# Current transformers

## Principle of operationof CT

• A current transformer is defined as “as an instrument transformer in which the secondary current is substantially proportional to the primary current (under normal conditions of operation) and differs in phase from it by an angle which is approximately zero for an appropriate direction of the connections.” This highlights the accuracy requirement of the current transformer” but also important is the isolating function, which means no matter what the system voltage the secondary circuit need be insulated only for a low voltage.
• The current transformer works on the principle of variable flux. In the “ideal” current transformer, secondary current would be exactly equal (when multiplied by the turn’s ratio) and opposite of the primary current. But, as in the voltage transformer, some of the primary current or the primary ampere-turns are utilized for magnetizing the core, thus leaving less than the actual primary ampere turns to be “transformed” into the secondary ampere-turns. This naturally introduces an error in the transformation. The error is classified into two-the current or ratio error and the phase error.
• CT is designed to minimize the errors using the best quality electrical steels for the core of the transformer. Both toroidal (round) and rectangular CT s are manufactured. The secondary current is usually smaller in magnitude than the primary current.
• The rated secondary current is commonly 5A or 1A, though lower currents such as 0.5A are not uncommon. It flows in the rated secondary load, usually called the burden, when the rated primary current flows in the primary winding.
• The primary winding can consist merely of the primary current conductor passing once through an aperture in the current transformer core or it may consist of two or more turns wound on the core together with the secondary winding.
• The primary and secondary currents are expressed as a ratio such as 100/5. With a 100/5 ratio CT, 100A flowing in the primary winding will result in 5A flowing in the secondary winding, provided the correct rated burden is connected to the secondary winding. Similarly, for lesser primary currents, the secondary currents are proportionately lower.
• It should be noted that a 100/5 CT would not fulfill the function of a 20/1 or a 10/0.5 CT as the ratio expresses the current rating of the CT, not merely the ratio of the primary to the secondary currents.
• The extent to which the secondary current magnitude differs from the calculated value expected by virtue of the CT ratio is defined by the (accuracy) “Class” of the CT. The greater the number used to define the class, the greater the permissible “current error” [the deviation in the secondary current from the calculated value].
• Except for the least accurate classes, the accuracy class also defines the permissible phase angle displacement between primary and secondary currents. This latter point is important with measuring instruments influenced both by magnitude of current and by the phase angle difference between the supply voltage and the load current, such as kWh meters, watt meter, var meters and power factor meters.
• Common burden ratings are 2.5, 5, 10, 15 and 30VA.
• Current transformers are usually either “measuring” or “protective” types, these descriptions being indicative of their functions.

## Some Definitions used for CT:

Rated primary current:

• The value of current which is to be transformed to a lower value. In CT parlance, the “load” of the CT refers to the primary current.

Rated secondary current:

• The current in the secondary circuit and on which the performance of the CT is based. Typical values of secondary current are 1 A or 5 A. In the case of transformer differential protection, secondary currents of 1/ root 3 A and 5/ root 3 A are also specified.

Rated burden:

• The apparent power of the secondary circuit in Volt-amperes expressed at the rated secondary current and at a specific power factor (0.8 for almost all standards)

Accuracy class:

• In the case of metering CT s, accuracy class is typically, 0.2, 0.5, 1 or 3.
• This means that the errors have to be within the limits specified in the standards for that particular accuracy class. The metering CT has to be accurate from 5% to 120% of the rated primary current, at 25% and 100% of the rated burden at the specified power factor.
• In the case of protection CT s, the CT s should pass both the ratio and phase errors at the specified accuracy class, usually 5P or 10P, as well as composite error at the accuracy limit factor of the CT.

Composite error:

• The rms value of the difference between the instantaneous primary current and the instantaneous secondary current multiplied by the turns ratio, under steady state conditions.

Accuracy limit factor:

• The value of primary current up to which the CT complies with composite error requirements. This is typically 5, 10 or 15, which means that the composite error of the CT has to be within specified limits at 5, 10 or 15 times the rated primary current.

Short time rating:

• The value of primary current (in kA) that the CT should be able to withstand both thermally and dynamically without damage to the windings, with the secondary circuit being short-circuited. The time specified is usually 1 or 3 seconds.

Instrument security factor (factor of security):

• This typically takes a value of less than 5 or less than 10 though it could be much higher if the ratio is very low. If the factor of security of the CT is 5, it means that the composite error of the metering CT at 5 times the rated primary current is equal to or greater than 10%. This means that heavy currents on the primary are not passed on to the secondary circuit and instruments are therefore protected. In the case of double ratio CT’s, FS is applicable for the lowest ratio only.

Class PS X CT:

• In balance systems of protection, CT s with a high degree of similarity in their characteristics is required. These requirements are met by Class PS (X) CT s. Their performance is defined in terms of a knee-point voltage (KPV), the magnetizing current (Imag) at the knee point voltage or 1/2 or 1/4 the knee-point voltage, and the resistance of the CT secondary winding corrected to 75C. Accuracy is defined in terms of the turn’s ratio.

Knee point voltage:

• That point on the magnetizing curve where an increase of 10% in the flux density (voltage) causes an increase of 50% in the magnetizing force (current).

Summation CT:

• When the currents in a number of feeders need not be individually metered but summated to a single meter or instrument, a summation current transformer can be used. The summation CT consists of two or more primary windings which are connected to the feeders to be summated, and a single secondary winding, which feeds a current proportional to the summated primary current. A typical ratio would be 5+5+5/ 5A, which means that three primary feeders of 5 are to be summated to a single 5A meter.

Core balance CT (CBCT):

• The CBCT, also known as a zero sequence CT, is used for earth leakage and earth fault protection. The concept is similar to the RVT. In the CBCT, the three core cable or three single cores of a three phase system pass through the inner diameter of the CT. When the system is fault free, no current flows in the secondary of the CBCT. When there is an earth fault, the residual current (zero phase sequence current) of the system flows through the secondary of the CBCT and this operates the relay. In order to design the CBCT, the inner diameter of the CT, the relay type, the relay setting and the primary operating current need to be furnished.

Interposing CT’s (ICT’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 transformers.

Rated primary current:

• The value of primary current which appears in the designation of the transformer and on which the performance of the current transformer is based.

Rated secondary current:

• The value of secondary current which appears in the designation of the transformer and on which the performance of the current transformer is based.

Rated transformation ratio:

• The ratio of the rated primary current to the rated secondary current.

Current error (ratio error):

• The error with a transformer introduces into the measurement of a current and which arises from the fact that actual transformation ratio is not equal to the rated transformer ratio. The current error expressed in percentage is given by the formula:
• Current error, percent = (Ka.Is-Ip) x 100 / Ip
• Where Ka= rated transformation ratio
• Ip= actual primary current
• Is= actual secondary current when Ip is flowing under the conditions of measurement

Phase displacement:

• The difference in phase between the primary and secondary current vectors, the direction of the vectors being so chosen that the angle is zero for the perfect transformer. The phase displacement is said to be positive when the secondary current vector leads the primary current vector. It is usually express in minutes

Rated output:

• The value of the apparent power (in volt-amperes at a specified power (factor) which the current transformer is intended to supply to the secondary circuit at the rated secondary current and with rated burden connected to it.

