_{1}

^{*}

Singly connected Hall plates with
*N* peripheral contacts can be mapped onto the upper half of the z-plane by a conformal transformation. Recently, Homentcovschi and Bercia derived the
*General Formula* for the electric field in this region. We present an alternative intuitive derivation based on conformal mapping arguments. Then we apply the
*General Formula* to complementary Hall plates, where contacts and insulating boundaries are swapped. The resistance matrix of the complementary device at reverse magnetic field is expressed in terms of the conductance matrix of the original device at non-reverse magnetic field. These findings are used to prove several symmetry properties of Hall plates and their complementary counterparts at arbitrary magnetic field.

The purpose of this work is to derive relations between Hall plates and their complementary counter-parts. A Hall plate is assumed to be a plane conductive region with thickness much smaller than its lateral dimensions. We discuss only singly-connected Hall plates without holes. An arbitrary number of at least two extended contacts are on the perimeter. Between neighboring contacts there is a unitary piece of insulating boundary. Thus the number of contacts equals the number of unitary pieces of insulating boundary. The complementary Hall plate comprises the very same conductive region, but contacts and insulating boundaries on the perimeter are swapped. A complementary Hall plate has the same number of contacts as the original Hall plate. If the contacts of the original Hall plate are small, the complementary Hall plate has large contacts. Therefore a high impedance original device will generally have a low impedance complementary device. The first question is if there is a more precise quantitative relation between both devices. The ultimate question is if we can obtain all electrical parameters of the complementary device from the parameters of the original Hall plate.

Several properties of complementary Hall plates at zero or weak applied magnetic field have already been studied in the past. We found the following ones.

Plain distributed resistive structures with complementary peripheral electrode geometries were studied at zero magnetic field in [

Van-der-Pauw measurement on Hall plates with 90˚ symmetry at zero magnetic field was discussed in [

In [

In [

Complementary Hall plates with three extended contacts on the perimeter were studied in [

In Section 2 we reconsider the General Formula of [

Recently Homentcovschi and Bercia found a closed form analytical solution for the electric field in the upper half plane with an arbitrary number of extended contacts on the real axis [

E ( z ) = E x − i E y = i L ( z ) ∑ n = 1 N c n z n − 1 with c n ∈ R (1a)

with

L ( z ) = ∏ n = 1 N ( z − a n ) 1 − γ ( z − b n ) γ (1b)

γ = 1 2 + θ H π with 0 < γ < 1 (1c)

In (1c) θ H is the Hall angle defined by tan ( θ H ) = μ H B ⊥ . We define the Hall mobility μ H > 0 for negative charge carriers and μ H < 0 for positive ones. This is opposite to the definition in [

In [

S ( w ) = S u − i S v = 1 ρ cos ( θ H ) exp ( − i θ H ) E ( w ) (2a)

(2a) is the complex notation of the general Ohm’s law in vector form

E = ρ S + ρ μ H S × B ⊥ (2b)

In many materials ρ and μ H do not depend on B ⊥ as long as B ⊥ is sufficiently small (see (11.19) in [

In (2a)， the conjugate complex of S ( w ) , E ( w ) point in the directions of the vectors S , E ,respectively. Thus, taking the conjugate of both sides of (2a) we obtain S by counter-clockwise rotation of E (for θ H > 0 ). All N − 2 output contacts must be horizontal and folded in exactly such a way that the outer boundary becomes a smooth parallelogram (without steps in its left and right edges). Only if all folded output contacts are horizontal they are compatible to homogeneous electric and current density fields in the parallelogram. Then the complex potential in the skewed parallelogram is given by (see (29) in [

F I ( w ) = I supply R sheet w 2 cos ( θ H ) 2 (3a)

The complex electric field in the parallelogram is (see left equalities in (25) in [

E ( w ) = E u − i E v = i d d w F I ( w ) = I supply R sheet i 2 cos ( θ H ) 2 (3b)

and the complex current density in the parallelogram is with (2a)

S ( w ) = S u − i S v = 1 2 t H I supply ( i + tan ( θ H ) ) (4)

E ( w ) and S ( w ) are homogeneous inside the parallelogram—they do not depend on w. The current flowing out of the Hall plate via the lower supply contact in

t H ∫ w = − 1 1 ( − 1 ) S v d w = I supply (5)

The supply voltage is given by

V supply = ∫ ϕ = 0 V supply d ϕ = ∫ B N A j ∇ ϕ ⋅ d w = − ∫ v = Im { B N } Im { A j } E v d v = I supply R sheet Im { A j − B N } 2 cos ( θ H ) 2 (6)

where ϕ is the electric potential inside the Hall effect region (the potential on the contacts is denoted by V 1 , V 2 , ⋯ , V N ). We are free to choose the ground potential and so we ground the N-th contact ( V N = 0 ⇒ Im { B N = 0 } ). The current flows out of the Hall plate through the lower contact N of the parallelogram, and it flows into the Hall plate through the upper contact j. The electric field in the upper half of the z-plane is given by

E ( z ) = E x − i E y = i d d z F I ( z ) = i d F I ( w ) d w d w d z = I supply R sheet i 2 cos ( θ H ) 2 d w d z (7)

whereby we used the fact that due to the conformal mapping between w-plane and z-plane it holds F I ( z ) = F I ( w ) . With the Schwartz-Christoffel formula [

E ( z ) = I supply R sheet i 2 cos ( θ H ) 2 K ( z − b 1 ) − ( π / 2 + θ H π ) ( z − a 2 ) − ( π / 2 − θ H π ) ( z − d 2 ) − ( − π π ) × ( z − b 2 ) − ( π / 2 + θ H π ) ⋯ ( z − a j ) − ( π / 2 − θ H π ) ( z − b j ) − ( π / 2 + θ H π ) ( z − a j + 1 ) − ( π / 2 − θ H π ) × ( z − d j + 1 ) − ( − π π ) ( z − b j + 1 ) − ( π / 2 + θ H π ) ⋯ ( z − a N ) − ( π / 2 − θ H π ) ( z − d N ) − ( − π π ) × ( z − b N ) − ( π / 2 + θ H π ) ( z − a 1 ) − ( π / 2 − θ H π ) ( z − d 1 ) − ( − π π ) = I supply R sheet i 2 cos ( θ H ) 2 K L ( z ) ∏ n = 1 n ≠ j N − 1 ( z − d n ) (8a)

hereby K is a scaling constant of the Schwartz-Christoffel mapping, contact N is the current drain contact at ground potential, and contact j is the current input contact at supply voltage. All output contacts—i.e., floating contacts with zero current—are folded. Points D n are inner pointed tips of these folded contacts inside the parallelogram in the w-plane. They are mapped onto the stagnation points d n on the real axis in the z-plane with a n < d n < b n for n = 2 , 3 , ⋯ , N − 1 ; ( n ≠ j ) . At contact 1, it holds a 1 < d 1 ∨ b 1 > d 1 . Hence, all d n are real numbers. K is a real number, because for points on the grounded contact N the product Π ( z − d n ) in the numerator of (8a) is a real number and L ( z ) is a negative real number (see

In the most general case currents are flowing through all contacts. Then the electric field is the following linear superposition:

E ( z ) = R sheet i 2 cos ( θ H ) 2 L ( z ) ∑ j = 1 N − 1 I j K j ∏ n = 1 n ≠ j N − 1 ( z − d n , j ) (8b)

whereby the real-valued scaling constant K j and the stagnation points d n , j depend on the number j of the contact at which current I j was supplied in

c N = 0 (9)

It means that for | z | → ∞ the electric field declines with the dominant term i c N − 1 z − 2 .

