Current Transformer Questions 2

[111R]

I'm attempting to fully understand CT operation and I'm getting stuck on a few seemingly simple concepts despite reading for quite a while. I realize this list is exhausting, so a few answers will still be helpful.

The excitation curve is a plot of secondary RMS exciting current vs secondary RMS exciting voltage.

1) I look at the accuracy classes of CTs and they are rated at C100, C200, C400, and C800. I found out that this means that a C800 CT can supply 800V into the secondary at 20x rated current (usually 5A*20=100A) within 10% accuracy. So, B1.0, B2.0, etc mean the same thing, but max burden impedance is given rather than secondary voltage. I am looking at PAGE 5 of the following PDF and I don't understand a few things:

http://www.idc-online.com/technical..._engineering/current_transformer_concepts.pdf

Why is this a C400 CT? 486V is quoted as the secondary voltage at 10A. How is 486V within 10% accuracy? It's way beyond the knee point. And, why 10A...shouldn't it be at 100A (20*5)? I guess I don't really understand what the excitation curve indicates. The knee point voltage on this curve seems to be around 200V. What does the knee point tell us?

2) The knee point is where the core saturates, correct? So, the terms excitation curve and saturation curve are synonymous? What about magnetization curve?

3) Is the linear section before the knee point the desirable area to stay within for accurate current monitoring?

4) The IEEE30 and IEEE45 refer to the angles of a line tangent to the knee point. They indicate the knee point voltage of the CT. There is also IEC 10/50 which is the point where a 10% increase in exciting voltage causes a 50% increase in exciting current. Since I'm in the US, I doubt I will use IEC often. How do I select between IEEE30 and IEEE45? Why is there a need for two? IEEE30 will give a higher knee point value, but is there an application for one over the other?

Thanks for your time.

 

[jghrist]

At 20 times rated primary current, the secondary current would be 100A (20·5) if there were no ratio error. 10A secondary exciting current will make the total secondary current 110A (100+10), for a error of 10% (110/100-1). With 10A exciting current, the voltage is 486V according to the excitation curve. The next lower standard rating from 486V is C400.

It is desirable to stay on the point of the excitation curve where the excitation current is low, keeping the raio error low.

The knee point is where the excitation current starts increase more quickly when the voltage (a function of secondary current and impedance) increases. There are different ways to measure the knee point, probably for historical reasons.

 

[randyman]

Relay class in IEEE world is defined as the available terminal voltage with rated burden at 20X rated current. The voltage required by the CT must be higher than the terminal voltage in order to overcome its own internal impedance ==> Vex = (Rs + Rb)*Is*20. This is only a definition. The fault level of 20X is only used to establish this definition. In reality, the fault level can be any number used in the above equation where Vex would be your required secondary exciting voltage.

And as jghirst mentioned, the 10A max exciting current represents 10% of 5A x 20.

The knee point is where the core enters saturation, and distortion of the secondary current from sinusoidal waveshape begins. Repeatable and predictable performance is assured at and below the knee point, not that you cannot work above the knee point - you can, just yields higher errors.

IEEE30 knee was one time referred to the point used for gapped cores. IEEE removed it from C57.13 since that standard really did not deal with gapped cores anyway.

IEC 10/50 is very close to the defined saturation voltage Vx refered to in C37.110, which is the intercept of the two major excitation slopes. Use IEEE45 as it is the only one defined in the current version of C57.13-2008 standard.

 

[111R]

Thanks, that was very helpful!

So, when looking at secondary exciting curves for CTs with a 5A secondary, the accuracy class voltage (C400=400V, etc.) should occur on the curve at 10A or below, correct? If not, then the CT cannot drive 100A into a 4 ohm burden at less than 10% error at that voltage...which is required for that classification?

With CT test equipment, there are settings for excitation voltage and current (Up to 2000V, 10A). If testing a CT with a 5A secondary, does the excitation current need to be set to 10A for the reasons stated above? If the CT is a C800 class, does the excitation voltage need to be above 800V? Or, is this unnecessary? I've seen test results that stop immediately after the knee point, so I'm a bit confused on how to select the correct excitation voltage and current.

Thanks again.

 

[DTR2011]

When field testing CT's. the test voltage should be set to a level much higher than expected. For example on a C800, the voltage is often taken to 1200V. The old rule of thumb of 1A is also popular. When testing, getting past the knee point an into saturation is desired.

Certainly the test voltage must be higher than expected (based on CT class). Consider that the results are typically plotted Log-Log. Sometimes a user may select different (above minimum) values for optimal presentation on the excitation current graph.

Ideally, there would be standard test voltages and currents as the excitation plots are a bit of a fingerprint and can be used when troubleshooting in the future. You can examine the manufacturers excitation plot and deduce the maximum test voltages & currents if necessary.

