Multiply-connected complementary Hall plates with extended contacts

We consider uniform plane Hall plates with an arbitrary number of holes exposed to a uniform perpendicular magnetic field of arbitrary strength. The plates have extended contacts on the outer boundary and on the boundaries of the holes. No symmetry is presumed. Pairs of complementary Hall plates are defined, where contacts and insulating segments on the boundaries are swapped. A unique stream function exists in these multiply-connected domains if the total current through each boundary vanishes. Then the voltages between neighbouring contacts and the currents into contacts of pairs of complementary Hall plates are linked in a peculiar way, which results in identical power density. Relations between the impedances of complementary multiply-connected Hall plates are derived. The prominent role of complementary symmetric Hall plates—with or without holes—is revealed. For arbitrary magnetic field, their resistance equals the one of a square without holes and with contacts fully covering two opposite edges.


Introduction
Common Hall plates are symmetric thin flat regions of a uniform semiconductor with four peripheral contacts.Textbooks often show Hall plates having shapes of clover leaves, Greek crosses, rectangles, or squares.Hall plates detect the magnetic field component orthogonal to the main chip surface.In the past decade the industry also began to offer Vertical Hall effect devices in large quantities [1].They detect magnetic field components parallel to the main chip surface.Their cross section into the depth of the substrate resembles the shape of traditional Hall plates: it is roughly rectangular with three, five, or even more contacts lined up on the accessible top surface only.Two or more contacts may be shorted by wires, and in some silicon technologies there may be an unavoidable big floating contact at the bottom of the Hall tub [2].It is also common to connect several Hall effect devices together (by wire) for reasons of symmetrization.This gives a big variety of Hall sensor arrangements, and therefore, one is interested in general rules on their properties and on their mutual relations.
For a good magnetic field sensor one usually avoids holes, discontinuities, and internal contacts in the Hall effect region.However, this is not always possible.Vertical Hall effect devices suffer from inhomogeneous doping profiles into the depth of the substrate.Also samples of newly discovered materials, like Graphen, are generally not perfectly homogeneous, because their fabrication process is not yet optimized.There are regions of lower and higher resistivity.One can model this as a homogeneous region comprising voids and highly conductive inclusions of varying sizes.Mathematically, this is a multiply-connected two-dimensional domain with holes having insulating segments or contacts on the boundaries.This is the motivation for the present work.
The electrical properties of Hall plates are fully described by their resistance matrix as a function of the applied magnetic field.Thereby, small contacts tend to give larger resistances than large contacts.
For every Hall plate we can define a complementary (or dual) one, which has the same shape (same boundaries, same holes), yet conducting and insulating segments on the boundaries are swapped.Thus, every high-ohmic Hall plate has a low-ohmic complementary Hall plate with the same number of contacts.For singly-connected Hall plates it is known how to compute the resistance matrix of a Hall plate from its complementary counterpart at reverse magnetic field [3].This is called the reverse magnetic field on complementary device (RMFoCD) theorem.Its proof was based on the General Formula for the electric field in singly-connected Hall plates, given in [4].For Hall plates with one hole and four extended contacts on the outer boundary, having two perpendicular mirror symmetries, the RMFoCD theorem still pertains, at least for weak applied magnetic field.This was proven recently by the method of Hall equivalent boundary current injection [5].The fact that the RMFoCD theorem holds in singly-and doubly-connected domains indicates that there must be a more fundamental reason for it than the General Formula (which holds only for singly-connected domains).In the absence of the Hall effect, the RMFoCD theorem degenerates to a simpler theorem [6].Its proof rests on the duality of the electric potential and the stream function: the partial derivatives of the potential give the electric field vector field, whereas the partial derivatives of the stream function give the current density vector field.Consequently, the potential is constant on the contacts, whereas the stream function is constant on the insulating segments.This suggests that in multiply-connected domains exposed to magnetic field the RMFoCD theorem might also be proven by the duality of the potential and the stream function.If all boundaries of a multiplyconnected uniform Hall plate are made of conducting segments (= contacts), and the potential is forced at some of them, the potential in the Hall plate and at the rest of the contacts does not depend on the applied magnetic field, i.e. there is no Hall effect [7].Conversely, if all boundaries of a multiply-connected uniform Hall plate are insulating, and current is forced through the Hall plate via point-sized contacts on the same boundary, the stream function, the current density, and the current streamlines in the Hall plate do not depend on the magnetic field, and the Hall effect is maximized [8,9].This suggests that in the general case of finite contacts being separated by finite insulating segments, the interplay between potential and stream function plays an important role, which we will study in this paper.
Section 2 starts with Ohm's law in the presence of the Hall effect.It introduces the resistance matrix of a multi-contact Hall plate, the electric potential, the stream function, and the boundary conditions in a multiply-connected 2D domain.Section 3 defines the complementary Hall plate.There we derive the RMFoCD theorem.In Sect. 4 we work out some consequences of this theorem.In Sect. 5 the analytical theory is checked by numerical simulations.

