Need Information about a Current Transformer

Hi,

I'm trying to find out information about a current transformer. Like what is the primary current and what is the secondary out. The CT is made of black plastic and is 2" round and =BE" thick and has a hole in it =BD" round. It has two wires for the secondary, #18 gage, one purple and one orange. On the label is the info below.

CAT # CST-2845 100 amp FERMITER Corp. 8-85 CLASS 2, (I think the 8-85 is the date made). INSUL. CLASS 105 INDOOR TYPE 1.

Thanks

Reply to
slc
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Most industry CT's are 5A secondary.

Reply to
scada

I took the CT apart and there is a 1 ohm resistor (about =BC watt) across the secondary. If I have 6 amps on the primary and I connect a volt meter across the secondary. What should the voltage be?

Reply to
slc

I'm trying to find out information about a current transformer. Like what is the primary current and what is the secondary out. The CT is made of black plastic and is 2" round and ¾" thick and has a hole in it ½" round. It has two wires for the secondary, #18 gage, one purple and one orange. On the label is the info below.

CAT # CST-2845 100 amp FERMITER Corp. 8-85 CLASS 2, (I think the 8-85 is the date made). INSUL. CLASS 105 INDOOR TYPE 1.

Thanks

Well, it's 100 amps to something. There are a number of 'standards' for secondary current, if the secondary connections are #18, then I think 1 amp is probably a reasonable guess. I suggest you test the thing by running an approximately known AC current to it and measuring the current in the secondary. That will confirm what the ratio is, it's almost certain to be something like 100A/1A, 100A / 0.5A or something like that.

You can generate the current by simply winding a few turns around the core, and using a lamp of known wattage. The 'equivalent current' is the actual current in your primary winding times the number of turns, if you simply run the wire through the centre, that is one turn. Important: Ensure that the secondary is not allowed to go open circuit, this can generate dangerously high voltages! On the contrary, these transformers are most happy with their secondary short circuited (minimum 'burden'), so your test is best conducted with a meter set to amps straight across the ssecondary leads, with no series resistor. Alternatively, if you have a LOW VALUE resistor you can use this and measure tne voltage across it due to the secondary current.

Class 2 refers to accuracy, 2% IIRC.

For a CT of that size, you're probably talking of a maximum burden of about

3VA, that means - for example - that if the ratio turns out to be 100A / 1A, then the maximum resistance you can have across the secondary if you're expecting 100A is 3 ohms, which will give a secondary voltage of 3 volts maximum.

Once again, to stress... never let the secondary go open circuit, and ensure that the secondary load is physically solid and securely terminated. An OC secondary on a CT can generate hundreds of volts, enough to kill.

Reply to
bruce varley

It depends on the turns ratio.

Don't you have a voltmeter with an "AC Volts" range? Just measure it! If it's a 10:1 CT, with your 6A through the primary hole, then there will be .6A flowing in the secondary, which, with the 1 ohm resistor, would drop .6 V, which is .6 * .6 = .36 watts, but since it's a 1/4 watt resistor, it's probably not 10:1. Probably a lot closer to 1000:1 - that would give you .006A, or 6 mA, with 6A through the primary, which would drop 6 mV across your 1 ohm resistor. So it's very probably something in between those two - time to get out the meter!

Good Luck! Rich

Reply to
Rich Grise

Certainly most bus-bar sizes tended to be 5A, but there are a lot of 1A units out there too. I haven't seen anything below 1A sec for a 100A primary.

Reply to
budgie

Ok, I put a 6 amp load thought the primary and measureed the secondary. The meter was set on "AC volts" and it read .020 So what would the rateing be, 100 amps to what?

Thnaks Kevin

Rich Grise wrote:

Reply to
slc

Since the burden resistor is 1 ohm, the current ratio would be 6 / .020 =

300:1 So, if you put 100A through a single turn primary, you would expect to read 0.3333 volts on the secondary. The current through the 1 ohm burden is 0.333A, and the burden dissipation is 0.1111W.

Hth

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 
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Reply to
DaveM

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I don't think the 1 ohm resistor is the actual burden. I believe it is there for protection, should you run the CT open circuited. Try the test with a shorted CT secondary using a clamp on ammeter, or a wired AC Ammeter directly connected to the secondary.

Reply to
scada

No, the 1-ohm resistor is the burden. It's across the secondary. CT secondary protection would likely be in the form of a series opposed pair of zeners. A CT secondary is not normally short-circuited, although it could be. The CT's used in power metering equipment are usually run directly into a 5A AC current meter, but even those are not a direct short circuit to the CT. Those types of CT are designated for 5A current meter connection... and the ratio will be expressed as something like 1000:5, meaning a primary current of 1000A will produce a 5A secondary current when connected to a 5A current meter. Cheers!!!

