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1 Radiated Emission and Susceptibility Tzong-Lin Wu, Ph.D. EMC Lab Department of Electrical Engineering National Taiwan University Differential-Mode v.s. Common-mode Currents

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1

Radiated Emission and Susceptibility

Tzong-Lin Wu, Ph.D.

EMC LabDepartment of Electrical EngineeringNational Taiwan University

Differential-Mode v.s. Common-mode Currents

2

Differential-Mode v.s. Common-mode Currents

0 1 0 2

0 0

00 0

,1 ,2

1 21 2

( ) ( )

1 2

1 2

?? ( ) ( )

( ) ( )

( ) ( )

j r j r

j r j r

j rj j

E E E

e eM I I Fr r

e eM I I Fr r

eM I e I e Fr

θ θ θ

β β

β β

ββ β

θ

θ

θ

− −

− +Δ − −Δ

−− Δ Δ

= +

= +

= ++ Δ −Δ

= +

M is a function of antenna type

F(θ) is the array factor

60 0 2 104

( ) sin

M j l j fl

F

η β ππθ θ

−⎧ = = ×⎪⎨⎪ =⎩

For Hertzian dipole

Differential-mode current emission model

A Question: Where is the maximum E field for the differential current ?

z

3

Differential-mode current emission model

Θ=90 is the maximum

Differential-mode current emission model

1

2

d

d

I II I== −

0 0 0/ 2 / 27,max

214

2 10 { }

1.316 10

j d j s j sDd

D

fI lE j e e edI f ls

d

β β βπ − −−

= × −

= ×

1. Reduce the loop size2. Reduce the current level

Can reduce the EMI

4

Differential-mode current emission for trapezoidal pulse train

EMI due to the differential-mode current is significant for higher frequencies.

An example

1m ribbon cable with 50mil distance carrying 20mA current at 30MHz

The EMI at 3m distance is about

40dBuV/m

It’s easy to over the FCC limit line

5

30cm

15cm12.5cm

5mm

30cm

15cm

Why the slot increase EMI ?

Common-mode Emission Model

1

2

c

c

I II I==

0 0 0/ 2 / 27,max

6

2 10 { }

1.257 10

j d j s j scd

c

fI lE j e e edI fld

β β βπ − −−

= × +

= ×

1. Reduce the trace length2. Reduce the current level

Can reduce the common-mode EMI

6

An Example

1m ribbon cable with 50mil distance carrying 7.96uA current at 30MHz

The EMI at 3m distance is about

40dBuV/m -> limit of FCC

Very small current will exceed the limit of FCC !

Common-mode current emission for trapezoidal pulse train

Common-mode EMI usually occursat low frequency range

7

Current Probe for common-mode current

Ampere’s Law andFaraday Law are both used inthe design

Calibration data:Zt = V / I (transfer impedance)

Why only common-mode current are Sensed in this probe?

Working principal ?

Current Probe Example

8

Examples for common-mode EMI

James L. Knighten et al., “Experimental Analysis of Common Mode Currents on Fiber Channel Cable Shields due to Skew Imbalance of Differential Signals Operating at 1.0625 Gb/s”, IEEE EMC Symposium, 1999

Lothar 0. Hoeft et al., “Spectral Analysis of Common Mode Currents on Fiber Channel Cable Shields due to Skew Imbalance of Differential Signals Operating at 1.0625 Gb/s”, IEEE EMC Symposium, 1998

James L. Knighten et al., “Effects of Device Variations on the EMI Potential of High Speed Digital Integrated Circuits”, IEEE EMC Symposium, 1997

Decomposition of the Common/Differential mode

9

Delay Skew effect on the Spectral components (1st harmonic)

Rise time is a minor effect on the spectral component

Skew effect on the Spectral components (1st harmonic)

10

Measurement setup for EMI and Common-mode current

Delay Skew design on PCB

11

Measurement Results for Common-mode current and their radiation

Fundamental Frequency

EMI increases about 9dB/decadeof the skew

Common-mode current v.s. EMI (measurement)

Fundamental Frequency

12

Higher harmonics

Delay Skew v.s. Common-mode current (measurement)

Questions:

• Why does 3rd harmonic not increase as skew increases ?

• Why does 2nd harmonic exist ?

Delay Skew v.s. Common-mode current

Increasing delay skew will increase the energy of 3rd harmonic.

But, impedance mismatch between trace on PCB and Cables increase the reflection coefficient.

Therefore the common-mode current of 3rd harmonic on the cable decreases.

13

About the Even harmonics2nd harmonic1st harmonic

Lack of symmetry of the waveform causes the even harmonics

Measurement setup for digital waveform

14

Digital waveform and its spectrum

Even harmonics

Asymmetry of the waveform may be resulted from

• Difference of the rise/fall times• Not 50% duty cycle (shorter or longer)

EMI for higher harmonics

15

Common-Mode EMI on the Ground Plane

A trace on a solid Ground plane

Common-Mode EMI on the Ground Plane

Equivalent Circuit

Common-mode Noise

It can be reduced by increasing the mutual inductance

How ?