The electromagnetic field problem in 2D space in the presence of the Hall effect
The flow of stationary electric current through most conductors is governed by Ohm's law E = ρJ , with E, J being the vectors of the electric field and the current density, respectively.The specific resistivity ρ is a positive real number, i.e.E and J are collinear, if the conductor is isotropic or if it is a crystal of cubic symmetry at zero mechanical stress like silicon [10].In the presence of the Hall effect, Ohm's law becomes in 3D space with the magnetic induction B applied by an external magnet to the Hall plate.× denotes the vector product, and μ H is the Hall mobility.In (1) we used the traditional sign convention with μ H > 0 for positive charge carriers (holes in p-doped semiconductors) and μ H < 0 for negative charge carriers (electrons in n-doped semiconductors).Let us assume only a single dominant type of charge carriers (holes or electrons as majority charge carriers).The right equation in (1) follows from the left one (see Appendix E in [11]).Let us limit the discussion to B = B z n z , with n z being the unit vector perpendicular to the plate.This implies E • B = 0 in (1).With (1) it is clear that J × B is perpendicular to J and lies in the thin Hall plate.Therefore, E lies also in the Hall plate and it is rotated against J by the Hall angle θ H = arctan(μ H B z ) in all locations in the Hall plate.In this work we assume uniform Hall plates at uniform magnetic field; therefore, ρ, μ H , B z , θ H do not vary in space.Equation ( 1) is a linear relation between the electric field vector and the current density vector.Therefore, also the respective macroscopic quantities, voltages and currents, are linearly related via resistance and conductance matrices (R ij ), (G ij ), V i is the electric potential at the i-th terminal, and I j is the current flowing into the j-th terminal.The terminals are numbered from 0 to N − 1 with V 0 = 0 (ground node at zero reference potential) and The zeroth terminal can be connected to a contact on the outer perimeter or on a hole boundary.A Hall plate may also extend to infinity-then it has no outer perimeter.Each terminal is connected to a dedicated contact having the same index.If a boundary has more than one contact, each contact has to be connected to its dedicated terminal.If a boundary has only one contact, within the scope of this paper it is not connected to a terminal.The reason will become apparent in Sect.2.3.Such contacts are called floating-they are not accessible, i.e. one cannot pass current through them or tap their electric potential.
The matrices (R ij ), (G ij ) with N −1 rows and columns are called definite matrices.Their determinants do not vanish.In fact they are positive definite, because the power dissipated in a resistive device such as a Hall plate is always positive, ( This also implies that these matrices can be inverted with (R ij ) = (G ij ) −1 .

The electric potential φ and the stream function ψ
From Faraday's law ∇ × E = −∂B/∂τ , with τ being the time, it follows in the stationary case, ∂/∂τ = 0, that the electric field is fully given by a scalar electric potential φ according to with ∇ = n x ∂/∂x + n y ∂/∂y being the nabla differential operator in 2D space.In the absence of free electric charges and for zero or homogeneous isotropic electric polarizability, Maxwell's third law becomes ∇ • E = 0, (the dot denoting the scalar product), which means ∇ 2 φ = 0. Hence, the electric potential φ is a solution of Laplace's differential equation in 2D space.
In the stationary case, Maxwell's first law is ∇ × H = J , with H being the magnetic field from the current flow (not from the external magnet).Thus, we may consider the current density to be generated by the H-field via the curl operator.For a two-dimensional current density in the (x, y)-plane of the Hall plate only the z-component of the H-vector is nonzero.This scalar function is commonly called the stream function ψ, defined by With this ansatz it follows ∇ • J = 0. Replacing the E-field in Faraday's law by (1) and expressing the J -field by the stream function with (5) gives, for static fields from the external magnet, Hereby we used n z • ∇ = 0 for flat domains in the (x, y)-plane and ∇ • J = 0. Hence, the stream function ψ is also a solution of Laplace's differential equation in 2D space, ∇ 2 ψ = 0. On a contour line of ψ it holds with ( 5) Hence, the stream function is constant along current streamlines.Since the current flows along insulating boundaries, the stream function is constant there, too.Entering (4) and ( 5 (9)