Typical CT test sets have different taps for various voltage / current combinations. This, more often, has to due with thermal limitations of the variac / step up transformers, while also keeping the equipment as light as possible.

 

[randyman]

Hello 111R,

There are two misnomers regarding relay class and how it relates to the characteristic curve, that you should be aware of;
1 - The class voltage must be at the knee
2 - The class voltage must be at 10A
Both statements are false, even though there are many who believe them to be true.

Just to be clear on the 10A voltage of secondary excitation, it has to be, as a minimum, the terminal voltage plus its coresponding internal drop.

Using your example of C400, let's say the winding resistance for this particular ratio happens to be 0.5 ohms. Assuming all impedances are purely resistive, the minimum secondary exciting voltage at 10A must be 450V to meet C400 class ==>(0.5 + 4.0)*5*20. This is a good rule of thumb approach. By the standard, the definition of B4.0 has 0.5PF so algebraically adding the values will yield something less than 450, but greater than 400V.

Hope this helps.

 

[scottf]

Randyman-

That's not quite correct. When you measure the secondary voltage, that drop across the Rct is already accounted for.

When we perform routine tests on relay class CTs, we apply the application accuracy limiting voltage (i.e. C400 = ALV of 400V) and confirm excitation current is less than 10 A.

We do not apply a voltage equal to the ALV plus the voltage drop across the Rct. The rating is only defined as the voltage able to be developed at the secondary terminals.

Also, it should be noted that the knee-point voltage (define it how you will) is generally around 80% of the accuracy limiting voltage, i.e. a C400 CT will generally have a knee-point voltage of around 320V.

With this said, the excitation current normally measured at the accuracy limiting voltage is typically in the range of 300 mA or so, i.e more manufacturers don't push the excitation current anywhere near the 10A limit.

 

[111R]

Great explanation...thanks again.

 

[111R]

Is it possible to use a metering CT (not for protection use) with a higher than needed ratio and set the relay to operate with a multiplier?

For example, if a line will never see a current in normal operating conditions above 500A, a 500:5 CT will be optimal. If a CT with a 1500:5 ratio (single ratio) is available, can you set the relay to multiply the input current by a factor of 3? So, when 500A is flowing through the line, the relay will translate the ~1.67A input from the secondary to mean 5A?

 

[davidbeach]

That would be far more complicated than necessary. You set the relay based on secondary amps. If you have a setting of x for a CTR of 100 you will have a setting of x/3 for a CTR of 300.

 

[randyman]

Hello Scottf,

I would like to rebute your not so quite correction if I may.

In order to measure the equivalent exciting current at the equivalent induction level of a given relay class, the Rs MUST be accounted for in the voltage calculation you excite the core to, which provides the effective error current, which for C class has to be 10A or less for 5A secondary.

We too perform relay class on all of our CTs, but at the equivalent operating level, not the secondary terminal voltage.

There is good explanation in C37.110-2007, section 4.4.1 and not so good explanation in C57.13-2008, section 8.1.7-10, but that is being rewritten.

In terms of relay class and secondary terminal voltage, then yes the Rs is already accounted for.

I think we are talking about the same thing, just wording it differently.

 

[scottf]

Randyman-

I work for an instrument transformer manufacturer (and have worked for other IT manufacturers as well).

During the factory routine testing for relay class CTs, the accuracy limiting voltage is applied to the secondary terminals and the excitation current is measured. If the Iexc is less than 10A, then the unit passes the C class. For example, a C800 rated CT has 800V applied to the secondary terminals and the Iexc should be less than 10A. In other words, the accuracy limiting voltage used for the definition of the class and the confirming of such through test is independent of the secondary DC resistance.

To my knowledge, all IT manufactures design and test relay class CTs in that fashion.

 

[111R]

David - I agree that it's probably more complicated than it's worth. But, I don't understand the setting. If you have CTR of 300, shouldn't the relay be set for 3X? The CT ratio is three times as high, so a given current will be reduced 3X as much compared to the CTR of 100. So, the relay should be set to multiply the input from the secondary of CTR=300 by a factor of three to have the same input current.

If you're in a bind and can live with the slightly lower accuracy in the linear portion, can you use a protection CT in place of a metering CT? Is the only concern the damage that can be done under a fault? Are metering CTs designed to saturate very quickly to prevent damage to metering relays during faults?

 

[davidbeach]

If you want the relay to pick up at a particular primary value, say 1200A, you divide that by the CTR and use that as the setting. So if you have a CTR of 100 you set the relay to pick up at 12A, but if you have a CTR of 300 you set the relay to pick up at 4A.

 

[randyman]

I hate to muddy the waters here as it would appear I am beating a dead horse, but I do want to stress one point which scottf and I seem to disagree on. I was debating whether or not I should even rebute this since we apparently are competitors, but since this forum is based on information exchange, that is what is important.

