How Resonant Structures Affect Power Distribution Networks and Create Emissions.

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Harold Tate
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1 How Resonant Structures Affect Power Distribution Networks and Create Emissions. Presented by Joanna McLellan January 17, Lots of people have the paradigm that adding capacitors is the Go-to solution for all EMC and Power Integrity problems. But mindlessly adding capacitors often adds to your troubles. The reason you might ask? They create new resonance structures. PDF copies of the presentation do not show the animations.

2 Where does the energy go? The energy moves back and forth as a current. PDF copies of the presentation do not show the animations.!2

3 Frequencies changes the Impedance At low frequencies At high frequencies 100 KHz < f < Fractional λ!3

4 Frequencies changes the Impedance At low frequencies At high frequencies 100 KHz < f < Fractional λ!4

5 Frequencies changes the Impedance At low frequencies At high frequencies 100 KHz < f < Fractional λ off!5

6 Frequencies changes the Impedance At low frequencies At high frequencies 100 KHz < f < Fractional λ off!6

7 Frequencies changes the Impedance!7

8 Frequencies changes the Impedance Impedance off Output impedance of a current switch mode power supply with a zener in the circuit. Note the movement of the resonance and antiresonance as the DC voltage changes. PDF copies of the presentation do not show the animations.!8

9 Frequencies changes the Impedance At low frequencies At high frequencies 100 KHz < f < Fractional λ!9

10 Let s take a look at two capacitors and the distances between them.!10

11 Let s take a look at two capacitors and the distances between them. 1 Frequency= There is one 2π L T C T resonance and one loop of current. L FR4 PCB 10 nh inch 400 nh meter L wire 15 nh inch 600 nh meter!11

12 Now let s take a look at an array of three capacitors and the distances between them.!12

13 Now let s take a look at an array of three capacitors and the distances between them.!13

14 Now let s take a look at an array of three capacitors and the distances between them.!14

15 Now let s take a look at an array of three capacitors and the distances between them.!15

16 Now let s take a look at an array of three capacitors and the distances between them. We now have three resonances.!16

17 But what happens as we add a fourth capacitor?!17

18 We get ten different resonant frequencies.!18

19 Hence adding lots of capacitors is not the answer to EMC problem mitigation. Because as capacitors are added you end up playing a game of Capacitor Whack-a-Mole.!19

20 Hence adding lots of capacitors is not the answer to EMC problem mitigation. So what do we do? We find and break the loop antennas.!20

21 This Switch Matrix failed EMC Testing Switch matrix inputs are very common. The microcontroller scans the rows and columns of the matrix to determine if a switch has been depressed. This product detected a switch depression due to RF excitation at 220 MHz.!21

22 Pin the tail on the resonant circuit. Row and column switch matrix inputs are very common. Can you find the 220 MHz resonant circuit?!22

23 Pin the tail on the resonant circuit. Here s a hint: with a higher value of R105 the EMC performance got worse.!23

24 Pin the tail on the resonant circuit. The traces are inductors, the open switch is a capacitor. The microcontroller input is also a capacitor and so are the off diodes.!24

25 Pin the tail on the resonant circuit. How do we make sure this is the right resonant circuit? We calculate the resonant frequency.!25

26 Pin the tail on the resonant circuit. First we need to know the capacitance of the Diode.!26

27 The Rohm Data sheet With the diode off the maximum capacitance is 3.5 pf.!27

28 The resonant frequency The total length to the switch through cables and PCB then back again is 20 inches. 1 = C T pf pf pf L T = 10nH inch 20inches=200nH C T = 2.4 pf Frequency center = 2π 1 200nH 2.4 pf 230 MHz!28

29 The resonant frequency The total length to the switch through cables and PCB then back again is 20 inches. Now that we have determined the offending circuit, how do we fix it?!29

30 Resistor are very Broadband The only DC current in the resistors is the input leakage current. The High Q tank is now a low Q tank and the emissions have been reduced by a substantial amount. But how much resistance do we need to pass?!30

31 We can calculate the amount of resistance required to pass the EMC requirement. In the words of Henry Ott, kill it dead The Schelkunoff small loop equation tells us: 20 Log E radiated1 E radiated 2 20 Log R 2 R 1 20 Log mω 60dB!31

32 We can calculate the amount of resistance required to pass the EMC requirement. In the words of Henry Ott, kill it dead 20 Log E radiated1 E radiated 2 20 Log R 2 R 1 20 Log mω 60dB!32

