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Home / Technical Articles / IEC and NEMA/IEEE ratings of current transformers (CTs) in medium voltage applications

Metering and protection purpose

First, let’s remind ourselves of the basics in a few sentences. That is something you must know. A current transformer (CT) is designed to produce a secondary current which is accurately proportional to the primary current. It consists of a single primary winding, which an external busbar or cable runs through, or it can have a single primary bar, brought out to two ends for termination.

IEC and NEMA ratings of current transformers (CTs) for a medium voltage switchgear
IEC and NEMA ratings of current transformers (CTs) for a medium voltage switchgear (photo credit: Energie Technik Becker GmbH)

A medium voltage current transformer can have up to three independent secondary winding sets. The entire current transformer assembly is encapsulated in resin, inside an insulated casing. Current transformers are used for metering or protection purposes.

The accuracy class and size depends on the individual application – for example, revenue metering would use high accuracy metering CTs.

Just to note, it’s very important to never leave the secondary winding of a CT open circuit. This creates extremely high voltages which pose a real danger to personnel.

Ok, let’s get on the IEC and later NEMA ratings of a current transformer. Some rating explanations have exercises and real examples, which I hope it will help for better understanding.

  1. IEC  ratings of current transformer
    1. Rated primary current
    2. Rated secondary current: Isr
    3. Transformer ratio: Kn
    4. Rated thermal short-time withstand current: Ith (kA)
    5. Overcurrent coefficient: Ksi
    6. Rated primary circuit voltage: Up (kV)
    7. Rated frequency
    8. Rated real output power (VA)
      1. Exercises
    9. Metering class CT
    10. Protection class CT
      1. Example
    11. Selection of current transformers
      1. Exercise to select appropriate CTs
        1. Exercise #1
        2. Exercise #2
        3. Exercise #3
  2. NEMA/IEEE ratings of current transformer
    1. Accuracy class
    2. Class rating
    3. Burden
    4. Examples

1. IEC Ratings

1.1 Rated primary current: Ipr (A)

The primary current rating of a CT must be greater than the expected maximum operating current it is monitoring.

Metering CT’s primary current rating should not exceed 1.5 times the maximum operating current. Protection CT’s primary current rating needs to be chosen so that the protection pick-up level is attained during a fault.

Standard values for Ipr are: 10, 12.5, 15, 20, 25, 30, 40, 50, 60, 75 A, and decimal multiples of these values (source: IEC 60044-1)

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1.2 Rated secondary current: Isr

The secondary current rating of a CT is either 1 A or 5 A. CTs with a 5 A secondary rating are becoming less common as more CT driven equipment becomes digital. For long secondary cable runs, CTs with 1 A secondary windings can minimize the transformer and secondary cable size.

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1.3 Transformer ratio: Kn

This is the ratio of secondary to primary winding turns: Kn = Ns/Np = Ipr/Isr

Current transformer nameplate
Figure 1 – Current transformer nameplate

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1.4 Rated thermal short-time withstand current: Ith (kA)

This is the highest level of rms primary fault current which the CT can endure, both thermally and dynamically, for 1 second without damage. When used in a medium voltage enclosure, the Ith rating should match the short-time withstand rating of the entire switchgear.

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1.5 Overcurrent coefficient: Ksi

This is the ratio of a CT’s short-time withstand current rating to its primary current rating:

Ksi = Ith/Ipr

This coefficient indicates how difficult it would be to manufacture a CT. A higher coefficient means a physically larger CT, which is more difficult to manufacture.

  • If Ksi < 100 it’s easy to manufacture
  • If Ksi 100 ~ 500 it’s difficult to manufacture, with certain limitations
  • If Ksi > 500 it’s  extremely difficult to manufacture

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1.6 Rated primary circuit voltage: Up (kV)

The primary circuit voltage rating indicates the level on insulation provided by the CT. If a ring type CT is installed around a cable or bushing, the insulation level can be provided by the cable or bushing.

Rated primary voltage
Upr (kV)
Suitable operation range
U (kV)
Power frequency withstand voltage
(kV) rms for 1 minute
Lightning impulse withstand voltage
(kV) peak, 1.2/50μs
7.233-7.22060
126-122875
17.510-17.53895
2412-2450125
3620-3670170

Source: IEC 62271-1


1.7 Rated frequency: fr (Hz)

This rating must match the system’s operating frequency. Standard frequencies are 50 Hz and 60 Hz. It’s very important to be cautios, because a 50 Hz CT can be used on a 60 Hz system, but a 60 Hz CT cannot be used on a 50 Hz system.