First, I too work for IT manufacturer, and in this business for 30+ years, having been associated with 3 very reputable suppliers in the US. The standard is quite clear on its definition of relay class and burden. And to your point scottf, the secondary resistance is part of the equation. And it will increase the measured secondary voltage in order to assure the rated secondary terminal voltage. The only way you have a situation like you describe is if,

a - you measure voltage across rated burden at 20X rated current by primary induced current,
b - you have superconductive secondary winding with zero winding resistance

neither are the case.

When referring to the Standard (C57.13-2008), it is clear from Figure 18 and the equations on the bottom of pg 48 that the excitation voltage capability (while maintaining the RCF limits) of a C class transformer must be increased by the voltage drop across the winding resistance.

RCF = Ist/Is = (Is + Ie)/Is = 1 + Ie/Is

Where Is = Vse/Zt

And Zt is defined explicitly as sqrt[(Rs + Rb)^2 + Xb^2)]

The required test voltage is very much dependent on the secondary winding resistance, and it is added to the secondary burden, not part of it. All suppliers I have been associated with have qualified their relay class CTs at the Vse point and NOT the terminal voltage.

So to your example of C800 class, if you tested a unit and measured 800V at 9.9A and passed it, I am afraid you shipped a bad unit.

Hopefuly enough said on this topic.

 

[scottf]

randyman...I suspect I may know who you are now

However, the figures you are referencing are for calculating CT performance from typical data and do not refer to routine test code. Figure 18 is mainly geared towards calculating performance based on field measurements and/or typical curve data, hence the inclusion of Rb and Xb. As you know, the common method for routine factory test confirmation of C class does not include application of an external burden.

I just pulled routine test reports from GE Somersworth, GE ITI, ABB Pinetops, Ritz (now AREVA) Waynesboro, and Ritz Lavonia and all match up to the statements I made above, i.e. a C800 rated CT has 800V applied across secondary and Iexc is measured.

What I believe you may be doing is confusing the requirements of the standard for classification and how relay performance calculations are done (where Rb, Rct are considered).

 

[randyman]

Hello Scottf,

I figured you would figure it out sooner ot later. I did talk to a few of your colleagues in Orlando this last week but did not get to meet you face to face. Maybe some other time??

It's funny you bring up routine testing because the standard does not say to test at Vse or terminal voltage BUT to test at the knee point and compare that to the published curve. I think as suppliers we may interpret the meaning a bit differently but accomplish the same end result. When I was with ITI (pre-GE days) we always tested at Vse. And quite honestly that point(Vse) makes the most technical sence to me. And yes it does come from an application perspective but the CT still has to satsify Vse < 10A to meet class, which is what the original question was relating to. I guess it really makes no difference what point you test at as long as the measured Iex is within the tolerances of your published curve. But I do think it is important for users to realize and understand what these values mean and how they are related to each other.

Cheers!!

 

[scottf]

Sorry I missed you Randy. The IEEE show is always a crazy time. I wasn't even in our booth half of the time.

I think most people fall back on section 6.4.1.4 "Secondary Terminal Voltage" where it lists the secondary terminal voltage as 800V for B8.0.

I think we probably both agree C57.13 needs a complete overhaul. This is 1 of many issues that are not clearly defined. My biggest pet peeve is listing mechanical short-time current as an RMS value, but that's another topic for another day

 

[111R]

I have a follow-up question regarding CTs. I found this article which is one of the best I've found for explaining CTs:

http://library.abb.com/GLOBAL/SCOT/...a922d51bf17c12570ae004b5518/$FILE/gan_pap.pdf

On page 4, it shows the output from a CT in saturation. I always (mistakenly) assumed that if a CT saturates at a secondary current of 10A (etc.), it will still output that current in saturation. But, this article just helped me realize that if a CT is in saturation, there is no longer a change in flux, so the output from the CT will drop to zero.

I still do not understand why the CT does not output any secondary current when the primary current waveform drops back below the saturation current level on the positive cycle and negative cycle. It seems like there should be two spikes on each half cycle of the sine wave. But, the CT does not output current again until the negative half of the cycle. Once a CT core is in saturation, does it stick that way with all poles aligned until an opposite voltage is applied...opposed to during normal operation when the CT secondary is proportional to the primary input?

With metering CTs, saturation would be reached immediately during a fault. The waveform would look like the one shown in the linked PDF, if not worse. Would this distorted waveform still operate a protection relay? As in, if you weren't concerned about the accurate measurement of fault current, but just wanted the relay to operate a contact beyond a certain current level (~5A secondary), would the metering CT still do it?

 

[111R]

Actually, I think my last question was answered in the article about the relay not functioning with severe CT saturation.