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Page 1: Current Transformers ALSTOM

GRIDTechnical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.

Page 2: Current Transformers ALSTOM

GRIDTechnical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.

Current Transformers

Page 3: Current Transformers ALSTOM

s3

Current Transformer Function

Reduce power system current to lower value for measurement.

Insulate secondary circuits from the primary.

Permit the use of standard current ratings for secondary equipment.

REMEMBER :

The relay performance DEPENDS on the C.T which drives it !

Page 4: Current Transformers ALSTOM

s4

Instrument Transformer Standards

IEC IEC 185:1987 CTs

IEC 44-6:1992 CTs

IEC 186:1987 VTs

EUROPEAN BS 7625 VTs

BS 7626 CTs

BS 7628 CT+VT

BRITISH BS 3938:1973 CTs

BS 3941:1975 VTs

AMERICAN ANSI C51.13.1978 CTs and VTs

CANADIAN CSA CAN3-C13-M83 CTs and VTs

AUSTRALIAN AS 1675-1986 CTs

Page 5: Current Transformers ALSTOM

s5

Current Transformer Construction

BAR PRIMARY WOUND PRIMARY

Primary

Secondary

Page 6: Current Transformers ALSTOM

s6

Bar Primary Type Current Transformer(s)

Primary Conductor

Ring Type Current Transformer

Primary Insulation

Core

Secondary Winding

Page 7: Current Transformers ALSTOM

s7

Typical Protection Bar-Primary Current Transformer

RELAY

1A ?1000A ?

1000 turns sec. ?

Insulation covered wire, giving inter-turn insulation & secondary to core insulation

Generator, or system voltage source

‘Feeder’ or ‘Bus-bar’ forming 1 turn of primary circuit

Insulation to stop flash-over from HV primary to core & secondary circuit

Laminated ‘strip’ wound steel toroidal core

Page 8: Current Transformers ALSTOM

s8

Polarity

P2P1

Is

S2S1

Ip

Inst. Current directions :-

P1 P2 S1 S2 Externally

Page 9: Current Transformers ALSTOM

s9

Flick Test

P1

S1

+

V

-

S2

P2

IsIp

FWD kick on application,

REV kick on removal of test lead.

Battery (6V) + to P1AVO +ve lead to S1

Page 10: Current Transformers ALSTOM

s10

Directional Protection

P2P1

S2

67

S1

IOP

VPOL

IOP = Current direction for operation

P2P1

S3S2

IFP

IFS

IFP

IFS

Operation for Forward Faults No Operation for Reverse Faults

Page 11: Current Transformers ALSTOM

s11

Differential Protection (1)

P2P1 IFP1

IFS

IFS1S2S1

P1P2

S1S2

R

IFP2

IFS2

Operation for Internal Faults

Page 12: Current Transformers ALSTOM

s12

Differential Protection (2)

P2P1 IFP

IFSS2S1

P1P2

S1S2

R

IFS

Stability for External Faults

Page 13: Current Transformers ALSTOM

s13

Differential Protection (3)

IFP

IFS

S2S1 S1S2

R

Maloperation due to Incorrect Connections

IFP

IFS

2IFS

Page 14: Current Transformers ALSTOM

s14

Basic Theory

Page 15: Current Transformers ALSTOM

s15

Basic Theory (1)

IP

IS

R

1 Primary Turn N Secondary Turns

For an ideal transformer :-

PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS

IP = N x IS

Page 16: Current Transformers ALSTOM

s16

Basic Theory (2)

IP

IS

RES

For IS to flow through R there must be some potential - ES = the E.M.F.

ES = IS x R

ES is produced by an alternating flux in the core.

