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Typical Hall plates for practical magnetic field sensing purposes are plane, simply-connected regions with peripheral contacts. Their output voltage is the sum of even and odd functions of the applied magnetic field. They are commonly called offset and Hall voltage. Contemporary smart Hall sensor circuits extract the Hall voltage via spinning current Hall probe schemes, thereby cancelling out the offset very efficiently. The magnetic field response of such Hall plates can be computed via the electric potential or via the stream function. Conversely, Hall plates with holes show new phenomena: 1) the stream function exists only for a limited class of multiply-connected domains, and 2) a sub-class of 1) behaves like a Hall/Anti-Hall bar configuration, i.e., no Hall voltage appears between any two points on the hole boundary if current contacts are on their outer boundary. The paper studies the requirements under which these effects occur. Canonical cases of simply and doubly connected domains are computed analytically. The focus is on 2D multiply-connected Hall plates where all boundaries are insulating and where all current contacts are point sized.

Conventional Hall plates have no holes. They are well understood, though the analytical calculation of the Hall output voltage remains challenging, particularly at large magnetic field [

Discontinuous Hall effect regions exhibit similar phenomena as multiply connected Hall Effect regions [

This paper gives a theory on the classical Hall effect in multiply-connected plane and schlicht regions bounded by Jordan curves [

One goal of the paper is similar to [

In fact the authors in [

In this paper we do not discuss quantum Hall effect phenomena, which are frequently mentioned in the context of multiply connected Hall effect regions [

Section 2 explains the “anti-Hall bar within a Hall bar” whose unexpected behavior bewildered several engineers with decades of experience in Hall sensor technology (including the author). Section 3 states the assumptions of the theory. Section 4 presents numerical simulations on a very general doubly connected device to give us an idea of the roles of symmetry, contacts size, and magnetic field strength. Section 5 develops a theory for Hall plates, where all boundaries are insulating. Sections 6-8 apply this theory to simply-connected, doubly-connected, and multiply-connected plane Hall plates without interior current sources. Section 9 explains the role of extended contacts in the limit of large applied magnetic field. Section 10 sums up the main achievements of the paper. In Appendix A we compute the current pattern in the unit disk with and without hole, if point current contacts are on the unit circle.

In [

( E x E y ) = ρ ( J x J y ) with ρ = ρ ( 1 μ H B a , z − μ H B a , z 1 ) (1)

All boundaries are insulating and a current of 1 Ampere is injected into the device at the left outer boundary in point A. The current is extracted at the opposite point B also being on the outer boundary. We are interested in the potential at points C, D, E, F on the vertical symmetry line with C and F on the outer boundary and D and E on the inner boundary. We are free to choose one point as reference point for the potential and so we ground D. For good accuracy a fine mesh with 1.9 million elements was used. The applied magnetic field was swept from μ H B a , z = − 1.28 to 1.28. For the largest magnetic field the current

streamlines are shown in grey color and the contours of the potential are drawn in color. The current streamlines are parallel in the upper and lower horizontal parts of the device, and there the contours of the potential are also straight lines tilted against the current streamlines by 90˚ minus the Hall angle. This is identical to simply connected Hall plates. In

In

point-sized contacts.

In this paper we deal only with plane Hall plates whose size in the (x, y)-plane is much larger than their small and homogeneous thickness t H in z-direction. All quantities are constant versus z-position: ∂ / ∂ z = 0 . A magnetic flux density B a is applied with external means to the Hall plate. It has a component only in z-direction B a = B a , z n z ( n z is the unit vector in z-direction) and it is homogeneous in the entire (x, y)-plane. It may be weak but it may also be very large. Then the classical Hall effect is described by

E = ρ J + ρ μ H J × B a (2a)

J = Ε − μ H Ε × B a + μ H 2 ( Ε ⋅ B a ) B a ρ ( 1 + μ H 2 | B a | 2 ) → Ε − μ H Ε × B a ρ ( 1 + μ H 2 B a , z 2 ) (2b)

with the specific ohmic resistivity ρ > 0 and the Hall mobility μ H > 0 . These two material parameters are assumed to be constant versus space and versus electric and magnetic fields. They are both assumed to be simple scalars. The positive mobility in (2a, b) is valid for electrons as majority carriers—for holes we have to use a negative mobility. We assume only a single dominant type of charge carrier. Equation (2b) follows from (2a) if B a is perpendicular to the thin Hall plate. Its derivation is given in [

We consider Hall plates with four contacts, the contacts having infinite conductivity. Via two supply contacts we force a current I supply through the Hall plate and at the other two sense contacts we tap an output voltage V out . Perfectly symmetric Hall plates have zero output voltage at zero magnetic field. However, in practice Hall plates always suffer from unavoidable asymmetries so that in general one has to account for a non-vanishing output voltage in the absence of applied magnetic field—this is called offset. If magnetic field is applied the output voltage will change. We decompose the output voltage into a sum of even and odd functions of the applied magnetic field. The even part is commonly called the offset and the odd part is the Hall voltage V H .

