Transformer Design

Question

I have transformer which is used for a sin-wave invert-er whose data are follows

**Primary Winding details : 250 Volt ,375 turns with 19 SWG copper .

Secondary Details : 7.5 Volt , 11 turns with two 10 SWG (two 10 SWG
parallel winded )**

Core Details : 7/2.5 inch or 43/2.5
Bobbin : Window area 5.2 cm ., thikness -6.7

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I tried to Design the same on a paper and i cannot get the similar design Values and got stuck pleas help me (sorry for my Bad English)

My Design as follows

Core Area = 1.52 8 sqrt of (O/P voltage * O/P Current)

Core Area = 1.52 x Sqrt 1000VA (not taken efficiency )
= 48.06 cm^2

TPV = 1/4.44 * CA * Bm * F // Bm ,flux maximum =1.3 ,Fre = 50Hz

Turn Per Volt = 1/ 4.44 * (10^-4 * 48.06) * 1.3 * 50 = .7290volt.

Primary winding current = (Secondary volt * Secondary Current) / primary volt
* 90% efficeancy
Primary winding current : (7.5*133.33) / (250 * .9) = 4.44 amp So 15 SWG wire taken as per table

( but in my real transformer taken 19 SWG how ?)

secondary current is 133 so how the copper wire select i stucked here

primary winding area : .7292 * 250 =180.turns (but in Real transformer has 375 turns wher was i wrong ?)

secondary winding turns = 1.04 * .7292 * 7.5 = 7 turns ( but in real transformer it is 10 swg or with double layer
11 winding)

Please help me to how to desgin are taken the transformer

Answer 1

I see: you are examing the transformer you bought and your calculations don't match. Volts per turn is 0.7292 V/turn. Let's ask how many turns are required for 250V? Answer is: \$N=\dfrac{250}{0.7292} \approx 343\$. The first mystery is solved.
The real transformer has 375 turns, \$\dfrac{375}{343}\cdot 1.3= 1.42\$, you are lucky, because if you buy from ebay you can get a transformer with \$B_{max}\$ at 1.9 Tesla.
The wires are too thin: maybe you have a diffent table, or they cheated on copper. If you see that the window has a lot of space left, then they cheated. If you see that window is full, then a thicker wire couldn't be used.