33 Do you add resistors or capacitors? It is better to De-Q the resonance with a resistor. Adding capacitors makes more resonances. Changing a capacitor value moves the center frequency. Let's not play hide and seek with the emissions. Add series resistors to series resonances. Add parallel resistor to parallel resonances.!33

34 Decoupling Circuits Power supply bulk charge well. Discreet decoupling capacitors. Interlayer sheet capacitance in the PCB. Discreet capacitance on the interposer. Interlayer sheet capacitance of the interposer. White space (unused) space on the die. Decoupling is a Bucket Brigade of charge from slow big buckets to fast small buckets.!34

35 The Bucket List Power supply bulk charge well. Discreet decoupling capacitors. Interlayer sheet capacitance in the PCB. Discreet capacitance on the interposer. Interlayer sheet capacitance of the interposer. White space (unused) space on the die.!35

36 Sizing the Buckets The size of each bucket is determined by the Current demand, Rate of charge drain, Voltage ripple requirements, and Electrical length to where the charge is needed.!36

37 Sizing the Buckets Each capacitor of the Bucket Brigade can form a resonant loop with another capacitor in the Bucket Brigade. This will cause ringing on the two resonating capacitors. The off-board communication bus drivers decoupling capacitors should be checked for ringing due to this issue. As again, we are playing Capacitor Whack-a-Mole.!37

38 I Peak Peak AC Current Estimation Current We estimate the average AC current demand as one third of the DC current demand. I = I DC demand AC average 3 I AC average Average Current tr tr Demand Time This estimate is from the work of Bruce Archambeault. tcycle Time to next pulse!38

39 Peak AC Current Estimation By equating the charge of these two waveforms we can determine the peak current demand. Q = I dt t 2 Q Pulse = t r 2t r I dt + I dt Q = I dt AC Average AC Average 0 t r t Cycle 0 t 1 Q Pulse = area RiseRight AngleTrangle + area Fall Right AngleTrangle Q Pulse = I Peak t r 2 + I Peak t r 2 = I Peak t r Q AC Average = area AC currrent demand rectangle = I AC Average t cycle I Peak t r = I AC Average t cycle I AC average = I DC demand 3 Reference: I Peak = t cycle t r I DC demand 3!39

40 I Peak Peak AC Current Estimation Current I Peak = t cycle t r I DC demand 3 I AC average Average Current Demand tr tr Time tcycle Time to next pulse!40

41 Peak AC Current Estimation We consider the voltage noise on VDD. t V ( t)= 1 C I ( t ) dt +V 0 0 ( ) then 2tr V ( 2t )= 1 r C I ( t ) dt +V 0 0 ( ) t r with Q Pulse = I dt + I dt 0 2t r t r = I Peak t r, I Peak = t cycle t r I DC demand 3 V ( 2t ) r = V DD V noise and V ( 0) = V DD than V DD V noise = Q Pulse C +V DD with V noise specified by the requirements, than V noise = Q Pulse C Q Pulse = t I cycle DC demand C = t I cycle DC demand Decoupling V V3 3 noise noise!41

42 I Peak Total Decoupling Distance to Decoupling Capacitance Estimation Capacitance Estimation Current C = t I cycle DC demand Decoupling V 3 noise I AC average Average Current Demand If physical limits do not allow compliance with this value, change the design. tr tr Time tcycle Time to next pulse!42

43 Total Decoupling Capacitance Estimation C = t I cycle DC demand Decoupling V 3 noise If physical limits do not allow compliance with this value, change the design.!43

44 Distance to Decoupling Capacitance Estimation We consider the speed pesky of slow propagation. speed of light. v= 1 µε = c µ r ε r t capacitor = t cycle 20 then d capacitor =t capacitor v= t capacitor µε = t capacitor c µ r ε r d capacitor = t cycle 20 c µ r ε r!44

I Peak Distance to Decoupling Capacitance Estimation C = t I cycle DC demand Decoupling V 3 noise Current I AC average Be sure to place this capacitance within d = t c cycle

45 I Peak Distance to Decoupling Capacitance Estimation C = t I cycle DC demand Decoupling V 3 noise Current I AC average Be sure to place this capacitance within d = t c cycle capacitor 20 µ ε r r to allow for 10 reflections. Average Current Demand If physical limits do not allow compliance with this value, change the design. tr tr Time tcycle Time to next pulse!45

46 Distance to Decoupling Capacitance Estimation Be sure to place this capacitance within d capacitor = t cycle 20 c µ r ε r If physical limits do not allow compliance with this value, change the design.!46

