Connect with us

Current transformer

Discussion in 'Electronic Design' started by Tim Williams, Dec 1, 2011.

Scroll to continue with content
  1. Tim Williams

    Tim Williams Guest

    Would've been nice if I had known a current transformer isn't a very good
    transformer. Instead of two terminals, it acts more like a single
    terminal, with your poor burden resistor caught in the middle. Ah well,
    such is basic research...

    Tim
     
  2. What?
     
  3. Let me help you....
    Here's the Jive Translation:

    Would've been supa' fine if ah' had knode some current transfo'ma' isn't
    some very baaaad transfo'mer. Ah be baaad... Instead uh two terminals,
    it acts mo'e likes some sin'le terminal, wid yo' poo' burden resisto'
    caught in de middle. What it is, Mama! Ah sheeit, such be basic
    research...

    Cheers
     
  4. admformeto

    admformeto Guest

    All transformers are current transformers, see the Faraday's lows of mutual
    induction.
    In practice it can be configured in current, voltage or power mode depending
    on application.



    Mathew Orman

    http://www.faster-than-light.us/
     
  5. Tim Williams

    Tim Williams Guest

    News to Faraday, my transformer doesn't transform. At least, not very
    well. At least, not within the first, about a microsecond.

    Simple example: go buy a bog standard Triad CT206, it rings like a bell at
    a rather low frequency (a MHz or so)!

    In fact, I'm pretty sure it's acting as a toroidial resonator, which isn't
    much heard of, helical resonators are more common. Same idea, but with
    four quadrants.

    Tim
     
  6. Your using it as an air core transformer. Didn't Telsa invent something
    similar ;)

    Cheers
     
  7. Tim Williams

    Tim Williams Guest

    It sure feels like it, but I'm pretty sure the core is still mu_r >> 100
    at whatever resonant frequencies these things are doing it at. I'd call
    it a ferrite-loaded helico-toroidial resonator, or something ungainly like
    that.

    Since L is large, that means F is small (~MHz), and the impedance is
    high -- sadly, the Q is also high, so the impedance (ESR, since it's a
    series resonant equivalent) at the feedpoint (i.e. burden R) is very low.

    It's very interesting to rotate the "primary" around the core and watch
    the nodes in the standing wave. I should see about making an air-cored
    version, driving it with RF and putting it in a low pressure neon
    environment or something.

    Tim
     
  8. Tim Williams

    Tim Williams Guest

    Indeed, effective parallel parasitic capacitance is a valuable concept.
    Sadly, it's just that, a concept -- the *actual* capacitance from end to
    end of, say, a solenoidal coil (i.e., as more advanced modelers call it, a
    helical resonator) is dramatically smaller than the turn-to-turn
    capacitance.

    Consider, if instead of a helix, you had a stack of rings. It's the same
    basic structure, except skewed by a turn, so the turns aren't turns,
    they're loops. Now ground ALL the rings, except for just the two ends.
    What is the capacitance between those two rings?

    The capacitance will not only be small due to distance, but almost
    entirely shielded by the turns inbetween them. When you unground them,
    all the intervening turns have their own capacitance, but it still doesn't
    even act as an ideal capacitive divider, because there is finite
    propagation delay along the structure (i.e., the speed of light) and
    because the interspersed turns have a comparable loading all their own
    (self capacitance to free space as well as "mutual capacitance" to
    adjecent turns).

    The same is true of the toroid, with the added boundary condition that the
    magnetic field must be equal at both ends -- in the helical resonator,
    they can be equal or opposite, allowing (N + 1) / 2 wave resonances;
    toroids only allow integer N. Speaking of which, it stands to reason that
    the bandwidth of this resonance should correspond to the evenness of the
    winding; if the leads are not at exactly 0 and 360 degrees (give or take
    the external reactance between them), the wave can be skewed by that many
    degrees, across the unwound portion of the core.

    Tim
     
  9. Tim Williams

    Tim Williams Guest

    Actually Jim, that's exactly the problem.

    A high impedance load will work perfectly.

    Note the comment about series resonance...

    You've been spending too much time on ICs where the transmission lines are
    approximately the resistivity of nichrome and the inductivity of mud ;-)

    Tim

    P.S. Tonight I also enjoy "cooking" with wine ;-)
     
  10. John S

    John S Guest

    Or you could just measure it.
     
