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Need Information about a Current Transformer

Discussion in 'Electronic Design' started by [email protected], Jan 9, 2006.

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  1. Guest

    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 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
     
  2. bruce varley

    bruce varley Guest

    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 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.
     
  3. scada

    scada Guest

    Most industry CT's are 5A secondary.
     
  4. Guest

    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
     
  5. DaveM

    DaveM Guest



    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
    the address)

    Never take a laxative and a sleeping pill at the same time!!
     
  6. scada

    scada Guest

    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.
     
  7. DaveM

    DaveM Guest

    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
    the address)

    Never take a laxative and a sleeping pill at the same time!!
     
  8. scada

    scada Guest

    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...
     
  9. ehsjr

    ehsjr Guest


    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
     
  10. scada

    scada Guest

    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...
     
  11. Glen Walpert

    Glen Walpert Guest

    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 :).
     
  12. scada

    scada Guest

    Exactly...
     
  13. ehsjr

    ehsjr Guest

    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
     
  14. scada

    scada Guest

    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...
     
  15. budgie

    budgie Guest

    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.
     
  16. ehsjr

    ehsjr Guest

    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

    <snip>
     
  17. scada

    scada Guest

    Remove the burden so that the secondary has no load, and you'll
    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.
     
  18. Rich Grise

    Rich Grise Guest

    No, the term "transformer" does not include "burden resistor", but that
    doesn't prevent anyone from packaging a resistor with their CT and selling
    int as a "CT with a built-in burden resistor included", albeit, with what
    little reading I've been done, that would limit the range of applications
    for a given part, and a cursory google search hasn't turned up any yet,
    but I haven't looked at page 2 of the results yet.

    Thanks,
    Rich
     
  19. Rich Grise

    Rich Grise Guest

    Well, just to prove a point, I finally found at least one: (two, actually)
    http://www.archenergy.com/products/mdl/applnote_ct_101002.pdf

    :)

    Cheers!
    Rich
     
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