# How to Size Current Transformers

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.

ABB Current

Transformer

Example of a CT specification: - avery 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 typical specification would be 5P10 15 VA.

## The IEC 60044 Method

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:

- K
_{ssc}- rated symmetrical short-circuit current factor - K
^{’}_{ssc}- effective symmetrical short-circuit current factor - K
_{td}- transient dimensioning factor

**Example IEC 60044 Calculation**

Consider a CT with the following specification and protection requirements:

- CT: 600/1 5P20 15 VA, R
_{ct}= 4 Ω - CT Leads: 6 mm
^{2}, 50 m long

- use R=2 ρ l /a to calculate = 0.0179 Ω/m - Relay: Siemens 7SJ45, K
_{td}= 1 - Short circuit current, I
_{scc max}= 30 kA

To find the lead resistance R_{leads} (two leads – supply, return) we can use the standard formulae for resistivity:

R_{leads} = 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:

R_{b} = 15 VA / 1 A^{2} = 15 Ω

R^{’}_{b} = R_{leads} + R_{relay} = 0.3 + 0.1 = 0.4 Ω

K^{’}_{scc} = K_{scc} (R_{ct} + R_{b})/(R_{ct} + 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.

*K*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 K

_{ssc}_{ssc}(i.e. K

_{ssc}= 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:

- R
_{ct}- secondary winding d.c. resistance at specified temperature - R
_{b}- rated resistive burden of the relay - R
^{’}_{b}- Rleads + Rrelay; this is thel connected burden

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 these conditions is a function of K^{’}_{scc} and the transient performance of the relay, K_{td}. The factor, K_{td} is supplied by the relay manufacturer. Correct functioning is achieved by ensuring the following is valid:

- I
_{ssc max}- maximum symmetrical short-circuit current -
I

_{pn}- 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 K
_{td} - 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:

is the connection lead length in mlis the resistivity in Ω mmρ^{2}m^{-1}(=0.0179 for copper)is the cross sectional area in mmA^{2}

## Other CT Sizing Methods and Requirements

### 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 voltageis 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 *U _{KN}* and the internal secondary resistance R

_{i}. To convert an IEC design the following can be used:

where:

Iis the nominal secondary current_{2N}

### 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.

Dear Steven,

It is a very informative technical article you shared here. Thanks for it. Could you please provide the reference clauses for the equations used for the calculations in IEC-60044. I could not find it in 60044.

Thanks,

Vinayan