Hall mobility µ H > 0, the magnetic flux density B ┴ perpendicular to the thin, plane Hall plate, the sheet resistance R sh , and the Hall-geometry factor 0< <1.The arctangent of µ H B ┴ is called the Hall angle.The electrical behaviour of such a 4C-device at zero magnetic field is described by an Equivalent Resistor Circuit (ERC) comprising six resistors with three different resistance values (Fig. 1b) [5].Irrespective of the size of the contacts, we can use some generalized van der Pauw technique to measure the sheet resistance and the resistances of the ERC [5].From the ERC, we can compute the input resistance R in between the supply contacts and the output resistance R out between the sense contacts.Thereby, the ratios of input and output resistances over sheet resistance R in / R sh , R out / R sh depend only on the lateral geometry of the Hall plate, i.e., its layout.We call them the effective numbers of squares of input and output Together with the sheet resistance these 3 DoF fully characterize the electrical behaviour of the Hall plate at zero magnetic field.Moreover, they also describe the behaviour of the Hall plate at arbitrary magnetic field.For small magnetic field, we even know an analytical relation [6]: with the incomplete elliptic integral of the first kind , the complete elliptic integral of the first kind K(k) = F (1, k), the complementary elliptic integral , and the modular lambda function L(y), defined by L(K'(k) / K(k)) = k 2 for 0 ≤ k ≤ 1 [5].There is a non-trivial symmetry [6].
In modern times, Hall plates are mostly operated in the spinning current Hall probe scheme, which reduces the offset error by 2 ½ decades (in silicon technology from 5 mT initial offset to 10 µT residual offset) [7,8].The scheme comprises various operating phases where the supply and sense contacts of the Hall plate are swapped (contact commutation).An appropriate sum or difference of output signals of individual operating phases cancels out offset errors while boosting the magnetic sensitivity.It also cancels out flicker noise, if the spinning frequency is chosen larger than twice the flicker noise corner frequency [9].For this scheme, it is convenient to use Hall plates with equal input and output resistances R in = R out or λ in = λ out .This can be readily done with layouts of 90° symmetry.The ERC has only two resistance values for the six resistors, because in Fig. (1b) we simply have to set R Df = R DP → R D .Then the generalized van der Pauw measurement has a simpler formula for λ in = λ out → λ and R sh (see (26) in [10]).The 2 DoF λ, R sh are linked to the ERC via ( ) with [10,11].In (4a,b) the minus-sign is used for and the plus-sign in the opposite case.The accuracy of (4a,b) is +/-0.02%.Again the number of squares λ depends only on the layout, and it fully determines the Hall-geometry factor at small magnetic field [12] (5a) with and c = 2.279, c 2 = 1.394, c 4 = 0.6699, c 6 = 0.4543.The accuracy of (5a) is -2%/+0%, and the accuracy of (5b) is -60ppm/+0.02%.Here the symmetry is between complementary devices Fig. (8) in [6]).
With these results, one can show that the ratio of the output signal over thermal noise is given by [6] (6) with Boltzmann's constant k b , the absolute temperature T, the effective noise bandwidth Δf, and the available Hall supply voltage V H sup = I in R in .At a given impedance R in the SNR gets largest for symmetric devices with [6] (7) Hence, in silicon technology at room temperature, a 500 Ω device operated at 2.5 V supply can achieve a maximum magnetic resolution of 566 nT in 1 kHz noise-equivalent bandwidth [6].This Hall plate needs to be quite thick (~45 µm) -alternatively one may connect 45 devices with 1 µm thickness in parallel (which can be accommodated in 150 µm x 150 µm chip area).
Recently, vertical Hall-effect devices (VHalls) have been attracting significant attention from the industry, because they allow one to measure the in-plane magnetic field components, i.e. the magnetic field components parallel to the chip surface (B x , B y ), whereas Hall plates -also called horizontal Hall-effect devices (HHalls) -respond to the magnetic field component orthogonal to the chip surface (B z ) [13,14].A vast plurality of topologies was proposed for VHalls, yet one important group of VHalls comprises a single Hall tub with only three contacts on one side and a single mirror symmetry Fig. (2) [15,16].It can also be viewed as a basic building block for more complex Hall-effect devices.Besides, this type of symmetry also applies to split-drain MAG-FETs [17].Therefore, we are interested in a theory of such devices analogous to the foregoing one for 4C-devices.And indeed, we will find many analogies and a few distinct differences between devices with 3 and 4 contacts.2).Vertical Hall-effect device with three contacts and a single mirror symmetry.

