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JOHN BALI 20th November 2011 10:31 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 10:37 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.

TonyTecson 20th November 2011 10:44 AM

are you winding on an EI core or torroid?

AndrewT 20th November 2011 10:46 AM


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


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

KatieandDad 20th November 2011 10:51 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 11:22 AM

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 11:57 AM

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 12:00 PM

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

KatieandDad 20th November 2011 12:07 PM

Looks fine to me.

AndrewT 20th November 2011 12:46 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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