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Help me make a transformer for inverter design

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nvd

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Inverter

I want to design an inverter. But I want to wind the transformer by myself.
Any rules for the number of turns of the primary and secondary?
Any suggestions?
 

Re: Inverter

what is the input and output of your inveter? u need to know at least the power or the current output of your inverter.
 

Re: Inverter

you can use following fomula to design your transformer:

L * Imax = N * A(cross-section area) * Bs(max flux density)

best regards





nvd said:
I want to design an inverter. But I want to wind the transformer by myself.
Any rules for the number of turns of the primary and secondary?
Any suggestions?
 

Re: Inverter

funster said:
you can use following fomula to design your transformer:

L * Imax = N * A(cross-section area) * Bs(max flux density)

Can you please explain what these symbols stand for (L,Imax...). I intend to design a 500W transformer. Regards
 

Re: Inverter

Not goning into details (everything will be as simplified as possible), I will give you here an example of calculating a transformer with 110V primary and 8V @ 5A secondary (typical "square" transformer):

If "a" is the cross-sectional area of the core in square inches and P is the secondary power (V*A), for the output power in our example of 40watts (8V * 5A) we have:
a=0.16*√P=0.16*√40=1.0 square inch
Plugging this value into equasion which will give us Turns per Volt (N/V) we have:
N/V=5/a=5/1.0=5 (turns per volt)
The primary winding is 110V; therefore the turns required on the primary are 110V*5=550 turn
The secondary is 8V, so turns required for the secondary winding: 8*5=40.

In practice transformers have losses. If you add, say, 10% turns to the secondary winding you will be covered for these loses. So, the secondary winding (in this eample) will have 44 turns ..

These equasions are based on a very conservative value of B (flux density) of 75,000 lines/square inch.

If you use this example for your calculations you probably will be wrong by 5-10% but in 1-off situation this accuracy is more than enough, especially, when you probably don't have detailed information on the core ..

Windings current rating:
Here I will give you another example on how to select copper wire AWG.
Again, not going into details, lets assume that we select 1000 circular mils as a safe density. So for your secondary current of 5A we will have 5*1000=5000 circular mils. From "General-Purpose Wire Table" (easy to find on the web: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html#c2 ) we can find the nearest AWG of the secondary winding as 13 gauge (wire size AWG=13).
In the previous example we had total power of 40VA and th primary voltage of 110V. In this case your prmary current will be 40/110≈0.36A.
Again using he same value of "circular mils" of 1000, from the same table you can select for 360 circular mils (0.36*1000) the AWG will be between 24 and 25 wire size, and I would suggest the value of 25.

Regards,
IanP
 
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