CT Accuracy Class

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Electriman

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I am considering CTs for three phase 12.47kv to 4.16kv and 12.47kV/480 V, 3.75 MVA transformers in my design. The CT ration is based on the nominal current. But the problem that I face is I don't know how to select CT's accuracy class (i.e. C200, C400, ...)? Can the accuracy class be different for primary and secondary or phase and neutral?
Please help me with that.
 
Electriman said:
I am considering CTs for three phase 12.47kv to 4.16kv and 12.47kV/480 V, 3.75 MVA transformers in my design. The CT ration is based on the nominal current. But the problem that I face is I don't know how to select CT's accuracy class (i.e. C200, C400, ...)? Can the accuracy class be different for primary and secondary or phase and neutral?
Please help me with that.
Click to expand...

not sure if I understand your question. CT classes are 0.1, 0.2, 0.5, 1 ..... A class 0.2 CT has error of 0.2%; i.e., ratio*(1?0.002)=primary/secondary.

What did you mean by "Can the accuracy class be different for primary and secondary or phase and neutral"?
 
LMAO said:
not sure if I understand your question. CT classes are 0.1, 0.2, 0.5, 1 ..... A class 0.2 CT has error of 0.2%; i.e., ratio*(1?0.002)=primary/secondary.

What did you mean by "Can the accuracy class be different for primary and secondary or phase and neutral"?
Click to expand...

There is another method for CT classification which is more commen and that is by using "C" letter and a number that comes after C. The existing C classes are C10, C20, C50, C100, C200, C400 and C800. I have seen that C200 and C400 are used more frequently than the rest. But I don't know the criteria to choose C200 over C100.

On the second question, for example if my secondary CT is 4500A/5A C200 MR and I select 50A/5A CT for that neutral which is grounded through a high resistance ground can I select C20 instead of C200 which is my phase CT class?
 
Keeping it basic.
Someone else can jump in to make it complicated.

A "CT metering class rating" is basically the ability of the CT to provide a linear output in excess of its nominal rating. This is very important when you are trying to respond to short circuit currents. To properly select the metering class you need to know the burden which it will be seeing, the amount of fault current you need to respond to, and the response curve of the CT.

The burden is the total impedance of the circuit, it is expressed in ohms and for the most part is only resistance. It includes the CT windings, the length and wire size of the conductors, and the protective device or meter.
The fault current is often chosen to be the amount of current flowing through the CT during a fault as opposed to the amount available to the system.
The CT curve needs to come from the CT manufacturer.

Yes, it is very common to have different metering classes on the primary and secondary sides of power transformers.
 
jim dungar said:
Keeping it basic.
Someone else can jump in to make it complicated.

A "CT metering class rating" is basically the ability of the CT to provide a linear output in excess of its nominal rating. This is very important when you are trying to respond to short circuit currents. To properly select the metering class you need to know the burden which it will be seeing, the amount of fault current you need to respond to, and the response curve of the CT.

The burden is the total impedance of the circuit, it is expressed in ohms and for the most part is only resistance. It includes the CT windings, the length and wire size of the conductors, and the protective device or meter.
The fault current is often chosen to be the amount of current flowing through the CT during a fault as opposed to the amount available to the system.
The CT curve needs to come from the CT manufacturer.

Yes, it is very common to have different metering classes on the primary and secondary sides of power transformers.
Click to expand...

Thanks Jim.

Please correct me if I am wrong. So what you are say is "C" rating is called "CT metering class rating".
I understood that selecting it has something to do with CT burden and CT curve. But I have never seen an example of how it is calculated. Do you have any example?

Second since my CTs are for protection not metering does it change anything?

Third, It seems going that route and calculating through that method seems extremely complicated is there any role of thumb that I can use?
 
Here's a simple tutorial sheet on CT accuracy that may be helpful. I split it into two files to meet the file size limit.
 

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jtinge said:
Here's a simple tutorial sheet on CT accuracy that may be helpful. I split it into two files to meet the file size limit.
Click to expand...

Thanks.

It says the number after C is the voltage that develops at 20 times of nominal current. But do you know if there is any easy way to justify picking C200 over C100?
 
Electriman said:
Thanks.

It says the number after C is the voltage that develops at 20 times of nominal current. But do you know if there is any easy way to justify picking C200 over C100?
Click to expand...