THE DEGREES OF FREEDOM AND THE EQUIVALENT RESISTOR CIRCUIT
shows a plane circular disk-shaped device without holes and with three arbitrary contacts on its perimeter.Obviously, its layout has 7 geometrical degrees of freedom α 1 , α 2 , ...α 6 plus the diameter, but we can scale the diameter to 1 and rotate the device so that α 6 → 0 without affecting its electrical properties, and then we end up with 5 DoF.We can further map the interior of the disk in the z -plane onto the upper half of the w' -plane by a Möbius transformation w' = (a 1 + a 2 z) / (1 + a 3 z) with 3 complex-valued parameters a 1 , a 2 , a 3 , which we are free to choose except for a 2 ≠ a 1 a 3 [18].Thus, from the 6 points Z 1 , Z 2 , ...Z 6 we can map 3 points at fixed positions in the w' -plane: in Fig. [19], so that only 3 points W 3 ', W 4 ', W 5 ' on the real axis of the w' -plane are free to describe the specific contact geometry.These 3 scalar parameters plus the sheet resistance give 4 electrical DoF, which fully describe the electrical behaviour of the device.On the other hand, we know from linear circuit theory that a resistive network with three terminals can be represented by its ERC, which has a resistor between each pair of contacts.This gives only 2+1 = 3 resistors Fig. (3c).Hence, we have 4 DoF W 3 ', W 4 ', W 5 ', R sh in the w' -plane, which are mapped to just 3 DoF R a , R b , R c in the ERC!In other words, for a given ERC we can freely choose one of the 4 DoF in the w' -plane.For example, we could choose a value for the sheet resistance and select appropriate three parameters W 3 ', W 4 ', W 5 ' to achieve any required ERC.Therefore, it is impossible to determine the sheet resistance from electrical measurements on a 3C-device at zero magnetic field.Finally, we increase the symmetry even further according to Fig. (5a).Now the device has a 120° symmetry and the layout in the z -plane has only one scalar parameter θ that affects the electrical behavior.After a Möbius-transformation the layout in the w' -plane has still one scalar parameter .The second parameter is not free -it follows from the first one .With the sheet resistance we have 2 electrical DoF, but the ERC has only a single resistance value (Fig. 5c).Thus, for a given electrical behaviour at zero magnetic field we can choose any arbitrary value for R sh and select k f according to the following equation.

(10a)
Therefore, R H / R sh and R D / R sh represent only a single DoF, and the 2 DoF of the ERC are covered by R H / R sh and R sh .The reason, why we can measure sheet resistance with van der Pauw technique is: the ratio R H / R D is accessible to electrical measurement and it depends only on the layout and not on the sheet resistance.From ( 6), ( 14), (15) in [10] and (A15) in [5] we derive the relation
If we reduce the symmetry according to Fig. (7a) the device has two orthogonal mirror symmetries and thus different input and output resistances.This gives 3 electrical DoF λ in , λ out , R sh (see (2a,b)).The ERC has three resistance values R H , R Df , R Dp which are related via the sheet resistance: From [5] we use (1a) to express R sh (1 / R H + 1 / (2R Df )), and then we insert (1b) and (2c) of [5] into (A14) of [5].Combining both results we can express R Df / R sf as a function of R Dp / R sh and R H / R sh : Consequently, the 3 electrical DoF can be represented by R Dp / R sh , R H / R sh , R sh .
We skip the discussion of 4C-devices with 4 and 5 DoF and conclude with the case of entirely asymmetric devices.The 4 contacts are defined by 8 arbitrary end-points, 3 of which we can map via Möbius-transformation onto defined points on the Re {w'}-axis.This gives 5 DoF of the layout plus the sheet resistance.So we end up with 6 electrical DoF, which matches the 6 resistances in the ERC [8].If we normalize these 6 resistances in the ERC by the sheet resistance, there must be one relation between them: the 6 th one follows out of the first 5 ones.In a weaker form this has already been mentioned in the appendix of the seminal paper by Van der Pauw [20].To sum up, in this section, we have shown two remarkable differences between 3C-and 4C-devices: (1) from mere electrical measurements on a 3C-device and without knowledge of the device geometry one cannot deduce the sheet resistance, however, for 4C-devices this works irrespective of the size of the contacts.(2) for a fixed shape of the Halleffect region and a fixed ERC (i.e.fixed electrical behavior) there is exactly one contact geometry and one sheet resistance in the case of 4C-devices, however, in the case of 3C-devices we will find infinitely many contact geometries and sheet resistances (whereby we assume plane devices, simply connected region, homogeneous and isotropic conductivity, constant thickness, and the contacts must be at the perimeter).

