# Loaded Transformer secondary current/voltage phase relationship?

#### kenc184

Joined Jan 11, 2018
3
Hi all,

This is probably not an incredibly smart question but I have to ask regardless!

If you consider a transformer to be two coupled inductors, and if you consider the secondary to be loaded with a resistor then you have the situation where The resistor and secondary winding share the same current AND the same voltage across their terminals. So while the load resistor insists on voltage and current being in phase (being a non reactive component), if the secondary is truly an inductor, it should force voltage and current to be in quadrature. Since both situations cannot be satisfied, and a simulation shows - somehow - that the resistor "wins", it means we have an inductor with voltage and current 180 degrees out of phase rather than the 90 degrees that physics tells us should exist.

What am I not getting here?

Thanks!

#### crutschow

Joined Mar 14, 2008
25,109
Transformers can be confusing and it took me awhile until I really understood how they worked.

You are not getting that the transformer inductance of a normal transformer does not appear in series with the load and is, in fact, transparent to the load.
The primary (and reflected secondary) coupled inductance act as one inductance (assuming perfect coupling between the primary and secondary windings) which establishes the magnetizing current and consequent magnetic flux from the applied primary voltage. That is the purpose of the inductance.

Here's the crux of the transformer operation.
Any secondary load current unbalances this magnetizing flux which then causes more primary current to flow to keep the flux in balance.
Thus any primary current is in phase with the secondary current, and there is ideally no phase-shift between the primary and secondary currents in the transformer.
The primary current thus always matches the secondary current times the turns ratio.

• tranzz4md

#### WBahn

Joined Mar 31, 2012
25,906
Hi all,

This is probably not an incredibly smart question but I have to ask regardless!

If you consider a transformer to be two coupled inductors, and if you consider the secondary to be loaded with a resistor then you have the situation where The resistor and secondary winding share the same current AND the same voltage across their terminals. So while the load resistor insists on voltage and current being in phase (being a non reactive component), if the secondary is truly an inductor, it should force voltage and current to be in quadrature. Since both situations cannot be satisfied, and a simulation shows - somehow - that the resistor "wins", it means we have an inductor with voltage and current 180 degrees out of phase rather than the 90 degrees that physics tells us should exist.

What am I not getting here?

Thanks!
Basically what is happening is that, as current is pulled from the secondary, the magnetic interactions between the primary current and the secondary current result in the inductance going away as energy is pulled from the field.

In fact, if you take this into account by analyzing the circuit's mutual inductance between the primary and secondary you discover that, to the source driving the primary, the combination of the transformer and the load on the secondary look like a scaled version of the load. The transformer essential does nothing but transforms the impedance by the square of the turns ratio. This is for an ideal transformer. Not surprisingly, real transformers have artifacts on top of this, but in many instances this is a very good approximation.

This has many useful application. One of them is in RF circuits where you can use an impedance matching transformer between two parts of a network so that each side sees the optimal impedance.

#### kenc184

Joined Jan 11, 2018
3
Thanks for the replies.

And yet the secondary winding has the physical form of an inductor, i.e. a wire wrapped around an iron (e.g.) core? I realize that the secondary winding has its voltage induced by the flux due to the current through the primary which is different to the applied voltage stimulus of a regular inductor but still it's hard to grasp that a wrapped wire which is the physical form of an inductor does not have the usual quadrature relationship of voltage and current when loaded with a resistor. This is non intuitive!

#### crutschow

Joined Mar 14, 2008
25,109
still it's hard to grasp that a wrapped wire which is the physical form of an inductor does not have the usual quadrature relationship of voltage and current when loaded with a resistor. This is non intuitive!
It's intuitive if you understand how the transformer actually works.
The secondary is not an isolated inductance but is part of the same magnetic circuit as the primary.
In a normal (ideal) transformer the primary and secondary magnetic circuits have the same magnetic path.
Thus whether the transformer has only one winding (as an autotransformer does) or multiple windings, there is only one magnetic circuit for all of them.
Therefore you can't look at the secondary as a separate inductance (unless the primary and secondary are loosely coupled, such as with an air-core transformer). That's where your intuition fails.
Until you understand now a transformer actually works, you will likely remain confused.

#### BR-549

Joined Sep 22, 2013
4,938
How bout thinking of a transformer as a coil with 2 quads. So the whole unit has 180.

