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

1. Nov 25, 2006

### 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?

Last edited: Nov 25, 2006
2. Nov 25, 2006

### 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.

3. Nov 25, 2006

### 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.

4. Nov 25, 2006

### Omegatron

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 switching on the DC is a changing step function current.

5. Nov 25, 2006

### abdo375

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.

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

6. Nov 25, 2006

### Averagesupernova

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?

7. Nov 25, 2006

### Omegatron

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, 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?

8. Nov 25, 2006

### 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.

9. Nov 25, 2006

### waht

Frequency of DC voltage = 0 Hz

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

10. Nov 25, 2006

### abdo375

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 .

11. Nov 25, 2006

### Omegatron

Why?
.

12. Nov 25, 2006

### 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.

13. Nov 25, 2006

### Averagesupernova

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.

14. Nov 26, 2006

### Staff: Mentor

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

15. Nov 27, 2006

### Gokul43201

Staff Emeritus
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.

16. Nov 27, 2006

### 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.

17. Nov 27, 2006

### Omegatron

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.

Last edited by a moderator: May 2, 2017
18. Nov 27, 2006

### es1

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.

Last edited: Nov 27, 2006
19. Nov 28, 2006

### Averagesupernova

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

20. Nov 28, 2006

### Staff: Mentor

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

Last edited: Nov 29, 2006