V out ( B a , z ) = V even ( B a , z ) + V odd ( B a , z ) (3a)

V even ( B a , z ) = V out ( B a , z ) + V out ( − B a , z ) 2 (3b)

V odd ( B a , z ) = V H ( B a , z ) = V out ( B a , z ) − V out ( − B a , z ) 2 (3c)

V out ( B a , z ) is measured at positive applied magnetic field and V out ( − B a , z ) is measured at negative applied magnetic field. With the principle of reverse magnetic field reciprocity (RMFR [

We can also measure the potential ϕ with a single point-sized probe at any location r in the Hall plate. Analogous to (3a-c) we define the even potential ϕ even and the odd potential ϕ odd (=Hall potential ϕ H )

ϕ ( r , B a , z ) = ϕ even ( r , B a , z ) + ϕ odd ( r , B a , z ) (4a)

ϕ even ( r , B a , z ) = ϕ ( r , B a , z ) + ϕ ( r , − B a , z ) 2 (4b)

ϕ odd ( r , B a , z ) = ϕ H ( r , B a , z ) = ϕ ( r , B a , z ) − ϕ ( r , − B a , z ) 2 (4c)

If the location r 1 is on the first sense contact and r 2 on the second sense contact then V H , 12 = ϕ H ( r 1 ) − ϕ H ( r 2 ) . Other authors define the Hall potential as the difference of potential with and without magnetic field, which differs from our definition by the magnetic field change of offset ( ϕ ( B a , z ) + ϕ ( − B a , z ) ) / 2 − ϕ ( B a , z = 0 ) . This difference is small of order O 2 ( B a , z ) for small magnetic field.

All odd functions vanish at zero applied magnetic field due to their definition.

V odd ( 0 ) = 0 , ϕ odd ( r , 0 ) = 0 (5)

We denote all even functions at zero applied magnetic field with an index 0

V out ( 0 ) = V even ( 0 ) = V out,0 (6a)

ϕ ( r , 0 ) = ϕ even ( r , 0 ) = ϕ 0 ( r ) (6b)

E ( r , 0 ) = E 0 ( r ) (6c)

J ( r , 0 ) = J 0 ( r ) (6d)

The Hall voltage depends on the supply current through the Hall plate, the magnetic field applied to the Hall plate, and on the material parameters and the geometry of the Hall plate. The literature on Hall plates casts this relation into the following form

V H = I supply R sheet G H tan θ H (7)

with the sheet resistance R sheet = ρ / t H , the Hall geometry factor G H , and the Hall angle θ H . In the entire Hall plate the electric field vectors E are rotated by the angle θ H against the current density vectors J , as can be seen in (2). From (2) we get tan θ H = μ H B a , z . Small magnetic field means θ H ≈ 0 ∘ , and large magnetic field means θ H → ± 90 ∘ depending on the polarity of the magnetic field. The Hall geometry factor summarizes all effects caused by the geometry of the Hall plate and the finite size of its contacts. It holds 0 ≤ G H ≤ 1 . If all four contacts are point-sized and peripheral G H = 1 (see also (19)). The larger the contacts, the smaller G H . Therefore G H describes the reduction of the magnetic sensitivity of a Hall plate caused by its contacts. The sense contacts shunt a part of the supply current away from the interior of the Hall plate so that not all the current is available for the Hall effect. And the supply contacts have a short circuiting effect on the Hall electric field which hinders the Hall voltage to develop unrestrained between the sense contacts. Due to an interplay between magnetic field and geometry at the contacts (see Section 9), in general G H is also a function of θ H . At large magnetic field G H → 1 regardless of the size of the contacts. That means at large magnetic field all contacts behave similar to point contacts. In this paper we will keep the definition (7) of the Hall geometry factor, but expand its range of validity to Hall plates with holes with − 1 ≤ G H ≤ 1 . Note that the Hall voltage is sampled between two points V H , 12 = ϕ H ( r 1 ) − ϕ H ( r 2 ) , and therefore the Hall geometry factor also relates to these two points G H = G H , 12 . However, we can ground one of these points and then the Hall geometry factor can be interpreted as a normalized Hall potential G H ( r 1 ) = ϕ H ( r 1 ) / ( I supply R sheet tan θ H ) for ϕ ( r 2 ) = 0 ⇒ ϕ H ( r 2 ) = 0 .

In

| tan θ H | > 1 . Then one may use a doubly connected Hall plate like in

If we make one current contact point sized, skip the large sense contacts, and sample the Hall potential only with point probes, the behavior is similar.

boundary. As we grounded an inner point, the Hall geometry factor for the outer points is not between 0 and 1 but between approximately −0.5 and 0.5 (only for this specific example)—therefore we plotted the Hall potential over tangent of Hall angle at 1 Ampere supply current and 1 Ohm sheet resistance. Anyhow, if one taps the output voltage between a point on the left outer boundary and a point on the right outer boundary its G H goes to 1 for large magnetic fields.