  11. Oppie

    Oppie Guest

    Consider yourself bitch slapped ( ;-P
     
  12. Joerg

    Joerg Guest

    There are a lot of parts that can transform themselves from
    multi-terminal parts into 1-terminal parts. Usually after some smoke has
    wafted off. In really bad cases they can turn into a 0-terminal part.
    Usually after a lot of smoke has wafted away, or after a loud bang.
     
  13. josephkk

    josephkk Guest

    Say what?

    ?-)
     
  14. Tim Williams

    Tim Williams Guest

    Oh, if you've got the right model and the system is reasonable (possibly,
    it must be non-dispersive), it even works all the way up to the first
    resonance. This is true of 1/4 and 1/2 wave resonators, in coax /
    stripline form, and still reasonable for microstrip (which is dispersive).
    Probably because the infinnitessimal circuit is a scaled down model of the
    same layout. Take a chunk of coax and it looks like series L and parallel
    C on almost any scale (limited by diameter, when higher modes take over);
    Z_o = sqrt(L/C) and F_res = 1 / (2*pi*sqrt(L*C)) are always true.

    I don't think it applies for helical and thin toroidial resonators,
    because the calculated effective parallel capacitance and inductance vary
    with frequency (going to extreme values at resonance, of course); and this
    is going to be a consequence of the infinnitessimal model, which involves
    mutual inductances and chains of capacitors, producing dispersive modes
    that are difficult to model in the same way.

    Tim
     
  15. Phil Allison

    Phil Allison Guest

    "vkj"
    ** For the simple reason that a "transformer" always converts voltages
    according to the turns ratio on the windings. When you feed an iron cored
    transformer prmary with a current, the resulting voltage drop is undefined
    and very non-linear.

    ** An inductor could be used to measure AC current (at a single frequency)
    as there is a simple ratio between current and voltage drop ( long as the
    inductive reactance is much bigger than the resistance). A secondary winding
    would enlarge the voltage to be measured - but that is no longer a *current
    transformer* at all and loses most of the advantages.

    The original idea of a current transformer was to allow safe and simple
    measurement of high AC supply currents with a standard moving iron ammeter.
    No rectifiers are needed, no high voltages involved and there is virtually
    no effect on the circuit being measured.


    .... Phil
     
  16. P E Schoen

    P E Schoen Guest

    "DJ Delorie" wrote in message
    Actually, the voltage would never get anywhere near that high, although it
    could get high enough to damage the insulation. For one thing, a 1000:1
    ratio is a 5000:5 CT (usually), which is fairly rare. Most systems run
    currents of 100 to 500 amps or so. If you do run current through a CT with
    an open secondary, it will just function as an inductor, and it will soon
    saturate at 60 Hz so the primary voltage drop will be only several volts.
    Thus the secondary will probably produce only several hundred to a thousand
    volts or so. However, this is often beyond the voltage rating of the control
    wiring and could cause breakdown of insulation, arcing, and other damage.
    And much higher spikes will be created when the power is switched on and
    off. This could be especially bad if the CT is in a PWM motor control line
    where the current may be switched at 20 kHz or so.
    Most CTs are shipped with a shorting jumper or spring clip on the secondary.
    And some are protected by a pair of high current anti-parallel diodes. But
    sometimes you may need more than 0.6V peak to drive the load, and since CTs
    sometimes operate at 2x to 5x nominal rating, that can be a lot of power to
    dissipate in the diodes.

    Paul
     
  17. Sylvia Else

    Sylvia Else Guest

    Only as long as the primary current doesn't drive the core into saturation.

    Sylvia.
     
  18. Phil Allison

    Phil Allison Guest

    "vkjerkoff"

    ** Then you are know nothing, fucking idiot.

    FFS go test one !!!!!!!!!!!!!

    Then PISS off to hell.
     
  19. Jasen Betts

    Jasen Betts Guest

    works fine with linear loads which draw sinusoidal current,
    doesn't work for phase control, bridge rectifiers feeding capacitors,
    or other active loads.
     
  20. josephkk

    josephkk Guest

    That is not all, maybe not even the most important. The flux in the core
    when operated properly in the CT mode is nearly zero, and if the secondary
    is open that is NOT true, and the core could easily saturate. BTW the
    relevant IEEE standards make a distinction between metering and
    protection/relaying CT and their performance properties in overload.

    ?-)
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-