current transformers

..........................................................................Collection Technique

Cahier technique no. 194

Current transformers:how to specify them

P. Fonti

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Cahier Technique Schneider Electric no. 194 / p.1

no. 194Current transformers:how to specify them

ECT 194 first issue, February 2000

Paola FONTI

INPG engineer (Institut National Polytechnique de Grenoble). Graduatein 1970.Joined Merlin Gerin in 1981 as consultant and head of the MediumVoltage Export engineering and design department. She is currentlyresponsible for the MV project completion and tender support groupfor Schneider Electric.


Lexicon

If: maximum through current crossing aprotected area.Is: current threshold setting.kn: nominal accuracy limit factor (ALF) of a CT(associated with its accuracy load).kr: real ALF of a CT associated with its real load.Pi: (=Rct In2). Internal losses of the CT at In.Pn: (=Rn In2). Accuracy power of the CT.Pr: (=Rr In2). Real load consumption of the CTat In.RL: wiring resistance.

Rp: protection relay resistance.ALF: accuracy limit factor.CT: current transformer.Overrating of a CT: selection of a CT whoseprimary In is greater than the In immediatelygreater than the load In.Matching, auxiliary or interposing CT: lowvoltage CTs installed at the secondary of themain CTs for correcting a ratio and/or the currentphase shift.SF: security factor.


Current transformers:how to specify them

Contents1 Introduction p. 42 Network disturbances and protections 2.1 Disturbances p. 7

2.2 Protections p. 73 Current transformers 3.1 Reminder of ferromagnetic transformers p. 9

3.2 Non-magnetic transformers p. 133.3 CT manufacturing and implementation p. 13

4 Choosing CTs according to protections 4.1 Choosing CT ALF according to protections p. 15and applications 4.2 Characterising CTs according to applications p. 17

4.3 Special case of differential protection p. 214.4 Distance protections p. 26

5 CT specification examples 5.1 Motor feeder protections p. 275.2 Transformer feeder protections p. 275.3 Transformers differential protection p. 285.4 Differential protection for busbars (87B) p. 28

6 Conclusion p. 31Bibliography p. 32

Electrical power management requires implementation of data processingunits able to monitor networks or equipment and, as applicable, to initiatethe appropriate actions...Data sent by current transformers are processed by protection, control andmonitoring units that send signals to operate switchgear and/or informationto a supervisory unit or to a central control room.The task, consisting of identifying and rating the current transformers andassociating them with the protection and/or metering units, has alwaysgiven rise to problems, both for electrical engineers (oversizing thecharacteristics) and for the manufacturer (random feasibility, excessivesizing, high costs).This document does not cover the technical demonstrations amply referredto in the literature (see Cahiers Techniques no. 164 and 170). Its purposeis to remind users of a few simple rules enabling the best possibledefinition of the secondary characteristics of a current transformer (CT)according to the protections and applications concerned.Practically, it offers constructive assistance for technicians who havereached a dead end:c either because they do not possess the necessary information,c or because the results of their design have led to current transformersthat cannot be manufactured by the potential suppliers.


The design of MV and HV electrical networks isa complex undertaking that must take intoaccount the needs to be satisfied, i.e.:c safety of people and equipment,c continuity of supply,c installation and operating costs.The designer uses the load points, theirsimultaneity coefficient and the above-mentioned

1 Introduction

criteria to draw up the single-line diagram for theinstallation (see fig. 1 ).He must then select the earthing system, definethe busways, calculate the fault currents, definethe protection system (discrimination, choice ofprotections, see Cahier Technique no. 174).Figure 2 shows an example of protectionschosen for the standby supply in figure 1.

Fig. 1 : example of a general single-wire diagram for a mine.

G

CrushingBank

63 kV

10 MVA2 MVA10 MVA

Stanby supply

5 kVLV LV

20 kV

Shaft no.1,level 300

5 kV

Processing Mills

Shaft no. 2, level 500


Fig. 2 : protections of a MV unit module.

Permanent insulation monitor

a49 Thermal image

46 Negative sequence

51V Voltage restrained O/C

32P Reverse active power

32Q Field loss(max. reactive power)

27 Undervoltage

59 Overvoltage

81 Over and under frequency

59N Zero sequence voltage

87T Differential

The protection plan must specify the operating ornon-operating conditions for all the protectionsduring a fault and during normal operation(transients). It must indicate the protectionsettings.However, the plan rarely indicates thecharacteristics of the protection input circuit andother data necessary to specify the currenttransformers (CT). This is because it is oftenvery hard for the designer to collect all thenecessary data.The consequences on industrial start-up can beserious: unsuitability, overrating, non-standardspecifications and high costs, addition of matchingCTs, last minute CT changes, postponement ofdelivery, commissioning, production times, etc.More serious still, incorrect definition can lead tomalfunctions in the protection channel causingdestruction of equipment or, worse still, dangerfor the operator.

A few examples:c Overestimation of the short-circuit current canlead to feasibility problems, overrating and highCT costs.c On the other hand, underestimation of theshort-circuit current can lead to failure to detectthe fault, thus destroying the equipment, placingthe operator in danger and generating operatingdowntime.c An output power or accuracy error can result ina malfunction or in failure to trip of the protectiondevices, thus destroying the equipment, placingthe operator in danger and generating operatingdowntime.c An error in defining the accuracy class of ametering winding will lead to incorrect energybilling and thus a loss of income for the electricalutility or the customer.c Etc.


The aim of this document is to provideassistance with defining current transformers.Before entering the heart of the subject,

Necessary information Abbreviations UnitsInsulation level Un kVNominal short-circuit current Isc kADuration (1 to 3 seconds) t sNominal primary current Ip ANumber of secondary windings (1 to 3)For each secondary winding:c what typec associated protection or metering and settingc output power Poutput VA(relay and wiring consumption)c accuracy factorv protection ALFv metering SFc Nominal secondary current (1 or 5 A) Is A

Fig. 3 : information necessary to specify a CT with a single primary.

a reminder is given of the informationnecessary to define a CT (see the table infigure 3 ).


A MV or HV electrical network is disturbed:c exceptionally by lightning overvoltages, bytemperature rises further to overloads orfollowing violent short-circuits between phases orphase-to-earth,c more frequently, and more naturally, byswitching overvoltages (e.g. capacitor energ-isation) or natural transient conditions (e.g. motorstarting or power transformer switching) resultingin high but temporary overcurrents.

The consequences of disturbancesMajor disturbances, such as short-circuitcurrents, can result in serious damage:c fatigue or deterioration of networkcomponents,c danger for people,c loss of supply and production, etc.

2 Network disturbances and protections

It is thus necessary to provide the relevantprotection devices with the right information toensure prompt action, as the greater thedamage, the longer and more costly the repairsand the heavier the losses.However, transient and normal disturbances area necessary evil and installations must bedesigned to withstand them. Furthermore, thecurrent transformer/protection pair(s) must notcause nuisance tripping.

Elimination of faultsPermanent monitoring of network electricalvalues by reliable and properly rated currenttransformers supplying protection relays allowsrapid isolation of the faulty area. These relaysmust ignore transient and normal disturbancesbut systematically trip when a destructive faulthas to be eliminated.

