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The electrical properties of magnetic sensing devices fabricated from anisotropic materials are not easily extracted. Here we present a method for determining the resistance matrix for an anisotropic device with multiple electrical contacts placed in a perpendicular magnetic field. By using the methods developed by Van der Pauw and Wasscher, the analysis for the anisotropic system is reduced to the equivalent problem for an isotropic sample, which can then be solved using methods developed previously. As a result, the method works in the case of structures with an arbitrary number of asymmetric extended contacts at large magnetic field strength. In addition to the extraction of nonisotropic resistivities, the resistance matrix can be used to analyze the Hall effect for anisotropic plates.

The electrical resistivity of materials is a very important property in the characterization of magnetic sensing devices. There is a large body of work devoted to the determination of the resistivity of isotropic materials dating back to the seminal work of van der Pauw [

Wasscher [

By using the affine transformation suggested by van der Pauw [

In the absence of a magnetic field, Versnel [

For the case of isotropic semiconductors, Homentcovschi and Bercia [

In this paper, the relationships relevant to the galvanomagnetic transport in two-dimensional anisotropic conductive structures (with multiple nonsymmetric contacts and at large magnetic field) are obtained as analytic formulas involving the equivalent isotropic resistivity, the Hall mobility and some one-variable integrals which can be evaluated numerically taking appropriate care of the point singularities at the ends of contacts in the original plane (see [

An important application of the resistance matrix is the determination of the Hall voltage in order to optimize the design of the Hall devices. In the last section, we present as an example the determination of the Hall voltage in a square nonisotropic semiconductor plate under the influence of an arbitrary magnetic field. We note that anisotropy generated by piezoresistive effects is not analyzed here. This subject was addressed in the excellent paper by Ausserlechner [

We restrict ourselves to the cubic, tetragonal, hexagonal trigonal and orthorhombic crystallographic symmetries for which the resistivity tensor is diagonal and has three resistivity components ρ 1 , ρ 2 , ρ 3 (along the x 1 = x , x 2 = y , x 3 = z axes). For the treatment of an anisotropic cubic sample, having the edges aligned with the principal axes of the resistivity tensor (with edge length l ), van der Pauw [

x ′ i = ρ i ρ x i , where ρ = ρ 1 ⋅ ρ 2 ⋅ ρ 3 3 (1)

and

l ′ i = ρ i ρ l ( i = 1,2,3 )

such that the cube D ˜ is transformed into an isotropic parallelepiped D ˜ ′ of resistivity ρ and dimensions l ′ 1 , l ′ 2 , l ′ 3 . The transformation (1) preserves the voltage and current, therefore the domains D ˜ and D ˜ ′ will have the same resistance R (see Miccoli et al. [

Consider in the physical domain a flat, anisotropic rectangular semiconductor sample with length l 1 , the width l 2 parallel to the directions of the principal resistivities ρ 1 , ρ 2 and thickness δ with its plane perpendicular to the principal resistivity ρ 3 .

D = { − l 1 / 2 < X < l 1 / 2 0 < Y < l 2

As shown in

D ′ = { − l ′ 1 / 2 < X ′ < l ′ 1 / 2 0 < Y ′ < l ′ 2

shown in

The complex variable sine-amplitude function

z = s n ( l ′ 1 Z ′ / ( 2 K ) , m )

conformally maps the interior of the rectangle with side lengths l ′ 1 and l ′ 2 in the ( Z ′ )-plane onto the upper half-plane Im ( z ) > 0 . Here, K ( m ) is the complete elliptical integral of the first kind of the parameter m and K ′ ( m ) = K ( 1 − m ) is its associate (another parameter used in connection to the elliptical integrals is the modulus k = m ). Since K ′ ( m ) / K ( m ) = 2 l ′ 2 / l ′ 1 we introduce the “nome” q ( m ) for the parameter m defined as follows:

q ( m ) ≡ exp ( − π K ′ ( m ) K ( m ) )

and the inverse nome q − 1 , which for our case yields,

q − 1 ( q ( m ) ) = m = q − 1 ( − 2 π l 2 l 1 ρ 2 ρ 1 ) (2)

