SANKAR NATHSANKAR NATH S.TECHNICAL ADVISOR at AL HASOUN AL ARABIAUpto Knee point, Secondary Current is proportional to Primary Current as per the Current Transformer Rating. To express in simple way, above Knee point, Secondary current remains more or less same irrespective of rise of Primary Current which will affect operation of Protective Relay. Saturation level of Current Transformer is selected as per the Protection system it is meant for. ShahbazShahbaz H.International Applications Engineer; Meggerand in saturation region your protection will not work; you should go through CT saturation studies. virgiliovirgilio R.Principal ConsultantThe knee-point voltage of a current transformer is critical in protective relay application because fault currents usually range from 20 to 30 times the rated currents of the current transformers. When the knee-point voltage of a current transformer is reached, the linearrelationship between the primary and secondary currents will no longer apply. Knee-point is defined as the voltage at which a 10% increase in applied voltage, will increase the magnetizing current by 50%. These could pose some problems in your relay settings and coordination. ZekiZeki A.R&D Manager @ ALCE Elektrik San. ve Tic. A.S.As the esteemed members have explained here is some more in depth info. For protection purposes there are 5P and 10P transformers according to IEC 61869-1 and IEC 60044-1. However the error values for 5P and 10P are more. These transformer secondaries almost operate above the saturation (above Vkp point). For more linear information (reflected to the secondary), that is for better error values, secondaries are designed with special Vkp, magnetizing current (Im) and Rct values.

The protection scheme engineers usually prefers to use such secondaries for differential protection of power transformers, generators and/or lines as they are more accurate. When such transformers are ordered, they are ordered inpairs (for each phase) and their Vkp values are chosen as close as possible. AntonioAntonio B.Director at ABS Representaciones 3533Lets state in a practical point of view.In a S/S with a 30 MVA 132/24 KV Transformer & 6 feeders 24 KV have the folowing details:Rated current for secondary main circuit of this transformer at 24 KV is 722 Amps, so 800/5 TC was selected. With 5P20 protection class CT, it means that linearity is granted up to 800 x 20= 16 KAmpsFeeders on the other hand are rated for 8 MVA: 192 Amps, 250/5 5 P20 was selected, Linear section is up to 250 x 20 = 5 KA.So see the difference?: if you had a 12 KA fault, feeder TC is saturated because 12 KA > 5 KA, providing false information about actual fault current, whereas Transformers main circuit CT is measuring properly because 12 KA 83.5 V it will not saturate so all OK

the other way to look at this is from the specification itselfwhen supplying max fault current into the connected burden, the VA isVAterminals = Vctterminals x Ifsecondary = 20.65 x 12.5 = 258.125 VA

5P20 15VA means it will handle up to 20 x 15 = 300 VA

so to your first question - can you use a CT with Vkp = 400 V?Depends! What is the Rct of that CT? you need to make sure that:Vkp > Vctterminals + (Ifsecondary x Rct)400 > 20.65 + (12.5 x Rct)i.e. Rct < (400 - 20.65) / 12.5 i.e. Rct < 30 OhmsI think it is fair to say that most CTs with Vkp = 400 V is most likely to be OK!!

so to second question - is it better or worse?Higher Vkp means more iron in the cross section of the CT core so its more expensive.So using a CT with higher than necessary Vkp capability works fine but costs you more than necessary.

third question - is it easier to make a CT with lower Vkp? Yes.5P20 15VA means that it will handle IbrahimIbrahim S.Testing & commissioning Project Manager at alfanar Co.Really i have enjoyed with that great discussion. but what i think is thatthe time is up for the OLD types CTs. Now we have to think for Optical CTas a much better solution. YogeshYogesh K.Executive Manager, Transmission and Distribution, Dar EngineeringI agree with Ibrahim. I liked all above discussion. But it will be more appropriate to start discussion about optical CTs and how it overcomes saturation problems and all other problems of conventional CTs. mohamedmohamed M.electrical engineer at NLSupervision Company A/SI think Regwaskey coil CT is overcome the knee point problem of old CT and its error is less than old CT FrancesFrances T.MV Engineering Manager at Plummers Industries Pty LtdEverything will be slow moving transition from Old type of Ct going to Optical CT due to some factors in terms market availability and cost between the two specially for small switchboard manufacturing, electrical suppliers which is critical for them for the total cost of their finished product, especially If they are participating in any bidding and still they can maintain their compititve cost, the old CT type was tested and proven in all ages and in very harsh environment that we have in mining industries, and in other electrical industries in remote areas. Thats how huge the gap that needs to be over came by the optical CT to prove that it can do much more than what the conventional ct can do. When it comes to design, it always demand flexibility and reliability in overall application.