The current density in the z-plane is given by (2a) if we replace w = u + i v by z = x + i y . Integration over the electric field and the current density gives the voltages between the contacts and the currents through the contacts. This was done in [

I n = ∑ k = 1 N N n , k c k (10a)

N n , k = ( − 1 ) N + n cos ( θ H ) 2 R sheet ∫ a n b n x k − 1 | L ( x ) | d x for n = 2 , ⋯ , N (10b)

N 1 , k = ( − 1 ) N + 1 cos ( θ H ) 2 R sheet ( ∫ − ∞ b 1 x k − 1 | L ( x ) | d x + ( − 1 ) N ∫ a 1 ∞ x k − 1 | L ( x ) | d x ) for k < N (10c)

N 1 , k = ( − 1 ) N + 1 cos ( θ H ) 2 R sheet ( ∫ max ( a 1 , − b 1 ) ∞ ( − 1 ) N x k − 1 | L ( x ) | + ( − x ) k − 1 | L ( − x ) | d x + { ∫ − b 1 a 1 ( − x ) k − 1 | L ( − x ) | d x ; − b 1 < a 1 ∫ a 1 − b 1 ( − 1 ) N x k − 1 | L ( x ) | d x ; a 1 < − b 1 ) for k = N (10d)

and to the potentials V n at the contacts

U n = V n − V n + 1 = ∑ k = 1 N M n , k c k for 1 ≤ n ≤ N with V N + 1 = V 1 (11a)

M n , k = ( − 1 ) N + n cos ( θ H ) ∫ b n a n + 1 x k − 1 | L ( x ) | d x (11b)

Note that N n , k has opposite sign than in [

∑ n = 1 N U n = 0 ⇒ ( ∑ n = 1 N M n , 1 ) c 1 + ( ∑ n = 1 N M n , 2 ) c 2 + ⋯ + ( ∑ n = 1 N M n , N − 1 ) c N − 1 + ( ∑ n = 1 N M n , N ) c N = 0 (12a)

In the appendix we prove with the calculus of residues that the sums in front of c 1 , c 2 , ⋯ , c N − 1 vanish, however, the sum in front of c N does not vanish. Therefore (12a) leads again to (9). Moreover, also the sum over all currents must vanish due to Kirchhoff’s current law I 1 + I 2 + ⋯ + I N = 0 .

∑ n = 1 N I n = 0 ⇒ ( ∑ n = 1 N N n , 1 ) c 1 + ( ∑ n = 1 N N n , 2 ) c 2 + ⋯ + ( ∑ n = 1 N N n , N − 1 ) c N − 1 + ( ∑ n = 1 N N n , N ) c N = 0 (12b)

Indeed, in the appendix we can show that the sums in front of all c n with n < N vanish.

From (12a) we see that we can skip the last equation in (11a) for n = N . With (9) it follows that M n , k is a square matrix with N − 1 columns and rows. With (12b) we can also skip one equation in (10a). Consequently, also N n , k is a square matrix with N − 1 columns and rows. Since contact N is the grounded supply contact we prefer to skip the last equation in (10a) for n = N . This leads to the matrix solution proposed in [

I = ( I 1 I 2 ⋮ I N − 1 ) = ( N 1 , 1 N 1 , 2 ⋯ N 1 , N − 1 N 2 , 1 N 2 , 2 ⋯ N 2 , N − 1 ⋮ ⋮ ⋱ ⋮ N N − 1 , 1 N N − 1 , 2 ⋯ N N − 1 , N − 1 ) ⋅ ( c 1 c 2 ⋮ c N − 1 ) = N ⋅ c (13a)

and

U = ( V 1 − V 2 V 2 − V 3 ⋮ V N − 1 − V N ) = ( M 1 , 1 M 1 , 2 ⋯ M 1 , N − 1 M 2 , 1 M 2 , 2 ⋯ M 2 , N − 1 ⋮ ⋮ ⋱ ⋮ M N − 1 , 1 M N − 1 , 2 ⋯ M N − 1 , N − 1 ) ⋅ ( c 1 c 2 ⋮ c N − 1 ) = M ⋅ c (13b)

with the ground node V N = 0 . Eliminating c in (13a) and (13b) gives

U = M ⋅ N − 1 ⋅ I (14)

Inserting (B7a) into (14) gives the contact potentials V = ( V 1 , V 2 , V 3 , ⋯ , V N − 1 ) T (the index T denotes the transpose of a vector or a matrix).

V = R ⋅ I (15a)

In (15a) the resistance matrix R is given by

R = Δ − 1 ⋅ M ⋅ N − 1 = ( ∑ l = 1 N − 1 M l , 1 ∑ l = 1 N − 1 M l , 2 ⋯ ∑ l = 1 N − 1 M l , N − 1 ∑ l = 2 N − 1 M l , 1 ∑ l = 2 N − 1 M l , 2 ⋯ ∑ l = 2 N − 1 M l , N − 1 ⋮ ⋮ ⋱ ⋮ M N − 1 , 1 M N − 1 , 2 ⋯ M N − 1 , N − 1 ) ⋅ ( N 1 , 1 N 1 , 2 ⋯ N 1 , N − 1 N 2 , 1 N 2 , 2 ⋯ N 2 , N − 1 ⋮ ⋮ ⋱ ⋮ N N − 1 , 1 N N − 1 , 2 ⋯ N N − 1 , N − 1 ) − 1 (15b)

In (15b) the matrix Δ is defined in Appendix B. The conductance matrix G is given by I = G ⋅ V . It follows from inversion of (15b). Note that (15b) is computationally not very efficient, because we need to compute two matrices, invert one, and multiply two matrices. Another method is to determine the points d n in (8a) by solving A n = B n on all folded contacts in the parallelogram. This was done in (41) and (42) in [

a ¯ n = b n and b ¯ n = a n + 1 for n = 1 , 2 , ⋯ , N (16)

whereby the quantities of the complementary device are denoted by an overbar. Note that in the complementary device all indices rotate by one instance to the right. If the device is mapped onto a circular Hall plate, the indices in the complementary circular device rotate one instance in positive mathematical sense (counterclockwise). If we walk along the boundary in the direction of increasing indices the conductive region is at the left hand side. Therefore, in

(16) states that in a complementary Hall plate a n and b n are swapped. Comparison with (1b) shows that L ( z ) remains unchanged if we additionally swap γ ↔ 1 − γ in the complementary device, which means that the applied magnetic field acting on the complementary Hall plate has opposite polarity. However, in the theory of Homentcovschi and Bercia we see that identical L ( z ) in the complementary Hall plate would give wrong boundary conditions, because the roles of insulating boundaries and contacts are swapped (see (6) in [

L ¯ ( z ) = exp ( − i π γ ¯ ) ∏ n = 1 N ( z − a ¯ n ) 1 − γ ¯ ( z − b ¯ n ) γ ¯ (17a)

where γ ¯ denotes the applied magnetic field on the complementary Hall plate

γ ¯ = 1 2 + θ H ¯ π with 0 < γ ¯ < 1 and tan θ H ¯ = μ H B ⊥ ¯ (17b)

