Current Transformer

[timm333]

On transformer primary/secondary feeders and motor feeder: should the primary of the CT be sized for 100-125% of the full load current of the feeder, or should it be more than 125% of the full load current. Thanks.

 

[templdl]

On transformer primary/secondary feeders and motor feeder: should the primary of the CT be sized for 100-125% of the full load current of the feeder, or should it be more than 125% of the full load current. Thanks.

You use the term 'current transformer but yet you post implies a control power transformer. But then again you are feeding a motor?

 

[timm333]

No I did not mean to say control power transformer. I meant to say current transformer for feeding over current relays and motor relays.

 

[templdl]

No I did not mean to say control power transformer. I meant to say current transformer for feeding over current relays and motor relays.

Oh, thanks.
I would target a standard CT. Remember that they come in standard ratios with a 5a secondary It doesn't appear as though you don't need metering accuracy just relay accuracy. Then select a primary value that would provide the secondary current that you are looking for to supply your relay I believe the relay should be considered first. The 125% should not be a consideration from what my experience is.

 

[JoeStillman]

A current transformer is a whole different animal from a power transformer. You don't size them for 125% of the load. The limiting factor is accuracy, not current carrying ability. Most commonly, they maintain their accuracy up to 20X nominal primary current. Choose a CT whose primary rating is closest to the expected full-load current. For a 100A load, you can safely use a 100:5 CT. You don't go up to a 125:5 just because its continuoous load.

An imprtant factor to consider is the available short-circuit current. It is possible to saturate the CT on short circuit, which means the primary:secondary ratio is no longer valid at high primary currents. If the available fault current is over 20X FLA, you may need to go up to the next higher class of CT, i.e. C100 to C200. Here is a good paper on choosing protective CT's.

 

[kingpb]

There are many factors in proper selection of CTs. It is very common to use ratios that are 1.5 - 2.5 times the rated full load current to avoid saturation of the CT. However depending on the application, saturation may not be a factor.

 

[GoldDigger]

There are many factors in proper selection of CTs. It is very common to use ratios that are 1.5 - 2.5 times the rated full load current to avoid saturation of the CT. However depending on the application, saturation may not be a factor.

As long as the secondary side is loaded heavily enough there will not be core saturation.
Core saturation of the unloaded CT at normal currents is actually desirable to prevent dangerous high voltages from being generated at the output.
At some point the secondary wire resistance itself will limit the secondary current to the point that core saturation might occur, but that would be with the CT size off by an order of magnitude or more.

 

[JoeStillman]

As long as the secondary side is loaded heavily enough there will not be core saturation.

Did you mean "not loaded too heavily"? Saturation occurs when secondary output voltage is too high, and output voltage increases with CT burden.

(Note to OP: the load on a CT is called a "burden" and not a "load" just to distinguish it from the way we think about power transformers.)

 

[GoldDigger]

Did you mean "not loaded too heavily"? Saturation occurs when secondary output voltage is too high, and output voltage increases with CT burden.

(Note to OP: the load on a CT is called a "burden" and not a "load" just to distinguish it from the way we think about power transformers.)

My impression is that as the resistance across the secondary is increased, the voltage rises. If it rises high enough, the current in the secondary will decrease and allow the core to saturate. I was referring to that when I said not loaded too lightly. In your terms, a CT requires the burden resistance to below a maximum value to work properly.
Short circuiting the secondary would be fine as long as you were still able to measure the secondary current accurately. But that would be a lower burden and higher "load" right?

I agree that the terminology has the potential to be confusing, specifically increased burden would be higher or lower resistance??
It think that we agree on what is happening, but are describing it in different ways.
To be totally unambiguous, one could just refer to resistance and have the meaning clear to everyone. (maybe)

 

[JoeStillman]

My impression is that as the resistance across the secondary is increased, the voltage rises. If it rises high enough, the current in the secondary will decrease and allow the core to saturate. I was referring to that when I said not loaded too lightly. In your terms, a CT requires the burden resistance to below a maximum value to work properly. )

I agree, sort of; the total secondary current never decreases. As secondary voltage increases, more current is wasted on excitingthe core, until the core is saturated and all the current goes to magnetization. After that, increasing primary current causes little or no increase in output voltage, although the total secondary current is always in proportion to the turns ratio. This is the area to the right of the knee on a CT excitation curve.

Short circuiting the secondary would be fine as long as you were still able to measure the secondary current accurately. But that would be a lower burden and higher "load" right?

A CT with a shorted output has zero external burden. The term "load" is not meaningful for a CT. The total secondary current (excitation plus output) is always proportional to the turns ratio. As long as the CT is not saturated, the output current is not proportional to the external resistance (burden). This is where our intuition about power transformers fails us. You would expect lower resistance to mean higher current, but it doesn't - its all about the turns ratio. A CT is a current source and a power transformer is a voltage source.

I agree that the terminology has the potential to be confusing, specifically increased burden would be higher or lower resistance??
It think that we agree on what is happening, but are describing it in different ways.
To be totally unambiguous, one could just refer to resistance and have the meaning clear to everyone. (maybe)

Increased burden would mean increased resistance. The total burden is (internal impedance)+(cable impedance)+(instrument impedance.) I have seen relay burdens given in VA or ohms. The "standard burdens" used to define protection class CT performance are given in ohms (resistance), milli-henries (inductance), ohms (total impedance) and VA (total power at 5 Amps). You can see these values in IEEE C57.13 or page 52 of the Buff Book. IEC classifies CT's a little differnetly. Here is a paper on that subject.

 

[GoldDigger]

I agree, sort of; the total secondary current never decreases. As secondary voltage increases, more current is wasted on excitingthe core, until the core is saturated and all the current goes to magnetization. After that, increasing primary current causes little or no increase in output voltage, although the total secondary current is always in proportion to the turns ratio. This is the area to the right of the knee on a CT excitation curve.

I have to disagree with you there. If you have an open circuit on the secondary, the secondary current will certainly no longer be proportional to the primary current divided by the turns ratio. And at some finite value of the burden the current will have to stop being proportional.
At some point in the process of increasing the burden the CT stops acting like a simple CT and starts looking like a saturated voltage transformer.
The transition from simple power/voltage transformer math to simple CT math goes through a region of complexity where neither simple approximation applies, although the formal solution is not too horrendous now that we have computers to plot the graphs.

Once you get past the point of core saturation, the remaining coupling between primary and secondary is simply that of two coils in air, which is pretty inefficient.

The saturation point of the core needs to be controlled to keep from developing excessive voltage on an open secondary. So you are correct that at some point increased primary current will cause little or no increase in output voltage.

 

[SG-1]

The only other consideration that has not been mentioned is that the CT should be sized to not satuturate during a fault, since your protective relays are depending on them.

 

[GoldDigger]

The only other consideration that has not been mentioned is that the CT should be sized to not satuturate during a fault, since your protective relays are depending on them.

Very important!
More accurately, not saturate until the secondary current is high enough to trip the protective relays. At some point the core will saturate during a bolted fault (and some other OCPD will handle that), but hopefully not under normal LRA or even normal starting current. With a safety factor for the trip characteristics.
For the OP, that will probably be a major consideration, and he needs to know the characteristics of the instrumentation that will be looking at the CT output and what it will be required to do.

 

[templdl]

Timm333,
I see you haven't chimed in here at all. Does any of the replies answers you question or apply to your OP?