2.1 Disturbances

2.2 ProtectionsDfinition of protectionsc Function of protectionsThe protection functions of a network areintended to monitor one or more parameters ofthe installation, for example: currents, voltage,temperature, frequency, etc.These values are permanently measured andcompared with setpoints or thresholds beyondwhich the situation is defined as abnormal anddangerous. When a fault occurs, the protectiondevice issues a tripping signal. Then, in order todurably isolate the faulty part, it preventsreclosing until the device has been repaired.It can also generate an alarm to informmaintenance personnel and enable them to takethe necessary action.c The technologiesWith the particularly rapid growth of electronicstechnology, protection relays, originallyelectromechanical, have become static devices:analogue electronic then digital electronic thanksto microprocessors. These devices performincreasingly sophisticated functions and are moreand more often referred to as processing units.

Digital technology is becoming widespread for allapplications (network components or protectedloads). This technology has an ample dataprocessing capacity, which allows control andmonitoring linked to protection functions andcommunication to a supervisory unit or acentralised control system.These units are normally supplied with astandard protection, control and indicationprogramme, thus enabling them to be usedwithout extra studies or programming. Onlyparameters have to be set on commissioning(e.g. the protection settings).They are designed to fully meet applicationneeds. All you have to do is select the versioncorresponding to the functions available for eachapplication. Some examples of applications are:v transformers,v generators,v capacitors,v motors,v substations,v etc.


These units incorporate meters such asammeters, voltmeters, wattmeters, kilowatt hourmeters, etc. with fewer wiring and compactspace requirements (see fig. 4 ). Furthermore,their reduced power requirements (lessdemanding than those of electromagneticrelays), requiring less powerful CTs, make themmore economic. When combined with protectionfunctions and control logic, they also displayalarm and operating messages.c The current transformersThe CT characteristics are defined according tothe technology selected for the processing unitand the scheduled functions (protection,metering, control and monitoring, indication).c ImplementationChoice of protection settings is very tricky. Itcalls for thorough knowledge of parameter levels

Fig. 4 : simplification and savings provided by a multifunctional digital system (protection - automation - metering)compared with the former electromagnetic technique.

according to whether the disturbance is normalor transient or caused by a fault that must beeliminated. Furthermore, it is commonknowledge that a short-circuit at one point of thenetwork may be detected right up to the source.The protection co-ordination study ensures thatonly the faulty part of the network is de-energised.The protections relating to short-circuits eitherphase-to-phase or phase-to-earth use one of thefollowing discrimination types, as applicable:v overcurrent,v time,v logic,v differential,v directional,(see Cahiers Techniques no. 174 and 113).

Multifunctional digitalsystem

Electromagnetictechnique


Instrument and protection CTsCurrent transformers are used to supplyinformation to the protective relays and/or current,power and energy metering instruments. Forthis purpose they must supply a secondarycurrent proportional to the primary current flowingthrough them and must be adapted to networkcharacteristics: voltage, frequency and current.They are defined by their ratio, power andaccuracy class. Their class (accuracy as afunction of CT load and of overcurrent) is chosenaccording to the application.c A protection CT must saturate sufficientlyhigh to allow a relatively accurate measurementof the fault current by the protection whoseoperating threshold can be very high. Currenttransformers are thus expected to have anAccuracy Limit Factor (ALF) that is usually fairlyhigh. Note that the associated relay must beable to withstand high overcurrents.c An instrument CT requires good accuracyaround the nominal current value. The meteringinstruments do not need to withstand currents ashigh as the protection relays. This is why theinstrument CTs, unlike the protection CTs,have the lowest possible Safety Factor (SF) inorder to protect these instruments through earliersaturation.

3 Current transformers

3.1 Reminders of ferromagnetic transformersc Some CTs have secondary windings dedicatedto protection and metering. These instrumentand protection CTs are governed by standardIEC 60044-1 (in France NF C 42-502).The matching of CTs with protection relays callsfor a thorough knowledge of CTs. The followingsection gives a few reminders of CTscorresponding to this use.

Characterisation of CTsc An example of a protection CT:v rated primary current: 200 A,v rated secondary current: 5 A.15 VA 5P 10

accuracy limit factor = 10accuracy class = 5Paccuracy power = 15 VA

Its accuracy load: Pn = 15 VAIts accuracy limit factor is ALF = 10For I = ALF. In, its accuracy is 5% (5P),(see fig. 5 )To simplify, for the protection CT given inexample, the ratio error is less than 5% at 10 In ,if the real load consumes 15 VA at In.However these data are not sufficient. Also, it isuseful to know the standard values.

Fig. 5 : example of the nameplate of a current transformer with two secondaries.

transformateur de courant - current transformer

n 919167117,5/38/95 kV 50 Hz Ith 25 kA 1 srapport bornes ratio terminals 150/5 1S1 - 1S2 150/5 2S1 - 2S2

type RCF 2 / Bnormestandard

Idyn 62,5 kA ext. %VA FS ou FLP 15 0,5 715 5P 10

2 221 625

classeclass

CEI - 185

Network voltage characteristicsRated insulation voltage: 17.5 kVPower frequency withstand voltage: 38 kV 1 mn 50HzImpulse withstand voltage: 95 kV peak

CT type

ApplicableCT standard

Safetyfactor (SF)

Accuracy limitfactor (ALF)

Accuracyclass

Accuracypower

1 primary circuit1 secondary circuit 1S1 - 1S21 secondary circuit 2S1 - 2S2

Ratio

NetworkcurrentcharacteristicIth : 25 kA/1 sIdyn : 62.5 kA peak

CT serial numberwith year ofmanufacture


c A few definitionsv Rated (nominal) primary current I1Defined by standards, it is chosen from thediscrete values: 10 - 12.5 - 15 - 20 - 25 - 30 -40 - 50 - 60 - 75 A and their decimal multiples.v Rated (nominal) secondary current I2Equals 1 or 5 A.v Ratio (I1 / I2)The primary and secondary currents arestandard, thus these values are discrete.v Accuracy loadLoad value on which the accuracy conditions arebased.v Rated (nominal) accuracy power PnExpressed in VA, it is the apparent powersupplied to the secondary circuit for the nominal(rated) secondary current and the accuracy load.The standard values are:1 - 2.5 - 5 - 10 - 15 - 30 VA.v Real power PrIn this Cahier Technique, it is the powercorresponding to the real load consumption ofthe CT at In.v Accuracy classThis class defines the error limits guaranteed onthe ratio and on the phase shift in specifiedpower and current conditions.For the nominal 5P and 10P classes, the table infigure 6 defines these limits.v Special accuracy classClass X is a class defined by British standardBS 3938. It must also be defined in the futurestandard IEC 60044-1 under the name of classPX. This class specifies the minimum value ofthe knee point voltage Vk of the CT.It also imposes a maximum value of Rct (CTsecondary winding resistance). Sometimes, itspecifies the maximum value of the magnetisingcurrent Io at knee point voltage.If we consider the magnetising curve V(Io) of theCT, the knee point voltage Vk is defined as thepoint on this curve from which a 10% increase in

voltage causes a 50% increase in themagnetising current Io.Class X corresponds to a better meteringaccuracy than classes 5P and even more so 10P(see fig. 7 ).It is always possible to find an equivalencebetween a CT defined in class X and a 5P CT orin some cases even a 10P CT (refer to CahierTechnique no. 195 which deals withequivalences).v Real accuracy factor (Fp or Kr)This is the ratio between the overcurrentcorresponding to the nominal error and the ratedcurrent of the CT when the real load is differentfrom the nominal load.v Accuracy limit factor (ALF or Kn)This is the ratio between the nominal overcurrent(e.g. 10 In) and the rated current (In).v Short time withstand currentExpressed in kA, this is the maximum current Iththat can be withstood for one second (when thesecondary is short-circuited). It represents the

Fig. 6 : errors on the module and the phase at nominal current according to standard IEC 60044-1.

