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Power XFMR magnetization currents in a "reverse" connection

  1. Apr 4, 2016 #1
    If my understanding is correct, with respect to dry-type building power transformers, due to the virtues of the transformer as a dumb machine it really doesn't care which side of itself is the primary versus the secondary. In other words a 480D-208Y tranformer can be correctly connected such that it steps up rather than steps down as they are designed to be used. The wrinkle I am told is the magnetization or inrush current; in a reversed application where a 208V 3ph connection is made on the primary side, giving a 480V three or four wire source on the secondary. It works great, except that where the magnetization current in the intended configuration is only a few times the full-load current of the transformer, in the reverse configuration it is something like a factor of seventeen.

    I can't recall for certain if what was said was specifically magnetization current or inrush current. The point was, one might have unforseen problems with any overcurrent device upstream of the transformer primary being able to handle such a sudden peak in current as such devices are commonly able to handle closer to six times the current without an instantaneous trip. I'm unclear on the dynamics of this sudden spike of current, is it only an issue when the transformer is first energized, or is it something that can be additive if say you were stepping up the load on a transformer in twenty percent increments. Another obvious implication lies in motor starting currents on the secondary side which I would imagine have their effects multiplied in some similar fashion. What are some effective ways of mitigating problems such as this?
  2. jcsd
  3. Apr 5, 2016 #2
    The idea of Inrush current is that the flux within the core of the transformer cannot change instantaneously. If you remember your ideal relations, the flux will be at its negative maximum when the input voltage is zero; obviously, this cannot be the case in reality. This means that the flux will be significantly higher than its steady-state maximum value when the transformer is energized. Most transformers are designed to saturate just above the steady-state value of flux, but, upon being energized, the flux will be much higher than that of the saturation value, leading to a very large current being drawn, or "rushing in." This current lasts until the flux "matches" the more ideal relations described by Faraday's Law.

    I would think the relation would apply regardless of one's naming convention.

    There is circuitry built into many transformers to try and mitigate the amplitude and duration of inrush current, but, at the end of the day, physics is physics, and one cannot overcome natural laws.

    Protective devices, ideally, should not trip breakers on the detection of inrush current as it is a normal part of transformer operation.
  4. Apr 7, 2016 #3
    Then it is correct to say that given a transformer already running at 5% of its rated full-load ampacity will not experience as dramatic a spike in flux if it is suddenly loaded to 95%, when compared to a de-energized transformer to which a 95% load is suddenly connected from an open circuit condition on the secondary. This was what I had suspected initially, if I am still incorrect please point out where.
    If there were such "circuitry" in something like a 300kVA transformer it would undoubtedly be very, very robust. I would imagine of course that in designing a transformer you would arrange the windings in such a way that would use physics and geometry to an advangage in order to minimize inrush. As was stated in the OP we are talking about conditions in using such a transformer in a somewhat non-traditional way.
    I'm well aware of how many common thermal-magnetic molded case power circuit breakers operate and there are many variables involved in trips. As I said per a general guideline one can expect a common T/M circuit breaker's instantaneous, or magnetic section, to have a sensitivity probably on the order of six times the thermal section's amp rating. To state the OP question on mitigation another way, if one were to run a continuous and small load on a transformer secondary would it help to reduce the large inrush currents should the secondary load be suddenly increased to a much larger percentage?

    Of course the obvious solution to the problem is to buy a purpose built step-up transformer instead of pressing a step-down transformer into walking on its hands. And if one were determined to use said transformer backwards one could always upsize the primary protection to probably 4x what it ought to be. Then upsize the conductor as well, to keep some semblance of safety, and also realize that all this bumps you up into an OCP frame size in a much higher cost bracket. With a budget target of something like as close to zero as possible it begs other options.
  5. Apr 7, 2016 #4

    jim hardy

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    Evergreen was headed in the right direction.

    Inrush to a transformer is a startup phenomenon.
    It depends completely on at what point in the AC cycle you close the switch that applies power.
    To understand it,, study the first half cycle.
    That is when you first establish magnetization in the core.

    Here's steady state operation, from a decent article at

    observe that at zero volts flux is always at a peak, as is magnetizing current.

    So - if you de-energize it, voltage current and flux all three fall to zero.

    If you re-energize it, you need to do it at a voltage peak so that flux starts from the correct value. Otherwise there'll be some sort of transient as flux and voltage re-establish their steady state relationship, 90° and centered on zero.

    Worst case is to re-energize on the zero crossing.
    That's because so long as voltage is positive , flux is increasing
    ... and if flux started from zero instead of its negative peak it has to go twice as high as normal.
    Most transformers don't have enough iron to carry that much flux,
    so they saturate, and the current goes sky high trying to push enough magnetizing current to produce that much flux in iron that's saturated.

    Inrush current looks like this as the voltage-flux relationship restores itself to normal.

    from another decent article here

    Fortunately Mother Nature made sinewaves spend most of their time near the peak, (what's sin30° ?)
    so statistically you only get that awful inrush current about once every five or ten switch closures.