Boundaries of holey Hall effect regions in 2D space
Finite Hall plates have an outer boundary, infinite Hall plates have no outer boundary.Moreover, a multiply-connected domain has an additional inner boundary for each hole.For plane Hall plates each boundary is a closed curve in the (x, y)-plane with a possible parametrization (x(s), y(s)) with s ≥ 0 being the arc-length starting at some arbitrary point on the boundary.In principle, one can walk along the boundary in two opposite directions.We define the positive direction (also called the directional sense) such that the Hall effect region is at the left-hand side of the boundary during this walk.For the outer boundary this means counter-clockwise direction when looking from z > 0 onto the (x, y)-plane, whereas for the holes it means clockwise direction.The tangential unit vector on this boundary is with ds = dx 2 + dy 2 , and with the prime denoting differentiation with respect to arc-length s.Then the normal unit vector on this boundary pointing away from the Hall effect region is Each closed boundary consists of one or more segments, which are contacts C m or insulating boundaries C m .The indices start with m = 0 on an arbitrary closed boundary and they increase in the directional senses of the closed boundaries.A closed boundary may consist of a single contact, or it may be entirely insulating, or it may consist of an equal number of contacts and insulating segments in an alternating sequence starting with a contact C m and followed in the directional sense by an insulating segment C m .

Existence of a unique stream function
So far we have tacitly assumed that a unique stream function exists, which is not always the case.From the ansatz (5) it follows that the divergence of the current density vanishes, ∇ • J = 0. We consider an arbitrary smooth closed curve ∂A in the 2D Hall effect region, which encircles the area A inside the Hall effect region.A is only a portion of the Hall effect region, and it may comprise complete holes but not fractions of holes.That means that the complete curve ∂A must lie in the Hall effect region without cutting through any holes and without cutting through the outer boundary of a finite Hall plate.Applying Gauss' integral law on A gives zero net current I ∂A flowing out of the closed curve ∂A, t H being the constant thickness of the Hall plate along the z-direction.The current I ∂A is linked to the circulation Γ ∂A specifying spiral current streamlines, Here we used the right equation in ( 1) with E • B = 0 and B = B z n z and μ H B z = tan θ H .In ( 13) the first integral on the right-hand side vanishes, because E • tds = −dφ (E is a conservative force field with ∇ × E = 0, therefore dφ is a full differential and consequently the integral over it along a closed curve vanishes).In the second integral we set • n and insert the left equation of (1).It follows The first integral on the right-hand side is proportional to the total current flowing out of the closed curve ∂A.With J × n z • n = J • t the second integral becomes again the circulation.Combining ( 14) with ( 13) finally gives Thus, a current pattern with I ∂A = 0 for any closed curve ∂A in the Hall effect region has zero circulation.Conversely, at non-vanishing magnetic field, θ H = 0, zero circulation means I ∂A = 0.Then, the sum of currents flowing into all contacts inside the closed curve ∂A vanishes.Consequently, the sum of currents flowing into all contacts of a closed boundary also vanishes-let us call this a neutral boundary.It is a pre-requisite for the theory of this work.If only one contact is on a boundary, let us call this a solitary contact.Solitary contacts may cover a portion of a closed boundary or the entire closed boundary.No current is allowed to flow into/out of solitary contacts.If we do not connect solitary contacts to terminals, we inhibit any current flow through them.This leads to the following rule for the theory developed in this work: Every contact is connected to a dedicated terminal, except for solitary contacts.( 16) Only the voltages and currents at the terminals are described by the matrices R, G.The solitary contacts have no effect on the numbers of rows and columns of R, G.

Boundary conditions for φ and ψ
Per definition, ideal contacts have infinite conductivity.Thus, the electric field is orthogonal to the contacts.With ( 4) and (10) this gives Hence, the electric potential φ is constant along each contact.Replacing the potentials in ( 17) by the stream function with (8) gives an oblique derivative boundary condition [12] with the normal derivative ∂ψ/∂n = n • ∇ψ and the tangential derivative ∂ψ/∂t = t • ∇ψ.On the insulating segment the normal current density has to vanish.With ( 5) and ( 11) this gives Consequently, the stream function ψ is constant along each insulating segment (see also the argument below ( 7)).Replacing the stream function in (19) by the potential with ( 9) again gives an oblique derivative boundary condition At the contact C m a current I m flows into the Hall effect region.With ( 5) and ( 11) this gives where R sheet = ρ/t H is its sheet resistance.
To sum up, if the potentials at the contacts are given, we compute the electric potential φ.It satisfies the Laplace equation with the following boundary conditions: φ is known on the contacts and it satisfies (20) at the insulating segments.If the currents at the contacts are given, we compute the stream function ψ.It also satisfies the Laplace equation and its boundary conditions are: ψ is known on the insulating segments, whereby we can choose ψ = 0 on C 0 (just like we can choose φ = 0 on C 0 ).On the contacts it holds (18).Thus, on the boundaries, φ behaves like ψ, if we swap all contacts and insulating segments and if we reverse the sign of the Hall angle θ H .    and a stream function ψ = v(x, y) will establish.The functions u(x, y) and v(x, y) are continuous, their partial derivatives exist, and they fulfil (8), (9).φ = u(x, y), ψ = v(x, y), with boundary conditions:

