小功率电源变压器设计与计算
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@wywa123
这个是我整理的,[图片]小功率电源变压器设计与计算
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!
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!
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@zhongpeng888
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=60kDCinmin=85v*1.414-20v=100vIpk=(2*Po)/DCinmin*Dmax =(2*10)/100*0.5 =0.4ALP=(DCinmin*Dmax*Ts)/Ipk =[100*0.5*(1/60000)]/0.4 =0.00208H =2.08mHNP=(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=0.491/2*(Imin+Ipk)*Dmax*DCinmin=(Po/n) Ipk=3*Imin1/2*(Ipk/3+Ipk)*0.49*100=10/0.85 Ipk=0.36ALp=(Dmax*DCinmin)/(f*Ipk) =(0.49*100)/(600000*0.36) =0.0023H=2.2mHNP=(LP*Ipk*10000)/(Bm*Ae) =(0.0023mH*0.36A*10000)/0.22*0.22 =171T完成!
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@zhongpeng888
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=60kDCinmin=85v*1.414-20v=100vIpk=(2*Po)/DCinmin*Dmax =(2*10)/100*0.5 =0.4ALP=(DCinmin*Dmax*Ts)/Ipk =[100*0.5*(1/60000)]/0.4 =0.00208H =2.08mHNP=(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=0.491/2*(Imin+Ipk)*Dmax*DCinmin=(Po/n) Ipk=3*Imin1/2*(Ipk/3+Ipk)*0.49*100=10/0.85 Ipk=0.36ALp=(Dmax*DCinmin)/(f*Ipk) =(0.49*100)/(600000*0.36) =0.0023H=2.2mHNP=(LP*Ipk*10000)/(Bm*Ae) =(0.0023mH*0.36A*10000)/0.22*0.22 =171T完成!
我有个产品输出Po=11W/Vo=40V,效率0.85,f=40K变压器骨架RM8(0.64平方厘米pc40铁氧体);输入电压AC90v-AC264v,用公式可以算出来变压器设计参数吗?可能会相差特别多,不知道有没有比较好的计算方法
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