lightning damage to electric vehicle charging systems · 5 1. abstract 6 this is a living document...

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Lightning Damage to Electric Vehicle 1

Charging systems 2

3 4




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1. Abstract 5

This is a living document and represents the current thinking of IEEE PES SPDC WG 3.6.14 on Electric 6 Vehicle Supply Equipment. Based on the SPDC’s experience of lightning events affecting electrical and 7 electronic systems, a series of scenarios are discussed. At present these are done on a mainly theoretical 8 basis, as there are not enough field reports of damage. 9

2. Introduction 10

There are 4 ways lightning could cause damage to the DC charging circuit: 11 • By Induction from a nearby strike (see clause 3.1) 12 • Through the effects of GCR (ground current rise see clause 3.2) 13 • Via flashover due to GPR (ground potential rise) and follow current (see clause 3.2) 14 • By a direct strike (see clause 4) 15

16 A direct lightning strike is the worst case, but unlikely to happen. So designing protection to meet this 17 threat may not be economic. So let’s consider the indirect effects in order. 18 19

3. Potential damage from indirect effects 20

3.1 Induction from a nearby strike 21

In this case we want to calculate the amount of current induced in the loop which represents the DC fast-22 charging circuit in Figure 1 caused by a lightning flash, represented by the long straight wire in Figure 1. 23

24

Figure 1 Representation of a lightning flash coupled to a loop 25

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To do that we need to first calculate the mutual inductance between the wire and the loop. To calculate the 27 mutual inductance M, we first need to know the magnetic flux through the rectangular loop. The magnetic 28




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field at a distance r away from the straight wire is B = μ0I / 2πr , using Ampere’s law. The total magnetic 29 flux ΦB through the loop can be obtained by summing over contributions from all differential area elements 30 dA =l dr: 31

32

Thus the mutual inductance is: 33

S

WSl

IM B ln

20

(1) 34

Let 35

btatpeak eeIti (2) 36

btatpeak beaeMI

dt

diMtV (3) 37

bsas

sabMIsV

peak

(4) 38

csL

sV

Z

sVsI

W

W

(5) 39

In equation 5, assume that Z is the lowest it can be, which is the impedance of the DC feed. Then in 40 equation 5, LW is the inductance of the wire, s = jω, and c = RW/LW. Substitution equation (4) in 41 equation (5) we have: 42

csbsas

sab

L

MIsI

W

peak

W )( (6) 43

The LaPlace transform of which is: 44

bcca

eabc

bc

be

ca

ae

L

MItI

ctbtat

W

peak

W (7) 45

To get an estimate of the size of IW(t), let’s put some numbers in equation 7. Let the height of the loop l be 46 the spacing between the DC feed wire and the ground return, and assume 0.1 m (the calculation is 47




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insensitive to this value – see equation 1). The worst-case cable length W is assumed to be 10 m, and 48 assume the lightning flash is S = 30 m away. Then from equation (1), where μ0 = 4π x 10-7 H/m 49

M = 5.75x10-9 H 50

Now assume that the feed cable is a 10 m length of #6 AWG wire, for which RW = 0.0013 ohms and LW = 51 15.5 μH. Let the lightning surge be the median surge from CIGRE TD 549 Table 3.5: 30 kA 5.5/75 [5.5 52 microseconds rise time /75 microseconds to half-peak]. Then a = 1x104, b = 8.1x105, and c = 84. Putting 53 these numbers into equation (7) we can make the plot shown in Figure 2. 54

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Figure 2 Current induced in a charging cable due to a typical 5.5/75 30 kA first flash 56

Since the induced current depends on di/dt, it is of interest to see what the induced current is for a 1.2/32 12 57 kA median second flash, which has a much faster rise time than the first flash. In this case a = 2.3x104 and 58 b = 4.78x106. Using these numbers for a and b, and otherwise the same assumptions as used for the median 59 first surge, we can use equation (7) to make the plot shown in Figure 3. Here we can see that the faster rise 60 time of the subsequent flash partially compensates for its lower peak current, relative to the first flash. But 61 the first flash is still the worst case. 62




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63

Figure 3 Current induced in a charging cable due to a typical 1.1/32 12 kA subsequent 64 flash 65

