Search This Blog

22 October 2021

Instrument Security Factor (ISF) of Current Transformer

For metering core, ISF is for protecting metering instruments like energy meters, transducers etc. These instruments are meant for measuring electrical quantities with high accuracy under normal operation. Measurement under fault condition is now required and insignificant due to short duration of faults. Metering instruments are designed to carry current upto a certain level. 

For example, an energy meter used with CT of 1000/1A ratio carries maximum 1A under full load. Therefore its input current carrying paths needs to be designed for current upto 150 to 200% of rated current (say 2A continuous) . For fault conditions, which will prevail for short duration, it should withstand upto 5A or 10A. But in case of severe fault of 40kA or 50kA, secondary current may go upto 40A or 50A. If meter is to be designed for such heavy currents its size and connections will be bulky. Its cost will go up and accuracy will go down. Therefore, CT core is designed to saturate at such heavy currents and maximum secondary current is specified as ISF.

ISF is ratio of rated instrument limit primary current to rated primary current. If ISF is specified as 5 for a 1000/1A CT, then CT metering core will saturate at 5000A primary and current in secondary will not go above 5A. Sometimes Auxiliary reactors are used to achieve ISF limit.

Equivalent circuit of CT:

IP = Primary current

IS = Secondary current

VS = Secondary voltage

ZE = Exciting impedance

IE = Exciting current

RS = Secondary resistance

XL = Leakage reactance

ZB = Burden impedance 

Effect of burden on ISF: ISF is specified for a certain burden. If actual burden is different from rated burden, ISF will change as below:

ISF at actual burden = ISF at rated burden x (Rated burden / Actual burden)

* Internal secondary winding resistance of CT to be added for calculating burden.

For example, for a CT with rated burden 20VA and ISF 5, if actual burden is 10VA, ISF will be =5 x (20/10) =10. I.e. maximum secondary current will be 10A.