Finally, if we replace the last big contact in

contacts, and in the case of large contacts it is the symmetry according to Haeusler [

If all boundaries are insulating the contacts must be point sized. Then it is better to use the stream function instead of the potential, because the boundary conditions specify the stream function and not the potential (see also Appendix A).

However, the stream function exists only in a limited class of topologies where the net current through every closed curve inside a multiply connected domain vanishes (regardless if the contacts are point sized or extended).

∮ C J ⋅ n d s = 0 ⇔ stream function exists (8)

with n being the unit vector perpendicular to the curve. Thereby the closed curve may also encircle holes. If a multiply-connected Hall plate has one current input contact and one current output contact and both are on the boundary of the same hole, a stream function exists. The same holds if both contacts are on the outer perimeter. If the current input contact is on the boundary of a different hole than the current output contact, no stream function exists. If the entire boundary of a hole is a single current contact, no stream function exists. This holds also in the limit of vanishing hole size. Topologies with no stream function will be focused on in the follow-up part II of this publication. A proof of (8) will become clear later (see (18)).

According to Maxwell’s first equation in the quasi-static case it holds

J = ∇ × H = ( ∂ H z / ∂ y − ∂ H y / ∂ z ∂ H x / ∂ z − ∂ H z / ∂ x ∂ H y / ∂ x − ∂ H x / ∂ y ) (9)

with the magnetic field H linked to the current flow. Note that H is the magnetic field that is generated by the current streamlines—it is not the magnetic field applied to the Hall plate μ 0 H ≠ B a . Most of the literature on the Hall effect tacitly assumes μ 0 | H | ≪ | B a | , i.e., the magnetic field caused by the current through the Hall plate is thought to be negligible against the externally applied magnetic field. With (9) the divergence of the current density vanishes

∇ ⋅ J = 0 (10)

because ∇ ⋅ ∇ × H = ∇ × ∇ ⋅ H = 0 . In a thin plane Hall plate it holds: 1) J z = 0 , 2) ∂ J x / ∂ z = ∂ J y / ∂ z = 0 , and 3) J x , J y generate no H x , H y within the conductive region. Therefore H z is the only non-zero component in the conductive region. It defines the stream function

ψ = − ρ H z (11)

In (11) the resistivity makes the dimensions of stream function and potential equal in order to combine them to the complex potential, and the minus sign makes this complex potential compliant with the Cauchy-Riemann differential equations in the case B a , z = 0 (see (20a, b) below).

In (9)-(11) we adopted a three-dimensional picture how J and H are linked. Due to the small thickness of the Hall plate several components of the vector fields vanish. The non-vanishing quantities are sufficient to define a consistent two-dimensional model where H z is a function of J x , J y . Such a 2D problem in the (x,y)-plane implicitly assumes ∂ / ∂ z = 0 for all quantities, and this implies an infinitely thick Hall plate t H → ∞ with J z = 0 . Therefore it assumes H x = H y = 0 —not only inside the Hall plate but everywhere. Then the point current contacts become line contacts parallel to the z-axis. It is important to keep in mind that in such a 2D problem − ψ / ρ is the magnetic field H z of the currents in an infinitely thick Hall plate, and not in the real, thin Hall plate (for the same current pattern J x , J y with J z = 0 the magnetic field in thick and thin Hall plates differs in z = 0 , which becomes apparent if we insert the current density into the law of Biot-Savart to compute the magnetic field:

H ( r ) = 1 4π ∫ A ′ ∫ z ′ J ( r ′ ) × r − r ′ | r − r ′ | 3 d A ′ d z ′ (12)

A ′ is the area in the (x, y)-plane. For the thick Hall plate the integration goes z ′ : − ∞ → ∞ whereas for the thin Hall plate it goes only z ′ : lim t H → 0 − t H / 2 → t H / 2 which is a simple multiplication by t H .

Since J is obtained from H z by spatial differentiation in (9), and H z is proportional to ψ in (11), the stream function ψ takes over the role of a 2D “potential” that can be used in a way similar to the potential ϕ . It holds

J = − ∇ × ψ n z ρ = − 1 ρ ∂ ψ ∂ y n x + 1 ρ ∂ ψ ∂ x n y (13)

n x , n y , n z are the Cartesian unit vectors. Note the different viewpoint: in (9) H is generated by J , yet in (13) J is generated by ψ . In fact both equations just describe how the quantities H , J , ψ are linked without causality: they do not differentiate between origin and reaction.