For more details on the computation of elliptic integrals using MATLAB see Batista [

s n ( K , m ) = 1 ; s n ( K + i K ′ , m ) = 1 / m

s n ( − a K , m ) = − s n ( a K , m ) ; | a | ≤ 1 (3)

s n ( a K + i K ′ ) = [ m s n ( a K , m ) ] − 1

From these relationships, we can calculate the extremities a 1 , b 1 of the contact A (similarly for B, C and D) in the z half-plane (see

Now we utilize the basic relationships for the upper half-plane Re ( z ) > 0 provided in Homentcovschi et al. [

∑ m = 1 3 B k m ( r ) V m = R s q ( i s o ) cos ( θ H ) ∑ m = 1 3 C k m ( r ) I m , ( k = 1 , 2 , 3 ) (4)

where

R s q ( i s o ) = ρ δ ′ = ρ ρ / ρ 3 / δ = ρ 1 ρ 2 δ

and

A k m ( r ) = ∫ b m a m + 1 | P b ( t ) P a ( t ) | γ t k − 1 P b ( t ) d t

B k m ( r ) = ∫ a m b m | P b ( t ) P a ( t ) | γ t k − 1 P b ( t ) d t

C k m ( r ) = − ∑ q = m 3 A k q ( r ) ( k , m = 1 , 2 , 3 ) .

where

P a ( x ) = ∏ j = 1 4 ( x − a j ) and P b ( x ) = ∏ j = 1 4 ( x − b j ) .

The basic relationships (4) together with the equation for the conservation of currents

I 1 + I 2 + I 3 + I 4 = 0 (5)

is a complete system for determining the electrical parameters of the device. The method can be easily extended for general positions of the contacts on the anisotropic rectangular sample.

The configuration studied now (the physical domain) is a flat anisotropic circular sample of radius r and thickness δ with its plane chosen perpendicular to the direction ρ 3 and having on the boundary four arbitrary perfectly conducting metallic contacts A , B , C , D (

{ X = r cos ϕ Y = r sin ϕ 0 ≤ ϕ ≤ 2 π (6)

According to the above transformation the anisotropic circular sample gives

{ X ′ = r ρ 1 / ρ cos ϕ Y ′ = r ρ 2 / ρ sin ϕ (7)

which is an electric equivalent isotropic elliptic sample (of resistivity ρ ) with semi-axes a = r ρ 1 / ρ and b = r ρ 2 / ρ , and thickness δ ′ = δ ρ 3 / ρ ; we assume that ρ 1 is larger than ρ 2 . By embedding the ellipse in the complex Z ′ plane we can write (

Z ′ = a cos ϕ + i b sin ϕ , 0 ≤ ϕ ≤ 2 π (8)

Since the elliptic sample is isotropic, it can be conformably mapped into the unit circle in the complex plane (z) (see

z = f ( Z ′ ) = m ⋅ s n ( 2 K ( m ) π sin − 1 Z ′ a 2 − b 2 , m ) (9)

where the parameter m of the elliptical integrals ( K , s n , c n , d n ) satisfies the Jacobi’s nome equation

q ( m ) ≡ exp ( − π K ′ ( m ) K ( m ) ) = ( ρ 1 1 / 2 − ρ 2 1 / 2 ρ 1 1 / 2 + ρ 2 1 / 2 ) 2 (10)

which determines the parameter as a function of ρ 1 / ρ 2 . On the circle | z | = 1 we obtain

z = exp ( i ψ ) = c n ( u , m ) d n ( u , m ) + i ( 1 − m ) s n ( u , m ) 1 − m s n 2 ( u , m ) (11)

where u = 2 K ( m ) ⋅ ϕ / π . The Formula (11) was used previously by Wasscher [

When the magnetic field B is normal to the circular disk the Hall-Ohm law introduces the effective resistance

ρ B = ρ ( 1 + μ H 2 B 2 ) = ρ / cos 2 θ H (12)

