current transformer no-load secondary voltage

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pnelspec

Member
Location
Australia
It only depends on CT's rated power (VAs). Theoretically, if CT is ideal, means unlimited power, and it able to push a secondary current towards an infinity impedance (no load on secondary), so in that case the voltage across secondary should be infinity as well.
In practical life, nothing is ideal. So, if your CT is small, let's say 5VA, it will probably fail to develop any dangerous voltages even when it's open.
BUT, if it is big, such as a few tens of VA, it can easy develop voltages of several kvolts.

How much exactly? You better not to try :).

Pol
 

pnelspec

Member
Location
Australia
For my estimation, at worse condition, it should be something like:
V = square root(maximum power*load impedance)
So, if the CT's power is 25VA and the load side impedance is 1MOhm=1000000 Ohm,
the maximum voltage is 5000V

Pol
 

gar

Senior Member
100311-0744 EST

electrics:

The maximum secondary open circuit voltage will greatly depend upon the saturation characteristics of the magnetic core and other factors of the magnetic circuit and the primary current. No easy way for you to calculate this voltage.

Also will depend upon the breakdown voltage, either internal or external.

At full load rating of the current transformer with your normal meter connected, if the maximum secondary voltage was 100 millivolts, then you could put a pair of back-to-back silicon diodes with adequate current capability across the current transformer secondary, and have the ability to remove the meter without excessive output voltage.

.
 

mivey

Senior Member
You could model it by having all of the impedances on the secondary side. Let rated current flow and a voltage will develope across the excitation shunt impedance. You would also need to account for transformer saturation.

There is no simple V = IZ formula.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
My theoretical mutterings:

In a normal transformer, the self inductance of the primary is high enough to limit current at the full primary voltage. This self inductance comes from the magnetic flux produced by the primary. Any secondary current acts to oppose the flux produced by the primary, forcing the primary to draw more current.

In a CT, the self inductance of the primary is very, very low. A CT can only work because we place a load in series with the primary circuit, and that load acts to limit current flow. In the ideal case, the actual voltage drop across the primary of the CT is a few mV.

The flux developed in the core of the CT is kept low because the current flowing on the secondary of the CT acts to oppose any flux developed by the primary.

The voltage developed in the secondary will always follow the standard transformer rule that the voltage developed in a wire by changing magnetic flux is equal to the rate change of the flux coupling that wire. Its just that in normal operation we arrange things so that the magnetic flux is quite low.

If you open circuit the secondary of the CT, then the core will almost certainly go into saturation, since you only have the primary current flow. If you know the core dimensions and the saturation flux density, then you will be able to estimate the total flux change per AC cycle, and from this you could estimate the 'average' voltage developed. Since the voltage developed is proportional to the rate of flux change, and we calculate the total flux change and the time period for that total flux change, we can calculate the average voltage.

However since the CT core is in saturation, we would expect that the flux will remain 'hard' at one polarity for most of the AC cycle, and then suddenly shift to the other polarity, with a very rapid transition. During this rapid transition, the rate change of flux is much higher than average, and the voltage developed in the secondary would be much higher than the average.

To answer the question, I guess what I would do is use the core characteristics to calculate the relationship between primary current and CT core flux (assuming an open secondary), use primary current to calculate core flux, and take the rate change of this calculated core flux to get secondary voltage.

-Jon
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
can anyone tell me how can we calculate the secondary no-load voltage of any current transformer?
How about a simple explanation?

Let's say we have a 400:5 CT, which is an 80:1 ratio, as a CT. With an open secondary, it acts more like a VT with a 1:80 ratio, so the theoretical potential is 80 times the line-to-neutral voltage.
 

jghrist

Senior Member
How about a simple explanation?

Let's say we have a 400:5 CT, which is an 80:1 ratio, as a CT. With an open secondary, it acts more like a VT with a 1:80 ratio, so the theoretical potential is 80 times the line-to-neutral voltage.

Except that you aren't putting line-to-neutral voltage across the primary, unless you are creating an intentional line-to-neutral fault through the CT.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Except that you aren't putting line-to-neutral voltage across the primary, unless you are creating an intentional line-to-neutral fault through the CT.
Agreed. It was a theoretical question, and that's the max I could figure.

What would you say is a realistic answer?

The concern to the CT itself is internal insulation damage, of course.
 
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