Defining J as the curl of − ( ψ / ρ ) n z implies that J is orthogonal to ∇ ψ

J ⋅ ∇ ψ = 1 ρ ( − ∂ ψ / ∂ y ∂ ψ / ∂ x ) ⋅ ( ∂ ψ / ∂ x ∂ ψ / ∂ y ) = 0 (14)

This explains the name stream function, because it is constant along each current streamline. With Maxwell’s second equation and (2a) we get

∇ × E = 0 = ρ ∇ × J + ρ μ H B a , z ∇ × ( J × n z ) (15)

∇ × ( J × n z ) = ( n z ⋅ ∇ ) J − n z ( ∇ ⋅ J ) = 0 because of n z ⋅ ∇ = 0 for 2D geometries and because of (10). With ρ ∇ × J = − ∇ × ( ∇ × ψ n z ) = − ∇ ( ∇ ⋅ n z ψ ) + ∇ 2 ψ n z = ∇ 2 ψ n z we conclude that the stream function must also fullfill Laplace’s equation ∇ 2 ψ = 0 . The boundary conditions at extended contacts are ∂ ψ / ∂ n = μ H B a , z ∂ ψ / ∂ t and at the insulating boundary the normal current density vanishes. Thus it holds ∂ ψ / ∂ t = 0 and therefore ψ is constant on insulating boundaries. This goes along with the notion that along any insulating boundary there flows one specific current streamline and the stream function is constant on current streamlines, thus, it is also constant along the insulating boundary.

Both ϕ and ψ are solutions of the Laplace equation, yet the Laplace equation is independent of the applied magnetic field. The magnetic field dependence of potential and current distribution is determined by the boundary conditions only. Hence, one might speculate that the volume is less important than the boundary for the Hall effect. After all, there is no Hall voltage in an infinitely large, unbounded Hall plate (cf.

If a Hall plate has only insulating boundaries and point-sized contacts where current is impressed, both the Laplace equation and the boundary conditions for the stream function do not contain the applied magnetic field any more. In such a Hall plate the stream function, the current density, and the current streamlines are constant versus applied magnetic field. For the case of a circular disk with peripheral point current contacts this was already stated in [

If we express the electric field in (2a) by the potential and the current density by the stream function this gives

∇ ϕ = ∇ × ψ n z − μ H B a , z ∇ ψ (16a)

At vanishing applied magnetic field (16a) reads

∇ ϕ 0 = ∇ × ψ n z (16b)

For the particular case of all-insulating boundaries we do not need an index 0 for ψ , because it is identical with and without applied magnetic field. From (16a, b) we get

∇ ϕ = ∇ ϕ 0 − μ H B a , z ∇ ψ ⇒ ϕ = ϕ 0 − μ H B a , z ( ψ − ψ g r o u n d ) (16c)

In the ground node it holds ϕ = ϕ 0 = 0 and therefore ψ = ψ ground . We are free to choose ψ ground and so we define ψ ground = 0 , which means ψ = 0 in the ground node. With this convention, with (4a), (16c), and with the fact that ψ is independent of B a , z the Hall potential becomes

ϕ H = − μ H B a , z ψ (17a)

Thus, the Hall potential is constant along current streamlines, and the Hall voltage between two points is simply proportional to the difference in stream function

V H , 12 = ϕ H 1 − ϕ H 2 = − μ H B a , z ( ψ 1 − ψ 2 ) (17b)

Another property of the stream function is its relation to the total current I 12 flowing across any contour (extruded into thickness direction) that starts at point 1 and ends at point 2. Using (13) we get

I 12 = t H ∫ 1 2 J n d s = t H n z ⋅ ∫ 1 2 J × t d s = − t H ρ ∫ 1 2 ∂ ψ ∂ x d x + ∂ ψ ∂ y d y = − t H ρ ∫ 1 2 d ψ = − ψ 2 − ψ 1 R sheet (18)

whereby we used J n = J ⋅ n , n = t × n z , n ⋅ n = t ⋅ t = 1 , and t is tangential to the path and points from point 1 to point 2. The current I 12 flowing across the contour connecting the two points 1, 2 is the difference in stream function at these two points divided by the sheet resistance. Inserting (18) into (17b) and comparing with (7) gives

G H , 12 = ϕ H 1 − ϕ H 2 I supply R sheet μ H B a , z = − I 12 I supply (19)

Thus, the Hall geometry factor between the points 1 and 2 is equal to the ratio of the current between the two points and the total current through the Hall plate. This holds for multiply connected conductive regions without extended contacts and without internal current sources (i.e., whenever a stream function exists). Obviously, 1) this ratio cannot exceed 1 and it is smaller if both points 1, 2 are in the interior of the Hall plate with peripheral contacts, 2) it is zero along any insulating boundary without current contacts between points 1 and 2, and 3) it is ±1 if points 1 and 2 are on the same boundary and a single point-sized supply contact is between them on this boundary.