Highest system voltage:

• The highest rms line to line voltage which can be sustained under normal operating conditions at any time and at any point on the system. It excludes temporary voltage variations due to fault condition and the sudden disconnection of large loads.

Rated insulation level:

• That combination of voltage values (power frequency and lightning impulse, or where applicable, lightning and switching impulse) which characterizes the insulation of a transformer with regard to its capability to withstand by dielectric stresses. For low voltage transformer the test voltage 4kV, at power-frequency, applied during 1 minute.

Rated short-time thermal current (Ith):

• The rms value of the primary current which the current transformer will withstand for a rated time, with their secondary winding short circuited without suffering harmful effects.

Rated dynamic current (Idyn):

• The peak value of the primary current which a current transformer will withstand, without being damaged electrically for mechanically by the resulting electromagnetic forces, the secondary winding being short-circuited.

Rated continuous thermal current (Un)

• The value of current which can be permitted to flow continuously in the primary winding, the secondary windings being connected to the rated burdens, without the temperature rise exceeding the specified values.

Instrument security factor (ISF or Fs):

• The ratio of rated instrument limits primary current to the rated primary current. The times that the primary current must be higher than the rated value, for the composite error of a measuring current transformer to be equal to or greater than 10%, the secondary burden being equal to the rated burden. The lower this number is, the more protected the connected instrument are against.

Sensitivity

• Sensitivity is defined as the lowest value of primary fault current, within the protected zone, which will cause the relay to operate. To provide fast operation on an in zone fault, the current transformer should have a ‘Knee Point Voltage’ at least twice the setting voltage of the relay.
• The ‘Knee Point Voltage’ (Vkp) is defined as the secondary voltage at which an increase of 10% produces an increase in magnetizing current of 50%. It is the secondary voltage above which the CT is near magnetic saturation.

Stability

• That quality whereby a protective system remains inoperative under all conditions other than those for which it is designed to operate, i.e. an in-zone fault Stability is defined as the ratio of the maximum through fault current at which the system is stable to nominal full load current. Good quality current transformers will produce linear output to the defined knee point voltage (Vkp).
• Vkp = 2If(Rs+Rp) for stability, where
• If = max through fault secondary current at stability limit
• Rs = CT secondary winding resistance
• Rp = loop lead resistance from CT to relay Transient Effects
• Balanced protective systems may use time lag or high speed armature relays. Where high speed relays are used, operation of the relay occurs in the transient region of fault current, which includes the d c asymmetrical component.
• The buildup of magnetic flux may therefore be high enough to preclude the possibility of avoiding the saturation region. The resulting transient instability can fortunately be overcome using some of the following  techniques.
• A) Relays incorporating capacitors to block the dc asymmetrical component
• B) Biased relays, where dc asymmetrical currents are compensated by anti phase coils.
• C) Stabilizing resistors in series with current operated relays, or in parallel with voltage operated relays. These limit the spill current (or voltage) to a maximum value below the setting value. For series resistors in current
operated armature relays.
• Rs = (Vkp/2) – (VA/Ir)
• Rs = value of stabilizing resistor in ohms
• Vkp = CT knee point voltage
• VA = relay burden (typically 3VA)
• Ir = relay setting current
• Note: The value of Rs varies with each fault setting. An adjustable.

Field Adjustment of Current Transformer Ratio:

• The ratio of current transformers can be field adjusted to fulfill the needs of the application.  Passing more secondary turns or more primary turns through the window will increase or decrease the turns ratio.
• Actual Turns Ratio = (Name Plate Ration- Secondary Turns Added) / Primary Turns.

## Types of Current transformers (CT’s)

### According to Construction of CT:

1)    Ring Core CT’s :

• There are available for measuring currents from 50 to 5000 amps, with windows (power conductor opening size) from 1″ to 8″ diameter.

2)    Split Core CT’s:

• There are available for measuring currents from 100 to 5000 amps, with windows in varying sizes from 1″ by 2″ to 13″ by 30″. Split core CT’s have one end removable so that the load conductor or bus bar does not have to be disconnected to install the CT.

3)    Wound Primary CT’s:

• There are designed to measure currents from 1 amp to 100 amps. Since the load current passes through primary windings in the CT, screw terminals are provided for the load and secondary conductors. Wound primary CT’s are available in ratios from 2.5:5 to 100:5

### 1) 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 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.
• This means that at these higher levels of primary current the secondary current is less than proportionate. The effect of this is to reduce the extent to which any measuring device connected to the CT secondary is subjected to current Overload.

2)  Protective CT:

• On the other hand the reverse is required of the protective type CT, the principal purpose of 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 more loads to be supplied by a voltage or power transformer where the devices are paralleled across the secondary winding.
• With a CT, an increase in the burden will result in an increase in the CT secondary output voltage. This is automatic and necessary to maintain the current to the correct magnitude. Conversely, a reduction in the burden will result in a reduction in the CT secondary output voltage.
• This rise in secondary voltage output with an increase in burden means that, theoretically, with infinite burden as is the case with the secondary load open circuit, an infinitely high voltage appears across the secondary terminals. For practical reasons this voltage is not infinitely high, but can be high enough to cause a breakdown in the insulation between primary and secondary windings or between either or both windings and the core. For this reason, primary current should never be allowed to flow with no load or with a high resistance load connected across the secondary winding.
• When 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 but that it also includes the burden imposed by the connecting cable and the resistance of the connections.
• If, for example, the resistance of the connecting cable and the connections is 0.1 ohm and the secondary rating of the CT is 5A, the burden of the cable and connections (RI2) is 0.1 x 5 x 5 = 2.5VA. This must be added to the burden(s) of the connected device(s) when determining whether the CT has an adequately large burden rating to supply the required device(s) and the burden imposed by the connections.
• Should the burden imposed on the CT secondary winding by the connected device(s) and the connections exceed the rated burden of the CT the CT may partly or fully saturate and therefore not have a secondary current adequately linear with the primary current.
• The burden imposed by a given resistance in ohms [such as the resistance of a connecting cable] is proportional to the square of the rated secondary current. Therefore, where long runs of cable between CT and the connected device(s) are involved, the use of a 1A secondary CT and a 1A device rather than 5A will result in a 25-fold reduction in the burden of the connecting cables and connections. All burden ratings and calculations are at rated secondary current.
• Because of the foregoing, when a relatively long [more than a very few meters] cable run is required to connect a CT to its burden [such as a remote ammeter] a calculation should be made to determine the cable burden. This is proportional to the “round trip” resistance, i.e. twice the resistance of the length of twin cable used. Cable tables provide information on the resistance values of different sizes of conductors at 20o C per unit length.
• The calculated resistance is then multiplied by the square of the CT secondary current rating [25 for 5A, 1 for 1A]. If the VA burden as calculated by this method and added to the rated burden(s) of the device(s) to be driven by the CT exceeds the CT burden rating, the cable size must be increased [to reduce the resistance and thus the burden] or a CT with a higher VA burden rating must be used, or a lower CT secondary current rating [with matching change in the current rating of the device(s) to be driven] should be substituted

