From U - Fachgebiet Hochspannungstechnik

Insulation Strength Characteristics 

Topics to be covered in the following: 

• Insulators under polluted conditions 

• Probability of flashover (Normal and Weibull distributions) 

• Behavior of parallel insulation 

• Coordination procedure: deterministic and statistical approach 

• Correction with altitude of installation 

• Clearances in air; "gap factors" 

Fachgebiet 

Hochspannungstechnik 

Overvoltage Protection and Insulation Coordination / Chapter 7 - 1 -

Probability of Disruptive Discharge of Insulation 

The breakdown process is statistical in nature to be taken into account, especially for 

impulse voltage stress! 

Non-self-restoring insulation 

• No method at present available for the determination of the probability of disruptive discharge 

• Therefore, it is assumed that the withstand probability changes from 0% to 100% at the value 

defining the withstand voltage. 

• Withstand voltage usually verified by application of a limited number of test voltages at standard 

withstand level with no disruptive breakdown allowed "Procedure A" of IEC 60006-1: 

[IEC 60060-1] 





The breakdown process is statistical in nature to be taken into account, especially for 

impulse voltage stress! 

Self-restoring insulation 

• Withstand capability can be evaluated by tests and be described in statistical terms. 

• Therefore, self-restoring insulation is typically described by the statistical withstand voltage 

corresponding to a withstand probability of 90%. 

• Withstand voltage verified by application of a limited number of test voltages at standard insulation 

level, allowing a certain number of discharges 

"Procedure B" of IEC 60060-1 "15/2-test" usually applied procedure in the "IEC world" 

"Procedure C" of IEC 60060-1 "3+9-test" 

See next three slides …. 





[IEC 60060-1] 





Comparison of Procedures B and C 

Only here both procedures are equivalent! 

[IEC 60071-2] 

Example: 

• equipment at the borderline, rated and tested at its U 10 

, has a 82% probability of passing the test in Procedure B 

• a better equipment, rated and tested at its U 5,5 

, has a 95% probability of passing the test in Procedure B 

• a worse equipment, rated and tested at its U 36 

, has only a 5% probability of passing the test in Procedure B; 

with Procedure C, its probability of passing would be higher, the 5% probability of passing would be given for 

equipment rated and tested at its U 63 

(see also next slide) 





Comparison of Procedures B and C 

Probability of passing the test: approx. 82 % 

at probability of breakdown of 10 % 

Probability of breakdown P(U) 

Probability of passing the test "15/2" 

Probability of passing the test "3+9" 

1,0 

0,9 

0,8 

0,7 

0,6 

0,5 

0,4 

0,3 

0,2 

0,1 

Pb(U) P(U) 

15/2 

3+9 

0,0 

-3 -2,5 -2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 

(U-U50)/Z 50 Test voltage referred to conventional deviation 





Comparison of Procedures B and C (IEC depiction) 

50 Equipment that has a 

5% probability of 

passing the 15/2-test, 

would have an approx. 

40% probability of 

passing the 3+9-test 

[IEC 60071-2] 





Completely different approaches: 

• Verification of withstand voltages (see slides before) 

• Evaluation of withstand voltages 

Determination of the probability function P = P(U), defined by the three following 

parameters in case of a Normal or Gaussian distribution: 

U 50 

… voltage under which the insulation has a 50% probability to flashover or to withstand 

Z … conventional deviation; Z = U 50 

– U 16 

U 0 

… truncation voltage (cannot be directly determined) 

IEC 60071-2: 

"For insulation co-ordination purposes, the up-and-down withstand method with 

seven impulses per group and at least eight groups is the preferred method of 

determining U 50 ". 





Up-and-down method (see HVT I, Ch. 5) 

Special case for determining U 50 

û 5 

∆U ≈ 3% of û 1 

û 4 

û 

û 3 

û 2 

û 1 

Count starts here 

breakdown 

no breakdown 

n 

General procedure see next slides… 





[IEC 60060-1] 

See slide before! 

Examples see next slides…. 





