How to fake a CT input

Discussion in 'General Electronics Chat' started by EB255GTX, Jun 23, 2016.

  1. EB255GTX

    Thread Starter Active Member

    Apr 30, 2011
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    Hi all,

    I think this is very simple but I have thought about it too much and confused myself....

    I need to test a project that has a CT input. For practicality reasons i would like to make a test fixture with knows that set the faked AC current.

    I thought this would be easy, just connect an AC/AC transformer type wall wart with a low voltage output to a potentiometer and connect the resulting variable AC voltage to the CT input, right?

    But then i started thinking about it ....There is a 1R burden resistor on the CT input, and the CT itself is a current source that drives a voltage across that resistor representative of the current.

    So if i connect a variable AC voltage in place of the CT, it has to be able to drive enough current to raise the voltage over that 1R resistor... but i wanted to use say 100k pots and have effectively tiny power used everywhere....but then again when the real CT is in place there is not significat power being dissipated - the burden resistor is a small SMD part!

    Where am i going off the rails on this thinking???
     
  2. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    Assuming CT expands to "Centre-Tapped" in this context (Granting that some elements of context allow for an expansion of "Current Transformer")

    Unless analysis of resistive divider behaviour in this regard is part of the exercise -- you may save yourself much 'agro' via use of a reactive (Spec inductive) divider:)

    Best regards
    HP

    PS
    Inasmuch as there seems to be a slight language barrier - a diagram would be most helpful!:)
     
  3. EB255GTX

    Thread Starter Active Member

    Apr 30, 2011
    54
    2
    thanks for the reply - "There is a 1R burden resistor on the CT input, and the CT itself is a current source that drives a voltage across that resistor representative of the current."

    it's definitely a Current Transformer :)

    I want to put a signal into points marked A and B that is adjustable with a potentiometer and looks like the CT signal.
     
    Last edited: Jun 23, 2016
  4. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Then you need an AC source capable of driving the one ohm resistor to the voltage level you require. I might suggest an audio amplifier IC may be ideal for this.
     
  5. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    Indeed!:)

    @EB255GTX -- Thanks for the clarification! -- FWIW I concur with @ErnieM --- It's merely a matter of impedance matching (q.v.):cool:

    Best regards
    HP
     
    ErnieM likes this.
  6. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    How much current are you expecting from the CT output winding?
    What is the frequency?
     
  7. EM Fields

    Member

    Jun 8, 2016
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    28
    Impedance matching doesn't matter as long as the source can drive the one ohm load to the desired voltage.
     
  8. EB255GTX

    Thread Starter Active Member

    Apr 30, 2011
    54
    2
    Thanks for the replies, after sleeping on it and reading the replies i really think my mind was making this more complicated than it is....

    the CT will push current though the burden to develop a voltage across it (about 70mA max, 50Hz). so i can fake the CT signal with a source that can do up to 70mA ..... for some reason i was thinking that a voltage source like a bench supply wouldn't work but if i set the voltage then control the current on the supply it should look to the CT input no different than the real thing :)
     
  9. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    --Emphasis added--

    ...Which, for the purposes at hand, is descriptive of a matched condition:)

    Very best regards
    HP:)
     
  10. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    Be advised that only the AC (i.e. time-varying) component of a 'signal' may be transformed -- hence pure DC is inapplicable to your circuit -- unless, of course, you are endeavoring to 'bias' a saturable core? (such being inconsistent with the attached schematic).

    'Voltage' and Current are not independently 'controllable' across and through a fixed impedance:confused:

    If, as it seems, your aim is simulation of a 'load condition' on the 'primary', please be advised that (Re: the transformer in question) the primary current will be much greater than the secondary current (as per Isec/Ipri = Npri/NSec) thus for a (conservatively) estimated turns ratio of 1:200, a secondary current of 70mA will require a primary current of 14 amps!:eek:

    Best regards
    HP:)
     
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