## Specification of CT:

1. RATIO: input / output current ratio
2. VA: total burden including pilot wires.
3. CLASS: Accuracy required for operation
4. DIMENSIONS: maximum & minimum limits
5. Specification of CT: (RATIO/VA BURDEN/ACCURACY CLASS /ACCURACY LIMIT FACTOR).
6. For example: 1600/5, 15VA 5P10
• Ratio: 1600/5, Burden: 15VA, Accuracy Class: 5P, ALF: 10

## Burden of CT:

• The CT burden is the maximum load (in VA) that can be applied to the CT secondary.
• 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 secondary circuit cable burden (cable burden = where I = CT secondary current, R = cable resistance per length, L = total length of the secondary circuit cable. If the proper size and short length of wire is used, cable burden can be ignored).
• The CT secondary circuit load shall not be more than the CT VA rating. If the load is less than the CT burden, all meters connected to the measuring CT should provide correct reading. So, in your example, there should not be any effect on Ammeter reading if you use a CT of either 5 VA or 15 VA (provided the proper size and short length of wire is used for the CT secondary side).
• 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 the maximum permissible CT error is only 0.5%.
• CT burden is the load imposed on CT secondary during operation.
• The burden is mentioned as x-VA for the CT.
• In the case of Measuring Current transformer, the burden depends on the connected meters and quantity of meters on the secondary i.e. no of Ammeters, KWh meters, Kvar meters, Kwh meters, transducers and also the connection cable burden (I2 x R x2 L) to metering shall be taken into account.
• where 2L is the to &fro distance of cable length L from CT to metering circuits,R=is the resistance of unit length of connecting cable, I=secondary current of CT
• Total burden of Measuring CT=Connecting cable Burden in VA + sum of Meters Burden in VA
• Note Meters burden can be obtained from manufacturer catalogue
• Selected CT burden shall be more than the calculated burden.
• In the case of Protection CTs the burden is calculated in the same way as above except the burden of individual protective relays burden shall be considered instead of meters. The connecting cable burden is calculated in the same way as metering CT
• Total burden of Protection CT=Connecting cable Burden in VA + sum of Protective relays Burden in VA.
• The external load applied to the secondary of a current transformer is called the “burden”. the burden can be expressed in two ways.
• The burden can be expressed as the total impedance in ohms of the circuit or the total volt-amperes and power factor at a specified value of current or voltage and frequency.
• Formerly, the practice was to express the burden in terms of volt-amperes and power factor, the volt-amperes being what would be consumed in the burden impedance at rated secondary current (in other words, rated secondary current squared times the burden impedance). Thus, a burden of 0.5-ohm impedance may be expressed also as “12.5 volt-amperes at 5 amperes,” if we assume the usual 5-ampere secondary rating. The volt ampere terminology is no longer standard, but it needs defining because it will be found in the literature and in old data.
• The individual devices may only be the current transformer, a short run of wire and a meter. In contrast, the circuit may have the current transformer, a lone run of wiring, a relay, a meter, an auxiliary current transformer and a transducer. While the latter configuration would not be used today, one may be required to make this calculation on an existing system.
•  All manufacturers can supply the burden of their individual devices. Although not used very often these days, induction disk over-current devices always gave the burden for the minimum tap setting. To determine the impedance of the actual tap setting being used, first square the ratio of minimum divide by the actual tap setting used and, second multiply this value by the minimum impedance.
• Suppose an impedance of 1.47 + 5.34j at the 1-amp tap. To apply the relay at the 4-amp tap the engineer would multiply the impedance at the 1-amp tap setting by (1/4)2. The impedance at the 4-amp tap would be 0.0919 + 0.3338j or 0.3462 Z at 96.4 power factor.
• The CT burden impedance decreases as the secondary current increases, because of saturation in the magnetic circuits of relays and other devices. Hence, a given burden may apply only for a particular value of secondary current. The old terminology of volt-amperes at 5 amperes is most confusing in this respect since it is not necessarily the actual volt amperes with 5 amperes flowing, but is what the volt-amperes would be at 5 amperes
• If there were no saturation. Manufacturer’s publications give impedance data for several values of over current for some relays for which such data are sometimes required. Otherwise, data are provided only for one value of CT secondary current.
•  If a publication does not clearly state for what value of current the burden applies, this information should be requested. Lacking such saturation data, one can obtain it easily by test. At high saturation, the impedance approaches the d-c resistance. Neglecting the reduction in impedance with saturation makes it appear that a CT will have more inaccuracy than it actually will have. Of course, if such apparently greater inaccuracy can be tolerated, further refinements in calculation are unnecessary. However, in some applications neglecting the effect of saturation will provide overly optimistic results; consequently, it is safer always to take this effect into account.
• It is usually sufficiently accurate to add series burden impedances arithmetically. The results will be slightly pessimistic, indicating slightly greater than actual CT ratio inaccuracy. But, if a given application is so borderline that vector addition of impedances is necessary to prove that the CTÕs will be suitable, such an application should be avoided.
• If the impedance at pickup of a tapped over current-relay coil is known for a given pickup tap, it can be estimated for pickup current for any other tap. The reactance of a tapped coil varies as the square of the coil turns, and the resistance varies approximately as the turns. At pickup, there is negligible saturation, and the resistance is small compared with the reactance. Therefore, it is usually sufficiently accurate to assume that the impedance varies as the square of the turns. The number of coil turns is inversely proportional to the pickup current, and therefore the impedance varies inversely approximately as the square of the pickup current.
• Whether CT is connected in wye or in delta, the burden impedance are always connected in wye. With wye-connected CTÕs the neutrals of the CTÕs and of the burdens are connected together, either directly or through a relay coil, except when a so-called zero phase-sequence-current shunt (to be described later) is used.
• It is seldom correct simply to add the impedance of series burdens to get the total, whenever two or more CTÕs are connected in such a way that their currents may add or subtract in some common portion of the secondary circuit. Instead, one must calculate the sum of the voltage drops and rises in the external circuit from one CT secondary terminal to the other for assumed values of secondary currents flowing in the various branches of the external circuit. The effective CT burden impedance for each combination of assumed currents is the calculated CT terminal voltage divided by the assumed CT secondary current. This effective impedance is the one to use, and it may be larger or smaller than the actual impedance which would apply if no other CTÕs were supplying current to the circuit.
• If the primary of an auxiliary CT is to be connected into the secondary of a CT whose accuracy is being studied, one must know the impedance of the auxiliary CT viewed from its primary with its secondary short-circuited. To this value of impedance must be added the impedance of the auxiliary CT burden as viewed from the primary side of the auxiliary CT; to obtain this impedance, multiply the actual burden impedance by the square of the ratio of primary to secondary turns of the auxiliary CT. It will become evident that, with an auxiliary CT that steps up the magnitude of its current from primary to secondary, very high burden impedances, when viewed from the primary, may result.
• Burden is depending on pilot lead length
 VA Applications 1 To 2 VA Moving iron ammeter 1 To 2.5VA Moving coil rectifier ammeter 2.5 To 5VA Electrodynamics instrument 3 To 5VA Maximum demand ammeter 1 To 2.5VA Recording ammeter or transducer
•  Burden (VA) of copper wires between instrument & current transformer for 1A and 5A secondary’s
 Cross Section (mm2) CT  1 Amp Secondary Burden in VA (Twin Wire) Distance 10 meter 20 meter 40 meter 60 meter 80 meter 100 meter 1.0 0.35 0.71 1.43 2.14 2.85 3.57 1.5 0.23 0.46 0.92 1.39 1.85 2.31 2.5 0.14 0.29 0.57 0.86 1.14 1.43 4.0 0.09 0.18 0.36 0.54 0.71 0.89 6.0 0.06 0.12 0.24 0.36 0.48 0.6

