CT Secondary Current Rating limited to 20 times the rated current

[D!NNy]

Why the current on the secondary side is limited to 20 times the rated current which is 5 * 20 = 100A

How and why we got to the number 20 TIMES?? is this manufacturer or IEEE C57.13

 

[jtinge]

Why the current on the secondary side is limited to 20 times the rated current which is 5 * 20 = 100A

How and why we got to the number 20 TIMES?? is this manufacturer or IEEE C57.13

Per ANSI, the CT ratio is selected to limit the maximum symmetrical fault current on the CT secondary winding to 100A to stay within the linear performance of the CT's accuracy curve. Thus 5A * 20 = 100A

 

[D!NNy]

Per ANSI, the CT ratio is selected to limit the maximum symmetrical fault current on the CT secondary winding to 100A to stay within the linear performance of the CT's accuracy curve. Thus 5A * 20 = 100A

Example if a CT of 600:5 has TR 120, with C200. 200 indicates the maximum voltage that can be appeared across the burden of the CT.
If burden is 1 ohm, for a fault current of 12000A, current on the secondary side is 100A and max voltage seen is 100V, so this doesnt saturate.
If burden is 1 ohm, for a fault current of 18000A, current on the secondary side is 150A and max voltage seen is 150V, so this doesnt saturate. However it exceeds the rated secondary current by 30 times (recommended limit is 20 times)....... why is it a problem though we are in the limit of voltage class C200???

Thanks

 

[jtinge]

Example if a CT of 600:5 has TR 120, with C200. 200 indicates the maximum voltage that can be appeared across the burden of the CT.
If burden is 1 ohm, for a fault current of 12000A, current on the secondary side is 100A and max voltage seen is 100V, so this doesnt saturate.
If burden is 1 ohm, for a fault current of 18000A, current on the secondary side is 150A and max voltage seen is 150V, so this doesnt saturate. However it exceeds the rated secondary current by 30 times (recommended limit is 20 times)....... why is it a problem though we are in the limit of voltage class C200???

Thanks

For the C200 CT, the 200 indicates that the CT can drive a 2 ohm burden up to 200 volts base on 20 times the 5A rated secondary current, and still be in the linear performance range of the CT. Over 20 times, linear performance isn't guaranteed, however, if the terminal voltage is below the knee point, the secondary current should still be linear.

5A x 20 x 2 ohms = 200V

Also, if the burden is less than 2 ohms, CT output will be linear up to the point where the terminal voltage is 200 volts.

For your example: 200 volts/(5A x 1 ohm) = 40, so the CT will be linear up to 40 times the rated secondary current.

 

[jtinge]

For the C200 CT, the 200 indicates that the CT can drive a 2 ohm burden up to 200 volts base on 20 times the 5A rated secondary current, and still be in the linear performance range of the CT. Over 20 times, linear performance isn't guaranteed, however, if the terminal voltage is below the knee point, the secondary current should still be linear.

5A x 20 x 2 ohms = 200V

Also, if the burden is less than 2 ohms, CT output will be linear up to the point where the terminal voltage is 200 volts.

For your example: 200 volts/(5A x 1 ohm) = 40, so the CT will be linear up to 40 times the rated secondary current.

While the above is true, I wouldn't recommended applying the CT's so that the available primary fault currrent would drive the CT secondary winding above the 20 times 5 A rated current. There are thermal limits to what the secondary CT windings can handle without damage, and the CT data sheets should indicate what the maximum ratings are.

 

[D!NNy]

For the C200 CT, the 200 indicates that the CT can drive a 2 ohm burden up to 200 volts base on 20 times the 5A rated secondary current, and still be in the linear performance range of the CT. Over 20 times, linear performance isn't guaranteed, however, if the terminal voltage is below the knee point, the secondary current should still be linear.

5A x 20 x 2 ohms = 200V

Also, if the burden is less than 2 ohms, CT output will be linear up to the point where the terminal voltage is 200 volts.

For your example: 200 volts/(5A x 1 ohm) = 40, so the CT will be linear up to 40 times the rated secondary current.

this is exactly what my understanding was however i kept hearing that rated secondary current is limited to 20 * 5amp = 100A.

 

[D!NNy]

While the above is true, I wouldn't recommended applying the CT's so that the available primary fault currrent would drive the CT secondary winding above the 20 times 5 A rated current. There are thermal limits to what the secondary CT windings can handle without damage, and the CT data sheets should indicate what the maximum ratings are.

so you are saying any excessive current on the secondary side of the ct over 20 times causes some thermal problems.

I agree more current more heat at some point it will deteriorate the insulation and breaks down. but every CT is different, different core size, different no of turns, different fault currents on the primary side ........ how could one number be fit for all CT's 20 times (though all the CT's have one thing in common, 5A or 1 A secondary current).
By chance can you guide me to a document of calculations that shows 20 times? apart from BH curves.

Just trying to understand
Thanks

 

[jtinge]

so you are saying any excessive current on the secondary side of the ct over 20 times causes some thermal problems.

I agree more current more heat at some point it will deteriorate the insulation and breaks down. but every CT is different, different core size, different no of turns, different fault currents on the primary side ........ how could one number be fit for all CT's 20 times (though all the CT's have one thing in common, 5A or 1 A secondary current).
By chance can you guide me to a document of calculations that shows 20 times? apart from BH curves.

