0
\$\begingroup\$

I've calculated the area of product needed for transformer, it's 71cm^4, but in core datasheets there is not value with cm^4 unit for example: https://www.tdk-electronics.tdk.com/inf/80/db/fer/e_65_32_27.pdf, am I missing something here? How can I choose a core for calculated area of product in cm^4 enter image description here

https://www.ti.com/lit/ml/slup126/slup126.pdf

\$\endgroup\$
  • 1
    \$\begingroup\$ I know very little about this topic but why would area be measured in \$ cm^4 \$? I don't think you're missing something - I think you've got something extra. \$\endgroup\$ – Transistor Jun 6 at 20:09
  • \$\begingroup\$ source added to question \$\endgroup\$ – Das D. Jun 6 at 20:14
  • 6
    \$\begingroup\$ The source mentions on page 8 the area product which is different to the "area of product" in your question. \$\endgroup\$ – Transistor Jun 6 at 20:16
  • 2
    \$\begingroup\$ This is a way to determine the power handled by a specific transformer core. The area product is the product of the core window area by the cross-sectional area. As both are expressed in \$cm^2\$ you have \$cm^4\$ at the end. This is a classic to design power transformers. Some manufacturers do include these data in their data-sheets (look at Magnetics for instance) but some not and you have to resort to a core listing giving this AP parameter. \$\endgroup\$ – Verbal Kint Jun 6 at 20:33
7
\$\begingroup\$

Now that we've got the formula and source reference we can see that the source mentions on page 8 the area product which is different to the "area of product" in your question. (If you want an "of" in there it would be the "product of areas" rather than the other way around.)

Core Selection: Size

A novice in the art of transformer design usually needs some guidance in making an initial estimate of the core size appropriate for the application requirements. One widely used method, with many variations, is based on the core Area Product, obtained by multiplying the core magnetic cross-section area by the window area available for the winding.

Just before the AP equation the article explains ...

Some core manufacturers no longer provide area product information on their data sheets, often substituting their own methodology to make an initial core size choice for various applications.

That means that you'd have to calculate it from the information given in a datasheet.


enter image description here

Going back to the datasheet we can see that the area of the centre core, AE is 20 × 27.4 = 540 mm2 = 5.4 cm2.

I don't know whether the winding window is in the same orientation or at right angles so I'll assume that it's in the same plane as AE. From the drawing we can calculate that the window area, AW = 12 × 27.4 = 324 mm2 = 3.24 cm2.

AW (often referred to as AN in datasheets) is the area that is filled by the cross sections of the wires:

enter image description here

AW = (22.2 × 2) × (12) = 533mm² = 5.33cm².

Since some of this area is occupied by the bobbin (a.k.a. coil former), it's always lower than calculated. That's why this area is given in the datasheets for various types of bobbins for the same core. As can be seen at p.4 in the datasheet, AW is given as 4.35cm² for the only bobbin. As stated above, winding window area is mostly given as AN in datasheets.

From those numbers we can calculate:

AP = AE × AW = 5.4 × 3.24 = 17.5 cm4 5.4 × 4.35 = 23.5cm4

Remember that I don't know what I'm talking about.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Aw is determined by the coil former. Because there's always a dead/unused zone for winding. So it's not the correct way to directly multiply the dimensions. \$\endgroup\$ – Rohat Kılıç Jun 6 at 21:09
  • \$\begingroup\$ Thanks for that. The article says that, "This formula is based on current density of 420 A/cm² in the windings, and assumes a window utilization of 40% copper." Would that make any difference to your comment? \$\endgroup\$ – Transistor Jun 6 at 21:19
  • \$\begingroup\$ I meant the way of the calculation of Aw in your answer. You multiplied the dimensions directly but it's not the way. I hope I could describe with my terrible English and understood what you said correctly. \$\endgroup\$ – Rohat Kılıç Jun 6 at 21:37
  • 1
    \$\begingroup\$ Can you correct my misunderstanding in your answer or by an add-on edit to mine? Your English is excellent in all your posts and better than many native writers here. \$\endgroup\$ – Transistor Jun 6 at 21:41
  • \$\begingroup\$ Thanks for your kind words. I've put the info about Aw in your answer here as this is more detailed than mine. \$\endgroup\$ – Rohat Kılıç Jun 8 at 5:13
3
\$\begingroup\$

As @Transistor stated, it's the Area Product, not Area of Product.

The value can be used to select a core for required power level.

It's the product of core's central area which is given as AE, and the winding window area which is given as AN (aka AW). Please note that, unlike how @Transistor has shown in his answer, the value of AN is determined entirely by the bobbin (i.e. coil former).

How can I choose a core for calculated area of product in cm^4

Today manufacturers publish a list of cores for various power ranges and topologies. But still the AP method is useful. If you have the value of AP and not have a list for cores/power levels then pick a core, find its AE and AN (or AW) from the datasheet and multiply them. If the result is larger than or equal to your required value then the core is suitable for your needs.

For example, let's check if E65 is suitable for you: AE=530mm²=5.3cm² and AN=4.35cm². AP=23.06cm⁴ which is much lower than your needs. So it's not suitable.

PS: I'd like to know which values are you used for AP calculation. 71cm⁴ is quite large to me.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I got 2kW power at output, so there is more than 2kW on transformer because of rectifier and filter losses so I took it as 2.5kW power (worst scenario), than K constant is 0.017 for bridge converters, I choose 200mT as deltaB(100 positive and 100 negative), and 30kHz for transformer operating frequency. \$\endgroup\$ – Das D. Jun 6 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.