 Cross Section (mm2) CT  5 Amp Secondary Burden in VA (Twin Wire) Distance 1 meter 2 meter 4 meter 6 meter 8 meter 10 meter 1.5 0.58 1.15 2.31 3.46 4.62 5.77 2.5 0.36 0.71 1.43 2.14 2.86 3.57 4.0 0.22 0.45 0.89 1.34 1.79 2.24 6.0 0.15 0.30 0.60 0.89 1.19 1.49 10.0 0.09 0.18 0.36 0.54 0.71 0.89

### Calculation of the CT burden:

• The actual burden is formed by the resistance of the pilot conductors and the protection relay(s).

### Resistance of a conductor

• The resistance of a conductor (with a constant cross-sectional area) can be calculated from the equation:
• R =ƿ*l/ A
• where ƿ  = resistivity of the conductor material (given typically at +20°C)
• l = length of the conductor
• A = cross-sectional area
• If the resistivity is given in μΩm, the length in meters and the area in mm2, the equation 1 will give the resistance directly in ohms.
• Resistivity: Copper 0.0178µΩm at 20 °C and 0.0216 µΩm at 75 °C

### 4- or 6-wire connection:

• If 6-wire connection is used, the total length of the wire, naturally, will be two times the distance between the CT and the relay.
•  However, in many cases a common return conductor is used (figure). Then, instead of multiplying the distance by two, a factor of 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.

• If, for example, the distance between the CT and the relay is 5 meters the total length is 2 x 5 m = 10 meter for 6-wire connection, but only 1.2 x 5 m = 6.0 meter when 4-wire connection is used.

### Burden of the relay:

• For example, the input impedance is less than 0.020 Ω for a 5 A input (i.e. burden less than 0.5 VA) and less than 0.100 Ω for a 1 A input (i.e. less than 0.1 VA).

### Example

• The distance between the CTs and the protection relay is 15 meters, 4 mm2 Cu conductors in 4-wire connection are used. The burden of the relay input is less than 20 mΩ (5 A inputs). Calculate the actual burden of the CT at 75°C:
• Solution:
• ƿ = 0.0216 µΩm (75°C)
• R = 0.0216 µΩm x (1.2 x 15 m) / 4 mm2 = 0.097 Ω
• Burden of CT = 0.097 Ω + 0.020 Ω = 0.117 Ω.
• Using CTs of burden values higher than required, is unscientific since it leads to inaccurate reading (meter) or inaccurate sensing of fault / reporting conditions.
• Basically, such high value of design burden extends saturation characteristics of CT core leading to likely damage to the meter connected across it under overload condition. e.g.When we expect security factor (ISF) to be 5, the secondary current should be restricted toless than 5 times in case primary current shoots to more than 5 times its rated value.
• In such an overload condition, the core of CT is desired to go into saturation, restricting the secondary current thus the meter is not damaged. However, when we ask for higher VA, core doesn’t go into saturation due to less load (ISF is much higher than desired) which may damage the meter.
• To understand the effect on Accuracy aspect, let’s take an example of a CT with specified burden of 15 VA, and the actual burden is 2.5 VA:15 VA CT with less than 5 ISF will have saturation voltage of 15 Volts (15/5×5), and actual burden of 2.5 VA the saturation voltage required shall be ( 2.5/5 x 5) 2.5 Volts against 15 Volts resulting ISF = 30 against required of 5.

## Accuracy Class of CT:

•  Accuracy is specified as a percentage of the range, and is given for the maximum burden as expressed in volt amperes. The total burden includes the input resistance of the meter and the loop resistance of the wire and connections between the current transformer and meter.
• Example: Burden = 2.0 VA. Maximum Voltage drop = 2.0 VA / 5 Amps = 0.400 Volts.
•  Maximum Resistance = Voltage / Current = 04.00 Volts / 5 Amps =0.080 Ohms.
• If the input resistance of the meter is 0.010 Ohms, then 0.070 Ohms is allowed for loop resistance of the wire, and connections between the current transformer and the meter. The length and gauge of the wire must be considered in order to avoid exceeding the maximum burden.
• If resistance in the 5 amp loop causes the burden to be exceeded, the current will drop. This will result in the meter reading low at higher current levels.Considering a core of certain fixed dimensions and magnetic materials with a secondary winding of say 200 turns (current ratio 200/1 turns ratio 1/200) and say it takes 2 amperes of the 200A primary current to magnetize the core, the error is therefore only 1% approximately. However considering a 50/1 CT with 50 secondary turns on the same core it still takes 2 amperes to magnetize to core. The error is then 4% approximately. To obtain a 1% accuracy on the 50/1 ring CT a much larger core and/or expensive core material is required
• As in all transformers, errors arise due to a proportion of the primary input current being used to magnetize the core and not transferred to the secondary winding. The proportion of the primary current used for this purpose determines the amount of error.
• The essence of good design of measuring current transformers is to ensure that the magnetizing current is low enough to ensure that the error specified for the accuracy class is not exceeded.
This is achieved by selecting suitable core materials and the appropriate cross-sectional area of core. Frequently in measuring currents of 50A and upwards, it is convenient and technically sound for the primary winding of a CT to have one turn only.
• In these most common cases the CT is supplied with a secondary winding only, the primary being the cable or bus bar of the main conductor which is passed through the CT aperture in the case of ring CTs (i .e. single primary turn) it should be noted that the lower the rated primary current the more difficult it is (and the more expensive it is) to achieve a given accuracy.

