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Assumption 1: An "ideal" transformer is said to have very large primary, secondary, and mutual reactance.. (self-inductance/mutual-inductance tending toward infinity), has a unity coupling coefficient (zero leakage flux), High or infinite magnetic permeability, absorbs zero real power (is lossless, 100% efficient).

Assumption 2: From a pure circuit analytical and mathematical standpoint, and without the "real" model elements, the infinite primary and secondary inductances in the "ideal" transformer will draw current when secondary load is not open, and zero current when the secondary load is open or tends to infinity.

Problem 1: How can the infinite reactance of either primary or secondary draw the current in assumption 2?

Problem 2: The secondary load gets transformed and appears in parallel to the primary inductance, so if the primary reactance is virtually open, why even put it in the circuit? what good does this do?...there are an infinite amount of parallel opens in any given circuit.

Thanks in advance!

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    \$\begingroup\$ An ideal transformer has no reactance. It just transforms impedances. So, your question reduces do, "in an ideal transformer, how can non-ideal properties of real transformers do stuff?" which makes no sense. \$\endgroup\$ – Phil Frost Oct 22 '14 at 17:59
  • \$\begingroup\$ Phil Frost, I want to keep it strictly theoretical and ideal. I don't want to complicate it with "real" behavior. So in terms of circuit analysis you can solve transformer circuits with mesh analysis...eventually it leads to a solution where analytically, current flows through the ideal part of the primary. The issue with this is that in assumption 1 above, this primary has infinite reactance. \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 18:45
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    \$\begingroup\$ An ideal transformer has no inductance, no leakage, no capacitance, perfect coupling, etc. There's nothing to appear in parallel. If you introduce non-ideal concepts (like series or parallel impedances) into an ideal model, of course it won't be strictly ideal anymore. That's the point of an ideal model. \$\endgroup\$ – Phil Frost Oct 22 '14 at 19:55
  • \$\begingroup\$ I think what you are saying is that my assumption 1 is incorrect. Can you confirm this? Thanks. \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 23:25
  • \$\begingroup\$ Yes, mostly. What you are describing (large primary, etc) are limits that approach an ideal transformer. The reason it's framed in language like that is ideal transformers aren't realizable, but if we are discussing purely theoretical concerns, all these problems go away. All an ideal transformer does is multiply impedances. Leakage flux, magnetizing inductance, etc, all these things don't exist in ideal transformers. \$\endgroup\$ – Phil Frost Oct 22 '14 at 23:33
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I think your confusion lies in your first assumption. An ideal transformer doesn't even have windings, because it can't exist. Thus, it doesn't make sense to consider inductance, or leakage, or less than perfect coupling. All of these issues don't exist. An ideal transformer simply multiplies impedances by some constant. Power in will equal power out exactly, but the voltage:current ratio will be altered according to the turns ratio of the transformer.

For example, it is impossible to measure any difference between a 50Ω resistor, and a 12.5Ω resistor seen through an ideal transformer with a 2:1 turns ratio. This holds true for any load, including complex impedances.

schematic

simulate this circuit – Schematic created using CircuitLab

Since an ideal transformer can't be realized, considering how it might work is a logical dead-end. It doesn't have to work because it is a purely theoretical concept used to simplify calculations.

The language you used in your first assumption is a description of the limiting case that defines an ideal transformer. Consider a simple transformer equivalent circuit:

schematic

simulate this circuit

Of course, we can make a more complicated equivalent circuit according to how accurately we wish to model the non-ideal effects of a real transformer, but this one will do to illustrate the point. Remember also that XFMR1 represents an ideal transformer.

As the real transformer's winding resistance approaches zero, then R2 approaches 0Ω. In the limiting case of an ideal transformer where there is no winding resistance, then we can replace R2 with a short.

Likewise, as the leakage inductance approaches zero, L2 approaches 0H, and can be replaced with a short in the limiting case.

As the primary inductance approaches infinity, we can replace L1 with an open in the limiting case.

And so it goes for all the non-ideal effects we might model in a transformer. The ideal transformer has an infinitely large core that never saturates. As such, the ideal transformer even works at DC. The ideal transformer's windings have no distributed capacitance. And so on. After you've hit these limits (or in practice, approached them sufficiently close for your application for their effects to become negligible), you are left with just the ideal transformer, XFMR1.

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  • \$\begingroup\$ Brilliant. I think this answers the question in full. Thank you again. Its interesting to note that the ideal transformer works for DC as you have mentioned. Very abstract. \$\endgroup\$ – XPTPCREWX Oct 23 '14 at 17:29
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How can the infinite reactance of either primary or secondary in an ideal transformer draw current?

For two coupled inductors, we have two coupled equations:

$$v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$$

$$v_2 = M \frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$

where \$M = k\sqrt{L_1L_2}\$ is the mutual inductance and \$k\$ is the coupling coefficient. Assume perfect coupling, \$k = 1\$, from this point forward.

Using phasor notation, the above equations are

$$V_1 = j\omega (L_1 I_1 + M I_2)$$

$$V_2 = j\omega (M I_1 + L_2 I_2)$$

Now, by (phasor) Ohm's Law, it must be the case that

$$V_2 = I_2Z_2 $$

where \$Z_2\$ is the impedance connected to the secondary.

It follows that

$$\frac{I_2}{I_1} = \frac{j\omega M}{Z_2 - j\omega L_2}$$

So, for finite \$L_1, L_2\$, the ratio of the secondary current to primary current is a function of frequency even when there is perfect coupling.