Figure 2 predicts that on the average a current of 10 – 20 A could flow on a charging cable due to 66 induction. An extreme lightning flash can have several times the peak current of a median flash, but it is 67 usually accompanied by a much slower rise time, so the net effect might not be a much greater induced 68 current. The actual induced current depends on many things, but it will likely be a few tens of amps. The 69 induced current will add to (or subtract from) the current due to ground potential rise (GPR), which we will 70 consider next. 71




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3.2 Ground potential rise (GPR) and ground current rise (GCR) 72

In the case of a uniform ground, the current density from a lightning flash spreads out from the point of 73 contact of the flash. The voltage created by the spreading flash current density decreases as the distance 74 from the flash increases, as illustrated in Figure 4. The result is that the ground potential at a point closer to 75 the flash, e.g. point A in Figure 4, is greater than the ground potential at a more remote point B, hence there 76 is a difference in potential due to a GPR between points B and A. 77

78

Figure 4 Spread of GPR from the source. Here V1 > V2 … >V6 79

Now referring to Figure 4, suppose a car is located at point A, and a charging system is located at point B. 80 When the car battery is being charged, point A and point B are connected by a wire having an impedance 81 Z. 82




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An equivalent circuit for this arrangement is shown in Figure 5. Here R1 is the resistance of the earth 83 between the flash striking point and point A, R2 is the resistance of the earth between the flash striking 84 point and point B, R3 is the earth resistance between point A and a remote ground, and R4 is the earth 85 resistance between point B and a remote ground. IF is the current due to the lightning flash. 86

R1

R3

V

R2

R4

Z

IF

I1

I3

I2

I4V3

V1 V2

V4

I5

87

Figure 5 Equivalent circuit for a car at point A and a charging system at point B 88

Analyzing the equivalent circuit, the current in the charging circuit, I5 due to a GPR can be expressed as: 89

2

4343

41325

RRRRZR

RRRRII

T

F (8) 90

Where RT = R1 + R2 + R3 + R4 91




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Let r0 be the radius of the area where the lightning flash strikes, let r1 be the distance of the car ground from 92 the lightning flash striking point, and let x be the distance of the charging system ground from the car 93 (points A and B respectively, in Figure 4). Then R1 is the resistance of the green area in Figure 6 between 94 r0 and r1; and R3 is the resistance of the yellow area between r1 and remote earth. 95

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Figure 6 Areas corresponding to resistances R1 and R3 97

98

Similarly R2 is the resistance of the gray area in Figure 7 between r0 and r1 + x; and R4 is the resistance of 99 the blue area between r1 + x and remote earth. 100

101

102




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Figure 7 Areas corresponding to resistances R2 and R4 103

Let ρ be the resistivity of the ground. Then using the guidance of IEEE Std.142:1991 104

12

11

2 10

1 Grr

R

(9) 105

2

10

22

11

2G

xrrR

(10) 106

3

1

32

1

2G

rR

(11) 107

4

1

42

1

2G

xrR

(12) 108

TT Gr

RRRRR

2

2

2 0

4321

(13) 109

Substituting equations (9) – (13) into equation (8) we get equation (14) 110

24343

41325

)(2

GGGGZG

GGGGII

T

F

(14) 111

The voltage ΔV = (V3 – V4) is also of interest. It is simply 112

ZIV 5 (15) 113

Let Z = RS + sLS, where RS is the sum of all resistances between the two ground points (circuit resistance, 114 resistance of the ground at A, and the resistance of the ground at B), and LS is the sum of all inductances. 115




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Referring to Figure 8, assume that the ground is ionized over the patterned area on the green, and that the 116 man’s feet are a foot apart. With these assumptions r0 is estimated to be 2.4 m (the calculation is not very 117 sensitive to this number). Fitting some typical numbers to equations (14) and (15), 118

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Figure 8 Used for estimating r0, assuming the man’s feet are 1 foot apart (courtesy CITEL) 120

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r1 is a variable for Figure 9, otherwise it is 30 m 122

x = 10 m 123

ρ = 300 ohm-m 124

RS = is a variable for Figure 10, otherwise it is 174 ohms (estimated total ground connection resistance, 125 calculated according to IEEE Std.142:1991) 126

LS = 15.5 μH (roughly the inductance of a #6 wire over a ground plane) 127

Let the waveform be the median first stroke 30 kA 5.5/75, for which Ipk = 30, a = 1.0x104, and b = 8.1x105 128

For the waveform assumed, the maximum value of sLS is 0.138 ohms at 8.9 kHz, very much less than RS. 129 So sLS can be neglected, and Z = RS. 130 131




