20 times rated secondary current for CT? 1

[sabap]

Why is the CT error specified in terms of 20 times rated sec current? What is the signifance of this number? Can be it be derived performing calculations on basic CT model?

Thanks in advance

 

[Skogsgurra]

If it is a protection CT (it most likely is) then the fault relays need correct measurement of current even in fault conditions. The deciscion which breaker to operate is done from impedance measurements and the impedance is calculated as U/I so a safe switch-off of the right branch needs that the CTs operate also under severe overcurrent. Such as during a short circuit fault.

 

[RalphChristie]

The composite error is defined as the rms value of the difference between the ideal secondary current and the actual secondary current, it includes current and phase errors and the effect of harmonics in the exciting current, usually expressed as a percentage of the primary current. This effect (error) is most noticable in the region approaching saturation of the CT core. (CT-performance is linear - or almost linear - up to the region before saturation. )
A protection CT is designed to maintain accuracy within specified limits of error up to a certain primary current. This current divided by the rated current is termed the accuracy limit factor. (ALF) Standard ALF-values are 5, 10, 15, 20 and 30.

In your case the ALF-value is 20. Thus, at 20 times rated current, you'll have a certain composite error, normally 5% or 10%
During a short-circuit is the fault current many times the normal rated current. Adequate protection depend on the effectiveness of the CT during this fault period.

See also standard BS 3938:1973


Regards

Ralph

 

[RalphChristie]

Also, the ALF-value is required when a CT is specified


Regards

Ralph

 

[scottf]

Ralph...your post is right on for BS/IEC/AS, etc... standards. In the IEEE/CSA world, all protection CTs have an accuracy limiting factor of 20. I.e. C400, C800. They specify that the CTs maintain a 10% accuracy to to 20 times rated current at a given burden. Instead of a burden listed in VA or ohms directly, they list the accuracy limiting voltage, i.e. 200, 400, 800 V. This can be transformed to a burden, i.e. 800V is the same as an 8 ohm burden, assuming a 5A rated secondary. 5A x 20 x 8 ohms = 800 V.