
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
1st June 2010, 06:21 PM  #1 
diyAudio Member
Join Date: Jul 2009
Location: West bengal/siliguri

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 Last edited by sekhar; 1st June 2010 at 06:28 PM. 
1st June 2010, 06:57 PM  #2 
diyAudio Member
Join Date: Apr 2006
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. 
2nd June 2010, 08:20 PM  #3 
diyAudio Member
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 
2nd June 2010, 10:14 PM  #4 
diyAudio Member
Join Date: Apr 2006
Location: Minnesota

I don't completely understand the above calculation, but it doesn't look right. Do you have the equation?

2nd June 2010, 10:37 PM  #5 
diyAudio Member
Join Date: Jul 2009
Location: West bengal/siliguri

ok give me ur email address i wil send the book in pdf version which i had been following

2nd June 2010, 11:23 PM  #6 
diyAudio Member
Join Date: Apr 2006
Location: Minnesota

I don't need the book. If you don't want to provide the formula, that's fine.

3rd June 2010, 09:12 AM  #7  
diyAudio Member
Join Date: Jul 2009
Location: West bengal/siliguri

dont be angry
Quote:
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 

3rd June 2010, 10:37 AM  #8 
diyAudio Member
Join Date: Mar 2005
Location: Kerala

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 
3rd June 2010, 10:54 AM  #9 
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

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. Sent from my desktop computer using a keyboard Last edited by AndrewT; 3rd June 2010 at 10:57 AM. 
3rd June 2010, 10:59 PM  #10 
diyAudio Member
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 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
500w Class D  drzhuang  Class D  52  24th April 2006 12:53 AM 
Selfoscillating HalfBridge 500W+500W by IR. Interested?  deflorator  Class D  0  10th October 2004 03:30 PM 
New To Site?  Need Help? 