47 Sizing the Decoupling Capacitor These equations will provide a minimum capacitance value at a C = t I cycle DC demand Decoupling V 3 noise maximum distance. But it is always best to use the biggest capacitor value available within the package size minus one step. d capacitor = t cycle 20 c µ r ε r!47

48 Sometime it Doesn t Work Out There is inductance between decoupling capacitors and the other capacitors on the rail. They will resonate at various frequencies. If any of these frequencies are at a harmonic of the current demand, there will be voltage noise on the capacitors and also on the driven outputs.!48

49 Power Plane Target Impedance For further information see Steve Sandler s YouTube Videos or his book:

50 Power Plane Target Impedance Z target = Allowed Voltage Ripple DynamicCurrent Often we do not know the dynamic current, but we can estimate it as half of the maximum current. Z target = Nominal Voltage %Allowed Ripple 50% MaximumCurrent Z target = 5V 5% 1.5A50% = 250mV 750mA =333mΩ!50

51 Decoupling Example 100Ω Impedance C1 Resonance 1µΩ Frequency 100K-1GHz A single 100nF capacitor has a self resonance frequency (SRF) of 16.4MHz.!51

52 100Ω Decoupling Example C1 & C2 Antiresonance C1 Only Impedance 1µΩ C2 Resonance C1 Resonance C1 & C2 Frequency 100K-1GHz By adding a 2.2µF capacitor we get a new resonance at 2.5MHz and an antiresonance at 4.5MHz!52

53 Decoupling Example 100Ω C1 & C2 Antiresonance C3 Zener Antiresonance Impedance 1µΩ C2 Resonance C1 Resonance C3 Zener Resonance Frequency 100K-1GHz By adding and an off zener diode we get a new resonance at 652MHz and an antiresonance at 234MHz.!53

54 100Ω New Antiresonance Decoupling Example C1 & C2 Antiresonance Impedance 1µΩ C2 Resonance C1 & C4 Resonance Deeper C3 Zener Resonance Frequency 100K-1GHz By adding a second 100nF capacitor we get a deeper resonance at 16.4MHz and 652MHz.!54

55 Decoupling Example 100Ω Moved zener Antiresonance Impedance 1µΩ Frequency 100K-1GHz The second 100nF capacitor also moved the zener s antiresonance to 314.5MHz.!55

56 Frequencies changes the Impedance!56

57 Decoupling Example 10Ω Impedance 1µΩ Frequency 100K-1GHz If the zener voltage is changing, the capacitance of the junction will change. This changes the frequency of the zener s antiresonance from 314MHz (off, 35pF) to 416MHz (on, 20pF).!57

58 Decoupling Example 10Ω Impedance 1µΩ Frequency 100K-1GHz If load current has frequency content at a antiresonance, large voltage ringing will result. This ringing has been known to reset microcontrollers and exceed maximum allowed voltages. It is also known as rogue waves.!58

59 Decoupling Example 10Ω Impedance 1µΩ Frequency 100K-1GHz If an external RF excitation is applied at an antiresonance frequency, the energy will be absorbed by the circuit and again rogue waves are created.!59

60 Networking is good for business, please send me a LinkedIn link request. And post a LinkedIn recommendation.

Thank you for allowing me to share this information with you today. Presented by Joanna McLellan 248-765-3599 Joanna Pirie McLellan 2019, All Rights Reserved JoannaEMC@icloud.

61 Thank you for allowing me to share this information with you today. Presented by Joanna McLellan Joanna Pirie McLellan 2019, All Rights Reserved No part of this presentation or any of its contents may be reproduced, copied, modified or adapted, without the prior written consent of the author, unless otherwise indicated for stand-alone materials. Respective copyrights are owned by their respective copywriter holders. Important: The information in this presentation is of a general nature, and should not be relied upon as individual professional advice. The Presenter makes no warranty or promise whatsoever that the information presented is complete, current, and accurate, and accepts no responsibility for any loss, damage, or personal injury that may result from the use of any material provided, whether orally or through any form of media, during the presentation. Presentations are intended for educational purposes only and do not replace independent professional judgment. Attendees should always make and rely upon their own separate inquiries prior to making any important decisions or taking any action on the basis of this information. Video and audio recording is prohibited.!61

ET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012

ET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012 1. In the single-phase rectifier shown below in Fig 1a., s = 1mH and I d = 10A. The input voltage v s has the pulse waveform shown