For me ot will take awhile for the said transition from conventional to optical CT. PatrickPatrick R.Sr Electrical Engineer at Infigen EnergyGreat discussion no marketing and sound engineering applications.My comment on optical ct's, they are great invention with many applications however it will take time before their presence in the field increases. Meanwhile we have a lot of ct's installed and we need to continue to teach those in our field about them and understand how they work. krishnakrishna R.HEAD-ELECTRICAL DESIGN at Gracetec GroupI enjoyed the reading very much and thanks for all contributors.esp for antonio for verbal clarity.Simply superb.Now about the lead burden.In low voltage 3P&N systems, apart from phase CTs, we have two neutral CTs, instead of one, one NCT with lead burden same as those for phases and another NCT with longer control cable length. With this arrangement how to fine tune the knee point voltage requirement. Will the average lead burden will do?Comments please.With regards,V.Krishna rao RodneyRodney H.Managing Director at Rod Hughes Consulting Pty LtdTop Contributor"Average" will obviously not do at all!You need the CTs to be able to produce the maximum voltage for the maximum output current associated with the maximum burden they are supplying.You may need to do a few calcs to check each possible scenario of which CT is pushing current through which burdens. krishnakrishna R.HEAD-ELECTRICAL DESIGN at Gracetec GroupMr.Rodney,Fantastic!Very convincing answer!!Thank you very much.Regards,V.Krishna Rao MeiruMeiru D.High Voltage Tester at SiemensHow about a Hall Effect current sensor? PrabakaranPrabakaran K.Electrical Design Engineer at Development Consultant private LimitedThank you for all... AdamAdam H.Substation Electrical Field Engineer at National GridThe CT Knee point voltage is directly related to the accuracy range of the CT and the level of current it can receive before saturation. A CT's accuracy is fairly linear and predictable (useable) until it reaches the knee point voltage. So our protection CT's must have a high enough knee point voltage to remain accurate in the expected fault levels of that circuit to enable the correct operation of protection.

At the same time for a metering CT we want this to be accurate in normal operating conditions and preferably saturate during a fault to prevent any confusing bills, so these would have a much lower knee point voltage but they are very accurate in the range their designed for.

In basic the knee point defines the range of primary current the CT will remain accurate for.

The relay can only work with what the CT can provide, so its essential to chose the correct class. If you have chosen the wrong CT class then obviously there could be problems RodneyRodney H.Managing Director at Rod Hughes Consulting Pty LtdTop Contributor@Adamthat is not totally correct.Kneepoint is only specified by class PX CTsKneepoint and Excitation current is then specified to define the magnetising curve so that we can make sure that CTs in differential applications - including Restricted Earth Fault differential - all perform the same way - i.e. having two CTs with the same Excitation current but on ewith a higher VKP means their slope will be different, or the same VKP but half the excitation current of the other, means that the slopes will cause a different amount of excitation current for a given voltage requirement, So if the CTs have different Vkp:Ie ratio, they will create a false differential current.

