Shielding

Introduction

Concept of shielding

Shielding of a metal shield (theory)

. . 20logi

t

ES E

E

For E field

For H field

. . 20logi

t

HS E

H

Do they be equal ?

Shielding of a metal shield (theory)

It depends. For plane wave, they are equal But for near field, they are not.

Shielding of a metal shield (theory)

0

22 / 2 /0 0

0 0

( )[1 ( ) ]

4

j tt j t t j ti

t

Ee e e e e

E

j

j

where :Skin depth

1. Plane wave assumption 2. Continuity of E and H at each boundary

How to derive?

11.1 Shielding effectiveness. (S.E.)

a. Definition

ˆ1 : . . 20 log

ˆ

ˆ2 : . . 20 log

ˆ

note:

1. If the incident field is an uniform plane wave,and the me

t

t

Eifor electric firld S E

E

Hifor magnetic field S E

H

tan

dium on

each side of the barrier are identical, 1 and 2 definitions are

identical.

2. For near fields and/or different medium, 1 and 2 are not

equivalent.

ˆ3. Definition 1 (for E field) is taken as s dard for 2 case.

-

1

( : )

.

(1) Re .

(2) :

(3)

: (1) & (2) . . , (3)

zattenuation

C Causes of shield

flection

Absorption e

Multiple reflection

Note will increase the S E of the barrier but will decrease

t

. . .

. . dB dB dB dB

he S E of barrier

S E R A M

11.2 Shielding effectiveness: far field sources

0

0

0

0

-

-

0

-

-

0

-1 1

11.2.1.

(1) :

ˆ ˆ

ˆˆ

ˆ ˆ

ˆˆ -

ˆ ˆ

j zi xi

j zii y

j zr r x

j zrr y

zx

Exact solution

For plane wave

E E e a

EH e a

E E e a

EH e a

E E e a

0

-110

-2 2 0 0 0 0

- 022 00 0

-

ˆˆ

ˆ ˆ

ˆˆ -

ˆ ˆ

zy

zx

zy

j zt t x

EH e a where

E E e a

EH e a

E E e a

0-

0

ˆˆ

j ztt y

j j j

jEH e aj

ˆ(2) , .

, , , .1 2

Exact solution

E is known and to determine the remaining amplitudesi

E E E and Er t

. . 0 1 2 1 2

1 2 1 2

0 0 0 0

1 2

B C at Z E E E E E E E Ei r i r

E E E Ei rH H H Hi r

at Z t E E E

- - - 1 2

- - - 1 2 - 1 2

0 0 0

2

0 -2- - 2203 . 1 ( )

40 0

t t tE e E e E et t

E E Et t ttH H H e e et

tj t

e e

-0

. 1/ , .

t j tj te e e

exact solution where j in barrier metal

0

0

0

-- - - -

4 Simplifications:

1. assume the barrier is a "good conductor".

- 1.

2. skin depth barrier thickness .

1

tt t j t j t

t

e e e e e

E

2

0 - -0

0

4 4

t ti

t

e e

E

0

10 10

0

10 10

2

-2- 20

0

5

. . : 20 log : , 20 log : .

4

6 The Multiple - reflection term is the middle term of (3).

. 20 log 20 log .

4

- 20 log 1 -

t

dB dB

t

dB dB

tj

dB

P S R e A

S E e M

M e e

- 2-2

( )

1 .

20 log 1 - 0

t

j tt

if t

where for good conductor

e e

11.2.2 Approximate solution

0

2

1

0

0

01

ˆ: 1 ( )

2

. Re ( )

0.

ˆ ˆ21

ˆ ˆ

ˆ 22

ˆ ˆ

i

t

Assumption a good conductor

t

a flection Loss E field

t

E

E

E

E

E

01

21

0

2

0 010

0

.

ˆ

3

ˆˆ ˆ4

ˆ

ˆˆ4 20 log 20 log 20 log

ˆ ˆ4 4

tt

i i

idB

t

the same as exact solution

E

In the absence of attenuation

EE E

E E E

ER

0

0

ˆ

:

ˆ2 0 1, 2

ˆ

2.

Note The transimission coefficient is very small

at the Boundary and is approximately

at the Boundary

ve

.1. ry little of E field is transmitted through the B

1

1 1 0 0

0

0

0

0

01 1 1 0

21 2

22 0

i i i

t

t t

E

H E

H E E

if

EH E

H E E

0

0

2

0

43

t t t

i i i

if

H H H

H H H

1

C. Absorption loss

is attenuated in the conductor.

Absorption factor

20log t t

dB

E

A Ae e

0

2

0

Note:

ˆ41.

