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
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