In Hall plates with all insulating boundaries and point sized contacts the current pattern does not change with applied magnetic field (see above). Therefore (19) implies that the Hall geometry factor is constant versus applied magnetic field. Consequently the Hall voltage is directly proportional to the applied magnetic field whenever the current through the Hall plate is kept constant. These devices are linear versus applied magnetic field—within the scope of our assumption that the material parameters ρ , μ H are constant. In engineering practice one may partly compensate for the magnetic field dependence of the material parameters with the magnetic field dependence of the Hall geometry factor in Hall plates with extended contacts [

If the integration path in (18) is a closed loop, points 1 and 2 are identical and this leads to (8). Hence, we see that in multiply-connected domains a stream function makes sense only in the absence of internal current sinks and sources.

The stream function jumps in point-sized current contacts according to (18), and so we must not use it there, because it would lead to self-contradictory predictions. Let us assume point 1 exactly at the current contact and point 2 somewhere inside the Hall plate. If we would erroneously apply (18), this gives ψ 2 = ψ 1 because obviously I 12 = 0 (both points are on a current streamline and therefore no current flows across this streamline). This would mean that the stream function is constant everywhere inside the Hall plate—which of course is wrong. If ψ is discontinuous in a point-sized current contact also ϕ H is discontinuous there according to (17a). Hence, also ϕ H makes no sense in point contacts. On the other hand we can sample the Hall voltage between two points on the same current streamline and it vanishes according to (17b). If the two points approach positive and negative supply contacts it still holds V H = 0 . Hence, the Hall voltage across supply contacts vanishes. The argument holds for point-sized supply contacts, but the Hall voltage also vanishes across extended supply contacts (this can be shown with the methods developed in [

In the absence of an applied magnetic field ( B z , a = 0 ) it holds

E x = ρ J x ⇒ ∂ ϕ 0 ∂ x = ∂ ψ ∂ y (20a)

E y = ρ J y ⇒ ∂ ϕ 0 ∂ y = − ∂ ψ ∂ x (20b)

These are the Cauchy-Riemann differential equations and therefore the functions F R 0 = ϕ 0 + i ψ and F I 0 = − ψ + i ϕ 0 are analytic functions, which are commonly called complex potentials. (20a, b) imply ∇ ϕ 0 ⋅ ∇ ψ = 0 , curves of constant ϕ 0 and curves of constant ψ are orthogonal. If we combine this with (17a) we arrive at the simple and well known fact that the gradient of the Hall potential is orthogonal to J 0 .

In the presence of an applied magnetic field E and J are not parallel, and therefore ϕ and ψ do not fulfill the Cauchy-Riemann differential equations. Nonetheless we can construct a complex potential whenever the stream function ψ exists. There are two possibilities: either the real or the imaginary part of the complex potential is equal to the electric potential ϕ , and the remainder is defined such as to satisfy the Cauchy-Riemann differential equations (enter (13), (16c), and J = J 0 into (2a)).

F R = − i F I = ϕ + i ( μ H B a , z ϕ + ( 1 + μ H 2 B a , z 2 ) ψ ) (21)

Thus, the complex number F I is equal to F R rotated by 90˚. (21) is valid for Hall plates with small and large contacts at applied magnetic field without internal current sources. For point sized contacts without internal current sources we can write (21) for positive and negative applied magnetic field with identical ψ . Subtracting both we get with (4b) and (17a)

F R , odd = F R ( B a , z ) − F R ( − B a , z ) 2 = ϕ H + i μ H B a , z ϕ even = μ H B a , z ( − ψ + i ϕ even ) (22)

(22) is again an analytical function. Since ψ does not depend on B a , z in the entire Hall plate also ϕ even is constant versus applied magnetic field because they have to fullfill the Cauchy-Riemann differential equations. Comparison of (4a) with (16c) and (17a) gives ϕ even = ϕ 0 . The contour lines of ψ and ϕ even = ϕ 0 are orthogonal. Contour lines of constant Hall potential are orthogonal to equipotential lines at zero applied magnetic field. Moreover, we have ϕ H = ϕ − ϕ even = ϕ − ϕ 0 , i.e., in this particular case the Hall potential is the change in potential with and without applied magnetic field. Finally we can write the complex potential in Hall plates with point contacts like this

F R = − i F I = ϕ 0 − μ H B a , z ψ + i ( μ H B a , z ϕ 0 + ψ ) . (23)

Hence, the complex potential has no quadratic dependence on applied magnetic field. This gives a simple relation between complex potentials with and without applied magnetic field.

F R = F R 0 ( 1 + i μ H B a , z ) = ( F R 0 / cos θ H ) exp ( i θ H ) . (24)

(24) is compatible with (C9) in [

E ( z ) = − d d z F R ( z ) = i d d z F I ( z ) = E 0 ( z ) cos θ H exp ( i θ H ) (25)

with z = x + i y . If we define a complex current density by J = J x − i J y it holds

J ( z ) = 1 ρ cos θ H exp ( − i θ H ) E ( z ) = 1 ρ E 0 ( z ) = J 0 ( z ) (26)

E 0 and J 0 are the complex electric field and the complex current density, respectively, at zero applied magnetic field. The leftmost equality in (26) is identical to (2a). If we take the conjugate of (26) we see that the current density is equal to the electric field rotated CCW by the Hall angle (and scaled in length and dimension).

