Is it possible to modify the ratio of a current transformer (CT)?

Product Line
3090SCCT

Environment
CT Ratio

Resolution
*Warning: Installation and maintenance of this device should only be performed by qualified, competent personnel that have appropriate training and experience with high voltage and current devices. Failure to follow the instructions / warnings in the Meter Installation Guide can result in permanent damage to the meter, severe injury or death.

Note: This is not a suggested solution. It is recommended that the appropriate CTs be purchased for each application.

The ratio assigned to a current transformer is based on the assumption that the primary conductor will only pass through the current transformer opening once. It is possible to reduce the rating by looping the primary conductor through the opening multiple times. The rating will be reduced by the times the primary conductors is looped through the CT. For example, a CT with the rating of 400:5A will be changed to 100:5A if four loops or turns are created by passing the primary conductor through the CT four times. See illustration and table below.

The ratio of the current transformer can also be modified by changing the number of secondary turns by either passing the secondary lead forward or backward through the current transformer opening.

Ip / Is = Ns / Np

Ip = primary amperage
Is = secondary amperage
Ns = number of secondary turns
Np = number of primary turns

400p / 5s = 80s / 1p

Note: One turn is dropped from the secondary as a ratio correction factor.

By adding secondary turns, while keeping the primary amperage the same will result in greater secondary output. For example, using the 400:5A current transformer, by adding 5 secondary turns it will now require 425 amps on the primary to maintain the 5 amp secondary output.

400p / 5s = 80s / 1p
Xp / 5s = 85s / 1p , where Xp equals the primary amperage.
Solving for Xp gives a requirement of 425 amps on the primary.

Similarly, subtracting 5 secondary turns will now require 375 amps on the primary to maintain the 5 amp secondary output.

400p / 5s = 80s / 1p
Xp / 5s = 75s / 1p , where Xp equals the primary amperage.
Solving for Xp gives a requirement of 375 amps on the primary.
Original Source: scribd/doc/16962202/Transformer-Turn-Ratio(no longer available)