How to calculate primary winding current

Discussion in 'General Electronics Chat' started by JStitzlein, Jun 12, 2011.

  1. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    I'm trying to calculate the primary current for the last example in the first section of transformers(worksheet). The equation I normally use works for both primary and secondary currents if the primary voltage and current are in phase.

    Ip = Np^2 * Vp
    ------------
    Ns^2 * Rload

    This equation works to give the secondary current in a situation where the primary current shift is from -90 to 0, but the primary is not accurate.

    How do i go about calculating the primary current?

    [​IMG]
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Are you ignoring the magnetizing current? In your model, this is the current drawn by the 1H primary inductance, which has only 377Ω reactance at 60Hz. That's very significant compared to your 1000Ω load.

    I also notice that the primary current seems to have a DC component. Unless your AC source is offset, this may be due to a transient coming from the starting value of the AC at T=0. If you increase the resistance in series with the primary to say 1Ω you may find that you can see the DC component decaying over say 100 cycles.
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    The phase difference in the secondary will be reflected back into the primary as an identical phase difference. Also, in the ideal transformer the secondary current is 180 out of phase with the primary. The CURRENT phase difference between the windings in the transformer will stay true to this no matter what the phase difference between voltage and current in the secondary.

    Unless the impedances are a required part of your calculation, you can get primary current simply by using the fraction of winding ratio times the secondary current. The impedance ratio is identical but uses the square of the turns ratio. Again any phase difference in the secondary voltage and current will be reflected back to the primary and it also will have the same phase relationship.
     
  4. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    I'm not ignoring the magnetizing current. I read the worksheet and I have no disagreements, the logic makes sense to me - a smaller inductance will produce a larger primary magnetizing current. I just want to know how to calculate it.

    For this particular example, you can calculate the primary current by Vp/377, which would give you the 54mA. A load resistance of 200 would give mismatched answers to a simulation for the primary.
     
  5. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    Why don't the usual equations work to calculate the primary current for this model?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As Adjuster points out there is an offset in the primary current arising from the transient condition. A complete solution involves a transient analysis - rather than the steady-state analysis you are probably assuming.

    Try setting the simulation source phase angle to something like 90° and you may note the change in the primary current offset.
     
  7. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    Thanks to all for the answers. I ran the simulation for a longer time and I saw that the answers were close. The primary current is still 5mA higher than the secondary, but it probably has something to do with the simulation that I don't know about.

    I can't believe that a link factor of .999 will result in a primary current that's 20mA higher.
     
    Last edited: Jun 12, 2011
  8. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    What happens in the case of a short for the load?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The primary side transient term will die away so long as you have a means of dissipating energy on the primary side - such as a series resistance. If you set the primary series resistance to say 1Ω then the transient term will vanish in a reasonable time frame - a few seconds perhaps.

    The primary side current is significantly higher compared with load current because the 1H magnetizing inductance takes a significantly higher current - as already pointed out earlier.

    If you short-circuit the secondary, the influence of the primary magnetizing current will be insignificant since in the absence of any limiting impedance the load (& hence source) current will be effectively infinite.
     
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