THE ERC OF A 3C-DEVICE WITH SINGLE MIRROR SYMMETRY
Here we compute the two resistances R d , R e of the device from Fig. (4a).Thereby, we look for operating conditions with symmetric potential distributions at zero magnetic field, which can be generated by only two contacts, because this leads to comparably simple, closed-form conformal transformations in terms of elliptic integrals.

Current Flow Across the Line of Mirror Symmetry
In Fig. (4a) the Re{z}-axis is the line of mirror symmetry.If we connect C 1 to +1V and C 3 to -1V, C 2 and the Re{z}axis will be at 0V and we need to study only the potential distribution in the lower half of the device with the two contacts C 1 , C 2 .Fig. (8) shows a sequence of conformal transformations, which map the semi-circular region onto a rectangle, with the contacts at opposite sides, so that we immediately know the resistance between them.The first transformation is [21]: only in an isotropic scaling factor w = (tan (α 2 / 2)) 2 w'.We symmetrize the contacts by the Möbius transformation: A final Schwartz-Christoffel transformation maps the upper half of the t-plane onto the interior of a rectangle in the q-plane (see also Fig. (3d) in [22]).

(14a)
sn(F(t, k), k)=t is the Jacobi sine-amplitude function.The aspect ratio of the rectangle gives the resistance between C 1 and C 2 in this operating mode

Fig. (8).
(a-d) Sequence of transformations that map the lower half of a circular device with current flow across its axis of single mirror symmetry onto a rectangle z → w → t → q with homogeneous current density.

Theory of Hall-Effect Devices Three Contacts
Open Physics Journal, 2018, Volume 4 21

Current Flow Along the Line of Mirror Symmetry
If we connect C

(16a)
The aspect ratio of the rectangle gives the resistance between C 1 and C 2
A final Schwartz-Christoffel transformation maps the upper half of the -plane onto the interior of a rectangle in the -plane.

The ERC and its Properties
Comparison of (14b) with the ERC in Fig. 4c gives Comparison of (16b) with the ERC in Fig. 4c gives Solving (17a,b) for the two resistances of the ERC gives Since R d , R e fully define the electrical behaviour of the device, (17a,b) imply that R sh λ d , R sh λ ed also describe the 2 electrical DoF.Thus, R sh λ d and R sh λ ed are independent of each other, and therefore λ d and λ ed are also independent of each other.Analogous to (18a) we can define (18c) so that any single one of the parameters λ d , λ ed , λ e can be expressed by the other two.
On the other hand, we can invert (13d) and (15d) to express two parameters out of W 1 , W 2 , W 3 by k d , k ed or by λ d , λ ed .

(19a) (19b)
We can obtain any arbitrary ERC by picking some arbitrary value for W 2 : then W 1 and W 3 follow from (19a,b).In particular, we may choose W 2 = -1.Then we can model all possible ERCs by W 1 (-1,0) and W 3 (-∞, -1).Thus, we have reduced the problem in three dimensions to a problem in two dimensions.Alternatively, we may also write . location and size of contact C 1 -so that we can obtain any arbitrary ERC with α 3 , i.e. with a fixed size of contact C 2 (the location of C 2 is given by symmetry).This has an important application for VHalls with three contacts: we can choose any convenient size for the center contact C 2 -by playing around with location and size of the outer contacts we still have all 2 DoF at our option.
With (18b) this means R e > 0. With the L.H.S. of (14b), (16b) and with (28) in [22] this also means which limits the allowed region in (k d , k ed ) -space to a narrow region.
If we swap isolating boundaries with contacts in Fig. (4a) we obtain the complementary device, which has also three contacts and a single mirror symmetry (see Figs. (11a, b)).Denoting the parameters of the complementary device by an overbar and the conjugate complex by an asterisk, we get the following relations.If a device has electrical symmetry λ d = 3λ ed , then also its complementary device has electrical symmetry .

