Current Transformer by Eazyword.com

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Edition: First, 2015

CURRENTTRAN F RMER



This page is left blank intentionally to dedicate to my parents



This Book is made with an intention to provide step by stepcomprehensive knowledge about correct selection of currenttransformer to those who’s working for an organization or whois pursuing degree in the electrical & electronics. Main focus is

put on the meaningful collaboration of practical as well as

theoretical knowledge. Aftermath of serious effort in thecollaboration of two distinct thoughts, we are able to generatea book with excellent features like explanation of various basicfactors of different verticals of current transformer in selectingthe correct one. These factors are filtered with the actualknowledge: needed at the time of specifying a CT as per yourrequirement. Various references had taken in composing thisbook; it requires lot of effort with free flow of time. Figures usedin this book are quite illustrative one, anyone can learn fromthat without going into detailed explanation of the same. Thisbook has been designed in a way to give better appreciation ofthe role played by current transformer in protecting variouselectrical equipment used in power system. An understandingof correct selection of current transformer will increasehealthiness of the system as well as performance .A better

performance increase the efficiency of the plant This bookmake you understand about the necessity of Currenttransformer. Simple calculations are included in order to makeyou easily understand the selection of current transformer atdifferent levels.

This book shall behave as an intermediate level between the practical & theoretical knowledge of current transformer. At theend of the book you will have an excellent knowledge of thebasic principles as well as associated trouble shooting duringselection & installation of current transformer. In addition, thisbook introduced a section named as interesting facts thatcovers the area of known facts what we usually forget toremind, these facts are based on basic theory of currenttransformer.We hope that you will gain a lot from this book and will helpyou to improve your professional career.

This book targets the following people who will find this bookuseful:

Electrical Engineers Design Engineers Project Engineers Instrumentation Engineers Electrical Technicians Field Technicians Electricians

PREFACE



Edition: First, 2015

CURRENTTRAN F RMER

Illustration & Production By:

Acknowledgment

The Technical Data Given in this book is for

information purpose. This Book is combined effort

of experience experts and based on Standards.

It is my great pleasure and honor to introduce this

Book. I am sure it will be used fruitfully by all

persons involved in the implementation of Current

Transformer

-Adhish Gupta

Electrical Engineer

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1. Introduction to basic theory

1.1

Let us understand current transformer1

1.2 Working & main constructional parts of current transformer 1

1.3 Ideal Transformer 2

1.4 C.T equivalent Circuit 3

1.5 Phasor Diagram 4

1.6 General principle of measuring current and voltage 4

1.7 Exciting Current 4

1.8 Why secondary of a CT never kept open? 5

3. Current transformer classification

3.1 Types of current transformer 9

3.2 Interesting facts 10

3.3 Remanence 12

3.4 How to correct remanance12

Contents

4. Amplitude and phase error calculation

4.1 Definition of amplitude and phase error 13

4.2 Ratio correction factor 13

4.3 Some useful formulae 15

4.4 Error reduction methods17

2. Current transformer importance in power

system

2.1 Why should we use in electrical system? 6

2.2 Result of incorrect definition of CT 6

2.3 How to specify the CT? 7

2.4 Definition of protection 7

2.5

CT class according to the application7




5. Terms to be used in specifying the CT

5.1 Useful terms in defining CT specification 18

5.1.1 Rated burden of CT 18

5.1.2 Actual burden of CT 18

5.1.3 Surge current Coefficient/Overcurrent Coefficient 20

5.1.4 Rated short-time thermal current (Ith) & Dynamic PeakValue(Idyn) 20

5.1.5 Rated frequency 22

5.1.6 Rated voltage of primary circuit (Upr) 22

5.1.7 Primary operating Current Ips) 22

5.1.8 Rated Primary Current of CT/Nominal Current (Ipn) 23

5.2 What is CT Burden? 18

5.3 Interesting facts18

5.4 4 or 6 wire connection 19


5.6 Difference between Dynamic peak Value and Rated Thermal short –

Circuit current 21


5.8 Secondary circuit characterstics 23 to 33

5.8.1 Rated Secondary Current (Isr) 23

5.8.2 Accuracy Classes27

5.8.3 Rated output 28

5.8.4 Safety Factor(SF)/Instrument Security Factor(ISF) 28

5.8.5 Accuracy limit Factor(ALF) 29

5.8.6 Knee Point Voltage 33

5.9 How to choose CT secondary output? 23


5.11 Protection Current Transformer 30 to 33

5.11.1 Definite time over current protection30

5.11.2 Inverse Definite Time Over current Protection 31

5.11.3 Differential Protection 31

5.11.3.1 Generator Differential 32

5.11.3.2 Motor Differential 32

5.11.3.3 Transformer Differential 33



6. How to do CT Selection

6.1 Why measuring CT is not advisable to use as protection 39


7.

Current transformer transient performance7.1 Introduction 41

7.2 Transient performance: Calculation 42

7.3 X/R ratio or time constant importance 42

7.4 How CT saturation does occur? 43

8. Current transformer testing & Commissioning

8.1 Equipment’s requires in testing 44

8.2 Checking & inspection 45

8.3 Test on Current Transformer

8.3.1 Type test 45

8.3.2 Routine test 45

8.4 Field tests to be performed before commissioning of current transformer 46

8.4.1 Insulation resistance test 46

8.4.2 Procedure of IR test 47

8.4.3

Polarity test47

8.4.4 Burden test 48

8.4.5 Protection CT magnetization curve test 49

8.4.6 Turns ration/Primary injection test 51

8.5 Advantage of IR test 46

8.6 Interview question: What happen if Transformer is given DC supply? 47

8.7 Factors which caused Burden error 48

8.8 Interview question: What will be the turns ratio required to match an

80Ω

source to a 320Ω

load?51

9. Terms &

Definitions

52 to 54

Did you know? Metering

function of current

transformer 54

10. Current transformer Summary

10.1 Functions of current transformer 55

10.2 Application of current transformer 56


10.4 Points to be remembered 57 to 58

Bibliography 59

Applicable standards 60



Introduction to

Basic TheoryC H A P T E R - 1

Ip

Let us understand what is current transformer

1

Working: A time varying voltage is applied to the primary winding which drives magnetic flux in the core and

induces a voltage in the secondary winding. The transformer draws an exciting current to maintain the flux in a core.

"Rate of change of flux linkage with respect to time is directly proportional to the induced EMF in aconductor or coil".

Faraday’s laws of Electromagnetic Induction

Main constructional parts of Current transformer

1. Primary winding of transformer – Alternating current through thePrimary winding produces a continually changing flux or alternating fluxwhich surrounds the Secondary winding, through magnetic core.

2. Magnetic Core of transformer – the magnetic flux produced by theprimary winding, will pass through this low reluctance path linked withsecondary winding and creates a closed magnetic circuit.

3. Secondary Winding of transformer – the flux, produced by primarywinding, passes through the core, will link with the secondary winding. IfSecondary winding circuit is closed, electric current start flowing throughit.

Hollow core

Primary Conductor

Primary Current, Ip

Secondary Winding

Ammeter

Is

Main Primary Conductor

Secondary Winding

Circuit Symbol

C.T

Is



For CTs Primary turns=1, so

PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS

Circuit Voltage Required:

E

S

= I

S

(Z

B

+ Z

CT

+ Z

L

) Volts

Where:-

I

S

=Secondary Current of C.T. (Amperes)

Z

B

=Connected External Burden (Ohms)

Z

CT

=C.T Winding Impedance (Ohms)

Z

L

=Lead Loop Resistance (Ohms)

E

s

=E.M.F induced across secondary for an ideal

transformer (volts)

V

t

=Secondary terminal voltage

For an ideal transformer

For IS to flow through Z there must be some potential, Es

ES is produced by an alternating flux in the core of CT.

I

P

= N x I

S

E n.1

E

s

= I

s

x Z E n.2

E

S

dØ

dt

Figure-1.0

Is

Z

Ip

Es

Where:-

N = Turns ration of turns between primary & secondary winding .Mostly, the primary of a CT is a straight

through bar or in other words we can say a single turn.

I

S

=Secondary Current of C.T. (Amperes)

For an ideal CT Circuit, impedance Z have Components as detailed below:

Figure-1.1

N

p

Ns

I

p

Is

E

k

CT

Z

B

V

t

V

t

= I

s

. Z

B

= E

k

- I

s

. Z

CT (Volts)

Eqn.3

AsZ

L

shall be negligible small, we can ignore it

2



Maximum Secondary Winding Voltage:

Where:-

E

k

= 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

ᴨ

x√2

= 3.14 x 1.414

C.T. Equivalent Circuit

E

K

= 4.44X BX AX fX N

(Volts)

E n.4

E

K

should be greater than

E

S

Example: 1

Sol:

Suppose a CT with Ratio 2000 / 5A is given having Max Flux Density = 1.6 T , Core C.S.A = 20 cm2 ,R S = 0.31Ω &

I MAX Primary = 40 kA. Then, find the maximum secondary burden permissible in terms of ohm if no saturation is to

occur.

Hence ,Maximum CONNECTED burden: 2.84 - 0.31 = 2.53 Ohms

Number of turns, N = 2000 / 5 = 400 Turns

Max. Sec. Current, IS MAX = 40,000 / 400 = 100 Amps

Using Eqn.4

VK = (4.44 x 1.6 x 20 x 50 x 400)/ 104 = 284 Volts

Therefore Maximum Burden = 284 / 100

= 2.84 Ohms

BURDEN

Please Refer Chapter-5, Page-18

for Definition

E

s

= Secondary excitation voltage

V

t

=

Secondary terminal voltage

I

p

= Primary rating of C.T

Z

CT

= C.T. secondary winding across the

C.T. terminals impedance in ohms (Rs+jXs) 3

Where:

I

e

= Secondary excitation current/Magnetizing current

I

s

=

Secondary currentZ

b

= Burden of relays in ohms

N = C.T. ratio

Z

e

= Secondary excitation impedance in ohms (Rm+jXm)

Figure-1.2

I

e

R

m

m

I

s

X

s

R

s

Z

b

I

p

V

t



1. Current transformer

For a Closed-circuited transformer the following is valid:

This equation gives current transformation in proportion to the primary and secondary turns. A current transformer isideally a short-circuited transformer where the secondary terminal voltage is zero and the magnetizing current isnegligible.

2. Voltage transformers

For a Voltage transformer in no load the following is valid:

This equation gives voltage transformation in proportion to the primary and secondary turns. A voltage transformer is ideally a transformer under no-load conditions where the load current is zero.

Phasor Diagram

General principles of measuring current and Voltage

I

p

X N

1

= I

s

X N

2

E

p

N

p

E

s

N

s

= E n.6

Where:

E

p

= Primary voltage

I

m

=

Magnetising currentE

s

= Secondary voltage

I

e

= Excitation current

= Flux

I

p

= Primary current

I

c

= Iron losses (hysteresis & eddy currents)

I

s

= Secondary current

Figure-1.3

Ip/N

Exciting

Current

In an ideal current transformer, the primary ampere-turns are equal to the secondary ampere-

turns. However, every core material requires some energy to produce the magnetic flux which

induces the secondary voltage necessary to deliver the secondary current. This energy isprovided by exciting current .Thus, in an actual current transformer, the secondary ampere-

turns are equal to the primary ampere-turns minus the exciting ampere-turns.

Eqn.5

4

I

P

= N x I

S

As per Eqn.1



If In CTs, the exciting current could be neglected, thetransformer should reproduce the primary current withouterrors and the following equation should apply to the primaryand secondary currents:

-

1

I

2

= I

1

N

2

I

e

E n.8

Simplified equivalent current transformer diagram converted

to the secondary side is shown in (Figure-1.5)

The current transformation requires a small amount of energy to magnetize the iron core that creates small energy

losses such as eddy currents, and heat caused by current flowing through the windings. Therefore, the secondary

current is not a perfect representation of the primary current. Hence Eqn.7 is modified to Eqn.8, where Ie represent

Excitin current.

