Ideal transformers can pass DC, but real ones can't

[Omegatron]

I only recently realized that ideal transformers can pass DC. Of course real ones can't. What's the difference? What is non-ideal about real transformers that prevents them from passing DC? Is it the fact that the flux is not linked perfectly between the two coils? Is it because the permeability of the core is not infinite? How does this change their electrical model?

 

[Averagesupernova]

How did you come to realize that ideal transformers can pass DC? I've never thought about it before. To be honest, I question it's validity.

 

[abdo375]

I agree with Averagesupernova, the whole idea behind the transformer is that when there is a fluctuating current (aka AC) in one coil the flux lines will cut the second coil and since the flux lines also fluctuates there will be an induced current in the second coil, but in case of an ideal DC current the flux lines are constant therefore there will be no current in the second coil.

 

[Omegatron]

I agree with Averagesupernova, the whole idea behind the transformer is that when there is a fluctuating current (aka AC) in one coil the flux lines will cut the second coil and since the flux lines also fluctuates there will be an induced current in the second coil,

I'm well aware that real transformers can't pass DC, but the ideal transformer — the mathematical model that engineers use — can. I'm wondering what property of real transformers prevents them from passing low frequencies well.

Also wondering if it would be possible with exotic materials (superconductors, infinite permeability cores, if such a thing could even exist) to create a DC-passing transformer.

but in case of an ideal DC current the flux lines are constant therefore there will be no current in the second coil.

But switching on the DC is a changing step function current.

 

[abdo375]

I'm well aware that real transformers can't pass DC, but the ideal transformer — the mathematical model that engineers use — can. I'm wondering what property of real transformers prevents them from passing low frequencies well.

There seems to be a mistake about the model your looking at, I believe that your looking at the model that is seen from the primary or secondary side of the transformer, not the complete exact model of the transformer, the model your looking at is constructed under the conditions that the flux lines out of coil one goes to coil two and produce electricity which will only happen in the case of AC, hence you can't apply the model your talking about to DC.

But switching on the DC is a changing step function current.

yes it is, but what we're talking about is the steady state analysis not the transit response of the transformer.

 

[Averagesupernova]

There seems to be a mistake about the model your looking at, I believe that your looking at the model that is seen from the primary or secondary side of the transformer, not the complete exact model of the transformer, the model your looking at is constructed under the conditions that the flux lines out of coil one goes to coil two and produce electricity which will only happen in the case of AC, hence you can't apply the model your talking about to DC.

Hence why I asked at how he arrived at his conclusion. Infinite inductance implies that not even DC will flow in a winding. Is this a clue as to how you've arrived at your conclusion Omegatron?

 

[Omegatron]

There seems to be a mistake about the model your looking at, I believe that your looking at the model that is seen from the primary or secondary side of the transformer, not the complete exact model of the transformer, the model your looking at is constructed under the conditions that the flux lines out of coil one goes to coil two and produce electricity which will only happen in the case of AC, hence you can't apply the model your talking about to DC.

I'm talking about the simple, idealized model of a transformer that engineers use; not a real transformer:

Vs/Ns = Vp/Np

There is no stipulation that it must be AC.

Please don't just say "transformers can't pass DC" without thinking about it. I'm very well aware that real transformers can't pass DC. That's what this whole question is about. I'm trying to start a discussion about the nitty gritty details of how a transformer works and why it can't pass DC.

Think of it this way. An ideal transformer can pass 60 Hz just as easily as 1 kHz, right? Real-life transformers are used in both power supplies and audio outputs, so this should be intuitively obvious. Transformers don't care about frequency. Ideal transformers have no losses, so an ideal transformer would pass any frequency without losses, even down to arbitrarily low frequencies, no? So it would pass AC with a period of several years just as readily as 60 Hz. AC with a period of several years is effectively DC.

yes it is, but what we're talking about is the steady state analysis not the transit response of the transformer.

Yes, but the step will always occur, and that is all you need to get the step magnetic flux and step secondary response. In a real transformer, the secondary would decrease exponentially after the step, right? Like a RC discharge curve. What about the transformer causes that decrease? Is it the magnetic field dissipating itself because of reluctance (magnetic circuit analogy to resistance)? Resistance in the wires?