ˆ

2. But primary transmission of H field occurs at the boundary 1,whereas

the primary transmission of the E field occurs at the boundary 2.

3. It means that "thick

t t

i i

E H

E H

" boundary has more effect on shielding against H

field than to the E field.

d. Multiple Reflection Loss

When t >>δ the multiple reflection loss may be important.

(1) for magnetic field

1.

in

inEinHt

H

HTHTH

ˆˆ

ˆ2

ˆˆˆˆˆˆ

0

0

1

inH

t1H

t 2H

boundary 1 boundary 2

2. The reflected wave is

3.

2 ˆ

ˆˆ

ˆ

1 ˆ

22

0

0

boundaryatHe

He

boundaryatHe

inrt

E

inrt

inrt

E

inrt

EHt HeTH ˆˆˆ 22

2

4.

5.

1 , )1(

ˆ

1ˆ

ˆˆˆˆ

21

42

1

321

t

rtE

rtE

rtEt

tttt

H

eleteeH

HHHH

421 1log20ˆ

ˆlog20

ˆ

ˆlog20

i

t

i

tdB

H

H

H

HSE

6.

2

20

0

22

20

0

ˆ20log 1

ˆ

ˆ20log 1

ˆ

t

dB

tj tβς

η ηM e

η η

η η e e

η η

same as exact solution

§11.3 Shielding effectiveness - near field sources

a. In the far field:

0

0

3 )3(

)2(

)1(

d

HE

orthogonalareHandE

b. Near field for Hertzian dipole

(1)

2 30

0 0

0

20

0

3

0 0

0

02

0

ˆˆ

ˆ 1

901

w

j j jr r rE

ZjH

r r

j

r

rr

(2)

(3)

0 0

0

ˆ 602

we

Zr

r

fieldnear fieldfar

3

0

H

E

0

r

sourceimpedance

highaisdipoleelectric

c. Near field for magnetic dipole (loop)

(1)

20

0

0

2 30

0 0

0 0

1ˆ

ˆˆ 1 1

90

w

jrE r

ZjH

r r r

r

(2)

(3)

23692ˆ

0

0

fZ

mw

fieldnear fieldfar

3

0

H

E

0

fieldnear

theinsourceimpedance

lowaisdipolemagnetic

d. Reflection loss for Electric source

(1) we know for plane wave

(2)

ˆ4

20logˆ4

ˆ20log

ˆ

ˆlog20

0

0

2

0

t

idB

E

ER

0

0

0

3 2

ˆusing

ˆ20log 322 10log

ˆ4

w

w r

dB

r

Z jr

ZR

f r

fieldnear

Shielding of a metal shield (Near Field)

060wZr

0

2369w

rZ

Electric dipole

Magnetic dipole

Shielding of a metal shield (Near Field)

3 2322 10log( )r

e

r

Rf r

2

14.57 10log( )rm

r

frR

For electric dipole

For magnetic dipole

Shielding of a metal shield (Near Field)

1. Reflection Loss increase as frequency decrease for E field 2. Reflection Loss decrease as frequency decrease for H field

Note: The Absorption loss also Decrease as frequency decrease for H-field.

How to solve the SE of magnetic field?

An example of S.E. for copper plate

Low Frequency, Magnetic Shielding

Two approaches

Low Frequency, Magnetic Shielding

Two factors that may degrade the ferromagnetic material: 1. Increasing frequency. 2. Increasing the field strength.

Larger than 4KHz, the u is the same with the steel

That’s why we use the steel as

The magnetic shielding material for Switch-mode Power Supply (20-100KHz).

Low Frequency, Magnetic Shielding

Phenomenon of the saturation of ferromagnetic materials

saturation

When increase the H-field

Low Frequency, Magnetic Shielding

Solution: Using multi-layer to reduce the effect of the saturation

The effect of aperture

In practice, the S.E. is limited by the necessary apertures and discontinuities.

20log2

SEd

The simplest formula

The modified formula

Ex: 20dB SE for 1GHz then, d < 1.6cm

The effect of mesh and Honeycomb

SE can reach 100dB

Why ?

The effect of Seams

Poor Better

Best

The effect of aperture orientation

Which one is better?

By this trick, the SE can be improved up to 10dB

The effect of image plane

Enclosure resonance

2 2 2150 ( / ) ( / ) ( / )F k l m h n w MHz

F ~ 212/l ~ 212/h ~ 212/w for equal square enclosure

Gasket and Contact Strip

Choosing concern: 1. Conductivity 2. Ease of mounting

Measurement of SE

Method 1: EN50147-1

. . .b Measurement of S E

Measurement of SE

Method two: Coaxial transmission line model