# Transformer theory

When you have no loads connected to the secondary of a transformer, what is the current in the primary?

When you start connecting loads to the secondary, what phenomenon is making the current increase in the primary?

Cannot find an easy way to explain this without drawing some schematic and transformer equivalent circuit, but

1) When there is no load, the current in primary is called magnetizing current, the core is magnetized,and there will be power loss associated. This core loss will be still there even when there is load connected. Also, the open circuit primary inductance is very high.

2) When there is load connected to the secondary, the primary impedance will be reduced and the primary current begin to increase. When you short the secondary, that is the least load and the least inductance on primary, so most likely the primary will be burned by extreme high current in the primary.

When you have no loads connected to the secondary of a transformer, what is the current in the primary?

When you start connecting loads to the secondary, what phenomenon is making the current increase in the primary?

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How is the loaded secondary decreasing the inductance of the primary?

You may just think of this way,

When you short the secondary, you remove that amount of turns from the primary, causing the primary inductance drops. If you have the same turns on primary and secondary (1:1 ratio), then when you short the secondary, you completely removed the primary turns and it will be a short circuit on the primary (here assuming the leakage inductance is zero).

So when you add load to the secondary, it partially "remove" the primary turns.

Not exactly so but I think this is an easy way to think about it.

How is the loaded secondary decreasing the inductance of the primary?

berkeman
Mentor
The load does not change the magnetizing inductance value.

The transformed load R appears in parallel with the magnetizing inductance Lm, and the leakage inductance Lk is in series before that parallel combination. So if you short out the secondary, that transforms to a short in parallel with Lm, leaving only Lk if you measure the inductance of the transformer with a shorted secondary.

This is much better than what I said.

Magnetizing inductance will not change (core and winding turns will not change).

The measured inductance on the primary will change, just like you add the load in parallel to the primary.

The load does not change the magnetizing inductance value.

The transformed load R appears in parallel with the magnetizing inductance Lm, and the leakage inductance Lk is in series before that parallel combination. So if you short out the secondary, that transforms to a short in parallel with Lm, leaving only Lk if you measure the inductance of the transformer with a shorted secondary.

berkeman
Mentor
This is much better than what I said.

Magnetizing inductance will not change (core and winding turns will not change).

The measured inductance on the primary will change, just like you add the load in parallel to the primary.

No, the measured inductance Lm will not change. The parallel impedance of the transformed load R and Lm will change, but Lm is not what is changing.

If you use an Impedance Analyzer like the HP 4194, and use its "Equivalent Circuit" feature, you will just see the transformed impedance changing as you vary the load resistance. The Lm and Lk values should not change with the load.

Here is what we measure the inductance.

We measure the inductance of primary with secondary open, call this open circuit inductance. For example 400uH.

Then we short the secondary, measure the inductance, we call this leakage inductance. For example 6uH.

If short the secondary with a resistor (normally don't do this), the inductance value will be lower than the open circuit, but higher than the short circuit, the value depends on the resistance.

No, the measured inductance Lm will not change. The parallel impedance of the transformed load R and Lm will change, but Lm is not what is changing.

If you use an Impedance Analyzer like the HP 4194, and use its "Equivalent Circuit" feature, you will just see the transformed impedance changing as you vary the load resistance. The Lm and Lk values should not change with the load.

berkeman
Mentor
Here is what we measure the inductance.

We measure the inductance of primary with secondary open, call this open circuit inductance. For example 400uH.

Then we short the secondary, measure the inductance, we call this leakage inductance. For example 6uH.

If short the secondary with a resistor (normally don't do this), the inductance value will be lower than the open circuit, but higher than the short circuit, the value depends on the resistance.

No. You are measuring with just a simple LCR meter, it sounds like. The Impedance is dropping as you lower the value of the load resistor; the magnetizing inductance Lm is not changing.

Do you have access to a Vector Impedance Meter or Impedance Analyzer?

We use HP4284A or Wayne Kerr PM3260A.

I have not done any measurement as you described.

No. You are measuring with just a simple LCR meter, it sounds like. The Impedance is dropping as you lower the value of the load resistor; the magnetizing inductance Lm is not changing.

Do you have access to a Vector Impedance Meter or Impedance Analyzer?

berkeman
Mentor
We use HP4284A or Wayne Kerr PM3260A.

I have not done any measurement as you described.

It looks like the Kerr PM3260A has an Impedance measurement mode. You could try that and calculate what the impedance should be with the transformed load in parallel with Lm (and that all in series with Lk) for a test transformer that you may have laying around. Probably you should include the winding resistances too, to get a good match.

It also looks like the PM3260A has a "series or parallel equivalent circuit" mode -- maybe check that out to see if it can give you values when there is a varying load resistor connected to the secondary of the test transformer.

jim hardy
Gold Member
Dearly Missed
In about 1880 they were figuring this stuff out. If you can find old engineering magazines it's fascinating to read their descriptions and explanations. Over the years the explanations become so concise and sparsely worded they lose their efficacy.

Okay we're talking sine-waves here....

You understand that magnetizing current establishes MMF and flux will circulate in the core in proportion to MMF(measured in amp-turns) divided by reluctance of core.

