

2022 Microchip Technology Inc. and its subsidiaries

DS20006355A-page 13

MIC2171

EQUATION 5-14:

Use a 12 µH primary inductance to overcome circuit

inefficiencies.

To complete the design, the inductance value of the

secondary winding must be calculated, so it will ensure

that the energy stored in the transformer during the

power switch on time will be completed discharged into

the output during the off time; this is necessary when

operating in discontinuous mode.

EQUATION 5-15:

Then:

EQUATION 5-16:

Finally, recalculate the transformer turns ratio to ensure

that it is less than the value found earlier by using

Equation 5-11.

EQUATION 5-17:

Then:

EQUATION 5-18:

This ratio is less than the ratio calculated in

Equation 5-11. When selecting the transformer, it is

necessary to know the primary peak current which

must be handled without saturating the transformer

core.

EQUATION 5-19:

So:

EQUATION 5-20:

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.

EQUATION 5-21:

Then:

3.22 0.76







0.5 1













2



2.5

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













OFF

2







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



Where:

Maximum secondary inductance

Power switch off time

–







2







2.5

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





a

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



a

12

3.6

-------



1.83

=







ON



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

=





–





–



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

=





2.04A

=











+



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



Where:

Output rectifier maximum peak

reverse voltage rating

a =

Transformer turns ratio (1.8)

Reverse voltage safety derating factor

(0.8)