Voltage measurement on a Current Transformer (CT)

[kenneth edmiston]

TL;DR

Coworker measured the output voltage of a zero sequence CT for a ground fault relay. This seems wrong.

I don’t know the ratio of the CT. I’m just looking for an explanation on this. I thought in order to measure output voltage you need to pass the current through a resistor and read the voltage drop.

A co worker put 400a through the CT and measured 130v. Something just seems off to me. Can you just hook up your DMM and test the output voltage? The intent was just to see if the CT was operating, not for a quantitative measurement.

But this seems dangerous. A modern voltmeter has an input impedance of around a megohm or maybe 10. Seems like the CT output voltage could get away from you quickly.

Any insight will help. Thanks!

 

[berkeman]

The CTs I work with (much lower currents, like 65A max) often have the "burden resistor" internal to the CT, so they present an output voltage to a high-impedance measuring device. The specs on the one I'm working with now are 333.33mVrms output into a minimum of 5k Ohms for a 5A CT current in the primary wire.

Can you post a link to the datasheet for that CT? Does it say anything about an internal burden resistor?

 

[berkeman]

Is it this kind of system?

https://zone4engineer.com/public/album_photo/4c/08/708a0e3b1be0670b10eb3fa0e01c30b8.PNG

 

[kenneth edmiston]

Yes. The manufacturer literature says to short “c” and “m” to prevent shock. It’s a 3 wire zero sequence CT for ground fault protection. The current from the CT energizes the trip coil. I’m just having a hard time reporting his 130v @ 400a to the customer because it seems wrong. The CT is a GE ground-break CAT: TGS0719 MOD 1

 

[Baluncore]

A Current Transformer should never be operated without a burden that will limit the voltage. The output of a CT is a current, not a voltage. The resistive burden converts the current to a voltage.

What burden was used during the 400 amp test that developed 130 volt ? Maybe there was an internal gas discharge limiter that prevented breakdown of the CT insulation.

 

[zoki85]

What's the turns ratio or the conversion ratio of your CT?

 

[kenneth edmiston]

http://www.electricalpartmanuals.co... Electric/Ground Fault Systems/GEI-86126H.pdf

 

[kenneth edmiston]

That is all I can find so far. I’m having very little luck getting any real information.

 

[zoki85]

In the photo one can find some data: "max continuous net primary current 3000 A.
That should be rated primary current of the CT as well.
In the installation instruction pdf the relay output rating is given as 5 A.
That should be CT rated secondary current as well.
Accordingly the CT ratio should be 3000:5 A

 

[Baluncore]

Accordingly the CT ratio should be 3000:5 A

The three phase current may be 3000 amp, but the trip current will probably be much less than that because it is the Earth leakage, not the continuous balanced current.

 

[zoki85]

The three phase current may be 3000 amp, but the trip current will probably be much less than that because it is the Earth leakage, not the continuous balanced current.

3000 A figure is for the ground break sensor and not very reliable as concerns CT.
Not so sure the trip current will be much less if the ground fault occurs in star connected configuration having neutral connected to ground. Tripping range is adjustable anyway and in time-current curve (Fig 4) continuous relay current is rather 1 A and not 5 A as I misread. CT's secondaries are usually rated for 1 A or 5 A

 

[kenneth edmiston]

Since there appears to be no burden resistor internally, and a digital voltmeter will have a very high input impeadance , I would say there is no way to get an accurate measurement of the CT output by measuring voltage on the secondary only. Current and voltage may not be proptional. It will however tell you that the CT is in fact operating, but it would be much better to measure output current. Since it’s a zero sequence and the only net mag field that should exist is an unbalanced current due to ground fault, the net primary current rating should be rated for a 3000a unbalanced current. There is also a separate neutral CT so the ZS will also pick up single phase-2 wire circuits. Output at 3000a should be 1 or 5 amps (unclear) so at 400 amps would be either .13a or .66a respectively. Does this sound correct? Thank you guys

 

[zoki85]

There must be a burden. If the sensor (ie. CT) is connected to relay, than it is the integral part of relay impedance after terminals "1" and "2" (see the schematics). Max burden (measured in VA) for relay class CT is actually expressed as the max secondary voltage if 20x CT rated current flows through the secondary. The CT accuracy class is compromised of 3 number-leter combination and can be found on the nameplate of CT. For instance, 3C100 means that CT is accurate within 3%, is of low leakage flux type, and if secondary rated 5A is passed by 100A, the secondary voltage must not exceed 100V (therefore max burden is 1Ω ). I don't why your "coworker" measured secondary volts instead of amps. Maybe he was confused by this. It shouldn't be difficult to hook up ammeter on the secondary side, disconnect the sensor from the relay, and while passing decent current through one primary conductor in window, take some measurements. You don't even to disconnect it but use good clamp-meter to give figures for orientation. And I'm not sure what the problem is really about when detailed testing procedure of the protective system with monitor panel is given in the pdf. You don't have a complete system or there is some problem with it?