THE MAGNETIC SENSITIVITY OF A 3C-DEVICE WITH SINGLE MIRROR SYMMETRY
A 3C-Hall-effect device can be operated in various operating modes (see Fig. (6) in [19]).For each operating mode one can find a spinning current scheme that cancels out offset errors perfectly, as long as the device is assumed to have electrical linearity 1 .First we show that even though the 3C-device may be entirely asymmetric like in Fig. (3), its 1 Electrical linearity means that the resistance values in the ERC do not depend on the potentials at the various nodes.However, the device may be magnetically nonlinear, which means that the resistance values in the ERC may well depend on the applied magnetic field -so we do not have to limit this discussion to small magnetic fields.
We use the principle of superposition as introduced in [24].Let us consider an asymmetric VHall device with three contacts.In a first operating phase ph 1 the device is supplied at its outer contacts (see Fig. (12a) and the ERC in Fig. (12b).We can use the ERC to compute the potentials and currents at the contacts of the device but we have to add an extra Hall term whenever a contact is left of the current flow through the device (and we subtract it, when the contact is right of the current flow2 ).We can think of two further operating phases ph 2 and ph 3 , where the current flows between an outer contact and the mid-contact.Phases ph 2 and ph 3 are chosen such that their superposition gives phase ph 1 (see Fig. 12c-12f).This means that the sum of currents into each contact in phases ph 2 and ph 3 must equal the current into the respective contact in phase ph 1 , and the sum of potentials at each contact in ph 2 and ph 3 must also equal the potential at the respective contact in ph 1 [25].With linear circuit theory we get the following potentials at contact C 1 .
Note that in ph 3 the Hall action of the VHall device tries to reduce the potential at C 3 by , but the ground wire ties C 3 to 0 V thereby lifting all other potentials in the device by .Superposition of ph 2 and ph 3 gives ph 1 :

(23d)
Inserting (23a-c) into (23d) gives .Since this is valid for all B ┴ we can differentiate the equation with respect to B ┴ and it follows S i (ph 2 ) = S i (ph 3 ).We repeat the same for contact C 2 . ( With (24a-d) it follows S i (ph 1 ) = S i (ph 3 ).Hence, the current related magnetic sensitivity S i is equal in all phases ph 1 , ph 2 , ph 3 .This holds for 3C-devices with and without symmetry.Thereby the magnetic field may even be strong, so that the resistances R a , R b .R c and the current related magnetic sensitivity S i become nonlinear functions of B ┴ -the principle of superposition requires only electrical linearity, not magnetic linearity.According to [19] we can define the Hallgeometry factor of 3C-Hall-effect devices by

Theory of Hall-Effect Devices Three Contacts
Open Physics Journal, 2018, Volume 4 27

The Weak Field Hall-Geometry Factor of a 3C-Device with Singe Mirror Symmetry
For small Hall angles one can use a perturbation approach, where the potential in the Hall-effect region is developed into powers of µ H B ┴ and only the lowest order term is used to compute the Hall output signal [26,27].The procedure is developed in detail in [22]: In a first step we compute the potential at zero magnetic field.Thereof, we get the electric field E p along the isolating boundaries.However, in the presence of a magnetic field, the Hall effect gives rise to an additional component of the electric field, which is normal to the isolating boundary E n = µ h B ┴ E p with the Hall angle .We take account of the component E n in a second step, where we tie all supply contacts to zero potential, and impose a perpendicular current density on the isolating boundaries J n = σ 0 E n , with σ 0 being the conductivity at zero magnetic field.As in the first step also in this second step we use the isotropic conductivity σ 0 throughout the Hall-effect region.The output contacts are at unknown potential V out and -depending on biasing conditions -usually no net current is flowing in or out of them.Solving the net current condition at the output contacts returns V out proportional to µ h B ┴ .The advantage of this method is that we can use symmetries of the device, whereas in the case of strong magnetic field these symmetries get lost.Another asset is that in the course of this calculation we can re-use conformal mappings of the ERC-computation in section 3.
From section 4.1 we know that the Hall-geometry factor is identical in all operating phases.So we choose current flow from contact C 1 to C 3 in a circular device with single mirror symmetry of Fig. (8), because it gives highly symmetric current flow lines.In step 1 this will give us the inhomogeneous electric field along the isolating boundaries Similar to [22] we make the ansatz for the potential in the -plane of Fig. (9d), whereby .This satisfies two boundary conditions and where V out is the change in potential at contact C 2 caused by the action of the small magnetic field.The net current into the output contact must vanish.Thus, (27a) (26) in (27a) and integration gives