Up-and-down method – Withstand procedure with m = 7 and n = 8 

û 5 

∆U ≈ 3% of û 1 

û 4 

û 

û 3 

û 2 

û 1 


group of 7 impulses with at least one disruptive discharge 

group of 7 impulses with no disruptive discharge 

n 





Up-and-down method – Discharge procedure with m = 7 and n = 8 

û 5 

∆U ≈ 3% of û 1 

û 4 

û 

û 3 

û 2 

û 1 


group of 7 impulses with no withstand 

group of 7 impulses with at least one withstand 

n 





Replacing the Normal (Gaussian) by a Weibull distribution 

Disruptive discharge probability described by a Gaussian cumulative frequency distribution: 

1 2 

1 

y 

2 

P( U ) = − 

e dy 

2π 

∫ 

where 

x 

−∞ 

x = ( U −U50) / Z 

U 50 

… 50% discharge voltage (P(U 50 

) = 0.5) 

Z … conventional deviation 

In order to reflect the real physical behavior, this function has to be truncated at 

U 0 

= U 50 

– 3Z or 

U 0 

= U 50 

– 4Z 

(No discharge can occur at voltages below U 0 

!) 






Arguments for replacing the Gaussian by the Weibull distribution: 

• the truncation value U 0 

is mathematically included in the Weibull expression; 

• the function is easily evaluated by pocket calculators; 

• the inverse function U = U(P) can be expressed mathematically and is easily evaluated 

by pocket calculators; 

• the modified Weibull expression is defined by the same parameters characterizing the 

truncated Gaussian expression: U 50 

, Z and U 0 

; 

• the disruptive discharge probability function of several identical insulations in parallel has 

the same expression as that of one insulation and its characteristics can be easily 

determined from those of the single insulation. 






General expression for Weibull distribution: 

⎛U 

−δ 

⎞ 

−⎜ ⎟ 

⎝ β ⎠ 

P( U ) = 1− e where 

γ 

δ … truncation value 

β … scale parameter 

γ … shape parameter 

Modification for description of discharge probability of an insulation with a truncated 

discharge probability: 

δ = 

U50 

− 

NZ 

β = 

γ 

NZ(ln 2) 

1 

− 

γ 

1 

⎛ 

⎞ 

⎛ 

⎞ 

⎜ γ 

U − U50 

+ NZ ⎟ ⎜ ( U − U50 

+ NZ )(ln 2) ⎟ 

−⎜ 

1 

⎟ 

−⎜ 

⎟ 

− 

NZ 

⎛U − U50 

+ NZ ⎞ 

⎜ γ 

− 

NZ (ln 2) 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎝ ⋅ ⎠ ⎝ ⎠ ⎝ NZ ⎠ 

P( U ) = 1− e = 1− e = 1− 

e 

γ 

(N may be 3 or 4) 

γ 

ln 2 

γ 

γ 

50 ⎛ U −U 

U −U 

50 ⎞ 

50 

⎛ U −U 

⎞ ⎛ ⎞ 

− ⎜1+ ⎟ ln 2 1+ 

⎜1+ 

⎟ 

⎝ NZ ⎠ 

⎜ ⎟ 

⎝ NZ ⎠ 

⎝ NZ ⎠ 

− ln 2 

( e ) 

= 1− e = 1− = 1− 

0.5 

γ 






P( U ) = 1− 

0.5 

Condition: 50 

⎛ U −U 

⎜1+ 

⎝ NZ 

50 

γ 

⎞ 

⎟ 

⎠ 

P( U − Z) = 0.16 

Solving the equation to γ: 

⎛ U 

⎜1+ 

⎝ 

−Z −U 

50 50 

ZN ⎠ 

1− 0.5 = 0.16 

⎛ 1 ⎞ 

⎜1− 

⎟ 

⎝ N ⎠ 

γ 

⎞ 

⎟ 

1− 0.5 = 0.16 

γ 

γ 

⎛ 1 ⎞ 

⎜1− 

⎟ 

N 

⎝ ⎠ 

0.5 = 1− 

0.16 

γ 

⎛ ⎛ 1 ⎞ 

⎜1− 

⎟ 

⎞ 

⎝ N ⎠ 

ln 0.5 = ln 1− 

0.16 

⎜ ⎟ 

⎝ ⎠ 

( ) 






