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JOHN BALI 20th November 2011 11:31 AM

Iron Core Transformer Formula
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


AndrewT 20th November 2011 11:37 AM

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.

AJT 20th November 2011 11:44 AM

are you winding on an EI core or torroid?

AndrewT 20th November 2011 11:46 AM


Originally Posted by JOHN BALI (Post 2788385)
This my formula for EI trafo for 50 Hertz frequency


Originally Posted by Tony (Post 2788396)
are you winding on an EI core or toroid?

KatieandDad 20th November 2011 11:51 AM


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

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.

AndrewT 20th November 2011 12:22 PM

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?

KatieandDad 20th November 2011 12:57 PM

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.

AndrewT 20th November 2011 01:00 PM

But, have I interpreted your formulae correctly and drawn the correct conclusions?

KatieandDad 20th November 2011 01:07 PM

Looks fine to me.

AndrewT 20th November 2011 01:46 PM

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

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