Now the new product L ¯ ( z ) E ¯ ( z ) has to fulfil the homogeneous boundary conditions Re { L ¯ ( z ) E ¯ ( z ) } = 0 on the entire real axis. Thereby E ¯ ( z ) is the electric field in the complementary Hall plate. Following the same arguments as in [

E ¯ ( z ) = E ¯ x − i E ¯ y = i L ¯ ( z ) ∑ n = 1 N − 1 c ¯ n z n − 1 with c ¯ n ∈ R (18)

If the magnetic field is reversed on the complementary Hall plate B ⊥ ¯ = − B ⊥ it holds γ ¯ = 1 − γ and this gives

L ¯ ( z ) | − B ⊥ = − exp ( i π γ ) ∏ n = 1 N ( z − a n ) 1 − γ ( z − b n ) γ = L ( z ) | − B ⊥ × ( − exp ( i π γ ) ) (19a)

Comparison of (19a) and (1b) gives the important result

| L ¯ ( z ) | − B ⊥ | = | L ( z ) | B ⊥ | (19b)

Computing the voltages gives with (11a) and

U ¯ n ( − B ⊥ ) = V ¯ n ( − B ⊥ ) − V ¯ n + 1 ( − B ⊥ ) = − ∫ contact n contact n + 1 d ϕ ¯ = − ∫ contact n contact n + 1 ∇ ϕ ¯ ⋅ d x = ∫ contact n contact n + 1 E ¯ x d x = ∫ b ¯ n a ¯ n + 1 Re { E ¯ ( z ) } d x = ∫ a n + 1 b n + 1 Re { i ∑ k = 1 N − 1 c ¯ k x k − 1 ( − 1 ) n exp ( − i π γ ) | L ( x ) | } d x = R sheet cos ( θ H ) ∑ k = 1 N − 1 N n + 1 , k ( B ⊥ ) c ¯ k ≡ ∑ k = 1 N − 1 M ¯ n , k ( − B ⊥ ) c ¯ k (20)

whereby the rightmost equation in (20) defines U ¯ ( − B ⊥ ) = M ¯ ( − B ⊥ ) ⋅ c ¯ ( − B ⊥ ) analogous to U ( B ⊥ ) = M ( B ⊥ ) ⋅ c ( B ⊥ ) in (13b). This means

M ¯ ( − B ⊥ ) = R sheet cos ( θ H ) N ↑ ( B ⊥ ) = R sheet cos ( θ H ) 1 ↑ ⋅ N ( B ⊥ ) (21a)

M ¯ n , k ( − B ⊥ ) = R sheet cos ( θ H ) N n + 1 , k ( B ⊥ ) (21b)

In (21a) we used the shift-up operation defined in Appendix B.

N ↑ = ( N ) m + 1 , n = ( N 2 , 1 N 2 , 2 ⋯ N 2 , N − 1 N 3 , 1 N 3 , 2 ⋯ N 3 , N − 1 ⋮ ⋮ ⋱ ⋮ N N − 1 , 1 N N − 1 , 2 ⋯ N N − 1 , N − 1 − ∑ l = 1 N − 1 N l , 1 − ∑ l = 1 N − 1 N l , 2 ⋯ − ∑ l = 1 N − 1 N l , N − 1 ) = ( N 2 , 1 N 2 , 2 ⋯ N 2 , N − 1 N 3 , 1 N 3 , 2 ⋯ N 3 , N − 1 ⋮ ⋮ ⋱ ⋮ N N − 1 , 1 N N − 1 , 2 ⋯ N N − 1 , N − 1 N N , 1 N N , 2 ⋯ N N , N − 1 ) (22)

In N ↑ all rows of N are shifted up once and the last row in N ↑ is equal to the negative sum of all rows in N . For the right equation in (22) we used (A5b) in Appendix A. We can get N from (21a) with (B10a)

N ( B a , z ) = cos ( θ H ) R sheet M ¯ ↓ ( − B a , z ) = cos ( θ H ) R sheet 1 ↓ ⋅ M ¯ ( − B a , z ) (23a)

M ↓ = ( M ) m − 1 , n = ( − ∑ l = 1 N − 1 M l , 1 − ∑ l = 1 N − 1 M l , 2 ⋯ − ∑ l = 1 N − 1 M l , N − 1 M 1 , 1 M 1 , 2 ⋯ M 1 , N − 1 ⋮ ⋮ ⋱ ⋮ M N − 3 , 1 M N − 3 , 2 ⋯ M N − 3 , N − 1 M N − 2 , 1 M N − 2 , 2 ⋯ M N − 2 , N − 1 ) = ( M N , 1 M N , 2 ⋯ M N , N − 1 M 1 , 1 M 1 , 2 ⋯ M 1 , N − 1 ⋮ ⋮ ⋱ ⋮ M N − 3 , 1 M N − 3 , 2 ⋯ M N − 3 , N − 1 M N − 2 , 1 M N − 2 , 2 ⋯ M N − 2 , N − 1 ) (23b)

with the definition M 0. l = M N . l . For the right equation in (23b) we used (A5a) in Appendix A. The currents flowing into the complementary Hall plate through the contacts are

I ¯ n ( − B ⊥ ) = t H ∫ a ¯ n b ¯ n S ¯ y d x = t H ∫ b n a n + 1 ( − 1 ) Im { cos ( θ ¯ H ) ρ exp ( − i θ ¯ H ) E ¯ ( z ) } d x = ( − 1 ) N + n + 1 cos ( θ H ) R sheet ∫ b n a n + 1 Im { i exp ( i θ H ) } ∑ k = 1 N − 1 c ¯ k x k − 1 | L ( x ) | d x = − cos ( θ H ) R sheet ∑ k = 1 N − 1 M n , k ( B ⊥ ) c ¯ k ≡ ∑ k = 1 N − 1 N ¯ n , k ( − B ⊥ ) c ¯ k (24)

here we used (11a). The rightmost equation in (24) defines I ¯ ( − B ⊥ ) = N ¯ ( − B ⊥ ) ⋅ c ¯ ( − B ⊥ ) analogous to I ( B ⊥ ) = N ( B ⊥ ) ⋅ c ( B ⊥ ) in (13a). This means

N ¯ n , k ( − B ⊥ ) = − cos ( θ H ) R s h e e t M n , k ( B ⊥ ) ⇔ N ¯ ( − B ⊥ ) = − cos ( θ H ) R s h e e t M ( B ⊥ ) (25)

for n = 1 , 2 , ⋯ , N − 1 (we do not need the case n = N ). (21) and (25) state that all elements of the resistance and conductance matrix of the complementary Hall plate at reverse magnetic field can be derived from the matrices M and N of the original Hall plate at non-reverse magnetic field. We call (21) and (25) the reverse magnetic field on complementary device theorem (RMFoCD).

With the RMFoCD theorem we can predict the behavior of complementary Hall plates from the original ones.

