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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.
Laplace Transform
(https://myelectrical.com/notes/entryid/173/laplacetransform)
Laplace transforms and their inverse are a mathematical
technique which allows us to solve differential equations, by
primarily using algebraic methods...
(http://myelectrical.com/Portals/0/SunBlogNuke/2/Windows-Live-Writer/How-to-Size-CurrentTransformers_E50F/ABBCT_2.jpg)
Robots - Interesting Videos
ABB Current
(https://myelectrical.com/notes/entryid/41/robot-folds-
Transformer
towels)
Example of a CT specification: - a very common specification for a protection class CT would be an accuracy
class 5P (1%), with rated accuracy limiting factors of 10 or 20. Typical burdens would be 5, 10, 15 or 20 VA. A
The robot folding towels post below was interesting enough
at the time to post a link. Recently I’ve come across a
couple of other interesting videos...
typical specification would be 5P10 15 VA.
Bows and Arrows
(https://myelectrical.com/notes/entryid/42/bows-andContents [hide]
arrows)
1. The IEC 60044 Method
It starts with me reading one of the Horrible History books
1. What the Manufacturer Wants
with my son (Groovy Greeks). Arrows were mentioned which
2. Connection Leads
lead to the discussion of the bodkin...
2. Other CT Sizing Methods and Requirements
1. BS 3938 and BS 7626
Understanding LV Circuit Breaker Fault Ratings
2. ANSI/IEEE C57.13
(https://myelectrical.com/notes/entryid/101/understandinglv-circuit-breaker-fault-ratings)
I think this post is going to be helpful to several of our
readers. While the IEC low voltage circuit breaker Standard
The IEC 60044 Method
[IEC 60947-2, Low voltage switchgear...
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:
- 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]
would be the CT accuracy class and the ‘P’ signifies a protection class CT.
Slightly more complicated is the effective factor,
K’scc.
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/181/operationalamplifier)
The fundamental component of any analogue computer is
Kssc - rated symmetrical short-circuit current factor
K’ssc
Operational Amplifier
the operational amplifier, or op amp. An operational
amplifier (often called an op-amp,) is a high...
Lighting - Lamps
Example IEC 60044 Calculation
(https://myelectrical.com/notes/entryid/187/lightinglamps)
Consider a CT with the following
Lamps are the essential part of any luminaire. These are the
specification and protection
light generating components. Since the advent of electrical
requirements:
lighting in the middle of the...
CT: 600/1 5P20 15 VA, Rct = 4
Frame Leakage Protection
Ω
(https://myelectrical.com/notes/entryid/218/frame-
CT Leads: 6 mm2, 50 m long
leakage-protection)
- use R=2 ρ l /a to calculate =
0.0179 Ω/m
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While not as popular as it once was, frame leakage
TOOLSSiemens
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Relay:
7SJ45, Ktd = 1
protection does still have some use in some
Short circuit current, Iscc max =
circumstances. In essence frame leakage is an earth fault...
30 kA
Differential protection, the good old days
(https://myelectrical.com/notes/entryid/90/differential-
To find the lead resistance Rleads (two
protection-the-good-old-days)
leads – supply, return) we can use the
This morning I was explaining how differential protection
standard formulae for resistivity:
works to a junior engineer. To give him something to read I
opened up the NPAG (Network Protection...
Rleads = 2 ρ l /a = 2 x 0.0175 x 50 / 6 =
0.3 Ω
Why a Sine Wave?
(https://myelectrical.com/notes/entryid/36/why-a-sine-
Numerical relays have low burdens,
wave)
typically 0.1 Ω (where possible the relay
I received this question by email a few weeks. First thoughts
manual should be consulted).
was that it is a product of the mathematics of rotating a
straight conductor in a magnetic...
Plugging everything into the equations:
How to Write an Electrical Note
Rb = 15 VA / 1 A2 = 15 Ω
(https://myelectrical.com/notes/entryid/149/how-to-writean-electrical-note)
R’b = Rleads + Rrelay = 0.3 + 0.1 = 0.4 Ω
Electrical notes are a collaborative collection of electrical
engineering information and educational material. Any
K’scc = Kscc (Rct + Rb)/(Rct + R’b)
registered user can add content. ...
(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
fault than
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
relay, Kcriteria.
the CT meets of
thethe
stability
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
Other CT Sizing Methods and
NOTES 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.
Try it free for 14 days
Keep translations in sync with
your code, save time and costs,
and scale globally fast.
Lokalise
Open
Knee voltage is defined as the point at which a 10% increase in voltage across the terminals, causes a 50%
increase in excitation current
Try it free for
14 days
Lokalise
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:
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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
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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)
(http://myelectricalengineering.co.uk)
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Thành Trần Thành
10 years ago
⚑
Hi Steven McFadyen
Thanks for your useful post, but it seems that how to size a neutral current transformer (for function 51G)
has not been included in your post, could you please add this?
34
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origin789
4 years ago
⚑
Hi Mr. Steven,
The formula, (rated secondary limiting emf), "Esl= Accuracy Limit Factor(CT rated burden in ohm+CT
secondary resistance in ohm) x CT secondary current" is correct or not?
and
CT required secondary emf=transient dimension factor x ALF x (CT secondary resistance in ohm + CT real
burden) x CT secondary current OK or not!
would you please answer!
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Rouhollah Emami
5 years ago
⚑
Hi Steven
can you share am link for a Siemens or GE-Schnider( micom series) protection relay which shows Ktd transient dimensioning factor ?
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Nilesh
5 years ago
⚑
For example , if we consider 300/5 A CT as per calculation for measuring purpose. so that means we
want to consider 100/5 A for each phase . Does it means any sense?
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Shivakumar Thayanithy
8 years ago
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is there a possibility to size ct's using calculation tools like ETAP?
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sureshkumar > Shivakumar Thayanithy
8 years ago
It is not possible in ETAP. We can do only manual calculation as per IEC and ANSI/IEEE.
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Steven McFadyen
8 years ago
> sureshkumar
A while ago, I also confirmed it is not possible. I thought about writing some external interface to the
ETAP database to do the calculations, but did not take it further. There is other software available,
CTDim from Siemens for example (although I have never used it).
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Jojo Alex
7 years ago
⚑
Can you provide the values for the above example with 5P10 CTs?
Also i have checked the IEC 60044-1 IEC standard, however the above calculation formulaes used above are
not mentioned in the IEC reference. Can you provide where is the calculation references.?
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Steven McFadyen
> Jojo Alex
5 years ago
You can Google it quiet easily.
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