Fig. 7 : voltages corresponding to different CT classes.

VS

VS210PVS15PVk

IS Io

X

Accuracy Current error for Phase shift for the Composite error for theclass the nominal current nominal current accuracy limit current as

as a % Minutes Centiradians a %5P 1 60 1.8 510P 3 10


thermal withstand of the CT to overcurrents (thestandard values are given by the standardsmentioned in the appendix).v CT rated voltageThis is the rated voltage to which the CT primaryis subjected. It is important to remember that theprimary is at HV potential and that one of theterminals of the secondary (which must never beopened) is normally earthed.Just as for any devices, a maximum withstandvoltage for one minute at power frequency and amaximum impulse voltage withstand are alsodefined. Their values are defined by thestandards.For example: for a rated voltage of 24 kV, the CTmust withstand 50 kV for 1 minute at 50 Hz and125 kV at the impulse voltage.c CT with several secondariesSome CTs may have several secondariesdedicated to protection or to metering.The most typical cases are CTs with 2secondaries, more rarely with 3 secondaries.Physically, these CTs group in the same mouldthe equivalent of 2 or 3 separate CTs that canhave different classes and ratios (see fig. 8 ).Influence of the load on the accuracy limitfactorRemember that the equivalent simplified diagramof the magnetic current transformer is shown infigure 9 .

Fig. 8 : manufacturing principle of a CT with 3secondaries (with 3 windings in the same mould).

S1

I1

S2 S3

Fig. 10 : operating points of the CT according to its load.

Fig. 9 : CT equivalent diagram.

V

I1

Io

IS Rct I2

R

Applied to this diagram, Ohms law lets uswrite: V = I2 (Rct + R), where:Rct : CT secondary winding resistanceR: load resistance including wiring,v if I2 = kn In and R = Rn = Pn / In2 ,Vn = kn In (Rct + Rn) (1) ( kn = nominal ALF)v if I2 = kn In and R = Rp = Pr / In2 ,Vr = kn In (Rct + Rp)On figure 10 we can see that although Rp is farsmaller than Rn, the CT saturation knee point isfar from being reached at the declared accuracylimit factor kn.The real accuracy limit factor corresponding tothe real load (protection + wiring) can becalculated. This is ALFr = kr for which thesaturation knee point voltage Vn is reached:Vn = kr In (Rct + Rp) (2)If Rp is less than Rn, kr is greater than kn(ALFr > ALF)

Vn

Vr

V (volts)

IonIor Io

Operating point of the CT when it is loaded at Pr and supplied by the same current kn In

Operating point of the CT when it is loaded at Pn and supplied by a current kn In


By combining the equations (1) and (2), we findthe following formula:

k kr r=+

+=

+

+ k R R

R R , or k P P

P Pnct n

ct pn

i n

i rwhere:Pi = Rct In2 = internal losses of the CT at InPn = Rn In2 = CT accuracy powerPr = Rp In2 = real load consumption of the CT at In.It is obvious that proper operation of aprotection relay is linked to the behaviour ofthe associated CT and to its real load and notto the behaviour of the CT associated with atheoretical nominal load.

Real needs enable us to determine the minimumaccuracy power to be chosen. Using a CT with aload Pr < Pn increases the ALF. Likewise, theALF increases more if the Rct (internal losses Pi)is low (see fig. 11 ).Calculation of the real ALF (kr) of a CT,associated with its real load, ensures that theright CT is chosen in all traditional cases.Note: for very demanding protections (e.g.differential protections), current transformers aremost often defined in class X. This class isalways defined according to the real load of theCT and to its own internal losses.

Fig. 11 : behaviour of the accuracy limit factor kr = f(Pr) of two CTs of 10 VA-5P20 with different internal losses(Rct) according to the real load connected to the secondary.

5 15 20100

0

40

20

30

50

60

70

80

10

kr

Pr

Pi = 2 VA

Pi = 5 VA


The output signal, delivered by the non-magnetictransformers (also known as ROGOWSKI coils)is a voltage proportional to the derivative of theprimary current.

(Lenz law: e n d= dt

)They do not saturate and their response is linear.Consequently, they can be used over widecurrent ranges: the only limitation is thedynamics and the linearity of the input circuit ofthe associated protection.The technology of the protection, control andmonitoring units connected to these non-magnetic transformers is of the digitalmicroprocessor type. This technology is able toprocess signals of very low amplitude.For a given non-magnetic transformer, in view ofthe linearity of the output signal, the nominalprimary current is replaced by a wide range, forexample 30 to 300 A.In addition to the advantage of linearity, the useof non-magnetic CTs reduces:c risks of error when choosing primary current atthe design stage of the installation,c the number of models to be managed. It alsominimises the delivery times.Today these transformers are seldom used. Astandard (IEC 60044-8) should define them.Schneider Electric has been using thesetransformers (see fig. 12 ) in association with theSepam protection, control, monitoring andmetering units since 1986.To specify them, all you have to do is indicate:

3.2 Non-magnetic transformers

1

2

3

4

56

7

1 - Primary winding2 - Dielectric screen3 - Dielectric insulation4 - Setting resistance

5 - Secondary winding6 - Secondary winding support7 - Magnetic shielding

Fig. 12 : cross-section of a non-magnetic transformerused in MV.

c the CT insulation level, defined just as for atraditional CT,c the rated thermal short-circuit current (Ith) andthe dynamic current (Idyn) set according to thesame rules as for the CTs,c the utilisation range (rated primary current andthe thermal current).

3.3 CT manufacturing and implementationCTs are industrial products designed accordingto standards. They are mass produced, therebyreducing costs and guaranteeing theircharacteristics.

Their live part is duplicate moulded in order tocomply with insulation, temperature rise andelectrodynamic withstand requirements. Thenumber of moulds corresponding to a standardrange is necessarily limited.

Furthermore, the switchgear and CTs are mostoften installed in panels that have been optimised,standardised and subjected to qualification tests.In this case, use of standard moulds is compulsoryas CTs perform other functions such as thebushings between the cable and the circuit-breaker compartments (see fig. 13 overleaf).Consequently, any modification in volume orshape of a CT results in major study, productionand test investments.


To solve special cases, without overcosts oradditional technical risks, it is therefore alwaysadvisable to look for solutions in order to fit themould of standard CTs. Panel structure is thusmaintained intact. These solutions are:

Fig. 13 : cross-sectional view of the panel and multifunctional CT.

c good rating of electrical characteristics: forexample avoid overrating in power and ALF,c use of CTs with two or three windings,c use of relays performing several protectionfunctions with the same current information.


Thorough knowledge of CTs, their possibilitiesand their limits is useful only when they areassociated with a specific protection relay whosecharacteristics and scope of action regarding themonitored current range are known.The protection relays installed on an electricalnetwork are defined in the protection plan.This plan specifies the position and setting of theselected protections. It also defines the positionof the CTs, their ratio and, more rarely, theirpower, accuracy and ALF. In point of fact,complete specification of CTs also requiresknowledge of:c the protection input impedance,

4 Choosing CTs according to protections and applications

c the wiring impedance,c the protection operating thresholds (normallytaken into account in the protection co-ordinationstudy).Today, most protections are of the digitaltechnology kind and are highly accurate.CT accuracy is thus a decisive factor.The type of protection also affects the requiredCT accuracy:c an overcurrent protection only takes thecurrent value into account,c a differential protection compares two currents,c an earth fault protection treats the sum of thethree phase currents.