γ 

⎛ 1 ⎞ 

⎜1− ⎟ ln 0.5 = ln 1− 

0.16 

⎝ N ⎠ 

( ) ( ) 

γ 

⎛ 1 ⎞ 

⎜1− ⎟ = 

⎝ N ⎠ 

( − ) 

ln ( 0.5) 

ln 1 0.16 

( − ) 

ln ( 0.5) 

γ 

⎛ 1 ⎞ ⎛ ln 1 0.16 ⎞ 

ln ⎜1− ⎟ = ln ⎜ ⎟ 

⎝ N ⎠ ⎝ ⎠ 

( − ) 

ln ( 0.5) 

⎛ 1 ⎞ ⎛ ln 1 0.16 ⎞ 

γ ln ⎜1− ⎟ = ln ⎜ ⎟ 

⎝ N ⎠ ⎝ ⎠ 

γ = 

( − ) 

ln ( 0.5) 

⎛ ln 1 0.16 

ln ⎜ 

⎝ 

⎛ 1 ⎞ 

ln ⎜1− 

⎟ 

⎝ N ⎠ 

⎞ 

⎟ 

⎠ 






γ = 

( − ) 

ln ( 0.5) 

⎛ ln 1 0.16 

ln ⎜ 

⎝ 

⎛ 1 ⎞ 

ln ⎜1− 

⎟ 

⎝ N ⎠ 

⎞ 

⎟ 

⎠ 

Assuming that U 0 

= U 50 

– 4Z N = 4 

( − ) 

ln ( 0.5) 

( − ) 

⎛ ln 1 0.16 ⎞ 

ln ⎜ 

⎟ 

γ = 

⎝ 

⎠ 

= 4,80 ≈ 5 

ln 1 0.25 

(reasonably accurate) 






γ 

50 U −U 

50 

⎛ U −U 

⎞ ⎛ ⎞ 

⎜1+ ⎟ ⎜1+ 

⎟ 

⎝ NZ ⎠ ⎝ 4Z 

⎠ 

P( U ) = 1− 0.5 = 1− 

0.5 

5 

With x = (U – U 50 

)/Z: 

⎛ ⎞ 

⎜1+ 

⎟ 

⎝ 4 ⎠ 

P ( U ) = 1 − 0.5 

x 

5 

Modified Weibull flashover probability 

Typical values for Z (if more accurate data are missing): 

For lightning impulses: Z = 0.03 U 50 [Z] = kV 

For switching impulses: Z = 0.06 U 50 

And for U 10 

, resulting from the distribution function: 

U 10 = U 50 – 1.3 Z see also chapter 3 






[IEC 60071-2] 





Many insulations in parallel 

Question: if the probability of flashover of one insulator at U is P(U), what is the probability 

P'(U) of M of these insulators connected in parallel to flashover? 

P(U) P 1 

(U) P 2 

(U) P 3 

(U) P 4 

(U) P 5 

(U) ........ P M 

(U) 

P‘(U) = ? 






Applying the rules of statistics: 

Probability of flashover of one single insulator: P(U) 

If many insulators of the same individual flashover probability are connected in parallel 

and one is looking for the probability of a flashover of one of them, this cannot be 

calculated by an addition of the individual probabilities. (Note: if this was the case, 100 

parallel insulators of 10% flashover probability (P(U) = 0.1) would have a probability of 

P'(U) = 10 to flashover, which is mathematical nonsense.) 

