CT Magnetizing current 2

[RalphChristie]

Hi

After a discussion I've realized I had a lack of knowledge on CTs - any specifically on specifying the magnetizing current of a Class PTX (X) CT, thus an CT been used in the IEC-world.
In a class X CT the performance is defined in terms of a knee-point voltage and the magnetizing current at the knee point voltage or 1/2 or 1/4 the knee-point voltage. Normally the client should specify the knee-point volatage and the magnetizing current. To obtain the required knee-point voltage is not an issue, but how would you define the magnetizing current? Thus, from the clients perspective,(or out of the field perspective) how should you choose the value of the magnetizing current? Any rule of thumb or formula to be used? I do understand Eskom (South African Electrical Utility) specify it as 4mA/turn - but do not know where the value is been obtained from. Any ideas from the CT-manufacturers? Scottf?

Thanks
Rgards
Ralph

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[scottf]

Simply stated, the magnetizing current at the knee-point voltage is another way of defining the secondary current magnitude error at the knee-point voltage. Think of the CT as a black box. Primary current comes in and secondary current comes out as a function of the turns ratio. The error in the secondary current magnitude is a function the primary current lost to excite the core, i.e excitation current.

As an example...

If you have a total secondary burden of 4 ohm, a 1A rated secondary current, and you want to ensure you have a certain CT error up to 40 times over current. (for simplicity I'm leaving out the adjustment for X/R ratio of systems, etc...)

So minimum knee-point voltage would be 40 * 1A * 4ohm = 160V

If you want to ensure a 1% accuracy at that point, then you would need to limit the excitation current to 40 x 1A x 1% = 0.4A

From a CT design perspective, the maximum excitation current spec of a class X/PX CT is rarely the controlling design parameter. For most specs, when you design the core/coil to meet the minimum knee-point voltage, the resulting excitation current is almost always well below what is specified by the customer.

In practice, most utility's Class X / PX ratings have been developed over time with input from CT manufacturers for each ratio/rating.

 

[randyman]

As part of Class X spec is also the secondary winding resistance. And when determining the desired operating knee voltage the interal IR drop of the CT must also be considered. So to add to Scott's exapmle, Vk > 40 X (Rs + 4 Ohms). I agree with Scott that the exciting current is seldom a design issue. But the limiting Rs can be a different matter. I find it interseting that IEEE has a class X modeled after the IEC Class X, but since its inception in 2008 I have yet to see a Class X specified to IEEE standards. What about you Scott - have you seen it in IEEE world??

 

[scottf]

Randy

No I've never seen it in IEEE market. I doubt many customers know it's there.

Yes...that's why I wrote 'total secondary burden'...was trying to keep it simple

Most class x specs either include a max Rct or min Vkp as a function of Rct.

 

[davidbeach]

Ralph, no answer to your question - I'm sure scottf did a far better job than I could have (spec'ing C800 seems to resolve all CT issues for me), but it's good see you sign in here again. Hopefully it won't be another few years before your next post.

 

[RalphChristie]

Scott: Thanks a lot for your answer. From a field perspective I thought I knew enough, I've tested a lot of CT in my career and I've used a lot of CT information to determine settings for protective schemes (for instance REF-protection when mentioning Class X CTs) - but never really thought about the magnetizing current, although I knew you should specify it on a new CT. When a new PIT (professional in training) asked me how do you specify the magnetizing current - I realized I actually do not know. Thank you again.

Randy: As indicated above, I was actually more interested in specifying the magnetizing current. However, from a future reference perspective, I think it is a valid inclusion to the thread. Most CT-literature (IEC-world) will include the way to determine the knee-point voltage in their documentation, but I was not able to track reference to the determination of the magnetizing current in any literature. (literature available to me) I phoned a few CT-manufacturers in South Africa, no-one could actually answer my question, most indicated that they use these days software programs to design CTs, and that they do not know how to do it manually. Thanks for your comments.

David: You are flattering me. Reason for my absence is actually two new family-members - I was blessed in my mid-years with two boys, one two and a half years old and one four months old. Family time is keeping me away from the computer. I do however peek in from time to time, but more as a spectator. But I must admit - if I need clarification or information, this forum is the first one that I would visit. (if I do not find answers from literature that is available to me or from manufacturers that I do know.) Maybe I can call myself an ex-addict?

Regards
Ralph

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[ScottyUK]

Ralph,

Congratulations on the additions to the family. Nice to see you posting here but you have definitely got your priorities right!

 

[veritas]

Ralph,

The formula for Ie = H*L/N

Ie = excitation current
H = magnetic field intensity
L = mean core length
N = turns ratio

As mentioned it is impractical to use this formula in general as the CT dimensions need to be known. However, I do employ it sometimes especially when there are space constraints such as indoor switchgear. I then find out from the manufacturer what core material is most likely to be used, e.g. M-4. Knowing the swbd dimensions I can calculate the largest size CT that can be accommodated in order to determine L.

Also, for a Class X CT, the flux density, B, at the kneepoint voltage is typically 1.4Tesla. From the B-H curve I can then determine H.

Note these are all approximations but should put you in the right ballpark.

I would caution against using a rule of thumb such as 4mA/turn as the magnetising current is usually less with the larger number of turns - and as stated above is dependant on core dimensions and material as well.

I do perform a more detailed Ie calculation when dealing with high Z diff applications, e.g. REF. The maximum excitation current at the stability voltage may be determined by considering the required DP (Degree of Protection). For example I strive to cover at least 90% of a trfr winding with an REF application. Let's say full-load amps for the winding is 218A.

Thus minimum operating primary current is 10% of 218A = 21.8A.

Let's say relay operating current is 20mA, CTR = 250/1 and there are 4 CT's in question (3 phase and one neutral). Iop = 21.8A = CTR*(Irelay + 4*Iexc)to get,

Iexc = (1/4)*(21.8/250 - 0.02) = 16.8mA. Thus at the kneepoint voltage, Ie = 2 * 16.8mA = 33.6mA.

Should this not be attainable, then some of the parameters may need to be massaged.

However, when it comes to low impedance diff applications, I am much more generous and typically specify 150mA for turns ratios <= 800 and 100mA above 800. Just a rule of thumb I've used over the years and has never been a problem.

 

[veritas]

Sorry, I meant to say 200mA for turns ratios <= 800.