Reverse magnetic field on a complementary Hall plate
In terms of the functions u, v this means whereby the Laplace equations are evaluated in the Hall effect region, i.e. in the domain where ρ and μ H are valid.Figure 1b shows the same portion of the Hall plate from Fig. 1a, but contacts and insulating segments are swapped and the polarity of the applied magnetic field is reversed, while its magnitude is the same.Thus, currents are now flowing through contacts C m , not through segments C m .All other quantities of this so-called complementary case are denoted by an overbar.Thus, currents I m flow through the new contacts C m , giving rise to an electric potential φ and a stream function ψ, both being solutions of the Laplace equation.The relations between φ and ψ are analogue to (8), ( 9), if we replace θ H → −θ H to account for the inverted polarity of the magnetic field.We define functions u(x, y), v(x, y) with Like φ, ψ also u, v are solutions of the Laplace equation.Analogous to Sect.2.4 the boundary conditions are (note the different sign of θ H due to inverted polarity of the magnetic field) In terms of the functions u, v this means Comparison of ( 26 R sq is the resistance through a square domain without holes, if two opposite edges are contacts (→ square resistance) [13].Replacing v m by v m in (22) and using (25) gives Equations ( 28) and (29) are the reverse magnetic field on complementary device (RMFoCD) theorem, which is visualized in Fig. 2. It says that knowing all voltages between neighbouring contacts and all currents into all contacts of a Hall plate exposed to a magnetic field, we also know the voltages between neighbouring contacts of the complementary device when it is exposed to the reverse magnetic field, provided that it is electrified by the specific currents I m of (28).The RMFoCD equation (28) states a specific biasing condition of the complementary Hall plate, under which its currents and voltages between neighbouring contacts are related to the ones of the original Hall plate.This theorem is identical to the one in [3], yet here it is proven for multiply-connected Hall plates with neutral boundaries whereas in [3] it was derived only for simply connected Hall plates.Yet, we should keep in mind that in (28) C m , C m+1 are neighbouring contacts, which are on the same boundary (Fig. 1b).Also, in (29) contacts C m−1 , C m are on the same boundary (Fig. 1a).

Implications of the RMFoCD theorem
Here we discuss the implications of the RMFoCD theorem on how the matrices R, G relate to R, G and on how the power densities in complementary Hall plates relate under the biasing conditions specified by the RMFoCD theorem.Equations ( 28), (29) refer to neighbouring contacts on the same boundary.Therefore, we have to distinguish how many boundaries have neighbouring contacts, i.e. how many boundaries have more than one contact.Single contacts per boundary are solitary contacts, which are not touched by ( 28), (29).In Sect.4.1 we start with the simpler case where only one boundary has more than one contact.In Sect.4.2 several boundaries have more than one contact.

... if a single boundary has more than one contacts
Due to the distinction between terminals (in the theory of electrical networks) and contacts (in electromagnetic field theory) we use different symbols for the potentials in this section, For the currents we use the symbols I m , I m .We group voltages and currents to vectors where the superscript T denotes the transpose.Thus, (2) reads V = RI and I = GV with the matrices R = (R ij ) and G = (G ij ).V m are the voltages between a terminal and the zeroth terminal (= ground node).However, in (28 ) and ( 29) there are voltages between neighbouring contacts on the same closed boundary.In this case we can define Note the very last element Here we used the Kronecker delta δ ij , which equals 1 if both indices are identical-otherwise it vanishes.In (29 which we call a shift-down operation (see also [3]).In matrix notation, (28) and (29) read Fig. 3. Hall plate with one exemplary hole in a circuit that guarantees neutral boundaries.The outer boundary #0 has N 0 contacts, the hole boundary has N 1 contacts.The restriction (42) for the currents at the hole boundary is It is implemented by the ideal transformers with unit turns ratios.At the terminals it holds V = R I , with the reduced matrix R being positive definite, yet not symmetric and not reverse magnetic field reciprocal In the left equation of (35) it holds V = R(B z )I with the resistance matrix evaluated at positive magnetic field (Fig. 1a).In the right equation of (35) it holds V = R(−B z )I with the resistance matrix evaluated at reverse magnetic field (Fig. 1b).Plugging the second into the first equation gives for arbitrary current vectors I, from which it follows The principle of reverse magnetic field reciprocity (RMFR) says G(−B z ) = (G(B z )) T [14,15].Introducing this into (37) gives whereby all matrices are evaluated at B z .In the derivation of (38) we used the identities Equation ( 38) is a consequence of the RMFoCD theorem: the definite conductance matrix of the complementary Hall plate is expressible in terms of the resistance matrix of the original Hall plate.This holds if only one boundary has more than two contacts.