I ( B ⊥ ) = N ( B ⊥ ) ⋅ M − 1 ( B ⊥ ) ⋅ U ( B ⊥ ) (26a)

Multiplying (26a) with 1 ↑ from left gives

1 ↑ ⋅ I ( B ⊥ ) = 1 ↑ ⋅ N ( B ⊥ ) ⋅ M − 1 ( B ⊥ ) ⋅ U ( B ⊥ ) (26b)

On the other hand, we can write for the complementary Hall plate at reverse magnetic field with (14)

U ¯ ( − B ⊥ ) = M ¯ ( − B ⊥ ) ⋅ N ¯ − 1 ( − B ⊥ ) ⋅ I ¯ ( − B ⊥ ) (27a)

With (21a) and (25) this is equivalent to

− cos ( θ H ) R sheet U ¯ ( − B ⊥ ) = 1 ↑ ⋅ N ( B ⊥ ) ⋅ M − 1 ( B ⊥ ) ⋅ R sheet cos ( θ H ) I ¯ ( − B ⊥ ) (27b)

(26b) and (27b) are sets of linear equations with identical coefficient matrix 1 ↑ ⋅ N ( B ⊥ ) ⋅ M − 1 ( B ⊥ ) . Thus, the vectors on the right hand sides must be equal, if the vectors on the left hand sides are equal—and vice versa. This means

U ¯ ( − B ⊥ ) = − R s q 1 ↑ ⋅ I ( B ⊥ ) ⇔ I ¯ ( − B ⊥ ) = 1 R s q U ( B ⊥ ) (28)

whereby we introduced the square resistance at magnetic field

R s q = R sheet 1 + μ H 2 B ⊥ 2 = R sheet / cos ( θ H ) (29)

The square resistance is the resistance of a square sample with contacts along two opposite edges (see [

Read (28): “if the left equation holds then the right equation follows and vice versa”. (28) means that at reverse magnetic field the complementary Hall plate has voltages between neighboring contacts, which are − R s q times the currents in the contacts of the original Hall plate, if the complementary Hall plate is supplied with currents, which are 1 / R s q times the voltages between neighboring contacts in the original Hall plate. This is shown in

We can express (28) also in terms of the resistance matrix of the complementary Hall plate at reverse magnetic field. It is defined as

V ¯ ( − B ⊥ ) = R ¯ ( − B ⊥ ) ⋅ I ¯ ( − B ⊥ ) (30a)

Now we express the left hand side in terms of parameters of the original Hall plate. With (B7c) and the left side of (28) we get

V ¯ ( − B ⊥ ) = Δ − 1 ⋅ U ¯ ( − B ⊥ ) = − R sheet cos ( θ H ) Δ − 1 ⋅ 1 ↑ ⋅ I ( B ⊥ ) (30b)

We also express the current on the right hand side of (30a) in terms of parameters of the original Hall plate. With the right side of (28) and (B7a) we get

I ¯ ( − B ⊥ ) = cos ( θ H ) R sheet U ( B ⊥ ) = cos ( θ H ) R sheet Δ ⋅ V ( B ⊥ ) (31a)

Inserting (30b), (31a) into (30a) and rearranging gives

I ( B ⊥ ) = − cos ( θ H ) 2 R sheet 2 ( 1 ↑ ) − 1 ⋅ Δ ⋅ R ¯ ( − B ⊥ ) ⋅ Δ ⋅ V ( B ⊥ ) (31b)

However, by definition (31b) can also be written as I ( B ⊥ ) = G ( B ⊥ ) ⋅ V ( B ⊥ ) . Comparison with (31b) gives the final result

R ¯ ( − B ⊥ ) = − R sheet 2 cos ( θ H ) 2 Δ − 1 ⋅ 1 ↑ ⋅ G ( B ⊥ ) ⋅ Δ − 1 (32a)

R ¯ m , l ( − B ⊥ ) = − R sheet 2 cos ( θ H ) 2 ∑ k = 1 N − 1 ∑ n = 1 N − 1 ( Δ − 1 ) m , n G n + 1 , k ( B ⊥ ) ( Δ − 1 ) k , l = − R sheet 2 cos ( θ H ) 2 ∑ k = 1 N − 1 ∑ n = m N − 1 G n + 1 , k ( B ⊥ ) ( Δ − 1 ) k , l = − R sheet 2 cos ( θ H ) 2 ∑ k = 1 l ∑ n = m N − 1 G n + 1 , k ( B ⊥ ) = R sheet 2 cos ( θ H ) 2 ∑ n = 1 m ∑ k = 1 l G n , k ( B ⊥ ) (32b)

In the last transformation we used the identity ∑ n = 1 m G n , k + ∑ n = m + 1 N G n , k = 0 . The sum over all N elements per row or column in the indefinite conductance matrix vanishes [

R ¯ ( − B ⊥ ) = R sheet 2 cos ( θ H ) 2 × ( G 1 , 1 G 1 , 1 + G 1 , 2 G 1 , 1 + G 1 , 2 + G 1 , 3 G 1 , 1 + G 2 , 1 G 1 , 1 + G 1 , 2 + G 2 , 1 + G 2 , 2 G 1 , 1 + G 1 , 2 + G 1 , 3 + G 2 , 1 + G 2 , 2 + G 2 , 3 G 1 , 1 + G 2 , 1 + G 3 , 1 G 1 , 1 + G 1 , 2 + G 2 , 1 + G 2 , 2 + G 3 , 1 + G 3 , 2 G 1 , 1 + G 1 , 2 + G 1 , 3 + G 2 , 1 + G 2 , 2 + G 2 , 3 + G 3 , 1 + G 3 , 2 + G 3 , 3 ) (32c)

whereby all G m , n of the original Hall plate are at non-reversed magnetic field G m , n = G m , n ( B ⊥ ) . Matrix inversion of both sides of (32a) gives with (B10a)

G ¯ ( − B ⊥ ) = − cos ( θ H ) 2 R sheet 2 Δ ⋅ R ( B ⊥ ) ⋅ 1 ↓ ⋅ Δ (33)

On the other hand we can multiply both sides of (32a) first with Δ and then with ( 1 ↑ ) − 1 from left and with Δ from right to get

G ( B ⊥ ) = − cos ( θ H ) 2 R sheet 2 1 ↓ ⋅ Δ ⋅ R ¯ ( − B ⊥ ) ⋅ Δ (34)

Matrix inversion of both sides of (34) gives with (B10a)

R ( B ⊥ ) = − R sheet 2 cos ( θ H ) 2 Δ − 1 ⋅ G ¯ ( − B ⊥ ) ⋅ Δ − 1 ⋅ 1 ↑ (35a)

With (B10c) and (B7d) we can write this

R m , l ( B ⊥ ) = R sheet 2 cos ( θ H ) 2 ∑ k = 1 N − 1 ∑ n = 1 N − 1 ( Δ − 1 ) m , n G ¯ n , k ( − B ⊥ ) ( Δ − 1 ) l , k = R sheet 2 cos ( θ H ) 2 ∑ n = m N − 1 ∑ k = l N − 1 G ¯ n , k ( − B ⊥ ) (35b)

In the introduction we have already stated that there is a relation between impedances of original and complementary Hall plates at vanishing magnetic field [

V supply = R m , m ( B ⊥ ) I supply (36)

In the complementary Hall plate we short all contacts left of a line through m-th and N-th contacts of the original device, and we ground all other contacts. Then we inject current at reverse magnetic field. It holds