The CT accuracy class must be equal to or greater than the required CT secondary terminal voltage.

For example, if you had standard 2.0 ohm burden the required secondary CT terminal voltage would be Vs=2.0 ohms x 5A X 20 = 200 Vrms.

If your available primary fault current drives the secondary CT current greater than 20 x 5A, then you would need to go to a higher class CT.

So in the above example, if you had a 600/5 ratio CT, and your primary fault current was 18kA, the CT secondary current would be 150 A, which is 30 X the 5A secondary CT current rating. In this condition the CT will saturate become non-linear. You would then need to go to a higher CT class.

There are other considerations that also come into play, but hopefully this will helps understand the importance of selecting the right CT accuracy class.
 
jtinge said:
The CT accuracy class must be equal to or greater than the required CT secondary terminal voltage.

For example, if you had standard 2.0 ohm burden the required secondary CT terminal voltage would be Vs=2.0 ohms x 5A X 20 = 200 Vrms.

If your available primary fault current drives the secondary CT current greater than 20 x 5A, then you would need to go to a higher class CT.

So in the above example, if you had a 600/5 ratio CT, and your primary fault current was 18kA, the CT secondary current would be 150 A, which is 30 X the 5A secondary CT current rating. In this condition the CT will saturate become non-linear. You would then need to go to a higher CT class.

There are other considerations that also come into play, but hopefully this will helps understand the importance of selecting the right CT accuracy class.
Click to expand...

Thank you for clarification.
Can I consider 2.0 ohm burden for all my CTs or is it different for each application considering that all I know the CT signal will go to a relay and that relay is a few feet away from the CT?
 
Electriman said:
Please correct me if I am wrong. So what you are say is "C" rating is called "CT metering class rating".
Click to expand...

No the C stands for 'calculated', it means basically means you can determine the performance of the CT based on its design instead of having to wait until it has been built. Most metering class CTs are going to be C type devices.
 
Electriman said:
Thank you for clarification.
Can I consider 2.0 ohm burden for all my CTs or is it different for each application considering that all I know the CT signal will go to a relay and that relay is a few feet away from the CT?
Click to expand...

Not necessarily. The standard CT burden is the load that will result in the rated secondary voltage Vs at 20 x 5A = 100A. So for a C200 CT, the standard burden will be 2 ohms, for a C400, it will be 4 ohms. The actual CT burden will be determined by the device connected to the CT and the impedance of the wire connecting the CT to the device. This total must be less than the CT standard burden.
 
jtinge said:
The CT accuracy class must be equal to or greater than the required CT secondary terminal voltage.

For example, if you had standard 2.0 ohm burden the required secondary CT terminal voltage would be Vs=2.0 ohms x 5A X 20 = 200 Vrms.

If your available primary fault current drives the secondary CT current greater than 20 x 5A, then you would need to go to a higher class CT.

So in the above example, if you had a 600/5 ratio CT, and your primary fault current was 18kA, the CT secondary current would be 150 A, which is 30 X the 5A secondary CT current rating. In this condition the CT will saturate become non-linear. You would then need to go to a higher CT class.

There are other considerations that also come into play, but hopefully this will helps understand the importance of selecting the right CT accuracy class.
Click to expand...

Perhaps you are getting class and ratio mixed. For your scenario above, you could use a higher ratio CT (1200/5) to drop the secondary current to 75 amps.

There are a few factors when determining ratio and class:

1.) You should find out what the available short circuit is. If it is 18kA, for example, you want to select a ratio that will reduce the secondary current to personnel and equipment in the event of the full fault. The rule of thumb is to keep it less than 100A, secondary; preferably below 75A. In this case, you would need no less than a 1200/5 ratio. The CT class has nothing to do with this.

2.) The class is determined by burden (or load). The load includes the relays and the CT cables and anything else in between the CT and the relays....like test switches, terminal blocks, etc. If the load is too high, you'll need a higher class CT. Typically, the older E-M relays have a high burden and require higher class CTs. Really long CT cables also requires higher class CTs. If you are using new microprocessor relays and short runs of CT cables (<300' of #10), then you should be able to use C200 class CTs. There are calcs that can help determine all this. Also, the class has nothing to do with CT ratio.
 
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