G. Ziegler. Digital Differential Protection

G. Ziegler, 10/2005 page 19

Differential Protection Symposium

Belo HorizonteNovember 2005

Id >

I

R0

k

1

k

2

I

Op

I

Res

Digital differential protection: Relay characteristic

Settings:

•Pick-up value Id >

•slope k

1

•slope k

2

with footing point I

R0

2

I

1

IIΣ

Res

I +==

2

I

1

I

Op

I +=Σ= I

∆I

Positive current polarity

G. Ziegler, 10/2005 page 20

Differential Protection Symposium

Belo HorizonteNovember 2005

B

k

kI

S

⋅

+

2

1

2

0

B

k

kI

S

⋅

−

+

2

1

2

0

2

0S

I

B

0S

I

[

]

100/%k

21

II

ΔI

+

=

21

III +=Σ

1

I

2

I

Internal

faults

slope

∆I / ΣI und I

1

/ I

2

diagram

(

)

BII

S

IIIkII

>+

−+>+

21

0

2121

1

2

G. Ziegler, 10/2005 page 21

Differential Protection Symposium

Belo HorizonteNovember 2005

Polar diagram of differential protection

k

−

+

−

1

k

+

−

1

2

I

=β

∆Ι

1

I

2

I

Protection

object

(k=0,5)













1

2

Re

I













1

2

Im

I

with

k

-1

1

or k

I

1

I

1

2

1

2

1

2

=β

>

β

β+

>

−

+

−⋅>+

2121

IIkII

-1

+1

Restraint area

β-Plane (remote/local current)

G. Ziegler, 10/2005 page 22

Differential Protection Symposium

Belo HorizonteNovember 2005

(

)

11

2

1

2

2121

I

B

I

1k

I

1

or

BIIkII

+













+⋅>+

++⋅>+

ϕj

1

2

e

I

B/I

1

= 0,3a1(Φ): k=0,3

a2(Φ): k=0,6

a3(Φ): k=0,8

Polar diagram of digital differential protection:

Basic pick-up value B > 0

120

150

180

210

240

270

300

0

330

30

60

90

510

ϕ

G. Ziegler, 10/2005 page 23

Differential Protection Symposium

Belo HorizonteNovember 2005

Composed current during

symmetrical 3-ph fault or load

Mixing

transformer

2

1

3

I

L1

I

L3

I

E

I

L1

I

L2

I

L3

I

L3

I

L3

2 ·I

L1

I

L1

I

L2

I

L3

7SD503: I

M

= 100 mA

7SS600: I

M

= 100 mA

7SD502: I

M

= 20 mA

⇒ composed current (vector sum)

3PhM

I3

w

1

I

−

⋅⋅=

L3

I4

L2

I3

L1

I5

E

I3

L3

I1

L1

I2

M

I

⋅+⋅+⋅=

I

M

Mixing transformer of composed current differential protection

G. Ziegler, 10/2005 page 24

Differential Protection Symposium

Belo HorizonteNovember 2005

Fault type

Composed current

Per unit composed current

related to 3-phase

symmetrical current

L1-EI

ML1-E

= 5 I

L1

I

L2

= I

L3

= 0

I

ML1-E

/ I

M

= 5 /

√ 3 = 2.9

L2-EI

ML2-E

= 3 I

L2

I

L1

= I

L3

= 0

I

ML2-E

/ I

M

= 3 /

√

3 = 1.73

L3-EI

ML3-E

= 4 I

L3

I

L1

= I

L2

= 0

I

ML3-E

/ I

M

= 4 /

√

3 = 2.3

L1-L2I

ML12

= 5 I

L1

+ 3 I

L2

I

L3

= 0

I

ML12

/ I

M

= 2 /

√

3 = 1.15

L2-L3I

ML23

= 3 I

L2

+ 4 I

L3

I

L1

= 0

I

ML23

/ I

M

= 1 /

√

3 = 0.58

L1-L3I

ML13

= 5 I

L1

+ 4 I

L3

I

L2

= 0

I

ML13

/ I

M

= 1 /

√

3 = 0.58

L1-L2-L3I

ML123

= 5 I

L1

+ 3I

L2

+ 4I

L3

|I

L1

| = |I

L2

| = |I

L3

|

I

ML123

/ I

M

=

√

3 /

√

3 = 1

Highest

sensitivity

Mixing transformer:

Pickup sensitivity of standard connection (I

M

= 2I

L1

+I

L3

+3I

E

)

G. Ziegler, 10/2005 page 25

Differential Protection Symposium

Belo HorizonteNovember 2005

I

F

L1

L2

L3

∆

I

A

B

Tripping

Tripping if k-setting < 0.6

Composed current differential protection

Behaviour during cross country fault (isolated/compensated network)

L1 internal und L3 external:

I

M-A

= 5⋅I

F

-4⋅I

F

= 1⋅I

F

und I

M-B

= + 4⋅I

F

∆I = |I

M-A

+ I

M-B

| = 5I

F

und ΣI = |I

M-A

| + |I

M-B

| = 5⋅I

F

∆I/ΣI = 5/5 = 1.0

L2 internal, L3 external

I

M-A

= -1⋅I

F

und I

M-B

= + 4⋅I

F

∆I = |I

M-A

+ I

M-B

| = 3 ⋅I

F

und ΣI = |I

M-A

| + |I

M-B

| = 5⋅I

F

∆I/ΣI = 3/5 = 0.6

G. Ziegler, 10/2005 page 26

Differential Protection Symposium

Belo HorizonteNovember 2005

21

32

1

3

I

M1

=2·I

L1

+ 1·I

L3

+ 3 ·I

E

=2 ·(–I

F

) + 1 ·(–I

F

)+ 3·0= –3·I

F

I

M2

=2·I

L1

+ 1·I

L3

+ 3 ·I

E

=2·I

F

+ 1 ·I

F

+ 3·3I

F

= +12 ·I

F

I

Op

=| I

M1

+ I

M2

| = 9 ·I

F

I

Res

=|I

M1

| +|I

M2

| = 15 ·I

F

k=9/15 = 0.6

Fault L2-E

I

Res

I

Op

k=0.5

Faults in other phases:

Fault L1-E: I

Op

= (3+12) ·I

F

= 15 ·I

F

,

I

Res

= (3+12) ·I

F

= 15 ·I

F

,

Fault L3-E: I

Op

= (0+12) ·I

F

= 12 ·I

F

,

I

Res

= (0+12) ·I

F

= 12 ·I

F

,

k=1

Composed current differential protection

Through current stabilisation with unsymmetrical earthing conditions

G. Ziegler, 10/2005 page 27

Differential Protection Symposium

Belo HorizonteNovember 2005

I

SC

I

SC

I

SC

I

SC

(

)

SC

i

CT

R

L

R

CT

E ⋅+=

−1

(

)

SC

i

CT

R

L

R

CT

E ⋅+=

−2

R

CT

R

CT

(

)

SC

i

CT

R

L

R

CT

E ⋅+⋅=

−

2

1

(

)

SC

i

CT

R

L

R

U ⋅+=

0=

R

U

Behaviour during external fault

with ideal current transformers

with CT saturation

U

R

High impedance differential protection: Principle

G. Ziegler, 10/2005 page 28

Differential Protection Symposium

Belo HorizonteNovember 2005

Given:n = 8 feeders

r

CT

= 600/1 A

U

KN

= 500 V

R

CT

= 4 Ohm

I

mR

= 30 mA (at relay pick-up value)

Pick-up sensitivity:

(

)

mR

In

Var

I

uppickR

I

CT

r

F

I ⋅++

−−

⋅=

−min

()

03.0805.002.0

1

600

min

⋅++⋅=

−F

I

(

)

%31186

min

⋅=

−

AI

F

R

L

= 3 Ohm (max.)

I

R-pick-up.

= 20 mA (fixed value)

R

= 10 kOhm

I

var

= 50 mA (at relay pick-up value)

Stability:

uppickR

I

SW

R

L

R

CT

r

throughF

I

−−

⋅

+

⋅<

−− max

02.0

43

000,10

1

600

max

⋅

+

⋅<

−−throuhF

I

n

IkA

throughF

I ⋅=<

−−

2817

max

High impedance differential protection:

Calculation example (busbarprotection)

G. Ziegler. Digital Differential Protection

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