... if several boundaries have more than one contacts
Suppose we have M closed boundaries, labelled from #0, #1, . . . (41) However, not all currents into all contacts are allowed to be chosen independently, because the existence of a stream function requires neutral boundaries: The sum of all currents flowing into the contacts of any closed boundary must vanish.We express the current into the first contact of each boundary (the one with lowest index) by the negative sum of all others on boundary #m: for m ∈ {1, 2, . . ., M − 1}.If we delete the currents into the first contacts of each boundary in the vector I, we get the reduced current vector I with only N −M rows for M ≥ 2. Applying the same procedure to V gives V .We call vectors reduced if they have only N −M rows, and matrices are reduced if they have N−M rows and columns, and we denote them by a prime.From V = RI we express all currents by I according to (42) and we delete the rows, which give potentials at the first contacts of each boundary.
What is left after this procedure is the set of equations V = R I with the reduced resistance matrix R .This matrix describes an electric system shown in Fig. 3. There, the current restrictions from (42) are implemented by the ideal transformers connected to the contacts of all boundaries except #0, which comprises the ground node C 0 .A closer inspection shows that the reduced resistance matrix is neither symmetric nor reverse magnetic field reciprocal, and some of its off-diagonal entries may become negative.However, the power dissipated in this system is P d = I T V = I T R I .Whenever such a Hall plate is electrified, i.e.I = 0, it must dissipate power, P d > 0. Therefore, R is positive definite, which means, that its inverse matrix exists.Consequently, it also holds I = G V with G = (R ) −1 .In (28) we need the vector U is a reduced vector with N − M entries.Most of its entries are subtractions of entries of the reduced vector V , yet, also the potentials of the first contacts of the boundaries come into play (like V 0 = 0 and V N0 in (43)).Thus, it holds D 1 has entries equal to 1 on its main diagonal and most entries equal to −1 adjacent above the main diagonal, while all other entries are zero.Thus, the determinant of D 1 is 1.The reduced matrix r has only M − 1 rows which do not vanish.There the entries are equal to R ij or to R ij − R ik , where i and k are equal to m−1 =0 N for m ∈ {1, 2, . . ., M − 1}, i.e. they refer to first contacts on boundaries.Translating (28) into vector notation gives In (29) we need the vector Similar to above it holds, r has again M − 1 rows which do not vanish, but these are different rows than in r .In these rows the entries are equal to R ij or to R ij − R ik , where i and k refer to first contacts on boundaries.Translating (29) into vector notation gives Plugging (48) into (45) one gets In the individual entries of the matrices one can again eliminate the negative polarity of the magnetic field by the RMFR principle.This gives (N − M ) 2 equations, which link the entries of the reduced matrix R with entries of the reduced matrix R .Yet, also entries R ij , R ij of the first contacts of the boundaries are involved, which are not part of the reduced resistance matrices.The conclusion is that one cannot compute all elements of the resistance matrix of such a general multiply-connected Hall plate from the resistance matrix of its complementary Hall plate.
with all R ij = R ij (Bz).Apparently, R and R are not symmetric and not reverse magnetic field reciprocal.Whenever a subtraction is used in one of their entries the element can also be negative.In r and r all involved R ij , R ij have an index 4 or 7, because C 4 , C 4 , C 7 , C 7 are the first contacts of closed boundaries #1 and #2.Inserting ( 50)-( 53) into (49) gives (N − M ) 2 = 36 equations, which can be solved for all entries of R , but not for the entries of r .

Impedance relations between complementary Hall plates
Suppose all contacts on a closed boundary are open (not connected to others) in the original Hall plate.Then all currents into these contacts vanish, and with (29) all contacts on this boundary in the complementary Hall plate will have identical potential.Therefore, we can short all these contacts.In general, there will be currents flowing out of some contacts C m and into others on this boundary.However, the sum of all currents through this boundary will vanish, because they are all shorted and not connected to an external current source.shorted to terminal T − .Current I supply is injected into T + and extracted at T − , and this will cause a voltage V supply = V (at T + ) − V (at T − ).Applying the first RMFoCD equation (28) to Fig. 4 gives