Solution: consider the probability of withstand: W(U) = 1 – P(U) 

The probability that a number of M insulators withstands at the same time can be 

calculated by multiplication according to the rules of statistics: 

W total 

(U) = W 1 

(U) · W 2 

(U) · W 3 

(U) · ….. · W M 

(U) = W(U) M = [1-P(U)] M 

The probability P'(U) that one out of M insulators flashes over is equal to the probability 

that not all insulators withstand at the same time, thus: 

[ P U ] 

P′ ( U ) = 1− 1 − ( ) M 






[ P U ] 

P′ ( U ) = 1− 1 − ( ) M 

⎛ ⎞ 

⎜1+ 

⎟ 

⎝ 4 ⎠ 

P ( U ) = 1 − 0.5 

x 

5 

5 

⎛ x ⎞ 

M ⎜1+ 

⎟ 

P ( U ) 1 0.5 

⎝ 4 ⎠ 

′ = − Flashover probability of M parallel insulations 

Introducing the normalized variable x M 

= (U – U 50M 

)/Z M 

: 

P′ ( U ) = 1− 

0.5 

⎛ 

⎜1+ 

⎝ 

x M 

4 

⎞ 

⎟ 

⎠ 

5 

Comparison of both equations yields: 

x 

x 

+ = M ⎛ ⎜ + 

⎞ ⎟ 

4 ⎝ 4 ⎠ 

M 5 

1 1 






x 

x 

+ = M ⎛ ⎜ + 

⎞ ⎟ 

4 ⎝ 4 ⎠ 

M 5 

1 1 

Replacing x and x M 

by their extended definitions: 

x = (U – U 50 

)/Z 

x M 

= (U – U 50M 

)/Z M 

and because: 

U 50 

– 4Z = U 50M 

– 4Z M 

= U 0 

Z 

M 

Z 

= U 

5 

50M 

= U50 − 4Z 

⎜1− 

5 

M 

⎛ 

⎝ 

1 

M 

⎞ 

⎟ 

⎠ 






Z 

M 

Z 

= U 

5 

50M 

= U50 − 4Z 

⎜1− 

5 

M 

⎛ 

⎝ 

1 

M 

⎞ 

⎟ 

⎠ 

Example 1: 

For M = 200: 

U 50(200) 

= U 50 

– 2.6 Z 

U 10(200) 

= U 50(200) 

– 1.3 Z 200 

= U 50 

– 3.1 Z 

Example 2: 

M = 100, U 50 

= 1600 kV, Z = 100 kV 

Z M 

= 39.8 kV, U 50M 

= 1359.2 kV 






Z 

M 

Z 

= U 

5 

50M 

= U50 − 4Z 

⎜1− 

5 

M 

⎛ 

⎝ 

1 

M 

⎞ 

⎟ 

⎠ 

Example 2, continued: 

[IEC 60071-2] 






Note: These values can 

directly be obtained from this 

equation: 

[ P U ] 

P′ ( U ) = 1− 1 − ( ) M 

see Example 1: 

U 50(200) 

= U 50 

– 2.6 Z 

[IEC 60071-2] 






… also relevant for apparatus design 

Grading capacitors 

Breaking chambers 

Closing resistors 

three parallel external insulations 

one external insulation 




Insulation Strength in Air 

Factors influencing the dielectric strength of the insulation: 

• magnitude, shape, duration and polarity of the applied voltage 

• electric field distribution in the insulation 

• homogeneous or non-homogeneous electric field 

• electrodes adjacent to the considered gap and their potential 

• type of insulation 

• gaseous 

• liquid 

• solid 

• combination of two or all of them 

• impurity content and the presence of local inhomogeneities 

• physical state of the insulation 

• temperature 

• pressure 

• other ambient conditions 

• mechanical stress 

• history of the insulation (aging, damage) 

• chemical effects 

• conductor surface effects 

Covered by equations U 50RP = f(d) 

where 

U 50RP … 50% probability 

breakdown voltage of 

a rod-plane-configuration 

d … gap spacing 

Covered by equations U 50 = f(U 50RP , K) 

where 

K … gap factor 

K is a factor indicating how much higher the 

electrical strength of a particular electrode 

configuration is in comparison with the rod-planeconfiguration 

(which gives least dielectric strength); 

factors K were experimentally found for standard 

switching impulse voltage stress 





Gap factors (Table G.1 of IEC 60071-2) 




