ES dØ dt

Page 17: Current Transformers ALSTOM

s17

Basic Theory (3)

IP

NP

NS

EKIS

ZBZCT

VO/P = ISZB = EK - ISZCT

Page 18: Current Transformers ALSTOM

s18

Basic Formulae

Circuit Voltage Required :

ES = IS (ZB + ZCT + ZL) Voltswhere :-

IS = Secondary Current of C.T. (Amperes)

ZB = Connected External Burden (Ohms)

ZCT = C.T Winding Impedance (Ohms)

ZL = Lead Loop Resistance (Ohms)

Require EK > ES

Page 19: Current Transformers ALSTOM

s19

Basic Formulae (1)

Maximum Secondary Winding Voltage :

EK = 4.44 x B A f N Volts …. 1

where :-

EK = Secondary Induced Volts

(knee-point voltage)

B = Flux Density (Tesla)

A = Core Cross-sectional Area

(square metres)

f = System Frequency (Hertz)

N = Number of Turns

Page 20: Current Transformers ALSTOM

s20

Simple Selection Example

Example Calculation :

C.T. Ratio = 2000 / 5A Max Flux Density = 1.6 T

RS = 0.31 Ohms Core C.S.A = 20 cm2

IMAX Primary = 40 kA

Find maximum secondary burden permissible if no saturation is to occur.

Solution : N = 2000 / 5 = 400 Turns

IS MAX = 40,000 / 400 = 100 Amps

From equation 1 the knee point voltage is :-

VK = (4.44 x 1.6 x 20 x 50 x 400) = 284 Volts

104

Therefore Maximum Burden = 284 / 100

= 2.84 Ohms

Hence Maximum CONNECTED burden :

2.84 - 0.31 = 2.53 Ohms

Page 21: Current Transformers ALSTOM

s21

Knee Point

Material : Cross

Flux Density (B)

Tesla (Wb/m2)

0 0.02 0.04 0.06 0.08 0.1 0.12

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

Magnetizing Force (H)Ampere-Turns / mm(Magnetic Field Strength : H, A/m)

Mag Curves

Page 22: Current Transformers ALSTOM

s22

ES

(Secondary E.M.F.)

(Magnetising Current) Ie

ES = 4.44 N f A B = Kv B

where :- Kv = 4.44 N f A

(B in Wb/m2 ; A in m2)

Ie = H . L = Ki . H

N where :-

Ki = L/N

L = mean magnetic path

in metres

H = amp. Turns / metre

Ie = Amps

Mag Curves Cont...

Page 23: Current Transformers ALSTOM

s23

Mult

iply

By K

V t

o o

bta

in

Volt

s

Multiply By Ki to obtain

Amps

B

H

KV x B = Volts (E)

Ki x H = Amps

Units : KV = E = 4.44 N f A (m2 x turns)

B

Ki = Ie = L (m / turns)

H N

Note : L = Mean Magnetic Path

Mean Magnetic Path = 2r

r = ro - ri + ri

2

ro

ri

r

Mag Curves Cont...

Page 24: Current Transformers ALSTOM

s24

Magnetisation Characteristic (1)

C.T Ratio 100/5 A connected to CDG11.

Relay Burden = 2 VA (at 10% rated I)

Problem :-

Calculate the necessary values of Kv and Ki to provide the

necessary output at ten times setting

(Assume maximum flux density before saturation = 1.6 Tesla)

Page 25: Current Transformers ALSTOM

s25

Magnetisation Characteristic (2)

(i) Bar Type Primary

CDG current setting = 10% = 0.5 A

Volts required to operate relay = 2 / 0.5 = 4 V

Therefore voltage required at 10 times setting = 10 x 4 = 40 V

Hence for a flux density of 1.6T the C.T. should have an output of at

least 40 V.

With bar primary, number of turns = 20.