4.1 Choosing CT ALF according to protectionsWhen choosing a CT out of the standard CTs, areminder is necessary of the relationship linkingthe nominal ALF (linked to Rn) and the real ALFlinked to the real load Rp:

kRR

RRn

p

n

n

p=

+

+=

+

+ k

RR

or k k RRr

ct

ctr n

ct

ct

A CT can supply several different protectionseither separate or grouped in a multi-protectionsystem (e.g. the Sepam). This leads us toexamine the protections determining CT sizing.

Definite time overcurrent protectionThe threshold Is (protection setting) can be set,for example, from 2 to 10 In of the CT if the CT Inis the application In.To ensure that the CT will not affect theoperating accuracy of the protection, it isnecessary to have no saturation up to the settingpoint. But it is usual to take a safety coefficientof 2 (see fig. 14 ).Thus the ALFr (kr) at real load will be:kr uuuuu 2 II

s

n

if Is = 10 In kr u 20

Example:200/5 CT - 10 VA-5P10,Load In: 160 AIs = 8 In of loadLet us verify if the proposed CT is suitable:

Operating point at kr InOperating point at 2 Is max

Operating point at Is max

V

2Vs

Vs

ImFig. 14 : operating points of the CT at maximumthreshold.

IIs

n

o

C f O/C of T

8 160200

6.4= =

the minimum recommended ALFr (kr) is thus:kr u 2 x 6.4 = 12.8.If the CT load and its internal resistance areknown with, for example:

Rct + Rp = 12 (Rct + Rn)

we obtain:kr = kn x 2 = 20 higher than the minimum valuerequired. Thus the CT is suitable.


Inverse time overcurrent protectionIf the aim is correct accuracy over the entirerelay inverse curve, it is necessary to know atwhat point it becomes a definite time curve. Formost relays it occurs at the minimum valuebetween 20 Is and 24 In (CT). So, assumingIscmax is the maximum short-circuit current, thesame reasoning as above, using a safetycoefficient of 1.5, gives:

krmin = minimum value between:

30 (CT), 36, and 1.5 max

(CT)s

n

sc

n

II

II

Directional current protectionThe rules, unless otherwise specified (refer toCahier Technique no. 181) are the same as forthe overcurrent protections.Note that for the three current protectionsdescribed above:c If several current protections are supplied bythe same CT, it is the one with the lowest curve(the shortest time delay) for high currents thatdetermines the sizing.c In difficult cases, the safety coefficient of 2 canbe lowered to 1.5.

Earth fault protectionAs the protection is supplied by the vector sumof the secondary currents of 3 CTs connected asper the Nicholson arrangement (see fig. 15 ), itis preferable to use identical CTs produced bythe same manufacturer. However, if there is aDC component (energisation of a transformer) orwhen a high current occurs, this arrangement(parallel-connection of the secondary of the 3CTs) will deliver a false earth fault current whichmay cause nuisance tripping of the protection.To give an example, with the 5P10 CTs, aprotection threshold of 10 % of CT In is a limitbelow which there is a risk of nuisance trippingfor definite time protections.The CT accuracy limit factor is given by the

expression: krh > X IIhs

n

.

The safety coefficient (X) is normally equal to 6(given by the relay manufacturers).

This corresponds to the fact that the CT(associated with the phase with the earth fault)must be able to develop a voltageVh = X Ihs (Rct + 2 RL + Rh).Note:c If a CT also supplies an overcurrent relay, Rhmust be replaced by Rh + Rp.c If the CTs are initially designed for overcurrentprotection, we recommend you check that theyare suitable for supplying earth fault protectiontoo. Thus, the krh of a 100/1 CT - 10 VA-5P10 isgiven by the expression:

k PR R R Rrh

n n

ct L p h

R

kct n=+

+ + +

/I2

2.

Bearing in mind that the impedance of the relayused depends on the setting of Ih (in this case0.1 A), the digital application gives:

Rh 1VA0.1A

100= ( ) =2,

krh 10 3+10

3+1+4+100 1.2= =

a value to be compared with the expression ofthe needed krh:

krh 6 0.11 0.6= = , the CT is suitable.

Ih >

Fig. 15 : the vector sum of the phase currents givesthe earth fault current.


If the short-circuit current is very high and if theprotection time delay is short, to avoid nuisancetripping it may be necessary to add a stabilisingresistance in series with the earth fault relay. Itspurpose will be examined in paragraph 4.3.In order to avoid difficulties that may arise withthe 3 CT arrangement, it is preferable, wheneverpossible, to use a toroidal CT around the 3phases (see fig. 16 ). Note that the 3 phasesmust be placed in the centre of the toroid toavoid local saturation of the magnetic material.Use of a toroidal CT allows very low operatingthresholds (a few amps) to be chosen.Differential protectionsIncreasingly used for the protection of transformers,rotating machines and busbars, they have theadvantage of being fast and independent fromthe other protections in terms of discrimination.These relays often operate during the transientfault current period. Just as for the earth faultprotections, an aperiodic DC component mayresult in transient saturation of the CTs and thusgenerate a false differential current. Sincedifferential protections are tricky to implement,manufacturers normally supply the necessaryinformation to design and install the CTs.

Fig. 16 : the toroid placed in (1) or (2) gives the sameinformation but the toroid placed in (1) also monitorsthe faults located upstream of the toroid (2).

1

1

2

23

4.2 Characterising CTs according to applicationsIn this context the term applications means thecomponents of the electrical network to beprotected: line incomers and feeders,transformers, busbars, generators, motors,capacitor banks, etc.Each component requires the use of severalprotections. An example is given by the diagramin figure 17 (overleaf) in which the protectionsare identified by their ANSI code (AmericanNational Standard Institute).However, to optimise choice of a CT, you needto know which protections it supplies, as well astheir settings, real impedance, short-circuitcurrent, etc.

The difficulty lies in the fact that these values arerarely known by the CT specifier when his choicehas to be finalised.

For this reason this section proposes to studythe various applications where values can bedefined by excess. They can therefore bechosen without risk as they will necessarily coverneeds. However, they will lead to a less restrictivedefinition which is more realistic than the oneproposed by relay suppliers in their catalogues.In point of fact, each supplier gives someguidelines for choice covering all the settingranges of his relay associated with CTs assumedto have high internal losses and also maximumwiring impedances.For the various applications, we shall list theprotection functions commonly used. The mostunfavourable protection will be chosen to sizethe current transformers.Note: Initially, the earth fault and differentialprotections will not be taken into account.

Conclusionsc It is the high setting of the overcurrentprotections that determines the minimum kr(real ALF) to be complied with.c It is the stability of the relay up againsttransient phenomena that determines the ALF orclass X for earth fault and differential protections.


a

M

M

M

27 27R

51G

150/6 kV22.5/30 MVAZsc = 11 % (30 MVA)YD 11

27I

27

87

27

51G

40

46

49

51

51V

3000/1

3000/5

100/5100/5

100/5

200/5300/5

600/546

6666

50

49

51

51N

50 51

51N

50t

49

46

51

49

50t

132 kWIn = 250 AId / In = 5Td = 2 sec

710 kWIn = 82.3 AId / In = 6Td = 5.2 sec

4500 kWIn = 491 AId = 5.8 InTd = 5 sec

2000 kVAIn = 192.45 A

6 kV/400/230 V2500 kVAZsc = 7.5 %DY11

400 kVAZsc = 4 %

51N51N

50t51

81

51G

315 A

87T

50 51150/1

300/1

300/1300/5

Fig. 17 : example of a protection plan.