• ### Fa=Fn X ( (Sin+Sn) / (Sin+Sa) )

• Fn = Rated Accuracy Limit Factor
• Sin = Internal Burden of CT secondary Coil
• Sn= Rated Burden of CT (in VA)
• Sa= Actual Burden of CT (in VA)
• Example: The internal secondary coil resistance of the CT(5P20) is 0.07 Ω, the secondary burden (including wires and relay) is 0.117 Ω and the CT is rated 300/5, 5P20, 10 VA. Calculate the actual accuracy limit factor.
• Fn = 20 (CT data 5P20),
• Sin = (5A)2 × 0.07 Ω =1.75 VA,
• Sn = 10 VA (from CT data),
• Sa = (5A)2 × 0.117 Ω = 2.925 VA
• Fa= 20 X ( (1.75+10) / (1.75+2.925) )
• ALF(Fa)= 50.3

### Accuracy Class of Metering CT:

•  Metering CTs In general, the following applies:
 Class Applications 0.1 To 0.2 Precision measurements 0.5 High grade kilowatt hour meters for commercial grade kilowatt hour meters 3 General industrial measurements 3 OR 5 Approximate measurements
 Protective System CT Secondary VA Class Per current for phase & earth fault 1A 2.5 10P20 Or 5P20 5A 7.5 10P20 Or 5P20 Unrestricted earth fault 1A 2.5 10P20 Or 5P20 5A 7.5 10P20 Or 5P20 Sensitive earth fault 1A or 5A Class PX use relay manufacturers formulae Distance protection 1A or 5A Class PX use relay manufacturers formulae Differential protection 1A or 5A Class PX use relay manufacturers formulae High impedance differential impedance 1A or 5A Class PX use relay manufacturers formulae High speed feeder protection 1A or 5A Class PX use relay manufacturers formulae Motor protection 1A or 5A 5 5P10

### Accuracy Class of Protection CT:

• In addition to the general specification required for CT design, protection CT’s requiring an Accuracy Limit Factor (ALF). This is the multiple of rated current up to which the CT will operate while complying with the accuracy class requirements.
• In general the following applies:
 Class Applications 10P5 Instantaneous over current relays & trip coils – 2.5VA 10P10 Thermal inverse time relays – 7.5VA 10P10 Low consumption Relay – 2.5VA 10P10/5 Inverse definite min. time relays (IDMT) over current 10P10 IDMT Earth fault relays with approximate time grading – 15VA 5P10 IDMT Earth fault relays with phase fault stability or accurate time grading required – 15VA
•  Accuracy Class: Metering Accuracy as per IEEE C37.20.2b-1994
 Ratio B0.1 B0.2 B0.5 B0.9 B1.8 Relaying Accuracy 50:5 1.2 2.4 – – – C or T10 75:5 1.2 2.4 – – – C or T10 100:5 1.2 2.4 – – – C or T10 150:5 0.6 1.2 2.4 – – C or T20 200:5 0.6 1.2 2.4 – – C or T20 300:5 0.6 1.2 2.4 2.4 – C or T20 400:5 0.3 0.6 1.2 1.2 2.4 C or T50 600:5 0.3 0.3 0.3 1.2 2.4 C or T50 800:5 0.3 0.3 0.3 0.3 1.2 C or T50 1200:5 0.3 0.3 0.3 0.3 0.3 C100 1500:5 0.3 0.3 0.3 0.3 0.3 C100 2000:5 0.3 0.3 0.3 0.3 0.3 C100 3000:5 0.3 0.3 0.3 0.3 0.3 C100 4000:5 0.3 0.3 0.3 0.3 0.3 C100

### Important of accuracy & phase angle

• Current error is an error that arises when the current value of the actual transformation ratio is not equal to rated transformation ratio.
• ### Current error (%) = {(Kn x Is – Ip) x 100}/Ip

• Kn = rated transformation ratio
• Ip = actual primary current
• Is = actual secondary current
• Example:
• In case of a 2000/5A class 1 5VA current transformer
• Kn = 2000/5 = 400 turn
• Ip = 2000A
• Is = 4.9A
• Current error = {(400 x 4.9 – 2000) x100}/2000 = -2%
• For protection class current transformer, the accuracy class is designed by the highest permissible percentage composite error at the accuracy limit primary current prescribed for the accuracy class concerned.
• Accuracy class includes: 5P, 10P
• Standard accuracy limit factor are: 5, 10, 15, 20, 30

### By phase angle

• Phase error is the difference in phase between primary & secondary current vectors, the direction of the vectors to be zero for a perfect transformer.
• You will experience a positive phase displacement when secondary current vector lead primary current vector.
• Unit of scale expressed in minutes / cent radians.
• Circular measure = (unit in radian) is the ratio of the distance measured along the arc to the radius.
• Angular measure = (unit in degree) is obtained by dividing the angle subtended at the center of a circle into 360 deg equal division known as “degrees”.
• Limits of current error and phase displacement for measuring current transformer (Classes 0.1 To 1)
 Accuracy Class +/- Percentage Current (Ratio) Error at % Rated Current +/- Phase Displacement at % Rated Current Minutes Centi radians 5 20 100 120 5 20 100 120 5 20 100 120 0.1 0.4 0.2 0.1 0.1 15 8 5 5 0.45 0.24 0.15 0.15 0.2 0.75 0.35 0.2 0.2 30 15 10 10 0.9 0.45 0.3 0.3 0.5 1.5 0.75 0.5 0.5 90 45 30 30 2.7 1.35 0.9 0.9 1.0 3 1.5 1 1 180 90 60 60 5.4 2.7 1.8 1.8
•  limits of current error and phase displacement for measuring current transformer For special application
 Accuracy Class +/- Percentage Current (Ratio) Error at % Rated Current +/- Phase Displacement at % Rated Current Minutes Centi radians 1 5 20 100 120 1 5 20 100 120 1 5 20 100 120 0.2S 0.75 0.35 0.2 0.2 0.2 30 15 10 10 10 0.9 0.4 0.3 0.3 0.3 0.5S 1.50 0.75 0.5 0.5 0.5 90 45 30 30 30 2.7 1.3 0.9 0.9 0.9
•  limits of current error for measuring current transformers (classes 3 and 5)
 Accuracy Class +/- Percentage Current (Ratio) Error at % Rated Current 50 120 3 3 3 5 5 5

## Class X Current Transformer:

• Class X current transformer is use in conjunction with high impedance circulating current differential protection relay, eg restricted earth fault relay. As illustrated in IEC60044-1, the class X current transformer is needed.
• The following illustrates the method to size a class X current transformer.
• Step 1: calculating knee point voltage Vkp
• ### Vkp = {2 x Ift (Rct+Rw)}/ k

• Vkp = required CT knee point voltage
• Ift = max transformer through fault in ampere
• Rct = CT secondary winding resistance in ohms
• Rw = loop impedance of pilot wire between CT and the
• K = CT transformation ratio
• Step 2: calculate Transformer through fault Ift
• ### Ift = (KVA x 1000)/ (1.732 x V x Impedance)

• KVA = transformer rating in kVA
• V = transformer secondary voltage
• Impedance = transformer impedance
• Step 3: How to obtain Rct
• To measure when CT is produce
• Step 4: How to obtain Rw
• This is the resistance of the pilot wire used to connect the 5th class X CT at the transformer star point to the relay
• In the LV switchboard. Please obtain this data from the Electrical contractor or consultant. We provide a table to
• Serve as a general guide on cable resistance.
• Example:
• Transformer Capacity: 2500kVA
• Transformer impedance: 6%
• Voltage system : 22kV / 415V 3phase 4 wire
• Current transformer ratio: 4000/5A
• Current transformer type: Class X PR10
• Current transformer Vkp : 185V
• Current transformer Rct  : 1.02½ (measured)
• Pilot wire resistance Rw : 25 meters using 6.0mm sq cable
• = 2 x 25 x 0.0032 = 0.16½
• Ift = (kVA x 1000) / (1.732 x V x impedance) = (2500 x 1000) / (1.732 x 415 x 0.06)= 57,968 (Say 58,000A)
• Vkp = {2 x Ift (Rct+Rw)} / k= {2 x 58000 (1.02+0.16)} / 800= 171.1½.

