question about CT's
- Thread starter steveng
- Start date
- Status
brantmacga
Señor Member
- Location
- Georgia
- Occupation
- Former Child
micromind
Senior Member
- Location
- Fernley, NV. (near Reno)
CT's are current transformers. They don't deal with voltage, only current. All of them have a ratio of primary (large) current to secondary (small) current. The secondary side is usually 5 amps. I've seen ratios all the way from 20:5 up to 6000:5. The advantage here is that a large current can be transmitted by a small wire.
They come in several different classes, depending on the amount of accuracy required. Obviously, a utilitys metering CT's are going to be very accurate, while a CT used to drive a simple ammeter can be a bit less so.
Polarity doesn't usually need to be observed unless they're used in conjunction with a voltage signal. The ratio can be 'changed' by the number of turns the primary wire passes through the core. For example, a 200:5 CT becomes a 100:5 if you pass the wire through the core twice in the same direction. If you go opposite directions, it'll always see 0 current. This is the principal behind 'differential protection' of transformers (usually huge ones), motors, generators (again, gigantic), and other equipment. The CT's measure the current on both sides, but opposite directions, if there's a difference, there's a fault.
CT's must always operate at near short circuit. I can't stress this enough; NEVER open the secondary circuit of any CT with current flowing in the primary. To do so will result in an extremely high voltage (several thousand volts) appearing across the secondary. Usually the CT will be destroyed instantly, sometimes spectacularly!
Sorry for the long-winded post, the above is only a small portion of what CT's can do.
They come in several different classes, depending on the amount of accuracy required. Obviously, a utilitys metering CT's are going to be very accurate, while a CT used to drive a simple ammeter can be a bit less so.
Polarity doesn't usually need to be observed unless they're used in conjunction with a voltage signal. The ratio can be 'changed' by the number of turns the primary wire passes through the core. For example, a 200:5 CT becomes a 100:5 if you pass the wire through the core twice in the same direction. If you go opposite directions, it'll always see 0 current. This is the principal behind 'differential protection' of transformers (usually huge ones), motors, generators (again, gigantic), and other equipment. The CT's measure the current on both sides, but opposite directions, if there's a difference, there's a fault.
CT's must always operate at near short circuit. I can't stress this enough; NEVER open the secondary circuit of any CT with current flowing in the primary. To do so will result in an extremely high voltage (several thousand volts) appearing across the secondary. Usually the CT will be destroyed instantly, sometimes spectacularly!
Sorry for the long-winded post, the above is only a small portion of what CT's can do.
They work on the same physics as all transformers, Sumation(NI) = 0 For a two winding xfm, I1( primary) X N1(primary turns) = -I2(secondary) X N2(secondary turns)
CT's are two winding two winding xfms. Lets take the case of a CT marked 500:5 with a 750kcmil going through. This is a 1 turn primary. The ratio is 500:5 or 100:1, so the secondary has 100 turns. The current ratio will be 100:1 - for 100A on the single turn primary, there will be 1 amp on the secondary.
Now about the voltage. Keeping inline with the above example: say you have a 5ohm secondary load. With 1A. the sec voltage is 5 V. The primary voltage is 5/100 = .05V. That's right, the 750kcmil going through the CT has a VD of .05V in 1".
Now what happens if you open circuit the secondary. Well, remember, the primary hs 100A going through it. The secondary voltage will rise until it has 1 amp going through it. Usually somewhere between 2000V and 480,000V the CT secondary shorts, burns up, cracks the case, DFW(engineering term) It's not good good to open circuit CT's.
That's the basics of how and why they work.
carl
Edited to add: Micro's post is excellent. Didn't see it before I posted.
CT's are two winding two winding xfms. Lets take the case of a CT marked 500:5 with a 750kcmil going through. This is a 1 turn primary. The ratio is 500:5 or 100:1, so the secondary has 100 turns. The current ratio will be 100:1 - for 100A on the single turn primary, there will be 1 amp on the secondary.
Now about the voltage. Keeping inline with the above example: say you have a 5ohm secondary load. With 1A. the sec voltage is 5 V. The primary voltage is 5/100 = .05V. That's right, the 750kcmil going through the CT has a VD of .05V in 1".
Now what happens if you open circuit the secondary. Well, remember, the primary hs 100A going through it. The secondary voltage will rise until it has 1 amp going through it. Usually somewhere between 2000V and 480,000V the CT secondary shorts, burns up, cracks the case, DFW(engineering term) It's not good good to open circuit CT's.
That's the basics of how and why they work.
carl
Edited to add: Micro's post is excellent. Didn't see it before I posted.
