1997

4-11

MIC2171

Micrel

4

(10)

L

0.5 f

V

T

P

PRI

SW

IN(min)

2

ON

2

OUT

≥

Where:

L

f

V

T

Then:

L

0.5

1 10

3.22

7.4 10

2.5

PRI

5

2

-6

2

≥

××

×

()

L

Use an 12

µH primary inductance to overcome circuit ineffi-

ciencies.

To complete the design the inductance value of the second-

ary is found which will guarantee that the energy stored in the

transformer during the power switch on time will be com-

pleted discharged into the output during the off-time. This is

necessary when operating in discontinuous-mode.

(11)

L

0.5 f

OFF

2

P

OUT

Where:

L

T

Then:

L

0.5

1 10

5.41

2.6 10

2.5

SEC

5

2

-6

2

≤

××

×

()

L

Finally, recalculate the transformer turns ratio to insure that

it is less than the value earlier found in equation (9).

(12)

a

≤

L

PRI

L

SEC

Then:

a

11.4

7.9

= 1.20

≤

This ratio is less than the ratio calculated in equation (9).

When specifying the transformer it is necessary to know the

primary peak current which must be withstood without satu-

rating the transformer core.

(13)

I

V

L

PRI

So:

I

=

3.22

7.6 10

12 H

PEAK(pri)

-6

××

µ

I

Now find the minimum reverse voltage requirement for the

output rectifier. This rectifier must have an average current

rating greater than the maximum output current of 0.5A.

(14)

V

V

V

()

F

Where:

V

reverse voltage rating

a = transformer turns ratio (1.2)

F

Then:

V

6.0 + 5.0

1.2

0.8

1.2

×

()

×

V

A 1N5817 will safely handle voltage and current require-

ments in this example.