Gap factors [Electra No. 29 (1973), 

pp. 29-44)] 

Electrode configuration 

Rod-plane 

K 

1.0 

Increasing dielectric strength 

Rod-structure (under) 

Conductor-plane 

Conductor-window 

Conductor-structure (under) 

Rod-rod (h = 6 m, under) 

Conductor-structure 

(over and laterally) 

Conductor-rope 

(under and laterally) 

Conductor-crossarm (end) 

Conductor-rod (h = 3 m, under) 

Conductor-rod (h = 6 m, under) 

1.05 

1.15 

1.20 

1.30 

1.30 

1.35 

1.40 

1.55 

1.65 

1.90 





Insulation response to power-frequency voltages (IEC60071-2, Annex G) 

U 

1.2 

50RP 

= 750⋅ 2 ⋅ ln(1 + 0.55 ⋅ d ) 

U 

with 

U 50RP 

… crest value in kV 

d in m; d ≤ 3 m 

= U 

50 50RP 

50 50RP 

exact for d < 1 m; conservative for 1 m ≤ d ≤ 2 m 

( 

2 

1.35 0.35 ) 

U = U K − K for d > 2 m 

≈ 300 kV/m (r.m.s. value) 

U 

≈ 0.9 ⋅U 

0 50 

(assuming U 0 

= U 50 

- 4Z and Z = 0.03 U 50 

) 

• influence of rain in an air gap negligible; but for insulators to be considered! 

• pollution for insulators to be considered! 

• altitude correction required! 





Insulation response to slow-front overvoltages (IEC60071-2, Annex G) 

U50RP = 1080⋅ln(0.46 ⋅ d + 1) 

with 

U 50RP 

… in kV; for positive polarity at most critical front-time (see Ch. 3) 

d in m; d ≤ 25 m 

U 

= ⋅d 

50RP 

500 

0.6 

with 

U 50RP 

… in kV; for positive polarity standard switching impulse voltage (see Ch. 3) 

d in m; d ≤ 25 m 

U 

= KU Note: for K ≥ 1.45 U 50neg. may become lower than U 50pos. 

50 50RP 






U 

≈ 0.75⋅ U (assuming U 0 

= U 50 

- 4Z and Z = 0.06 U 50 

) 

0 50 

• influence of rain in an air gap negligible; but for insulators to be considered! 







For phase-to-phase insulation similar gap factors as for phase-to-earth insulation 

can be applied. 

But: the influence of negative and positive components has to be taken into 

account by a factor α: 

peak negative component 

α 

= sum of peak negative and positive components 

[IEC 60071-2] 





Insulation response to fast-front overvoltages (IEC60071-2, Annex G) 

U 

= ⋅d 

50RP 

530 

i.e. linear increase with gap spacing 

with 

U 50RP 

… in kV; for positive polarity 

d in m; d ≤ 10 m 

Note: for negative LI voltages, dielectric 

strength is higher and increases nonlinearly 

with gap spacing! 

The gap factors K (found for SI voltages!) cannot be directly applied. 

From experimental investigations: 

U = K ⋅U 

+ 

50 ff 50RP 

K 

+ ff 

= 0.74 + 0.26K 

with 

K + ff 

… fast-front overvoltage gap factor 

for positive polarity 

K …... gap factor for SI voltage 

according to tables 





Insulation response to fast-front overvoltages (IEC60071-2, Annex G) 

Estimation of negative line insulator flashover voltage (in order to determine lightning 

overvoltages impinging on a substation: 

U 

= ⋅d 

50 neg, line insulator 

700 

• influence of insulators to be considered particularly for range II! 

