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I was working on the SMPS transformer and wanted to wind my own. For that I found the initial step of power calculation of the SMPS transformer. It stated to find the power using the products of area given as:

AwAc=(PoDcma)/(fBKt)

a) Considering an EE type core, the area of the window is taken to the area shown. I am not sure if the area is taken for only one window(Aw) or both? If one side then why? Should it be both added together.

enter image description here

Image link.

  1. the link above also tells that we need to decrease the area of window by some around due to bobbin.

I found one of the transformers in the 1000W inverter. Where the bobbin size almost made the window area negligible. But still the transformer successfully worked at its full power. How is it doing that. Shouldn't it's power capacity decrease as window area gets small?

enter image description here

Kindly help me in the following issues and kindly help in case I am missing some concepts.

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The window area is the area of the section where the wires' cross-sectional areas sit.

Should it be both added together.

There are no multiple window areas. There's only one window area and it's generally given in the core (and accessories) datasheet as AN or AW. The following is from TDK's ETD39 datasheet:

enter image description here

The image you nicked from the linked question doesn't correctly show the window area (I had edited @Transistor's answer there, BTW). The area marked with red diagonal lines is the maximum you can get, theoretically. But practically it's going to be less because there's a bobbin. That's why you should always check the datasheets.

I found one of the transformers in the 1000kW inverter. Where the bobbin size almost made the window area negligible. But still the transformer successfully worked at its full power. How is it doing that. Shouldn't it's power capacity decrease as window area gets small?

1000 kW means, if we assume the output voltage is 250 Vrms and the load is pure resistive or power-factor-corrected, 4 kArms. This current level is so high that there's no way you can carry it with those thin wires (Even if we assume you meant 1 kW and made a typo, still the above comment holds). Despite the fact that the transformer may belong to an HF inverter, the skin effect wouldn't allow the designer to use those thin wires to carry 4 kArms.


Before everything else you need to understand the magnetisation (flux swing, energy transfer, etc) and the relation with frequency at first. I admit, winding your own transformer and taking measurements on it is a very good way. The area-product formula which is available in different forms everywhere is based on some assumptions which may or may not apply (fully or partially) to your needs. Yes, it gives a good starting point but there are many other things to consider.

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  • \$\begingroup\$ This was found in a high frequency inverter. Secondly it is possible that the frequency used in the inverter was very high that made it for such a power. But my main question was regarding the window size in this case. Its very less as compared to actual given in the datasheet due to the bobin and winding but still its able to work well. Almost 5 wires are combined and used as a single one which gives high current capability \$\endgroup\$
    – kam1212
    Oct 19, 2023 at 18:34
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    \$\begingroup\$ @kam1212 Yes, the area product formulas are just an approximation with a number of assumptions because they don't take cooling, mounting etc into account. So it gives you a starting point but you may end up with a larger transformer/core after your tests and measurements. For example, if you rely on convection cooling (no fan or forced air) then for wire selection you may start with a current density of J = 420A/cm². And you'll notice that the wires will never fit in :) \$\endgroup\$ Oct 19, 2023 at 18:52
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    \$\begingroup\$ @kam1212 It's the inverse of the current density (1/J in my previous comment). EU prefer J = amp per millimetre squared, US prefer D = circular mils per ampere. \$1 \ cmil \approx 5 \ 10^{-4} \ mm^2\$. It's basically the same thing but in different forms. 420 A/cm² in my previous comment corresponds to ~500 cmil/A which sounds aggressive. Using a smaller number (e.g. 500 or lower) brings you thicker wires (or, taking the skin effect into account, thinner wires but a higher number of strands) therefore less copper resistance and copper loss, but requires larger space (window area). If your ... \$\endgroup\$ Oct 19, 2023 at 19:38
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    \$\begingroup\$ ... system employs good cooling (forced air, etc) you can select higher numbers (750 or greater), so use thinner wires and less space (window area). But the copper losses will be higher. You can rely on your cooling but the efficiency will be lower. \$\endgroup\$ Oct 19, 2023 at 19:40
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    \$\begingroup\$ @kam1212 CORRECTION The diameter of one strand can't be higher than twice the skin depth (d in mm, not in cm) of copper at the maximum operating frequency \$d_{mm}=72/\sqrt{f_{Hz}}\$. Make sure you use enough number of strands to meet the required (or selected) cross-sectional area which can be obtained from \$D_{cma}\$ (or J). \$\endgroup\$ Oct 20, 2023 at 6:27

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