CT Design principles
The key parameter for the design of current transformers is the maximum flux density (Bm). The maximum allowable value of the maximum flux density is determined by the core material. The flux in the core starts saturating when this value is exceeded. In other words, the induced voltage (Es) in the secondary winding will not increase proportionately when the magnetising current (Ie) increases. This is the main factor which determines the accuracy of the current transformer.
With the core materials commonly used for CT manufacturing, the following values of maximum flux density (Bm) is used for the design:
10P CT – Accuracy limit induced voltage corresponds to Bm of 1.9 T
5P CT – Accuracy limit induced voltage corresponds to Bm of 1.6 T
PX CT – Knee point induced voltage corresponds to Bm of 1.4 T
The above design factors for flux density are interesting. From an accuracy point of view, a Class PX CT is equivalent to, say, a Class 5P CT—provided it is operated at 1.4 T instead of the designed value of 1.6 T. This can be achieved either by operating a given CT at a reduced burden or at a reduced primary fault current. In other words, we need to maintain the ‘effective accuracy limit factor’ at the ‘desired’ value of the maximum flux density.
Refer to ‘Schneider Electric Cahier Technique No 195’ for details of the conversion from one Class to another!
Knee point voltage (Ek)
We now know the difference between the design of Class PX and Class P CTs.
The obvious question is, what is the need to specify a Class PX CT when we can purchase a cheaper Class P CT and tailor it to our accuracy requirement?
Good question. However, the CTs used in differential protection need to have the same saturation characteristics. This cannot be assured for Class P current transformers. In the case of IEC Class P specifications, the manufacturer is not obliged to provide the CT characteristics.
Class PX facilitates specification of CT characteristics in a very clever way. Let us first understand how this is done. The CT characteristics for a Class PX CT is automatically specified by the specification of the ‘knee point voltage (Ek)’ and the corresponding magnetising current (Ik).
The definition of the knee point voltage as per AS 600044.1 is reproduced below:
“Rated knee point e.m.f. (Ek): That minimum sinusoidal e.m.f. (r.m.s.) at rated power frequency when applied to the secondary terminals of the transformer, all other terminals being open-circuited, which when increased by 10% causes the r.m.s. exciting current to increase by no more than 50%.”
Are you confused? You are not alone! There is a saying that, “a picture is worth a thousand words”. It is very true in this case. The picture shown in Figure 2 makes things clear. In Figure 2, the knee point voltage ‘Ek’ and the corresponding magnetising current ‘Ik’ are specified as a part of Class PX CT specification.
Figure 2. CT Knee point voltage
In Figure 2, the knee point voltage ‘Ek’ is 300 V and the corresponding magnetising current ‘Ik’ is 0.04 A. This corresponds to point ‘A’ on the CT characteristics. The characteristics is assumed to be linear between point ‘A’ and the origin. Let us now locate a point ‘B’ on the plot. This is done by increasing magnetising current by 50% of ‘Ik’, that is 1.5 x 0.04 = 0.06 A, and increasing the induced voltage increased by 10% of ‘Ek’, that is 1.1 x 300 = 330 V. As per the definition of knee point voltage, the specified CT must have an induced voltage of less than or equal to 330 V at Ik = 0.06 A.
Consider another point on the curve in Figure 2, say Es = 150 V and Ie = 0.02 A. If the magnetising current is increased from 0.02 A to 0.03 A (an increase of 50%), the corresponding voltage increases from 150 V to 235 V. This voltage is greater than 165 V (150 V + 10% of 150 V). Hence, Es = 150 V is below the knee point voltage.
In practice, the knee point voltage and the CT characteristics are established by a simple test. A variable voltage source (Variac) is connected to the secondary winding, and the primary winding is kept open. The voltage is increased starting from 0 V in steps. The value of voltage and the corresponding current are noted. When the voltage reaches the specified knee point voltage, the voltage is increased by 10% and the corresponding increase in current is noted. If the current increase is more than 50%, then the knee point has been reached.
The notations ‘Ek’ and ‘Ik’ are used in IEC / Australian standards. The corresponding notations for Class X CTs in the IEEE standards are ‘Vk’ and ‘Ik’. So, it is important not to get confused with these notations.
Very informative and clearly laid out. Thank you.
Thanks Alpana.
Regards,
Sesha
Hi
For clarity
The FIRST and FOREMOST criteria for diff prot is to make sure the relay will NOT OPERATE for a through fault even with one CT saturated.
That means you must get the the CT kneepoint Ek correct FIRST!
Assume CT2 is saturated.
