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Old 27th December 2011, 03:08 PM   #71
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The Power transformer we are going to compute specifies 12 volts output with 750ma current (my Example) converting milliamperes to ampere will give us. .75a , multiplying voltage by current gives us wattage or power.

12x.75 = 9 watts

after having solved for the wattage of the transformer, which is 9 watts, look at Tables 1, where 9 watts is listed, follow the table horizontally and pick up the date for the center leg, thickness and turns per volt.

Center leg - 3/4
thickness - 3/4
turns per volt - 10

the nest step is to compute for the number of turns for the the primary and secondary coils respectively. the number of turns per volts in the primary is equal to the number of turns per volts in the secondary.

Primary coil - 220 volt (base our voltage rating in our country which is the PHILIPPINES)
Secondary coil - 12 volts
Number of turns (primary) = 220x10 = 2200 turns
Number of turns (secondary) = 12x10 = 120 turns

having computed for the number of turns in the primary and secondary coils. determine the sizes of wire for these coils in the next step . before computing for the guage or size of wire (AWG) must find out the current capacities of wires and their corresponding sizes. this data is provided in table 2 .

the power formula is used again in solving for the currents of wires in the primary and secondary. dividing the watts over voltage gives us the current of the wire:

I = w/e

I secondary = 9watts/12volts = .75 a
I primary = 9watts/220volts = .04a

after having solved for the currents of wires pick up the corresponding sizes of wires from table 2. choose the closest current listed in the table.

Size of wires (Primary) - #35
Size of wires (Secondary) - #23

in case of a multi-voltage secondary, the higher voltage should be taken for computation. if the transformer has multiple secondary coils, the wattage of each coil is computed and then summed up. in other words, the wattage of the primary is equal to the total wattage of all secondary coils.

if the secondary coil is only center-taped, half of the wattage and current is taken to compute for the wattage of that coil.

the primary is 220-0 volts, the secondary is 12-0-12 volts
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File Type: jpg transformer coil table 2.JPG (144.9 KB, 139 views)
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Old 2nd January 2012, 11:42 AM   #72
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Quote:
Originally Posted by tranqspogi View Post
the primary is 220-0 volts, the secondary is 12-0-12 volts
Thanks tranqspogi, its useful for me...
our primary is the same that is good but I guess the frequency is different.
Right here in Indonesia that is 50Hz, so my trafo need more turn for same T.

Btw I want some opinion about the "red paper" that I attach,
I want to use it, or maybe you can tell me what paper did you use to make your transformer? or some picture of it

Regards
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Old 3rd January 2012, 12:01 AM   #73
kbeist is offline kbeist  United States
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Thank you tranqspogi for the great information and excellent examples.
But I would like to know where I might be able to access the tables that you have pictures of??? I am sure the internet has them somewhere, but it would be bennificial if you could direct some of us laymen to the proper website.
Thank you for your knowlege.
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Old 3rd January 2012, 03:15 AM   #74
AJT is offline AJT  Philippines
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Quote:
Originally Posted by JOHN BALI View Post
Thanks Tony

I find this paper

Maybe it is can be used...
my mother often use it in the kitchen to make some cake
btw, now is mother's day here in my country
Happy mother's day for my mom
wish you all the best

Then if we make trafo that run cool (below 1T)
why we need so expensive insulator material?
But if we run the trafo more than 1T
that good insulating material must be used

So if I go 0.8-0.9T & I use this paper I guess it just okay
but if not okay please tell me...
The trafo that I made earlier even not using inter layer insulator
you can use that paper.....make sure it is dry.....
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Old 4th January 2012, 02:37 PM   #75
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ok...
now I have better feeling about it

I'll be back with my latest traffo build
& a few more formula
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Old 5th January 2012, 07:34 AM   #76
Riff is offline Riff  Indonesia
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Quote:
Originally Posted by JOHN BALI View Post
Does someone know the formula
to get the best numbers of turn to this kind of transformer/trafo?
I do have the formula, but maybe some other formula to compare with

This my formula for EI trafo for 50 Hertz frequency

primary winding
n = 45 * V / A

secondary winding
n = 49.5 * V / A

n = number of winding turn
V = voltage input or output (volt)
A = core area (cm square)

Maybe someone help to add another formula here, please...
Wanted, trafo DIYers

Thanks
salam kenal....saya arif dr jkt,
bli Nyoman pnya skema smps 500w ga?
tentunya sparepartnya yg mudah dicari di jkt.
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Old 5th January 2012, 07:54 AM   #77
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Riff This is an English speaking forum. Please post an English translation with your post.

Silahkan posting terjemahan bahasa Inggris.
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Old 5th January 2012, 09:44 AM   #78
AndrewT is offline AndrewT  Scotland
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Post79 shows a translation from english that appears, very unusually, to be much shorter than the english version.
Is there something missing from the translation?
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Old 5th January 2012, 09:50 AM   #79
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Hi Andrew the only part that was translated from English was "Please Post an English translation with your post" I find that the translation engines work best with simple sentences

The translation engine did not do a particularly good job of translating the post I referred to though, so whether my message gets through I am uncertain, but putting the Indonesian back through (always a good test) returns "Please post the English translation" So I guess google simplified it



Tony.
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Old 5th January 2012, 10:45 AM   #80
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This question has been asked many times and answered almost as many times.

During my HND training we were taught the following formula.

Turns/Volt = (1 x 10e9) / (K * FF * B * f * Ai)

Where FF is Form Factor - 1.11 for Sine Wave, 1.00 for Square Wave and 1.10 for Triangular Wave.
K = 4
B = Max Flux Density in mT. (780mT is good practice but anything up to 1350mT)
Ai = Area of magnetic CORE in square millimeters
f = Frequency

And for Ai:

Ai = (Square Root W) / K

Where W is 110% of the TOTAL secondary power
K = 0.0086 for EI, 0.015 for TU and 0.035 for Ferrite type A13
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