http://bbs.dianyuan.com/topic/36555
内容都是这里贴来的
AC输入:85-265V
输出功率:10瓦 n=0.85
查磁芯规格F=60KHZ时宽电压10W选EE19合适,
查得Ae=0.22平方厘米 Bm=0.22T
例1:
设Dmax=0.5 f=60k
DCinmin=85v*1.414-20v=100v
Ipk=(2*Po)/DCinmin*Dmax
=(2*10)/100*0.5
=0.4A
LP =(DCinmin*Dmax*Ts)/Ipk
=[100*0.5*(1/60000)]/0.4
=0.00208H
=2.08mH
NP =(LP*Ipk)/(Ae*Bm)
=0.00208*0.4/0.22*0.22
=172T
例2:
Pin=Po/n =10/0.85=11.76W
Ts=1/60000=16.7us
ton=Dmax*Ts=0.5*16.7=8.33
Np=(DCinmin*ton)/Ae*Bm
=100*8.33/0.22*0.22
=172T
Is=Pin/DCinmin=11.76/100=0.12A
Iave=(Is*Ts)/ton
=0.12*16.7/8.33=0.24A
Imin=Iave/2=0.24/2=0.12A
Ipk=3*Imin=0.12*3=0.36A
LP=(DCinmin*ton)/Ipk
=100*0.00000833/0.36
=0.0023H=2.3mH
例3:
Vf反射电压
VmosMOS管耐压 设600V留150V裕量
DCinmax=ACinmax*1.414-20
=265*1.414-20=355V
Vf=Vmos-DCinmax-150v
=600-355-150=95V
DCinmin*Dmax=Vf*(1-Dmax)
100*Dmax=95*(1-Dmax)
Dmax=0.49
1/2*(Imin+Ipk)*Dmax*DCinmin=(Po/n)
Ipk=3*Imin
1/2*(Ipk/3+Ipk)*0.49*100=10/0.85
Ipk=0.36A
Lp=(Dmax*DCinmin)/(f*Ipk)
=(0.49*100)/(600000*0.36)
=0.0023H=2.2mH
NP=(LP*Ipk*10000)/(Bm*Ae)
=(0.0023mH*0.36A*10000)/0.22*0.22
=171T
完成!
我想问一下,在上面所有的计算中,求LP的时候他都是用的Ipk,也就是原边电流的最大值,在电流连续方式的情况下,是不是应该用△IP而不应该用IPK啊
