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How to Size Current Transformers

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How to Size Current Transformers
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How to Size Current Transformers
By Steven McFadyen (/user-profile/userid/3)on November 7th, 2011
The correct sizing of current transformers is required to ensure satisfactory operation of measuring instruments and
protection relays. Several methods exist to size current transformers. This note will look at several methods, with particular
attention being paid to protection class CTs sized in IEC 60044, which is adopted internationally.
110 or 230 Volts
(https://myelectrical.com/notes/entryid/22/110-or-230volts)
I've been considering a blog on the 110 or 230 Volt issue for
a while. While browsing the Internet I came across a great
summary by Borat over at engineering...
(http://myelectrical.com/Portals/0/SunBlogNuke/2/Windows-Live-Writer/How-to-Size-CurrentTransformers_E50F/ABBCT_2.jpg)
Introduction to Current Transformers
ABB Current
(https://myelectrical.com/notes/entryid/209/introduction-
Transformer
to-current-transformers)
Current transformers (CTs) are used to convert high level
Example of a CT specification: - a very common specification for a protection class CT would be an accuracy
currents to a smaller more reasonable level for use as
class 5P (1%), with rated accuracy limiting factors of 10 or 20. Typical burdens would be 5, 10, 15 or 20 VA. A
inputs to protection relays and metering...
typical specification would be 5P10 15 VA.
How Electrical Circuits Work
(https://myelectrical.com/notes/entryid/139/howContents [hide]
electrical-circuits-work)
1. The IEC 60044 Method
If you have no idea how electrical circuits work, or what
1. What the Manufacturer Wants
people mean then they talk about volts and amps, hopefully
2. Connection Leads
I can shed a bit light. I’m intending...
2. Other CT Sizing Methods and Requirements
1. BS 3938 and BS 7626
Electrical Engineering
2. ANSI/IEEE C57.13
(https://myelectrical.com/notes/entryid/4/electricalengineering)
Electrical engineering is a field that covers a wide variety of
sub-fields, including electricity and electronics. It is a field
The IEC 60044 Method
that goes back to the...
IEC 60044 specifies the requirements for protection CTs (in addition to measuring CT's, VTs and electronic sensors).
The key to CT dimensioning under the standard, is the symmetrical short circuit current and transient dimensioning factors:
Kssc
- rated symmetrical short-circuit current factor
K’ssc
- effective symmetrical short-circuit current factor
Ktd
- transient dimensioning factor
The factor Kssc is relatively easy to understand and relates to the liner portion
of a CT characteristic. The voltage and current across a CT are linear only
up till a certain value (normally specified as a multiple of the nominal rating),
after which the CT will saturate and the curve will level off. A CT rated at say
5P20 will stay linear to approximately 20 times its nominal current. This
linear limit is the Kssc (i.e. Kssc = 20). As a reminder, the 5 [in the 5P20]
Slightly more complicated is the effective factor,
(https://myelectrical.com/notes/entryid/219/introductionto-cathodic-protection)
If two dissimilar metals are touching and an external
conducting path exists, corrosion of one the metals can take
place. Moisture or other materials...
Resistors
Example IEC 60044 Calculation
(https://myelectrical.com/notes/entryid/195/resistors)
Resistors are electronic components that oppose the flow of
Consider a CT with the following
current. Manufactured in various types and ranges they
specification and protection
have a wide application to electronics...
requirements:
would be the CT accuracy class and the ‘P’ signifies a protection class CT.
K’scc.
Introduction to Cathodic Protection
This is a calculated
value which takes into account the burden (resistance) of the relay,
resistance of the CT windings and resistance of the leads:
https://myelectrical.com/notes/entryid/99/how-to-size-current-transformers
How to Calculate Motor Starting Time
CT: 600/1 5P20 15 VA, Rct = 4
(https://myelectrical.com/notes/entryid/107/how-to-
Ω
calculate-motor-starting-time)
CT Leads: 6 mm2, 50 m long
- use R=2 ρ l /a to calculate =
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1/25/23, 2:23 PM
How to Size Current Transformers
0.0179 Ω/m
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Request to look at induction motor starting time have come
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Relay:
Siemens 7SJ45, Ktd = 1
up a few times on the site. Hopefully in this post, I give you
Short circuit current, Iscc max =
guys some idea on how to calculate...
30 kA
IEC Reference Designations
(https://myelectrical.com/notes/entryid/24/iec-reference-
To find the lead resistance Rleads (two
designations)
leads – supply, return) we can use the
The IEC publishes a series of documents and rules
standard formulae for resistivity:
governing the preparation of documents, drawings and the
referencing of equipment. Depending on country...
