如图这几个器件是干嘛用的?正常的话一个二极管就OK了吗?
PFC升压二极管两端并两个串联的二极管的作用
全部回复(12)
正序查看
倒序查看
现在还没有回复呢,说说你的想法
@米老鼠
降低损耗用,主要是吸收主二极管反向恢复损耗,并近似无损的传送到输出端
1. Saturation core L206
is design to block the reverse recovery current of D229 and provide ZCS at the same time.
2. C236,D230 and D231 stores the
reverse energy while D229 off period
and then recycle it to Bulk at the
Q218,9 off moment.
3. C237, R282 and D232 bypasses L205’s current a small
period to let L206 saturated and then take over it’s current.
4. L206 Turn ratio Tr
= N2/N1 about 0.1, the
reverse recovery current is given as Irr=IL205 *Tr/(1-Tr).
For example, Tr=0.1, then Irr=IL205*0.1/(1-0.1)=(0.111)IL205
0
回复
提示
@gbfdyx
1.SaturationcoreL206 isdesigntoblockthereverserecoverycurrentofD229andprovideZCSatthesametime.2.C236,D230andD231storesthereverseenergywhileD229offperiodandthenrecycleittoBulkatthe Q218,9offmoment.3.C237,R282andD232bypassesL205’scurrentasmallperiodtoletL206saturatedandthentakeoverit’scurrent.4.L206TurnratioTr=N2/N1about0.1,thereverserecoverycurrentisgivenasIrr=IL205*Tr/(1-Tr). Forexample,Tr=0.1,thenIrr=IL205*0.1/(1-0.1)=(0.111)IL205
解释得非常好!
0
回复
提示
@gbfdyx
1.SaturationcoreL206 isdesigntoblockthereverserecoverycurrentofD229andprovideZCSatthesametime.2.C236,D230andD231storesthereverseenergywhileD229offperiodandthenrecycleittoBulkatthe Q218,9offmoment.3.C237,R282andD232bypassesL205’scurrentasmallperiodtoletL206saturatedandthentakeoverit’scurrent.4.L206TurnratioTr=N2/N1about0.1,thereverserecoverycurrentisgivenasIrr=IL205*Tr/(1-Tr). Forexample,Tr=0.1,thenIrr=IL205*0.1/(1-0.1)=(0.111)IL205
这是什么资料上面的?
0
回复
提示
@gbfdyx
1.SaturationcoreL206 isdesigntoblockthereverserecoverycurrentofD229andprovideZCSatthesametime.2.C236,D230andD231storesthereverseenergywhileD229offperiodandthenrecycleittoBulkatthe Q218,9offmoment.3.C237,R282andD232bypassesL205’scurrentasmallperiodtoletL206saturatedandthentakeoverit’scurrent.4.L206TurnratioTr=N2/N1about0.1,thereverserecoverycurrentisgivenasIrr=IL205*Tr/(1-Tr). Forexample,Tr=0.1,thenIrr=IL205*0.1/(1-0.1)=(0.111)IL205
这个位号都对得上?还是你改的?
是出自什么资料上的吗?
0
回复
提示