Just trying to understand
Thanks

ABB has an instrumentation transformer guide that has a nice discussion of overload and short circuit capacities of CT's, pg 24 in link below.

https://library.e.abb.com/public/e2462bd7f816437ac1256f9a007629cf/ITTechInfoAppGuide.pdf

Other manufacturers have similar guides you can find on the web.

 

[D!NNy]

ABB has an instrumentation transformer guide that has a nice discussion of overload and short circuit capacities of CT's, pg 24 in link below.

https://library.e.abb.com/public/e2462bd7f816437ac1256f9a007629cf/ITTechInfoAppGuide.pdf

Other manufacturers have similar guides you can find on the web.

The IEEE Guides for Loading recognize that transformer insulationcan withstand a considerable degree of overheating for ashort time without severe deterioration. For example, in the eventof a line short circuit, the fault current may easily be fifty timesthe rated current of the current transformer, but will probablyflow for not over one second. IEEE C57.13 permits a 250°Ctemperature for this very short time (compared to 95°C averagecontinuous winding temperature). All current transformers areassigned a one-second thermal current limit which denotes howmuch current they will carry (usually denoted in terms of timesnormal rated current) for one second. For durations up to fiveseconds, current transformers will carry currents lower than this.The transformer will carry more current for a time less than onesecond according to the same rule, up to the mechanical currentlimit (which is also given for standard current transformers.) Atthis limit of current, the electro-mechanical forces tending toseparate the primary and secondary coils, become high enough todamage the transformer.

Based on this and figure 20 in the link
u could let the current flow up to 50 times for less than 1 second.
upto 20 times for 10 seconds .........

in most cases ur protection devices trips and leave the circuit before ct's saturation time comes.

 

[GoldDigger]

...

in most cases ur protection devices trips and leave the circuit before ct's saturation time comes.

A minor misstatement there.
Saturation will occur essentially instantaneously (less than one cycle). It is the damaging effects of overcurrent in terms of heating of the CT that will be prevented by the I-T curve of the OCPD in most cases.

 

[D!NNy]

A minor misstatement there.
Saturation will occur essentially instantaneously (less than one cycle). It is the damaging effects of overcurrent in terms of heating of the CT that will be prevented by the I-T curve of the OCPD in most cases.

question 1:
I am confused according the ABB article those curves say that it could go as high as 50 times for less than one second!!!!!! you are saying in less than a cycle it will gets saturated??????

question 2:
If the secondary circuit of the current transformer is accidentallyopened, the flux density will become very high and even if thecircuit is immediately closed again, the core may be left with permanentmagnetization.

2A why the flux density goes really High??????
2B what does it mean by permanent magnetization:::: lets say CT is open circuited momentarily while bus is energized and then Bus is de-energized if the CT is left open circuited, will the core be still producing constant magnetic field on the secondary winding? if the CT's are shorted will we see contact secondary current circulating in the secondary winding???

 

[GoldDigger]

question 1:
I am confused according the ABB article those curves say that it could go as high as 50 times for less than one second!!!!!! you are saying in less than a cycle it will gets saturated??????

question 2:
If the secondary circuit of the current transformer is accidentallyopened, the flux density will become very high and even if thecircuit is immediately closed again, the core may be left with permanentmagnetization.

2A why the flux density goes really High??????
2B what does it mean by permanent magnetization:::: lets say CT is open circuited momentarily while bus is energized and then Bus is de-energized if the CT is left open circuited, will the core be still producing constant magnetic field on the secondary winding? if the CT's are shorted will we see contact secondary current circulating in the secondary winding???

1. If the burden on the secondary is high enough that the current in the secondary cannot do its job of balancing the flux generated by the current in the primary then the primary can saturate. That can happen in a fraction of a cycle.
As long as the load resistance on the secondary is low enough the primary current will not saturate the core, but the primary and secondary current will both heat the CT, which can cause damage if it reaches too high a temperature.

2. The residual field left by an even such as you describe, then just as can happen when you put a screwdriver into a degaussing coil and open the switch at the wrong time, you can be left with a magnetized core (screwdriver). This is a static magnetic field which cannot induce any current in anything. It can however, make it easier for the core to saturate when exposed to an alternating magnetic field.
Consider this: I insert a bar magnet into a coil of copper wire and wait a few seconds. Will opening or closing a switch located between the coil terminals cause any current to flow? No. The magnetic field affecting the wire is constant. Only a time-varying or moving magnetic field can induce current in a stationary wire loop.

 

[D!NNy]

1. If the burden on the secondary is high enough that the current in the secondary cannot do its job of balancing the flux generated by the current in the primary then the primary can saturate. That can happen in a fraction of a cycle.
As long as the load resistance on the secondary is low enough the primary current will not saturate the core, but the primary and secondary current will both heat the CT, which can cause damage if it reaches too high a temperature.

Got it

2. The residual field left by an even such as you describe, then just as can happen when you put a screwdriver into a degaussing coil and open the switch at the wrong time, you can be left with a magnetized core (screwdriver). This is a static magnetic field which cannot induce any current in anything. It can however, make it easier for the core to saturate when exposed to an alternating magnetic field.
Consider this: I insert a bar magnet into a coil of copper wire and wait a few seconds. Will opening or closing a switch located between the coil terminals cause any current to flow? No. The magnetic field affecting the wire is constant. Only a time-varying or moving magnetic field can induce current in a stationary wire loop.

Thanks for educating me, apparently i completely forgot the Faraday's law....... if there is no change in glux no emf to drive current in the loop.

Thank you very much all