CT(current transformer) ratings

[panthripu]

I would like to know , If I have a CT in our electrial system with ration 500/1A. Does it mean that if the current is more than 500A , it would not funtion properly. If no , upto what current can it work without any problem ?

 

[templdl]

Are you sure it's 500:1? They would normally be 500:5.
The problem that you would have when running over 500a is the core of the CT would go into saturation. The accuracy is based upon a max of 500a.
If you need to increase the ratio often times the secondary winding is wrapped around the window if you have a window type CT. Depending upon the direction of the wrap you can either increase or decrease the ratio. The number of wraps then determines the ratio.
But, you must have a meter that corresponds to that ratio. Most Ammeter movements are 5a full scale with the dial actually marked to show the primary amps.

 

[Joethemechanic]

Will the manufacturer set parameters for the resistance of the secondary circuit conductors? Such as for long runs between the CTs and the meters? Or is total circuit resistance just adjusted with a pot?

 

[iwire]

Will the manufacturer set parameters for the resistance of the secondary circuit conductors? Such as for long runs between the CTs and the meters? Or is total circuit resistance just adjusted with a pot?

Voltage is irrelevant, the meter works on current.

 

[jim dungar]

I would like to know , If I have a CT in our electrial system with ration 500/1A. Does it mean that if the current is more than 500A , it would not funtion properly. If no , upto what current can it work without any problem ?

Many utility systems and even some process control use CT's with 1A output. Every manufacturer of electronic overcurrent relays have options of either 1A or 5A inputs.

Now to the OP.

A CT is just like any other transformer it will do what you want until it overheats and fails.
As the primary current rises the magnetic core of the CT begins to saturate which means its output stops being linear and thus. In your case this might mean something like:
500A in = 1A out
1000A = 2A out
1500A = 2.2A out
2000A = 2.3A out

How much overload the CT can handle is determined by how much steel is in the core. For short circuit protection, I often choose CT's that can accurately handle 15-20X their rating. Physically small CT's may only be rated for 2X.

 

[Joethemechanic]

Voltage is irrelevant, the meter works on current.

Oh yeah I understand that Bob, but as overall circuit resistance increases current should fall. I've just never had to worry about it, so I never really thought about it much. Most every time I installed a CT it was the doughnut type. And it was always for a generator or monitoring some motor on a machine, so probably the longest run was probably 50 feet, and that would be 100 feet of conductor.

So my added resistance from the conductors would have been insignificant.


But there has got to be a limit, has anyone ever had a long long run where that limit was significant?


And if we really think of it, amp meters are really volt meters. Picture a simple dc shunted analog ammeter. It is just measuring voltage drop over a fixed resistance. It's a volt meter, it's just speaking in amps

When you guys are dealing with CTs it's mostly for utility metering, and ground fault devices isn't it?

 

[iwire]

Joe, the voltage will increase as needed to maintain the current, the current is the same in every point of the circuit at any given moment.

There are likely limits but nothing you would run into in real life.

 

[Joethemechanic]

There are likely limits but nothing you would run into in real life.

Bob, that was the part I was wondering about. If there ever had to be compensation made. But a piece of 14 gauge wire with 5 amps max flowing, is going to have very little losses.

 

[jim dungar]

Bob, that was the part I was wondering about. If there ever had to be compensation made. But a piece of 14 gauge wire with 5 amps max flowing, is going to have very little losses.

Depending on the purpose for measuring the current, the meter (load) and the CT, I would probably not recommend a 50ft run of #14.

 

[Joethemechanic]

Depending on the purpose for measuring the current, the meter (load) and the CT, I would probably not recommend a 50ft run of #14.

Most everything I use a ct for is less than critical. I never need super accuracy. And I'll bet most of the circuits have been around 6 feet long total (3 feet from CT to meter)

What would you recommend for a 50 foot run? The time I had to go out 50 feet, I think I used some 14 SO cord because I wanted to sit in the office trailer and watch my ammeters where it was warm and I could surf the net while waiting for something to happen.

 

[Open Neutral]

Oh yeah I understand that Bob, but as overall circuit resistance increases current should fall.

In a word, no....

The CT emulates a constant current supply.

Given a 500:5, with 100 amps being measured.

If you have a {say} 10 ohm loop, it sources 1 amp. If you have a 20 Ohm loop, it sources 1 amp. If you have a 136.876 ohm loop, it sources 1 amp.

Where you are asking for it is the following issue.....

For each of the above loops, what will the output voltage be at the CT terminals?

I*R = E
1 amp * 10 ohms = 10 volts
1 * 20 = 20 V
1 * 136.876 = 136.876 volts.