Let us consider an original
Applying the second RMFoCD equation (29) to Fig. 4 gives Combining ( 55) and (56) gives a more general impedance relation between multiply-connected complementary Hall plates than in Section 5 of [3], If one knows the full matrix R, it is possible to compute V supply (−B z )/I supply (−B z ) = f mn (R) with f mn (R) being a function of the entries of R. On the other hand, it is easy to show With (57) this gives the following N − M relations between R and R, Applying the principle of RMFR to the left side of (58 Hence, (57) also holds if the magnetic field has identical polarity in both Fig. 4a, b.Moreover, if the Hall plate is identical to its complementary counterpart, it follows One geometry, which fulfils this so-called complementary symmetry and which has all contacts open, is a singly-connected square domain with two contacts C 0 , C 1 on opposite edges.Hence, we have proven that R sq from (28 ) is indeed the magneto-resistance between opposite edges of a square [13].
If not all but the two power supply contacts C m , C n in the original Hall plate are open, one gets a class of symmetric multiply-connected Hall plates, which are identical to their complementary counterparts and which are electrified in RMFoCD biasing conditions (see Fig. 5).The derivations of (57) and (59) still hold.Thus, the remarkable result is that Hall plates with fairly complex symmetric geometries have a simple magneto-resistance equal to R sq .In Fig. 5a and b the outer boundary and the quadrupel of holes have 90 • symmetry with respect to their coinciding centres.All contacts are placed such that they become insulating boundaries after 90 • rotation, and vice versa.The contacts in the holes are all open and shorted, alternatingly.Figure 5a has highly symmetric holes and contacts.Without holes and inner contacts the square Hall plate has the resistance R sq , and with the holes and the contacts on the hole boundaries it stays the same.This high symmetry is broken in Fig. 5b, but the resistance still stays the same.In Fig. 5c the triple-connected domain and its outer boundary have 180 • symmetry, but not the hole boundaries.We first have to rotate the original Hall plate 90 • counter-clockwise and then flip it (together with the magnetic field) around the vertical axis in the drawing plane to get the complementary Hall plate.Finally, we have to invert the sign of the indicated current to establish the exact RMFoCD operating condition of (28).All these manipulations do not alter the resistance, and therefore it is again equal to R sq .The example in Fig. 5c shows that (59) is also valid for Hall plates with flipped complementary symmetry; then the top view of the original Hall plate is identical to the bottom view of the complementary Hall plate.In the last example in Fig. 5d the impedance R sq is measured between the contacts of the hole.
In [7] the prominent role of R sq is explained in the context of diagonal-symmetric Hall plates, where every contact becomes an insulating segment after being mirrored on the symmetry axis, and vice versa (like in Fig. 5c).For singly-connected Hall plates, diagonal and complementary symmetry are identical, but for multiply-connected Hall plates complementary symmetry is more general than diagonal (e.g.Fig. 5a, b have no diagonal symmetry).In particular, diagonal symmetry does not explain, which contacts are open or shorted: even though Fig. 5c has diagonal symmetry, the contacts in only one hole are shorted.
It is noteworthy that half of the total boundary of complementary symmetric Hall plates is covered by contacts, and the indicated resistances are proportional to 1/ cos θ H = 1 + μ 2 H B 2 z (exponent 1/2).In [9] it was proven that multiply-connected Hall plates with all of their boundaries covered by contacts have a resistance matrix with all its entries being proportional to 1/(cos θ H ) 2 = 1 + μ 2 H B 2 z (exponent 1)-they are all even functions of B z and therefore no Hall output signal can be tapped.If such a Hall plate is our complementary one, the right-hand side of (58) becomes constant from the magnetic field.Then the original multiply-connected Hall plate has only point-sized contacts, with the expressions on the left-hand side of (58) being constant versus magnetic field (∝ (1 + μ 2 H B 2 z ) 0 , exponent 0).This is consistent with [8], where it was proven that for point-sized contacts and neutral boundaries the even part of the electric potential, (φ(B z ) + φ(−B z ))/2, does not depend on the magnetic field.To sum up, the magneto-resistance effect tends to grow with the size of the contacts while the Hall effect tends to diminish with the size of the contacts.

Power and power density in complementary devices
Figure 6 shows the E and J vectors in a test point of the original Hall plate at applied magnetic field B z and E and J in the same test point of the complementary Hall plate at reverse magnetic field −B z , if the RMFoCD biasing conditions (28) and (29) hold.Then, the curves u(x, y) = const are isopotential lines (E(Bz) is perpendicular to the curves), and they are also current streamlines (J (−Bz) is tangential to the curves).Conversely, the curves v(x, y) = const are isopotential lines (E(−Bz) is perpendicular to the curves), and they are also current streamlines (J (Bz) is tangential to the curves).The isocurves u(x, y) = const and v(x, y) = const intersect at an angle π/2 − θ H .With u = u, v = v from (27) and with (22), (24) it holds Surprisingly, the power density field is identical in the original Hall plate and in the complementary one if the RMFoCD biasing conditions (28), (29) hold, The equality of power densities in complementary Hall plates at RMFoCD biasing conditions is quite extraordinary: it holds for arbitrary magnetic field everywhere in the Hall domains.All Hall plates, which are related by conformal transformations, have the same resistance matrix, i.e. the same electric behaviour at their terminals [16].However, their power densities are different.Conversely, in general, complementary Hall plates have different resistance matrices, yet there is an infinite set of biasing conditions, which gives identical power density in pairs of complementary Hall plates.With (63) also the total power is identical, (65) The power through boundary #m in the complementary device at reverse magnetic field is whereby we used (28).In (65) it holds V −1 = V Nm−1 , and in (66) it holds V Nm = V 0 .Term-wise comparison of (65) and (66) shows their identity, P d,#m = P d,#m .