The diagram shows that not all the equivalent primary current passes through the secondary circuit. Part of it

consumed by the core, which means that the primary current is not reproduced exactly. The relation between

the currents is stated in Eqn.8 (Figure-1.6)

Thus, primary current contains two components:

• An exciting current Ie

, which magnetizes the core and supplies the eddy current and hysteresis losses, etc.

• A remaining primary current component (I

1

-I

e

), which is available for transformation to secondary current in the

inverse ratio of turns.

I

2

=

I

1

1

N

2

X

E n.7

It is necessary to keep secondary of CT either shorted or connected in

series with low resistance coil such as current coil of wattmeter, coil of

ammeter etc. If it is left open, then current through secondary becomes

zero hence the ampere turns produced by secondary which is generally

oppose primary ampere turns becomes zero. As there is no counter m.m.f

to oppose primary m.m.f (ampere turns), this lead the production of high

flux in the core results in excessive core losses. This may damage the

insulation of windings or nearby personnel. Current transformer generally

works at a low flux density. The core usually made up of very good metal

to give a small magnetizing current. When it is open circuit the secondary

impedance now becomes infinite and the core deeply saturates.

As the AC wave moves from positive half cycle to the negative half cycle,the rate of change of flux dø/dt is so great that very high voltage isinduced in the secondary winding

It is usual practice to ground the CT on the secondary side to avoid adanger of shock to operator.

secondaryOf CT never kept open?

Why

BURDEN

I

s

I

p

Figure-1.4

I

2

N

1

/N

2

) x I

1

I

2

I

e

Figure-1.5

5



To make power system more sensitive towards detecting even the smallest fault protection relays are used which isolate

the fault by tripping CBs. Current transformer is responsible to give fault current related feedback signal to respective relayto operate when secondary current of CT exceeds the threshold value of relay (Generated Feedback signal of transformeris in form of small current which is ranges from 0A to 5 A).

Major disturbances, such as short-circuit currents, canresult in serious damage:

a. fatigue or deterioration of network components,

b. danger for people,

c. Loss of supply and production, etc.

Why should we use CT in electrical System?

Permanent monitoring of network electrical parameters by reliable and properly selected current transformerssupplying protection relays allows rapid isolation of the faulty area. These relays must ignore transient and normal

disturbances but systematically trip when a destructive fault has to be eliminated. Thus, correct selection of Currenttransformer is necessary otherwise it can lead to malfunctions in the protection channel causing destruction ofequipment and create danger to the operator.

Incorrect definition can lead to malfunctions in the protection, for examples:

1. Overestimation of the short-circuit current can lead to feasibility problems, overratingand high CT costs.

2. On the other hand, under-estimation of the short-circuit current can lead to failure todetect the fault, thus destroying the equipment, placing the operator in danger andgenerating operating downtime.

3. An output power or accuracy error can result in a malfunction or in failure to trip of theprotection devices, thus destroying the equipment, placing the operator in danger andgenerating operating downtime.

4. An error in defining the accuracy class of a metering winding will lead to incorrect

energy billing and thus a loss of income for the electrical utility or the customer

Result of Incorrect definition of CT

C H A P T E R - 2

Current transformer

Importance in power system

6



Protection Class Instrument Class

A “protection” CT must saturate sufficiently high to allowa relatively accurate measurement of the fault current bythe protection whose operating threshold can be veryhigh. Current transformers are thus expected to have an Accuracy Limit Factor (ALF) that is usually fairly High.Note that the associated “relay” must be Able towithstand high over currents.

An “instrument” CT requires good accuracy around thenominal current value. The metering instruments do notneed to withstand currents as high as the protectionrelays. This is why the “instrument” CTs, unlike the“protection” CTs, have the lowest possible Safety Factor(SF) in order to protect these instruments through earlierSaturation.

a. Insulation Level(KV)

b. Short-Circuit Current Withstand Capacity(KA)

c. Short-Circuit Duration(S)

d. Nominal Primary Current(A)

e. Number of secondary Winding

f. Associated Protection & Metering with Secondaryg. Output Power(Including relay and wire

Consumption)(VA)

h. Accuracy Classa. ALF(Protection)b. SF(Metering)

i. Nominal Secondary Current(A)

How to specify the CT?

Necessary informationrequired to specify the CTwith single primary are:

Importance of CTCT is defined by ratio, powerand accuracy class. CT class(accuracy as a function of CTload and of over current) ischosen according to theapplication.

Current transformers are used to supply informationto the protective relays and/or current, power andenergy metering “instruments”. For this purpose theymust supply a secondary current proportional to theprimary current flowing through them and must beadapted to network characteristics: voltage,frequency and current.

Characterization of CTs

CT class according to the application

Definition of protection

The protection functions of a network are intended to monitor one or more parameters of the installation, forexample: currents, voltage, temperature, frequency, etc. These values are permanently measured andcompared with set points or thresholds beyond which the situation is defined as abnormal and dangerous. When a fault occurs, the protection device issues a tripping signal. Then, in order to durably isolate the faulty part, itprevents reclosing until the device has been repaired. It can also generate an alarm to inform maintenance personneland enable them to take the necessary action.

Table.-2.0

7



Interesting Facts

It should be remembered when two or more devices are required to be connected at thesecondary terminal of a CT, Devices must be connected in series across the winding. This isexactly the opposite of the method used to connect two or more loads to be supplied by avoltage or power transformer where the devices are paralleled across the secondary winding.

CT Rating Plate indicates the following details, an

example of representation of protection CT

Rated primary current: 200 A,Rated secondary current: 5 A.

15 VA 5P 10 Accuracy limit factor = 10 Accuracy class = 5P Accuracy power = 15 VAIts accuracy load: Pn = 15 VAIts accuracy limit factor is ALF = 10For I = ALF X In, its accuracy is 5% (5P)

Example: 2

For the protection CT given in theExample, the ratio error is less than 5% at 10 In,if the real load consumes 15 VA at In.

OrGiven CT is rated for 15VA Burden and will nothave more than 5% error at 10 times of ratedcurrent.

8



Current transformers are classified based on construction and application. Classifications according to both are explained

below:

Types of Current transformers

1. Current transformers are classified In accordance with their nature of construction

C H A P T E R - 3

Current transformer

Classification

Figure-3.0

Figure-3.1

Figure-3.2

These CT‟s can be used whereprimary current lies between 50 to

5000 amps. It has an opening in

centre to accommodate primary

conductor through it.

1) Ring Core CT’s:

These CT‟s can be used whereprimary current lies between 100

to 5000 amps. Split core CT‟s

have one end removable so that

the Primary conductor need not to

be disconnected to install the CT.

2) Split Core CT’s:

These CT‟s can be used

where primary current lies

between 1 to 100 amps.

Since the load current

passes through primary

windings in the CT, screw

terminals are provided for

the load and secondary

conductors.

3) Wound Primary CT’s:

9



2. Current transformers are classified In accordance with their nature of application

.

The principal requirements of a measuring CT are that, for primary

currents up to 120% or 125% of the rated current, its secondary

current is proportional to its primary current to a degree of accuracy

as defined by its “Class” and, in the case of the more accurate

types, that a specified maximum phase angle displacement is not

exceeded. A desirable characteristic of a measuring CT is that

it should “saturate” when the primary current exceeds the

percentage of rated current specified as the upper limit to

which the accuracy provisions apply . These CT‟s require high

accuracy, a low burden (output) and a low saturation voltage

a. Measuring Current Transformer

The principal purpose of Protective Current transformer is toprovide a secondary current proportional to the primary current

when it is several, or many, times the rated primary current. Themeasure of this characteristic is known as the “Accuracy LimitFactor” (A.L.F.). A protection type CT with an A.L.F. of 10 will produce aproportional current in the secondary winding (subject to theallowable current error) with primary currents up to a maximum of10 times the rated current.

b. Protective Current Transformer

Interesting Facts

Ring type (or rectangular type) CTs are normally preferred over other types of CTs. becausethey are simple in construction, mechanically stronger and cheaper. In a ring type/bar primarytype CTs the working ampere-turns are determined by the primary current and thereforenecessarily, the accuracy that can be offered with these CTs becomes progressively inferior as

the rated primary current decreases.-If higher accuracy and burdens are required for CTs of low primary current wound types CTs are used.

Transformer differential relayscompare the phase and magnitude of

the current entering one winding ofthe transformer with that leaving viathe other winding(s). Any difference inphase or magnitude between themeasured quantities will causecurrent to flow through the operatewinding of the relay. If this currentexceeds the relay setting, tripping ofthe transformer circuit breakers willbe initiated.

c. Interposing Transformer

Figure-3.5

Figure-3.3

Figure-3.4

10



To enable a comparison to be made, the differential scheme should be arranged so that the relay will see rated currentwhen the full load current flows in the protected circuit. In order to achieve this, the line current transformers must bematched to the normal full load current of the transformer. Where this is not the case it is necessary to use an auxiliaryinterposing current transformer to provide amplitude correction. The connection of the line CTs should compensate forany phase shift arising across the transformer. Alternatively the necessary phase correction may also be provided by theuse of an interposing CT.

Thus, the main function of an interposing CT is to balance the currents supplied to the relay where therewould otherwise be an imbalance due to the ratios of the main CTs. Interposing CTs are equipped with a

wide range of taps that can be selected by the user to achieve the balance required.

d. Core Balance CTs

An earth fault relay, connected to the secondary winding, is energized only when there is residual current in the primary

system. The advantage in using this method of earth fault protection lies in the fact that only one CT core is used in

place of three phase CT‟s whose secondary windings are residually connected. In this way the CT magnetizing current

at relay operation is reduced by approximately three-to-one, an important consideration in sensitive earth fault relays

where a low effective setting is required.

Core Balance CTs are special CTs used to detect Earth faults & usually used for Restricted Earth Fault Protection.

It is a ring type CT through which the cables carrying current of all the three phases (R, Y & B) are passed through.

Under normal operating conditions, summation of current through the three phases shall be equal to zero. In event of a

fault (as shown in Figure-3.6), the summation of the current shall no longer remain zero (zero sequence current shall flow

during earth fault) & thus the fault can be detected.

Core-balance transformers are normally mounted over a cable at a point close up to the cable gland of switchgear or

other apparatus. Physically split cores („slip-over‟ types) are normally available for applications in which the cables are

already made up, as on existing switchgear.

These are ring type CTs & suitable for themeasurement of the sum of three phase currents in a3-phase cable. Under normal operating conditions thissum is zero. In the event of an earth-fault the sum ofthe current is equal to the zero sequence current. It isnecessary to specify leakage current to be detectedalong with minimum setting of the relay and size ofcable at the time of ordering CT.

I s flows only when there is an earth-fault Ia+I

b+I

c ≠0

Figure-3.6

Earth

Fault

Relay

Core

11

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NORMAL HYSTERSIS CURVE HYSTERSIS CURVE WITH REMANENCE

12

Flux

Density

Magnetizing Force

(A) (B)

Figure-3.7

The remanence can be corrected by demagnetizing the current transformer.

This is accomplished by applying a suitable variable alternating voltage to the

secondary, with initial magnitude sufficient to force the flux density above the

saturation point, and then decreasing the applied voltage slowly and

continuously to zero. If there is any reason to suspect that a current transformer

has been subjected recently to heavy currents, possibly involving a large DC

component, it should be demagnetized before being used for any test requiring

accurate current measurement

?correct

How to

remanance

Remanence Remanent flux can be set up in the core of a current transformer under operating or test conditions. During operatingconditions, remanent flux can be left in the core while the primary current is interrupted and the flux density in the core ofthe transformer is high. Remanence flux may be left in the core of transformer during clearing of fault current, testing suchas resistance or continuity measurements.

The remanent flux in the core depends on many factors such as

Magnitude of primary current, Impedance of the secondary circuit

Amplitude and time constant of any offset transient.

