hi , Friends i am very happy to see all replies but i really didn't understand i mean need a formula for example if i use 22 guage primary then what should be the guage for secondary is there any accurate formula i visted page http://www.powerstream.com/Wire_Size.htm here i found
Maximum amps for power transmission = 0.92 and
Maximum amps for chassis wiring = 7
i am surprised what is actually 22 guage wire can bear in a transformer i mean is it 7 or 0.92
and if i used 12 guage wire for secondary then for it is it 41 amps or 9.3 amps
friends i really need a formula like for guage of primary wire , guage of secondary should be . i am not good in electronics . help me friends thanks

 

Transformers can handle some maximum amount of power. While there are some losses, the power in the primary is about the same as power in the secondary. Here is the article in out Ebook that discusses transformers - http://www.allaboutcircuits.com/vol_2/chpt_9/2.html.

Notice that you have to know if the transformer is stepping up the voltage, or stepping it down. If it is stepping down, then current in the secondary will be greater than in the primary so the product of voltage and current will be the same. In that case, the secondary wire gauge will be larger than the primary.

For a good answer to your question, we need to know the transformer power, it's primary voltage, and the secondary voltage.

 

hi , thanks a lot for replying well detail for transformer is
primary voltage = 220
secondary voltage = 24
frequency = 50/60 hz
now problem is that i don't know guage ratio i know formula for turn ratio like
Vp/Vs = Np/Ns
but for guage i have nothing i visited page http://www.powerstream.com/Wire_Size.htm
where Maximum amps for chassis wiring are different and Maximum amps for power transmission are different i am totally confused i mean i my transformer is ideal and i used 22 guage magnetic wire for primary then amp it can bear according to table is 0.92 and thus
p= 220 * 0.92= 202.4 va and for secondary i have 24 volts thus i used 12 guage magnetic wire then
p= 24*9.3 = 223 va
is it right what i am calculating or totally wrong
last time i opened a good transformer i measured the gauge of primary it was 21 then i calculated va rating of transformer and it was
p= 1.2 * 220 = 264 va which is totally wrong rating mentioned on it was 1000 va . help me .

i need a good formula by using which i can wind a transformer like if primary is 22 guage what should be the guage of secondary . and if i use 24 guage primary then guage of secondary and so on ??????? help me friends please. thanks

 

hi , thanks a lot for replying well detail for transformer is
primary voltage = 220
secondary voltage = 24
frequency = 50/60 hz
now problem is that i don't know guage ratio i know formula for turn ratio like
Vp/Vs = Np/Ns
but for guage i have nothing i visited page http://www.powerstream.com/Wire_Size.htm
where Maximum amps for chassis wiring are different and Maximum amps for power transmission are different i am totally confused i mean i my transformer is ideal and i used 22 guage magnetic wire for primary then amp it can bear according to table is 0.92 and thus
p= 220 * 0.92= 202.4 va and for secondary i have 24 volts thus i used 12 guage magnetic wire then
p= 24*9.3 = 223 va
is it right what i am calculating or totally wrong
last time i opened a good transformer i measured the gauge of primary it was 21 then i calculated va rating of transformer and it was
p= 1.2 * 220 = 264 va which is totally wrong rating mentioned on it was 1000 va . help me .

i need a good formula by using which i can wind a transformer like if primary is 22 guage what should be the guage of secondary . and if i use 24 guage primary then guage of secondary and so on ??????? help me friends please. thanks

 

Hello,

There is no real formula for a gauge ratio.
You should use a wire gauge value that can handle a larger current then your maximum calulated current.
For more information on transformers see :
http://www.educypedia.be/electronics/electricitytransformers.htm

Greeting,
Bertus

 

22 gauge would be limited closer to .9 A---not 7 A. And 12 gauge is rated for 9A, not 41 A. 7 Amps would require 13 AWG, and 41 Amps would require 5 AWG.

A safe formula that might help is AWGsize = 50 - 10*log(700*I) where I is current, of course. Round your answer down to the next whole number (e.g., 5.42 to 5).

Likewise, if you already have an AWG wire size but would like to know its current carrying capacity (safe), I = [10^((50-AWG)/10)] / 700. You could just look at the tables, but math is more fun.

As you can see, the formula and the AWG scale are logarithmic. I don't think you will find a formula as simple as the turns ratio formulas.

For your second power calculation, I think you used 9.3 A because that is what 12 AWG is rated for. But Is/Ip = Np/Ns = Vp/Vs, and I got 8.43 amps which in turn gives a power rating of 202.4 VA. Stick with the lower calculation, not the 223 VA. 12 AWG winding is still good for 8.4 A, though.

 

See also: http://www.pupman.com/listarchives/1998/April/msg00222.html

As has been noted, the wire need simply be large enough to carry the maximum expected current. It is a good idea to go larger than needed.

 

Oh, and please refrain from posting multiple identical topics. Thank you.