### Accuracy Limit Factor:

• Accuracy limit Factor is defined as the multiple of rated primary current up to which the transformer will comply with the requirements of ‘Composite Error’ . Composite Error is the deviation from an ideal CT (as in Current Error), but takes account of harmonics in the secondary current caused by non-linear magnetic conditions through the cycle at higher flux densities.
• Standard Accuracy Limit Factors are 5, 10, 15, 20 and 30. The electrical requirements of a protection current transformer can therefore be defined as :
• Selection of Accuracy Class & Limit Factor.
• Class 5P and 10P protective current transformers are generally used in over current and unrestricted earth leakage protection. With the exception of simple trip relays, the protective device usually has an intentional time delay, thereby ensuring that the severe effect of transients has passed before the relay is called to operate. Protection Current Transformers used for such applications are normally working under steady state conditions Three examples of such protection is shown. In some systems, it may be sufficient to simply detect a fault and isolate that circuit. However, in more discriminating schemes, it is necessary to ensure that a phase to phase fault does not operate the earth fault relay.

## Common Characteristics of CT’s

### 1)    Frequency affects a C/T only

• Because the lines of flux generated by the primary current begin to appear as DC as the frequency gets very low; a C/T needs the AC CYCLE changes to induce the secondary current. With anyone’s toroidal C/T, you will experience a drop in accuracy as 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.

### 2)    Below 60 Hz, the accuracy will be affected by the drop in frequency and voltage:

• with Instrument Transformers CT’s having the maximum acknowledged accuracy of 0.3% ANSI Rating, you will experience a drop in accuracy at 9 Hz to 5%; at 6 Hz it might be 7.5% of full scale. A Split Core unit might have double the inaccuracy, or more (for example, a 1% Split Core being used at 9 Hz will experience an accuracy rating of 33% – {.3%/5% is as 1%/X or X = 5/.3 = 16.7 x 2}. Remember, it is difficult to come up with test equipment with enough power to test full scale at unusual frequencies. The lesson here is to take the most accurate C/T you can if you are running in lower frequencies than 60 Hz.

### 3)    Exercising the C/T beyond its current rating for short periods is not usually a problem;

• Each CIT has a Thermal Rating Factor (if not published, then you must assume it is 1.0). This is a “continuous thermal current rating factor”. The Instrument Transformer model 5A (page 5, Section 2) has a factor of 1.33 at 300C. This means this particular C/T can be operated at 133% of its primary rated current CONTINUOUSLY without overheating (a 200:5 can thus be operated at 200 x 1.33 or 266 primary amps continuously). Other CIT’s have thermal rating factors of 1.5 and 2.0 etc. On a momentary basis, any CIT will usually operate at 64 times its primary current rating for 1 second; 150 time its current rating for 1 cycle.

### 4)    Above 60 Hz, a CIT becomes conversely more accurate up to about 4000 Hz.

• Above this, you must examine the wave shape carefully because it causes the core to saturate. 400 Hz is the published limit with some manufacturers; there is usually no problem with accuracy or heat or saturation at this frequency.4-20 mA DC Transmitters
• For all such transmitters, an independent, stable prime power is a requisite for published operational accuracy and characteristics.
• The internal transmitter of the device usually will not operate below 85 volts (43 Hz)
• Frequency response with a constant 120V 60 Hz Prime Power starts to fall off at 20 Hz; by 9 Hz it will be off by 5% Full Scale. At 6 Hz it will be off by 7.5% etc.

### 5)    P/T’s and Frequency:

• The ratio of voltage to frequency is important to a P/T (but not to a C/T). It must remain constant, or the P/T will overheat. Lesson: do not powers a P/T from a variable frequency drive unless this ratio can be made constant.Do not create a “current loop” by connecting the shield cylinder to ground from both ends. Current flowing in this loop will also be measured by the CT.
• The following are considerations which need to be made in order to properly select a current transformer.
• Indoors Or Out Door: Determine if the transformer is going to be subjected to the elements or not. Indoor transformers are usually less costly than outdoor transformers. Obviously, if the current transformer is going to be enclosed in an outdoor enclosure, it need not be rated for outdoor use. This is a common costly error in judgment when selecting current transformers.
• What do you need: 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 a 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 a 1% accuracy, you will want a current transformer with a 0.3 to 0.6 accuracy.
• You must keep in mind that the accuracy ratings are based on rated primary current flowing and per ANSI standards may be doubled (0.3 becomes 0.6%) when 10% primary current flows. As mentioned earlier, the rated accuracies are at stated burdens. You must take into consideration not only the burden of the load (instrument) but you must consider the total burden. The total burden 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.
• Voltage Class: You must know what the voltage is in the circuit to be monitored. This will determine what the voltage class of the current transformer must be as explained earlier.
• Primary Conductor: If you have selected a current transformer with a window you must know the number, type and size of the primary conductor(s) in order to select a window size which will accommodate the primary conductors.
• Application: The variety of applications of current transformers seems to be limited only by ones imagination. As new electronic equipment evolves and plays a greater role in the generation, control and application of electrical energy, new demands will be placed upon current transformer manufacturers and designers to provide new products to meet these needs
• Safety: For personnel and equipment safety and measurement accuracy, current measurements on conductors at high voltage should be made only with a conducting shield cylinder placed inside the CT aperture. There should be a low electrical impedance connection from one end only to a reliable local ground.
• An inner insulating cylinder of adequate voltage isolation should be between the shield cylinder and the conductor at high voltage. Any leakage, induced or breakdown current between the high voltage conductor and the ground shield will substantially pass to local ground rather than through the signal cable to signal ground.
• CT output signal termination: The CT output coaxial cable should preferably be terminated in 50 ohms.  CT characteristics are guaranteed only when CT is terminated in 50 ohms.  The termination should present sufficient power dissipation capability.  When CT output is terminated in 50 ohms, its sensitivity is half that when terminated in a high-impedance load.

## CT Consideration:

### Application & Limitation of CT:

• Increasing the number of primary turns can only decrease the turn’s ratio.  A current transformer with a 50 to 5 turns ratio can be changed to a 25 to 5 turns ratio by passing the primary twice through the window.
• The turn’s ratio can be either increased or decreased by wrapping wire from the secondary through the window of the current transformer.
• When using the secondary of a current transformer to change the turns ratio, the right hand rule of magnetic fields comes into play.  Wrapping the white lead or the X1 lead from the H1 side of the transformer through the window to the H2 side will decrease the turns ratio.  Wrapping this wire from the H2 side to the H1 side will increase the turns ratio.
• Using the black or X2 lead as the adjustment method will do the opposite of the X1(white) lead.  Wrapping from the H1 to the H2 side will increase the turns ratio, and wrapping from the H2 to the H1 side will decrease the turns ratio.
• Increasing the turns ratio with the secondary wire, turns on the secondary are essentially increased. A 50 to 5 current transformer will have a 55 to 5 ratio when adding a single secondary turn.
• Decreasing the turns ratio with the secondary wire, turns on the secondary are essentially decreased.  A 50 to 5 current transformer will have a 45 to 5 ratio when adding a single secondary turn.
• Decreasing the turns ratio with the primary, accuracy and VA burden ratings are the same as the original configuration.
• Increasing the turns ratio with the secondary will improve the accuracy and burden rating.
• Decreasing the turns ratio with the secondary will worsen the accuracy and burden rating.

