CT Ratio

The CT should be chosen so that its maximum amperage is about 2 times the motor full load current.

I don't know what your voltage is, but as an example, a 200KW motor on 480V would draw about 322A. For this, I would probably use a 600:5 CT. There is some leeway; 500:5 or 750:5 might also be acceptable.

CT primary should have atleast 20% over load capacity, to avoid the CT SATURATION.

How does that contradict post 1?

wrong!

Easiest way to solve your requirement is to email Bill @ <email removed>

He will set you up with what you need to monitor your amps. Tell him Jim sent you and he will take care of you.

Jim

CR4 Admin - email address removed

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The first information you need is the motor's FLA (Full Load Amperes) to size your CT. As suggested the CT's input Amperes should be at 120% of FLA. The input amperes of a CT is the large number of the CT ratio ie; 600:5, 600 is the input amperes to the load, 5 is the output amperes to the meter or overload protection. Your FLA will be on the motor name plate, or should be in the motor's specification paperwork. If the FLA can't be found, then it can be approximated from motor KW and Voltage.

It would have been a good idea to search CR4 for your answer, as this is the answer I gave to an almost identical question in May 2011.. but you didn't so here it is again!

What is the max amps/load in the circuit?

It is standard practice to have a meter or amp chart recorder working at the mid point, that is working on 2.5amps into the meter/recorder and the scale would be to suit, if the FSD is 5amps. For example;

Load = 50 amps. (substitute your motor running load here)

CT = 200:5. (this CT gives you a multiplication factor of 40.. 200 divided by 5 = 40 remember this number)

Amps(load) divided by CT ratio multiplied by No turns thru the CT

50amps/(200/5 =40) x No of turns

50/40x1 = 1.25 amps at the meter.

Now add another turn... (number of turns refers to the number of times the load carrying cable/conductor passes through the CT)

50/40x2 = 2.50 amps at the meter.

Every thing you said it true, but you just confused the hell out of the OP :~)

Posts #1 and #2 are correct particularly for the primary side. It is important to account for the burden or load the output / secondary of the CT will be driving. Many CT's, especially those rated at 5A output will only handle a 0.5Ω load. That includes the wiring and connectors! So if you meeter is a distance from the CT you must use an AWG ≤ 12 to connect your 5A output CT.

Your load may be expecting 1Arms max.

Maybe you could post a bit more information on what the CT is driving? The signal preamp boards in switch control systems used 1000:1 at the input to read 5A CT signals and then terminated that into small value resistor read by low voltage electronics.

Tonado is correct on all counts not just on the primary side, his statement saying that a CT 600:5 is exact as if the RC is 322amps then the CT output will be 2.68amp.

I don't follow what you are saying...

It is important to account for the burden or load the output / secondary of the CT will be driving. Many CT's, especially those rated at 5A output will only handle a 0.5ohm load. That includes the wiring and connectors! So if you meeter is a distance from the CT you must use an AWG ±12 to connect your 5A output CT.

I've never seen a CT in situ.... miles away from its signal receiver.. have you?

It is common practice to place the CT near, or as near as one can to a motor controller or the control unit that needs the CT output, yes?

OK if the CT is not mounted directly to your electronics use low AWG wire to connect it. If you terminate your 5A CT with a 1Ω resistor and it is rated to drive no more than 0.5Ω you will saturate your CT and have some what less than accurate measurements.

Reg, I am STILL not sure of what you are trying to say!

surely this would be ANOTHER topic, and not totally relevant to the original question where the OP only asked for a formula to calculate the correct size of CT.

I have NEVER seen a CT mounted/connect further than 4 feet away from the item of equipment that requires the CT output..... are you referring to what one should do if that distance is longer??

I'm a little surprised that no one has been specific to point out that CT's for metering are not the same as CT's for protection, so it is equally important to select the correct type depending upon what the OP requires them to do

the difference is????

I assume that you do know the difference, but for anyone that does not know, the main differences between metering & protection CT's are accuracy and saturation level

Metering CT's are normally more accurate, but saturate shortly after their rating, however protection CT's can give linear outputs up to 20X their rating.

Using protection CT's to drive meters can lead to damage during short circuits, since the CT's can drive very high currents into the meter, So I would say that the correct selection of CT type is equally as important as the selection of rating, to ensure the OP gets a system that works as expected.

Soeb, my apologies, I did miss your point.

I asked what is the difference to give you the opportunity to explain in full to the OP, which is what you should have done in the first instance.

It is a good point to remember that some posts are from folks for who English is NOT their first language.. his name.. ikram, is a bit of a give away there, so he would either not understand your reply or just accept it, which is worse, as you've not given a FULL answer, more so a statement without explanation.

Yes, I do know the difference, and thanks for reiterating for me.

So please don't think I was "having a go", I just wanted you to fully explain your statement to the OP.

have you missed reading post # 2 Craig?

Birch,

In my previous job I was designing distributed control systems for the electric utility market. The automated switches at the top of the power poles have CT's and the control systems are typically > 20ft away!

The burden on the CT is critical in the design of the CT but since I was stuck with the CT provided the way it is connected and loaded is CRITICAL.

Gentlemen on and all.. I do believe the point of the Original Poster's question is to provide him with the formula to calculate the size of CT, not offer advice on HOW he should use the CT.

Actually Brich, the question was "Can anybody tell me that how to choose CT for a specific load?"

To choose the correct CT it is critical to know what the CT is driving.

the Op said....

Can anybody tell me that how to choose CT for a specific load? He then continued by stating that his specific load is... I have a 200kw induction motor, which ratio CT should be installed?. Then he asked... Is there a formula or calculation that determines the CT ratio for a load?

So please don't take part of the question and answer it.. answer it all.

For me he wants a CT for a specific load of a motor rated at 200Kw and how does he size the thing correctly.. do you agree or disagree with that??

I see no reference to what he is using it for.. so lets not second guess what he wants and KISS.

Did you not say in an earlier post that in your opinion I had confused the OP with my answer.. that being the case, then think how he feels now and his state of confusion after your replies. IMHO!