( I ¯ 1 ( − B ⊥ ) ⋮ I ¯ m − 1 ( − B ⊥ ) I ¯ m ( − B ⊥ ) ⋮ I ¯ N − 1 ( − B ⊥ ) ) = G ¯ ( − B ⊥ ) ⋅ ( 0 ⋮ 0 V ¯ supply ( − B ⊥ ) ⋮ V ¯ supply ( − B ⊥ ) ) (37)

From which we get

I ¯ supply ( − B ⊥ ) = ∑ n = m N − 1 I ¯ n ( − B ⊥ ) = ∑ n = m N − 1 ∑ k = m N − 1 G ¯ n , k ( − B ⊥ ) V ¯ supply ( − B ⊥ ) (38)

Inserting (35b) into (38) gives

I ¯ supply ( − B ⊥ ) = cos ( θ H ) 2 R sheet 2 R m , m ( B ⊥ ) V ¯ supply ( − B ⊥ ) (39)

Comparison of (39) with (36) gives the result

I ¯ supply ( − B ⊥ ) V ¯ supply ( − B ⊥ ) = cos ( θ H ) 2 R sheet 2 V supply ( B ⊥ ) I supply ( B ⊥ ) (40a)

which we can re-write with (29)

( 1 R s q V supply ( B ⊥ ) I supply ( B ⊥ ) ) × ( 1 R s q V ¯ supply ( − B ⊥ ) I ¯ supply ( − B ⊥ ) ) = 1 (40b)

It means: The product of the numbers of squares of both resistances is equal to 1. Thereby the number of squares of a resistance is defined as the ratio of this resistance over the square resistance—not over the sheet resistance (see (29)). (40b) also means that the product of both measured resistances is equal to the square of the square resistance:

V supply ( B ⊥ ) I supply ( B ⊥ ) × V ¯ supply ( − B ⊥ ) I ¯ supply ( − B ⊥ ) = R s q 2 = R sheet 2 ( 1 + μ H 2 B ⊥ 2 ) (40c)

This opens up a way to measure the sheet resistance and the Hall mobility of an unknown material. Note that we are free to choose contact m in

It seems possible to extend this scheme for R m , l ( B ⊥ ) with m ≠ l in (36), but we do not pursue this further.

Let us consider general Hall plates with three peripheral contacts. Current is injected into contact 1, voltage is tapped between contact 2 and 3, whereby contact 3 is the current sink and ground node at zero volts (see

If such Hall plates have a single mirror symmetry, and if we supply the original and the complementary Hall plate with the same supply voltage, their output voltages are identical. This was proven for weak magnetic field in [

( I supply 0 ) = ( G 11 G 12 G 21 G 22 ) ⋅ ( V supply V 2 ) (41a)

which gives the output voltage

V 2 = − G 21 G 22 V supply (41b)

At zero magnetic field the voltage V 2 does not vanish—this is called the offset voltage. Therefore V 2 is of little use in practical sensor applications. However, if one subtracts the voltage V 2 at positive and negative magnetic field, this procedure cancels out the offset voltage and renders the so-called Hall voltage V 2 H .

V 2 H = V 2 ( B ⊥ ) − V 2 ( − B ⊥ ) 2 (42)

In other words, the voltage V 2 may be decomposed into even and odd functions versus B ⊥ and the odd one is the Hall voltage (see also [

G m , n ( − B ⊥ ) = G n , m ( B ⊥ ) for 1 ≤ m , n ≤ N − 1 (43)

Using (43) in (42) we get

V 2 H = G 12 − G 21 2 G 22 V supply (44)

where the G m , n refer to non-reversed magnetic fields G m , n = G m , n ( B ⊥ ) . For the complementary Hall plate in

( V supply V ¯ 2 ) = ( R ¯ 11 ( − B ⊥ ) R ¯ 12 ( − B ⊥ ) R ¯ 21 ( − B ⊥ ) R ¯ 22 ( − B ⊥ ) ) ⋅ ( I ¯ supply 0 ) (45a)

where the quantities with overbar are at reverse magnetic field. It follows

V ¯ 2 ( − B ⊥ ) = R ¯ 21 ( − B ⊥ ) R ¯ 11 ( − B ⊥ ) V supply = G 11 + G 21 G 11 V supply (45b)

For the right equation in (45b) we used (32c). The Hall voltage for the complementary Hall plate is given by

V ¯ 2 H ( − B ⊥ ) = G 21 − G 12 2 G 11 V supply (45c)

where V ¯ 2 H is the Hall voltage of the complementary device at reverse magnetic field, yet G 11 , G 12 , G 21 are the conductances of the original device at non-reverse magnetic field. Thus, for asymmetric Hall plates neither the voltages nor the Hall voltages are identical for the original and for the complementary devices: V 2 ≠ V ¯ 2 and V 2 H ≠ V ¯ 2 H . However, if the original Hall plate has a single mirror symmetry with the symmetry line through the grounded reference contact 3, it holds G 1 , 1 = G 2 , 2 . Then it follows

V 2 ( B ⊥ ) + V ¯ 2 ( − B ⊥ ) = V supply and V ¯ 2 H ( − B ⊥ ) = − V 2 H ( B ⊥ ) for G 1 , 1 = G 2 , 2 (46)

Note that the complementary device also has a single mirror symmetry, but this symmetry line does not go through the grounded reference contact 3 ¯ . Swapping the roles of original and complementary Hall plate we conclude that (46) holds also if the symmetry line does not go through the grounded reference contact. Thus, our proof is complete: Hall plates with three contacts and single mirror symmetry have identical Hall output voltages as their complementary Hall plates when they are supplied by voltage sources—and this holds for all Hall angles.

Let us consider Hall plates with four peripheral contacts and two perpendicular mirror symmetries. Moreover, the contacts should be such that the resistance between two non-neighboring contacts equals the resistance between the other two non-neighboring contacts. All devices with 90˚ symmetry belong to this group of Hall plates. However, also rectangular shapes with properly chosen contact sizes have such properties. Some popular shapes are shown in

If we supply the original and the complementary Hall plate with the same supply voltage, their output voltages are identical. This was proven for weak magnetic field in [

The conductance matrix of a Hall plate with 90˚ symmetry has the following symmetry.

G = ( G 11 G 12 − G 11 − G 12 − G 21 G 21 G 11 G 12 − G 11 − G 12 − G 21 G 21 G 11 ) (47)

This can be proven by applying general currents and voltages to the Hall plate and rotating them by 90˚ and 180˚. Due to the symmetry of the device the same currents will give the same voltages. This gives 9 equations, which can be solved for the 9 elements of G . The solution has the structure of (47). For the original Hall plate in

( 0 I supply 0 ) = G ⋅ ( V 1 V supply V 3 ) (48a)

which gives the output voltage

V 1 − V 3 = − G 12 + G 21 2 G 11 + G 12 + G 21 V supply (48b)

For the complementary Hall plate in

( V ¯ 1 V supply V ¯ 3 ) = R ¯ ( − B ⊥ ) ⋅ ( 0 I ¯ supply 0 ) (49a)

where the quantities with overbar are at reverse magnetic field. With (32c) the resistance matrix of the complementary device at reverse magnetic field is