(27b)
The boundary conditions on the left and on the right edge of the rectangle in the -plane are Positive J n on the left edge = 0 means that current flows into the Hall-effect region, whereas positive J n on the right edge = 1/λ d means that current flows out of the Hall-effect region.Introducing the ansatz ( 26) into (28a-c) and making a Fourier series expansion gives the unknown coefficients a n , b n in (27b) We insert (29a,b) into (27b), use J n = µ H B ┴ J p , and reverse the sequence of summation and integration, thereby using .

(30)
In (30) we only have to determine the transformation of current density from q-plane in Fig. (8d) to -plane in Fig. (9d), which is detailed in Appendix A. The subtraction in (30) means that the current flowing along the boundary between the supply contacts C 1 and C 3 reduces the Hall output signal.Finally, the weak magnetic field limit of the geometry factor of a 3C-Hall-effect device with single mirror symmetry is given by (31) with the abbreviations L d = L(λ d ), L ed = L(λ ed ).Equation (31) is the core result of this work.It is valid for any operation mode, where current I in flows into one contact and out of another contact and voltage V out is tapped at the third contact.λ d , λ ed are the 2 DoF of the layout and with (17a,b) they can be expressed by ratios of resistances of the ERC over the sheet resistance.Thus, (31) gives the low field limit of the Hall-geometry factor as a function of purely electrical parameters R d / R sh , R e / R sh , irrespective of the geometry of the device.

Discussion of Magnetic Sensitivity and SNR of 3C-Devices
With numerical integration it is straightforward to plot versus its 2 DoF in the allowed region 0 ≤ λ ed ≤ λ d (cf. (20a)) as shown in Fig. (13).There the black solid curve on the surface represents for electrically symmetric devices with R d = R e which means λ d = λ e = 3λ ed .Obviously, → 0 for small λ d , λ ed , i.e. large contacts.On the other hand, with (21) we know that for point-sized contacts at arbitrary position we are located at the far end of the black solid curve.There → 1 [19].If we plot the points on the black solid curve versus λ ed we get the same plot as in Similar to 4C-devices, we note also for 3C-devices a symmetry of the Hall-geometry factor.For devices with electrical symmetry λ d = 3λ ed numerical inspection suggests the following conjecture

(32)
The physical significance is also the same: it links the Hall geometry factor (λ ed ) of a device with the Hall geometry factor of its complementary device from (22e).Therefore we only need to know the Hall geometry factor for small contacts with because we can obtain its values for large contacts with the above symmetry relation.A simple approximation is (33) with an accuracy of -2.3%/+1.7%.In [19] we showed that at a given impedance level the signal-to-noise ratio (SNR) of a 3C-Hall-effect device is proportional to .This parameter is plotted in Fig. ( 14) for all possible devices.Similar to 4C-devices we note a clear maximum and this maximum occurs for devices with electrical symmetry: for and .Such devices with circular shape may have various contact sizes, such as α 1 = 30 ). 3 Meanwhile it was proven rigorously that any Hall device with three contacts and at least one mirror symmetry has the same SNR as its complementary device [30]./sqrt(λ ed λ d )

Theory of Hall-Effect Devices Three Contacts
Open Physics Journal, 2018, Volume 4 31

THE VERTICAL HALL-EFFECT DEVICE WITH THREE CONTACTS
Such a vertical Hall-effect device is shown in Fig. (2).The Hall-effect region is a tub with contacts at its top side.In silicon technology, the tub may be a CMOS n-well, a deeper high-voltage CMOS n-well or an epitaxial layer, and the ohmic contacts are made by shallow n + source/drain diffusion.We neglect the depth of the contacts and the inhomogeneous doping profile, and we assume a rectangular cross-section of the Hall-tub.Then we can apply our theory to clarify, if it is possible to optimize such devices despite the limitation that all contacts have to be on the top side.
To this end we simply have to find a mapping from the rectangular cross-section of Fig. (2) to the half-plane geometry in Fig. (4b).This is shown in Figs.(15a-c).There we draw a scaled rectangular device in the q-plane and transform it to the t-plane similar to (14a).