The next thing then is that VKP and hence class PX has no concern about current accuracy. It only worries about the same magnetisation performance. You can see this that class PX specifies Winding Ratio, where as class P specifies current ratio and an accuracy at rated burden and current.The difference is the Winding Ratio is the exact number of turnsCurrent ratio means that there may be a different number of turns needed to compensate for the excitation current in order to achieve a specific accuracy

Kneepoint does not define the range of primary current accuracy since the Excitation current is dependant on secondary system burden. i.e. even at say half rated current, if the secondary burden, including the CT winding resistance, means that applying Ohm's Law Ix(Rct + Rburden) is greater than the CT Vkp, then the CT will saturate, even at less than rated current. alirezaalireza K.electrical engineer manager at kwpafor old generation of the relay protection it's very important to know where is the knee point of the CT's, because the relay is located in secondary side of the CT's and after CT saturation the relay couldn't understanding the fault strength.so the relay face to maloperation.but in new version of the relay it's not important for relay operation(if the relay have CT sat diagnostic).and its still important for CT because in this cases the ct face with over heating and maybe cause some problem for CT. RodneyRodney H.Managing Director at Rod Hughes Consulting Pty LtdTop Contributorit is not true that "new version of the relay it's (CT kneepoint voltage) not important for relay operation".(sorry but I can't read the name of the previous post author)

We must remember that saturation means the CT stops producing current output.so for the first, say, 2 to 3 milliseconds the output follows the expected sinusoid path, then drops straight to zero and remaisn there for the rest of that half cycle till it comes back out of saturation. In the next half cycle something similar happens. It is actually quite complex but imagine the first half cycle like this when the fault starts/|___and then the next half cycle the same but sloping down in the negative direction\|

Some electromechanical relays - such as high impedance circulating current Merz-Price connection differential relays - are specifically designed to operate when the CT does saturate like this because the fault condition is associated with the high impedance path through the relay and will HEAVILY saturate by design. (Please make sure you have evidence from the manufacturer of electronic HiZ relays being able to perform in the same way for heavily saturated waveform ... I haven't seen any proof myself yet)

CT saturation detectors will identify this weird saw tooth shape and stabilise the protection to usually NOT operate because it doesn't have the full sinusoid waveform - ESPECIALLY in numeric relays where they quite often tend to just work off the 50 Hz component of the waveform.

So please make sure that the CTs are specifically designed not to saturate:in the case of non-differential applications, P class CTS need to have a capability of supplying the maximum fault current into the burden at a certain accuracyi.e.for P class 1000/5 10VA 5P20 will provide 5% accuracy for 20x 1000A primary with a 10VA rated burden connected. Note this does not specify the kneepoint point voltage although you can infer the kneepoint voltage is not less than a derived value.

for PX class with diff applications, we need the CT to be able to drive maximum through fault current around the CTs path even with one CT saturated based on the formuilaVkp> 2 x Ifmax x (Rct + Rl)so a 5A 200/1 0.5PX400 R0.1 means it has EXACTLY 200 turns on the secondary (hence inferring a nominal but not specific accuracy 1000/5 current ratio) has a kneepoint defined at 400V and 0.5A excitation current with a CT secondary winding resistance of 0.1ohms alirezaalireza K.electrical engineer manager at kwpaHI,thank you rodney.I really enjoyed when i read your comments.but my article is really about differential protection and i think my idea its correct even when i read your comments. RodneyRodney H.Managing Director at Rod Hughes Consulting Pty LtdTop ContributorHi Alirezacertainly diff protection we tend to use PX CTs so that does mean we are interested in kneepoint voltage.Modern numerical relays though do not use circulating current connections - each CT has its own input terminals to the relay so it "looks" more "like" a general P class application in that respect.We also may have more options to help stabilise the scheme with bias of CT saturation detection, but we are still really wanting the CTs to perform with the same Vkp-Ie characteristic to minimise false differential currents so we are still very much interested in Vkp - the slope of a CT with Vkp=400V and Ie=0.05A Ie is very different to one with 600V/0.02A alirezaalireza K.electrical engineer manager at kwpaHI Rodneythank you for your attentions.I should tell you all of your comments are true.but my idea say 'when you use numerical relay with CT sat detection you shouldn't worry about mal-operation of the relay (I do some tests in ABB factory for proving this about ten years ago).and my friend, my old comments did not tell us,'we use CT in knee point area for protection' if you read it carefully you can see I didn't recommend it because of CTs.