Ek = 4.44 B A f N

= 4.44 x 1.6 x A x 10-4 x 20 x 50

A = 40 / 0.71 = 56.3 cm2

Assume a stacking factor of 0.92

=> Total C.S.A. = 61.2 cm2

Assume I.D. = 18cm, O.D. = 30cm & Depth = 10.2cm

KV = AN / 45 = (56.3 x 20) / 45 = 25

Ki = L / N = 24TT / 20 = 3.77 cm / turn

Page 26: Current Transformers ALSTOM

s26

Magnetisation Characteristic (3)

(ii) Wound Type Primary

Assume primary has 5 turns

- Therefore secondary has 100 turns

Ek = 4.44 B A f N

= 4.44 x 1.6 x A x 10-4 x 100 x 50

A = 40 / 3.55 = 11.26 cm2

Assume a stacking factor of 0.92

=> Total C.S.A. = 12.24 cm2

Assume I.D. = 18cm, O.D. = 30cm &

Depth = 2.04cm

KV = AN / 45 = (11.26 x 100) / 45 = 25

Ki = L / N = 24TT / 100 = 0.754 cm/turn

Page 27: Current Transformers ALSTOM

s27

Low Reactance Design

With evenly distributed winding the leakage reactance is very low and usually ignored.

Thus ZCT ~ RCT

Page 28: Current Transformers ALSTOM

s28

Knee-Point Voltage Definition

Iek

+10% Vk

+50% Iek

Vk

Exciting Current (Ie)

Exci

tin

g V

olt

age

(VS)

Page 29: Current Transformers ALSTOM

s29

C.T. Equivalent Circuit

ZbN

P1

ZCT

S1

VtZe

IsIp

Ip/NIe

Es

Ip = Primary rating of C.T. Ie = Secondary excitation current

N = C.T. ratio Is = Secondary current

Zb = Burden of relays in ohms Es = Secondary excitation voltage

(r+jx) Vt = Secondary terminal voltage

ZCT = C.T. secondary winding across the C.T. terminalsimpedance in ohms (r+jx)

Ze = Secondary excitationimpedance in ohms (r+jx)

Page 30: Current Transformers ALSTOM

s30

Phasor Diagram

IcEp

Ip/N

Ie

IsIe

Im

Es

Ep = Primary voltage Im = Magnetising current

Es = Secondary voltage Ie = Excitation current

= Flux Ip = Primary current

Ic = Iron losses (hysteresis & Is = Secondary currenteddy currents)

Page 31: Current Transformers ALSTOM

s31

Saturation

Page 32: Current Transformers ALSTOM

s32

Steady State Saturation (1)

100A

100/11A

E 1 ohm

E =1V

100A

100/11A

E10ohm E =

10V

100A

100/11A

E100 ohm

E =100V

100A

100/11A

E1000 ohm

E = ?

Page 33: Current Transformers ALSTOM

s33

Steady State Saturation (2)

Time

+V

-V

0VA

Assume :- Zero residual flux

Switch on at point A

Page 34: Current Transformers ALSTOM

s34

Steady State Saturation (3)

Time

+V

-V

0VC

Assume :- Zero residual flux

Switch on at point C

Page 35: Current Transformers ALSTOM

s35

Steady State Saturation (4)

Time

+V

-V

0VB

Assume :- Zero residual flux

Switch on at point B

Page 36: Current Transformers ALSTOM

s36

Steady State Saturation (5)

Mag CoreSaturation

+V

-V

0V

Mag CoreSaturation

Time

Page 37: Current Transformers ALSTOM

s37

Steady State Saturation (6)

Time

+V

-V

0V

+V

-V

0V

Mag CoreSaturation

Mag CoreSaturation

Output lost due to steady state saturation

Actual Output Voltage

Prospective Output Voltage

Page 38: Current Transformers ALSTOM

s38

Transient Saturation

L1R1

Z1

i1

v = VM sin (wt + )

v = VM sin (wt + )

TRANSIENT STATESTEADY

e . ) - ( sin - ) - (wt sin

e . ) - ( sin Z

V ) - (wt sin

Z

V i

1L / t1R-11

1L / t1R-

1

M

1

M1

ˆˆ

1

M1

1

11-

12

11

Z

V

R

wL tan

L w R Z -: where

ˆ

22

Page 39: Current Transformers ALSTOM

s39

Transient Saturation : Resistive Burden

FLUX

CURRENT

Primary Current Secondary

Current

Actual Flux

Mag Current

Required Flux

ØSAT

0

0 10 20 30 40 50 60 70 80

M

Page 40: Current Transformers ALSTOM

s40

CT Types

Page 41: Current Transformers ALSTOM

s41

Current Transformer Function

Two basic groups of C.T.

