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Acceptable relay input 1

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farale

Electrical
Apr 7, 2018
36
Dear All,

Many digital relays accept 0 to 1 A in their inputs for functions 50, 51, 37, etc.
But what is maximum acceptable input value for a relay?
As you know, current transformers saturate in case of short circuit. So they limit input current of relays.
To select current transformer, we need to know maximum accepatble current value of relays. So it will be possible to select ALF'.
thanks
 
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Check the manual. Usually they accept a very high value for period a fault will be on a system. You For continuous current carry capability, most of the digital relays accept 15 A for 5A relays. SEL's 421 will take 1250 A for one cycle and 500 A for a second. You will never get close to that on any system with properly tapped CTs.

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If you can't explain it to a six year old, you don't understand it yourself.
 
Your most important parameter for what you're trying to achieve is your current transformer ratio. 1A or 5A and current rating of primary winding. Then comes your ALF and the rest.
 
What you need is load current, first. Size the CT based off that, first. That will put you into the 0-1A range, secondary. If you're dealing with IEC or European standards, they use 0-1A CT's, secondary. USA uses 0-5A CT's, secondary. As for saturation, that's a different issue. If you want the saturation voltage, Vs, in the U.S. standards (ANSI) it's the voltage, minus the drop across the secondary windings, at 20X rated secondary current that one would find on a CT excitation curve. That gives a 10% error and the CT is considered fully saturated for practical purposes. Alternatively, one could use the criteria that the saturation voltage must be below 1/2 the C rating voltage of the CT for the max symmetrical fault and calculate the Vs, given the max symmetrical fault and the CT burden. The ratings of the relay for 1250A, as HamburgerHelper pointed out, are for transient or withstand ratings of the relay. You don't really need those, because you're practically not going to reach those values for a sustained period.
 
Yo121,

CT's with 5A secondaries are as common as dirt over here, especially in older installations.
 
What FreddyNurk posted. The specifications of the relay will contain that information. Check the manual.
 
What is the model number of the relay?
Here is an example from an SEL-351 relay.
The 1A current inputs will take 3A continuous, 100A for 1 second, and 250A for 1 cycle.
Maybe someone here can explain what "Linear to 20 A symmetrical" means in this context.
sel_vlsxbg.png
 
The maximum current value on the D side of the ADC would be 20A. For higher currents it would just flat top.
 
davidbeach, that's what I would have expected as well, the difference between physically being able to withstand the current versus the means to measure the current. I'd imagine that above 20 x the rated current, there's probably not a lot of resolution needed to work out that you'd want it to trip.

EDMS Australia
 
JG2828,

I believe it has to do with the size of the registers used for measuring the currents. I don't think you are even allowed to set pickups above 20x amp rating in SEL relays. Most CTs at least on the transmission and distribution level are guaranteed no more than 10% error at 20 times rated current at rated burden for symmetrical currents. For example, a C800 CT is designed to handle 5 x 20A = 100A at 8 ohm burden with no more than 10% error. CTs can handle more than 20x symmetrical if the burden is less than rated but SEL probably just picked the size of the registers as what they are to use rated values. In my opinion, someone is using bad CT ratios if you are ever approach 20x rated current on the secondary.

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If you can't explain it to a six year old, you don't understand it yourself.
 
Thanks davidbeach, FreddyNurk, HamburgerHelper,
That makes sense. I wasn't thinking about the digital side of things.
 
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