In the review process of this paper the author became aware of [

Riemann’s mapping theorem says that all simply connected bounded plane domains can be mapped onto the unit disk with a conformal transformation. Such a mapping is described by an analytical function and it preserves angles. The potentials at corresponding points are identical and also the currents into corresponding contacts are identical. The Hall angle between E and J is also identical, but E and J themselves are generally not identical. If we are only interested in the potential, we can limit the discussion to the circular unit disk of

I 12 = t H n z ⋅ ∫ 1 2 J × d s = t H n z ⋅ n y × n x ∫ x = − 1 x = 1 J y d x = − t H ∫ x = − 1 x = 1 J y d x = − I supply (27)

The Hall geometry factor between any two points inside the disk is smaller than 1. It can be computed in closed form (see Appendix A, compare also Figures 5 and 6 in [

These are plane domains with the shape of a ring. After Riemann’s mapping theorem, a ring of general shape can always be mapped onto the annular region between two concentric circles [

First we assume that all current contacts are on the outer circle as shown in

current partitioning between left and right, which depends solely on the spacing of contacts (see Appendix A).

With the principle of RMFR [

Since the current streamlines do not change with applied magnetic field, we can cut the Hall plate of

cut we have two disjunct Hall plates and we have to ground both of them. If we ground them along the cut line we re-establish original continuity of Hall potential across the cut line as it was in the original device before cutting. But this means that the two parts of the hole are also tied to the very same Hall potential, which gives zero Hall voltage between two points of the original hole boundary. On the left boundary of the left device the Hall geometry factor is 1 because it is left of the current flow and we grounded the opposite boundary, and on the right boundary of the right device it is −1 because it is right of the current flow and we grounded the opposite boundary. The current through the left device is ζ I supply and the current through the right device is ( 1 − ζ ) I supply with ζ = 1 / 6 denoting the current splitting in

As long as all current contacts are on boundaries and the net current into each boundary is zero, the problem can be described by a stream function. In the case of point contacts this stream function is constant versus applied magnetic field, as in the previous sections. If several hole boundaries have identical value of the stream function, there is no Hall voltage between them. An example is shown in

hole boundary until the current through the straight black lines L1, L2 became equal. Thus there is a single streamline (in red color) being split in two parts by the first hole, then joining up again to a single streamline and being split up again in two parts by the second hole. Therefore, the Hall potential is identical on the boundaries of the square and triangular holes. On the outer boundary the Hall potential is also constant but not equal to the one on the hole boundaries.

In

This behavior can be studied analytically by conformal mapping of the z-plane in

d w d z = 2 i π exp ( i θ H ) cos θ H ( z + 1 ) θ H / π − 1 / 2 ( z − 1 ) − θ H / π − 1 / 2 (28)

the upper half z-plane is mapped onto the strip of infinite length in the w-plane. The strip is slanted by the Hall angle, and the contact is between w = − 1 + i × 0 and w = 1 + i × 0 . Due to the slanting the electric potential is constant on horizontal lines. This gives a homogeneous electric field in negative v-direction. The current density is homogeneous and parallel to the long sides of the strip. Therefore the complex potential is

F I ( w ) = I supply R sheet w 2 cos 2 θ H (29)

It holds F I ( w ) = F I ( z ) if w and z are linked by (28). With (25) it follows

E ( z ) = i d F I d z = i d F I d w d w d z = − I supply R sheet 1 + i tan θ H π ( z + 1 ) θ H / π − 1 / 2 ( z − 1 ) − θ H / π − 1 / 2 (30)

and finally we get the current density with (26)

J ( z ) = − I supply 1 π t H ( z + 1 ) θ H / π − 1 / 2 ( z − 1 ) − θ H / π − 1 / 2 (31)

For infinite positive applied magnetic field θ H → π / 2 and J ( z ) → − I supply π − 1 t H − 1 ( z − 1 ) − 1 . This is identical to a negative point current source in z = 1 + i × 0 . The current density is constant on semi-circles around this point and it points towards z = 1 + i × 0 . At infinite negative applied magnetic field θ H → − π / 2 . Then the center of the semi-circles moves to z = − 1 + i × 0 . The numerical example in

This paper answers the question “Under which circumstances does the Hall voltage—tapped between two contacts on a boundary in a multiply-connected Hall plate—vanish?”. This question was kindled by the Hall/Anti-Hall bar configurations in [

The author gratefully acknowledges the vivid and fruitful discussions with Prof. Dieter Süss of University of Vienna and Dr. William J. Bruno of New Mexico Consortium, Los Alamos, NM.

The author declares no conflicts of interest regarding the publication of this paper.