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1.8 Rated real output power (VA)

The maximum power a CT secondary can deliver, to guarantee its accuracy and performance. The total sum VA (including cable, connectors and load) must not exceed the rated real output power of the CT. Standard values are: 1, 2.5, 5, 10, 15 VA.

Cable burden can be calculated the following way: VAcable = k × L/S, where:

  • k = 0.44 for 5 A secondary, = 0.0176 for 1 A secondary
  • L = total feed/return length of cable (m)
  • S = cross sectional area of copper cable (mm2)

Metering instrument burden:

  • Metering instrument (digital) = 1 VA (approx.)
  • Metering instrument (electromagnetic or induction) = 3 VA (approx.)
  • Transducer (self powered) = 3 VA (approx.)

Protection instrument burden:

  • Protection instrument (digital) = 1 VA (approx.)
  • Protection instrument (electromagnetic overcurrent) = 3-10 VA (approx.)
Medium voltage current transformers
Figure 2 – Medium voltage current transformers

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1.8.1 Exercises

Exercise #1 – A CT with a 1 A secondary is connected to an electromagnetic ammeter located 10 m away, using 2.5 mm2 copper cable.

Calculate the minimum required VA rating of the CT.

  • VAcable = k × L/S = 0.0176 × 20/2.5 = 0.14 VA
  • VAammeter = 3 VA
  • VAtotal = 0.14 + 3 = 3.14 VA

The total burden is 3.14 VA. Use a 5 VA CT.


Exercise #2 – A CT with a 5 A secondary is connected to a digital protection relay located 2 m away, using 1.5 mm2 copper cable.

Calculate the minimum required VA rating of the CT.

  • VAcable = k × L/S = 0.44 × 4/1.5 = 1.17 VA
  • VAammeter = 1 VA
  • VAtotal = 1.17 + 1 = 2.17 VA

The total burden is 2.17 VA. Use a 2.5 VA CT.

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1.9 Metering class

A metering class indicates the accuracy of the CT secondary current at 5 to 125% of rated primary current. Above this level, the CT starts to saturate and the secondary current is clipped to protect the inputs of a connected metering instrument.

  • General metering CT would use a metering class CL 0.5 – 1.0
  • Revenue metering CT would use a metering class CL 0.2 – 0.5
Operating range for metering class current transformer
Figure 3 – Operating range for metering class current transformer

Where:

  1. Saturation
  2. Linear operating range, at accuracy class tolerance

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1.10 Protection class CT

A protection class CT provides a linear transformation of the primary to secondary current at high overload levels. This characteristic makes them suitable for use with overcurrent protection relays.

A relay trip setting is normally 10~15 times the maximum load current and this level should fall on the linear part of the CT secondary current curve. If a CT saturates before the relay trip level is reached, the fault will remain undetected, leading to equipment damage and serious danger to personnel.

The most commonly used protection class is a 5PX, where X is the accuracy limit factor (ALF) or multiplication factor of the rated primary current. The secondary current is +/-1% accurate at rated primary current and +/-5% accurate at X times rated primary current.

Typical protection class CT ratings are 5P10, 5P15, 5P20.

Operating range for protection class current transformer
Figure 4 – Operating range for protection class current transformer

Where:

  1. Saturation
  2. Linear operating range, at accuracy class tolerance
  3. Ideal protection setting trip zone 50%~100% ALF

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1.10.1 Example

A 200/1 A CT has a protection class rating of 5P15. The secondary current is guaranteed to be linear up to 15 times the rated primary current. The secondary current will be 1 A (+/-1%) at 200 A primary current and 15 A (+/-5%) at 3000 A primary current.

For guaranteed operation, any overcurrent trip setting should be between 7.5 ~ 15 A secondary current.

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1.11 Selection of current transformers

The main considerations for selecting a CT are the primary and secondary current ratio, real output power rating (VA) and accuracy class. Secondary selection considerations are rated primary voltage, frequency and thermal short-time withstand current.


1.11.1 Primary and secondary current ratio

Rated primary current: Ipr (A)

SourceRated primary current Ipr (A)
Incomer from transformerIpr ≥ 1.0-1.25 of nominal source current
Feeder to transformerIpr ≥ 1.0-1.25 of transformer’s rated primary current
Feeder to motorIpr ≥ 1.0-1.5 of motor full load current
Feeder to capacitor bankIpr ≥ 1.3-1.5 of nominal capacitor current

Rated secondary current: Isr (A)

  • Use 1 A and 5 A for local installation
  • Use 1 A for remote installation

1.11.2 Real output power (VA)

The real output rating of the CT must be the next highest nominal size above the expected total burden on the CT secondary. Total burden is the sum of output cable, connectors and instruments.


