Generation of Equivalent Circuit fromFinite Element Model Using Model Order Reduction
Yuki Sato and Hajime Igarashi, Hokkaido University, Sapporo, Japan
Generation of Equivalent Circuit

Introduction
Finite element method (FEM) has been widely used to develop and design
electromagnetic devices.

Reduced admittance functionq

We want to develop and design the electromagnetic devices
considering the control and driving circuits connected to FE model.

Y (s0 + σ ) = k ∞ +FEM

Analytical solution
for simple geometry
Admittance function:

Vectorfitting[1]1jFoster CircuitkjIm(I)_FEM= R j + jωL j

R∞

R1Rq

L∞

L1Lq

Current [A]60000

under the condition of |pj|>2πfmax

40000
20000
000100200300

400

500

600

700800900

1000Frequency [Hz]Foster
Realization

Numerical Results
R0R1Rq

L0L1LqLq

R1R0j

j =1λ j =0-60000

1L1L0k = ∑l rµν-40000

pp

jt

80000

qq

1

λ-20000

a s + La s + a
Y ( s) =b s + Lb s + 1

It is not easy toapply theCauer Realization
complicated objects.

jjq

j∞−2πf max + jω − p j

Zj =120000Re(I)_FEM∑σ − p,p=jwhen σ=2πfmax+jωFrequency characteristic :100000kj

111
1=+++L+Z ∞ Z1 Z 2Zq

Equivalent circuits using Rational Polynomial Approximation
Maxwell equations

jλj

j =1λ j ≠0

The electromagnetic devices are often modeled as an equivalentcircuit for design of the control and driving circuit.

− l rµνt

k=Rq

Coil windings model
Skin depth

y[mm]

z[mm]

70.06

6

1.53

0.66

Skin depth[mm]PurposeWe propose a novel method to generate the equivalent circuit of theelectromagnetic devices using model order reduction.

0.17
0.17

0.9

0.06

5432
10.03
0.9

x[mm]

x[mm]

1.53

0.9

1.53

0100

1000

10000

1000001000000

Frequency [Hz]Padé approximation via the Lanczos process[2]Analysis condition

Laplace-transformed discrete Maxwell equationss Nx + K x = b v

N, K∈Rn×n, b, l, x∈Rni = lt xv : voltage, i : current

Conductivity Relatively Permeability Maximum frequency Number of elements
Number of nodes(tetrahedral elements)
κ[S/m]
fmax[MHz]µr
11298201
52077
5．76×107

Impedance with respect to frequency

Y( s ) = l t (K + sN) −1 b0.06

A=−(K+s0N)−1N
r=(K+s0N)-1bY( s 0 +σ ) = l t (I - σA) −1 r

Spectraldecomposition of A

Y(s0 +σ ) = l t (I -σSΛS-1)−1r =nfg

∑1−iσλi ii=1

This formulation would be unsuitable for real uses because of heavy
computational burden in solution of the eigenvalue problem.
Admittance function

Neumann series expansion(Y ( s 0 +σ ) = l t (I - σA ) −1 r

Im(Z)_q=5

Im(Z)_q=2Re(Z)_q=5Re(Z)_q=2

Re(Z)_FEM

Im(Z)_FEM

Circuit parameters
0.5

0.04

0.4

0.03

0.3

0.02

0.2

0.01

0.1

0100

1000

10000

q52q52R1[Ω] R2[Ω] R3[Ω] R4[Ω] R5[Ω]1.03e-2 5.69
6.82
141
58.3
1.11e-2
3.51---------L1[H]
L2[H]
L3[H]
L4[H]
L5[H]
6.97e-8 8.02e-7 7.25e-7 1.31e-6 1.79e-6
6.89e-8 3.18e-7----------

01000000

100000Frequency [Hz]Inductor model coupled with DC-DC converter∞) ∑mσY( s 0 + σ ) = l t I + σ A + σ 2 A 2 + L r =

0.6

0.05

Im(Z) [ΩΩ]s → s0+σRe(Z) [ΩΩ]Admittance function

iiz[mm]y[mm]Analysis condition

i=00.60

Lanczos method

Number ofelements
117548

1.35
0.45
0.17

Reduced admittance function
Y ( s0 + σ ) =Spectral decomposition of Tq∞∑ l r (e T e )σtt ii
1 q 1i =0= l t re 1t ( I − σ Tq ) −1 e1−1Yq (s0 + σ ) = l t re1t (I − σSq Λ qS-1q ) e1=Tq∈Rq×q : Tridiagonal matrix (n>>q)

ql t rµ jν jj =1j∑ 1 − σλ0.9

0.06
0.9x[mm]x[mm]1.30 1.35

0.9

1.35

LFEM
EDriving condition
R0C0Currents in the case of E=0.3 V

Lanczos method

Tridiagonal matrix

α 1ρ 2Tq =  0 M 0

obtain Avn and A wn.

β2
α2
ρ30L

β3OOOOOO

L0

ρq0 M 0 βq α q Avn = −(K + s0 N)−1 Nvn{}

tAt wn = − (K + s0 N) −1 N wn

References[1] B. Gustavsen, A. Semlyen, “Rational Approximation of Frequency Domain Responses by Vector

Fitting,” IEEE Trans. Power Delivery, vol. 14, no. 3, pp. 1052-1061, 1999.
[2] P. Feldmann and R. A. Freund, “Efficient Linear Circuit Analysis by Padé Approximation via the Lanczos
Process,” IEEE Trans. Computer-Aided Design, vol. 14, no. 5, pp. 639–649, May 1995.

Duty factor
0.9

25

520

Current [A]4

Current [A]---------algorithm--------0) Set ρ1=||r||2, η1=||l||2, v1=r/ρ1, w1=l/ η1, v0=w0=0 and δ0=0For n=1,2…,q do1) Compute δn=wntvn2) Set αn= wntAvn/ δn, β n= ηnδn/δn-1, γn= ρnδn/δn-1
We need to solve the3) Set v = Avn-αnvn- β nvn-1, w = Atwn-αnwn- γnwn-1
following equations to4) Set ρn+1=||v||2, ηn+1=||w||2, vn+1 = v/ρn+1 , wn+1 = w/ηn+1t

DC voltage Driving
E [V]frequency
0.3V, 1.2 V 1MHz

Currents in the case of E=1.2 V6We can obtain tridiagonal matrix Tq whose eigenvalues correspond to thesignificant eigenvalues of A.

Number ofnodes
20985

32FEM
staticPresent

100.0E+00

5.0E-06

1.0E-05

1.5E-05

Time[sec.]

1510FEMstatic5Present

2.0E-05

00.0E+00

5.0E-06

1.0E-05

1.5E-05

2.0E-05

Time[sec.]

Computational time for generation of equivalent circuit
Coil Windings model Inductor for DC-DC converter model
40 min. (q=5)
11 min. (q=5)
We use Xeon W5590/3.2GHz(12GB RAM)

Circuit analysis VS FE analysis
Coil Windings model
DC-DC converter model
FEM*Present
FEM(E=0.3V) FEM(E=1.2V)
Present
230 min. less than 1 sec.
240 min.
360 min.
less than 1 sec.
*the elapsed time of field computations by FEM at 13 sampling frequencies