STEP 1: Define the system specifications Vline

min = 100 V Vline

max = 240 V fL = 60 Hz Po = 7 W Eff = 75%

P¿=Po

E ff

= 70.75

=9.33W

K L(5 )=Po(5)

Po

=5.67

×100=80%

K L(12)=Po (12)

Po

=1.47

×100=20%

STEP 2: Determine DC link capacitor and DC link voltage range Dch = 0.2

CDC=(3×10−6 ) (7W )=21μF

V DCmin=√2 ∙(V line

min)2−P¿ ∙ (1−D ch )

CDC ∙ f L

=√2 (100 )2−(9.33 ) (1−0.2 )(21×10−6 ) (60 )

V DCmin=118.643V

V DCmax=√2V line

max=(√2 ) (240 )

V DCmax=339.411V

STEP 3: Determine maximum duty ratio (Dmax)

V DSnom=(600 ) (0.7 )=420V

V RO=V DSnom−V dc

max=420−339.411=80.859V

V RO=Dmax

1−Dmax

∙V DCmin

80.859=Dmax

1−Dmax

(118.643 )

Dmax=0.4045

STEP 4: Determine transformer primary inductance (Lm) KRF = 0.5 fS = 67 KHz

Lm=(V dc

min ∙ Dmax)2

2 ∙P ¿∙ f S ∙K RF

=[ (118.643) (0.4045 ) ]2

(2 ) (9.33 ) (67×103 ) (0.5 )=3.684×10−3H

IEDC=P¿

(V DCmin) ( Dmax )

= 9.33(118.543 ) (0.4045 )

=194.411×10−3 A

∆ I=(V DC¿¿min) (Dmax )

(l¿¿m)(f s)¿¿

∆ I=(118.643 )(0.4045)

(3.684×10−3 )(67×103)∆ I=0.1944313136I dc

peak=I EDC+( ∆I /2 )2

I dcpeak=194.411×10−3+( 0.19443131362 )

2

I dcpeak=0.2916267327

I dsrms=√3 ( IEDC)

2+( ∆I2 )

2 Dmax

3

I dsrms=0.1296459243

V dcCCM=( I

√2lmfsPin− 1

V RC )−1

V dcCCM=424.8762256V

STEP 5: Choose the proper FPS considering the input power and current limit

I ¿=I dspeak+ I ds

peak (0.12)I ¿>326.621mA

STEP 6: Determine the proper core and the minimum primary turnsk

N pmin=

Lm I ¿

B sa+Ae

×106

N pmin=

(3.684×10−3 )(326.621×10−3)(0.35 )(20.1)

×106

N pmin=171 turns

STEP 7: Determine the number of turns for each output

STEP 8: Determine the wire diameter for each windingK

STEP 9: Choose the rectifier diode in the secondary sideK

STEP 10: Determine the output capacitorK

STEP 11: Design RCD snubberK

STEP 12: Design feedback control loopK