here is the new inverter transformer designed by me & with a lot guidance from tony(Antonio) in my previous project & TONY THANKS AGAIN FOR THE GUIDANCE
calculation
1.secondary current = 500/12=41.66~42A
2.so a CT transformer should have 21 + 21 A secondary rating . I did it 25 + 25A just for safety sake ......
3.assuming current density = 3.8A/mm square (I am nt very sure about this current density, i have read current density being used = 5A/mm square,6A/mm sqr & so on but just to be on the safe side took 3.8A/mm sq)
4.so the VA rating stands at = 24 X 25= 600VA
5.secondary turn = 17 + 17 (which should have been 18 + 18 my mistake)
6.wire gauge = 11SWG(cal = 25/3.8=6.61mm sq )
7.primary current = 600/240 X 0.85 =2.94 A
8.wire size = 2.94/3.8=0.77mm sq (using 19SWG= 0.81mm sq)
9.turn per volt = 1.5
10.I have multiple taps on the primary side- 220V=330 turns(+ 5%),240V=360 turns(+ 5%),270V=405 turns(+ 5%)
11.the transformer uses a EI core with following spec
12.area under bobbin = 5 sq inch
13.width of central arm = 2inch
14.width of the core =2.5 inch
15.window area = 3sq inch
Here Are the pic kindly comment- & i need advise about the current density I have read different current density being used which one to use wat is the standard ??????
I have made the transformer it works very well no heating , no humming.
one more Question I have calculated for 500W how much do u think this transformer can take up happily.......
& if somebody wants detailed calculations I can post it
regards
sekhar
calculation
1.secondary current = 500/12=41.66~42A
2.so a CT transformer should have 21 + 21 A secondary rating . I did it 25 + 25A just for safety sake ......
3.assuming current density = 3.8A/mm square (I am nt very sure about this current density, i have read current density being used = 5A/mm square,6A/mm sqr & so on but just to be on the safe side took 3.8A/mm sq)
4.so the VA rating stands at = 24 X 25= 600VA
5.secondary turn = 17 + 17 (which should have been 18 + 18 my mistake)
6.wire gauge = 11SWG(cal = 25/3.8=6.61mm sq )
7.primary current = 600/240 X 0.85 =2.94 A
8.wire size = 2.94/3.8=0.77mm sq (using 19SWG= 0.81mm sq)
9.turn per volt = 1.5
10.I have multiple taps on the primary side- 220V=330 turns(+ 5%),240V=360 turns(+ 5%),270V=405 turns(+ 5%)
11.the transformer uses a EI core with following spec
12.area under bobbin = 5 sq inch
13.width of central arm = 2inch
14.width of the core =2.5 inch
15.window area = 3sq inch
Here Are the pic kindly comment- & i need advise about the current density I have read different current density being used which one to use wat is the standard ??????
I have made the transformer it works very well no heating , no humming.
one more Question I have calculated for 500W how much do u think this transformer can take up happily.......
& if somebody wants detailed calculations I can post it
regards
sekhar
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Last edited:
The current density you are using is reasonable.
I assume you are using a line frequency of 50 or 60 Hz. How did you determine the number of turns needed? Also note that the RMS current will be greater than the average current because the current will be large narrow pulses.
I assume you are using a line frequency of 50 or 60 Hz. How did you determine the number of turns needed? Also note that the RMS current will be greater than the average current because the current will be large narrow pulses.
dont be angry
hey it is nt that i dont want to provide the formula but i dont know how well i can do it so i was planning on giving the complete book to u so that u can use it
well neways
ur equation - primary voltage X 100000000/ 4 X line frequency x Number of magnetic line passing(84000 - 65000) X cross section of core = Number of turn for a given voltage for a specific core size.
core size = root of VA/5.58
hope it helps but taking the book was a better choice
regards
sekhar
I don't need the book. If you don't want to provide the formula, that's fine.
hey it is nt that i dont want to provide the formula but i dont know how well i can do it so i was planning on giving the complete book to u so that u can use it
well neways
ur equation - primary voltage X 100000000/ 4 X line frequency x Number of magnetic line passing(84000 - 65000) X cross section of core = Number of turn for a given voltage for a specific core size.
core size = root of VA/5.58
hope it helps but taking the book was a better choice
regards
sekhar
mistype7.primary current = 600/240 X 0.85 =2.94 A
Pri I = 600/240 / 0.85 = 2.94Aac
Is the estimate of efficiency at maximum load really as bad as 85%?
That means you input 706W and output 600W and 106W is dissipated as heat.
A toroid of this VA would be expected to be between 96% and 97% efficient.
I would guess that an EI would have very approximately double the losses, i.e. instead of 3 to 4% assume 6 to 8% giving efficiency ~93%
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fr square wave also i will use the same power, only thing is the t/v will be change it will be around 1.1 times higher.
one thing I am using 8+8 =16 2n3055 transistor in the drive is the number of transistor enough for this given transformer
regards
sekhar
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