Since the impedance of the secondary circuit is generally fixed, the magnitude of remanent flux is governed by themagnitude of the symmetrical component of the primary current and the magnitude of the offset transient prior to the primary current interruption.

Maximum remanent flux can be obtained under conditions whereby the primary current is interrupted while thetransformer is in a saturated state.

When the current transformer is next energized, the flux changes required will start from the remanent value. If the required

change is in the direction to add to the remanent flux, a large part of the cycle may find the current transformer saturated.

When this occurs, much of the primary current is required for excitation and secondary output is significantly reduced and

distorted on alternate half cycles.

When excitation is removed during high magnitude fault events, this remnant flux can be quite high. The remnant flux

essentially shifts the normal operating flux of CT and will require either more or less exciting current .During a subsequent

fault; this remnant flux can push the core deeper into saturation.

This phenomenon is illustrated in Figure-3.7

Remnant

flux

The flux in the core of a CT is a function of both the excitation voltage and the magnetic properties of the

core itself. When excitation is removed from the CT, same of the magnetic domains retain a degree of

orientation relative to the magnetic field that was applied to the core. This is known as remnant flux .



C H A P T E R - 4

Amplitude and phase

Error calculations

The difference in phase (phase error) between the primary and secondary current vector must be zero for the perfect

transformer, therefore the quality of material is so chosen that the phase difference comes out be minimum. The phase

displacement (phase error) is said to be positive when the secondary current vector LEADS the primary and the phase

displacement (phase error) is said to be negative when secondary current vector LAGS the primary current vector.

Phase Error: Phase error is defined as difference in phase angle between the primary and secondary

Definition of amplitude error & phase error

Amplitude Error: The error which a transformer introduces into measurement of a current arises from the fact that the

actual transformation ration is not equal to the rated transformation ratio. (For a CT it can be termed as Current Error)

13

Ratio Correction Factor:

The Ratio Correction Factor (RCF) is the ratio between the true ratio & marked ratio.

RCF=

True ratio

Marked Ratio

True ratio=

RMS Primary Current

RMS Secondary Current

Marked ratio=

Rated Primary Current

(As mentioned on nameplate of CT)

Rated Secondary Current

(As mentioned on nameplate of CT)

PACF=

True Power Factor, Cosø

Measured Power Factor, Cos(ø-

β)

E n.9

E n.10

E n.11

E n.12

The factor by which the reading of a watt meter or the registration of a watt hour meter must be multiplied to

correct for the effect of ratio error and phase angle is the Transformer Correction Factor (TCF)

Where:

ø = angle of lag of load current behind load voltage

cos ø = Power Factor , Cosine of angle between the voltage and current.

TCF= RCF X PACF E n.13



Thus, Reading of wattmeter must be multiply with 0.9945 to get correct reading

Interesting Facts

The phase error is generally not significant where as amplitude of the current isimportant, but phase error matters significantly when the CT is used in measuringpower, when voltage and current signals are multiplied together. Accordingly,applications requiring accurate power measurements should use a CT with low phaseerror. For highest accuracy, a non-opening nickel iron alloy toroidal core providesthe most inductance, and therefore the least error.

Sol:

If a CT with RCF of 1.0020 has a phase angle error = +15‟ and is used for measuring a load whose power factor is0.500 lagging, determine its phase angle correction factor, PACF?

Example: 3

RCF=1.0020

The primary current lags the line voltage by an angle whose cosine equals the power factor.

cos-1 (0.500) = 60° = θ

Or

cos θ = cos 60° = 0.500

The secondary current leads the primary current by 15'. Therefore, the primary currentactually lags the primary voltage by 59° 45'.

θ = 60° = 59° 60'

(θ - β) = 59° 60' - 0° 15' = 59° 45’

Thus,

cos (θ - β) = cos 59° 45' = 0.5038

Using Eqn.12

Using Eqn.13

= 0.9925

0.500

0.5038

=

Cosø

Cos (ø-β)

PACF =

TCF = 1.0020 x 0.9925

=0.9945

14



1. The secondary induced voltage Esi can be calculated

2. The inductive flux density necessary for inducing the voltage Esi can be calculated from

3. The exciting current, Ie necessary for producing the magnetic flux B.

4. Phase angle between the induced voltage Esi.

5. Current errorThe error with a transformer introduces into the measurement of a current and which arisesfrom the fact that actual transformation ratio is not equal to the rated transformer ratio. Thecurrent error expressed in percentage is given by the formula:

Where

f =

Frequency in Hz,

A

j

= Core Are in MM2,

N

2

=Number of secondary turns,

B =Magnetic Flux in Tesla,

All current transformers include an inherent phase shift relative to the current measurement. This phase shift introduces anerror in the power measurement.

B =

π X √2 X f X

j

X N

2

E

si

Where

Z =Impedance, Ω

R

i

=Winding Resistance, Ω

R

b

=Load Resistance, Ω

X

b

=Load Inductance, Ω

Ø =

X

b

R

i

+ R

b

)

Eqn.18

Eqn.19

Where

H =

Exciting force in At/m,

L = Length of magnetic path in Mtr ,

N

2

=Number of secondary turns,Eqn.17

e

=

H X L

N

2

Eqn.16

15

Some useful formulae

Current error ( ) =

(K

a

X I

s

- I

p

)

I

p

X 100

E

si

= I

2

X Z

(Volts)

E n.14

(R

i

+ R

b

) ²+ X

b

²

= E n.15

As per Eqn.4



.

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.

1 VA ammeter is connected to the secondary of CT (2000/5A) measuring current of 4.9A when primary currentflowing through it is 2000A, then find out how much current error will be there if CT rated Burden is given as 15VA?

Example: 4

As mentioned above,Primary Current, Ip= 2000ASecondary Current, Is=4.9 AThus, Transformation Ratio, Kn= 2000/5 =400

Using Eqn.19

Current error (%) = (400 x 4.9 – 2000) x100/2000 = -2%

Here, negative sign figure represents secondary current is lower than the rated nominal value.

Sol:

Where

K

a

= rated transformation ratio

I

p

=actual primary current (A)

I

s

= actual secondary current when Ip is flowing under the conditions of measurement (A)

ACCURACY CLASS CURRENT ERROR AT

RATED CURRENT (%)

PHASE DISPLACEMENT AT

RATED

CURRENT(Minutes)

COMPOSITE ERROR

RATEDACCURACY LIMIT

PRIMARY CURRENT(%)

5P ±1 ±60 5

10P ±3 - 10

15P ±5 - 15

ERROR LIMITS OF PROTECTION CT

Table-4.0

16



6. Composite Errors(εc):

Under-Steady Conditions, the R.M.S value of the difference between the instantaneous values of theprimary current, and the instantaneous values of the actual secondary current multiplied by the ratedcurrent transformation ratio, the positive signs of the primary and secondary current corresponding tothe convention for terminal marking is generally expressed as percentage of the primary currentaccording to the following formula;

-

T

∫

c

=

100

I

p

(k

n

i

s

i

p

)² d

0

Where

k

n

= Rated transformation ratio

I

p

= R.M.S value of primary current (A)

T =

Duration of one cycle (Sec.)

i

p

= Instantaneous value of primary current (A)

i

s

= Instantaneous value of Secondary current (A)

ErrorreductionCan be done by:

1 Using better quality magnetic material

2 Shortening the mean magnetic path

3 Reducing the flux density in the core

17

E n.20



Useful terms in defining CT specifications

1. Rated Burden of CT (As printed in Name plate of CT)

2. Real Burden or actual burden of CT

P

n

= I

n

²

X R

n

E n.21

Where,

P

n

= Rated Burden of CT, VA

I

n

= Rated Current, A

R

n

= Resistance of CT Winding, Ω

CT burden (VA) is the load imposed on CT

secondary during operation.

or As per IEC 60044-1

The burden is usually expressed as the apparent

power in volt-amperes absorbed at a specific power

factor and at the rated secondary current.

WhatCT Burden ? Is

C H A P T E R - 5

Terms to be used in

Specifying CT

Where,

P

r

= Actual Burden of CT,VA

I

n

= Rated Current, A

R

p

= Resistance of CT Winding & Cable

used Between CT and Relay, Ω

P

r

= I

n

²

X R

p

Interesting Facts

CT performance is characterized by:

1. Turns ratio,2. Turns ratio error (ratio correction factor),3. Saturation voltage,4. Phase angle error, and5. Rated secondary circuit load (burden).

18

E n.22



4- or 6-wire connectionIf 6-wire connection is used, the total length of the wire, naturally, will be two times the distance between the CT and therelay. However, in many cases a common return conductor is used .Then, instead of multiplying the distance by 2, a factorof 1.2 is typically used. This rule only applies to the 3-phase connection.

The factor 1.2 allows for a situation, where up to 20% of the electrical conductor length, including terminal resistances, uses

6-wire connection and at least 80% 4-wire connection.

6 wire CT connection 4 wire CT connection

Suppose, the distance between the CT and the relay is 10 meters. Then, total length is 2 x 10 m = 20 meter for6-wire connection whereas for 4-wire connection total length is 1.2 x 10=12 meter

19

Sol:

Burden of CT = 0.0648+0.015 = 0.0798 Ω

Resistivity of Cu Conductor, ρ=0.0216µΩm (75 ⁰ C)

Using Eqn.23

R = 0.0216 µΩ m X (1.2 X10)4 Sq.MM

Hence,R=0.0648 Ω

Resistance, R =

(ρ X L)/

Consumption of Various Devices are

given in the manufacture’s technical

data sheet

The distance between the CTs and the protection relay is 10Mtr. 4Sq.MM CU Conductors in the 4-wire connectionare used. The burden of the relay input is less than 15 mΩ (5A input).Calculate the actual burden of the CT at 75 ⁰ C.

Example: 4

E n.23

Figure-5.0



3. Surge Current Coefficient/Over current Coefficient

4. Rated short-time thermal current (Ith) & Dynamic Peak Value(Idyn)

Standard R.M.S values of current, expressed in KA, are:

Scale order of

Ksi

Manufacturing

Ksi<100 Standard

100<Ksi<300Sometime difficult forsome secondarycharacteristics

100<ksi<400 Difficult

400<Ksi<500Limited to somesecondarycharacteristics

Ksi>500 Very often impossible

K

si

= (I

th

/I

n

)

6.3

–

8

–

10

–

12.5

–

16

–

20

–

25

–

31.5

–

40

–

50

–

63

–

80

–

100

Where,

K

si

=

Surge Current Coefficient

I

th

= Rated thermal short-circuit current

(Generally, R.M.S Value of installation’s Max.Short-circuit Current and duration of this isgenerally taken to be 1s)

I

n

=Nominal Current of CT at primary side

A high “Ksi” lead to over-dimensioning of primary

winding cross sections. This will limit the number ofwindings in the primary coils, in turn limit induced EMFof the CT.

CT manufacturing becomes difficult at this stage.

Knowing over current

coefficient allows us to know

whether a CT will be easy to

manufacture or not.

Lower the Surge Current Factor Ksi higher will be

the feasibility of current transformer to get

manufactured

This is the maximum Primary current, which the transformer can withstand keeping secondary winding shortcircuited for a period of one second, without reaching a temperature that would be disastrous to the insulation,e.g. 250 ºC for oil immersed transformers.

Rated Thermal short-Circuit Current

Table-5.0

Each CT Must be able to thermally and dynamically withstand the defined short-

circuit current in defined duration (usually, 1sec is preferred) passing through its

primary current circuit until the fault is effectively broken.

For easier production we can:

Reduce the secondary characteristics as far as possible.

Over-rate the primary rated current.

Reduce the Short-circuit current duration/Surge Current Factor (Subjectedto the agreement of protection engineer, in normal cases don’t allow to go

beyond 0.8 s).