Ausserlechner, U. (2019) The Classical Hall Effect in Multiply-Connected Plane Regions Part I: Topologies with Stream Function. Journal of Applied Mathematics and Physics, 7, 1968-1996. https://doi.org/10.4236/jamp.2019.79136

Here we compute the current pattern for the Hall plate with insulating hole boundary and point-sized supply contacts at the outer boundary (see

We use polar coordinates ( r , φ ) . The current input contact is located in r = 1 , φ = − α , the current output is in r = 1 , φ = α with 0 < α < π . The annular conductive region is in r 1 ≤ r ≤ 1 with 0 < r 1 < 1 . For the stream function ψ we make the ansatz (see [

ψ = ( A 0 ln r + B 0 ) ( C 0 φ + 1 ) + ∑ k = 1 ∞ ( A k r k + B k r − k ) ( C k sin k φ + cos k φ ) . (A1)

From periodicity requirement ψ ( r , φ ) = ψ ( r , φ + 2 π ) ∀ r it follows C 0 = 0 . The outer boundary is split up in a left segment φ ∈ ( − π , − α ) ∪ ( α , π ) and a right segment φ ∈ ( − α , α ) . In both segments the stream function has to be constant according to (18), because no current flows through the boundary. According to our rule (see (16c)) we set its value at the grounded segment to zero: ψ right = 0 . At the left segment its value is ψ left . With (18) we get

− ψ right − ψ left R sheet = lim ε → 0 t H n z ⋅ ∫ φ = − α − ε − α + ε − I supply 2 ε t H n r × n φ d φ = − I supply . (A2)

n r and n φ are the unit vectors in radial and azimuthal directions, respectively. On the inner insulating boundary without current contacts it has some constant value ψ = ψ hole ∀ φ , which leads to B k = − A k r 1 2 k ∀ k ≥ 1 . Thus, ψ is given on all boundaries (Dirichlet problem) and it is an even function in φ . Therefore C k = 0 ∀ k ≥ 1 .

With ∇ ⋅ E = 0 it follows from Gauss’ theorem ∫ φ = − π π ∂ ϕ / ∂ r r d φ = 0 for r 1 < r < 1 . With (16c) it follows at arbitrary applied magnetic field: ∫ φ = − π π ∂ ψ / ∂ r r d φ = 0 . With (A1) this gives A 0 = 0 . Thereby it is allowed to differentiate the ansatz (A1) term-wise because with A k obtained below in (A4b) the term-wise differentiated series converges uniformly in r 1 < r < 1 [

B 0 is obtained from computing the mean of ψ on the outer perimeter

∫ φ = − π π ψ ( r = 1 ) d φ = 2 π B 0 = 2 ( π − α ) ψ l e f t = − 2 ( π − α ) I supply R sheet . (A3)

and A k is obtained by the Fourier series expansion of ψ on the outer perimeter

∫ φ = − π π ψ ( r = 1 ) cos k φ d φ = ∫ φ = − π π ( 1 − r 1 2 k ) A k cos 2 k φ d φ = ∫ φ = − π − α + ∫ φ = α π ( − 1 ) I supply R sheet cos k φ d φ (A4a)

⇒ A k = I supply R sheet α π sin k α k α 2 1 − r 1 2 k . (A4b)

Finally, the stream function (with and without applied magnetic field) is given by

ψ = I supply R sheet [ α π ( 1 + 2 ∑ k = 1 ∞ r k − r 1 2 k r − k 1 − r 1 2 k sin k α k α cos k φ ) − 1 ] . (A5a)

The series can be summed up if the hole vanishes

ψ | r 1 = 0 = I supply R sheet 1 π ( α − π + arctan r sin ( α − φ ) 1 − r cos ( α − φ ) + arctan r sin ( α + φ ) 1 − r cos ( α + φ ) ) . (A5b)

The stream function is finite everywhere, and it is continuous except in the current contacts where it jumps only along the boundary. Now we compute the current that flows at the RHS of the hole. The current density on the x-axis is J = ρ − 1 ∂ ψ / ∂ r n φ . With (18) and

I 12 = t H n z ⋅ ∫ 1 2 J × d s = t H n z ⋅ ∫ x = r 1 x = 1 J y n y × n x d x = − t H ∫ x = r 1 x = 1 J y d x = − I supply ( 1 − α π ) . (A6)

Point 2 is the ground node in ( r , φ ) = ( 1 , 0 ) and point 1 is on the hole boundary in ( r , φ ) = ( r 1 , 0 ) . We used the sum ∑ 1 ∞ k − 1 sin α k = ( π − α ) / 2 for 0 < α ≤ π . With (19) the Hall geometry factor for one probe on the hole boundary against grounded point 2 is

G H , 12 = 1 − α / π (A7)

Thus, the Hall geometry factor on the hole boundary is 1 for α → 0 ∘ , 0.5 for α = 90 ∘ , and 0 for α → 180 ∘ . This symmetry agrees with common sense. Interestingly, the current I 12 at the RHS of the hole is independent on the size of the hole r 1 . This also holds for the Hall voltage measured between a point on the hole and a point on the outer boundary.