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For this set of variables, Figure 9 has two plots. The first shows the surge current that could flow in the 132 charging wire due to a GCR as the distance of the lightning strike from the car is varied – up to 200 amps in 133 this case for a total ground resistance of 174 ohms (calculated according to IEEE Std.142:1991). The 134 second plot shows the GPR voltage that can be developed across the sum of the ground resistances and the 135 resistance of the charging circuit. The voltage across RS can reach 30 kV, if the lightning flash is close to 136 the charging circuit. 137

138

Figure 9 Variation in peak wire current, and voltage across Rs 139 for Rs = 174 ohms and ground spacing = 10 m 140

141




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Figure 10 is a plot of equation (14) as Rs is varied, with r1 = 30 m and ground spacing = 10 m.. In this case 142 for a reasonable value of Rs the current in the charging circuit could reach 100 A. The voltage across Rs 143 doesn’t vary much with as Rs is varied. For example in the present case the voltage across Rs goes from 144 5200 V to 6000 V as Rs goes from 20 ohms to infinity. 145

146 Figure 10. Variation in peak wire current with total ground resistance Rs, 147

for r1 = 30 m and ground spacing = 10 m. 148

The peak currents in Figure 9 and Figure 10 were calculated for a median lightning flash, and could be 149 more or less than that shown, depending (among other things) on the peak current of the lightning flash and 150 the distance of the lightning strike from the car. 151

4. Lightning direct strike 152

Although it would be a rare event, it’s interesting to see what would happen to the DC feed wire if it was 153 hit by a direct strike. 154

From [B1] for copper wire 155

42 024.0 dtIA (16) 156

Where d = diameter of the copper wire in mils. For a double exponential lightning surge of time-to-half 157 peak = τ, 158




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272.0 peakIA (17) 159

25.0230 peakId (18) 160

From Wikipedia 161

5log86.1936

dAWG (19) 162

Knowing Ipeak and τ, the largest diameter of a wire that would fuse is given by equation (18), and the 163 corresponding AWG by equation (19). Thus any AWG with a number higher than that calculated from 164 eauation (19) would fuse by the chosen lightning flash. For example for a median 30 kA 5.5/75 flash, any 165 AWG with a number equal to or higher than 19 would fuse. Simerly a 200kA 10/350 would fuse any wire 166 with an AWG equal to or higher than 7. The feed wire will likely have a low enough AWG that it would 167 survive a direct lightning flash, but the system would need to be protected against it. 168

5. Damage potential 169

Both the induced current and the GCR current due to GPR depend on many things, including the 170 waveshape of the flash, the distance of the flash from the charging cable, and the resistances of the grounds. 171 As an estimate, the combined peak induction current and peak GCR current could be in the range 172 50 – 100 A. There is a possibility that this current adds to, or subtracts from the normal battery charging 173 current. If it subtracts from the normal charging current, and if it is greater than the charging current, then 174 reverse current can flow through the battery, which will damage it1. This case is most probable if the 175 charging cable is connected to the car, but not running charging current (for example, if the battery is fully 176 charged; or the charger is not turned on). 177

Another possibility is that the voltage due to a GPR causes insulation breakdown, which if it happened in 178 the charging circuit could allow power follow current to flow. In this case damage would most certainly 179 occur. A high voltage could also cause damage to any electronics associated with the charging circuit. 180

Without sufficient field reports of damage for guidance, it is hard to determine the most prominent cause of 181 damage due to lightning. The IEEE PES SPDC WG3.6.14 is developing a Guide to cover this subject. 182

1 For example, this warning from a lithium battery manufacturer: Do not reverse the positive (+) and negative (-) terminals when charging. Otherwise, the battery pack will be reverse-charged, abnormal chemical reactions will occur, and the excessively high current will cause damage, overheating, smoke emission, bursting, and/or fire.




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Annex A 183

(informative) 184

Bibliography 185

[B1] Kenneth C. Chen, Larry K. Warne, Yau T. Lin, Robert L. Kinzel, Johnathon D. Hu®, Michael B. 186 McLean, Mark W. Jenkins, and Brian M. Rutherford, CONDUCTOR FUSING AND GAPPING FOR 187 BOND WIRES, Progress In Electromagnetics Research M, Vol. 31, 199-214, 2013 188

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lightning damage to electric vehicle charging systems · 5 1. abstract 6 this is a living document...

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