1.11.3 Class type

Use a metering class CT for metering and indication. A higher class CT gives greater accuracy between the primary and secondary currents.

Use a 5PX protection class CT for current based protection relay inputs. The ALF must be selected so that the relay trip point lies on the linear part of the secondary current curve, between 50% and 100% of the ALF.

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1.11.4 Exercise

Select appropriate CTs for the following transformer incomer and feeder circuits.

Example transformer incomer and feeder for selection of appropriate CTs
Figure 5 – Example transformer incomer and feeder for selection of appropriate CTs

Where:

1. Transformer Incomer:

  • MV/MV transformer (TXR1): 5 MVA, 36/11 kV, 10% Z
  • Instantaneous overcurrent trip setting = 15 × In for digital protection relay (OC1) driven off CT1-2
  • Electromagnetic ammeter (A) is driven off CT1-1

2.Transformer Feeder:

  • MV/LV transformer (TXR2): 2 MVA, 11/0.4 kV, 5% Z
  • Instantaneous overcurrent trip setting = 10 × In for digital protection relay (OC2) driven off CT2

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Exercise 1 – Metering CT1-1 for transformer incomer circuit:

Step 1 – Calculate transformer TXR1 nominal secondary current: In (A)

  • In = S/(√3 × U) = 5000/(√3 × 11) = 262 A
  • The secondary current for TXR1 is 262 A

Step 2 – Calculated max. expected short circuit current at CT1 installation: Isc (A)

  • Ignoring any power cable or busbar impedances:
  • Isc = In × 100/Z = 262 × 100/10 = 2620 A
  • The maximum expected short circuit current at CT1 is 2620 A

Step 3 – Select metering CT1-1 ratings:

  • Primary rated current: Ipr = (1.0-1.25) × In = (1.0-1.25) × 262 A
    Use a rating of 300 A

  • Secondary rated current: Isr
    Use a rating of 1 A

  • Short-time withstand rating: Ith ≥ Isc
    Use a rating of 10 kA

  • Primary circuit voltage: Up ≥ U
    Use a rating of 12 kV

  • Real output power: Typically > 3 VA for electromagnetic type meter
    Use 5 VA (this allows 2 VA for cable burden, etc.)

  • Accuracy Class
    Use Class 1.0 (common class for general metering)

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Exercise 2 –  Protection CT1-2 for transformer incomer circuit:

Step 1 – Select ratings common to both the metering and protection CTs

  • Primary/secondary rated current: Use 300/1 A
  • Short-time withstand rating [Ith]: Use 10 kA rating
  • Primary circuit voltage [Up]: Use 12 kV rating

Step 2 – Select real output power

  • Real output power: typically > 1VA for digital type protection relay
  • Use 2.5 VA (this allows 1.5 VA for cable burden, etc.)

Step 3 – Calculate protection class 5PX

  • The instantaneous trip current level of protection relay OC1 is set to 15 × In.
  • ITRIP = 15 × 262 = 3930 A (primary current)

Note: In most digital protection relays, the trip current levels are set with respect to the secondary current. In this case

  • ISEC = 3900/300 × 1 = 13.1 A
  • The instantaneous trip current level for the CT secondary is 13.1 A

The trip current level should fall between 100 to 50% of the accuracy limit factor (ALF). Using an ALF of 10 (5P10), the trip current level of 3930 A falls outside the range 100% to 50% ALF, so a 5P10 protection class CT is not suitable.

  • 100%(ALF) = 1.0 × 10 × 300 = 3000 A
  • 50%(ALF) = 0.5 × 10 × 300 = 1500 A

We may notice that 1500 ≤ 3930 ≥ 3000 A. Using an ALF of 15 (5P15), the trip current level of 3930 A falls within the range 100% to 50% ALF so a 5P15 protection class CT is suitable.