Applications with typical protections

Examination of the protections normallyimplemented for the various applications(see fig. 18 ), shows that it is always the highthreshold overcurrent protection which is the CTsizing factor. The exception to this rule is,however, the motors protected by fuses forwhich the locked rotor protection is the mostrestrictive.To optimise the CTs, the settings of theseprotections need to be taken into account.c Line incomers and feedersThe high threshold overcurrent protection isnormally set between 3 i Is i 10 In, hence a

krm u 2 IIs

n

with at maximum a krm = 20.

If an inverse time protection is used:krmin = minimum value between:

30 (CT), 36, and 1.5 max

(CT)s

n

sc

n

II

II

Fig. 18 : protections implemented by application, in the most common cases.

c Generator incomerThere is no reason to select a setting > 7 In,given the relatively low short-circuit current of agenerator,hence a krm u 14.c Transformer incomerIf we take the example of the protectionsdownstream of a HV/MV transformer, the settingthreshold must be lower, for example, than 70 %of the Isc at the secondary side, i.e. as an initialapproximation:

Isc = 0.7 In2 100Zsc

.

If we apply the rule: krm u 2 IIs

n

,

it becomes: krm u 1.4 100Zsc.

This krm is maximised; in reality it is thediscrimination study of the downstream networkthat sets the Is and thus determines the krm.

ANSI Typical protections Applicationscode Line Generator Trans- Trans- Capacitor Motor feeder with

incomer incomer former former feeder circuit- contactorand feeder incomer feeder breaker + fuses

37 Undercurrent c c46 Unbalance c c c49 Thermal image c c (c) c c c50 Instantaneous high c c

threshold overcurrent50N Instantaneous c

earth fault51N Time delayed earth fault c c c c c c c51N1 Star-star capacitor c

unbalance51LR Locked rotor / c c

too long start51V Voltage restrained c

overcurrent

51-1 Time delayed low c c c c cthreshold overcurrent

51-2 Time delayed high c c c cthreshold overcurrent

66 Number of starts c c67 Directional overcurrent c c c67N Directional earth fault c c c


c Transformer feederThe operating current of the high thresholdovercurrent protection (Is), placed upstream ofthe transformer, must be higher than the currentobserved at the primary (Isct) on a short-circuit atthe transformer secondary terminals.As a first approximation we can state that:

Isct i In1 100Zscwhere

In1 = Pn3 Un1

= nominal primary current,

Pn = nominal power of the power transformer,Un1 = primary voltage.In fact, the upstream source impedance alsohelps to limit Isct. We can thus be certain that:

Is i In1 100ZscIf the general rule krm u 2 II

s

n

applies,

we obtain: krm u 2 IIn

n

1

100Zsc

.

The typical values for short-circuit impedancesof power transformers can range from 4 % forsmall transformers to 20 % for the largest ones(see fig. 19 ).This would result in requiring minimum krmranging from 10 I

In

n

1 for large transformers to

50 IIn

n

1 for the smallest ones.

These values may be too difficult to obtain forlow rated CTs particularly when their thermalwithstand is high (e.g. 40 kA, 1 s).Therefore, in difficult cases, the problem offeasibility could be solved by overrating the CTprimary or by using a reduced coefficient(1.5 instead of 2) as described in paragraph 4.1,which gives:

krm u 1.5 IIn

n

1

100Zsc

.

If you choose to overrate a CT, you must ensurethat the setting of the transformer thermal

protection is still possible. Otherwise you mustplan this protection downstream of thetransformer.c Capacitor feederThe time delayed high threshold recommendedby capacitor suppliers is 3 In (0.3 s).If we take Ismax = 5 In ; krm uuuuu 10.c Motor feederThe high threshold must be set above thestarting current, which is always i 7 In.If we take Ismax = 8 In ; krm uuuuu 16.If the short-circuit protection is provided by fuses,then the locked rotor / too long start protectionis the most restrictive. Its setting is between 2.5and 4 In ; krm uuuuu 8.Note: From the above, we have assumed thatCT In = application In. If this is not the case, theresulting krm must be multiplied by the ratio:

II

n

n

(application) (CT) .

The table in figure 20 summarises the minimumALF to be complied with according to theapplications.

Fig. 19 : typical short-circuit impedances for powertransformers.

Transformer rating Zsc(MVA) (%)0.5 40.63 40.8 41 52.5 55 610 820 1030 1240 1380 18160 20


Fig. 20 : real ALF (kr) required for overcurrent according to the applications.

kr mini for definite kr mini for inversetime overcurrent time overcurrent

c Is is known, all applications

kr = 2 IIs

n

between: 30 (CT), 36, and 1.5 max

(CT)s

n

sc

n

II

II

c Is is unknown

Transformer feeders kr = 2 IIn1

n sc

100Z

Not concerned

Transformer incomers, kr = 20 (by excess) between: 30 (CT), 36, and 1.5 max

(CT)s

n

sc

n

II

II

,

Line incomers and feeders if no definite time on a second thresholdCapacitor feeders kr = 10 (by excess) Not concernedCircuit-breaker motor feeders kr = 16 Not concernedContactor with fuse motor feeders kr = 8Generator incomers kr = 14* kr = 14*

In this table: Is is the setting current of the overcurrent threshold for which the response time isshortest for the high currents,In is the CT nominal primary current,In1 is the power transformer nominal current,(*) = general cases.

4.3 Special case of differential protectionAlthough the manufacturers of relays dedicatedto differential protection impose the necessaryCT secondary characteristics required for properoperation, it is useful, for understanding andavoiding errors, to possess minimum knowledgeof this type of protection.A differential protection monitors an area limitedby CTs which measure incoming and outgoingcurrents. If the outgoing currents are notconsistent with the incoming currents, this isnormally because a fault has occurred in theprotected area.We shall examine in turn high impedance, pilotwire, percentage differential and low impedancedifferential protections with their respective CTrequirements.

According to the type of protection and its use,relay manufacturers have had to apply a varietyof principles varying in complexity in order toguarantee the stability of their relays againsttransient phenomena likely to cause nuisancetripping of this protection.

High impedance differential protectionThis type of protection is normally used forprotection of motors, generators, busbars aswell as for restricted earth fault protection oftransformers.c GeneralThis type of protection is used to protect an areawith the same voltage level.


Fig. 21 : high impedance differential principle.

In healthy conditions, the input current ie isidentical to the output current is and thus thedifferential current id = 0 (see fig. 21 )A high fault current can flow through themonitored area and cause CT saturation, hencethe risk of nuisance tripping of the protection(non stability). The stability of the relay isobtained by connecting it in series with astabilising resistance Rst.This resistance is calculated in such a way thatthe derived current in the differential circuit(Rst + Rp) cannot reach the relay settingthreshold when the maximum through currentsaturates a CT with its DC component. Thisresults in:

(1) (Rst + Rp) u (Rct + 2 RL) IIssc

rwhereIssc = maximum through current observed at theCT secondary,Ir = relay secondary setting current.Rst may vary between a few ohms and a fewhundred ohms (exceptionally it may be greaterthan 1000 ohms).For the relay to operate properly at Ir if a faultoccurs in the area, the knee point voltage Vkmust be greater than:2 (Rst + Rp + Rct + 2 RL) Ir .As a rule Rct + 2 RL are negligible compared withRst + Rp thus:(2) Vk u 2 Ir (Rst + Rp).By combining the equations (1) and (2) we find:(3) Vk uuuuu 2 Issc (Rct + 2 RL).These equations show that Rst (expression 1)and Vk have greater values when Rct is high.A high value stabilising resistance generateshigh overvoltages at the CT secondary.Therefore when overvoltages in excess of3000 V are anticipated, a protection by a non-linear resistance (ZnO) is added.The result of these observations is that the CTsare optimised if Rct and Vk are as low as possibleand if the through current (seen from the CTsecondary, i.e. Iscc) is defined without excess.Whatever the application where the highimpedance differential is used, all the CTs musthave:v the same ratio,v the same magnetising curve(same minimum Vk),v the same maximum Rct,and comply with expression (3) ; for Vk, since therelevant CTs are not at the same distance fromthe relay, take the maximum RL.For this protection, a maximum value of themagnetising current Io must also be defined atVk /2 according to the required sensitivity.