20

Eqn.24



Suppose we have following figures for a particular project

Short-Circuit VA = 250MVA

Operational Voltage = 11KV

Rated Primary Current (In)= 20 A

Using Eqn.24

According to the Table-5.0,CT is quite difficult to manufacture

If short-circuit time limited to 0.8 sec rather than 1 sec, then = 9622 X √0.8 =8606A

Rated Thermal short – Circuit

currentDynamic peak Value

This is the maximum (R.M.S) Primarycurrent, which the transformer can

withstand keeping secondary windingshort circuited for a period of onesecond, without reaching a temperaturethat would be disastrous to theinsulation.

The peak value of the primary currentwhich a current transformer will

withstand, without being damagedelectrically or mechanically by theresulting electromagnetic forces, the

secondary winding being short-circuited.

Difference between Dynamic peak Value and Rated Thermal short – Circuit current

Interesting FactsDynamic peak Current (Idyn) is always greater than rated thermal short-circuit current, as per

electrical standards, value of Idyn

in terms of Ith

is stated given below.

IEC 50Hz 2.5 X Ith

IEC 60Hz 2.6 X Ith

ANSI/IEEE 60Hz 2.7 X Ith

Table-5.1

21

The short-time current for periodsother than one second Ix can becalculated by using the followingformula:

Where,

X=the actual time in seconds

Eqn.25

I

x

=

√x

I

t

Ksi = 9622/20 = 481.1 ≈ 480

Effect of reducing short circuit current duration / surge current Factor

Example: 5

250 X 10 ³ KVA

11 X 1.732KV = 9622 A Rated thermal short-circuit current (Ith)=

Now, Ksi = 8606/20 ≈ 430

This transformer would be quite easier to produce than a transformer with K si = 480.It can be noticed that

by reducing the short-circuit duration, the value of Ksi reduces which increases the chance of CT getting

manufactured.



5. Rated Frequency: It is the frequency of installation

6. Rated Voltage of the primary Circuit(Upr): This defines the insulation level of Equipment

Rated CT Voltage > Rated Installation Voltage

7. Primary operating Current(Ips):

It is the actual current of equipment flows when it is connected to its respective load. We can calculate the exact

value of current for incomers & feeders by using the given below formulas.

Generally we choose the rated voltagebased on the duty voltage, Us, according tothe figure:

Generator Incomer I

ps

= S / (1.732 X U)

Motor Feeder I

ps

= P / (1.732 X U X Cosø X ŋ)

Because Frequency affects a CT only because the lines

of flux generated by the primary current begin to appear

as DC as the frequency gets very low; a CT needs the AC

CYCLE changes to induce the secondary current. With

toroidal CT, you will experience a drop in accuracy asthe frequency goes down from 60 Hz.

Therefore, CT installed on a 60Hz network cannot be

used for 50Hz network for the same level of accuracy

as its accuracy goes down as induced secondary

current reduces.

CT defined for 50 Hz can be

installed on a 60 Hz network with

the same level of accuracy.

However, the opposite is not true.

Why?

Figure-5.0

Eqn.26

Where,

S = Apparent power in kVA

U

= Primary operating Voltage in KV

P = Active power of motor in KW

Q =

Reactive power of capacitors in Kvars I

ps

= Primary operating Current in A

Incomer Cubicle: I

ps

= S / (1.732 X U)

Transformer Feeder I

ps

= S / (1.732 X U)

Capacitor Feeder I

ps

= (1.3 X Q) / (1.732 X U)

Eqn.27

Eqn.28

22**Ŋ=Motor Efficiency

**In Case of Capacitor Feeder, 1.3 is a derating coefficient to take account of temperature rise due to capacitor harmonics.



8. Rated Primary Current of CT/Nominal Current(Ipn):

Rated Current (I pn ) > Operating Current (I ps )

Standardized Values:

a. Current Transformer must be able to withstand more than the rated current on a constant basiswhich is normally expressed as 120%,150% and 200% of the rated primary current and this iscalled extended current rating.

b. In the case of an ambient temperature greater than 40 ⁰ C for the CT ,the CT’s nominal Current

(ipn) must be greater than Ips multiplied by the derating factor corresponding to type of feeder

or Cubicle.As a general rule derating factor is of 1% Ipn per degree above 40 ⁰ C.

c. Temperature rise depends on three parameters:

1. Rated current

2. Ambient temperature

3. Feeder type or cubicle type and its IP(Protection index)

Secondary Circuit Characteristics according to IEC standards

9. Rated Secondary Current (Isr):

In general Case:

a. For local use, Isr = 5A

b. For remote use, Isr = 1 A

10-12.5-15-20-25-30-40-50-60-75 and their multiples and factors

Using 5A for a remote application is not

forbidden but leads to an increase in

transformer dimension and cable section

(Line loss= I X R ^2)

The standard CT secondary current ratings are 1A & 5A, the selection is basedon the lead burden used for connecting the CT to meters/Relays.5A CT can beused where Current Transformer & protective‟s device are located within sameSwitchgear Panel where as 1A CT is preferred if CT leads goes out of theSwitchgear.

See example: 6 for better understanding

How to

Choose

CT secondary

output?

23



** Cable resistance increase as cable length increase, Increase in the resistance will increased the burden on CT

**R

p

or R

ct

= Resistance of CT winding & cable between CT and relay

Let’s consider 11kV/220KV transformer for transferring power from generator (at 11KV) (not shown) to Power grid

(220KV).Suppose CT-A & CT-B is used for GT Differential protection. Distance between the CT-A and 87 GT differential

protection relay of 1VA is 15Mtr whereas distance between CT-B and 87 GT differential protection relay is 40 Mtr.

Example: 6 SELECTION CRITERIA IN CHOOSING CT 1A or 5A SECONDARY OUTPUT

Figure-5.1

CT-A

CT-B

POWER GRID

GENERATOR

Rp or Rct ≤ 5 Ω, statesthat the burden imposed

by CT winding andconnecting lead betweenCT & relay should be lessthan 5 Ω.

Rp or Rct ≤ 2 Ω, states thatthe burden imposed by CTwinding and connectinglead between CT & relayshould be less than 2 Ω.

87 GT Differential

Protections

Relay

TRANSFORMER

24



Suppose a. 4Sq.MM CU Conductors cable is suggested to connect 87 GT differentials relay from CT-A and CT-B


Distance between the CT-A and 87 GT differential protection relay is 15Mtr Therefore,

Rp= 0.0216 µΩ m X (1.2 X15)4 Sq.MM

Burden of CT due to connecting cable, Rp=0.0972 Ω Burden due to relay= 1VA

Burden in terms of VA = (5) ^2 X 0.972=2.43VA

Total Burden in terms of VA=1VA+2.43VA=3.43VA

Burden of CT due to connecting cable, Rp=0.0972 Ω

Burden due to relay= 1VA

Burden in terms of VA = (1) ^2 X 0.0972=0.0972VA

Total Burden in terms of VA=1VA+0.0972VA=1.0972VA


Distance between the CT-B and 87 GT differential protection relay is 40 MtrTherefore,

Rp= 0.0216 µΩ m X (1.2 X40) 4 Sq.MM

Hence,

Burden of CT due to connecting cable, Rp=0.2592Ω

Case-1

Case-2

4- or 6-wire connectionRefer page-19Considering 4 wire system, 1.2 factor isintroduced in calculating Rp.

Selected CT specification

5VA Burden5P Accuracy Class

10 ALF

2500/5 1/500 turn ratio

Total Burden imposed on CT by connecting lead (4 Sq.MM) of length 15Mtr and relay is calculated as follows:

With 5A secondary output CT

Selected CT specification2VA Burden

5P Accuracy Class

10 ALF

1200/1 1/1200 turn ratio


Total Burden imposed on CT by connecting lead (4 Sq.MM) of length 40Mtr and relay is calculated as follows:

25

If we make comparison between 2500/5 & 1200/1 CT in selection, we will find that the 2500/5 CT is quite preferable as 5Asecondary CT is much cheaper than 1A secondary CT regardless of burden imposed of 5A secondary CT is higher than 1Asecondary CT because 5A secondary CT have 500 turns in its secondary that make it smaller in size as compare to 1 Asecondary which is expensive as well as require large area to accommodate in electrical panels.Hence in this case we can go for 5VA CT 2500/5 for 15 Mtr Connecting lead length.

CONCLUSION:




Burden in terms of VA = (5) ^2 X 0.2592=6.48 VA

Total Burden in terms of VA=1VA+6.48 VA=7.48 VA


Burden in terms of VA = (1) ^2 X 0.2592=0.2592 VATotal Burden in terms of VA=1VA+0.2592VA=1.2592 VA

Case-1

Burden due to connecting lead with CTsecondary of 5A is comparably low, in additionto this, cost saving is another factor in choosing5A secondary output CT over 1 A secondaryoutput.

CT with 5A secondary output Selected

Case-2

Burden due to connecting lead with CTsecondary of 1A is comparably low, in additionto this, cost saving is another factor in choosing1A secondary output CT over 5 A secondaryoutput. As the distance between the CT-B and

Relay is more than that of CT-A and Relay,therefore burden increases which is in turnincreases the cost of CT with 5A SecondaryOutput in comparison to 1A CT secondaryoutput..

CT with 1A secondary output Selected


Selected CT specification10 VA Burden

5P Accuracy Class

10 ALF

2500/5 1/500 turn ratio

Selected CT specification2VA Burden

5P Accuracy Class

10 ALF

1200 1/1200 turn ratio


26

If we make comparison between 2500/5 & 1200/1 CT in selection, we will find that the1200/1 CT is quite preferable as 1Asecondary CT is comparable cheaper than 5A secondary CT because 1A secondary CT burden is quite lower than 5Asecondary CT. Hence, we can say with the increase of distance beyond 30 Mtr between CT and relay ,1A secondary CT ismore preferable than 5A secondary because burden imposed on CT due to the connecting length drastically increasedfor a distance greater than 30 mtr.

CONCLUSION:

COMPARISON:



10. Accuracy Class:

Accuracy Class implies how accurate the CT is. It tells the percentage error of the CT. For example, an Accuracy Class 1.0 CT means the error will be +/- 1 %. That means, say, you have a 100/5A CT andwhen you pass 100A through the primary of this CT, the secondary current can be anywhere between4.95A to 5.05A There is always some difference in expected value and actual value of output of Currenttransformer due to Current Error and Phase Angle Error, because primary current of current transformerhas to contribute the excitation component of CT core. Accuracy class of current transformer is thehighest permissible percentage composite error at rated current. The standard accuracy classes ofcurrent transformer as per IS – 2705/IEC 60044-1 are 0.1, 0.2, 0.5, 1, 3 & 5 for metering CT. Here inthe protection Current Transformer , 5P means 5% 10P means 10% and 15P means 15% error and ′P′stands for Protection.

Class For burdens

Limits of errors

Application at % rated

currentRatio error %

Phase

displacement

minutes

0.125-100% ofrated burden

5 0.4 15Laboratory

20 0.20 8

100 0.1 5

120 0.1 5

0.2

25-100% ofrated burden

<15 VA1VA-100%

5 0.75 30Precisionrevenuemetering

20 0.35 15

100 0.2 10

120 0.2 10

0.2S**

25-100% ofrated burden

<15 VA

1VA-100%

1 0.75 30

Precisionrevenuemetering

5 0.35 15

20 0.2 10

100 0.2 10

120 0.2 10

0.525-100% ofrated burden

5 1.5 90Standard

commercialmetering

20 0.75 45

100 0.5 30

120 0.5 30

0.5S**25-100% ofrated burden

1 1.5 90

Precisionrevenuemetering

5 0.75 45

20 0.5 30

100 0.5 30

120 0.5 30

1.0 25-100% ofrated burden

5 3.0 180Industrial

grademeters

20 1.5 90100 1.0 60

120 1.0 60

3.0 50-100%50 3.0 -

Instruments120 3.0 -

5.0 50-100%50 5.0 -

Instruments120 5.0 -

5P and5PR ***

100%100 1.0 60

Protection ALF x In 5 2 -

10P and10PR ***

100%100 3.0 -

Protection ALF x In 10 2 -

PX **** Ek , Ie, Rct 5) - - - Protection

Accuracy classes accordingto IEC 60044-1

Table-5.2

A 5P10 CT means 10 times rated current has a maximum error of 5% and only 1% at nominal current whereas 10P15 CT means 15

times rated current has a maximum error of 10% and 3% at nominal current.