• less influence from long insulators without metallic parts (long rod, composite, station 

post) than for cap-and-pin insulators 


• virtually no influence of rain neither for air gaps nor for insulators 

Conventional deviation: 

Z ≈ 0.03·U 50 for air gaps and positive polarity 

Z ≈ 0.05·U 50 for air gaps and negative polarity 

Z ≈ (0.05 … 0.09)·U 50 across insulators 





Independent from the theoretical and empirical background given so far, IEC 60071-2 

offers tables on minimum clearances in air (Annex A). Not all values of these tables 

can be derived from above equations, as they additionally take into account 

• withstand values instead of U 50 

-values 

• feasibility 

• economy 

• experience 

• average influence of environmental conditions (pollution, rain, insects, …) 





[IEC 60071-2] 




Procedure for Insulation Coordination in Four Steps 

Next slide 

Flow chart of IEC 60071-1 

(Figure 1) 

we are here! 

Sorry, no time this year 

[IEC 60071-1] 





Performance criterion IEC 60071-2, Cl. 3.2 

According to definition 3.22 of IEC 71-1, the performance criterion to be required from the 

insulation in service is the acceptable failure rate (R a 

). 

The performance of the insulation in a system is judged on the basis of the number of 

insulation failures during service. Faults in different parts of the network can have different 

consequences. For example, in a meshed system a permanent line fault or an unsuccessful 

reclosure due to slow-front surges is not as severe as a busbar fault or corresponding faults 

in a radial network. Therefore, acceptable failure rates in a network can vary from point to 

point depending on the consequences of a failure at each of these points. 

Examples for acceptable failure rates can be drawn from fault statistics covering the existing 

systems and from design projects where statistics have been taken into account. For 

apparatus, acceptable failure rates R a 

due to overvoltages are in the range of 0.001/year 

up to 0.004/year depending on the repair times. For overhead lines acceptable failure rates 

due to lightning vary in the range of 0.1/100 km/year up to 20/100 km/year (the greatest 

number being for distribution lines). Corresponding figures for acceptable failure rates due to 

switching overvoltages lie in the range 0.01 to 0.001 per operation. Values for 

acceptable failure rates should be in these orders of magnitude. 





Altitude correction 

In general, withstand or breakdown voltages must be corrected for air density (pressure, 

temperature) and absolute humidity. 

Temperature and absolute humidity tend to cancel out each other. Thus correction is 

mainly required for pressure, which has its strongest influence in the altitude of 

installation. 

Therefore, in the procedure of insulation coordination, an altitude correction must be 

performed in the step from the coordination withstand voltage U cw 

to the required 

withstand voltage U rw 

. 

− 

H 

Air density vs. altitude: 8150 

δ = e (regression of experimental data) 

where H … altitude above sea level in m 

Voltage correction depends on voltage shape (the kind of pre-discharges), thus a voltagedependant 

factor m is introduced: 

k = e 

H 

−m 

8150 





Altitude correction 

m = 1 for LI voltage 

m = acc. to Figure 9 for SI voltage 

m = 1 for short-time alternating voltage 

m = 0.5 for long-time alternating 

voltage and tests under pollution 

Final altitude correction factor: 

K 

a 

1 

= = 

k 

e 

H 

m 

8150 

[IEC 60071-2] 

approx. 1.3% per 100 m (for m = 1) 






(Figure 1) 

we are here! 

[IEC 60071-1] 





From U cw 

U rw 

Urw = Ka ⋅ Ks ⋅Ucw 

where K a 

… altitude correction factor 

K s 

… safety factor, taking into account: 

• differences in equipment assembly 

• dispersion in product quality 

• quality of installation 

• aging effects 

• other unknown influences 

Internal insulation: 

• no altitude correction (K a = 1) 

• K s = 1.15 

External insulation: 

• K a = f(m,H) = exp(m·H/8150) 

• K s = 1.05 





Example (for m = 1) 

2000 

From U cw 

U rw 

1500 

1000 

U/kV 

External insulation: U rw 

= 1.05·exp(H/8150)·1000 kV 

Internal insulation: U rw 

= 1.15·1000 kV = 1150 kV 

Assumption: U cw 

= 1000 kV 

≈ 

0 1000 2000 3000 H/m 4000 






(Figure 1) 

we arrived here! 

[IEC 60071-1] 




Insulation Coordination 

For calculation examples, see IEC 60071-2, Annex H! 




Insulation Coordination

From U - Fachgebiet Hochspannungstechnik

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