Ek of CT1 must be at least sufficient to drive the secondary max fault current Ifmax through the saturated CT2
Ohm’s Law then says
Ek ≥ Is1max x (Rct1 + Rl1 + Rl2 + Rct2)
which simplifies to approximately
Ek ≥ 2 x Is1max x (Rct + Rl)
(you can see that the same equation applies for Ek of CT2 if CT1 is saturated)
Now we can consider the relay setting for that through fault condition where the relay MUST NOT operate!
Again, Ohm’s Law says if CT2 is saturated then the voltage at the relay is
Vs = Is1max x (Rl2 + Rct2)
So the effective relay setting Vr must be greater than Vs by a “margin”
If the relay is a voltage setting relay e.g. MFAC14, then the relay setting Vr must be greater than the calculated Vs.
No stabilising resistor is required as the relay is inherently a high impedance (MFAC14 operating current at any setting is ~20 mA).
If the relay is a current setting relay e.g. MCAG14 , then the relay setting is Id and we need the stabilising resistance.
The total minimum impedance is therefore
Rtot > Vs/Id
Rs = Rtot – Rrelay
If we assume the relay is zero impedance we can simplify to the correct version of Eq 8 as:
Rs > Vs/Id
It is just a Ohm’s Law mathematical coincidence that the Vr_minimum is half of the Ek_minimum
But it is not necessarily true that Vr_actual is half of Ek_actual
Hence Eq 8 is not truly correct that Rs is based on half the kneepoint voltage
The kneepoint voltage could be “5 times the minimum” which would result in a much larger than necessary Rs
That would mean the CT waveform for an internal fault is driven much harder and faster into saturation than necessary making it harder for the relay to operate.
It is really
Rs ≥ (Ek_min/2) / (Id)
I discuss this in my own technical reference site:
https://ideology.atlassian.net/l/c/6v3TpqKv
Hello Rod,
Thanks for your comments. I have updated the blog and included a note regarding the CT resistance.
Regards,
Sesha
I have to strongly disagree!
It is not feasible, and should not be done, to “specify Class P CTs for differential protection”
I agree that modern numerical relays provide some great features and settings that make things a lot easier.
Whilst the Ohm’s Law consideration of the CTs and settings for a Merz-Price Circulating Current High Impedance Differential do not apply in the same way to modern “low impedance relays” (i.e. relays with individual CT inputs not paralleled to the other CTs), the underlying principle of preventing unwanted diff relay operation for a through fault must still be applied.
That cannot be achieved by simply specifying P class CTs .
P class defines what you can connect to the CT in terms of total burden to achieve a certain accuracy at the ALF and rated burden.
PX class defines the construction of the CT to achieve a certain dynamic performance of a certain magnetisation curve and internal Rct.
You could have two CTs with the same P class specification but with VERY different Ek, Ie and Rct.
Consequently for the same through fault current connected to a very low burden such as a modern diff relay, at a particular “required” output current of the two CTs, the terminal voltage could be VERY different.
That means the Excitation current will be very different.
That means the real output current of each CT will be different … and that is not what we want fro a through fault as that means a false differential current calculation.
Yes, the relay settings MAY provide sufficient bias to compensate for differences of P class CTs.
But you have no idea what that difference is going to be when you specify P class CTs
You will ONLY know what the difference is when you test the actual P class CTs to determine their actual Ek, Ie and Rct
The real point of specifying PX is that we really need to know Vk, Ie and Rct anyway for diff applications and so we can ensure the dynamic performance based on the slope Vk/Ie is the same so that the “false differential current” is minimised
Hello Rod,
Thanks for your comments.
The aim of these blogs is to provide an in-depth exposure to CT fundamentals, theory and CT design. Based on the information provided, the protection engineer can make judicious decisions as per site requirements.
The motivation for including the Class P specifications for differential protection in this blog are as below:
1. The Schneider SEPAM manuals recommend the use of 5P20 Class P CTs for their microprocessor based differential protection. I have included a link to SEPAM manual information in this blog.
2. I had the personal experience of tailoring the Class P CT for differential protection. The site had specified and installed a 22kV Class P CT erroneously for transformer differential protection. This was discovered at the time of commissioning the project. Replacing the CT was not an option. I had to research the CT design details to tailor the Class P CT on the site for differential protection. Please refer to my blog on ‘CT Specifications – Gold Report’ for details.
Regards,
Sesha
Very informative
Rodney Huges response adds the value
Very detailed information, like the way it has been presented.
Easy to follow through with calculations.
Give a good insight on the IEEE & IEC CT sizing. Thank you.
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