current transformers

[dan55]

I am trying to determine if a current transformer is working or not. should there be voltage at the leads that terminate at the meter.

 

[winnie]

_DO NOT DISCONNECT THE CURRENT TRANSFORMER FROM ITS METER_

Sorry for shouting. There should be little if any voltage at the terminals of an operating current transformer.

A current transformer operates so that its secondary _current_ (the current flowing through its output terminals) is proportional to its primary current (the current flowing in the conductor in the hole). This proportional current flows through a low impedance meter (at most a few ohms) and is detected.

The voltage at the terminals will be (by ohms law) the product of the proportional current and the 'burden resistance'. Depending upon the meter design, the 'burden resistance' might be only a few milliohms, in which case the terminal voltage will only be a few millivolts.

If you were to open the secondary circuit, the current will _try_ to remain constant, and the voltage across the open connection will rise to lethal levels; an operating current transformer must always remain connected to a suitable load, or its output must be _shorted_ to prevent excessive voltage.

You can use a clamp on current meter to measure the current flowing in the secondary circuit, and thus determine if the current transformer is operating.

-Jon

 

[charlietuna]

good advice, but understand most new current transformers have resistors across the secondary to prevent high voltage buildup.

 

[djohns6]

good advice, but understand most new current transformers have resistors across the secondary to prevent high voltage buildup.

That's a VERY broad statement and I disagree .

 

[zog]

Me too, better to assume they arent there and NEVER DISCONECT CT's with the system energized.

 

[72.5kv]

An open circuited CT can expose you to voltages in the kilovolt levels. As mention in the other posted use a clamp on meter to read the current on the secondary. If it is a must that you test these ct, take them out of service to conduct your test. As mention before under no circumstance should the ct be opened circuited during the test.

Standard secondary current should be 5Amp but in some new designs 1Amp. at full primary current ie 800:5 or 1600:5

 

[djohns6]

If it's an 800:5 or 1600:5 CT , why would the secondary current only be 1 amp at full primary ?

 

[beanland]

CT Check

CT Check

If you want to check if the CT is working, use a clamp on current meter to measure the current in the CT secondary lead. You may need an aux 5A CT that plugs into a DMM if the currents are really low.

A C800 class CT will saturate open-circuited at over 800V.

 

[roger]

That's a VERY broad statement and I disagree .

And so do I.

As an aside, there is nothing guaranteeing that the resistor is intact even if this is a newer CT.

Roger

 

[LarryFine]

If it's an 800:5 or 1600:5 CT , why would the secondary current only be 1 amp at full primary ?

I'm not sure I understand your question, but . . .

When a CT is operated with the secondary open, it behaves like a PT. A 400:5 CT has an 80:1 current ratio, but a 1:80 PT.

That can damage the CT as well as surrounding equipment and people.

 

[djohns6]

I'm not sure I understand your question, but . . .

When a CT is operated with the secondary open, it behaves like a PT. A 400:5 CT has an 80:1 current ratio, but a 1:80 PT.

That can damage the CT as well as surrounding equipment and people.

A previous post stated that an 800:5 CT might have 1 amp secondary current at full primary load ( 800 amps ) . I was asking for an explanation of that . :smile:

BTW , I've seen the firerworks that an open CT can produce . Luckily I was about 30 feet away .

 

[weressl]

A previous post stated that an 800:5 CT might have 1 amp secondary current at full primary load ( 800 amps ) . I was asking for an explanation of that . :smile:

BTW , I've seen the firerworks that an open CT can produce . Luckily I was about 30 feet away .

I think what he meant to say is that there are CT's that have 800:1, in other words CT's with a 1A secondary rating.

Having said that, the 1A rating in the US is not a standard CT rating like the universally accepted 5A is/was. Rather manufacturers of formerly 'solid state' and now 'electronic' trip devices, either integral or separate, have developed proprietary current sensors with lover current rating that are more suitable to interface with electronic circuitry.

 

[djohns6]

I think what he meant to say is that there are CT's that have 800:1, in other words CT's with a 1A secondary rating.

Having said that, the 1A rating in the US is not a standard CT rating like the universally accepted 5A is/was. Rather manufacturers of formerly 'solid state' and now 'electronic' trip devices, either integral or separate, have developed proprietary current sensors with lover current rating that are more suitable to interface with electronic circuitry.

Ok , now I understand . I deal with CT's every day and was confused by the post . Thanks .

 

[dereckbc]

Not to over do the subject but current and voltage are inversely proportional. I have worked in the utility biz as a sub station engineer earlier in my career and a standard CT used in 345 KV substations is 600:1 . Ok what that means is the voltage ratio is 1:600 so if you were to open a CT on a active line, and it has been done, the voltage can go to 207 MV. I knew a few relay techs and substation switch operators that accidently opened them. Only one survived if that is what you want to call it.

 

[LarryFine]

Not to over do the subject but current and voltage are inversely proportional. I have worked in the utility biz as a sub station engineer earlier in my career and a standard CT used in 345 KV substations is 600:1 . Ok what that means is the voltage ratio is 1:600 . . .

Didn't I say that? :smile:

When a CT is operated with the secondary open, it behaves like a PT. A 400:5 CT has an 80:1 current ratio, but a 1:80 PT (ratio).

 

[roger]

Didn't I say that? :smile:

Oh, and like you have never posted something someone else had. :grin:

Roger

 

[weressl]

Not to over do the subject but current and voltage are inversely proportional. I have worked in the utility biz as a sub station engineer earlier in my career and a standard CT used in 345 KV substations is 600:1 . Ok what that means is the voltage ratio is 1:600 so if you were to open a CT on a active line, and it has been done, the voltage can go to 207 MV. I knew a few relay techs and substation switch operators that accidently opened them. Only one survived if that is what you want to call it.

Sorry, that voltage ratio is just plain false.

 

[winnie]

Actually, I believe that the voltage ratio is correct.

The error is the assumption applying the _full_ system voltage would ever be 'dropped' by the primary, which is the input of that voltage ratio.

Consider just the CT donut, with no secondary windings, sitting on a conductor. This is essentially a _large_ bead inductor. At any given current level, there will be a certain voltage drop across that inductor.

Now wrap your secondary coil around the donut. The maximum possible secondary voltage when open circuit will be the primary voltage drop times the turns ratio. Current flow in the secondary will _reduce_ this voltage drop.

-Jon

 

[beanland]

Are you sure?

Are you sure?

I have worked in the utility biz as a sub station engineer earlier in my career and a standard CT used in 345 KV substations is 600:1 . Ok what that means is the voltage ratio is 1:600 so if you were to open a CT on a active line, and it has been done, the voltage can go to 207 MV. I knew a few relay techs and substation switch operators that accidently opened them. Only one survived if that is what you want to call it.

I have worked in the utility business for 35 years as a substation engineer. An open circuited CT will rise to a voltage based on its saturation curve. Voltages of 800V to 1000V are common. Unless the CT faults to the 345kV line, which would blow it totally away, it cannot ever generate 207MV.

CTs in the USA are commonly 600:5, 1200:5, 2000:5, etc. A PT (potential transformer) may have a ratio of 600:1 (typical for 115-138kV lines) but that is a totally different issue. If you unground an energized PT, you will cause the "ground" to rise to line potential. Totally different situation.

I am concerned that a "moderator" can make this large an error.

 

[LarryFine]

Oh, and like you have never posted something someone else had. :grin:

Well . . . :roll:



This is different: it's me!

I am concerned that a "moderator" can make this large an error.

So I'm excused, since I'm not a member of The Mod Squad.


Added: But I have read that the voltage can rise well above the insulation damage point.