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Transformer's core size

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julian403

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We know that the mains voltage is 110 sen (2 Π 60 t)

And by the faraday's law:

\[V = - \frac{ d}{ d t} {\int}_{S} \vec{B} . d \vec{S}\]

\[B\] is proportional to current. \[B = \mu \frac{N I}{L}\] where L is the magnetic path.

\[V = - \frac{ d}{ d t} {\int}_{S} \mu \frac{N I}{L}. d \vec{S}\]

And as the magnetic field in the core is constants:

\[V = - \frac{ d}{ d t} \mu \frac{N I}{L} A\]

The only staff which can changes is the current, so:

\[V = - \mu \frac{N }{L} A \frac{ d I}{ d t}\]

So, in the transformer there is:

\[{V}_{1} = - \mu \frac{{N}_{1} }{L} A \frac{ d{I}_{1}}{d t}\]

\[{V}_{2} = - \mu \frac{{N}_{2} }{L} A \frac{ d {I}_{2}}{d t}\]

And here is my question, the core is made of silicon steel which \[{\mu}_{r} \] ≈ 14000 and the saturation magnetici field is 2 [T].

So B can not be bigger than 2[T]

\[110 sen (2 \pi 60 t) = - 2 \pi 60 \mu \frac{{N}_{1} {I}_{1}}{L} A sen (2 \pi 60 t)\]

And \[B =\mu \frac{{N}_{1} {I}_{1}}{L} \] ≦ 2[T]


\[110 sen (2 \pi 60 t) = - 2 \pi 60 (2[T]) A sen (2 \pi 60 t)\]

I that way A must be \[0.14 [{m}^{2}]\] and that's bigger that the comertial transformer. Why?
 

For 60Hz laminated cores I simply use the formula:

gif.gi.gif​
 
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Yes, well what I want it's how not just apply an expresion. If you see I start with maxwell's equation. And I got an expression where the potential diference is proportional to B and A but B can't be bigger than the saturation values of the material which for electric silicion is 2[T]. So, in my thinking way there is something wrong.
 

julian403 said:
We know that the mains voltage is 110 sen (2 Π 60 t)

And by the faraday's law:

\[V = - \frac{ d}{ d t} {\int}_{S} \vec{B} . d \vec{S}\]

\[B\] is proportional to current. \[B = \mu \frac{N I}{L}\] where L is the magnetic path.
Click to expand...

My understanding is that B is proportional to volt-time over area.
 
Why? Is it a valid assumption?
Click to expand...
Yes because there is a coil with ferromagnetic core.

B is proportional to the magnetising current, not the load current, the two add vectorially...
Click to expand...
I is the magnetic current, because \[B = \mu \frac{I}{L]}\]

the quies
 

Ok, I'm not sure if I got the meaning ;-), but perhaps it is where the julian's assumption fails. The µ parameter should't properly be understood as a constant, but as being a function of the current B.
 

julian403 said:
Yes because there is a coil with ferromagnetic core.
Click to expand...

The magnetic field in the transformer core cannot be considered constant (over time) or over cross sectional area (this may be more acceptable for high flux density).

I do not understand the logic.

Also, how does the core knows that this is magnetising current and that is a load current and this is going to create a magnetic field and ....

Current is just a flow of electrons and you cannot distinguish between the magnetising current and a load current...
 

I'm not sure if I got the meaning , but perhaps it is where the julian's assumption fails. The µ parameter should't properly be understood as a constant, but as being a function of the current B.
Click to expand...

Well if we consider the resistors losses \[R\] and the dispersal flow in the coils \[L\], where \[u(t) = 100 \sqrt{2} sen (\omega t)\]

the equation is

\[u(t) = i(t) R +L \frac{d i(t)}{dt} - \frac{\mu N A}{L} \frac{d i(t)}{dt}\]

if you solve that you will get: \[i(t) = -k cos (\omega t) + f sen (\omega t)\]

where k and F are constants and \[k cos (\omega t)\] is the curretnt which generate the magnetic flux.

For ferromagnetic materials we should't consider it magnetic permeability as linear.
Click to expand...
That's what I means, there is a point that B does not grow more, the saturation point which for electric silicon is 2[T].
 

andre_teprom suggested a rule of thumb to estimate the core size for a particular transformer power. That's o.k. from an engineering viewpoint, in fact you'll find power (VA) numbers listed along with core sizes in transformer catalogs.

Nevertheless I would go a step back to emphasize the design method. Useable peak induction for a 50 or 60 Hz transformer is lower than 2 T, e.g. 1.2 to 1.6 depending on the core material and acceptable core losses. For a given core cross section, peak induction and frequency, you next calculate turns per volt. Then determine the minimal wire gauge for the specified current and check if the winding fits the coil former window. If not, use a large core.
 
In a choke all the current is mag current, in a Tx there are opposing currents, supply side and load side, the difference is Imag, and is measurable...
 

you are saying that. For the faraday's law: \[fem [V] = - \frac{d}{dt} {\int}_{S} \vec{B} . d\vec{S}\]

The B module can't be bigger than 15 [T] and it can't changes faster than \[2 \pi 60[Hz]\] (because that the line's frecuency). So the only thing we can modify to have a big voltage (fem) is the core sizes (A)

\[\dst fem [V] = - A \frac{d}{dt} 1.5[T] cos (2 \pi 60[Hz] t)= - 2 * \pi * 60[Hz] * 1.5[T]* A sen (2 \pi 60[Hz] t )\]

So if we want to have 20 [V] on the secondary \[20 = 2 * \pi * 60[Hz] * 1.5[T] * A\] so \[A = 0.035 [{m}^{2}]\]

And that a big core compare to the comertials
 

FvM said:
... Useable peak induction for a 50 or 60 Hz transformer is lower than 2 T, e.g. 1.2 to 1.6 depending on the core material ...
Click to expand...

In addition, it is wise to be conservative in design and there are so wide differences in the mu for transformer sheets. High silicon high mu steel is rather hard and unpopular with people who do the stamping. It is also tricky to do any spot testing...
 

So the only thing we can modify to have a big voltage (fem) is the core sizes (A)
Click to expand...
To be honest, I don't understand your calculations at first sight, but I don't see number of turns at all. Thus I guess, they are somehow wrong.

it is wise to be conservative in design.
Click to expand...
My rule of thumb when I was designing mains transformers is 42 turns*cm²/V for 50 Hz, based on Bmax of about 1.2T and 90 percent core fill factor. Double number of turns for audio output transformers (25 Hz lower cut-off frequency).
 

the formulae assume a single turn I assume (or are per turn if you like)
Click to expand...
Apparently. But the OP doesn't seem to consider that a winding can have more than one turn. Otherwise he won't deduce huge core cross section numbers from voltage specifications.
 

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