On the hole boundary the sum in (A5a) vanishes and we get

ψ ( r = r 1 , φ = 0 ) = ψ hole = I supply R sheet ( α π − 1 ) . (A8)

On the right segment of the outer boundary we have in accordance with (A2)

ψ ( r = 1 , | φ | < α ) = ψ right = 0 (A9a)

and on the left segment of the outer boundary we have

ψ ( r = 1 , φ ∉ [ − α , α ] ) = ψ left = − I supply R sheet . (A9b)

The Hall geometry factor between points on left and right segment of the outer boundary is simply 1, independent of the hole and its size.

We call the path along which the specific current streamline flows, which separates the annulus into left and right branches of current flow around the hole, the separation curve. It meets the inner circle at a right angle and it meets the outer circle in the current contact at the angle α . This is obvious if we note that the current flows isotropically out of the current contact (which is proven below) and the fraction α / π takes the left detour around the hole. If we want to compute the separation curve, we have to solve ψ ( r , φ ) = ψ hole which gives

∑ k = 1 ∞ r k − r 1 2 k r − k 1 − r 1 2 k sin k α k cos k φ = 0 . (A10)

We used this in

With ψ we know the Hall potential ϕ H according to (17a), but not the potential ϕ = ϕ 0 + ϕ H . The potential at zero applied magnetic field is obtained by integrating the Cauchy-Riemann differential equations (20a, b).

ϕ 0 = ∂ ∂ φ ∫ ψ r d r = I supply R sheet − 2 α π ∑ k = 1 ∞ r k + r 1 2 k r − k 1 − r 1 2 k sin k α k α sin k φ . (A11a)

ϕ 0 is finite except in the current contacts. It is zero on the x-axis and odd in φ and α . If the ring degenerates to a disk r 1 → 0 we can sum it up in terms of elementary functions

ϕ 0 | r 1 = 0 = I supply R sheet 1 2π ln 1 + r 2 − 2 r cos ( φ − α ) 1 + r 2 − 2 r cos ( φ + α ) . (A11b)

For r 1 → 0 the streamline ψ = ω I supply R sheet with − 1 ≤ ω ≤ 0 is a circle with radius sin α / | sin ( α + ω π ) | having its center at x = sin ω π / sin ( α + ω π ) and y = 0 . This can be shown by entering the points on this circle into (A5b), eliminating y, and differentiating w.r.t. x, which gives zero. Hence, along this circle the stream function is constant. The circular stream function crosses the x-axis in x = cos ( − α / 2 + ω π / 2 ) / cos ( α / 2 + ω π / 2 ) .

We draw a small circle of radius δ r → 0 around the negative current contact in r ( − ) = ( cos α , sin α ) . At points within this circle we compute the current density with (A5b) and (13). A series of the current density in powers of δ r

has the dominant term J = − I supply ( r − r ( − ) ) / ( π t H | r − r ( − ) | 2 ) + O ( | r − r ( − ) | ) 0 . This

proves that the current flows isotropically into the negative supply contact. If two current streamlines are given by ψ 1 = ω 1 I supply R sheet and ψ 2 = ω 2 I supply R sheet a current | ω 1 − ω 2 | I supply flows between them according to (18), and in the current contact the two streamlines define an aperture angle of | ω 1 − ω 2 | π .

A first streamline ψ = ( − 1 + α / π ) I supply R sheet goes through the origin (compare with (A5b)). Then it holds ω = − 1 + α / π and the radius of this circle is equal to 1 / ( 2 cos α ) . A second streamline is for ω = − α / π which has infinite radius. This is a vertical line through the current contacts. Comparison of first and second streamlines says that the current flowing left of the center is equal to the current flowing right of the current contacts.

The complex potential in the disk without hole can be computed more elegantly without recurring to series developments. Thereby one writes down the current density J in the upper half plane when a current enters the half plane

in x = − 1 : J = I supply π − 1 t H − 1 ( ( x + 1 ) n x + y n y ) / ( ( x + 1 ) 2 + y 2 ) . Thus, the potential at zero magnetic field is ϕ 0 = − ρ I supply π − 1 t H − 1 ln ( x + 1 ) 2 + y 2 . Then the complex

potential is obtained as the analytic function whose real part is ϕ 0 . This gives ϕ 0 + i ψ = − ρ I supply π − 1 t H − 1 ln ( z + 1 ) with z = x + i y . With the mapping z = − i cot ( α / 2 ) ( w − 1 ) / ( w + 1 ) the upper half of the z-plane is mapped inside the unit disk in the w-plane with the current input at w = exp ( − i α ) . Subtracting the contribution of the outgoing current at x = 1 one gets the complex potential at zero magnetic field

ϕ 0 + i ψ = I supply R sheet π ( ln ( − i tan ( α / 2 ) z − 1 z + 1 − 1 ) − ln ( − i tan ( α / 2 ) z − 1 z + 1 + 1 ) ) (A12)

which matches (A5b, A11b).