A B

Ie

id

i'e i's

Is

Protection(Rp)

Rst

Application(monitored area)

For the relay to detect a current Ir, a voltageVs = Vk /2 must be developed at the terminals ofthe parallel-connected CTs; for this purpose, theminimum primary current Irms really detected bythe relay will be Irms = n(Ir + Io), wheren = CT ratio and = number of parallel-connected CTs (theremay be many of them on a busbar protection!).c Application to the motor differential protectionThe maximum through current for which themotor must remain insensitive is in this case themotor starting current:Issc = Ist (seen at the secondary).If you do not know this current Ist, you know that:Ist < 7 In motor

c Application to the generator differentialprotectionThe maximum through current is in this case theshort-circuit current supplied by this generatoronly.If you know the generator subtransient reactanceX %, the following will be taken:

I Issc n 100X

= ;

If this value is not known, X % = 15 will betakenNote: the peak voltage at the CT secondary mustbe calculated using:Issc maxi = Igenerator + Isc network.c Application to the busbar differentialprotection (see fig. 22 ).In this case, the through current is equal toswitchboard Isc :Issc = switchboard Isc seen from theCT secondary.c Application to the restricted earth faultdifferential protection of transformers (REF)v In the case of figure 23a , this protectiondetects the insulation faults at the transformersecondary windings and up to the CTs locateddownstream.


1

4 5

2 3

Rst87B

Fig. 22 : high impedance busbar differentialprotection.

Fig. 23 : restricted earth fault protection of thesecondary [a] or primary [b] windings of a transformer.

a

b

Rst

Rp

RstRp

Pilot wires

RL

Line differential87L

Line differential87L

RL

v In the case of figure 23b , this protectiondetects the insulation faults at the transformerprimary and advantageously improves the typicalearth fault protection which is sensitive totransformer violent inrush currents and to thethrough currents resulting from a downstreamasymmetrical short-circuit.In this case also, we shall calculate Rst and Vkbased on the maximum through current in theCTs for a fault outside the protected area. As afirst approximation, we can say that this currentis less than the current limited by the transformer

impedance, i.e. Ithrough sctn

PU

=

3,

where Psct P 100Zn sc

= , (transformer short-circuitpower); if we know the upstream short-circuitpower (Pu), a more accurate value can becalculated replacing

Psct with P

Psct

sct

P P

u

u

( )+( ) .

The through fault current must then be convertedinto Issc seen at the CT secondary.

Line or cable differential protection with pilotwires (see fig. 24 ).A relay of this type is installed at each end of thecable or the line.On the pilot wires, each relay reproduces avoltage which is the image of the sum:a I1 + b I2 + c I3 + d Ih .If the two voltages are different, both relays trip.NB:Note that the coefficients a, b, c and d aredifferent to ensure that all fault types correspondto a sum other than zero. Consequently, theoperating threshold of a two-phase fault or of aphase-to-earth fault is slightly different accordingto the faulty phase.In this case also, class X CTs are required, andeach manufacturer gives an empirical formula forthe minimum knee point voltage Vk.

Fig. 24 : line or cable differential protection with pilot wires.


An example of the minimum knee point voltagerequired:Vk mini = 0,5 N kt In (Rct + X RL)where N, kt and X are constants associated withthe relay response time, its sensitivity and itstype of connection.Another example:

Vk mini = 50In

+ If (Rct + 2 RL)where:In is the CT nominal secondary current (1 or 5 A),If is the through short-circuit current, seen at theCT secondary.The stability of this relay is achieved both bycomplying with the required knee point voltageand by a percentage differential operatingthreshold which will increase with the throughcurrent by use of retaining windings.The CTs at the ends of the line must have thesame ratio and must comply with the minimumVk and maximum Io specified by themanufacturer. However, their magnetisingcurves and Rct do not need to be identical.

Percentage biased differential protection fortransformersThe term percentage differential stems from thefact that the operating threshold increases withthe through current.Simple comparison of the currents in eachupstream phase with the currents in the samedownstream phases is not suitable fortransformer differential protection.This is because:v the upstream and downstream currents of apower transformer do not have the sameamplitude or the same phase angle,v when the transformer is energised, itsmagnetising current is only seen upstream,v the presence of an earthing generator in theprotected area (e.g. earthing the transformerneutral) can trip the protection, while the fault is,for example, located on a downstream feeder.

c Precautions to be taken to solve theseproblems:The aim is to ensure that the relay seesupstream and downstream currents of the sameamplitude and in phase during normal operatingconditions. This can be achieved by intelligentchoice of CT ratio and connections.

Matching CTs are used for this purpose andoften help to eliminate the zero-sequence currentthat could trip the protection when an earth faultoccurs outside the protected area.However, most of the new digital relays are ableto perform internally, by parameter setting, theadjustments necessary to reset the currents,thus considerably simplifying theirimplementation.Furthermore, it must be noted that all thetransformer differential relays are immunised tothe 2nd order harmonic blocking their operationwhen the transformer is energised.c CT voltage VkIn 99 % of cases, a class X is requested.The minimum knee point voltage is imposed anddepends on the resistance of the secondarywinding Rct of the CT and of its real load Rr.More complex specifications are sometimesmentioned, which include the X/R ratio of thenetwork or the magnetising current of the powertransformer. However, faced with the problemsthat users have in obtaining all these parameters,relay suppliers sometimes provide simplifiedempirical formulas which lead to a slightoversizing.Examples of minimum knee point voltageimposed for the Sepam 2000 D02 (SchneiderElectric):Vk mini = A Ib (Rct + 2 RL)where:2 RL = total resistance of the secondary wiring,Rct = CT secondary winding resistance,Ib = power transformer nominal current seen atthe CT secondary,A = constant depending on transformer power:Some suppliers take into account the throughcurrent, for example:

Vk u 43

3 R RL pIf

ctR + +( )( )on the star side of the power transformer,and Vk u 4 If (Rct + 2 RL + Rp)on the delta side of the power transformer.The through current will be defined in the sameway as for the restricted earth fault protection.Note:Use of matching CTs leads to differentexpressions of the knee point voltage for themain CTs which must take into account the extraload that they represent.