16

Common-Mode EMI on the Ground Plane

Reducing trace HeightTo increase Mgs

Common-Mode EMI on the Ground Plane

Adding another return pathto decrease I2

Guard trace

Shunt trace

17

Common-Mode EMI on the Ground Plane

Resonant effect byImage Plane

Radiated Susceptibility

Interconnection filter Field generation antennaIncoming mains power filter

Area of uniform field

(1.5m×1.5m)

Field generation equipment Chamber penetration cables

3 mInterconnecting cables

0.8m

18

Test levels for radiated immunity( 80 MHz to 1000 MHz)

Level Test field strengthV/m

1 12 33 10x Special

NOTE – x is an open test level. This levelmay be given in the product specification.The signal is 80 % amplitude modulatedwith 1 kHz sinewave to simulate actualtreats.

Radiated Susceptibility

a.Modelling a two conductor line to determine the terminal voltages induced by incident eletromagnetic field.

8-2 Simple Susceptibility Models for wires and PCB Lands

SR+

_

+

_ LR

L y

x

VS VLS

Ei

Hi

Incident wave

x

y

z

S

The problem is:

predicting the V and V given the magnitude,polarization ,and direction of

the incident uniform plane wave(E , H )

_

L

i i

19

.Only two components of incident wave contribute to the induced voltage.b

SSR E iy

izH⊗− LR

equivalent Tx line model

SSR

XL+

( )_V X

XC

( )I x x+

LR( )I xS XI

ii yE E=

ii zH H= ( )induce

0

0

where per-unit-length

L= ( )

( )

n

n

sr

C sr

μπ ω

πε

ω

=for parallel-wire line

( ) ( )

0 00

. , = ?(1)by the Faraday's law the incident magnetic field

H will induce the emf in the loop.

emf =

=- -

per-unit-length so

s x s x

iz

iz

ssi i

z zys

cV I

j B d S

j H dS j x H dy

ω

ωμ ωμ=

=

∫ ∫

i

i

( ) 00

urce at x , -

(2)The incident E induces a voltage between the two conductors

,which induces a displacement current in the per-unit-length capacitance C.

( )

s is x zy

iy

emfV j H dyx

Is x

ωμ=

= =

∴ =

0-

s iyy

j C E dyω=∫

20

x

00

.Derive the t -line equations

V( ) V( ) ( ) V( )

( ) ( ) V( ) ( ) Dividing x , and x 0

dV( ) ( ) V ( )

d ( )

S

s iS zy

d

x x x j L xI x x x

I x x I x j C x x x I x x

x j LI x x j H dydxI x jdx

ω

ω

ω ωμ

ω

=

+ − = − −

+ − = − + +→

⇒ + = − = +

+

0V( ) ( ) -

s iS yy

C x I x j C E dyω=

= = ∫

0

. if the tx-line is electrically short = Lthe , C can be ignored

e λ

SR VS

+

_ VL

+

_LR

0izj H Aωμ i

- iyj CE Aω i

0 0 00

0

total source

V

- - -

s i i iS z z zy

s i i iS y y yy

L j H dy j H S L j H A

I L j C E dy j CE S L j CE A

ωμ ωμ ωμ

ω ω ω

=

=

= ≅ =

= ≅ =

i i i

i i i0S L λ

0S L λ

area

21

0

0

f. It is easy to compute the induced voltage V and V

V

V

g.An example:

S L

S S Li iS z y

S L S L

L S Li iL z y

S L S L

R R Rj LSH j CLSER R R R

R R Rj LSH j CLSER R R R

ωμ ω

ωμ ω

= −+ +

= − −+ +

50Ω 150Ω

1m

50milVS

+

−VL

+

8 7 30

(1)only magnetic field induce voltage source.

(2)V 2 10 4 10 1 1.27 1070

26.6

ii

S n

Ej LSH A j m

j mv

ωμ π π − −= = × × × × × × × ×

=

10

100

vEi mf MHz

=

=

(3)

50Ω 150ΩVS

+

26.6j mv

50 V 26.6 6.6550 150

150 V 26.6 2050 150

S

L

j mv j mv

j mv j mv

= =+

= − = −+

22

Radiated Susceptibility for a transmission line

Example (1):

What kind of noise will couple to the transmission line ?

Radiated Susceptibility for a transmission line

Voltage noise source

23

Radiated Susceptibility for a transmission line

Example 2

Radiated Susceptibility for a transmission line

Example 3. Current Injection Technique in Conducted Susceptibility Test

Can this technique replace the R.S. test ?“Investigation of the Bulk Current Injection Technique by Comparison to Induced Currents from Radiated Electromagnetic Fields”, IEEE EMC Symposium, p412 – p417, 1996

24

Radiated Susceptibility for a transmission line

Testing probes and their working principle

Test Plate for generating uniform EM wave

Radiated Susceptibility for a transmission line

25

Radiated Susceptibility for a transmission line

Uniform field check

Radiated Susceptibility for a transmission line

Three different cases

26

Radiated Susceptibility for a transmission line

Calculated by previous theory

Measured under Parallel Plate

Measured under BCI technique

They are quite consistent below100MHz

1 Ohm case

Radiated Susceptibility for a transmission line50 Ohm Case270 Ohm case

Summary: BCI technique can be used in low frequency range to complement thehigh frequency radiation susceptibility test