27

http://www.electrical4u.com/electrical-transformer/current-transformer.php





11. Rated Output:

The standardized Values of Rated output are:

12. Safety Factor(SF)/Instrument Security Factor(ISF):

.

.

2.5- 5- 10- 15- 30 VA

1. For classes 0.1 – 0.2 – 0.5 – 1, the current error and phase

displacement at rated frequency shall not exceed the values given in

Table-5.2 when the secondary burden is any value from 25-100% of the

rated burden.

2. For classes 0.2S – 0.5S, the current error and phase displacement at

rated frequency shall not exceed the values given in Table-5.2 when the

secondary burden is any value from 25-100% of the rated burden

ISF or Instrument Security Factor of current transformer is defined as the ratio between the limit primary

current (I

pl

) to the rated primary current (I

pn

).

The instrument limit primary current of metering CT is the value primary current beyond which CT core becomessaturated

Safety Factor (SF) is

defined for metering

devices only.

SF = I

pl

/I

pn

The rated Instrument Security Factor (SF) indicates the over current as a multiple of the rated current at which themetering core will saturate. It is thus limiting the secondary current to SF times the rated current. The safety of themetering equipment is greatest when the value of SF is small. Typical SF factors are 5 or 10. It is a maximum valueand only valid at rated burden.

To protect the instruments and meters from being damaged by high currents during fault conditions, a meteringcore must be saturated typically between 5 and 20 times the rated current. Normally energy meters have the lowestwithstand capability, typically 5 to 20 times rated current.

Definition of ISF

Eqn.29

Interesting Facts An ammeter is usually guaranteed to support a short time current of 10 In

i.e. 50 A for a 5 A device.To ensure that the appliance is not destroyed during fault occurrence in the primary, the

current transformer must be able to saturate before 10 In in the secondary.

For this reason, a SF of 5 is sufficient .

28



13. Accuracy Limit Factor(ALF)

Using Eqn.27, 28 & 29 , We can get

Internal Losses of the CT at In S

in

=R

ct

X I

n

² CT Accuracy Power S

n

= R

n

X

I

n

² Real Load Consumption of the CT at In S

a

= R

p

X

I

n

²

Eqn.30

Eqn.31

Eqn.32

F

a

≈

F

n

X

S

in

+

S

in

+

S

a

S

n

Eqn.33

Eqn.34F

a ≈ F

n

X

R

ct

+

R

ct

+

R

p

R

n

Where,

F

n

= Actual Accuracy Limit Factor

F

n

=Rated Accuracy Limit Factor

S

in

= Internal Burden of the CT secondary Coil in VA

S

n

=

Rated Burden of the CT in VA

S

a

= Actual Burden of the CT in VA

Definition of ALF

ALF usually defined for the protection class current transformer, ALF can be stated as the ratio ofaccuracy limit primary current to the rated primary current

29



.

For protection current transformer

The ratio of accuracy limit primary current to the rated primary current is called Accuracy Limit Factor of current transformer.

Recommended value of ALF for different protections is stated below:

1. Definite time over current protection:

Relay will function perfectly if

ALF > 2 X (I

r

/I

sn

)

For a transformer outgoing , therewill usually be a high instantaneous set- point set on 14 Ir maximum which meansthat the necessary effective ALF is > 28.

For a motor feeder , we will generallyhave a high set-point set on 8 Irmaximum which means that the

necessary effective ALF > 16.

Suppose we have selected the below given CT for our purposeCT is rated as 300/5, 5P20, 10 VA

Suppose internal secondary coil resistance of the CT is 0.07 Ω and the secondary Side burden (includ ingwires and relay) is 0.117 Ω

Let us find out whether we have choosen right CT or not?

Example: 7

Using Eqn.30,31,32

F

n

= 20 (CT data 5P20),

S

in

= (5A)^2 × 0.07 Ω =1.75 VA

S

n

= 10 VA (from CT data),

S

a

=

(5A)^2 × 0.117 Ω = 2.925 VA

Using Eqn.34

Sol:

Hence, Fa > Fn

It shows we have chosen right CT

1.75+ 2.9225= 50.3

a

≈ 20 X1.75+10

Eqn.35

Where,

I

r

=Relay Threshold Setting/Relay Set point

I

sn

= Rated secondary current of CT in A

30



2. Inverse Definite Time Over current Protection

For inverse definite time overcurrent protection, recommended value of ALF is stated below:

3. Differential Protection :

For differential Protection, many manufactures recommend Class X CT‟s with Vk given below

Fault current calculation in determining the knee point voltage for differential protection

ALF > 20 X Ir

Where,

V

k

=

knee-point voltage in volts

a = coefficient which refers to the asymmetrical configuration or Saturation Factor

R

ct

= maximum resistance of the secondary winding in Ohms

R

b

= resistance of the loop (i.e. the return line) in Ohms

R

r

= resistance of the relay outside the differential part of the circuit in ohms

I

f

= the maximum fault current value measured by the CT in the secondary circuit for a fault outside the zone

to be protected in Amps (If=Isc/k n)

I

cc =

short-circuit current of the primary circuit

K

n

= CT ratio

V

k

≤ a X I

f

(R

ct

+ R

b

+ R

r

)

Eqn.36

Where,

I

sc sec

= CT secondary short-circuit current

I

r2

= Instantaneous high setting threshold for the module

Special Cases:1. If max. Short-Circuit Current is greater than or equal to 10 Ir

2.

If max. Short-Circuit Current is less than 10 Ir

3.

If the protection device has an instantaneous high threshold

ALF > 20

X

( I

r

/I

sn

)

ALF > 2

X

(

Isc Sec

./I

sn

)

ALF > 2

X

(I

r2

/I

sn

)

Eqn.37

Eqn.38

Eqn.39

Eqn.40

31



For Generator Differential:

1. If Isc is known,

2. If Isc is not Known, then

3. If rated Current as well as Short-circuit current is not known.

For Motor Differential:

1. If Isc is known,

2. If Isc is not Known,

3. If Rated Current as well as Short-circuit current is not known.

Fault Current I

f

=7 times of Rated current of generator / K

n

Fault Current I

f

=7 times of CT Sec. Rated current / K

n

Fault Current I

f

=I

sc

/K

n

Fault Current I

f

=7 times of Rated current of Motor / K

n

Fault Current I

f

=7 times of CT Sec. Rated current / K

n

Where,

I

sc

= Short circuit current of generator

Eqn.42

Eqn.43

Figure-5.3

Eqn.44

Where,

I

sc

= Short circuit current of Motor

Eqn.45

Eqn.46

Fault Current, If

=I

sc

/K

n

Eqn.41

Figure-5.2

32



For Transformer Differential:

14. Knee Point Voltage:

The main purpose of calculating the knee point voltage is to identify the core saturation characteristics.

This is mainly done on protection CT's. Knee Point Voltage signifies saturation level of CT core. Knee

point calculation is done for CTs used for differential protection and REF.

The Knee Point Voltage of CT varies from Relay to Relay as per their implementation for different Protection

Fault Current I

f

=20 times of CT sec. Rated Current

V k is defined as the point on curvefrom which a 10% increase involtage causes a 50% increase inthe magnetizing current Ie.

Eqn.47

Figure-5.4

Figure-5.5

Why would protective CT Require larger cross

section than measuring CT for same VA

Ratings?

Because of small ratio error for large currents. Betweenthe points A (Ankle Point) and B (Knee point) the

characteristics is linear. Working range of protective CT

lies between the point A and B as a protective CT is

expected to transform primary current linearly on to the

secondary for relatively large range of current about 20

times the full load current. Whereas measuring CT

usually operates at a point around the ankle point as the

measuring CT is expected to be accurate up to 120%

rated current only.Therefore, if the same material is used

for both the CTs,it is quite obvious that for the same VAratings, protective CT would require large cross section

and this would be bigger in size.

I n t e r e s t i n g F a c t s

S e c o n d a r y v o l t a g

e

Exciting Current

B

A Figure-5.6

33



Similarly

C H A P T E R - 6

How to do

CT Selection

Let‟s start with an example, this will make easy to understand the selection criteria for CT

Objective:

Selection of 600/1 Current transformer specification of a bus bar differential protection for Line feeder using

relay MICOM 746(An Areva Product).

Example: 8

Required parameters to specify CT:

i) Resistance/KM of selected cable which is used to connect CT to Relay.

ii) Length of cable required to connect the CT to Relay.

iii) Max. Sec. Winding Resistance (Rct)= 0.005 per primary tap/sec tap

iv) Fault Current of Internal Circuit or External Circuit

v) Class of CT(Here We are using CT for protection purpose, therefore PX Class is chosen)

Points to be remembered:

1. Lead Resistance ,R l = 2 X single lead resistance for single phase

= 1 X single lead resistance for three phases

2. Knee Point Voltage for External Fault is

3. Knee Point Voltage for Internal Fault is

Wiring characteristics - Calculation of rated resistive burden

Cable Cross section 4 mm² - Copper As per Manufacturer‟s Cable

Catalogue

Resistance per km 5.9 Ω/km As per Manufacturer‟s Cable

Catalogue

Tentative Distance 95 MtrDistance between the CT & Relay

(Assumed)

Maximum length (lead) for

single phase fault current2 X 95 Mtr


three phase fault Current

1 x 95 Mtr

Total resistance for single

phase fault current shall be 5.9 x 2 x 95 / 1000 =1.121Ω

V

k

≥

0.5I

fmax.

X

(R

ct

+2 R

l

)

Eqn.48

V

k ≥

I

fmax int.

X (R

ct

+2 R

l

) Eqn.49

34



Single Line Diagram:

Total resistance for three


Max. Sec. Winding

Resistance(Rct)0.005 X 600/1 =3Ω 600/1 represents CT ratio

Legends

A- 220KV Feeder

B- 220 KV Busbar

C- 220KV Grid line A B C

Load Flow

Fault occurs here is

called internal faults

Fault occurs here is called External faults, with If=12.583 KA

current

B

B

C C

A A Figure-6.0

For If=11.58KA

35



Calculation of Knee Point Voltage:

The rated knee-point voltage Vk required by the relay with safety factor for maximum external /Internal faultCurrent is

From case (1) and case (2), one can conclude that maximum value of Vk requirement shall be > 68.7236 V.The Value is divided by 600 to connect the fault current equivalent to secondary circuit.

Therefore Selected CT Specification is:

Case:1Vk (For External Fault) using Eqn.45 > 0.5 x 12583 x 1 x (3 + 0.5605) / 600

>37.335 V

Case:2 Vk(For Internal Fault) using Eqn.46 > 11581 x (3 + 0.5605) x 1 / 600 >68.723 V

Rated primary current600 A

Assumed

Rated secondary current 1 A Assumed

Accuracy classPX

As per protectionclass

Maximum secondary winding resistance (Rct)3Ω

As permanufacture

catalogue

Maximum exciting current (Ie) at the rated

knee-point(Vk/2) 30 mA As per standard

Rated knee-point voltage (Vk)

100V

Nearest value of

68.7236 which canbe manufactured

Example: 9

Objective:

Selection of Current transformer specification for energy metering for Line feeder using relay MICOM 746(An

Areva Product).

Required parameters to specify CT:

vi) Resistance of selected cable which is used to connect CT to Relay.

vii) Length of cable required to connect the CT to Relay.

viii) Max. Sec. Winding Resistance (Rct)= 0.005 per primary tap/sec tap

ix) Fault Current of Internal Circuit or External Circuit

x) Class of CT(Here We are using CT 0.2 class is chosen for energy meter as per table no.5.2 )

In case of selection of CT for metering purpose, we do not need to calculate knee point voltage as “protection”CT operating threshold usually kept very high to allow a relatively accurate measurement of the fault

current.Therefore, there is need to calculate knee point voltage in case of protection CT. Same is not true forCT selected for Metering purpose. The metering instruments do not need to withstand currents as high as the protection relays. This is why the “instrument” CTs, unlike the “protection” CTs, have the lowest possible SafetyFactor (SF) in order to protect these instruments from earlier Saturation.