 

[abdo375]

Ok let's start from the beginning, where did "Vs/Ns = Vp/Np" come from;

e_{p}=N_{p}\frac{d\varnothing}{dt}
e_{s}=N_{s}\frac{d\varnothing}{dt}

now in an idealized case where there will be no losses

e_{p}=N_{p}\frac{d\varnothing}{dt}=e_{s}=N_{s}\frac{d\varnothing}{dt}

this will result in:

\frac{e_{p}}{e_{s}} = \frac{N_{p}}{N_{s}}

but if you look at the original law you will see that there must be

\frac{d\varnothing}{dt}

and it must have a value for the e_{p} and e_{s} to exist, because without the variation in the flux the coils will just act like normal wires.

So as you can see the law assumes implicitly that there is a change in the flux, and the only reason frequency doesn't matter in an ideal transformer is because it's the same on both sides of the transformer which is not the case in an non-ideal transformer in which the losses is a function of the frequency applied.

So to sum it up frequency doesn't matter in an ideal transformer unless it's zero, but it does matter in a non ideal transformer.

 

[waht]

Frequency of DC voltage = 0 Hz

plug it in into some of the models and you get 1/0 which blows everything up.

 

[abdo375]

Yes, but the step will always occur, and that is all you need to get the step magnetic flux and step secondary response. In a real transformer, the secondary would decrease exponentially after the step, right? Like a RC discharge curve. What about the transformer causes that decrease? Is it the magnetic field dissipating itself because of reluctance (magnetic circuit analogy to resistance)? Resistance in the wires?

Now if your talking about the step response in an ideal transformer the voltage on both coils will simply tends to \infty and will be dispated in whatever load is connected to the secondary coil, now what will happen if a non-ideal transformer is applied to a unit-step input, well aside of it bursting into flames it will discharge in the load , winding resistance and some will be lost in the leakage flux and eddy currents .

 

[Omegatron]

which is not the case in an non-ideal transformer in which the losses is a function of the frequency applied.

Why?
.

 

[abdo375]

Basically there are two frequency dependent losses, the magnetic loss (eddy current loss + hysteresis loss) and the Leakage flux, both are a function of frequency, I can't recall there exact equations so you can Google it, if you want to know the exact reason as to why they are frequency dependent, I advice you to get a book about the microscopic view of magnetic materials, I recommend kasap's electronic material and devices for you, because it will be a very long post to explain on the forum.

 

[Averagesupernova]

I'm talking about the simple, idealized model of a transformer that engineers use; not a real transformer:

Vs/Ns = Vp/Np

There is no stipulation that it must be AC.

Don't be so sure. Consider the following:
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Vs/Ns = Vp/Np
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And then consider this:
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vs/ns = vp/np
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You may not have known it, but lower case indicates AC and upper case indicates DC.
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I also think you need to evaluate what "Ideal" actually is. Other than when a conductor is supposed to have resistance most devices modeled as ideal are superconductors. For example, a capacitor has zero Equivalent Series Resistance in the ideal world. In the real world, capacitors have a non-zero resistance. In the real world, transformer windings are NOT superconducting but ideally the windings would have zero resistance. The laminations of the core would also be superconductors. However, the ideal transformer does not necessarily have infinite inductance and zero magnetizing current. So if you are going off of the formula vs/ns = vp/np to conclude that the ideal transformers work on DC then you are sadly mistaken.

 

[berkeman]

The difference is that the OP's ideal transformer doesn't saturate, and can pass infinite flux and current.

 

[Gokul43201]

Unless there's some kind of funky non-linear resistance in series with the primary windings, you're going to have the turn on current rise like 1-Ie^{-t/\tau}. This will give you a decaying Ve^{-t/\tau} voltage in the secondary so long as you don't saturate the core.

 

[marcusl]

I vote with abdo375; transformers require a flux change to induce an emf in the secondary and induce a current. What a superconducting transformer can do that a normal one can't is sustain that current after the flux has stopped changing. Current in a normal transformer will decay away with time constant L/R, where R is the effective resistance of the circuit including all loss mechanisms. The superconducting current will (theoretically) persist for ever. It's not quite DC, though--the primary current had to change at some time to produce the secondary current.

 

[Omegatron]

So if you are going off of the formula vs/ns = vp/np to conclude that the ideal transformers work on DC then you are sadly mistaken.

Nevermind. I don't even know why I bother coming to this forum.

Not only do I immediately get shot down every time I post anything speculative (without actually reading what I've written or thinking about it or being the least bit imaginative), I get attitude about it, too.

