Iron Core Transformer Formula

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Do you know the formula?
to get the best numbers of turn to this kind of transformer / trafo?
I have over simplified formula, but maybe some other formula to compare with

This my formula for EI trafo for 50 Hertz frequency

primary winding
n = 45 * V / A

secondary winding
n = 49.5 * V / A

n = number of winding turn
V = voltage input or output (volt)
A = core area (cm square)

Maybe someone help to add another formula here, please...
I want to make right trafo this time

Thanks
 
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the secondary shows a +10% correction for loaded output voltage.
I suspect this is a guesstimated value for regulation of the final transformer build.
Regulation can vary from <=3% to >=30%. A fixed turns ratio correction of +10% is not appropriate to all transformers.
 
This is OVERSIMPLIFIED.

The important thing to learn is the magnetic properties of the transformer that you are trying to wind.

The first formula is

A = Square Root (W)
.......----------------...= mm2
................K

Where W is the Secondary Power + 10%

K = 0.0086 for standard E & I laminated sheet
0.0150 for T & U laminations.

Once you've got the correct iron work for the transformer.

Turns / Volt = 1 x 10e9
..................------------
...................K x B x f x A

Where K = 4.44 for a SINE wave, 4.00 for a SQUARE wave.

B = Max Flux Density - 780mT is good for reliabilty, hoever this can be pushed to 1350mT

f = Frequency 50Hz or 60Hz (Unless you are design SMPS then its kHz)

A = The Core Area as discussed above.


Once you know the Turns / Volt its a simple case of working out the Primary and Secondary Turns.
 
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Is the first formula?
A = sqrt(W) / K
where W is the input power in Watts and K is 0.0086 for EI.
The answer in mm^2 (sqmm)

Is the second formula?
Turns/Volt = 10^9 / [ K' * B * f * A]
where K' = 4.44 for sinewave, B = 780mT for EI to stay cool and avoid the saturation knee on high mains voltage, f = mains frequency, A = core area in sqmm

Inserting values into that first equation gives the same answer as: Watts = 31 * [Core Area (in square in)]^2
using 100W and 1160sqmm = 1.8sqin

Inserting values into the second formula for 4.44, 780mT, 50Hz, 1160sqmm,
gives 5T/V
resulting in 1145T for 230Vac @ 100VA

Whereas, John's first formula result in 892T for 230Vac with 1160sqmm. This seems to confirm that Johns' first formula has assumed B=1000mT = 1T

Have I got this correct?
 
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1000mT is fine for commercial design.

If you've got the core material I would reduce the Magnetising Force (B).

You can ALWAYS increase the Turns / Volt, NEVER reduce them.

780mT is what the MoD employ. It also allows for Mains Transients of over 100%, not that the windings wouldn't take that for a short time anyway.
For SMPS designs the B figure needs to be MUCH lower.
 
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I have only done it once so far, but would testing for the knee in the B vs V curve give confidence in having guessed at the correct identification of the core material?

I found that plotting for Vac from 10Vac to 260Vac showed that the tiny 3VA transformer was being run at far too high a voltage. 160Vac would be far cooler than running it at a nominal 240Vac, even though it is rated as 220/240Vac. Lies!! or the cheapskates thought I couldn't care about being burnt by the surface of the EI core
 
Am I correct in guessing that the extra 10% for Turns on the secondary is related to a transformer regulation of 10%?

In which case that +10% should be replaced by the target regulation.

I can see that turns ratio primary:secondary is an easier way to get to secondary turns. But, knowing the primary T/V is useful for many subsequent design calculations.
 
That is right Andrew (your post #12)

The complete formula is:

B = 50/f x 45 / (Afe x N) x U, where

f = frequency in Hz;
U = rms value of voltage in V;
Afe = magnetic core cross-section in cm²;
B = flux density amplitude in T;
N = number of turns.

Source: Vacuumschmelze data on GOSS c-cores.
Please note that for other core configurations the value "45" might be slightly different.
 
Could you express a simple way to use or modify John's second equation.

or is
Once you know the Turns / Volt its a simple case of working out the Primary and Secondary Turns.
the simplest way to proceed. These turns ratios give winding emf. Fully loaded voltage values require knowledge of the final transformer regulation.
Can you see a way to address this?
 
Transformers working towards core saturation will be more "regulating" than transformers working at or below 1T.

i trade "regulation" for low exciting current and lower operating temperatures.....

equations are fine, they are mere guides as to how to proceed with traffo buildup....design is iterative process....

what happens if you find out that winding primary you end up with 5.5 layers? you can either use a bigger wire if space permits, use a smaller wire, make it 6 full layers and adjust the turns, or use smaller number of turns with just 5 layers.............or you can wind a screen winding.......

there are many design choices one can make, as long as you stay below the knee of the magnetization curves, you are able to do any of these things....

bottomline, whatever figure that comes out of the formula does not mean you have to stick to it, it never happens that way in the real world.....
 
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Thanks to all reply,
now I can recalculate every things, every thinks too :):)

my formula is OVERSIMPLIFIED, yes
because I like simple things but works nice

So that is the complete formula.
I just remember the 45 * V/A but don't know where its come from
for the 49.5, I get from 45 * 10%, just for minimize the looses from the cooper wire
so its for regulation too

Then for the wire to be used to winding
from the book that I was read about 9 years ago
(don't have any note just from memories in my head)
get this
Allowable currents to flow in the cooper wire per area (cm²) symbol S
S = 3A to 5A per cm²
for 100VA transformer I use 0.25 to 0.3mm diameter cooper wire for primariy (220volt)
for secondary I use as thick as possible from winding window left
Is this formula okay or not?

Thanks
 
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