36



Therefore Selected CT Specification is:

Rated primary current 600 A

Rated secondary current 1 A

Rated Output 15VA

Accuracy Class 0.2

Safety Factor/Security

Factor

5

Points to be remembered:

1. Lead Resistance ,R l = 2 X single lead resistance for single phase

= 1 X single lead resistance for three phases

Wiring characteristics - Calculation of rated resistive burden

Cable Cross section 4 mm² - Copper As per Manufacturer‟s Cable

Catalogue

Resistance per km 5.9 Ω/km As per Manufacturer‟s Cable

Catalogue

Tentative Distance 90 MtrDistance between the CT &

Relay(Assumed)


single phase fault current

2 X 90 Mtr

Total resistance for single


Equivalent Consumption in VA I²R = 1² X 1.062 = 1.062VA

Secondary current is 1 A as

600/1 CT is used in this example

Measuring devices characteristics - Calculation of apparatus consumption

Energy meter and Multi function meter are supposed to be connected to current transformers.

The total consumption is supposed to be approximately 6 VA (each of 3 VA).

Selected rated output

Thus, the actual burden is about 1.062 + 6 = 7.062 VA. Selected rated output for CT is 15 VA (> Actual Burden)

While considering the application of a CT it should be remembered that the total burden imposed on the

secondary winding is not only the sum of the burden(s) of the individual device (s) connected to the winding butalso includes the burden imposed by the connecting cable and the resistance of the connections.

This factor indicates that selected CT

metering core must not saturated when

current ≤5 times the rated current.

For a instance, Suppose one Current

Transformer has rating 100/1A and SF

is 2 and another Current Transformer

has same rating with SF 5.It means, in

first CT, the metering core would be

saturated at 2 X100 or 200 A,whereas second CT, core will be

saturated at 5X100 or 500A.Means

whatever may be the primary current

of both CTs, secondary current will not

increase further after 200A & 500A of

primary current of the CTs

respectively

38







Why measuring CT

is not advisable to

use as protection

CT?

A metering CT is made to be very accurate at expected load

current levels where as a protection CT is made to be accurate atfault current level (which is often significantly above load current).If measuring CT used as a protection CT, it is likely to saturate soonand it will not give actual fault current feedback.

Measurement of current

The output required from a current transformer depends on the application and the type of load connectedto it:

1. Metering equipment or instruments, like kW, kVar, Amp instruments or kWh or kVArh meters, aremeasuring under normal load conditions. These metering cores require high accuracy, a lowburden (output) and a low saturation voltage. They operate in the range of 5-120% of rated currentaccording to accuracy classes:

a. - 0.2 or 0.5 as per IEC

b.

- 0.15 or 0.3 or 0.6 as per IEEE

2. For protection relays and disturbance recorders information about a primary disturbance must betransferred to the secondary side. Measurement at fault conditions in the over current rangerequires lower accuracy, but a high capability to transform high fault currents to allow protectionrelays to measure and disconnect the fault.

a. Typical relay classes are 5P, 10P, PR, PX or TP (IEC) or C 100-800 (IEEE)Remember:To protect the instruments and meters from being damaged by high currents during fault conditions, ametering core must be saturated typically between 5 and 20 times the rated current. Normally energy

meters have the lowest withstand capability, typically 5 to 20 times rated current.

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 voltage at

low flux density.

it operate only at ankle point

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.

it operate from ankle point to knee point

Ankle Point

Knee Point

39



Eqn.47

Eqn.46

Number of turns, N = 2000 / 5 = 400 Turns

Max. Current to flow at secondary of CT, IS MAX = 40,000 / 400 = 100 Amps

Maximum Secondary Winding Voltage:

E

K

= 4.44 x B x A x f x N Volts

where :-E

K

= 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

UsingEqn.50

E

K

= (4.44 x 1.6 x 20 x 50 x 400)/100² = 284.16 Volts

Therefore,

Maximum Burden = E

k

/ IS

MAX

= 284 / 100 = 2.84 Ohms

Hence, Maximum burden that can be connected to CT is:

2.84 - 0.31 = 2.53 Ohms

Example: 10

SOL

Find maximum secondary burden permissible if no saturation is to occur.Details of CT are given below:

C.T. Ratio = 2000 / 5A Max Flux Density = 1.6 TR S = 0.31 Ohms Core C.S.A = 20 cm2

I MAX Primary = 40 kA

I N T E

R E S T I N G

F A C T S In current transformer design, the core characteristics must be carefully selected because excitation

current Ie essentially subtracts from the metered current and affects the ration and phase angle of the

output current.Higher the Value of excitation current or core loss the will be the current

The excitation current determines the maximum accuracy that can be achieved with current transformer

We can say that,

Permeability of core of material high core loss low Excitation current small

Current error small

Eqn.50

40



C H A P T E R - 7

Current Transformer

Transient Performance

Large asymmetrical primary fault currents with a decaying

dc component.

Residual magnetism left in the core from an earlier

asymmetrical fault, or field-testing, if the CT has not been

demagnetized properly.

Large connected burden combined with high magnitudes of

primary fault currents.

CT output is impacted drastically when the CT operates in the nonlinear region of its excitation characteristic. Operation in

this region is initiated by:

Short-circuit current has DC component

flowing for several tens of milliseconds.

DC components cause saturation of the

CT core resulting in error. DC component

is a problem for transformer as the DC

component tends to create more flux in the

core that adds to the main flux driven by

AC voltage. Therefore, longer the system

time constant, the more likely the CT is tosaturate

OFFSET FAULT CURRENT: CT secondary waveform when severe AC saturation occurs

Figure-7.0

DC

AC

Saturation due to DC transient distorts AC

waveform output

In practical application the DC current is not

sustained but decays with a time

constant(X/R).Where X is inductance in henry

and R is resistance in Ω of the pri mary power

system

DC Components produces almost constant

magnetic flux in comparison with magnetic flux produced by 50Hz.

ø

I

p

I

S

t

t

t

It is clear that to produce the DC offset along with an AC waveform without distortion, production of much larger steel

core flux required to normalize the effect of flux produced by the DC transient/offset. As a result, CT Size increases

drastically whereas CT‟s dimension will be a limiting factor. Therefore, installation of such CT is impractical.

41

Introduction



I (t) =V

m

sin(

ω

t + Ѳ- ø)

Z

)

-(R/L)t

(t) =V

m

(

sin (

ω

t + Ѳ- ø)- sin( Ѳ- ø)

Z

Eqn.53

Where,

Z =

R

2

+

ω

L

2

ø= tan

1

ω

L

R

Eqn.54

Eqn.55

V

s

(t)=V

m

sin(wt +Ѳ)

From the circuit analysis, we know that t→∞

We know that at initial condition (t0), I (0) =0 e

-(R/L)t And there is thumb print

Eqn.51

Eqn.52

After math analysis, we can get

Figure-7.1

42

Transient Performance: Calculations

X=ωL

Eqn.56

Power factor = Cos ø =

tan

1

)

R

Cos(

X Eqn.57

If p.f=1,then the impedance only has resistance

If p.f=0, then the impedance only has reactance

p.f increases as X/R decreases or vice-versa

0

1

0.5

Pure reactive or inductive

Pure Resistive

Figure-7.2

X/R ratio or time constant importance:

Responsible for the decaying DC component in fault current. DC component produces almost constant magnetic

flux in comparison with magnetic flux produced by 50HZ sine wave.



For a transformer in service supplying normal load

.The flux requirement is very small and therefore

exciting current is very small and secondary current

is almost a faithful reproduction of the primary

current. Suppose, a short-circuit occurs at the

primary side. During short-circuit condition, CT

primary current increases results in an increased

secondary current.

The increased secondary current results in a higher

voltage drop across the CT winding resistance and

connected burden of the CT and results in higher

excitation voltage.

Higher excitation voltage creates more flux which is

enough to cause saturation of the transformer core.

Hysteresis loop becomes negligible for this high level

of excitation.

where :-

K

s

= Saturation Factor

V

s

= Saturation Voltage of CT (Volt)

I

sec

= Sec. Current at the CT

Z

s

= Total Sec. Impedance of CT circuit

f = System frequency

X = System reactance

R = System resistance

How

How

DoesCT saturation

Occurs?

K

s

=

V

s

I

sec

x Z

s

Eqn.56

X/R

K

s

-1 x ln

ᴨ

f

T

s

=

-X/R

) - (

Eqn.57

43

To Avoid Saturation, CT shall have enough capacity to develop the following voltage:

Vs = X I

f

(R

ct

+ R

b

+ R

r

)

R

X

1+

(

)

Eqn.58

X/R ratio is important because it determines the peak asymmetrical fault current. X/R ratio is

responsible for the decaying DC component in fault current.

As per C37-110.1996, primary current asymmetry and CT saturation due to DC offset current

component are the primary component while performing CT calculation whereas symmetrical

primary current is no longer acceptable while performing CT calculation.

With the increase of X/R ratio the difference between the min-fault current and the steady state fault

current increases. This means that for highly reactive fault path the current measured by CT in the

first few cycles is significantly smaller than the actual fault current. As right after a fault occurs, the

current waveform is no longer sine wave. Instead, it can be represented by the sum of a sine wave

and a decaying exponential. This decaying exponential or DC wave added to sine wave (as shown in

Figure-7.0

) causes the current to reach a much larger value than that of the sine wave alone and thusbecomes asymmetrical current waveform(Combination of sine wave and decaying exponential

waveform).



Equipment requires in performing Testing on CT are:

Insulation tester1. Polarity Tester2.

Multimeter3. Variac (0-2KV )4.

2.5Sq.MM

connecting wire 6.Primary

injection Kit 5.

C H A P T E R -8

Current Transformer

Testing & Commissioning

44



.

Mechanical Checking and Visual

Inspection before CT testing

1. Verify nameplate & Ratings mentioned on nameplate. It should be in accordance with theapproved drawings and specifications.

2. Inspect for any physical damage/ defects and mechanical condition.3. Verify correct connection of transformers with system requirements.4. Verify that adequate clearances exist between primary and secondary circuit wiring.5. Verify tightness of accessible bolted electrical connections by calibrated torque-wrench method.6. Verify that all required grounding and shorting connection has been done.7. Verify all shorting blocks are in correct position, either grounding or open as required.8. Verify single point grounding of each core done properly. Grounding point shall be nearer to the

CT location. However grounding shall be at relay point in case of several CT secondariesconnected together like differential protection.

C h e c k i n g &

I n s

p e c t i o n

TESTS ON CURRENT TRANSFORMER

TYPE TEST ROUTINE TESTType tests are intended to verify compliance

with the requirements laid down in the relevant

Standards for a given type of equipment. Type

tests are carried out on a sample of such

equipment or on such parts of equipments

manufactured to the same or similar design.

They shall be carried out on the initiative of the

manufacturer.

Routine tests are intended to detect faults in

materials and workmanship. They are carried

out on every equipment after its assembly or on

each transport unit. Routine Tests on

equipments are normally undertaken at the

manufacturer's premises

Routine Tests on equipments are not intended

to be repeated on site.

Verification of terminal markings and

polarities.

Short-time current test

Temperature rise test

Impulse voltage test

Power frequency voltage withstand test on

Primary

Overvoltage inerturn test

Error measurement

Verification of terminal markings and

polarities.

High voltage power frequency

withstand test on Secondary

Overvoltage inerturn test

Determination of errors and accuracy

class

Type tests are: Routine tests are:

45

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1. Insulation Resistance (IR) test

Field tests to be performed before Commissioning of current transformer

The insulation resistance (IR) test is also commonly known as a Megger test.This test is a spot insulation test

which uses an applied DC voltage (typically 250Vdc, 500Vdc or 1,000Vdc for low voltage equipment <600V and2,500Vdc or 5,000Vdc for high voltage equipment) to measure insulation resistance in either kΩ, MΩ or GΩ. Themeasured resistance is intended to indicate the condition of the insulation or dielectric between twoconductive parts.

