AN-8024

APPLICATION NOTE

© 2009 Fairchild Semiconductor Corporation

www.fairchildsemi.com

Rev. 1.0.1 • 9/18/09

5

(Design Example)

EEL-19 core is selected, whose

effective cross-sectional area is 25mm

saturation flux density as 0.3 T, the minimum number

of turns for the primary side is obtained as:

min

6

6

6

10

900 10

1.2

10

144

0.3 25

MLIM

P

SAT

e

LI

N

BA

−

⋅

=×

×⋅

=×

=

⋅

[STEP-7] Determine the Number of Turns for Each

Winding

Figure 6 shows the simplified diagram of the transformer.

First, calculate the turn ratio (n) between the primary side

and the secondary side from the reflected output voltage,

determined in step 3, as:

RO

P

SO

F

V

N

n

NV

V

==

+

(13)

supply is determined as:

*

1

DD

FA

AS

OF

VV

NN

VV

+

=⋅

+

(14

)

avoid the over-voltage protection condition during the peak

load operation.

Figure 6. Simplified Transformer Diagram

(Design

Example)

Assuming the diode forward-

voltage drop is 0.5V, the turn ratio is obtained as:

100

18.18

50.5

RO

P

SO

F

V

N

n

NV

V

==

=

=

++

min

8,

146

SP

S

P

NN

n N

N

== ⋅

=

>

*

as 15V, the number of turns for the

auxiliary winding is obtained as:

*

15 1.2

824

50.5

+

+

=

⋅=

⋅ =

++

DD

FA

AS

OF

VV

NN

VV

[STEP-8] Determine the Wire Diameter for Each Winding

Based on the RMS Current of Winding

The maximum RMS current of the secondary winding is

obtained as:

1

RMS

RMS

MAX

SEC

DS

MAX

D

In I

D

−

=⋅

(15)

The current density is typically 3~5A/mm

long (>1m). When the wire is short with a small number of

turns, a current density of 5~10A/mm

Avoid using wire with a diameter larger than 1mm to avoid

severe eddy current losses as well as to make winding easier.

For high-current output, it is better to use parallel windings

with multiple strands of thinner wire to minimize skin effect.

(Design Example)

The RMS current of primary-side

winding is obtained from step 4 as 0.36A. The RMS

current of secondary-side winding is calculated as:

1

10.47

18.18 0.36

6.9

0.47

RMS

RMS

MAX

SEC

DS

MAX

D

In I

D

A

−

=⋅

−

=⋅

=

0.3mm (5A/mm

wires are selected for primary and secondary windings,

respectively.

[STEP-9] Choose the Rectifier Diode in the Secondary

Side Based on the Voltage and Current Ratings

The maximum reverse voltage and the RMS current of the

rectifier diode are obtained as:

MAX

IN

DO

O

V

VV

n

=+

(16)

1

RMS

RMS

MAX

DO

DS

MAX

D

In I

D

−

=⋅

(17)