B Are silicon iron sheets really necessary for a commercial transformer?

[Steve Rogers]

TL;DR Summary

Does a transformer with its primary coil and secondary coil intertwined really need a core of iron?

Hi, everyone. I just finished studying the principle on which a transformer works. It relies on Faraday's law of induction. And my high school physics book uses the following picture for illustration:

Roughly speaking, the core of iron is used to pass the magnetic field built by the primary coil to the secondary coil. But I found an interesting video clip on youtube that really confuses me:

Build an electric transformer (DIY)

In this video, a man tries to make a transformer from scratch. The design for his transformer is basically the same as that for a commercial transformer, that is, the primary coil and the secondary coil are wound coaxially. And I'm wondering if the silicon iron sheets he used are that indispensable. After all, the primary coil can now pass its magnetic field on its own. Does anyone have an idea? Thank you.

 

[anorlunda]

https://duckduckgo.com/?q=air+core+transformer

Do that search for air core transformer. You'll see many hits.

The qualifier "commercial" is too vague to be meaningful.

But even better, you should learn about autotransformers, where a portion of the primary and secondary coils are shared. Below is a depiction of an autotransformer with three secondaries.

 

[Steve Rogers]

Thank you. Yeah, maybe you got a point, but I used the term "commercial" to distinguish the transformer in the clip from the one in physics textbooks. And I really want to know what role the silicon iron sheets play in a "commercial" setting. Do they make the transformer work better? If not, can we remove these sheets?

 

[Vanadium 50]

Do they make the transformer work better? If not, can we remove these sheets?

Do you really think people are buying these steel sheets just for fun? That they aren't buying them because they need them?

The steel directs the magnetic flux into the secondary. Seems like a good reason.

 

[Gordianus]

An iron core greatly enhances the magnetic coupling between primary and secondary, making the transformer more "ideal". Air core transformers are difficult to design and too bulky.
Simple drawings make you believe the iron core might be redundant.

 

[DaveE]

The iron is used in sheets so that currents won't be generated in the core material, just like they are in the windings. The insulation created between the sheets interrupts the current path that is concentric with the windings. Without this you'll have a very hot and inefficient transformer.

Other core materials deal with this in different ways. Powdered iron has lots of these interstitial gaps to prevent eddy currents in the core. Ferrites are highly resistive by nature since the ferromagnetic atoms are contained within the ceramics.

 

[sophiecentaur]

Summary:: Does a transformer with its primary coil and secondary coil intertwined really need a core of iron?

And I'm wondering if the silicon iron sheets he used are that indispensable.

In the past, there was a very common terms for your "sheets" and that was "laminations". If you use that term in your searches, you may find a large number of hits.
TBH, I have never come across the term you are using.

 

[rude man]

Post 6 is the right answer. Without laminations (i.e. with a solid version of the same core with the same core material) you would have Eddy currents. You should convince yourself that the laminations are arranged such as to prevent these currents by the electrical isolation between laminations

 

[alan123hk]

There is a more intuitive explanation regarding the use of laminated structures to reduce eddy current losses. Although this explanation does not use strict mathematical proofs, I personally think that this explanation is convincing and worthy of reference.

Please consider a square iron core, as shown on the left below, and then cut the iron core into four smaller parts, as shown on the right below. First, since the magnetic flux density does not change, the area of each small part is only one-fourth of the original, so the EMF of each small part is only one-fourth of the original.

The trick now is how to consider the effective resistance of each small part. Obviously, the effective resistance length of each small part is reduced to half, but its effective resistance cross-sectional area is also reduced to half, that is, we can approximate that its effective resistance has not changed.

Assuming that the uncut EMF is 1V and the effective resistance is 1 ohm, so the original power loss is 1W. Now the total power loss after cutting is reduced to $~4 ~\frac {(\frac {1} {4} V)^2} {1Ω} = \frac {1} {4} W ~$, that is equal to one quarter of the original.