 

[kenneth edmiston]

Zoki - The only problem is that there are holes in my understanding. The question was more about helping me learn the relationship between E and I in a CT. The coworker and job really isn’t relevant. The GF system is malfunctioning, he checked the CT voltage to verify the condition of the CT. Just seemed odd. I would’ve just measured the current. But It doesn’t seem like it should be that confusing if a CT is just another transformer. You don’t induce current, you induce EMF so the secondary current should be dictated by the resistor and induced voltage. (Turns, impedance, flux etc) Refering to it as an “impedance transformer” also throws me off and I’d like to understand why. Reading online is helpful, but I like to get opinions from others. People on here seem to have a handle on things. The more the better. Thanks for the help

 

[Baluncore]

You don’t induce current, you induce EMF so the secondary current should be dictated by the resistor and induced voltage.

Wrong. A current transformer transforms current to current.
You must have a resistor to convert the output current to a voltage.

The CT is not connected across the voltage like a normal transformer. Primary current is set externally, voltage across primary is unknown. Power in = power out. Secondary current is primary current reduced by CT ratio.

 

[kenneth edmiston]

Baluncore — never looked at it that way. That correction will help.

The difference in a power transformer and a current transformer being the voltage on the primary, there is no transformed voltage in a ct. But a voltage must still be present across the secondary for a current to flow.There is a piece I’m missing. So in the primary, the ct will pick up the magnetic effect of the current but has no known voltage. But the primary still has a power value. The primary current is transformed in the secondary winding, and the voltage will be a function of the current and impeadance of the circuit to maintain that power value? What dictates the voltage across the winding when secondary is open and impeadace is very high? Can’t be I2R because the voltage jumps so high. My brain is twisted, please bear with me.

 

[zoki85]

Zoki - The only problem is that there are holes in my understanding. The question was more about helping me learn the relationship between E and I in a CT. The coworker and job really isn’t relevant. The GF system is malfunctioning, he checked the CT voltage to verify the condition of the CT. Just seemed odd. I would’ve just measured the current. But It doesn’t seem like it should be that confusing if a CT is just another transformer. You don’t induce current, you induce EMF so the secondary current should be dictated by the resistor and induced voltage. (Turns, impedance, flux etc) Refering to it as an “impedance transformer” also throws me off and I’d like to understand why. Reading online is helpful, but I like to get opinions from others. People on here seem to have a handle on things. The more the better. Thanks for the help

All transformers, based on transformer action via magnetic link, work on the same principle: variable magnetic flux, traversing simultaneously primary and secondary windings induce corresponding EMFs.
Now, the obtained EMFs can be used in different ways in regard how the transformer terminals are connected to the power source and load. Normally, for this reason potential transformers and current transformers differ in constructive details. Still they are two sides of the same medal (sort to speak)
. I think it would be a waste of time elaborating on transformer theory and writing down equations when you can find a lot of relevant materials and video tutorials on the internet. Regarding your statement that "GF system is malfunctioning", my feeling, based on experience, that problem is somewhere else. It's not likely CT became defective. The relay module box can frequently be the problem. Check the module inputs and outputs independently of the sensor signal.

 

[Baluncore]

The primary current is transformed in the secondary winding, and the voltage will be a function of the current and impeadance of the circuit to maintain that power value?

The input current is known. The current ratio of the transformer decides lower secondary current. The product of secondary current by burden resistance determines voltage on the secondary. That secondary voltage is transformed (or reflected) back to the primary input as a lower input voltage.

What dictates the voltage across the winding when secondary is open and impeadace is very high? Can’t be I2R because the voltage jumps so high. My brain is twisted, please bear with me.

With an open circuit, or high impedance burden, the secondary voltage is set by the insulation breakdown voltage of the secondary.

I suspect the 130 volts measured was the breakdown voltage of a protective device such as an internal varistor, or of an inert gas discharge device.