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#21 |
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diyAudio Moderator
Join Date: May 2003
Location: Palatiw, Pasig City
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from the available winding area, i allow 80% for copper, 20% for insulation materials....
then from the 80% percent, i allot half of that for primary and the other half for the secondary..... you choose your wire size this way....even if your formula tells you to use gage 14 say, if space permits, nothing stops you from using #13.... if you can not get the desired wire size fit into the window, then you use the next bigger core size.....
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#22 | |
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Account disabled at member's request
Join Date: Apr 2009
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Quote:
There are professional software programs providing you with a complete winding scheme (number of turns, gauges, interleaving thickness a.s.o) given a particular core size and voltage/current requirements, and these programs are accurate (transformers are physical things; you can calculate everything). I have seen this in the "real world", just a pity that I can't afford something like that (the program I have seen "at work" is actually not for sale but is available on a lease basis). The fact that we wind our transformers on an iterate basis has more to do with not having this software; the equations I use are perfectly accurate and actually the same as in these software programs. Doing the calculations by hand just takes more time; the outcome however is equally accurate. |
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#23 | |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Quote:
0.3mm diameter and 5A/sqmm gives 353mAac times 230Vac gives 81VA before accounting for efficiency. I'd guess at ~70 to 75VA for 0.3mm diameter and 5A/sqmm. 0.25mm diameter and 3A/sqmm gives 147mAac times 230Vac gives 32.4VA. Reduce to ~25VA for output rating. If my sums are correct then your primary wire looks too thin. Can someone confirm where I have gone wrong? Last edited by AndrewT; 22nd November 2011 at 10:08 AM. |
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#24 |
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Account disabled at member's request
Join Date: Apr 2009
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Essentially right.
3A / mm² is for quality transformers, especially when temperature is a factor. 5A / mm² is for economy, where temperature, strayfields are not important. Primary current is determined by secondary power (load), not by what the core might be able to deliver. For example when I calculate a transformer which must supply 300 watts of secondary power, and I want B not exceeding 1T, I know a need a "600" watts core because of the larger required winding space. Wire diameter of the primary however can be calculated for 300 watts (no problem with the momentary higher inrush current at start up). |
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#25 | |
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diyAudio Moderator
Join Date: May 2003
Location: Palatiw, Pasig City
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Quote:
unfortunately, materials available to you is not available to me vis-a-vis cores, you know your core manufacturers, i do not.... you deal with fresh virgin materials, i use scrapped traffos and cores.... i build traffos that work, are not too hot to touch and doesn't burn out in use, i must be doing things right.....
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http://www.elab.ph/forum/index.php?topic=32688.0 |
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#26 |
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Account disabled at member's request
Join Date: Apr 2009
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I understand Tony, of course.
But be it your or my world, the same formulas apply. Important to know is what core material we deal with; the difference between 0.35 and 0.5 mm is physically obvious. When the laminations show an oxyde coating you can be pretty sure that the material is heat annealed. So we have already 4 possibilities, covering almost all options. For these options the important data are easily to find on the www, also for you. I would not rely on RDH3, 4 or whatever; mostly out of date. Good luck with your projects. |
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#27 |
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diyAudio Member
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I know I am late for this discussion, but here are the formulas I used back in high school, when quality ready-made toroids were not available. I did design a number of hand made transformers with these formulas, and all worked reliably. These formulas are (1) oversimplified, (2) very conservative, made for poor quality iron, and (3) designed for 50Hz.
Power(W) = (Core-Area)^2; Core-Area is in cm^2 Turns-per-volt = (50..70) / (Core-Area); Core-Area is in cm^2 Primary turns = Primary voltage * Turns-per-volt Secondary turns = Secondary voltage * Turns-per-volt * 1.2 Wire diameter = 0.7 * sqrt(Current), Current is in Amps Looks like these are is agreement with the (complicated) formulas above. |
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#28 |
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diyAudio Member
Join Date: Jun 2010
Location: Bali
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Yes its should be mm²
![]() but I gonna confused with this image from my real world no 1 = must be fake, 5A * 64v = 320VA to small for that power dimension 3" * 2.5" * 1.75" (stack) primary wire = 0.25mm diameter secondary wire = 0.7mm diameter no 2 = maybe its okay, 5A * 64v = 320VA, or maybe 5 * 32 = 160VA? dimension 3.75" * 3.25" * 2.15" primary wire = 0.5mm diameter secondary wire = 1.2mm diameter no 3 = was used in 30watt amp so its maybe quite good transformer, even the iron core is black in color, anyone know what is this core name? all trafo for 220 volt ac input the AWG table make me more confuse, from the shop that I bought wire they ask me in mm diameter not AWG's. Can anyone show me the right AWG to metric conversion. then for 0.25mm 0.5mm or 0.7mm diameter how much currents can flow safely? Please tell me the actual reliable VA for no 1 & 2 ? so my mind can be clear I'm also read this http://www.diyaudio.com/forums/power...ffo-build.html very good traffo making, but tube is out of my range Sir Tony, that winding machine is a DIY tool? so cool Thanks Last edited by JOHN BALI; 23rd November 2011 at 07:44 AM. |
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#29 |
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Account disabled at member's request
Join Date: Apr 2009
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#30 |
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diyAudio Member
Join Date: Jun 2010
Location: Bali
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ok, but sometimes I miss the right choice
too much result Maybe some experience of winding from you? I already made some rewinding of trafo EI & toroids too. some customize too but not sure about it work not sure the ratting itself So please give me your secret formula?
Last edited by JOHN BALI; 23rd November 2011 at 08:33 AM. |
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