Measurement C.T.s

Limits well defined

Protection C.T.s

Operation over wide range of currents

Note : They have DIFFERENT characteristics

Page 42: Current Transformers ALSTOM

s42

Measuring C.T.s

B

Protection C.T.

Measuring C.T.

H

Measuring C.T.s

Require good accuracy up to approx 120% rated current.

Require low saturation level to protect instruments, thus use nickel iron alloy core with low exciting current and knee point at low flux density.

Protection C.T.s

Accuracy not as important as above.

Require accuracy up to many times rated current, thus use grain orientated silicon steel with high saturation flux density.

Page 43: Current Transformers ALSTOM

s43

CT Errors

Page 44: Current Transformers ALSTOM

s44

Current Transformer Errors

ZS

ZbZe

IsIp

Ie

Es

Ip/N

IeIq

Ie

Es

Ep

Ic’

Ip/N

Ie

IsIs

Page 45: Current Transformers ALSTOM

s45

Current Transformer Error

Transformer Error vs Primary Current

Error

RatedCurren

t

Primary Current

Phase Error

Current Error

Errors can be reduced by :-

1. Using better quality magnetic material

2. Shortening the mean magnetic path

3. Reducing the flux density in the core

Page 46: Current Transformers ALSTOM

s46

Current Transformer Errors

Current Error Definition.

Error in magnitude of the secondary current, expressed as a percentage, given by :-

Current error % = 100 (kn IS - Ip)

Ip

kn = Rated Transmission Ratio

Ip = Actual Primary Current

Is = Actual Secondary Current

Current Error is :-

+ve : When secondary current is HIGHER than the rated

nominal value.

-ve : When secondary current is LOWER than the rated

nominal value.

Page 47: Current Transformers ALSTOM

s47

Current Transformer Errors

Phase Error Definition.

The displacement in phase between the primary and secondary current vectors, the direction of the vectors being chosen so the angle is zero for a perfect transformer.

Phase Error is :-

+ve : When secondary current vector LEADS the

primary current vector.

-ve : When secondary current vector LAGS the

primary current vector.

Page 48: Current Transformers ALSTOM

s48

Current Transformer Ratings

Page 49: Current Transformers ALSTOM

s49

Current Transformer Ratings (1)

Rated Burden

Value of burden upon which accuracy claims are based Usually expressed in VA Preferred values :-

2.5, 5, 7.5, 10, 15, 30 VA

Continuous Rated Current

Usually rated primary current

Short Time Rated Current

Usually specified for 0.5, 1, 2 or 3 secs No harmful effects Usually specified with the secondary shorted

Rated Secondary Current

Commonly 1, 2 or 5 Amps

Page 50: Current Transformers ALSTOM

s50

Current Transformer Ratings (2)

Rated Dynamic Current

Ratio of :-

IPEAK : IRATED

(IPEAK = Maximum current C.T. can withstand without suffering any damage).

Accuracy Limit Factor - A.L.F. (or Saturation Factor)

Ratio of :-

IPRIMARY : IRATED

up to which the C.T. rated accuracy is maintained.

e.g. 200 / 1A C.T. with an A.L.F. = 5 will maintain itsaccuracy for IPRIMARY < 5 x 200 = 1000 Amps

Page 51: Current Transformers ALSTOM

s51

Choice of Ratio

Clearly, the primary rating

IP normal current in the circuit

if thermal (continuous) rating is not to be exceeded.

Secondary rating is usually 1 or 5 Amps (0.5 and 2 Amp are also used).

If secondary wiring route length is greater than 30 metres - 1 Amp secondaries are preferable.

A practical maximum ratio is 3000 / 1.

