AN1673 APPLICATION NOTE

18/24

The design of the winding starts calculating the minimum requested primary turns number.

For turns number calculation:

[16]

2

(*) In the case AP is defined by the AP1,

∆B is calculated by the formula [17].

[17]

where:

Instead, in the case that AP should by AP2,

∆Bmax will be fixed by the designer (up to 0.400Tesla).

Choose primary turns number Np = 60

The turn's (n = Np/Ns) ratio will define the secondary winding turns number.

[18]

; n = 10

Where: duty=0.5;

Choose n = 12*; Secondary turns Ns = 5**

* The ratio n can be significantly higher than [18] because the Lmag effect

** The secondary winding is realized with Ns = 5+5 turns because of the center-tapped solution.

In order to limit skin effect (and the inherent power losses), the Litz wire solution has been adopted for the

windings:

As the primary and secondary windings have been defined, we need fix the inductances of the transformer.

The formula [19] allows calculating, with good approximation, the value of the primary inductance in func-

tion of ferrite set and the gap length, the ferrite's set parameters and the primary turns number.

[19]

Where:

l_gap is the gap size (in mm).

The calculated primary inductance will be in mH. In our case the L=0.85mH.

N

P

V

in min

4

⋅

8B

∆

ƒ

sw

Ae

⋅⋅

⋅

------------------------------------------

=

B

∆

Pcv

kh ƒ

sw

Ke ƒ

sw

2

⋅

+

⋅

----------------------------------------------------







1

2.4

--------

0.230T

==

Pcv

Pt

2Ve

⋅

---------------

=

n

V

in min

⋅

V

o

V

ƒ

+

--------------------------------------

=

L

10

1

–

µ

0

------------

le

µ

r

Ae

⋅

-----------------

0.1 l_gap

⋅

Ae

0.2 l_gap

π

4

---

⋅⋅

+



--------------------------------------------------------------

⋅











⋅









–

Np

2

⋅

=