    It really is that simple.

    old jim
  6. Apr 8, 2016 #5
    I never realized there was such an element of probability to inrush currents. Re-reading evergreen's post it's now more clear to me what was said about flux at its state of negative maximum.

    I expect the behaviors of current can get pretty complicated in a three phase arrangement and I will do some further reading before I come back with any more questions on what novel ways power transformers are built in order to minimize the chances they would violenty disassemble themselves in an out of the ordinary inrush condition.

    EDIT: Further recollection, it might have been said to me that phenomena involving winding distance from the core is to blame for a backwards connected transformer to experience a higher inrush characteristic, such that the primary winding now exists in a physical configuration where the secondary winding ought to be, drastically changing the characteristics of the flux inside the coils. Not 100% on this it just popped into my head.
    Last edited: Apr 8, 2016
  7. Apr 8, 2016 #6

    jim hardy

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    I always say-
    Use what you already know to figure things out.
    A thought experiment that turns out wrong teaches us something, and we don't have to admit we were dumb enough to even consider it.
    So let's run a thought experiment .
    At risk of looking dumb -
    I'll make this absurd claim about your friend's wild sounding "Inrush" statement :
    it's because of the lower winding resistance in low voltage winding.

    Think about it - when core saturates winding resistance is what limits inrush current.

    Now -
    in order to have equal power loss in high and low voltage windings during normal operation, what would a designer chose for ratio of their resistances?
    Arithmetic tells us
    Ihi2Rhi = Ilo2Rlo
    Rlo/RIhi = (Ihi/Ilo)2
    Let's just assume a 3::1 turns ratio , if the numbers make sense we could try to derive an equation.
    If Ilo = 3Ihi (turns ratio)
    Rlo = Rhi/9
    Aha a winding's resistance isn't in proportion to its voltage but its voltage 2
    the secondary has fewer ohms per volt than does the primary, by the turns ratio.
    So its lower resistance is less effective at limiting inrush current.

    so my thought experiment says rated voltage applied (at the zero crossing) to a transformer's secondary instead of its primary should give inrush that's larger than that winding's rated current by a factor about equal to turns ratio.
    Consider a 3::1 transformer ..
    If inrush to primary is 500% rated, inrush to secondary should be about 1500% rated.

    Sound logical ?

    My logic could be way off - please test it yourself.
    And if it's right, doubtless one of you sharp younger guys can summarize it more concisely.

    Was it Lord Kelvin said to effect "To be able to describe a phenomenon is only a claim to meager knowledge of it. When we can put a number on it we are beginning to understand."

    I never before considered this question, so check me.

    old jim
    Last edited: Apr 9, 2016
  8. Apr 11, 2016 #7
    Theoretically, there will be no inrush at all if the transformer is already loaded since it is a phenomenon experienced only at startup. In reality there may be some slight inrush current associated with extreme load fluctuations but not enough to elicit any consideration.

    Yes, but I don't see how one could, or would, do such a thing.
  9. Apr 11, 2016 #8

    jim hardy

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    Mr Evergreen is saying the right thing about your question.

    Let me turn it just a litle bit -
    Keep your apples separate from the oranges.
    Inrush current to the transformer is magnetizing current not load current. Inrush is over once the transformer has established its equilibrium voltage-flux relationship e = -ndΦ/dt .
    Inrush current to a LOAD is quite a different thing and just propagates through the transformer like any other load current.

    Here's your circuit model of a transformer ,from Wikipedia i think
    It has at its heart the ideal transformer in middle which can neither saturate nor overload and needs no magnetizing current .
    Magnetizing current flows down into Xm . If we want our model to approximate a real transformer, Xm must have a B-H saturation curve like a real transformer.
    Load current flows through the ideal transformer not Xm. Inrush to a load affects Xm only by causing voltage drop across Rp and Xp.

    I hope that's some help. That transformer model is a handy thinking-aid.
  10. Apr 11, 2016 #9
    Jim Hardy always has the best of explanations. They're always understandable to us more simple-minded and less experienced. He and anorlunda are the "power-houses" when it comes to anything power related on this forum. I have found their insights vastly useful.

    I have some questions about the calculation of power/load flow that may just be perfect for our resident experts. I'm only waiting until I know the subject well enough to converse intelligently... it shouldn't be too long now.
  11. Apr 11, 2016 #10
    Ah, yes... I should have differentiated between Load inrush and Transformer inrush. Any transient or fault type current flow "outside" the transformer will simply propagate through the machine since, as the OP has pointed out, the transformer is a "stupid" machine.

    To further illuminate the OP's question, using the secondary winding as the "high side" to step-up the voltage would yield the same inrush as if you'd connected it as intended. This is due to the consideration given to the number of turns. If you use the side with fewer turns but higher voltage, it would, ideally, produce the same inrush as if you'd used the side with more turns, but smaller voltage. Now this ideal relationship is not too useful, but, unless you're considering an overloaded transformer, it's useful for this case.

    Be sure if you use the transformer model to swap the values of the Rp and Xp with Rs and Xs and use 1/a rather than a.
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