If larger primary ratings are required (e.g. for large generators), can use 20 Amp secondary together with interposing C.T.

e.g. 5000 / 20 - 20 / 1

Page 52: Current Transformers ALSTOM

s52

Current Transformer Designation

Class “P”

Specified in terms of :-

i) Rated burdenii) Class (5P or 10P)iii) Accuracy limit factor (A.L.F.)

Example :-

15 VA 10P 20

To convert VA and A.L.F. into useful volts

Vuseful VA x ALF

IN

Page 53: Current Transformers ALSTOM

s53

BS 3938

Classes :- 5P, 10P. ‘X’

Designation (Classes 5P, 10P)

(Rated VA) (Class) (ALF)

Multiple of rated current (IN) up to which declared

accuracy will be maintained with rated burden connected.

5P or 10P.

Value of burden in VA on which accuracy claims are based.

(Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA)

ZB = rated burden in ohms

= Rated VA IN

2

Page 54: Current Transformers ALSTOM

s54

Protection Current Transformers

Table 3 - Limits of Error for Accuracy Class 5P and 10P

Phase Displacement atrated primary current

AccuracyClass

Current Error atrated primarycurrent (%) Minutes Centiradians

Composite Error(%) at rated

accuracy limitprimary current

5P

10P

1

3

60 1.8 5

10

Page 55: Current Transformers ALSTOM

s55

Interposing CT

Page 56: Current Transformers ALSTOM

s56

Interposing CT

NP NS ZB

ZCT

LINECT

Burden presented to line CT

= ZCT + ZB x NP2

NS2

Page 57: Current Transformers ALSTOM

s57

‘Seen’ by main ct :- 0.1 + (1)2 (2 x 0.5 + 0.4 + 1) = 0.196 (5)

Burden on main ct :- I2R = 25 x 0.196 = 4.9VA

Burden on a main ct of required ratio :-

Total connected burden = 2 x 0.5 + 1 = 2Connected VA = I2

R = 2

The I/P ct consumption was about 3VA.

NEG. 1A

0.10.1 0.4 1VA @ 1A 1.0

R

0.55A

500/5

0.5

R

1.0500/1

Page 58: Current Transformers ALSTOM

s58

Current Transformer Designation

Page 59: Current Transformers ALSTOM

s59

Current Transformer Designation

Class “X”

Specified in terms of :-

i) Rated Primary Current

ii) Turns Ratio (max. error = 0.25%)

iii) Knee Point Voltage

iv) Mag Current (at specified voltage)

v) Secondary Resistance (at 75°C)

Page 60: Current Transformers ALSTOM

s60

Choice of Current Transformer

Instantaneous Overcurrent Relays Class P Specification

A.L.F. = 5 usually sufficient

For high settings (5 - 15 times C.T rating) A.L.F. = relay setting

IDMT Overcurrent Relays Generally Class 10P

Class 5P where grading is critical

Note : A.L.F. X V.A < 150

Differential Protection Class X Specification

Protection relies on balanced C.T output

Page 61: Current Transformers ALSTOM

s61

Selection Example

Page 62: Current Transformers ALSTOM

s62

Burden on Current Transformers

RCT

RCT

RCT

IF

IFRLRL RLRL

Rr Rr RrRr

RCT

RCT

RCT

IF

IF RLRL RLRL

Rr Rr RrRr

1. Overcurrent : RCT + RL + Rr 2. Earth : RCT + 2RL + 2Rr

Page 63: Current Transformers ALSTOM

s63

Overcurrent Relay VK Check

Assume values : If max = 7226 A RCT = 0.26 C.T = 1000 / 5 A Rr = 0.02

7.5 VA 10P 20 RL = 0.15

Check to see if VK is large enough :

Required voltage = VS = IF (RCT + Rr + RL)

= 7226 x 5 (0.26 + 0.02 + 0.15) = 36.13 x 0.43 = 15.54 Volts 1000

Current transformer VK approximates to :-

VK ~ VA x ALF + RCT x IN x ALF In

= 7.5 x 20 + 0.26 x 5 x 20 = 56 Volts 5

VK > VS therefore C.T VK is adequate

Page 64: Current Transformers ALSTOM

s64

Earth Fault Relay VK Check

Assume values : As per overcurrent.