R ¯ ( − B ⊥ ) = R sheet 2 cos ( θ H ) 2 ( G 1 , 1 G 1 , 1 + G 1 , 2 − G 2 , 1 G 1 , 1 + G 2 , 1 2 G 1 , 1 + G 1 , 2 + G 2 , 1 G 1 , 1 + G 1 , 2 − G 1 , 2 G 1 , 1 + G 2 , 1 G 1 , 1 ) (49b)

which gives the output voltage

V ¯ 1 − V ¯ 3 = G 12 − G 21 2 G 11 + G 12 + G 21 V supply (49c)

The output voltage of the complementary Hall plate at reverse magnetic field in (49c) has equal magnitude and opposite sign as the output voltage of the original Hall plate at non-reverse magnetic field in (48b), i.e., both devices have identical output voltages at identical magnetic field. In general the supply currents are different

I supply = 2 G 11 2 + G 12 2 + G 21 2 + 2 G 11 ( G 12 + G 21 ) 2 G 11 + G 12 + G 21 V supply (50a)

I ¯ supply = 1 2 G 11 + G 12 + G 21 cos ( θ H ) 2 R sheet 2 V supply (50b)

If the Hall plates have no symmetry the complementary Hall plate will have different Hall output voltage than the original device when both are supplied by the same voltage source.

Finally we compute the cross-resistances R x , R ¯ x of original and complementary Hall plates. We define

R x = V 1 I 1 + I 3 for V 1 = V 3 and V 2 = V 4 = 0 (51)

From I = G ⋅ V it follows with the 90˚ symmetry constraints (47)

R x = − 1 2 ( G 1 , 2 + G 2 , 1 ) (52)

From I ¯ ( − B ⊥ ) = G ¯ ( − B ⊥ ) ⋅ V ¯ ( − B ⊥ ) it follows with G ¯ ( − B ⊥ ) = R ¯ − 1 ( − B ⊥ ) and (32c) and (47)

R ¯ x = − 1 2 ( G 1 , 2 + G 2 , 1 ) R sheet 2 cos ( θ H ) 2 (53)

Note that (52) and (53) are independent of the polarity of the magnetic field, because G 1 , 2 ( B ⊥ ) + G 2 , 1 ( B ⊥ ) = G 1 , 2 ( B ⊥ ) + G 1 , 2 ( − B ⊥ ) according to the RMFR principle [

2 R x R ¯ x = R s q = R sheet 1 + μ H 2 B ⊥ 2 (54)

(54) is more general than (13a) in [

We gave a simple derivation of the electric field in the infinite upper half plane with N contacts on the real axis. The original formula has N real coefficients, but we could show that the N-th one vanishes. Based on these results the resistance and conductance matrices of an N-contact Hall plate can be expressed in terms of two system matrices M , N with N-1 rows and columns. Then we applied this theory to complementary Hall plates where all contacts and insulating boundaries are swapped. It turned out that the resistance matrix of the complementary Hall plate at reverse magnetic field can be expressed by the conductance matrix of the original Hall plate at non-reverse magnetic field. This Reverse Magnetic Field on Complementary Device theorem (RMFoCD) was applied to various circuits of Hall plates to prove their symmetry properties. Hall plates with four contacts and 90˚ symmetry have identical Hall output voltage as their complementary counterparts, if they are supplied by the same voltage. Hall plates with three contacts and a singly mirror symmetry also have identical Hall output voltage as their complementary counterparts, if they are supplied by the same voltage. For several resistances between contacts or group of contacts in the original and complementary devices, it was shown that their product relates to the square of the square resistance. These findings can be used to determine the resistivity and the magnitude of the Hall mobility of Hall plates with extended contacts as functions of the applied magnetic field. This is an alternative method to van-der-Pauw’s method. With due diligence, one might find more relations between Hall plates and their complementary Hall plates with less or no symmetry at all.

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

Ausserlechner, U. (2019) Relations between Hall Plates with Complementary Contact Geometries. Journal of Applied Mathematics and Physics, 7, 2836-2867. https://doi.org/10.4236/jamp.2019.711195

From (1b) we get TableA1. With (10b-d) and (11b) this gives

∫ x = − ∞ + ∞ x k − 1 L ( x ) d x = − R sheet cos ( θ H ) 2 ∑ l = 1 N N l , k + ( tan ( θ H ) − i ) ∑ l = 1 N M l , k (A1)

For the integral in (A1) we use the calculus of residues [

lim ε → 0 ∫ s . c . n z k − 1 L ( z ) d z = lim ε → 0 ∫ ϑ = π 0 ε k − 1 exp ( i ( k − 1 ) ϑ ) ε exp ( i ϑ ) i d ϑ ε 1 − γ exp ( i ( 1 − γ ) ϑ ) ( ∏ m = 1 m ≠ n N ( a n + ε exp ( i ϑ ) − a m ) 1 − γ ) ( ∏ m = 1 N ( a n + ε exp ( i ϑ ) − b m ) γ ) = lim ε → 0 ε k − 1 + γ = 0 for k = 1 , 2 , ⋯ , N ; γ > 0 (A2)

where we used z = a n + ε exp ( i ϑ ) . For the integral along the large semi-circle (denoted by l.s.c.) we get with z = R exp ( i ϑ ) and for 1 ≤ k ≤ N

lim R → ∞ ∫ l . s . c . z k − 1 L ( z ) d z = lim R → ∞ ∫ ϑ = 0 π R k − 1 exp ( i ( k − 1 ) ϑ ) R exp ( i ϑ ) i d ϑ L ( R exp ( i ϑ ) ) = i ∫ ϑ = 0 π lim R → ∞ R k − N exp ( i ( k − N ) ϑ ) d ϑ = i π δ k , N (A3)

The integral along the closed contour is proportional to the sum of all enclosed residues. However, this sum is zero, because we cut out all singular points. Therefore we get with (A1), (A2), and (A3)

0 = ∫ x = − ∞ + ∞ x k − 1 L ( x ) d x + lim R → ∞ ∫ l . s . c . z k − 1 L ( z ) d z + ∑ n = 1 N lim ε → 0 ∫ s . c . n z k − 1 L ( z ) d z = − R sheet cos ( θ H ) 2 ∑ l = 1 N N l , k + ( tan ( θ H ) − i ) ∑ l = 1 N M l , k + i π δ k , N (A4)

The imaginary part of both sides in (A4) gives

∑ l = 1 N M l , k = π δ k , N (A5a)

with Kronecker’s delta δ k , N . With (A5a) the real part of both sides in (A4) gives

∑ l = 1 N N l , k = π sin ( 2 θ H ) 2 R sheet δ k , N (A5b)

Inserting (A5a) into (12a) and (A5b) into (12b) gives (9).

