Got one more question guys to complete the above formuli you's have kindly simplified for me. That's wire thickness for primary and secondary windings... Any help would be appreciated. Thanks
It's all about wire resistance and losses, which you want to both minimize and distribute evenly between windings.
Power dissipation is I² * R;
Assuming an ideal transformer, the current is indirect proportional to the turns ratio: Ip/Is = Ns/Np;
Ip = primary current [A]
Is = secondary current [A]
Np = primary number of turns
Ns = secondary number of turns
For equal losses in the primary and secondary windings, you want Ip² * Rp = Is² * Rs;
If you do the basic math, you'll find that this criterion is met with:
Ap = As * Ns/Np;
Ap = Cross-sectional wire area for the primary winding
As = Cross-sectional wire area for the secondary winding
Hence the diameter of the wire used for the primary winding is:
dp = ds * sqrt(Ns/Np);
It just so happens that this means, that the total cross-sectional area of the primary winding should be about equal to that of the sum of all secondary windings.
The resistance of a wire is calculated: R = rho * l / A;
rho = electrical resistivity [ohm*m] (depending on wire material)
l = lenght of the wire [m]
A = cross-sectional area of the wire [m²]
see
https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivityExample:
If you want two windings, say 10 turns for the primary and 1000 for the secondary, then you know the current ratio, i.e. primary current will be 100 times the secondary current.
Now you have to figure out what the maximum wire diameter is that fits the space. Assume that space is e.g. 20mm wide and 14mm high (just some random numbers not related to any particular core!), then you will subtract a few millimeters of the height to leave room for insulation wrappers plus a little spare, so let's say you take 4mm for that and then 10mm are left. That is a cross-sectional area of 20mm * 10mm = 200mm². That means 100mm² for each winding.
Primary side: 100mm² / 10 turns = 10mm²;
Secondary side: 100mm² / 1000 turns = 0.1mm²;
For calculating the wire diameters, do not use the common formula A = d² * Pi/4, as there is a fill factor that can never be 100% with round wires. Simply use d² instead, thus d = sqrt(A);
Now the associated diameters are 3.16mm and 0.316mm.
In practice, you might use 3mm for the primary and 0.3mm for the secondary in this example.