Rleads = 2 ρ l /a = 2 x 0.0175 x 50 / 6 =
0.3 Ω
Surface Treatment – Ladders, Trays and Baskets
(https://myelectrical.com/notes/entryid/186/surface-
Numerical relays have low burdens,
treatment-ladders-trays-and-baskets)
typically 0.1 Ω (where possible the relay
Steel ladders, trays and baskets form the backbone of cable
manual should be consulted).
containment systems. Often these items need some form of
surface treatment to prevent corrosion...
Plugging everything into the equations:
Microsoft OneNote
Rb = 15 VA / 1 A2 = 15 Ω
(https://myelectrical.com/notes/entryid/32/microsoftonenote)
R’b = Rleads + Rrelay = 0.3 + 0.1 = 0.4 Ω
A couple of months ago I came Microsoft's OneNote and
downloaded the 60 day free trail. Since then I have been
K’scc = Kscc (Rct + Rb)/(Rct + R’b)
using it regularly and now have a full license...
(http://myelectrical.com/Portals/0/SunBlogNuke/2/WindowsLiveWriter/HowtoSizeCurrentTransformers_FC74/7345944727c6cb5e8f2423756b28f0bb_2.png)
= 20 (4 + 15 )/ (4 + 0.4) = 86.4
Rct - secondary winding d.c. resistance at specified temperature
Required K’scc > 1 x 30000/600 = 50
Rb - rated resistive burden of the relay
R’b - Rleads + Rrelay; this is thel connected burden
In this case the effective K’scc of 86.4 is
CTs need to be able to supply the required current to drive the relays during transient
faultthan
conditions.
The ability
theand
CT
greater
the required
K’scc ofof50
and relay to operate under these conditions is a function of K’scc and the transient performance
thestability
relay, Kcriteria.
the CT meetsofthe
td. The factor,
Ktd is supplied by the relay manufacturer. Correct functioning is achieved by ensuring the following is valid:
(http://myelectrical.com/Portals/0/SunBlogNuke/2/WindowsLiveWriter/HowtoSizeCurrentTransformers_FC74/c176f53a1ee1cd44331dd80e552366ff_2.png)
Issc max - maximum symmetrical short-circuit current
Ipn - CT rated primary current
That it. Once you have confirmed the above is ok, you know your CT is ok.
What the Manufacturer Wants
There is a slight complication in the manufacturers know their relays better than we (or the IEC) do. As general advice, you
should always refer to the manufacturers information:
firstly it is the only way to get the factor Ktd
secondly manufacturers sometimes have additional requirements; for example Siemens’ overcurrent, motor
protection, line differential (non-pilot)and transformer differential are good to go with the above, while their line
differential (pilot wire) and distance relays require the above and have additional limitations on K’scc
Connection Leads
In the sizing of protection transformers, the resistance (burden) of the connection leads can have a considerable effect. In
calculations, the resistance of the connection leads can be estimated from:
where:
l is the connection lead length in m
ρ is the resistivity in Ω mm2 m-1 (=0.0179 for copper)
A is the cross sectional area in mm2
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How to Size Current Transformers
Other CT Sizing Methods and
Requirements
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BS 3938 and BS 7626
BS 3938 and BS 7626 are older British Standards which deal the the specification and sizing of current transformers. Both
of these have been withdrawn and are superseded by the IEC 6044 standard.
The standards adopted the concept of knee voltage and it is still common to find knee voltage voltage being used as a CT
sizing parameter.
Knee voltage is defined as the point at which a 10% increase in voltage across the terminals, causes a 50%
increase in excitation current
Utilising the British Standards, CTs were defined by the knee point voltage UKN and the internal secondary resistance Ri. To
convert an IEC design the following can be used:
where: I2N is the nominal secondary current
ANSI/IEEE C57.13
The IEEE standard C57.13 covers the requirements for CT sizing in the North American markets.
Class C of the standard defines CTs by their secondary terminal voltage at 20 times nominal current (for which the ratio error
shall not exceed 10%). Standard classes are C100, C200, C400 and C800 for 5 A nominal secondary current.
This terminal voltage can be calculated from the IEC data as follows:
with
and
If anyone has any questions, comments or suggestions on how to improve the post, please add them below.
Current Transformers (https://myelectrical.com/notes?tag=current+transformers), Equipment Ratings
(https://myelectrical.com/notes?tag=equipment+ratings), Protection (https://myelectrical.com/notes/tag/protection)
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Steven McFadyen (/user-profile/userid/3)
Steven has over twenty five years experience working on some of the largest construction projects.
He has a deep technical understanding of electrical engineering and is keen to share this
knowledge.
About the author(/user-profile/userid/3)
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