Numerical check of the theory
The off-diagonal elements account for the Hall effect, the fraction in front of the matrix accounts for the magneto-resistance effect at strong magnetic field.The mesh had 833536 triangular elements.In the original Hall plate the resulting potentials at the contacts are (in units of V) The potentials at the contacts of the complementary Hall plate at reverse magnetic field (θ H = −27 • ) were computed with FEM (in units of V), whereby C 0 was grounded.
The potential of the small hole was 1.3586381 V.It has no corresponding row and column in the resistance matrix, because it is a solitary contact.The voltages between neighbouring contacts in the complementary Hall plate are (in units of V) which is in accordance with (29)-the discrepancies are less than 3 × 10 −6 .The potential of the small hole and V 4 are not predicted by the RMFoCD theorem.Conversely, the voltages between neighbouring contacts on each boundary in (72) are precisely predicted by the RMFoCD theorem.Figure 7c superposes the potential of Fig. 7a and the current streamlines of Fig. 7b. Figure 7d superposes the potential of Fig. 7b and the current streamlines of Fig. 7a.Apparently in both cases the streamlines go along the contours of the potentials of the complementary Hall plate at reverse magnetic field, because these potentials are also the stream functions of the original Hall plate (up to a constant factor cos θ H , see ( 22), (24), ( 27)).The power dissipated in both cases is 22.2929 W. Integration over the power density E • J gives the same value as the sum over V i I i , both for the original and the complementary devices (the discrepancies are less than 1.2 ppm).The power through boundary #0 is 17.48 W, and through boundary #1 it is 4.81 W, both in the original and in the complementary Hall plate (discrepancy is only 1.5 ppm).Figure 7e and f shows the power densities in the original and in the complementary Hall plate in a logarithmic colour scale.Both are identical, in accordance with Sect.4.4.I also checked that the subtraction of both gives zero, except for some round-off errors ± ≈ 10 −4 dBW near the vertices of the contacts.I computed the definite conductance matrices with FEM via G ij = I i /V j by setting the potential at the j-th contact to 1 V and grounding all other contacts.which is consistent with R(−B z ) in (74) in view of (50).

Conclusion
This paper discussed multiply-connected uniform plane Hall plates with numerous extended contacts exposed to a homogeneous perpendicular magnetic field of arbitrary magnitude.Inter-relations between such Hall plates and their complementary (or dual) counterparts were studied.Thereby, a complementary (or dual) Hall plate is defined by the same geometry and the same material properties as the original one, yet contacts and insulating segments are swapped.If the sum of currents into all contacts of a closed boundary is zero (neutral boundary), and if this holds for all closed boundaries, the current pattern can be described by a unique stream function.Then the duality between the stream function and the electric potential leads to the following relations: The voltages between neighbouring contacts of the complementary Hall plate at reverse magnetic field are proportional to the currents into the original Hall plate (29), if the currents into the complementary Hall plate are proportional to the voltages between neighbouring contacts of the original Hall plate (28), the constants of proportionality being −R sq and

Figure
Figure 1a shows a portion of a Hall plate with a boundary comprising contacts C m−1 , C m , C m+1 and interleaving insulating segments C m−1 , C m .The directional sense of the boundary goes from right to left.Magnetic field B z > 0 is applied.Each of these contacts is connected to a terminal.Potentials are applied to the terminals, which give rise to currents flowing through the contacts.Finally, a potential φ = u(x, y)

Fig. 1 .
Fig.1.Portions of a Hall plate with segments Cm, Cm on one of its boundaries.Contacts and insulating segments are swapped in the two complementary cases.Also, a magnetic field with opposite polarity (but same magnitude) is applied to the two complementary Hall plates