The resistance matrix determined in the Homentcovschi et al. [

R s q ( e f ) cos ( θ H ) = ρ δ ′ cos θ H = ( ρ 1 ⋅ ρ 2 ⋅ ρ 3 ) 1 / 3 δ ( ρ 3 / ( ρ 1 ⋅ ρ 2 ⋅ ρ 3 ) 1 / 3 ) 1 / 2 cos θ H ≡ ( ρ 1 ⋅ ρ 2 ) 1 / 2 δ cos θ H (13)

The basic relationships (19) from the paper [

∑ j = 1 3 B p j ( c ) V j = R s q ( i s o ) cos θ H ∑ j = 1 3 C p j ( c ) I j , ( p = 1 , 2 , 3 ) . (14)

Here, R s q ( i s o ) = ( ρ 1 ⋅ ρ 2 ) 1 / 2 / δ is the sheet resistance in the physical domain, and

A p j ( c ) = ∫ β j α j + 1 h ( t ) d t sin ( ( t − β 4 ) / 2 ) sin ( ( t − β p ) / 2 )

B p j ( c ) = ∫ α j β j h ( t ) d t sin ( ( t − β 4 ) / 2 ) sin ( ( t − β p ) / 2 ) (15)

C p j ( c ) = − ∑ q = j 3 A p q ( c ) .

where p , j = 1 , 2 , 3 and we define

h ( t ) = ∏ j = 1 4 | sin ( ( t − β j ) / 2 ) sin ( ( t − α j ) / 2 ) | γ , γ = 1 2 − θ H π . (16)

The compatibility relationships (14) require the conservation equation for the currents for closure (5). In both cases considered in this section, the voltages of the contacts A , B , C are V 1 , V 2 , V 3 and the terminal currents are I 1 , I 2 , I 3 , respectively; the contact D is grounded ( V 4 = 0 ) and the current I 4 is obtained from (5).

The basic system in the case of half-plane (4) and (14) for the unit circle) contains the matrices B = [ B p j ] , ℂ = [ C p j ] , the vector of terminal potentials v = [ V 1 , V 2 , V 3 ] T , the vector of terminal currents i = [ I 1 , I 2 , I 3 ] T and the physical parameters R s q and θ H . These are the relevant relationships to the galvanomagnetic transport in two-dimensional conductive structures with an arbitrary number of asymmetric extended contacts on the boundary in the case of small or large magnetic field. The basic system can be written using matrix notation as

B ⋅ v = R s q ( i s o ) cos ( θ H ) ℂ ⋅ i (17)

hence, we can define the Resistance Matrix as

ℝ ( R s q , θ H ) = R s q ( i s o ) cos ( θ H ) B − 1 ⋅ ℂ (18)

It is convenient to extract R s q ( i s o ) by defining the geometry matrix G ,

ℝ ( R s q ( i s o ) , θ H ) = R s q ( i s o ) G ( θ H ) , (19)

where

G ( θ H ) = ( cos ( θ H ) ) − 1 B − 1 ℂ

which only depends on the location of the extremities of the contacts and the parameter θ H = arctan ( μ H B ) . We note that the two physical parameters μ H and B appear in all of the results only as their product, which yields the Hall angle θ H . The reverse-magnetic-field reciprocity theorem given by Cornils and Paul [

[ V A V B V C ] = ℝ ( R s q ( i s o ) , θ H ) [ I A I B I C ] (20)

where

ℝ ( R s q ( i s o ) , θ H ) = [ R A D , D A ( R s q ( i s o ) , θ H ) R B D , D A ( R s q ( i s o ) , θ H ) R C D , D A ( R s q ( i s o ) , θ H ) R B D , D A ( R s q ( i s o ) , − θ H ) R B D , D B ( R s q ( i s o ) , θ H ) R C D , D B ( R s q ( i s o ) , θ H ) R C D , D A ( R s q ( i s o ) , − θ H ) R C D , D B ( R s q ( i s o ) , − θ H ) R C D , D C ( R s q ( i s o ) , θ H ) ] (21)

We denoted R i j , k l = ( V l − V k ) / ( I i j ) , where I i j is the current which flows into the Hall plate at contact i and out at contact j. The off-diagonal elements in (21) are pairwise equal apart from the change of sign of their second arguments. Hence, there are six independent resistance functions of θ H . Cornils et al. [