## Testing & Ratio Modification of Current transformers

### Installing of CT:

•  Window type CT’s should be mounted with the H1 side of the window towards the power source. The X1 secondary terminal is the polarity terminal (Figure 3).The polarity marks of a current transformer indicate that when a primary current enters at the polarity mark (H1) of the primary, a current in phase with the primary current and proportional to it in magnitude will leave the polarity terminal of the secondary (X1).

• Normally CT’s should not be installed on “Hot” services. The power should be disconnected when the CT’s are installed. Many times this is not possible because of critical loads such as computers, laboratories, etc. that cannot be shut down. Split core CT’s should not be installed on “Hot” un insulated bus bars under any conditions.

## Recommendations for Installation of CT:

### 1.    The primary current must be centered in the CT aperture.

• Un-centered current may cause errors in current measurement. When the current to be measured is at high voltage, capacitive coupling between the high voltage conductor and the CT must be minimized.  This becomes a critical issue when a low-sensitivity CT is used.  In this context, CTs with less than 0.5 V/A in high-impedance output are considered “low sensitivity”.

### 2.    The CT couples with the primary current conductor in two modes:

• Magnetic coupling, which measures the current.  This is the only desirable coupling.
• Capacitive coupling with the conductor high voltage, which is undesired coupling.

### 3.    Magnetic coupling and the capacitive coupling can be identified:

• The CT output resulting from magnetic coupling changes polarity when the current direction changes.
• The CT output resulting from capacitive coupling does not change when the current direction changes.
• Therefore, to identify the signal caused by unwanted capacitive coupling, compare the CT output when
• the current conductor passes through CT in one direction, then in the other direction:  The output signal is
• the sum from magnetic coupling and capacitive coupling: the signal from magnetic coupling has changed
• polarity, while the signal from capacitive coupling has not changed polarity.

### 4.    To minimize unwanted capacitive coupling:

• Install common-mode filters on the CT output cable.  To realize simply a common-mode filter, use a
• ferrite (or better: nanocrystalline) core and pass the coaxial cable 6 to 8 times through the core.  It will
• constitute an excellent common-mode filter.
• Install a cylindrical shield between the current carrying conductor and the CT.  The shield must be
• grounded with a low-impedance grounding wire.  The shield must be grounded on one side only.  If it
• were grounded on two sides, it would constitute a one-turn short around the CT (to be avoided!)
• When possible,  maximize the “good” signal from magnetic coupling, by  using the most sensitive
• possible CT.  To determine the most sensitive model which can be used, take into consideration:

### 5.    The CT I x t product must be higher than the primary pulse charge.

• Higher sensitivity CTs also have higher droop.  The CT output signal droop must be acceptable in consideration of the duration of the signal to observe.  CT output does not droop when the current is nil, in-between pulses.
• Short pulses (<50ns) peak current can be up to 4 times the CT maximum current. SMA and BNC connectors can withstand repetitive 3000 volts peak for short time. If the CT output signal is too high, attenuators can be used.

## Primary/Secondary Turns Ratio Modification:

•  The nameplate current ratio of the current transformer is based on the condition that the primary conductor will be passed once through the transformer opening. If necessary, this rating can be reduced in even multiples by looping this conductor two or more times through the opening.
• A transformer having a rating of 300 amperes will be changed to 75 amperes if four loops or turns are made with the primary cable as illustrated.
• The ratio of the current transformer can be also modified by altering the number of secondary turns by forward or back-winding the secondary lead through the window of the current transformer.
• By adding secondary turns, the same primary amperage will result in a decrease in secondary output.
• By subtracting secondary turns, the same primary amperage will result in greater secondary output. Again using the 300:5 example, adding two secondary turns will require 310 amps on the primary to maintain the 5 amp secondary output or 62/1p = 310p/5s.
• Subtracting two secondary turns will only require 290 amps on the primary to maintain the 5 amp secondary output or 58s/5p = 290p/5s. The ratio modifications are achieved in the following manner:
• To add secondary turns, the white lead should be wound through the CT from the side opposite the polarity mark.
• To subtract turns, the white lead should be wound through the CT from the same side as the polarity mark.

### How to make Modifications in Primary Turns Ratio of CT:

• The ratio of the current transformer can be modified by adding more primary turns to the transformer. By adding primary turns, the current required to maintain five amps on the secondary is reduced.
• ### Ka = Kn X (Nn/Na)

• Ka= Actual Turns Ration.
• Kn=Name Plate T/C Ratio.
• Nn=Name Plate Number of Primary Turns.
• Na=Actual Number of Primary Turns.
• Example: 100:5 Current Transformers.
1. Primary Turns=1Nos:

•  Ka= (100/5) X (1/1) = 100:5
1. Primary Turns=2Nos:

• Ka=(100/5) X (1/2) = 50:5
1. Primary Turns= 4Nos:

• Ka= (100/5) X (1/4) = 25:5

• ### Formula  : Ip/Is = Ns/Np

• Ip = Primary Current
Is = Secondary Current
Np = Number of Primary Turns
Ns = Number of Secondary Turns
• Example: A 300:5 Current Transformer.
• The ratio of the current transformer can be modified by altering the number of secondary turns by forward or back winding the secondary lead through the window of the current transformer.
• By adding secondary turns, the same primary current will result in a decrease in secondary output. By subtracting secondary turns, the same primary current will result in greater secondary output.
• Again using the 300:5 example adding five secondary turns will require 325 amps on the primary to maintain the 5 amp secondary output or:  325 p / 5s = 65s / 1p
• Deducting 5 secondary turns will only require 275 amps on the primary to maintain the 5 amp secondary output or: 275p / 5s = 55s / 1p
• The above ratio modifications are achieved in the following manner:

CURRENT TRANSFORMERS RATIO MODIFICATION

 CT RATIO NUMBER OF PRIMARY TURNS MODIFIED RATIO 100:5A 2 50:5A 200:5A 2 100:5A 300:5A 2 150:5A 100:5A 3 33.3:5A 200:5A 3 66.6:5A 300:5A 3 100:5A 100:5A 4 25:5A 200:5A 4 50:5A 300:5A 4 75:5A
• A primary turn is the number of times the primary conductor passes through the CT’s window. The main advantage of this ratio modification is you maintain the accuracy and burden capabilities of the higher ratio. The higher the primary rating the better the accuracy and burden rating.
• You can make smaller ratio modification adjustments by using additive or subtractive secondary turns.
•  For example, if you have a CT with a ratio of 100:5A. By adding one additive secondary turn the ratio modification is 105:5A, by adding on subtractive secondary turn the ratio modification is 95:5A.
• Subtractive secondary turns are achieved by placing the “X1″ lead through the window from the H1 side and out the H2 side. Additive secondary turns are achieved by placing the “X1″ lead through the window from the H2 and out the H1 side.
• So, when there is only one primary turn each secondary turn modifies the primary rating by 5 amperes. If there is more than one primary turn each secondary turn value is changed (i.e. 5A divided by 2 primary turns = 2.5A).
•  The following table illustrates the effects of different combinations of primary and secondary turns:
 CT RATIO 100:5A PRIMARY TURNS SECONDARY TURNS RATIO ADJUSTMENT 1 -0- 100:5A 1 1+ 105:5A 1 1- 95:5A 2 -0- 50:5A 2 1+ 52.5:5A 2 2- 45.0:5A 3 -0- 33.3:5A 3 1+ 34.97:5A 3 1- 31.63:5A

## Ratio Correction Factor Curve of CT:

• The term correction factor is defined as that factor by which the marked (or nameplate) ratio of a current transformer must be multiplied to obtain the true ratio.
• The ratio errors of current transformers used for relaying are such that, for a given magnitude of primary current, the secondary current is less than the marked ratio would indicate; hence, the ratio-correction factor is greater than 1.0.
• A ratio-correction-factor curve is a curve of the ratio-correction factor plotted against multiples of rated primary orSecondary current for a given constant burden.
• Such curves give the most accurate results because the only errors involved in their use are the slight differences in accuracy between CT having the same nameplate ratings, owing to manufacturer’s tolerances. Usually, a family of such curves is provided for different typical values of burden.
• To use ratio-correction-factor curves, one must calculate the CT burden for each value of secondary current for which he wants to know the CT accuracy. Owing to variation in burden with secondary current because of saturation, no single RCF curve will apply for all currents because these curves are plotted for constant burdens; instead, one must use the applicable curve, or interpolate between curves, for each different value of secondary current.
• In this way, one can calculate the primary currents for various assumed values of secondary current; or, for a given primary current, he can determine, by trial and error, what the secondary current will be.
• The difference between the actual burden power factor and the power factor for which the RCF curves are drawn may be neglected because the difference in CT error will be negligible. Ratio-correction-factor curves are drawn for burden power factors approximately like those usually encountered in relay applications, and hence there is usually not much discrepancy.
• Any application should be avoided where successful relay operation depends on such small margins in CT accuracy that differences in burden power factor would be of any consequence.
• Extrapolations should not be made beyond the secondary current or burden values for which the RCF curves are drawn, or else unreliable results will be obtained.
• Ratio-correction-factor curves are considered standard application data and are furnished by the manufacturers for all types of current transformers.

## Testing of CT

• A number of routine and type tests have to be conducted on CT s before they can meet the standards specified above. The tests can be classified as :
• Accuracy tests to determine whether the errors of the CT are within specified limits.
• Dielectric insulation tests such as power frequency withstand voltage test on primary and secondary windings for one minute, inter-turn insulation test at power frequency voltage, impulse tests with 1.2u/50 wave, and partial discharge tests (for voltage >=6.6kv) to determine whether the discharge is below the specified limits.
• Temperature rise tests.
• Short time current tests.
• Verification of terminal markings and polarity.

### Polarity Test:

• In situations where the secondary bushing identification is not available or when a transformer has been rewound, it may be necessary to determine the transformer polarity by test. The following procedure can be used.
• The H1 (left-hand) primary bushing and the left-hand secondary bushing are temporarily jumpered together and a test voltage is applied to the transformer primary. The resultant voltage is measured between the right-hand bushings.
• If the measured voltage is greater than the applied voltage, the transformer is Additive Polarity because the polarity is such that the secondary voltage is being added to the applied primary voltage. If, however, the measured voltage across the right-hand bushings is less than the applied primary voltage, the transformer is Subtractive Polarity.
• Note: For safety and to avoid the possibility of damaging the secondary insulation, the test voltage applied to the Primary should be at a reduced voltage and should not exceed the rated secondary voltage.
• In the example below, if the T.C is actually rated 480 – 120 volts, the transformer ratio is 4:1 (480 / 120 = 4).
• Applying a test voltage of 120 volts to the primary will result in a secondary voltage of 30 volts (120 / 4 = 30). If transformer is subtractive polarity, the voltmeter will read 90 volts (120 – 30 = 90). If the voltmeter reads 150 volts, the transformer is additive polarity (120 + 30 = 150).The red arrows indicate the relative magnitude and direction of the primary and secondary voltages.

### Ratio Test

•  The ratio is defined as the number of turns in the secondary as compared to the number of turns in the primary.
• Apply one volt per turn to the secondary of the CT under test. Raise voltage slowly while observing meters. When one volt per turn has been reached on the secondary voltmeter, one volt should appear on the primary meter.
•  If the CT saturates before one volt per turn is reached, apply a smaller voltage which is a convenient fraction of one volt per turn. (e.g. .5v per turn). The Primary voltmeter should read the chosen fraction of a volt.
•  If a multi-ratio CT is being tested, the selector switch can be placed in the “External Meter” position. The primary voltmeter can be used to read the voltage between taps on the secondary winding while a known voltage per turn is applied to the winding, either between taps or to the full winding.
• CAUTION: TO PROTECT AGAINST INSULATION FAILURE, DO NO EXCEED MORE THAN 1000 VOLTS ON ANY OF THE SECONDARY WINDINGS OF THE CT UNDER TEST.
• Leads should be connected to the test set EXT VOLTS binding posts only when the selector switch is in the EXT METER CONNECTION position.

### Saturation Test

•  IEEE defines saturation as “the point where the tangent is at 45􀀨 to the secondary exciting amperes.” (See Figures).
• With the test set secondary binding posts X1 and X2 connected to the CT secondary and the H1 and H2 binding posts connected to the CT primary, increase output observing the ammeter and secondary voltmeter. Increase voltage until a small increase in voltage causes a large increase in current. Most CT’s will saturate at 1 amp or less and 600 volts or less.
• Note: It may be necessary to plot a curve to detect the saturation point. See Figures and ANSI/IEEE C57.13 for illustrations of typical curves for Class C transformers.

## Advantages of using a CT having 1A Secondary

• The standard CT secondary current ratings are 1A & 5A,The selection is based on the lead burden used for connecting the CT to meters/Relays.5A CT can be used where Current Transformer & protective’s device are located within same Switchgear Panel.
• 1A CT is preferred if CT leads goes out of the Switchgear.
• For Example if CT is located in Switch Yard & CT leads have to be taken to relay panels located in control room which can be away.1A CT is preferred to reduce the load burden. For CT with very High lead length, CT with Secondary current rating of 0.5 Amp can be used.
• In large Generator Circuits, where primary rated current is of the order of few kilo-amperes only,5A CTs are used, 1A CTs are not preferred since the turns rations becomes very high & CT becomes unwieldy.