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 
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Reply to
DaveM

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If the 1 ohm is the burdon, then when a secondary load (Ammeter, relay, etc...) is connected, then some of the current would flow through the resistor and some through the load, wouldn't it! Or am I wrong in my thinking? Or does one have to remove the resistor when the CT is put into service? I never worked with a CT that could be energized without a load connected to it's secondary.

Thanks...

Reply to
scada

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CT secondaries drive current devices, not to measure potential. A PT relay is used to monitor potential (voltage). A relay wired in series with a CT's secondary is a "Current Relay" and is of extremely low impeadance! That impeadance must be within the design of the CT burden. Such relays are used typically as overcurrent protection for equipment.

Thanks...

Reply to
scada

A CT with *nothing* connected to its output other than the burden resistor would be almost useless. (I suppose you could measure the temperature in the burden and do something with that.) So, for it to be included in a circuit and serve a useful purpose, there would have to be some use of the current on the secondary, other than through the burden.

In the case we are discussing, connect an 11 megohm volt meter across the CT secondary. It will introduce an undetectable error - the meter won't be able to display a difference that small. (Do the math - 11000000/11000001 )

Or get a worse indicator - say a 1 mA meter movement with 87 ohms resistance. It will introduce an error of less that

2 percent, indicating 296.59 amps when 300 amps is the primary current.

If the CT is used to switch a relay, there would be a high impedance connected to the CT, so the additional load wouldn't matter.

Ed

Reply to
ehsjr

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Exactly...

Reply to
scada

In power distribution applications CTs are current in, current out devices which *never* have a built in burden resistor, so that they can deliver current to any number of loads in series within the CT burden rating. CTs designed to be incorporated into electronic equipment *always* have a built in burden resistor, they are current in, voltage out devices suitable for driving high impedance loads only. Woe to he who confuses the two :-).

Reply to
Glen Walpert

I said, "in the case we are discussing". That's a CT with a built in one ohm burden. (And it has a 300:1 ratio, with 6 amps on the primary.) What extremely low impedance relay are you driving directly with that? Post a reference to the relay, please.

And for your consideration, CT are certainly used in circuits to develop a potential which represents primary current and is measured. They convert a current in the primary to a voltage across the burden in the secondary. Note that that is not their only use. But to indicate that they are not used in that manner is just plain wrong.

Ed

Reply to
ehsjr

CT's convert primary currents to secondary currents, never to secondary voltages! Internally to a device connected to this secondary, often an electronic current relay, can be a resistor which a potential is measured from - the Burden. I have never seen a Burden included within a CT. Such a device would not be a CT, but rather a "Current to Potential" device of sorts.

I suspect in this example, the actual Burden is well below the 1-ohm resistance (very common), therefore that 1-ohm is negligible. The only true way to test the CT ratio in question would be to use it in a test circuit of known primary current, and measure the secondary current with an ammeter.

Thanks...

Reply to
scada

That is correct, no argument. That's why removeable-case current relays have internal shorting switches.

That's the question! A CT with a built in burden? A CT outputs a current, period! That current often goes to a series connected string of ammeters, protection relays, load measuring devices, etc... How can that be with the purposed scenario of an internal Burden resistor? I simply suggest that this device is not a "CT" as known, perhaps some special device. I can only assume the resistor is there for protection should the CT be energized with no secondary load, it could clamp the secondary voltage to a safe value.

I won't argue the point further, and I don't agree with theory that the CT has an internal Burden.. However I welcome and defend your right to disagree.

Reply to
scada

What "secondary voltages"?? Read it again. You do not seem to realize that current in the secondary creates a potential across the burden to which it is connected. You are arguing with a point that was not made. Nevertheless, let me make it: Remove the burden so that the secondary has no load, and you'll have a very high voltage across the secondary when the primary is carrying appreciable current. It is precisely because a CT

*can* convert a primary current into a secondary voltage that a burden is required.

Setting that aside - the discussion is of a CT with a built-in

1 ohm burden. Such a CT produces a voltage across the burden. That voltage is available to the "outside world" via the output terminals of the CT, and is directly proportional to primary current. It can be used by the attached electronics in various ways, including measurements and driving a relay through some high input impedance circuit element, like a comparator. The OP told us that, with a voltmeter attached to it, and 6 amps in the primary, he measured .02 volts at the output terminals.

You seem to think that the CT being discussed can drive a relay directly. I would like to see a reference to that relay - can you provide one? You also mentioned connecting an ammeter across it - what would you expect the ammeter to display?

Ed

Reply to
ehsjr

with the appropriate correction for the presence of the 1 ohm shunt. While the ammeter resistance may be way below 1 ohm, the resultant ratio calculation will be off. Furthermore (not directed at you) the O/P's application of the ratio info will similarly need to take into account any fixed secondary shunt impedance.

Reply to
budgie

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