In the case of point-wise contacts the geometry of the problem is described by one parameter such that, in this case, we need to measure two resistances, namely the two used by van der Pauw (in the case of point-like contacts) R A B , C D and R B C , A D . By taking I A = − I B = I and I C = 0 , we obtain from Equation (20),

R A B , C D ( 0 ) = [ V D − V C ] / I A B = R C D , D B ( R s q ( i s o ) , 0 ) − R C D , D A ( R s q ( i s o ) , 0 ) (22)

Hence,

R A B , C D ( R s q ( i s o ) , 0 ) = R s q ( i s o ) [ G 23 ( 0 ) − G 13 ( 0 ) ] . (23)

Similarly,

R B C , D A ( R s q ( i s o ) , 0 ) = R B D , D A ( R s q ( i s o ) , 0 ) − R C D , D A ( R s q ( i s o ) , 0 )

and

R B C , D A ( R s q ( i s o ) , 0 ) = R s q ( i s o ) [ G 12 ( 0 ) − G 13 ( 0 ) ] (24)

In the case of finite contacts (non point-like contacts) these formulas will provide corrections for the sheet resistance.

To define the Hall voltage, we use

R B D , A C ( R s q ( i s o ) , θ H ) = R C D , D B ( R s q ( i s o ) , − θ H ) − R B D , D A ( R s q ( i s o ) , θ H )

from relationship (21), we have,

R B D , A C ( R s q ( i s o ) , θ H ) = R s q ( i s o ) [ G 32 ( θ H ) − G 12 ( θ H ) ] (25)

This formula can be used for determining the Hall voltage for a certain biasing.

In [

In the case of the method based on (11), we analyzed the images on the unit isotropic disk of the uniform mesh on the original anisotropic sample.

The geometry of the problem is similar to that considered by Versnel: A circular disk of radius r and thickness d containing four perfectly conducting contacts of the same size. The two lines, each of which connects the midpoints of the two opposite contacts are orthogonal to each other. The position of these lines is determined by the angle ϕ with respect to the direction ρ 1 . The size considered for all contacts equals 3˚. We can write

R 1 ( ϕ ) = R B C , D A ( 0 ) = R s q ( i s o ) [ G 12 ( 0 ) − G 13 ( 0 ) ]

R 2 ( ϕ ) = R A B , C D ( 0 ) = R s q ( i s o ) [ G 23 ( 0 ) − G 13 ( 0 ) ]

how to determine the specific resistivities of the anisotropic semiconductor from the measured values R 1 and R 2 . The same method is also described in the paper by Nader and Kouba [

As an application of the present results, we show how the resistance matrix can be used to determine the Hall voltage in an anisotropic sample with four terminals. The example chosen is a square semiconductor plate (as described in II.A) with four equal metallic terminals A = Q E ¯ , B = F R ¯ , C = S G ¯ , D = H P ¯ of length 0.1l, where l is the length of each side ( l = 1 here).

The resistance matrices for anisotropic Hall plates (square and circular) were developed by considering finite size, nonsymmetric, perfectly conducting, contacts on the boundary. By using the Van der Pauw transforms, the anisotropic problem is reduced to the isotropic case. Subsequently, by using the Wasscher transformation to the canonical domain, the problems are then solved by applying the methods presented in Homentcovschi et al. [

As application examples, the methodology is applied to study the influence of finite contact dimensions on the determination of specific anisotropic resistivities, and to the study of the Hall effect for a square anisotropic plate with finite contacts in an arbitrary magnetic field.

The authors wish to thank the reviewer who provided detailed comments which improved the manuscript.

The authors declare no conflicts of interest regarding the publication of this paper.

Homentcovschi, D., Oprea, R. and Murray, B.T. (2021) Resistance Matrix for an Anisotropic Hall Plate with Multiple Extended Asymmetric Contacts on the Boundary. Journal of Applied Mathematics and Physics, 9, 1911-1925. https://doi.org/10.4236/jamp.2021.98125