As the frequency tends to zero, the ratio tends to zero. As the frequency becomes arbitrarily large, the ratio tends to

$$\frac{I_2}{I_1} \rightarrow -\sqrt{\frac{L_1}{L_2}} = -\frac{N_1}{N_2}$$

Now, keeping the ratio \$\sqrt{\frac{L_1}{L_2}}\$ constant while allowing both \$L_1\$ and \$L_2\$ to become arbitrarily large, we have

$$\frac{I_2}{I_1} = \frac{j\omega M}{Z_2 - j\omega L_2} \rightarrow \frac{j\omega M}{-j\omega L_2} = -\sqrt{\frac{L_1}{L_2}} = -\frac{N_1}{N_2}$$

The point is this: Even though the individual reactances go to infinity as the individual inductances go to infinity, the reactances 'cancel out' leaving the well known result true at any frequency.

In other words, the answer to your question is found by taking the limit as the inductances go to infinity and observing that the frequency dependent reactances in the numerator and denominator become a frequency independent, non-zero ratio.

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  • \$\begingroup\$ Excellent this is definitely along the lines of what I am asking. Much appreciated. \$\endgroup\$ – XPTPCREWX Oct 23 '14 at 17:26
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Your question isn't completely clear, because assumption 2 says two things. I'll address it both ways.

In the case where the secondary load is open there is no current. So problem 1 isn't a problem in that case. In the case where the secondary load is not open, there will be current, but now the reactance is no longer infinite.

For problem 2, I would say that the secondary winding is a real component, and as such we would want to have a model for it. It's a way of saying, "I see that other coil there, and I've determined that it effect on the model is vanishingly small, and here's how." It's true there are other "parallel opens" (like the air leakage across a length of wire), but those don't appear as a component. We'd get into an infinite effort accounting for those, and worse, an infinitely long philosophical discussion might erupt :)

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  • \$\begingroup\$ Can you clarify what reactance you are talking about in your first paragraph when you say ,"but now the reactance is no longer infinite." \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 18:42
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  1. It doesn't. Current flows through the windings, but not through the reactance. that is why it is shown in parallel, not in series (leakage reactance is shown in series, but an ideal transformer doesn't have any leakage).

  2. If you don't put any reactance in the circuit then what do you have? Zero reactance, a short circuit! Any reactance less than infinity will cause some current to bypass the load.

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  • \$\begingroup\$ Would you mind clarifying what you mean by "windings" and "reactance" in response 1? i.e. what constitutes winding and which reactance. Thanks. \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 18:40
  • \$\begingroup\$ The physical windings are the actual coils in the transformer. They can be considered as consisting of various ideal components, each of which represents a particular property of the transformer. The ideal winding is just a winding (it has no internal inductance or resistance, since these properties have been separated out). Load current is drawn through the (ideal) winding, not through the inductor in parallel with it. So assumption 2: is simply wrong - applying a secondary load does not cause current to pass through the 'infinite primary and secondary inductances'. \$\endgroup\$ – Bruce Abbott Oct 22 '14 at 20:22
  • \$\begingroup\$ Thanks for your response but I'm not entirely sure I follow. Allow me to explain my confusion. 1.) You mention that "The ideal winding is just a winding (it has no internal inductance or resistance, since these properties have been separated out)" but clearly, if you read any EE text it states the properties of an ideal transformer as those in my assumption 1 above. Are these text books wrong? 2.) You also mention that, "Load current is drawn through the (ideal) winding, not through the inductor in parallel with it."..which parallel inductor are you referring to? Thanks. \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 23:39
  • \$\begingroup\$ The parallel inductor is a virtual component which represents the inductance of the winding (leaving an 'ideal' winding whose impedance is a reflection of the load, in parallel with its inductance). In an ideal transformer the winding inductance is infinite, so it has infinite reactance and any current flowing through it will generate infinite voltage. Therefore the phrase 'infinite primary and secondary inductances will draw current' is nonsensical. Which EE text says that this happens? \$\endgroup\$ – Bruce Abbott Oct 23 '14 at 6:37
  • \$\begingroup\$ Thanks for your clarification. I mentioned that "assumption 1" was taken from the EE texts. I think you might be confusing that with "problem 1" (which was my problematic understanding). \$\endgroup\$ – XPTPCREWX Oct 23 '14 at 17:36
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Equivalent circuit of a transformer: -

enter image description here

The area in the above circuit that appears to show a transformer with Es on the input and Ep on the output is, in fact, an ideal power converter such that: -

\$E_P\times I_P = E_S\times I_S\$ and further...

\$E_S = \dfrac{N_S}{N_P}\times E_P\$

Ignoring the small series secondary components (\$R_S\$ and \$X_S\$) Any load impedance on the secondary will appear on the input side to the ideal power converter as: -

Referred impedance of secondary onto primary \$= (\dfrac{N_P}{N_S})^2\times Z_{secondary}\$

So, if the magnetization inductive reactance (\$X_M\$) is infinite and the core loss (\$R_C\$) is also infinite then, apart from the small primary series components (\$R_P\$ and \$X_P\$), the impedance seen at the real primary input is secondary impedance multiplied by turns ratio squared.

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  • \$\begingroup\$ Andy aka, the Xp quantity in your third equation isn't the primary leakage reactance in the above model you provided, is it? This is also Xp. \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 18:37
  • \$\begingroup\$ @XPTPCREWX - you are quite correct and I shall correct this. \$\endgroup\$ – Andy aka Oct 22 '14 at 18:43
  • \$\begingroup\$ Andy aka, I believe that in equation 3, Xs should be Zload. Can you confirm this? \$\endgroup\$ – XPTPCREWX Oct 22 '14 at 23:31
  • \$\begingroup\$ @XPTPCREWX Again you are quite correct!! \$\endgroup\$ – Andy aka Oct 23 '14 at 9:58

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