That flux induces a voltage in primary which opposes current flow and we call that voltage "counter-EMF".

Same flux also induces a voltage in secondary because it's wrapped around same core.

If current is allowed to flow in secondary , that secondary current establishes a second MMF which opposes the primary MMF.
Since secondary MMF opposes primary MMF, flux decreases, meaning less counter-emf, so more current rushes into primary because applied voltage is no longer so vigorously pushed back against by counter-emf.
That increased primary current re-establishes flux to a level that makes enough counter-emf to stabilize primary current.

This was described in an 1880-ish magazine as "A beautifuly self regulating system". The experimenters were overjoyed that no external device was required to limit primary current - if you keep the reluctance of core low it'll regulate itself.

I hope this word picture helps you. There's plenty of folks here who can do a better job with the equations that I could.

It is often helpful to retrace our predecessors' footsteps.

If you get interested in antiquities of EE, Sylvanus P Thompson's 1897 book "Dynamo Electric Machinery" is reprinted and is available through Amazon.
https://www.amazon.com/dp/B005YQWUF8/?tag=pfamazon01-20

It has fascinating descriptions of many electrical machines. I used it a lot in my career - Thompson was a master explainer. His "Calculus Made Simple"(1917 i think) never went out of print and is in college bookstores today..
Someplace i have the 1901 version.

old jim

jambaugh
Gold Member
Some things to remember when parsing cases. An open circuit is an "infinite load" since R=infinity. So high load equates to the primary acting more like a simple inductor. If the ferromagnetic core is assumed to be ideal (completely coupling the primary and secondary) then it will make the inductance very high (idealized case = infinite). The primary acts as if it too were infinite load under any varying current.

A zero load on the secondary (short circuit) with no internal resistance gives you a superconducting electromagnet... effectively a very high diamagnetic core. The induced currents in the secondary will cancel out all the B field generated by the primary (in the idealized case).
The primary acts as if it has zero inductance since it is unable to induce a B field. It behaves as if it too were short circuited.

Thank you for the thorough explanations... Exactly what I was looking for. I was having a hard time finding explanations like this...

So the primary current flow (with an open secondary) can be calculated by figuring out the inductive reactance of the entire transformer and adding that to the wire resistance of the primary... and then applying ohm's law?

In the second case (with a shorted secondary), we can do the same thing but will find that the inductive reactance is significantly smaller... correct?

berkeman
Mentor
Thank you for the thorough explanations... Exactly what I was looking for. I was having a hard time finding explanations like this...

So the primary current flow (with an open secondary) can be calculated by figuring out the inductive reactance of the entire transformer and adding that to the wire resistance of the primary... and then applying ohm's law?

In the second case (with a shorted secondary), we can do the same thing but will find that the inductive reactance is significantly smaller... correct?

Correct on both counts. FOIWATER
Gold Member
I would like to associate myself with all the comments made by Jim Hardy, and thank him for the links.

Not sure about the open circuit one, because there is a magnetizing current for the core that can not be recovered.

If it is a pure inductor, then ohm's law applies. The energy goes into the inductor and then given back by the inductor. But there is a core, so there is energy loss (core loss).

Thank you for the thorough explanations... Exactly what I was looking for. I was having a hard time finding explanations like this...

So the primary current flow (with an open secondary) can be calculated by figuring out the inductive reactance of the entire transformer and adding that to the wire resistance of the primary... and then applying ohm's law?

In the second case (with a shorted secondary), we can do the same thing but will find that the inductive reactance is significantly smaller... correct?

jambaugh
Gold Member
Not sure about the open circuit one, because there is a magnetizing current for the core that can not be recovered.

If it is a pure inductor, then ohm's law applies. The energy goes into the inductor and then given back by the inductor. But there is a core, so there is energy loss (core loss).

Very late addendum: Even without a core the unshielded inductor acts as an antenna and you will have losses. But the point is to start with an idealized case, understand the asymptotic behavior and then (especially in resolving infinities which crop up) apply that understanding to realistic cases.

When answering questions like this I like to use the circuit model on slide 28 of this presentation. It's not perfect but it works fine for most applications.

For example it will pretty easily predict the secondary open circuit and short circuit behavior of a transformer.

How is the loaded secondary decreasing the inductance of the primary?

$v_1(t), i_1(t)$ is the primary voltage and current and $v_2(t), i_2(t)$ is for the secondary. voltage polarities are defined positive (+) for the "dotted" terminals and defined as positive current going into the same dotted terminal. then ignoring the core loss and copper loss, the ideal transformer has volt-amp characteristic as:

$$v_1(t) = L_1 \frac{di_1(t)}{dt} \ + \ M \frac{di_2(t)}{dt}$$
$$v_2(t) = M \frac{di_1(t)}{dt} \ + \ L_2 \frac{di_2(t)}{dt}$$

where M is the mutual inductance and is no greater than $\sqrt{L_1 L_2}$.

you can convert that into a phasor or Laplace domain expression. then you can connect a load to the secondary and see what the volt-amp relationship is for the primary. undergraduate electrical engineering level.