Fig. 2 .
Fig. 2. RMFoCD theorem: an exemplary Hall plate has four contacts on the outer boundary and three contacts on the hole boundary.It is electrified by currents I 0 , . . ., I 6 with neutral boundaries, i.e.I 0 + I 1 + I 2 + I 3 = 0 and I 4 + I 5 + I 6 = 0, and it responds with voltages U 0 , . . ., U 6 between neighbouring contacts.If the complementary Hall plate is electrified by currents Im = Um/Rsq, and if it is exposed to the same magnetic field with reverse polarity, it responds with voltagesV 1 − V 0 = RsqI 1 , V 2 − V 1 = RsqI 2 , V 3 − V 2 = RsqI 3 , V 5 − V 4 = RsqI 5 ,V 6 − V 5 =RsqI 6 .The currents into the first contacts per boundary are indicated in red colour (Color figure online) Fig. 2. RMFoCD theorem: an exemplary Hall plate has four contacts on the outer boundary and three contacts on the hole boundary.It is electrified by currents I 0 , . . ., I 6 with neutral boundaries, i.e.I 0 + I 1 + I 2 + I 3 = 0 and I 4 + I 5 + I 6 = 0, and it responds with voltages U 0 , . . ., U 6 between neighbouring contacts.If the complementary Hall plate is electrified by currents Im = Um/Rsq, and if it is exposed to the same magnetic field with reverse polarity, it responds with voltagesV 1 − V 0 = RsqI 1 , V 2 − V 1 = RsqI 2 , V 3 − V 2 = RsqI 3 , V 5 − V 4 = RsqI 5 ,V 6 − V 5 =RsqI 6 .The currents into the first contacts per boundary are indicated in red colour (Color figure online) ) with (23) shows: If we set u m = u m ∀m, the differential equations and the boundary conditions for u and u are identical.Likewise, if we set v m = v m ∀m, the differential equations and the boundary conditions for v and v are identical.Consequently, it holds u(x, y) = u(x, y), v(x, y) = v(x, y) (27) in the Hall effect region and on its boundaries.Replacing u m by u m in (24) and replacing u m by φ(on C m ) gives

#M − 1 ,
where the m-th boundary has N m ≥ 2 contacts and N m insulating segments, m ∈ {0, 1, . . ., M − 1}.Thus it holds M −1 m=0 N m = N, (40) and the matrices R, G, R, G have N −1 rows and columns.We label the contacts C m consecutively, on boundary #0: C 0 , C 1 , . . ., C N0−1 , . . . on boundary #m: Example: Let us consider a Hall plate with four contacts on the outer boundary, three contacts on the boundary of a first hole, and two contacts on the boundary of a second hole.This means M = 3, N 0 = 4, N 1 = 3, N 2 = 2, N = 9.The reduced matrices with N − M = 6 rows and columns are

1 =0NFig. 4 .
Fig. 4. Impedance relation between complementary Hall plates at reverse magnetic field.Both Hall plates are energized by a current source

Fig. 5 .
Fig. 5. Hall plates with resistance Rsq to the indicated current flow.The colour coding gives the electric potentials (red means positive, blue means negative) and the grey lines are the current streamlines.These plots were obtained by FEM simulation for θ H = 3.3 • (Color figure online)

Fig. 6 .
Fig. 6.Electric field and current density in the original Hall plate at Bz and in the complementary Hall plate at −Bz if the RMFoCD biasing conditions (28) and (29) hold.Then it also holds u = u, v = v, see (27)

Figure 7
Figure7shows a finite Hall plate with two holes.The outer boundary has four contacts, the first hole has three contacts, the second hole has no contact in Fig.7a and asingle floating contact in the complementary case of Fig. 7b.The Hall angle is θ H = 27 • and the square resistance is R sq = 1Ω, hence, the sheet resistance is R sheet = cos θ H = 0.891007Ω.Currents I 1 = 1 A, I 2 = 2 A, I 3 = −3 A are injected in the outer boundary.Contact C 0 is grounded.No current flows into the ground node, I 0 = −I 1 −I 2 −I 3 = 0 A. Currents I 5 = 0 and I 6 = 1 A are injected into the contacts of the first hole.Hence, I 4 = −I 5 − I 6 = −1 A flows into the first contact C 4 of this hole.The potential and the current streamlines were computed with a finite element simulation (FEM) method, implemented in the commercial program code COMSOL Multiphysics.The 2D application mode 'Conductive Media DC (emdc)' was chosen.The various segments are straight lines, circular arcs, and stretched circular arcs.The geometry is specified in Fig. 7c.The simple model has only one physical parameter, the anisotropic conductivity tensor σ, which follows immediately from (1),

Fig. 7 .
Fig. 7. Exemplary Hall plate with two holes for a numerical check of the RMFoCD theorem (28), (29).Results of finite element simulations for θ H = ±27 • and Rsq = 1Ω.Figures a and b show the electric potential in colour code and the current streamlines in grey.Figures c and d superpose the electric potential of one of the figures (a) and (b) with the streamlines of the other one of these figures.Figure e shows the power density in the original Hall plate and figure f shows the power density in the complementary Hall plate (in units of dBW).Obviously, both plots are identical.Colour code: red means largest value, green is the median and blue the lowest value (Color figure online)