(34a)
The parameter k is given by the aspect ratio of the Hall-tub.

(34b)
d and r are the depth and the length of the rectangular Hall-tub (see Fig. 15a).The mapping of the contacts is where r a is the length of the center contact, r b is the spacing between center and outer contacts, and r c is the length of the outer contacts (see Fig. 15a).A degenerate Möbius transform  Summarizing these findings we may assume r, d, r a and derive r b , r c for optimum SNR:

 
For (39a) we used (34b,c), for (39b) we used (38) and (39a), for (39c) we used (34d) and (39b), and for (39d) we used (34e) and (39c).For real-valued F(w,k) with 0 ≤ k ≤ 1 it must hold 0 ≤ w ≤ 1.Hence, from (39d) we obtain a maximum allowed: The meaning of ( 40) is: if we assume a given aspect ratio d / r of the Hall tub, we can only realize optimized devices (i.e.devices with electrical symmetry and with ) if r a / r is small enough.For practical reasons r a must be larger than the feature size of the semiconductor technology.Fig. (16) shows a plot of the R.H.S. of (40) versus d and r.Obviously, for small d we must use very small r a .For fixed d the length r a can be largest for r → ∞, namely ln(4/3)d/π 0.092d.On the other hand, if one can use the entire chip as Hall-tub, d/r is large, and this means .Thus, even for deep Hall-tubs the length of the center contact must be less than 5% of the device length to achieve -and for shallow Hall-tubs it must be even shorter.With respect to minimum feature size r b and r c are less critical than r a , because from (39c,d) it follows r b > r a and r c > r a .In practice, one should take care that too small r a and r b gives too large electric field, which leads to velocity saturation, electrical nonlinearity, temperature gradients and finally to reduced magnetic sensitivity and to poor residual offset at the output of the spinning scheme.For a device with electrical symmetry but with we have to replace in (40) by  , and this requires even smaller r a /r.

Example:
Please note the usefulness of an analytical treatment in this context.Not only does it prove, that a device with contacts only on one side of the rectangular Hall-region can still have electrical symmetry despite its degenerate geometry -it also shows, that this is possible for all rectangular aspect ratios and what the penalty is, that one has to pay for shallow Hall-tubs.It would have been painstaking to find the optimum of this problem with 4 DoF (r a /r, r b /r, r c /r, d/r) by purely numerical methods.

Fig. (17)
. FEM-simulation of a 3C-VHall-device with d = 5µm, r = 20µm, r a = 0.45µm, r b = 0.633548µm, r c = 6.78451µm operated in phases 1 and 2. In both phases the voltages at the input contacts and the magnetic sensitivities are identical and match with our analytical theory up to better than 649 ppm and 40 ppm, respectively: , .The color coding denotes the electric potential and the grey lines are current streamlines.

CONCLUSION
In this paper, we gave an analytical theory of Hall effect devices with three contacts and a single mirror symmetry.This class of devices is of considerable practical relevance, because it includes split-drain MAG-FETs and many Vertical Hall effect devices.The Equivalent Resistor Circuit (ERC) at vanishing magnetic field has three resistors with  We aim at a device with length = 20 µm.The Hall-tub is 5 µm deep.With (40) we may choose r a = 0.45 µm and from (39c,d) it follows r b = 0.633548 µm, r c = 6.78451 µm.The results of a finite element (FEM) simulation on this device are shown in Fig. (17).The FEM model used a conductivity of 1 S/m and a sheet resistance of 1 Ω at zero magnetic field.The complete conductivity tensor with the Hall-effect was 2 ).The mesh had 1.9 million elements and this gave 3.8 million equations.At zero magnetic field the FEM simulation gave a resistance between the outer contacts of 1.154369 Ω (phase 1), and between the center contact and the right contact it was 1.15395 Ω (phase 2).This matches up to 287 ppm and 649 ppm with the analytical formulae (37a,b).Next, the Hall-output voltage was computed for phases 1 and 2 in the limit of vanishing magnetic field.In both cases they agree up to 40 ppm with our analytical formula (31).In both phases the device was supplied by 1 A. Then the Hall-geometry factors are identical up to 0.1 ppm and the supply voltage matches up to 363 ppm for low and high magnetic fields in the range µ H B ┴ = 0.01, 0.02, 0.05, 0.1, ... ,5. two different resistance values, but the device geometry has three parameters (two for the layout and one for the thickness).Hence, a 3C-device has 2 electrical DoF and 3 geometrical DoF.As a consequence, one cannot determine the sheet resistance by electrical measurements on a 3C-device.This is a striking contrast to devices with four contacts, whose Hall-output voltage is a unique function of the ERC for fixed input current and Hall angle, whereas the Halloutput voltage of 3C-devices is not defined by the ERC alone -in addition one needs the sheet resistance.We also gave an analytical formula for the weak magnetic field limit of the Hall-geometry factor as a function of the 2 DoFs of the device layout.Various properties of this Hall-geometry factor of 3C-devices were discussed.Numerical values were given in tabular form and some of them were checked by finite element simulations.The maximum signal-to-noise ratio (SNR) at given impedance level is obtained for optimum devices, where the resistance between any two contacts equals times the sheet resistance.It was also shown that VHalls with three contacts can be optimum for arbitrary depths and lengths of their tubs, however, their center contact has to be smaller than 4.6% of the tub length and smaller than 9.2% of the tub depth.

