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Old 1st June 2010, 06:21 PM   #1
sekhar is offline sekhar  India
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Location: West bengal/siliguri
Default 500W or more

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
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Last edited by sekhar; 1st June 2010 at 06:28 PM.
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Old 1st June 2010, 06:57 PM   #2
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Location: Minnesota
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.
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Old 2nd June 2010, 08:20 PM   #3
sekhar is offline sekhar  India
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Join Date: Jul 2009
Location: West bengal/siliguri
calculation for no OF turns
240 X 100000000/4 X 50 X 65000 X 5= TURNS AT 240V
SUBSEQUENTLY THE TURN PER VOLT CAN CALCULATED
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Old 2nd June 2010, 10:14 PM   #4
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I don't completely understand the above calculation, but it doesn't look right. Do you have the equation?
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Old 2nd June 2010, 10:37 PM   #5
sekhar is offline sekhar  India
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ok give me ur email address i wil send the book in pdf version which i had been following
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Old 2nd June 2010, 11:23 PM   #6
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I don't need the book. If you don't want to provide the formula, that's fine.
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Old 3rd June 2010, 09:12 AM   #7
sekhar is offline sekhar  India
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Join Date: Jul 2009
Location: West bengal/siliguri
Default dont be angry

Quote:
Originally Posted by sawreyrw View Post
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
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Old 3rd June 2010, 10:37 AM   #8
rajudp is offline rajudp  India
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This transformer can handle 735w on sinewave. if you want you can use it for more power but not continues.

and for center taped winding i use current on each side is calculation as (total current/1.41) A in your case it is 42/1.41=29A
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Old 3rd June 2010, 10:54 AM   #9
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by sekhar View Post
7.primary current = 600/240 X 0.85 =2.94 A
mistype
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%
__________________
regards Andrew T.

Last edited by AndrewT; 3rd June 2010 at 10:57 AM.
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Old 3rd June 2010, 10:59 PM   #10
sekhar is offline sekhar  India
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Join Date: Jul 2009
Location: West bengal/siliguri
i am going to use it for a square wave or pwm based inverter............ then how much load will it take ???????????????????????/



regards
sekhar
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