Open Circuit Transformer

I would like to ask, why in the real world open-circuit transformers have non-zero value for its primary current?

What are the physical effects that cause the primary current to be non-zero.

Dear intensified
Hi
It is simple . because we have some dissipations . such as copper dissipation and iron dissipation .
Best Wishes
Goldsmith

Intensified,

I would like to ask, why in the real world open-circuit transformers have non-zero value for its primary current?

What are the physical effects that cause the primary current to be non-zero. .

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I assume you are referring to a transformer with a primary and secondary. It is physically imposisible for the current in the primary to be complete zero unless there is a open in the windings. Usually the primary is designed with enough inductance so that the current is limited when there is no load on the secondary. When the secondary is driving a load, a negative impedance is reflected back to the primary and current increases.

Ratch

When the secondary is driving a load, a negative impedance is reflected back to the primary and current increases.

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Dear Ratch
Hi
I'm disagree with you about it . why negative resistance ? each transformer is an electromagnetic circuit that we can model it to an electrical circuit . there is no negative resistance . core is like a wire to carrying magnetic flux !
Best Regards
Goldsmith

Hello, if you leave a transformer open, the current is determined by the primary inductance. As the permeability of the core isn't infinite, the inductance isn't infinite. In other words, you need a certain H-field to get certain magnetization (B-field), you can find this in the material's hysteresis curve (BH-curve). As H = I*n/(magnetic path length), you need some current to get certain B-field in the core.

You may know "core flux" = B*Acore. The core flux is directly related to the Volt*second product for each turn.

One can reduce the primary current by giving it more turns (so reducing the peak flux density in the core and reducing the core loss). However, more copper means more resistive loss. As the winding area is limited, one has to find a balance between full load loss and no load current (and loss).

As mains transformers operate at high flux density (close to saturation), you can't derive the no load current based on low voltage inductance measurement of the primary winding. When you measure the no load primary current, you will see that it contains harmonics (so the shape of the current is not sinusoidal, though the driving voltage is).

goldsmith,

I'm disagree with you about it . why negative resistance ? each transformer is an electromagnetic circuit that we can model it to an electrical circuit . there is no negative resistance . core is like a wire to carrying magnetic flux !

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What happens in the secondary influences the primary. This is called a "reflection". I said negative impedance, not negative resistance. That implies both LCR from the secondary get reflected back to counter the primary's impedance in a negative manner. Reflection is a legitimate analysis tool. see the link below.

Ratch

http://www.vias.org/basicradio/basic_radio_06_07.html

What are the physical effects that cause the primary current to be non-zero.

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I guess you mean "What are the physical effects that cause a (copper) coil current to be non-zero"... right?

Even with zero resistance, zero hysteresis and zero eddy current loss, you still have magnetizing current. You need to polarize the magnetic domains. This proces stores energy (that comes from the source providing voltage and current. When the polarization becomes zero again, the energy is given back to the source. Therefore current and voltage are out of phase.

In a practical core, there is "friction" that causes the hysteresis effect (with its associated loss). In addition you have the eddy current loss in the core, and depending on size and frequency, in the windings.

Attached is transformer equivalent model. With no load, a good quality transformer looks like a high value inductor caused by the Xm inductance reactance. The current should be near 90 degs in relationship to input voltage phase. Resistive losses due to primary winding and magnetic core losses move the current phase relationship lower and more real power is dissiplated as heating. Cheap transformer skimp on turns per volt and core area so and higher no load losses.

For a given core area there is a sweet spot of turns per volt to keep the no load Lm high and still keep the winding resistance loss and core losses low over the total power load range. The larger the core, the lower the turns per volt.

Dear Ratch
Hi
As i asked from some of my friends , the behavior of transformer can't be considered as a negative impedance . negative impedance , means that when the voltage is increasing , the current decreasing , too , but at transformers we won't have this , and i didn't see it at some of famous text books that i have seen ever . ( definition of negative impedance for transformers . )
I got your meaning and i know that what you're trying to say , but we can't justify it's behavior as a negative impedance , because however there is an electrical isolation between input and out put , but we will have an electromagnetic circuit , ( you can model it to a simple circuit) , thus your consideration can't be correct .
Best Regards
Goldsmith

goldsmith,

As i asked from some of my friends , the behavior of transformer can't be considered as a negative impedance .

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Perhaps it is time to acquire some additonal friends.

negative impedance , means that when the voltage is increasing , the current decreasing , too ,

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That is only if the impedance is totally negative. However, a negative impedance can be used to cancel part of the positive impedance so that the voltage and current still increase together, but at a different rate.

but at transformers we won't have this , and i didn't see it at some of famous text books that i have seen ever . ( definition of negative impedance for transformers . )

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If the secondary influences the primary, reflective impedance will be present.

I got your meaning and i know that what you're trying to say , but we can't justify it's behavior as a negative impedance , because however there is an electrical isolation between input and out put , but we will have an electromagnetic circuit , ( you can model it to a simple circuit) , thus your consideration can't be correct .

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Did you really understand what I said? Electrical isolation means nothing when the transfer of energy is done through magnetic fields. Suppose we have a power transformer with a light load on the secondary. The primary will have a fairly large impedance because it does not supply its full power. Now suppose we load the secondary so that a large amount of power is needed. What happens? The secondary reflects a negative impedance to the primary so that the primary impedance becomes less. That does not mean that the primary is completely swamped with a negative impedance. The smaller primary impedance allows the primary to increase its current and thereby supply the secondary with more power. I am including a couple of pages from a textbook that describes this.

Ratch

Hi,

Transformer primary circuit is just like a normal coil. Adding a secondary coil with mutual magnetic flux makes it a transformer. Adding load to the secondary coil affects the mutual magnetic flux and thus the behavior of the primary coil. So an open transformer is seen as primary coil and transformer with load gets affected by the load to the primary side too.

You can google "transformer equivalent circuit" to see how this looks in electrical point of view.

In my opinion, the "reflected impedance" approach isn't but a way to make simple things complicated. I never met this concept during my studies and meant to understand transformers well without it up to now.

I understand, that some textbooks apparently like "reflected impedance". Personally, I don't find the concept instructive or visual. But that's probably a matter of taste or just a "school" in engineering theory, thus I don't want to fight the concept. I simply assume, that the text book calculations are consistent so far, it looks like as this is the case.

In my view, the simple idea of "transformed load impedance" (a positive quantity for real world loads!) is better suited to describe transformer operation. It seems to me, that after many words, the text book passage about reflected load ends up with an expression similar to transformed positive load impedance.

You should however understand, that statements involving "negative impedance" related to passive electronic networks sound confusing at least for people with a regular engineering background and bring up contradiction. I don't see the term "negative impedance" founded by the quoted literature, by the way. You may be beating a dead horse in this point.

In other words, you don't need to teach Goldsmith about transformers. Just consider, that there are reasons why he had difficulties to understand your previous contribution.

FvM,

You should however understand, that statements involving "negative impedance" related to passive electronic networks sound confusing at least for people with a regular engineering background and bring up contradiction. I don't see the term "negative impedance" founded by the quoted literature, by the way. You may be beating a dead horse in this point.

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Although it works for me, perhaps "negative impedance" may not be the best way to describe what goes on in a transformer. I do like the concept of "reflective impedance", however.

Ratch