How to Size Current Transformers

How to Size Current Transformers

One common method used internationally is IEC 60044 and this is what we will be concentrating

on.

The key to CT dimensioning is symmetrical short circuit current and transient dimensioning

factors:

• Kssc - rated symmetrical short

• K’ssc - effective symmetrical short

• 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),

curve will level off. A CT rated at say 5P20 will stay linear to approx

nominal current. This linear limit is the

would be the CT accuracy class and the ‘P’ signifies a protection class CT.

Slightly more complicated is the effective factor,

into account the burden (resistance) of the relay, resistance of the CT windings and resistance of

the leads:

• Rct - secondary winding d.c. resistance at specified temperature

• Rb - rated resistive burden of the relay

• R’b - Rleads + Rrelay;

CTs need to be able to supply the required current to drive the relays during transient fault

conditions. The ability of the CT and relay to operate under thes

How to Size Current Transformers



rated symmetrical short-circuit current factor

effective symmetrical short-circuit current factor

transient dimensioning factor

is relatively easy to understand and relates to the liner portion of a CT

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

A CT rated at say 5P20 will stay linear to approximately 20 times its

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, K’scc. This is a calculated value which takes


secondary winding d.c. resistance at specified temperature

rated resistive burden of the relay

this is thel connected burden


The ability of the CT and relay to operate under these conditions is a function of



is relatively easy to understand and relates to the liner portion of a CT

The voltage and current across a CT are linear only up till a certain value

after which the CT will saturate and the

imately 20 times its

As a reminder, the 5 [in the 5P20]


This is a calculated value which takes



e conditions is a function of

K’scc and the transient performance of the relay, K

manufacturer. Correct functioning is achieved by ensuring the following is valid:

• Issc max - maximum symmetrical short

• 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

Time for an example but before that, there is a slight complication in the manufacturers know

their relays better than we (or the IEC) do.

information – firstly it is the only way to get the factor K

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 re

K’scc.

Example Calculation

Now for the promised example:

• CT: 600/1 5P20 15 VA R

• CT Leads: 6 mm2, 50 m long (use R=2

• Relay: Siemens 7SJ45, K

• Iscc max = 30 kA

and the transient performance of the relay, Ktd. The factor, Ktd is supplied by the relay

Correct functioning is achieved by ensuring the following is valid:

maximum symmetrical short-circuit current

CT rated primary current

Once you have confirmed the above is ok, you know your CT is ok.

What the Manufacturer Wants


their relays better than we (or the IEC) do. You should always refer to the manufacturers

firstly it is the only way to get the factor Ktd. Secondly manufacturers sometimes


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

Example Calculation

Now for the promised example:

600/1 5P20 15 VA Rct = 4 Ω (the VA is the output rating of the CT)

, 50 m long (use R=2 ρ l /a to calculate = 0.0179 Ω

Relay: Siemens 7SJ45, Ktd = 1

is supplied by the relay

Correct functioning is achieved by ensuring the following is valid:

Once you have confirmed the above is ok, you know your CT is ok.


You should always refer to the manufacturers

ly manufacturers sometimes


pilot)and transformer differential are good to go with the above, while their line

lays require the above and have additional limitations on

(the VA is the output rating of the CT)

Ω/m)

To find the lead resistance Rleads (two leads – supply, return) we can use the standard formulae

for resistivity:

Rleads = 2 ρ l /a = 2 x 0.0175 x 50 / 6 = 0.3 Ω

Numerical relays have low burdens, typically 0.1 Ω (where possible the relay manual should be

consulted). Plugging everything into the equations:

Rb = 15 VA / 1 A2 = 15 Ω

R’b = Rleads + Rrelay = 0.3 + 0.1 = 0.4 Ω

K’scc = Kscc (Rct + Rb)/(Rct + R’

b) = 20 (4 + 15 )/ (4 + 0.4) = 86.4

Required K’ scc > 1 x 30000/600 = 50

In this case the effective K’scc of 86.4 is greater than the required K’

scc of 50 and the CT meets

the stability criteria.

How to Size Current Transformers

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