Higher the resistance, better the condition of the insulation. Ideally, the insulation resistance would be infinite,but as no insulators are perfect, leakage currents through the dielectric will ensure that a finite (though high)resistance value is measured.

Higher the value of resistance indicates the healthy condition of

insulation. Usually, Resistance value should be in MΩ or GΩ.

It is a non-destructive test as DC voltages do not cause harmful and/or cumulative

effects on insulation materials and provided the voltage is below the breakdown

voltage of the insulation, does not deteriorate the insulation. IR test voltages are

all set within the safe test voltage for most (if not all) insulation materials.

Advantage

Of the IR test

46

The insulation resistance test is applied before the first highvoltage energizing off the equipment, during routinemaintenance and after repair before re-commissioning.

Nominal Circuit

Voltage (Vac)

Test Voltage

(Vdc)

Insulation

Resistance (MΩ)

250 500 25

600 1000 100

1000 1000 100

2500 1000 500

5000 2500 1000

8000 2500 2000

15000 2500 5000

25000 5000 20000

34500 and above 15000 100000

Minimum values for IR tests from manufacture’s catalogue Table-8.0

Nominal

Circuit

Voltage(Vac)

Test

Voltage

(Vdc)

Insulation

Resistance

(MΩ)

Extra low

Voltage250 ≥0.5

Up to 500V 500 ≥1.0

Above

500V1000 ≥1.0

Minimum values for IR tests as per IEC

60364-6 [1] Table 6A

Table-8.1



2. Polarity Test

SECONDARYPRIMARY

TEST 1Primary winding to core

Temporary short for the duration of the test

TEST 2Secondary winding to core

Figure-8.0

MΩ 500V

Insulation resistance tests between each winding and

earth

Temporary short for the duration of the test

MΩ 500V

SECONDARYWINDING

PRIMARY

WINDING

Figure-8.1

Insulation resistance tests between windings

Insulation resistance of the order of several hundred/thousand mega ohms to infinity is necessary between phase

conductor and earth, between two or more phase conductors, between phase conductors and the neutral

conductor, between two conducting parts separated by insulation.

Polarity test is conducted to confirm the polarity marking on the CT primary and secondary and verify

it is matching with concerned drawing. Moreover it gives an idea, how to connect the secondary‟s tomake the protection like directional, differential and metering function properly.

-Ve+Ve

Battery 9V

P1 P2

S2S1

- +

Connect battery –ve terminal to the current transformer P2primary terminal. This arrangement will cause current to flowfrom P1 to P2 when +ve terminal is connected to P1 until theprimary is saturated. If the polarities are correct, a momentarycurrent will flow from S1 to S2.

Closing and opening the battery switch frequently which isconnected at the primary side will give deflection ingalvanometer. Correct polarity of CT shall be confirmed onlyif the galvanometer pointer is moving +ve direction whileclosing and –ve direction while opening for correct polarity.

How to testCT Polarity?

Isolate CT secondary from the load and put the galvanometer across the secondary of CT as shown in

Figure-8.2

Figure-8.2

Procedure for IR test of current transformer

InterviewQuestion

No back E.M.F is induced while DC supply is given to transformer in the transformer primary

winding. Importance of Back E.M.F is that it limits the current drawn by the machine. In theabsence of back E.M.F transformer starts drawing excessive currents that leads to the

burning of primary winding.

What happen if Transformer is given DC supply?

47



3. Burden Test

Burden test is to ensure the connected burden to CT is within the rated burden which is mentioned on thenameplate.

Injected the rated secondary current of the CT, from CT terminals towards load side by isolating the CTsecondary with all connected load and observe the voltage drop across the injection points. The burdenVA can be calculated as

Burden VA = Voltage drop x rated CT sec. Current

Burden testing of metering class CT involves measuring the secondary current of the CT while introducingadditional burden in graduated steps up to the rated burden of the CT and assuming constant customerload during the duration of the test. If the measured secondary current remains constant during theduration of the test then the CT is deemed to have passed the burden test as rated . Thus, this test ensuresCT‟s ability to feed the rated burden as mentioned on nameplate. If the Connected load is changing duringthe burden test then the secondary current will correspondingly change giving a false indication of burdenerror . This is a major limitation of burden testing

Hence when a burden error is detected a tightening of the screws in that circuit will usually correct theerror. Otherwise, a search for and elimination of the high burden component will be necessary.

Burden test is conducted to check CT‟s abilit to deliver current.

The calculated burden should be less than rated CT burden.Limits:

Eqn.59

Note: Ammeter selector switch should be at respective phase during test.

High impedance relays shall be shorted during the test.

Whenever a burden error is detected we should firsttightening the screws in circuit, if the problem persist thenwe need to remove extra burden on CT

BurdenERRORCaused by:

Long secondary wires

Wrongly connection

Loose connections

Corroded connections

48



4. Protection CT Magnetization Curve Test

It is necessary to test the characteristics of a CT before it is put into operation, since the Results produced bythe relays and meters depend on how well the CT behaves under normal and fault conditions.

This test shall be conducted before ratio test and after secondary resistance and polarity test, since residualmagnetism left in the core due to DC test (polarity, resistance), which leads additional error in ratio test. Themeters used for this test shall be having true RMS measurement.

Demagnetisation

Before starting the test, demagnetize the core by Injecting voltage on secondary terminals and increasevoltage gradually till there is considerable increment in current with a small voltage increment. Now startdecreasing the voltage to zero, the rate at which increased.

Magnetisation test

Now increase the voltage and monitor the excitation current up to the CT reaching near to saturation point.

Record the reading of voltage and current at several points. Plot the curve and evaluate the V

k

and I

mg

from the graph

Suppose we have selected Protection Current transformer 5P 10 for some application in which CT secondarywire resistance is 0.1Ω.

Errors in CΤ happened because of excitation current, therefore, in order to check CT functioning is correct or not, it

is essential to plot excitation curve.

The magnetization current I

e

of a CT depends on the cross section, length of the magnetic circuit, the number of

turns in the windings, and the magnetic characteristics of the material.

From Eqn.2

Voltage across the magnetization impedance, E

s

, is directly proportional to the secondary current. I

s

, when the

primary current and therefore the secondary current is increased, these currents reach a point where the core

commences to saturate and the magnetization current becomes sufficiently high to produce an excessive error.

While investigating the behavior of a CT, the excitation current should be measured at various values of voltage,

this test is also known as secondary injection test.

E

s

= I

s

x Z

Objective:

Study on current transformer-Protection CT magnetization curve test

Example: 11

49





5. Turns ratio Tests/Primary Injection Test

This test is to ensure the turn‟s ratio of CT at all taps. The primary current of minimum of 25% rated primarycurrent to be injected on primary side of CT with secondary‟s shorted and the secondary current can bemeasured and recorded for all cores.

The ratio test is performed by comparing voltage applied to the secondary winding to the resulting voltage

produced on the primary winding.

As an example, if one volt per turn is applied to the secondary winding, the voltage present on the primary

winding would be one volt. More specifically, if 120 Volts were applied to the secondary of 600/5 CT (120:1),

one volt would be present on the primary winding.

Alternative method to perform Turn Ratio test

To protect against insulation failure, do not exceed more than 1000 volts on any of the

secondary windings of the CT under test

Primary injection test is done forensuring that CT respected coreare properly connected and thereshould be no mix up of phase.Hence, Phase identification can bedone through primary injection test.

Figure-8.4

R, Y and B Temporary shorted

Primary Injection

=

V

1

V

2

N

1

N

2

Hence, turns ratio is 2

InterviewQuestionWhat will be the turns ratio required to match an 80 Ω source to a 320 Ω load?

V

1 X

I

1

=V

2 X

I

2

Condition for matching impedance is Eqn.60

X 1

V

1

R

1

= V

2

V

2

R

2

Now solve Eqn.54=

V

1

V

2

)

2

R

1

R

2

Eqn.61

=

V

1

V

2

)

2

320

80 =2

51



1. Instrument transformer: A transformer intended to supply measuring instruments, meters, relays and other

similar apparatus.

2. Current Transformer: A current transformer in which the secondary current, in normal conditions of use ,is

substantially proportional to the primary current and differs in phase from it by an angle which is approximately

zero for an appropriate direction of connections.

3. Primary Winding: The winding through which the current to be transformed flows.

4. Secondary Winding: the winding which supplies the current circuits of measuring instruments, relays or similar

apparatus.

5. Rated Primary current: The value of primary current on which the performance of the transformer is based.

6. Rated Secondary current: The value of secondary current on which the performance of the transformer based.

7. Actual Transformation Ratio: The ratio of primary current to secondary current.

8. Current Error Ratio: The error which a transformer introduces into measurement of a current and which arises

from the fact that the actual transformation ration is not equal to the rated transformation ratio

9. Phase Displacement: Difference in phase between the primary and secondary current vectors, the directions ofvectors being so chosen that the angle is zero for prefect transformer.

Phase error is said to be positive when the secondary current vector leads the primary current vector. It is usually

expressed in minutes or centiradians.

10. Accuracy Class: A designation assigned to current transformer the error of which remain within specified limits

under prescribed conditions of use.

11. Burden: the impedance of secondary circuit in ohms and power-factor. The burden is usually expressed as the

apparent power in volt-amperes absorbed at a specific power factor and at the rated secondary current.

Current error ( ) =

(K

a

X I

s

- I

p

)

I

p

X 100

Where

K

a

= rated transformation ratio

I

p

=actual primary current (A)

I

s

= actual secondary current when Ip is flowing under the conditions of measurement (A)

52

C H A P T E R -9

Terms and

Definitions as per standard

Eqn.62



12. Rated Burden: The value of apparent power (in Volt-ampere at specific power factor) which the transformer intended

to supply to the secondary circuit at the rated secondary current and with rated burden connected to it.

13. Highest voltage for equipment: the highest r.m.s phase to phase voltage for which a transformer is designed in

respect of its insulation.

14. Rated Insulation Level: the combination of voltage values which characterizes the insulation of a transformer with

regard to its capability to withstand dielectric stresses.

15. Rated short-time thermal current(Ith): the r.ms value of the primary current which a transformer will withstand for

one second without suffering harmful effects, the secondary winding being short-circuited.

16. Rated Dynamic current(Idyn): the peak value of primary current which a transformer will withstand without being

damaged electrically or mechanically by resulting electromagnetic forces, the secondary winding being short-circuited

17. Rated continuous thermal current(Icth):the value of current which can be permitted to flow continuously in the

primary winding ,the secondary winding being connected to the rated burden, without the temperature rise exceeding

the values specified.

18. Exciting Current: the r.m.s value of the current taken by the secondary winding of current transformer, when a

sinusoidal voltage of rated frequency is applied to the secondary terminals, the primary and any other windings

winding being open-circuited.

19. Secondary winding resistance (Rct): Secondary winding d.c resistance in ohm connected to 75⁰C or such other

temperature as may be specified.

20. Measuring Current Transformer: a current transformer intended to supply indicating instruments, integrating meters

and similar apparatus.

21. Rated instrument limit primary current(IPL):the value of the minimum primary current at which the composite

error of the measuring current transformer is equal to or greater than 10%,the secondary burden being equal to the

rated burden.

22. Instrument Security Factor(IFS): the ratio of rated instrument limit primary current to the rated primary current

23. Secondary limiting E.M.F: the product of instrument security factor FS, the rated secondary and the vectorial sum of

the rated burden and the impedance of the secondary winding.

