AN1262 APPLICATION NOTE

10/42

(that is a larger input capacitor) is of help. Some iterations, involving a recheck of the points mentioned in "Pre-

liminary Calculations - step 2", may be necessary.

If no solution can be found, either some specification should be relaxed or the power handled by the converter

should be derated.

9

FLYBACK TRANSFORMER DESIGN

primary-to-secondary turns ratio (n) are still to be defined.

The primary inductance will be chosen so that the converter is operated on the boundary between DCM and

(4)

(5)

With the complete set of specification, the transformer design can start with the selection of the magnetic core

material and geometry.

Table 9. Ferrite Materials selection

As to the magnetic material, a standard soft ferrite for power applications (gapped core-set with bobbin) is the

usual choice: the switching frequency is not so high thus special grades for high frequency operation are not

required. Table 9 shows some suitable materials.

The geometry will be usually a popular E or E-derived type. Other configurations, such as RM or PQ cores, are not

recommended because they are inherently high leakage geometries, since they result in narrower and thicker wind-

ings. Consider that minimizing leakage inductance is one of the major tasks in the design of a flyback transformer.

Among the various shapes and styles offered by manufacturers the most suitable one will be selected with technical

and economic considerations. Table 10 shows some possible choices with the relevant data useful for the design.

Grade

Saturation flux density

[T]

Manufacturer

B2

0.36

THOMSON

3C85

0.33

PHILIPS

N67

0.38

EPCOS (ex S+M)

PC30

0.39

TDK

F44

0.4

MMG

Lp

V

inm i n

V

DS on

)x

()

–

() D

X

⋅

[]

2

2f

sw

P

inT

⋅⋅

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

=

n

V

R

V

out

V

F

+

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

=

1.15 10

5

–

B

2.26

∆

f

sw

1.11

⋅⋅

⋅

=

1.54 10

7

–

B

2.62

∆

f

sw

1.54

⋅⋅

⋅

=

8.53 10

7

–

B

2.54

∆

f

sw

1.36

⋅⋅

⋅

=

1.59 10

6

–

B

2.58

∆

f

sw

1.32

⋅⋅

⋅

=

2.39 10

6

–

B

2.23

∆

f

sw

1.26

⋅⋅

⋅

=