CT Excitation and Knee Point

[BigJohn1]

I'm getting into CT testing, and am having trouble understanding the value of a knee-point excitation test.

The majority of the tests I've done are on 600V CTs between 500-5000A. The knee point values are usually in the mA range at a couple of hundred volts.

My understanding is the ANSI knee is the stable 45 degree slope just before the saturation of the CT core, correct? And this saturation point is the point at which an increase in the primary current would not produce a reliable increase in secondary current, correct?

So, if in actual operation, the saturation point of the secondary is often tens of amps, at very low voltage, how does my knee-point test result at high voltage and low current relate to this?

Am I to simply take it on face-value that it should match the manufacturers published excitation curve? Because any of the curves I'm finding have excitation values much higher than indicated by testing.

Thanks for any help.

 

[oldfieldguy]

How are you testing this?

I use a variable transformer and a voltmeter and an ammeter (or a CT test set with these features), injecting voltage into the secondary winding and monitoring current. You can plot these values, but for example, a CT with a C800 accuracy class can be expected to have an excitation voltage above 800 volts.

As you step the voltage up from zero, you will see small increments of current increase. As you approach saturation voltage, corresponding small voltage increases will result in larger current increases. When you reach saturation, a small voltage increase will result in a large current increase.

If you plot these on log-log paper, you can take a 45-degree line and move it to a point tangent to the plotted curve. This is the 'knee' voltage.

I've tested hundreds of CT's in the 600-volt class and have only found a few which did not meet manufacturer specifications.

You can also do this test with varying resistances (burdens) in the secondary circuit, pushing primary current, but this requires large test equipment. I prefer secondary injection.

old field guy

 

[ScottyUK]

BigJohn,

Think of the CT equivalent circuit as a perfect source plus a resistance. When a fault occurs and the secondary current in tens of amps - perhaps hundreds of amps if the secondary is 5A nominal - the series resistance of the windings will drop a lot of voltage according to Ohm's Law. The voltage produced by the 'perfect' source and the terminal voltage are not the same value. When you do a mag curve there is (very close to) zero current flowing through the series resistance, and you measure the full internal source voltage.