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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 1

2 Differential-Mode v.s. Common-mode Currents E = E + E θ θ,1 θ,2 e e = M I + I F jβ01 r jβ02 r? ( 1 2 ) ( θ ) r1 r2 jβ0( r+δ) jβ0( r Δ) e e = M( I1 + I2 ) F( θ ) r+δ r Δ jβ0r e jβ0δ jβ0δ = M ( I1e + I2e ) F( θ ) r M is a function of antenna type F(θ) is the array factor ηβ = = π 4π F( θ) = sinθ M j l j2 10 fl For Hertzian dipole Differential-mode current emission model A Question: Where is the maximum E field for the differential current? z θ x 2

3 Differential-mode current emission model Θ=90 is the maximum Differential-mode current emission model I I 1 2 = I d = I d fi l Ed,max = j e e e d 7 D jβ0d jβ0s /2 jβ0s 2π 10 { /2 } = ID f ls d 1. Reduce the loop size 2. Reduce the current level Can reduce the EMI 3

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 4

5 Why the slot increase EMI? 15cm 15cm 12.5cm 5mm 30cm 30cm Common-mode Emission Model I I 1 2 = I = I c c fi l Ed,max = j e d e + e 6 Ic fl = d 7 c jβ0d jβ0s /2 jβ0s 2π 10 { /2 } 1. Reduce the trace length 2. Reduce the current level Can reduce the common-mode EMI 5

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 occurs at low frequency range 6

7 Current Probe for common-mode current Working principal? Ampere s Law and Faraday Law are both used in the design Calibration data: Z t = V / I (transfer impedance) Why only common-mode current are Sensed in this probe? Current Probe Example 7

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 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 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 8

9 Delay Skew effect on the Spectral components (1 st harmonic) Rise time is a minor effect on the spectral component Skew effect on the Spectral components (1 st harmonic) 9

10 Measurement setup for EMI and Common-mode current Delay Skew design on PCB 10

11 Measurement Results for Common-mode current and their radiation Fundamental Frequency EMI increases about 9dB/decade of the skew Common-mode current v.s. EMI (measurement) Fundamental Frequency 11

12 Delay Skew v.s. Common-mode current (measurement) Higher harmonics Questions: Why does 3 rd harmonic not increase as skew increases? Why does 2 nd 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 3 rd harmonic on the cable decreases. 12

13 About the Even harmonics 1st harmonic 2nd harmonic Lack of symmetry of the waveform causes the even harmonics Measurement setup for digital waveform 13

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 14

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? 15

16 Common-Mode EMI on the Ground Plane Reducing trace Height To increase Mgs Common-Mode EMI on the Ground Plane Adding another return path to decrease I 2 Guard trace Shunt trace 16

17 Common-Mode EMI on the Ground Plane Resonant effect by Image Plane Radiated Susceptibility Interconnection filter Incoming mains power filter Field generation antenna Area of uniform field (1.5m 1.5m) 0.8m Field generation equipment 3 m Interconnecting cables Chamber penetration cables 17

18 Radiated Susceptibility Test levels for radiated immunity ( 80 MHz to 1000 MHz) Level Test field strength V/m x Special NOTE x is an open test level. This level may be given in the product specification. The signal is 80 % amplitude modulated with 1 khz sinewave to simulate actual treats. 8-2 Simple Susceptibility Models for wires and PCB Lands a.modelling a two conductor line to determine the terminal voltages induced by incident eletromagnetic field. y z x y L Ei Incident wave RS + _ VS S VL + _ RL Hi The problem is: x predicting the V S and V L given the magnitude,polarization,and direction of the incident uniform plane wave(e i, H i) _ 18

19 b.only two components of incident wave contribute to the induced voltage. RS S E i y i H z RL E i = E H i = H i y i z ( induce) equivalent Tx line model I ( x+ x) RS S I ( x) L X + V ( X ) _ C X I S X RL where per-unit-length μ 0 s L= n( ) π rω C = πε 0 s n( ) rω for parallel-wire line cv. s( x), I s( x) =? (1)by the Faraday's law the incident magnetic field i H will induce the emf in the loop. z i emf = jω B i ds s z i =-jωμ H ids = - jωμ x H dy i 0 z 0 s y= 0 emf s i per-unit-length source at x, Vs( x) = = - jωμ 0 H dy 0 z x y= i (2)The incident E induces a voltage between the two conductors y,which induces a displacement current in the per-unit-length capacitance C. s Is i ( x) = - j ω C E dy y= 0 y s z 19

20 d.derive the tx-line equations V( x+ x) V( x) = jωl xi ( x) V( x) x I ( x+ x) I ( x) = jωc xv( x+ x) + I S ( x) x Dividing x, and x 0 dv( x) s i + jωli ( x) = V S( x) =+ jωμ0 H dy y 0 z dx = d I ( x) + jω dx C V( x s i ) = I S( x ) = - j ω C E dy y= 0 y e. if the tx-line is electrically short = L λ 0 the, C can be ignored RS + VS _ i jωμ 0H i A z - jωce i y i A + VL _ RL total source s i i i VSL= jωμ 0 H dy jωμ 0H isil= jωμ 0H ia y= 0 s i i i ISL= - jωc E dy - jωce isil= - jωce ia y= 0 z z z S L λ 0 y y y S L λ 0 area 20

21 f. It is easy to compute the induced voltage V S and VL RS i RSRL i VS = jωμ 0LSH jωclse z y RS+ RL RS+ RL RL i RSRL i VL = jωμ 0LSH jωclse z y RS+ RL RS+ RL g.an example: 50Ω + VS 50mil + VL 150Ω Ei = 10 v m 1m f = 100MHz (1)only magnetic field induce voltage source. E (2)V = j26.6mv i i S = jωμ 0LSH A= j π π m n (3) j26.6mv 50Ω + VS 150Ω 50 VS = j26.6mv = j6.65mv VL = j26.6mv = j20mv

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 22

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,

24 Radiated Susceptibility for a transmission line Testing probes and their working principle Radiated Susceptibility for a transmission line Test Plate for generating uniform EM wave 24

25 Radiated Susceptibility for a transmission line Uniform field check Radiated Susceptibility for a transmission line Three different cases 25

26 Radiated Susceptibility for a transmission line 1 Ohm case They are quite consistent below 100MHz Measured under BCI technique Measured under Parallel Plate Calculated by previous theory Radiated Susceptibility for a transmission line 270 Ohm case 50 Ohm Case Summary: BCI technique can be used in low frequency range to complement the high frequency radiation susceptibility test 26

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