To conclude, the stability of this protection isensured by:c the threshold which increases with the throughcurrent (restraint system),c the right choice of CT knee point voltage Vk ,c a system ensuring immunity to 2nd orderharmonics generated by inrush currents,c the most sophisticated relays are also immuneto 5th order harmonics which occur during powertransformer overexcitation (saturation).Low impedance differential protectionThis protection is used for busbar differentialprotection. It is very costly and space-consuming, as it requires a large number of

87BB1

87BB3

87BGlobal

87BB2

87BB4

B1

B1 C1.2

C1.3 C2.4

D4D3D2D1

C3.4

B3

B2

B4

B3

B2

B4Global

CMTD1

CMTD2

CMC1.3

CMC1.2/C3.4

CMC2.4

CMTD3

CMTD4

CMT = current metering and transfer (for feeders D1, D2, D3 and D4)CM = current metering (for crossing and in-line couplers C1.2, C1.3, C2.4 and C3.4)1 = transfer and metering2 = protection

1

2

modules and matching CTs which need one ormore cubicles according to switchboard size(see fig. 25 ).In the case of a double busbar switchboard, theprotection must be continually informed on theposition of the transfer switches in order to directthe currents of each feeder and incomer to therelay associated with monitoring of the busbar onwhich this feeder or incomer is connected.The CTs associated with this sophisticatedprotection may have different ratios. Theirsecondaries are also defined as class X in mostcases. However, as saturation can be tolerated,knee point voltage requirements are less severethan for high impedance differential protection.

Fig. 25 : example of a low impedance differential protection for a double busbar.


4.4 Distance protectionsThese protections, extremely common in highvoltage, are increasingly used for very longmedium voltage lines, as pilot wires do not needto be installed (see fig. 26 ).The formula generally used for CTs defined asclass X is as follows:

Vk u If 1 XR

R R 2 Rp ct L+ + +( )

Besides the usual terms already defined, thefollowing parameters can also be found:XR : reactance/resistance ratio between the

source and a three-phase short-circuit occurringat the end of the protected area.If is in this case equal to the three-phase short-circuit current at the end of the monitored area,seen at the secondary side of the upstream CT.Rp : relay resistance.

In many cases, information regarding the line tobe protected (cross-section, length) is non-existent or impossible to obtain before theswitchboard is delivered. However, thecalculation example in figure 27 shows theconsiderable difference betweenCT characteristics according to line length.Between 1 and 12 km, there is a ratio of 10between the characteristics.

This type of relay is always used for very longlines. It would not be reasonable, withoutinformation, to resign ourselves to using for If thevalue Isc at the head of the line. The example infigure 27 shows that the short-circuit currentdrops from 26.2 kA to 13.4 kA for a 2 km lineonly, then to 3.8 kA for a 12 km line.Knowledge, even approximate, of line length isan important factor in optimising CTs.

Fig. 26 : distance protections: at each end of the line a relay monitors 80 % of the line with instantaneousoperation.

L

RL

RL

Monitored area

Monitored area

21

21

Fig. 27 : calculation of CT knee point voltage Vk for distance protection relays, for various line lengths, showing theadvantage of using the Isc as a reference at the end of the line to define these CTs.

Source Line CalculationsIsc Rs Xs L Rlllll Xlllll Xt / Rt Zt Isc If (A) Vk

(km) (kA) (V)26.2 0.015 0.727 1 0.12 0.338 7.925 1.073 17.75 29.59 (264.1 x Rct) + 108.2826.2 0.015 0.727 2 0.2396 0.675 5.518 1.425 13.37 22.28 (145.23 x Rct) + 59.5426.2 0.015 0.727 5 0.599 1.688 3.936 2.492 7.646 12.74 (62.90 x Rct) + 25.7926.2 0.015 0.727 12 1.4376 4.051 3.29 4.994 3.815 6.358 (27.28 x Rct) + 11.1826.2 0.015 0.727 24 2.8752 8.102 3.055 9.29 2.051 3.418 (13.86 x Rct) + 5.6826.2 0.015 0.727 40 4.792 13.5 2.961 15.02 1.268 2.114 (8.37 x Rct) + 3.43

U = 33 kVCT primary side: 600CT secondary side: 1Rp = 0.36 2 RL = 0.05

Xt = Xs + XlRt = Rs + Rl

Vk u If 1 XR

R R 2 Rtt

p ct L+

+ +( )

Line cross-section = 150 mm2Rl = 0.12 /kmXl = 0.388 /kmRs , Xs = source resistanceand reactance


We shall deal in an incomplete but educationalmanner with two examples of specifications

5 CT specification examples

5.1 Motor feeder protections

For this application the functions are, forexample:v overcurrent,v thermal image,v unbalance.With electromagnetic relays, serial-connected inthe CT secondary, the minimum specificationoften encountered is 20 VA-5P30.With multifunction digital relays, the specificationis often 5 VA-5P20 it is superabundant.

The minimum ALF is 2 8 (motor) (CT)

n

n

II

,

i.e. kr u16 if In (motor) = In (CT).

But taking account of a motor In (200 A) for a300/1A CT: 16 x(300/200) = 12.The relay consumption is for example 0.025 VA(Sepam 2000) and 0.05 VA for wiring (6 m in 2.52);the 5 VA-5P20 CT has internal losses of 2 VA.Let us calculate kr :

kr 20 2 + 5

2 + 0.075 67.5= =

a value far greater than 12!A 2.5 VA-5P10 CT (where Pi = 1.5 VA) is morethan sufficient. Its kr is:

kr 10 1.5 + 2.5

1.5 + 0.075 25= =

5.2 Transformer feeder protectionsThis is the high threshold overcurrent protectionwhich sizes the CT (refer to paragraph 4.2):kr u 1.5

IIn

n

1

100Zsc

where In1 = Inominal of the transformer primaryand In = Inominal of the CT primary.Let us take the example of a 1 MVA transformer;Zsc = 5 %; Uprimary = 22 kV, hence In1 = 26.2 A.This gives, when In = 30 A, a minimum kr of 26.

Taking into account that the thermal withstand ofthe requested CT is 50 kA-1s the CT cannotbe manufactured. In actual fact the problemsstart as soon as Ith / In > 500, and in this case50000/30 = 1666!Faced with this kind of problem, the CT primarycan be overrated.In view of their characteristics, figure 28 showsthe overratings of CTs able to match the requiredALF and satisfy CT feasibility.

concerning conventional protections and twoexamples concerning differential protections.

Transformer U = 22 kV CT characteristicsPower Zsc Isc max In TFO In ALF required(MVA) (%) (kA) (A) (A) (Isc / In x 1.5)0.5 4 0.3 13 40 12.30.63 4 0.4 17 40 15.50.8 4 0.5 21 40 19.71 5 0.5 26 50 15.72.5 5 1.3 66 100 19.75 6 2.2 131 200 16.410 8 3.3 262 300 16.420 10 5.2 525 600 13.130 12 6.6 787 1000 9.840 13 8.1 1050 1500 8.180 16 13.1 2099 2500 7.9

Fig. 28 : standard CT for a 22 kV transformer feeder.


In this example, the relay used is a Sepam2000 D02 (Schneider Electric). This relay doesnot need an interposing CT (see fig. 29 ).The minimum knee point voltage Vk required isgiven by the formula:Vk = A Ib (Rct + 2 RL)where:Ib = power transformer nominal current at theCT secondary,Rct = CT secondary winding resistance,RL = resistance of a conductor linking the CT andthe relay,A = constant depending on transformer power:c 30 for 2 MVA < Pn < 14 MVA,c 24 for 15 MVA < Pn < 39 MVA,c 16 for 40 MVA < Pn < 70 MVA.Let us take an examplePn = 50 MVA hence: A = 16,I1 = 600 A U1 = 63 kV In1 = 1 A,I2 = 3000 A U2 = 11 kV In2 = 1 A,

I IIbn

11

1

P3 U

0.764 A,n2

= ( )

=

I IIbn

22

2

P3 U

0.875 A,n2

= ( )

=

where:I1 = nominal current of the CT at the primary sideof the power transformer,I2 = nominal current of the CT at the secondaryside of the power transformer,In1 = nominal current of the secondary windingsof the CT located at the primary side of thepower transformer,

In2 = nominal current of the secondary windingsof the CT located at the secondary side of thepower transformer.