COMPETING INTERESTS
The author declares that there are no competing interests.

CONSENT FOR PUBLICATION
Not applicable.

APPENDIX A
Here we compute the integrals in (30).We start with on the straight line .The mapping transforms the current density with in the interval 0 ≤ u ≤ 1 (cf.Fig. 8d).Input voltage and current are linked via (17a) V in = 2λ ed R sh I in .With (16a) we have in -1 / k d ≤ ≤ -1 .There it holds

(A1)
where we used the substitution y = -x.Thus, we get

(A2)
Next, we change the integration variable du = (du / dt)(dt / dw)dw.For du/dt we use (14a) in the interval

Theory of Hall-Effect Devices Three Contacts
Open Physics Journal, 2018, Volume 4 37 Plugging (A5) and (A9) into (30), we get with (25) in the limit of small magnetic fields

(A10)
where we used the abbreviation .With [28,29] we get . The rest of (A10) can be integrated in parts with the same formulae by Prudnikov.J o is solved by partial integration.With (52a,b) in [22] we have Inserting (B2a,b) into (B1a) gives

(B3)
With (A17) and (A18) from [5] this leads to For λ ed = 0 we set L ed = 1 + dL ed with small dL ed < 0. We develop the second summand in the integrand in (31) into powers of dL ed and integrate.The result goes to and so it vanishes for λ ed = 0.The first summand in the integral in (31) can be integrated in parts The integral at the R.H.S. of (B5) is finite at L ed = 1 and .So the integral in (31) remains finite for λ ed = 0, but the numerator .Therefore it holds .
With (19c,d) we see that λ ed = 0 means α 1 = 0 while α 2 , α 3 are arbitrary.This means that contacts C 1 and C 3 touch and the Hall signal disappears.
Fig. (1).(a) Traditional 4C-Hall-plate with current streamlines at strong magnetic field µ H B ┴ = 1 pointing out of the drawing plane.(b) Equivalent Resistor Circuit (ERC) of the same Hall plate at zero magnetic field.
Fig. (4b).With the sheet resistance this gives 3 DoF (k 12 , k 23 , R sh ), however, the ERC has only two resistance values (R d , R e ).Again we may freely choose one out of k 12 , k 23 , R sh to obtain any given ERC.

Fig. ( 4
Fig. (4).(a) 3C-HHall with a single mirror symmetry axis Re{z}.(b) Its conformal transformation on the upper half of the w' -plane has 3 electrical DoF k 12 , k 23 , R sh .(c) Its ERC has only 2 electrical DoF R d , R e .
6 resistors, but with the high symmetry we have only two resistance values R H , 2R D Fig. (6bc) [10].

2
Fig. (9).(a-d) Sequence of transformations that map the lower half of a circular device with current flow along its axis of single mirror symmetry onto a rectangle z → w → → with homogeneous current density.
2 < -1/3 and with -∞ ≤ W 3 ≤ -3.Since this device must also have electrical symmetry, the ERC leads to R d = R e .With (18a,b,c) it follows λ d = λ e = 3λ ed .A device without geometrical symmetry may still have electrical symmetry R d = R e with λ d = λ e = 3λ ed .With (14b), (16b) this means K'(k d ) / K(k d ) = 12K(k ed ) / K'(k ed ).This modular equation of degree 12 can be solved in two steps by K

Fig. ( 10 ).
Fig. (10).All possible parameter-sets α 1 , α 2 , α 3 for circular 3C-devices with single mirror symmetry from Fig. (4), that have electrical symmetry R d = R e .The straight solid black line denotes devices with geometrical 120°-symmetry.The rest of the surface denotes devices without geometrical symmetry but with electrical symmetry.The surface is symmetric: each device right of this line has a complementary device left of this line where contacts and isolating boundaries are swapped.Two examples of such devices are shown and their respective locations on the surface are indicated.The dashed lines denote devices with constant λ ed and λ d and thus identical (i.e.constant Hall output signal at fixed supply current, see (31)).