Note For earth fault applications require to be able to pass 10 x relay setting.

Check to see if VK is large enough : VK = 56 Volts

Total load connected = 2RL + RCT + 2Rr

= 2 x 0.15 + 0.26 + 2 x 0.02

Maximum secondary current= 56 = 93.33A

0.6

Typical earth fault setting = 30% IN

= 1.5A

Therefore C.T can provide > 60 x setting

C.T VK is adequate

Page 65: Current Transformers ALSTOM

s65

Voltage Transformers

Page 66: Current Transformers ALSTOM

s66

Voltage Transformers

Provides isolation from high voltages

Must operate in the linear region to prevent accuracy problems - Do not over specify VT

Must be capable of driving the burden, specified by relay manufacturer

Protection class VT will suffice

Page 67: Current Transformers ALSTOM

s67

Typical Working Points on a B-H Curve

Flux Density‘B’

Magnetising ForceAT/m

Cross C.T.’s & Power Transformers

Saturation

V.T.’s

Protection C.T. (at full load)‘H’

1000 2000 3000

1.6

1.0

0.5

Tesla

Page 68: Current Transformers ALSTOM

s68

Types of Voltage Transformers

Two main basic types are available:

Electromechanical VT`s Similar to a power transformer May not be economical above 132kV

Capacitor VT`s (CVT) Used at high voltages Main difference is that CVT has a capacitor

divider on the front end.

Page 69: Current Transformers ALSTOM

s69

Electromagnetic Voltage Transformer

ZB(burden)

NP / NS

= Kn

VP

RP

IP

LP

LM

IM IC

Re VSEP = ES

Ie

RS

IS

LS

Page 70: Current Transformers ALSTOM

s70

Basic Circuit of a Capacitor V.T.

C1

C2

VP

VC2 Vi

L

T

VS

ZB

Page 71: Current Transformers ALSTOM

s71

Ferro-resonance

The exciting impedance of auxiliary transformer T and the capacitance of the potential divider form a resonant circuit.

May oscillate at a sub normal frequency

Resonant frequency close to one-third value of system frequency

Manifests itself as a rise in output voltage, r.m.s. value being 25 to 50 per cent above normal value

Use resistive burden to dampen the effect

Page 72: Current Transformers ALSTOM

s72

VT Earthing

Primary Earthing Earth at neutral point

Required for phase-ground measurement at relay

Secondary Earthing

Required for safety

Earth at neutral point

When no neutral available - earth yellow phase (VERY COMMON)

No relevance for protection operation

Page 73: Current Transformers ALSTOM

s73

VT Construction

5 Limb Used when zero sequence measurement is

required (primary must also be earthed)

Three Single Phase Used when zero sequence measurement is

required (primary must also be earthed)

3 Limb Used where no zero sequence measurement is

required

V Connected (Open Delta) No yellow phase

Cost effective

Two phase-phase voltages

No ground fault measurement

Page 74: Current Transformers ALSTOM

s74

VT Connections

a b c

a b cA B C N

da dn

a b c n

V ConnectedBroken Delta

Page 75: Current Transformers ALSTOM

s75

VT Construction - Residual

Used to detect earthfault Useful where current operated protection cannot be

used

Connect all secondary windings in series

Sometimes referred to as `Broken Delta` Residual Voltage is 3 times zero sequence voltage

VT must be 5 Limb or 3 single phase units

Primary winding must be earthed

Page 76: Current Transformers ALSTOM

s76

Voltage Factors Vf

Vf is the upper limit of operating voltage.

Important for correct relay operation.

Earthfaults cause displacement of system neutral, particularly in the case of unearthed or impedance earthed systems.

Page 77: Current Transformers ALSTOM

s77

Protection of VT’s

H.R.C. Fuses on primary side

Fuses may not have sufficient interrupting capability

Use MCB

Page 78: Current Transformers ALSTOM

GRIDTechnical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.