Now... Suppose the loop resistance is high, really high, oh 999999 ohms. What will the voltage be then?

Never over-circuit a current transformer secondary; there will be an arc that may will damage it, or you....

 

[Joethemechanic]

So you are saying it tends to act like a constant current device like a series street lighting transformer or a welding machine? I got to think about that for a bit.


About the open circuit, been there, seen it done, and seen the damage (all against my advise of "You better shunt that before you forget and turn it on")


BTW, I am not an apprentice,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

I'm just new to this forum,,,,,,,,,

 

[Besoeker]

In a word, no....

The CT emulates a constant current supply.

Given a 500:5, with 100 amps being measured.

If you have a {say} 10 ohm loop, it sources 1 amp. If you have a 20 Ohm loop, it sources 1 amp. If you have a 136.876 ohm loop, it sources 1 amp.

Unless you exceed the VA rating of the CT.
We generally use either 5VA or 15VA depending on the application.

 

[Open Neutral]

Unless you exceed the VA rating of the CT.
We generally use either 5VA or 15VA depending on the application.

I'd say "until you exceed" but YMMV. My unspoken point was: short of falling off the end of the specs, the loop resistance is irrelevant to calibration.

Joe; I often write a response not just on my guess as to your specific knowledge; but to fill out the picture for someone reading the thread in the future.

{I also proofread badly, you grasped I mean 'OPEN-circuit' in that post.}

By the way, conventional D'Arsonval meter movements are usually thought of as current meters, with resistors either internally or externally making a "voltmeter" out of it.

 

[Joethemechanic]

Joe; I often write a response not just on my guess as to your specific knowledge; but to fill out the picture for someone reading the thread in the future.

{I also proofread badly, you grasped I mean 'OPEN-circuit' in that post.}

Yes I often do the same thing. I would like to apologize for my snippy reply. I have some things external to this forum causing me a bit of stress. It seems to be causing me to be a bit hypersensitive at times.

 

[templdl]

Open Neutral,
You hit it on the head regardiing the ammeter which you described it well. Ammeter movements also include a shunt which bypassed most of the current form the meter movement allowing enough to deflect the movement. If the movement is based upon a full scale deflection of 5a then the meter face is marked based upin the CT ratio. As such am ammeter should in so much be a short circuit being that it is in series with the circuit, that is it should have no significant resistance.
And you said the voltmeter is basically an ammeter with a resistor is series which will allow just enough current to deflect the movement.
And I agree that you never leave a CT open circuited when the circuit being metered is energized. If you do the voltage across the open circuit will go through the roof, it will most certainly skyrocket. Think it as the primary of a transformer with the secondary being the conductor running through the window or bar and lets say the primary as the donut with 500 turns. Talk about one hell of a step up transformer a CT can turn into when left open circuited.

 

[kingpb]

Saturation of the CT core will occur and the accuracy of the CT will become severely diminished and the waveform distorted by harmonics resulting in lower than expected current based on the CT ratio. If used for protection, loss of coordination could result.

 

[Besoeker]

Saturation of the CT core will occur and the accuracy of the CT will become severely diminished and the waveform distorted by harmonics resulting in lower than expected current based on the CT ratio. If used for protection, loss of coordination could result.

I routinely deal with power electronics with non-linear waveforms and have generally not found this to be a problem.
We typically use ACCTs to measure the AC into a rectifier. The output from the ACCTs is rectified and fed to a burden resistor where it then give a transformed down version of the current in the rectifier power circuit. Been doing that for a lot of years......
It's safer than using a shunt because the output is isolated from the power circuits and cheaper and more robust that Hall effect transducers.

 

[kingpb]

I routinely deal with power electronics with non-linear waveforms and have generally not found this to be a problem.
We typically use ACCTs to measure the AC into a rectifier. The output from the ACCTs is rectified and fed to a burden resistor where it then give a transformed down version of the current in the rectifier power circuit. Been doing that for a lot of years......
It's safer than using a shunt because the output is isolated from the power circuits and cheaper and more robust that Hall effect transducers.

My response is based on the OP, which seemed to be electrical power system related. Power electronics, I will leave to you.

 

[Besoeker]

My response is based on the OP, which seemed to be electrical power system related. Power electronics, I will leave to you.

That's fair enough.
I was just giving you and example of where CTs have not been rendered inaccurate but non-sinusoidal current waveforms.
But actually, with the ever increasing proliferation of electronic equipment in the domestic and commercial sectors, currents are not very often without harmonic content. And sometimes, quite a lot of it.