Calculating CTsWe assume that the relay is located in thedownstream switchboard, resulting in a wiring2 RL of 1000 m for upstream CTs and 10 m fordownstream CTs.c CTs at the primary side of the powertransformerIf the wiring is 2.5 mm2 (i.e. 8 per km):2 RL = 8 x 1000/1000 = 8 .This gives:Vk > 16 x 0.764 (Rct + 8)Vk > 12.2 Rct + 98c CTs at the secondary side of the powertransformerIf the wiring is 2.5 mm2 (i.e. 8 per km):2 RL = 8 x 10/1000 = 0.08 .This gives:Vk > 16 x 0.875 (Rct + 0.08)Vk > 14 Rct + 1.12

In the example: choose In1 = 50 instead of 30 A.The minimum kr moves from 26 to 15.7.The ratio Ith / In becomes 50000/50 = 1000.It is still higher than 500, but such a CT is nowfeasible.

Fig. 29 : transformer differential protection principle.

87T

I1 / In1 I2 / In2

5.4 Differential protection for busbars (87B)For 87B busbar differential protection (seefig. 30 ), the relay used is a Sepam 100 LD(Schneider Electric). The minimum knee pointvoltage Vk required for this relay is given by:Vk u 2 If (Rct + 2 RL)

where:If = maximum through current at the CT secondary,Rct = CT secondary winding resistance,2 RL = resistance of the wiring loop between theCT and the relay.

Schneider Electric, manufacturer of protectionCTs and MV panels, offers standard2.5 VA-5P20 CTs suitable for electronic anddigital protection and which consume less than0.5 VA with a wiring resistance of 2 RL < 0.1 .

5.3 Transformer differential protection


Calculating 2 RL2 RL = (2 L/S)Loop length: 2 L = 45 m,Wiring cross-section: S = 2,5 mm2,where = 1.8 x 10-8,2 RL = 0.324

Calculating If

I I IIf scn

n =

2

1

I1n = CT nominal primary current,I2n = CT nominal secondary current,Isc = short-circuit current at switchboard level,I1n = 1250 AI2n = 1 AIsc = 25 kAIf = 20 ANow Vk can be determined:Vk > 2 x 20 x (Rt + 0.32),i.e.Vk > 40 Rct + 13After consultation, the proposed CT has an Rct of6 and a Vkr equal to 270 V.It is suitable as:40 x 6 + 13 = 252.96 V < 270 V.

Calculating Rst

Rst V2

Rkr

p= IIr = setting current,Rp = relay resistance,Vk = minimum knee point voltage required.

We choose:Ir = 5 % I2n = 0.05 AIn this case we can consider that Rp = 0Rst = 2530

Calculating Vpeak

Vp 2 2 V V V kr s kr= ( )Vs = (Rs + Rp) Isscwhere:Issc = maximum fault current seen at the CTsecondary: in this case Issc = IrVkr = real value of the knee point voltage of theCT (270 V),Vs = 50600 V,we find: Vp = 10426 V.Vp > 3000 V, a surge limiter is necessary.

Calculating the fault current Id really detectedId = Ir + Io m,where:Io = magnetising current at Vk / 2 (data given bythe CT manufacturer),m = number of CTs per phase used for busbarprotection, in this case = 5,Io = 0.006 A,We find: Id = 0.08 A, i.e. 100 A on the primaryside.We thus observe that earth faults will alsobe detected in a satisfactory manner since inthis network the earth fault current is limited to300 A.

Rst87B

Fig. 30 : busbar differential diagram.


This section has given only a few examples ofCT specification (ALF or Vk) according to theapplications.Their complete and optimised specification

Fig. 31 : information to be provided by each contributor for defining CTs.

requires active and co-ordinated participation ofmany contributors. The main informationrequired for each protection is given in the tablein figure 31 .

Information Typical Differential protections Contributorsto be provided protections

High impedance differential % Pilotprotection (see note) differen- wire

tialOvercurrent BB Motor Gen. Earth Transfo. Line51 + 51 N 87B 87M 87G 87N 87T 87L

Switchboard c c c c c c Networkreal Isc designerMaximum through c c c c c ccurrentMaximum earth c c c c c c cfault current Ihlif earth faultdetectionTransformer c cratingTransformer Zsc c cVector group of cthe powertransformerCT ratio c c c c c c c Protection

plandesigner

Relay manufacturer c c c c c c cand typeRelay setting c c c c ccurrent IrMotor starting c Motorcurrent manufacturerGenerator c Generatorsubtransient manufacturershort- circuitreactanceDistance between c c c c c c c SwitchboardCT and relay designerCross-section of c c c c c c cwiring used(or value of RL)

Note:When consulting the class X CT suppliers, you MUST ask for all the values of minimum Vk, maximum Rctand maximum Io as these are essential for completing the study.Besides the minimum Vk, the value of real Vk is also necessary to calculate peak voltage, when highimpedance differential relays are used.


6 Conclusion

Current transformers are essential links betweenelectrical busways and the devices protecting theMV and HV network components.

Their precise definition and optimisation are notsimple and require thorough understanding oftheir operation and close co-operation of manycontributors. As a rule the choice of a CT iseasier when it is associated with conventionalprotections. However, when differentialprotections (class X) are chosen, they must beexamined with care and require goodcommunication with the CT manufacturer.

However, as we have shown in this document, itis possible to specify CTs by excess according toprotections and applications. This solution is ameans of avoiding many problems relating tosafety, costs and lead times.

CT technology must also be taken into accountas it can offer advantages, for example:c standard CTs are optimised and available,c CTs with several secondaries offer space andcost savings,c multifunctional CTs, used in standard panels,allow cost savings.If, despite all the precautions taken, problemsarise, there is always a solution. This is thepurpose of Cahier Technique no. 195 whichhighlights the traps (most common errors) andthe possible solutions.Finally, this Cahier Technique demonstrates theextreme care that installation designers musttake when defining CTs, in particular whendrawing up the protection plan and thediscrimination study.


Bibliography

Standardsc IEC 60185: Current transformers -Characteristicsc IEC 60044-1: Instrument transformers - Part 1:Current transformers (replaces IEC 185).c IEC 60044-8: Instrument transformers - Part 8:Electronic current transformers.c NF C 42-502: Measuring instruments. Currenttransformers. Characteristics.c BS 3938 (replaced by BS 7626): Specificationfor current transformers.

Schneider Electrics Cahiers Techniquesc Protection des machines et des rseauxindustriels HT.P. ROCCIA, Cahier Technique no. 113.c Current transformer for HV protection.M. ORLHAC, Cahier Technique no. 164.c Protection of industrial and tertiary MVnetworks.A. SASTRE, Cahier Technique no. 174.c Directional protection equipment.P. BERTRAND, Cahier Technique no. 181.c Dynamic stability of industrial electricalnetworks.B. DE METZ NOBLAT and G. JEANJEAN,Cahier Technique no. 185.c Current Transformers: specification errors andsolutions.P. FONTI, Cahier Technique no. 195.

Various worksc Guide de lingnierie lectriqueELECTRA - 07.86c Protection des rseaux lectriquesCh. PREVE - Ed. Hermes - 06.98

Schneider Electric Direction Scientifique et Technique,Service Communication TechniqueF-38050 Grenoble cedex 9Fax: (33) 04 76 57 98 60

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