  • 100%(ALF) = 1.0 × 15 × 300 = 4500 A
  • 50%(ALF) = 0.5 × 15 × 300 = 2250 A

We may notice that 2250 ≤ 3930 ≤ 4500 A. Use protection class 5P15

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Exercise 3 –  Protection CT2 for transformer feeder circuit:

Step 1 – Calculate transformer TXR2 nominal primary current: In (A)

  • In = S/(√3 × U) = 2000/(√3 × 11) = 105 A
  • The primary current for TXR2 is 105 A

Step 2 – Calculated maximum expected short circuit current at CT2 installation: Isc (A)

  • Ignoring any power cable or busbar impedances
  • Isc = In × 100/Z = 105 × 100/5 = 2100 A
  • The maximum expected short circuit current at CT2 is 2100 A

Step 3 – Select protection CT2 ratings

  • Primary rated current Ipr = (1.0 – 1.25) × In = (1.0 – 1.25) × 105
    Use a rating of 150 A

  • Secondary rated current Isr
    Use a rating of 1 A

  • Short-time withstand rating, Ith ≥ Isc
    Use a rating of 10 kA

  • Primary circuit voltage Up ≥ U
    Use a ratings of 12 kV

  • Real output power: Typically > 1 VA for digital type protection relay.
    Use 2.5 VA (this allows 1.5 VA for cable burden, etc.)

Step 4 – Calculate protection class 5PX

  • The instantaneous trip current level of protection relay OC2 is set to 10 × In
  • ITRIP = 10 × 105 = 1050 A (primary current)

Note: In most digital protection relays, the trip current levels are set with respect to the secondary current. In this case

  • ISEC = 3900/300 × 1 = 13.1 A
  • The instantaneous trip current level for the CT secondary is 7 A

The trip current level should fall between 100 to 50% of the accuracy limit factor (ALF). Using an ALF of 10 (5P10), the trip current level of 1050 A falls within the range of 100% to 50% ALF so a 5P10 protection class CT is suitable.

  • 100%(ALF) = 1.0 × 10 × 150 = 1500 A
  • 50%(ALF) = 0.5 × 10 × 150 = 750 A
  • We may notice that 750 ≤ 1050 ≤ 1500 A
  • Use protection class 5P10

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2. NEMA/IEEE Ratings

These ratings are typically used for current transformers manufactured or used in North American installations. As well as a stated primary to secondary nominal current ratio, the device also carries an overall accuracy rating in the format.

AC-CR-BU

Where:

  • AC = accuracy class
  • CR = class rating
  • BU = maximum burden (ohms)

2.1 Accuracy class

Designates the accuracy of the secondary current with respect to the primary rated current. This accuracy is only guaranteed provided the maximum burden is not exceeded.

Accuracy classTolerance at 100% primary current
1.2±1.2 %
0.6±0.6 %
0.5±0.5 %
0.3±0.3 %

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2.2 Class rating

Designates the intended application of the device.

  • B = for metering applications
  • H = for protection applications. The CT secondary accuracy is guaranteed at 5 to 20 times the nominal primary rated current

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2.3 Burden

The maximum load allowed to be connected to the current transformer secondary, to guarantee the accuracy class. The maximum burden includes secondary cable/wire, connectors and the load.

The following table converts burden in ohms to VA, for a 5 A secondary.

Ω0.040.060.080.120.160.200.240.280.320.360.400.480.560.640.720.80
VA11.523456789101214161820

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tabela

2.4 Examples

0.5-B-0.1

This example indicates a current transformer with an accuracy of ±0.5%, and a maximum allowable secondary burden of 0.1 Ω (or 2.5 VA on a 5 A secondary CT). This is a metering class rated current transformer.


1.2-H-0.2

This example indicates a current transformer with an accuracy of ±1.2%, and a maximum allowable secondary burden of 0.2 Ω (or 5 VA on a 5 A secondary CT). This is a protection class rated current transformer.

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Sources:

  1. Medium Voltage Application Guide by Aucom
  2. Electric Power Substations Engineering By James C. Burke
  3. Selection of current transformers and wire sizing in substations – Sethuraman Ganesan; ABB Inc.

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author-pic

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.

5 Comments


  1. Dang Tuan Khanh
    Nov 04, 2020

    Can you explain for me this:

    Why do we use 1 A (Isr of CT) for remote installation instead of 5 A ?

    Thanks a lot !


  2. Geoffrey Okyere Broni
    May 02, 2020

    Primary rated current Ipr for CT2 = (1.0 – 1.25) × In = (1.0 – 1.25) × 105
    This gives a value between 105A and 131.25A. Kindly help me understand why you used 150A for Ipr


    • Karthikeyan
      Nov 03, 2020

      Selected next standard size for CT


  3. Sandeep Sharma
    Jan 22, 2020

    Really practical information, really like
    Please mail wright up on MV/LV Switchgears Panel Designing.


  4. Md shariful islam
    Jan 20, 2020

    I’m md shariful islam I’m working qatar MHM- al muftah company

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