Location of test point x | L ( x ) / | L ( x ) | at B ⊥ | L ¯ ( x ) / | L ¯ ( x ) | at − B ⊥ |
---|---|---|

x < b 1 | ( − 1 ) N | ( − 1 ) N + 1 exp ( i π γ ) |

b 1 < x < a 2 | ( − 1 ) N exp ( − i π γ ) | ( − 1 ) N + 1 |

a 2 < x < b 2 | ( − 1 ) N − 1 | ( − 1 ) N exp ( i π γ ) |

b 2 < x < a 3 | ( − 1 ) N − 1 exp ( − i π γ ) | ( − 1 ) N |

⋮ | ⋮ | ⋮ |

a N < x < b N | −1 | exp ( i π γ ) |

b N < x < a N + 1 = a 1 | − exp ( − i π γ ) | 1 |

a 1 < x | 1 | − exp ( i π γ ) |

Here we summarize the mathematics to describe Hall plates with numerous contacts via impedance and conductance matrices. Let us consider a Hall plate with N contacts on the perimeter (see FigureB1). Due to Kirchhoff’s nodal current law it holds

∑ l = 1 N I l = 0 (B1a)

which gives

I N = − ∑ l = 1 N − 1 I l (B1b)

It is common practice to define the potential on one contact as reference potential and to set it equal to zero volts (=ground). We use the potential on the N-th contact as reference potential:

V N = 0 (B2)

The voltages between neighboring contacts are defined as

U l = V l − V l + 1 for 1 ≤ l ≤ N − 1 . (B3a)

According to FigureB1 the N-th voltage is

U N = V N − V 1 = − ∑ l = 1 N − 1 U l (B3b)

which gives

∑ l = 1 N U l = 0 (B3c)

According to FigureB1 the loop currents J l are linked to the currents I l via

I l = J l − J l − 1 for 2 ≤ l ≤ N (B4a)

with the first current as

I 1 = J 1 − J N (B4b)

If we add the same current to all loop currents this will not change the currents I l . Therefore, we are free to choose one loop current. We define

J N = 0 (B5a)

With (B1a) it follows

∑ l = 1 N J l = 0 (B5b)

Thus, the loop current J m flows from terminal m through the Hall plate towards terminal m + 1, and the voltage U m drops between the same terminals. This definition is in accordance with [

I = ( I 1 , I 2 , ⋯ , I N − 1 ) T (B6a)

V = ( V 1 , V 2 , ⋯ , V N − 1 ) T (B6b)

U = ( U 1 , U 2 , ⋯ , U N − 1 ) T (B6c)

J = ( J 1 , J 2 , ⋯ , J N − 1 ) T (B6d)

whereby the index T denotes the transpose. Then, we can write (B3a) as a matrix formula

U = Δ ⋅ V with Δ m , n = δ m , n − δ m + 1 , n , 1 ≤ m , n ≤ N − 1 (B7a)

Δ = ( 1 − 1 0 ⋯ 0 0 0 1 − 1 ⋯ 0 0 0 0 1 ⋯ 0 0 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 − 1 0 0 0 ⋯ 0 1 ) (B7b)

Δ has N − 1 rows and columns. In (B7a) we used the Kronecker delta δ m , n ,which equals 1 if both indices are identical—otherwise it vanishes. The inverse of (B7a) is given by

V = Δ − 1 ⋅ U with ( Δ − 1 ) m , n = { 1 for n ≥ m 0 for n < m (B7c)

Δ − 1 = ( 1 1 1 ⋯ 1 1 0 1 1 ⋯ 1 1 0 0 1 ⋯ 1 1 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 1 0 0 0 ⋯ 0 1 ) (B7d)

We define shift-up and shift-down manipulations of vectors:

I ↑ = ( I 2 I 3 ⋮ I N − 1 − ∑ l = 1 N − 1 I l ) and I ↓ = ( − ∑ l = 1 N − 1 I l I 1 ⋮ I N − 3 I N − 2 ) (B8)

With these manipulations we shift all elements by one place, kick top or bottom elements out, and receive a new element, which is equal to the negative sum of all former elements. Due to (B1b) the new element is equal to I N ,and the manipulations (B8) do not destroy information. We can apply shift-up and shift-down to arbitrary vectors. In particular, we can use U instead of I . We can shift-up and shift-down also J and V but there the new element − V 1 − V 2 − ⋯ − V N − 1 is not equal to V N (the same applies to J ). This is a consequence of our definitions (B2, B5a), which differ from (B1b, B3b). We could also redefine the reference potential V N analogous (B1b) but this would give more complicated matrices in (B7b) and (B7d). We can describe the shift-up procedure by a matrix multiplication:

I ↑ = ( I 2 I 3 ⋮ I N − 1 − ∑ l = 1 N − 1 I l ) = 1 ↑ ⋅ I = ( 0 1 0 ⋯ 0 0 0 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 1 − 1 − 1 − 1 ⋯ − 1 ) ⋅ ( I 1 I 2 ⋮ I N − 2 I N − 1 ) (B9a)

where 1 is the ( N − 1 ) × ( N − 1 ) identity matrix, and 1 ↑ is obtained from 1 by shifting all rows up once and setting the elements of the bottom row equal to the negative sum of all elements per column in the original matrix 1 . In other words, we apply the shift-up procedure not only to vectors but also to matrices. The shift-down manipulation is also equivalent to a matrix multiplication:

I ↓ = ( − ∑ l = 1 N − 1 I l I 1 ⋮ I N − 3 I N − 2 ) = 1 ↓ ⋅ I = ( − 1 − 1 − 1 ⋯ − 1 − 1 1 0 0 ⋯ 0 0 0 1 0 ⋯ 0 0 0 0 1 ⋯ 0 0 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 0 ) ⋅ ( I 1 I 2 ⋮ I N − 2 I N − 1 ) (B9b)

It holds

( 1 ↓ ) − 1 = 1 ↑ (B10a)

Therefore, shift-up and shift-down annihilate each other

1 ↑ ⋅ I ↓ = 1 ↑ ⋅ 1 ↓ ⋅ I = 1 ⋅ I = I (B10b)

However, we are not allowed to reverse the order: ( 1 ↓ ) − 1 ≠ ( 1 − 1 ) ↓ . Moreover, it holds

Δ − 1 ⋅ 1 ↑ = − ( Δ − 1 ) T (B10c)

We can also write (B4a), (B4b) with (B5a) as a matrix relation

I = Δ T ⋅ J (B11)

Throughout this paper we use the following definitions of resistance matrix R ,impedance matrix Z ,and conductance matrix G = Y :

V = R ⋅ I (B12a)

U = Z ⋅ J (B12b)

I = G ⋅ V = Y ⋅ V (B12c)

All these matrices have N-1 rows and columns, whereby N is the number of contacts of the Hall plate. In [

R − 1 = G = Y (B13a)

but Z − 1 ≠ Y because Z ≠ R . The correct relation between Z − 1 and Y is obtained by inserting (B7a) and (B11) into (B12b):

Z = Δ ⋅ R ⋅ Δ T (B14a)

Z − 1 = ( Δ T ) − 1 ⋅ Y ⋅ Δ − 1 (B14b)

For the complementary Hall plate at reverse magnetic field it follows

Z ¯ ( − B ⊥ ) = Δ ⋅ R ¯ ( − B ⊥ ) ⋅ Δ T = R sheet 2 cos ( θ H ) 2 1 ↑ ⋅ G ( B ⊥ ) ⋅ ( 1 ↑ ) T (B15a)

where we used (32a), (B10c), and (B14a). This can be re-written as

Z ¯ m , n ( − B ⊥ ) = R sheet 2 cos ( θ H ) 2 Y m + 1 , n + 1 ( B ⊥ ) (B15b)

with Y m , n = G m , n . At zero magnetic field (B15b) differs slightly from (21) in [

Here is another argument which proves that Z ¯ m , l ( − B ⊥ ) is not proportional to Y m + 1 , l ( B ⊥ ) as claimed for B ⊥ = 0 in (21) of [

In Appendix C we check our formulae against finite element simulations to guarantee their correctness. Moreover, we emphasize that all our equations work only if the following rules are applied:

1) The contacts are labeled in ascending order 1 to N.

