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 Thanks 
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. 
are you winding on an EI core or torroid?

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

Looks fine to me.

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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