RCC实例计算二

RCC实例计算

OUT :DC 5.5V 0.5A Vd=1V
Po : (5.5+1)×0.5=3.25W{(Vout+Vd)×A)}
∩ :0.7
Pi :3.25÷0.7=4.64W

INPU:AC 85-264
DCmin 85 ×1.20=102V
DCmax 264×1.41=372V

F :50KHz
Ts :1÷50KHz=20us 1÷F
D :0.45
Ton :20×0.45=9us Ts×D
Toff :20-9 =11us

磁蕊 EE 13
Ae :17.1平方MM
△B :0.3T
Ipk :2×4.64÷0.45÷102=0.2A 2×Po÷D÷Vin dc min

Lp :102×9÷0.2=4590uH Vin dc min×D÷Ipk
N :102×9÷{(5.5+1)×11}=12.8 Vin dc min×Ton{(Vout+Vd)×Toff}
Np :102×9÷(0.3×17.1)=178.9T Vin dc min×Ton÷(△B ×Ae)
Ns :178.9÷12.8=13.9 NP÷N
反馈绕组请按开关管类型选择:
1.三级管取5V驱动电压. (IN AC min)
2.MOSFET取10V驱动电压. (IN AC min)

验证
85V时:
NP :取179T
NS :取 14T
LP :4.59(实际变压器初级电感量)

D :(5.5+1)÷(14÷179×102+5.5+1)=0.4489 (Vout+Vd)÷(NS÷NP×Vin min dc+Vout+Vd)
Ipk :2×4.64÷(102×0.4489)=0.20 (2×Pin)÷(vin min dc×D)
Vds :372+179÷14×(5.5+1)+120=575 Vin max dc+NP÷NS×(Vout+Vd)+Vlk(保留范围值)
△B :4590×0.2÷179÷17.1=0.299 Lp×Ipk÷Np÷Ae
Vd :14÷179×372+5.5+1=35.6 Ns÷Np×Vin max dc +Vout+Vd
F :102×102×0.4489×0.4489÷4.64÷2÷4590=49.2KHz Vin max dc平方×D平方÷Pin÷2÷Lp(mH)= KHz