2) If we walk along the perimeter of the Hall plate in the direction of rising contact labels, the conductive region lies at the left hand side.

3) The N-th contact is grounded and the N-th loop current vanishes.

4) The complementary Hall plate is obtained from the original one, by exchanging all contacts with insulating boundaries and vice versa, whereby the contacts of the complementary device are shifted into the direction of rising contact labels against the respective contacts of the original device (compare

Here we give numerical examples, where we compare the results of our theory with finite element simulations (FEM). The FEM was conducted with the 2D conduction model emdc of COMSOL MULTIPHYSICS with the following conductivity matrix:

( E x E y ) = ρ ( S x S y ) with ρ = ρ ( 1 μ H B ⊥ − μ H B ⊥ 1 ) (C1)

We use realistic values for silicon at room temperature with a phosphorus doping of 2 × 10^{16}/cm^{3}: μ H = 0.11 / T and ρ = 0.002924 Ω ⋅ m . The thickness of the Hall plates was assumed to be t H = 0.9 μ m . This gives a sheet resistance R sheet = 3248.89 Ω at zero magnetic field. We assume that ρ and μ H are constant with B ⊥ . We consider semi-infinite Hall plates according to

The first example has the following parameters: N = 4, a 1 = 10 2 μ m , a 2 = − 10 π μ m , a 3 = − 10 μ m , a 4 = 10 / 3 μ m , b 1 = − 50 μ m , b 2 = − 20 μ m , b 3 = 0 μ m , b 4 = 10 / 11 μ m . The Hall angle was θ H = π / 6 = 30 ∘ ,which means a strong magnetic flux density B ⊥ = 5.24864 T . With (10b), (10c), and (11b) we get the following system matrices

N = ( − 2.068966952 × 10 10 14226.81805 − 29.30524564 4.920031176 × 10 10 − 1.129662241 × 10 6 26.53209581 − 1.376581519 × 10 11 389605.9384 − 2.518997254 ) (C2a)

M = ( − 5.923279984 × 10 13 2.370993550 × 10 9 − 97557.41642 2.715479249 × 10 14 − 4.330425950 × 10 9 72482.27809 − 4.514465398 × 10 14 − 5.196538114 × 10 8 − 1171.994832 ) (C2b)

(A5a) and (A5b) also hold for k = 1, 2, 3, and 4. If we inject the currents (in units of amps)

I = ( 0.001 0 − 0.001 / 7 ) T (C3a)

i.e., I 1 = 1 mA , I 2 = 0 mA ,and I 3 = − 1 / 7 mA ,we get from (13a)

c = ( − 6.868102436 × 10 − 16 − 8.294336001 × 10 − 10 − 3.404135684 × 10 − 5 ) T (C3b)

With (C3b) we may compute the electric field inside the Hall region (TableC1). From (13b) and (B7c) we get (in units of volts)

U = ( 1.395086802 0.9379037967 0.7809727339 ) T (C4a)

V = V theory = ( 3.113963333 1.718876531 0.7809727339 ) T (C4b)

With (15b) we get the resistance matrix (in units of Ohms)

R = ( 3443.026651 3032.326444 2303.443227 2131.222364 5315.718944 2886.420836 1310.983587 1756.075838 3710.075970 ) (C5)

In the FEM model the upper half of the z-plane was modeled as a semi-disk with 40 cm diameter. The FEM simulation used Lagrange multipliers. The mesh had 10.14 million elements, the equation system had 20.29 million degrees of freedom. The solution time was 2 hours 30 minutes on a desktop computer with Intel Core i7-4930K CPU, ASRock X79 Extreme 6 main-board, 64 GB RAM, and 3.4 GHz clock cycle. The solution of the potentials was (in units of volts)

V FEM = ( 3.11298544 1.718219614 0.780452059 ) T (C6a)

and the difference between (C6a) and (C4b) is

V theory − V FEM = ( 0.000977893 0.000656917 0.000520675 ) T (C6b)

The main reason for the difference between the FEM result and the analytical theory seems to come from the finite radius of the semi-disk (FigureC1).

Next we check

I ¯ = cos ( θ H ) U / R sheet = ( 371.789 × 10 − 6 249.972 × 10 − 6 208.037 × 10 − 6 ) T (C7a)

into the contacts of the complementary Hall plate, and the sign of the magnetic field was reversed θ ¯ H = − π / 6 = − 30 ∘ . With (24) we get the coefficients

Test points (x, y): | (−55, 1) | (−55, 10) | (−40, 3) | (3, 0) | (5, 7) | (13, 0) | (16, 13) |
---|---|---|---|---|---|---|---|

Original Hall Plate at Positive Magnetic Field in | |||||||

FEM: E_{x} | 7507.1 | 23185.4 | 41859.8 | 321585.8 | −51865.8 | −297705.8 | −50851.7 |

FEM: E_{y} | 48065.3 | 15976.0 | −20193.3 | −185668.8 | −101472.4 | 171882.0 | 1135.6 |

THEORY: E_{x} | 7515.7 | 23193.6 | 41851.7 | 321543.9 | −51875.3 | −297592.8 | −50856.6 |

THEORY: E_{y} | 48094.4 | 15974.8 | −20191.5 | −185643.4 | −101481.8 | 171815.3 | 1136.3 |

| E THEORY − E FEM | | E THEORY | | 0.062% | 0.029% | 0.018% | 0.013% | 0.012% | 0.038% | 0.010% |

Complementary Hall Plate at Negative Magnetic Field in | |||||||

Ex (FEM) | −45404.8 | −25425.5 | −3437.9 | 0.0 | 113789.4 | 0.0 | 24436.3 |

Ey (FEM) | −17535.4 | 12097.9 | 46326.1 | 371068.5 | 5803.1 | −343419.0 | −44599.2 |

Ex (THEORY) | −45408.8 | −25431.4 | −3439.5 | 0.0 | 113823.5 | 0.0 | 24,444.3 |

Ey(THEORY) | −17538.4 | 12098.9 | 46340.4 | 371286.9 | 5815.6 | −343630.6 | −44,611.3 |

| E THEORY − E FEM | | E THEORY | | 0.010% | 0.021% | 0.031% | 0.059% | 0.032% | 0.062% | 0.028% |

c ¯ = ( 6.868102436 × 10 − 16 8.294336001 × 10 − 10 3.404135684 × 10 − 5 ) T = − c (C7b)

Inserting (C7b) into (18) gives the electric field in the complementary Hall plate (see TableC1). According to

U ¯ theory = − ( R sheet / cos ( θ H ) ) 1 ↑ ⋅ 1 = ( 0 0.53592768 3.21556607 ) T (C8a)

The FEM simulation returned the potentials

V ¯ FEM = ( 3.75095275 3.750968675 3.214946154 ) T (C8b)

from which we get (in volts)

U ¯ FEM = ( − 0.000015925 0.536022521 3.214946154 ) T (C8c)

The discrepancy is only (in volts)

U ¯ theory − U ¯ FEM = ( 0.000015925 − 0.000094843 0.000619917 ) T (C8d)