AN2528

Flyback stage design

7/21

Equation 7

Hence

Equation 8

From here we can now calculate the peak current on primary.

Equation 9

To keep the transformer size very small and to get a very effective cost solution, we prefer to

slightly increase the minimum working frequency in order to decrease the primary

inductance.

In order to have a 15 mH inductance and to keep an EF20 core, a lot of turns are needed on

the primary side. This can generate either not enough space on the EF20 core to

accomodate such a high number of windings or the remaining space is not large enough to

ensure good design. These considerations might induce designing a smaller primary

inductance value accepting a higher switching frequency.

There is no contraindication in using a smaller primary inductance which leads to a higher

minimum switching frequency and theoretically also to a higher maximum frequency.

However the maximum switching frequency is then limited not only by the inductance value,

but also by the L6565 PWM driver. When using an L6565, the internal blanking time limits

the minimum off-time and, in turn, the maximum switching frequency. To better understand

this phenomenon, please refer to the L6565 datasheet and to the next paragraphs.

After bench tests and fine tuning we used a transformer with the following specs:

Equation 10

Equation 11

The part number of the transformer is CSM 2010-104 from Cramer.

In the next Section 3, we see from bench verification that the real minimum working

frequency is 50 kHz even if the inductance is 7.5 mH but with a peak current of about

250 mA.

P

IN

1.25

P

OUT

•

1

2

---

L

P

I

2

P

•

•

T

s

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

=

1

2

---

V

2

dcmin

•

T

2

onmax

•

L

P

T

S

•

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

=

=

L

P

V

2

dcmin

T

2

onmax

•

2.5

T

S

•

P

OUT

•

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

14.7mH

=

=

I

P

V

dcmin

T

onmax

•

L

P

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

143mA

≈

=

L

P

7.5mH

=

N

p

N

s

-------

23.8

=

N

p

N

aux

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

18.87

=