Last edited:
REEngineer
Member
How to take in saturation effect when selecting a CT?
How to take in saturation effect when selecting a CT?
Assume the following situation:
We have a 11KV busbar and there are just a few motor loads connected to it. Rated current of the busbar would be about 50-100A. But short circuit level is nearly 10-12KA.
The above arrangement is not unique. I guess there are many situations in which short circuit current exceeds 20*In(=rated current) enormously. We may have currents up to 200*In or so.
What I seek is a reliable procedure to help me trade off between reliability and price of the CT. Selecting a 3000:5 CT for a rated current of 40A seems to be too much expensive. On the other hand, what if a fault occurs and saturation of a low ratio CT prevents relays from tripping?
I tried IEEE C37.110:1996. But it was of no real help as it had emphasized on eliminating saturation. I'd like to know what happens to relays(uP particularly)when CTs are heavily saturated.
Any help/recommendation/link would be highly appreciated.
How to take in saturation effect when selecting a CT?
Assume the following situation:
We have a 11KV busbar and there are just a few motor loads connected to it. Rated current of the busbar would be about 50-100A. But short circuit level is nearly 10-12KA.
The above arrangement is not unique. I guess there are many situations in which short circuit current exceeds 20*In(=rated current) enormously. We may have currents up to 200*In or so.
What I seek is a reliable procedure to help me trade off between reliability and price of the CT. Selecting a 3000:5 CT for a rated current of 40A seems to be too much expensive. On the other hand, what if a fault occurs and saturation of a low ratio CT prevents relays from tripping?
I tried IEEE C37.110:1996. But it was of no real help as it had emphasized on eliminating saturation. I'd like to know what happens to relays(uP particularly)when CTs are heavily saturated.
Any help/recommendation/link would be highly appreciated.
Current Transformers
Current Transformers
You do not need a 3000:5 CT At most you would want to install a CT with a 100:5 ratio. I have done start up electrical at probably 15 power plants over the years. The CT accuracy is chosen for it's purpose, like metering CT's for the Protective Relaying in the switchyard. Large motors may have a less accurate relay, but are less expensive. I, or another electrical engineer would always run ratio tests and saturation tests on these CT's, and would purposely run them until we found the saturation point, and then bleed off the saturation. That's no big deal - easy - and I recommend you do that. I have found ( once only - but it only takes 1 time ) a factory CT in a 4160 200HP motor starter that was not the ratio it was marked. I do not remember the particulars, but it was either a lot larger or a lot smaller than it was suppose to be. The meters on the cabinet door, or the Multi-Lin protection would not have operated.
You are worried about the short circuit current - it should never be available long enough to be of any problem.
A really inexpensive way would be to have current sensing relays trip the motor. I have seen these years ago, and set them to required values. Not sure anymore who made them. But these were a pretty inexpensive solution for what they were used. They were accurate enough - we used them to alarm on overamp conditions - they could just as well trip the motor. Hope this helps.
Current Transformers
You do not need a 3000:5 CT At most you would want to install a CT with a 100:5 ratio. I have done start up electrical at probably 15 power plants over the years. The CT accuracy is chosen for it's purpose, like metering CT's for the Protective Relaying in the switchyard. Large motors may have a less accurate relay, but are less expensive. I, or another electrical engineer would always run ratio tests and saturation tests on these CT's, and would purposely run them until we found the saturation point, and then bleed off the saturation. That's no big deal - easy - and I recommend you do that. I have found ( once only - but it only takes 1 time ) a factory CT in a 4160 200HP motor starter that was not the ratio it was marked. I do not remember the particulars, but it was either a lot larger or a lot smaller than it was suppose to be. The meters on the cabinet door, or the Multi-Lin protection would not have operated.
You are worried about the short circuit current - it should never be available long enough to be of any problem.
A really inexpensive way would be to have current sensing relays trip the motor. I have seen these years ago, and set them to required values. Not sure anymore who made them. But these were a pretty inexpensive solution for what they were used. They were accurate enough - we used them to alarm on overamp conditions - they could just as well trip the motor. Hope this helps.
Ct Selection
Ct Selection
You can select the CT depend on short circuit current.
in your case even Ratio can be 100 / 1 or 100 / 5 . short circuit with stand depend on it's class.
Example : if you select protection class 5P10 or 5P20 Means it will give Only 5 % error on 10 or 20 times rated current ( of 100 A)
Ct Selection
You can select the CT depend on short circuit current.
in your case even Ratio can be 100 / 1 or 100 / 5 . short circuit with stand depend on it's class.
Example : if you select protection class 5P10 or 5P20 Means it will give Only 5 % error on 10 or 20 times rated current ( of 100 A)
- Status