24. Protective current transformer: a current transformer intended to supply protective relay

25. Rated accuracy limit primary current: the ratio of the rated accuracy limit primary current to the rated primary

current

26. Accuracy limit factor: the ratio of the rated accuracy limits primary current to the rated primary current.

Measuring Current transformer

Protective Current transformer

53



27. Secondary limiting e.m.f: the product of accuracy limits factor, the rated secondary current and the vectorial sum

of the rated burden and the impedance of the secondary winding.

28. Saturation Flux: the peak value of flux which would exist in a core in the transition from the non-saturated to the

fully saturated condition and deemed to be that point on the B-H characteristics for the core concerned at which

10% increase in B causes H to be increased by 50%.

29. Remanent flux: the value of flux which would remain in the core 3 Min after interruption of an exciting current of

sufficient magnitude to induce the saturation flux.

30. Remanence Factor(Kr): the ratio Kr=100 X (Saturation Flux / Remanent flux),expressed in %

31. Rated knee point E.M.F: the minimum sinusoidal e.m.f( r.m.s) at rated frequency when applied to the secondary

terminal of the transformer ,all other terminal being open-circuited, which when increased by 10% causes the r.m.s

exciting current to increase by no more than 50%.Note: Actual knee point e.m.f > rated knee point e.m.f

32. Turns ratio error: the difference between the rated and actual turns ratios, expressed in %

33. Dimensioning Factor (K x ):a factor assigned by the purchaser to indicate the multiple rated secondary current

(Isn) occurring under power system fault conditions, inclusive of safety factor ,up to which the transformer is

required to meet performance requirements.

Turn ratio Error ( ) = (Actual turns ratio – Rated turns ratio)

X 100

Rated Turns ratio

Eqn.63

Metering Functions of a

Current Transformer

Current across the secondary winding of

a transformer is directly proportional to

the current across the primary winding.

Direct measurement of current acrossthe secondary winding allows an indirect

calculation of the current conducted

across the primary winding.

Measurements may be used for

calculating energy usage in power

supply billing.

54



A current transformer can perform following functions

Monitoring Functions of a Current Transformer

Current measurements across the secondary winding may be used to monitor and

display the current conducted along a circuit or power line to check for spikes and

drops.

Control Functions of a Current Transformer

Measurements from the secondary winding of a current transformer may trigger

switches in controllers such as ground fault circuit interrupter (GFCI) circuit breakers

when measurements exceed allowed limits.

Protection Functions of a Current Transformer

Metering, monitoring, and control equipment used with AC power supplies can be

damaged by high currents. Transformers can step-down current so that appliance

circuitry is protected. For this reason, current transformers are often termed

instrument transformers.

Relaying Functions of a Current Transformer

Power grid transformers may be used to increase or decrease the current of the

power supply. A step-up transformer decreases the current and increases the

voltage of power from an energy plant generator before transmission over long

distance power lines. This reduces the costs of transmission by permitting use of

smaller power lines with less energy loss. Step-down transformers increase current

at the point of use.

Current Transformers Perform Critical Roles in the Use of Electricity

Without current transformers, long distance transmission of electricity would not be

cost-effective. Home use of AC power would not be safe. The metering, monitoring,

relaying, control and protection functions of current transformers enable the essential

role electricity plays in high technology societies.

1

2

3

4

5

C H A P T E R -10

Current Transformer

Summary

55

http://www.electronics-manufacturers.com/products/electrical-electronic-components/transformer/current-transformer/#_blank

http://www.usbr.gov/pmts/client_service/recent/studytransformers.pdf#_blank

http://www.usbr.gov/pmts/client_service/recent/studytransformers.pdf#_blank

http://www.electronics-manufacturers.com/products/electrical-electronic-components/transformer/current-transformer/#_blank



Applications of Current Transformer

Measuring CT

The principal requirements of a measuring CT are that, for primary currents up to 120%

or 125% of the rated current, its secondary current is proportional to its primary current toa degree of accuracy as defined by its “Class” (See table 5.2) and, in the case of the more

accurate types, that a specified maximum phase angle displacement is not exceeded.

A desirable characteristic of a measuring CT is that it should “saturate” when the

primary current exceeds the percentage of rated current specified as the upper limit

to which the accuracy provisions apply.

6

7

Protection CT

On the other hand the reverse is required for the protective type CT, the principal purposeof which is to provide a secondary current proportional to the primary current when

it is several, or many, times the rated primary current. The measure of this

characteristic is known as the “Accuracy Limit Factor” (A.L.F.).

A protection type CT with an A.L.F. of 10 will produce a proportional current in the

secondary winding (subject to the allowable current error) with primary currents up to a

maximum of 10 times the rated current.

It should be remembered when using a CT that where there are two or more devices

to be operated by the secondary winding, they must be connected in series across

the winding. This is exactly the opposite of the method used to connect two or moreloads to be supplied by a voltage or power transformer where the devices are

paralleled across the secondary winding.

I

N T E R E S T I N G F A C

T S Comparison face Power Transformer Current transformer

No. of cores One core Two core or more

No. of primary turns At least more than oneturns

One turns( often 2 turns)

Rating Power MVA VA

Changing of secondaryvoltage

Using tap changer Voltage reduction

Load Connection Parallel Series

Power Transformer V/S Current Transformer

56



Points to be remembered

Frequency affects

At very low frequency, lines of flux generated by the primary current begin to appear as DC;

a C/T needs the AC CYCLE changes to induce the secondary current. With toroidal C/T,you will experience a drop in accuracy if the frequency goes down from 60 Hz. One can

manufacture a C/T with an exotic metal core that is not quite as affected as the silicon

grain oriented steel most commonly used, but the improvement would be questionable and

at high cost.

Instrument Safety Factor for Measuring CT

The rated Instrument Security Factor (FS) indicates the over current as a multiple of the

rated current at which the metering core will saturate. It is thus limiting the secondarycurrent to FS times the rated current. The safety of the metering equipment is greatestwhen the value of FS is small. Typical FS factors are 5 or 10. It is a maximum value andonly valid at rated burden.To protect the instruments and meters from being damaged by high currents duringfault conditions, a metering core must be saturated typically between 5 and 20 timesthe rated current. Normally energy meters have the lowest withstand capability,typically 5 to 20 times rated current.

Accuracy limit factor for Protection CT

Require accuracy up to many times rated current, thus use grain orientated silicon steel

with high saturation flux density that makes Protection CT to operate from ankle point to

knee point

CT Selection

If you want an indication, the first thing you need to know is what degree of accuracy is

required. For example, if you simply want to know if a motor is lightly or overloaded, a panel

meter with 2 to 3% accuracy will likely suit your needs. In that case the current

transformer needs to be only 0.6 to 1.2% accurate. On the other hand, if you are going

to drive a switchboard type instrument with 1% accuracy then you need a currenttransformer with a 0.3 to 0.6 accuracy.

You must keep in mind that the accuracy ratings are based on rated primary current

flowing. As mentioned earlier, the rated accuracies are at stated burdens. You must take

into consideration total burden which includes the burden of the current transformers

secondary winding, the burden of the leads connecting the secondary to the load, and of

course, the burden of the load itself. The current transformer must be able to support the

total burden and to provide the accuracy required at that burden. If you are going to drive a

relay you must know what relay accuracy the relay will require.

8

9

10

11

57



17

Points to be remembered

Interposing CT’s

Interposing CT‟s are used when the ratio of transformation is very high. It is also

used to correct for phase displacement for differential protection of transformer.

Core balance CT (CBCT)

The CBCT, also known as a zero sequence CT, is used for earth leakage and earth faultprotection.In the CBCT, the three core cable or three single cores of a three phasesystem pass through the inner diameter of the CT.When the system is fault free, no current flows in the secondary of the CBCT. Whenthere is an earth fault, the residual current (zero phase sequence current) of the systemflows through the secondary of the CBCT and this operates the relay. In order to designthe CBCT, the inner diameter of the CT, the relay type, the relay setting and the primary

operating current need to be furnished

CT Secondary Load

The CT secondary load = Sum of the VA’s of all the loads (ammeter, watt meter,transducer etc.) connected in series to the CT secondary circuit + the CT secondarycircuit cable burden.

CT Accuracy

Accuracy of a CT is another parameter which is also specified with CT class. For example,if a measuring CT class is 0.5M (or 0.5B10), the accuracy is 99.5% for the CT, and themaximum permissible CT error is only 0.5%.

12

13

14

15CT Magnetization

As in all transformers, errors arise due to a proportion of the primary input current beingused to magnetize the core and not transferred to the secondary winding. The proportion ofthe primary current used for this purpose determines the amount of error. Thus, essence ofgood design of measuring current transformers is to ensure that the magnetizing currentis low enough to ensure that the error specified for the accuracy class is notexceeded. This is achieved by selecting suitable core materials and the appropriate cross-sectional area of core.

16Standard CT burdens are defined in IEEE Std. C57-13-1993

Metering burdens are B0.1, B0.2, B0.5, B0.9, and B1.8 where each number represents thetotal impedance at a 0.9 power factor. VA for each burden is 2.5, 5.0, 12.5, 22.5, and 45.Relay burdens are B1, B2, B4 and B8 where each number represents the total impedanceat a 0.5 power factor

58



BIBLIOGRAPHY

59

General Requirements for Transformers, Regulators, and Reactors Publ. C57.11 1948;

American Standard Requirements, Terminology, and Test Code for Instrument

Transformers Publ. C57.13-1954; and Guide for Loading and Operation of Instrument

Transformers Publ. C57.33, American Standards Assoc., Inc., 70 East 45th St., NewYork 17, N. Y.

Application Guide for Grounding of Instrument Transformer Secondary Circuits and

Cases Publ. 52, American Institute of Electrical Engineers, 33 West 39th St., New York18,

ASA C57.23, see Reference 1.

ÒA Simple Method for the Determination of Bushing-Current-TransformerCharacteristics by S. D. Moreton, AIEE Trans., 62 (1943), pp. 581-585. Discussions,

pp. 948-952. ÒA Simple Method for Determination of Ratio Error and Phase Angle in Current

Transformers by E. C. Wentz, AIEE Trans., 60 (1941), pp. 949-954. Discussions, p.1369.ÒA Proposed Method for the Determination of Current-Transformer Errors,Ó by G.

Camilli and R. L. Ten Broeck, AIEE Trans., 59 (1940), pp. 547-550. Discussions, pp.1138-1140.

Overcurrent Performance of Bushing-Type Current Transformers by C. A. Woods, Jr.,and S. A. Bottonari, AIEE Trans., 59 (1940), pp. 554-560. Discussions, pp. 1140-1144.

Computation of Accuracy of Current Transformers by A. T. Sinks, AIEE Trans., 59

(1940), pp. 663-668. Discussions, pp. 1252-1253. ASA C57.13, see Reference 1.

Current Transformers and Relays for High-Speed Differential Protection, withParticular Reference to Offset Transient Currents by W. K. Sonnemann and E. C. Wentz,AIEE Trans., 59 (1940), pp. 481-488. Discussions, p. 1144.

Transient Characteristics of Current Transformers during Faults, Óby C. Concordia,,

C. N. Weygandt,, and H. 3. Shott, AIEE Trans., 61 (1942), pp. 280-285. Discussions, pp. 469-470

Transient Characteristics of Current Transformers during Faults, Part II,Ó by F. S. Rothe

and C. Concordia, AIEE Trans., 66 (1947), pp. 731-734.

The Effect of Current-Transformer Residual Magnetism on Balanced-Current or

Differential Relays by H.T. Seeley, AIEE Trans., 62 (1943), pp. 164-168. Discussions,

p. 384.

Peak Voltages Induced by Accelerated Flux Reversals in Reactor Cores Operating above

Saturation Density by Theodore Specht and E. C. Wentz, AIEE Trans., 65 (1946), pp. 254-263.

Overvoltage’s in Saturable Series De vices,Ó by A. Boyajian and G. Camilli, AIEE Trans.,70

(1951), pp. 1845-1851. Discussions, pp. 1952-1853.



Applicable standards

IEC

IEC 185:1987 CTs

IEC 44-6:1992 CTs

EUROPEAN

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

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