#### crutschow

Joined Mar 14, 2008
25,109
How bout thinking of a transformer as a coil with 2 quads. So the whole unit has 180.
What? #### nsaspook

Joined Aug 27, 2009
7,344
Thanks for the replies.

And yet the secondary winding has the physical form of an inductor, i.e. a wire wrapped around an iron (e.g.) core? I realize that the secondary winding has its voltage induced by the flux due to the current through the primary which is different to the applied voltage stimulus of a regular inductor but still it's hard to grasp that a wrapped wire which is the physical form of an inductor does not have the usual quadrature relationship of voltage and current when loaded with a resistor. This is non intuitive!
Maybe it's useful to think of the transformer as a mechanical circuit of two gears. In both systems we have a physical system that transforms energy properties as a system. The coupled inductors of the transformer are physically (by mutual inductance) connected by magnetic fields just like the coupled gears are physically connected by teeth into a single system. So the transformer doesn't just store energy in a magnetic field and release it back to the source in an isolated inductor, it passes energy using that magnetic field from input to output in a mainly electrically transparent manner to the load.

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#### kenc184

Joined Jan 11, 2018
3
I understand the nature of a transformer regarding stepping up, stepping down, the relationship to the turns ratio. I understand that the power delivered to the load must come form somewhere and clearly that's the source driving the primary. I understand the fact that secondary loads look like loads either divided by or multiplied by the square of the turns ratio depending on whether we are stepping up or stepping down.

I also understand that if we presuppose power being dissipated by the load then that same power must be provided by the secondary (via the primary). Given the connection of the load and the secondary this dictates that the voltage and current in the secondary must be in antiphase.

I get all of that.

I just need to understand the fact that the secondary cannot be treated as what it physically appears to be - an inductor - which it clearly cannot given my previous paragraph.

#### crutschow

Joined Mar 14, 2008
25,109
I just need to understand the fact that the secondary cannot be treated as what it physically appears to be - an inductor
It is physically an inductor but you need to recognize that you can't separate the secondary magnetic circuit from the primary.
They share the same core.

#### nsaspook

Joined Aug 27, 2009
7,344
I understand the nature of a transformer regarding stepping up, stepping down, the relationship to the turns ratio. I understand that the power delivered to the load must come form somewhere and clearly that's the source driving the primary. I understand the fact that secondary loads look like loads either divided by or multiplied by the square of the turns ratio depending on whether we are stepping up or stepping down.

I also understand that if we presuppose power being dissipated by the load then that same power must be provided by the secondary (via the primary). Given the connection of the load and the secondary this dictates that the voltage and current in the secondary must be in antiphase.

I get all of that.

I just need to understand the fact that the secondary cannot be treated as what it physically appears to be - an inductor - which it clearly cannot given my previous paragraph.
It's like all of us have telling you, a transformer is not just A inductor. A transformer uses the property of inductance to transfer energy with a magnetic circuit. We have Faraday's law of induction AND Hopkinson's law.
The concept of a "magnetic circuit" exploits a one-to-one correspondence between the equations of the magnetic field in an unsaturated ferromagnetic material to that of an electrical circuit. Using this concept the magnetic fields of complex devices such as transformers can be quickly solved using the methods and techniques developed for electrical circuits.

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#### WBahn

Joined Mar 31, 2012
25,906
I just need to understand the fact that the secondary cannot be treated as what it physically appears to be - an inductor - which it clearly cannot given my previous paragraph.
You do NOT just have two coils each with an inductance. You have two coils, each with a self inductance, and then with a mutual inductance between the two coils. Your circuit analysis must take the mutual inductance into account. It is the mutual inductance that provides the mechanism for energy transfer between the two parts of the circuit.

Just as a the voltage across an inductor is proportional to the time rate of change of the current in that inductor, it is also proportional to the time rate of change of the current in the other inductor, as well. For an ideal transformer, the mutual inductance is the geometric mean of the individual inductances.

#### crutschow

Joined Mar 14, 2008
25,109
For an ideal transformer, the mutual inductance is the geometric mean of the individual inductances.
But note, that inductance does not appear in series with the output load as the TS is concerned about.

#### BR-549

Joined Sep 22, 2013
4,938
"I just need to understand the fact that the secondary cannot be treated as what it physically appears to be - an inductor - which it clearly cannot given my previous paragraph."

The reason is because the secondary coil was not excited by current(which is what you have studied)......it was excited by flux.

• nsaspook