C 2 _ 4 . 1 .
Fig. (12).(a, c, e) Asymmetric 3C-VHall in operating phases ph 1 , ph 2 , ph 3 .(b, d, f) Its respective ERCs.The dashed lines in (a, c, e) denote the global current flow in the Hall-effect regions.The potentials at the contacts are computed by the ERCs in (b,d,f) and the extra termsshown in (a, c, e), which have to be added to the potentials at the indicated contacts to account for the Hall effect.
is constant for all phases ph 1 , ph 2 , ph 3 , too.If we know and the ERC, we know the Hall output signals in all operating conditions.


Fig. (8d) this electric field is simple to compute and homogeneous.It becomes inhomogeneous via the transformations q → t → w.For step 2 we look at the complete original device in Fig. (4) with its symmetry.There we note that with C 1 and C 3 at zero volts and the current J n imposed on all isolating boundaries the current flow of the Hall reaction will not flow across the real axis.So we can use the lower semi-circular region of Fig. (9a) to compute the Hall output voltage V out .It is easier to use the w-plane in Fig. (9b), where we already have an expression for J n in the intervals and from step 1.We can transform this current via w → → onto the rectangle in Fig. (9d), where J n is impressed on the boundaries and .To sum up: the homogeneous current J p along the isolating boundary in the q-plane is transformed into the -plane via q → t → w → → where it defines the boundary condition J n = µ H B ┴ J p on parts of the isolating boundary.The solution of the potential in the -plane finally gives the Hall output voltage.

Fig. ( 18 )
Fig.(18) of[19].Hence, our analytical formula (31) is consistent with results from numerical simulations on 120°s ymmetric 3C-HHalls.Interestingly, in Fig.(13a) the black solid curve does not lie on the crest of the surface: for fixed λ d the function has its maximum for λ ed > λ d / 3, thus, not for symmetric devices.At fixed λ d the function goes to zero for small λ ed (which means that the spacing between contacts C 1 and C 2 gets small) and for λ ed → λ d (which means that contacts C 1 and C 3 get small while C 2 remains finite) (see Appendix B).

Fig. ( 13
Fig. (13).The weak-field Hall-geometry factor of devices with three contacts and a single mirror symmetry versus its 2 DoF of the layout.In (a) the 2 DoF are λ ed , λ d in (b) the 2 DoF are the resistances of the ERC normalized to the sheet resistance R d /R sh , R e /R sh and in (c) the 2 DoF are the resistances between two contacts normalized to the sheet resistance R C1 → C2 /R sh , R C1 → C3 /R sh .The black solid curves denote devices with electrical symmetry R d = R e (i.e.λ d = λ e = 3λ ed and R C1 → C2 = R C1 → C3 ).R e 2R d means the parallel connection of R e and 2R d .

Fig. ( 14 Table 1 .
Fig. (14).forall 3C-devices with single mirror symmetry versus its 2 DoF λ d , λ ed .The black solid curve denotes devices with electrical symmetry R d = R e (i.e.λ d = λ e = 3λ ed ).This curve goes right over the peak of the surface at with .

Fig. ( 15
Fig. (15).(a) Vertical Hall-effect device with three contacts and single mirror symmetry in the z-plane.(b) The same device in the normalized q-plane.(c) Its conformal transformation on the upper half of the t-plane.(d) Its final transformation on the lower half of the w' -plane: there it is rotated by 180° to Fig. (4b) for W a ' = W 4 ' = k 23 and W c ' = W 6 ' = k 12 .

Fig. ( 16
Fig. (16).Minimum required r a of an optimum 3C-VHall device of Fig. (15a) with .For a given depth d and length r of the Hall-tub the length r a of the center contact has to stay below the respective curve (see (40)).