"The ever-elusive ideal transformer would perform this function over an infinitely wide frequency range (including DC) and with zero loss. Even though such an element is physically unrealizable, it is nonetheless a useful starting point for constructing models of real transformers" - The Design of CMOS Radio-Frequency Integrated Circuits (ISBN 0521835399)

"It would be useful at this stage to introduce the concept of the ideal DC-to-DC transformer (a transformer that would provide a DC output for a DC input). Although this may seem a strange concept at first, consider the conventional ideal transformer model. The ideal transformer would pass all frequencies from DC upward in both directions, with no power loss. It would provide any required voltage or current ratio, and would have infinite galvanic isolation between input and output. It would also provide the same performance for all windings. Clearly such an ideal device does not exist." - Switchmode Power Supply Handbook (ISBN 0-07-005330-8)

"The ideal transformer transforms direct voltage, i.e., DC voltages on primary and secondary sides are related by turns ratio. This is not a surprising result because for the ideal transformer, we have assumed infinite core material permeability with linear (non-saturating) characteristics permitting core flux to rise without limit under a DC voltage application." - Transformer Engineering: Design and Practice (ISBN 0-8247-5653-3)

"For ac voltages and currents, an ideal transformer gives results that are within a few percent of those of the corresponding actual power transformer. But for dc voltages and currents, an ideal transformer gives incorrect results. The reason is that an ideal transformer will transform dc voltages and currents while an actual transformer will not." - Schaum's Outline of Theory and Problems of Basic Circuit Analysis (ISBN 0-07-047824-4)

"Ideal Transformer: The ideal transformer is a lossless transformer. It differs from the perfect transformer as it will pass DC and all frequencies. It is of course unrealizable in the 'real world'; however it serves as a building block for building complex, non-ideal transformers." http://www.beigebag.com/case_xfrmer_1.htm"

"It is a fact of life that the physical device called the “transformer” cannot operate properly in a dc circuit. The ideal transformer, because it is ideal, is defined to work according to equation (2) equally well for dc or ac voltages and currents, or any combination of the two." http://www.eecs.utoledo.edu/~rking/circuits/IdealTrans.pdf"

"However, the diminishing impedance value of these inductors as the frequency is lowered implies that there’s a lower frequency bound to which this will work. As a consequence, the coupled-inductor model doesn’t simulate an ideal transformer at zero hertz (dc)." http://www.elecdesign.com/Articles/ArticleID/6204/6204.html"

 

[es1]

I only recently realized that ideal transformers can pass DC. Of course real ones can't. What's the difference?

So, I think your answer, based on the quotes you gave, is:

The ideal transformer passes DC because they said it can.

i.e. they wanted a mathmatical model of a transformer that passes voltages at all frequencies so they just created such a model. This is no different from using an ideal voltage controlled voltage source with a constant of proportionality. It just seems they perfer to label their vcvs as 'ideal transformer'. Fine with me, it's their book. :)

A real transformer doesn't pass DC because the coupling is done via magnetic field on finite coils made of real metals placed a measurable distance apart, not via arbitrary math. But of course you knew that... So I am not sure where the confusion is or what the debate is about.

 

[Averagesupernova]

Nevermind. I don't even know why I bother coming to this forum.

Not only do I immediately get shot down every time I post anything speculative (without actually reading what I've written or thinking about it or being the least bit imaginative), I get attitude about it, too.

Now hold on just a sec. Nowhere in this thread have I said it isn't possible. I said I question it. In fact, I asked you this:

How did you come to realize that ideal transformers can pass DC?

I threw the ball in your court. You finally have come with some reference. My comment about the formula vs/ns = vs/ns still holds true. Lower case refers to AC. You specifically said:

There is no stipulation that it must be AC.

and I called you on it. I also asked you:

Infinite inductance implies that not even DC will flow in a winding. Is this a clue as to how you've arrived at your conclusion Omegatron?

You didn't reply to it. Is it because you don't know? Because if it is, please say so. There is nothing wrong with not knowing, but don't come on here spouting about whatever with no reference of significance until your fifth post and then cry about your idea getting shot down.
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I feel as I should rephrase my comment about infinite inductance not allowing DC to flow. Of course DC can flow, but, DC is not really considered DC if we start with zero current and an open switch in the circuit and then close the switch. The infinite inductance in the circuit will prevent current flow in this case.

 

[berkeman]

I'd like to tie off this thread and lock it. It's not going anywhere useful anymore.