请教LC滤波器的设计
我想用lc滤波器将峰值300v的spwm波平滑成正弦波,请问如何设计?
spwm方波的频率为1.6k,
请教LC滤波器的设计
全部回复(8)
正序查看
倒序查看
现在还没有回复呢，说说你的想法
let me give u a hand
lc filter's transfer function is a two order system,so u can achieve it's corner frequency which should be 1/7~1/10 of your spwm frequency,
here your spwm frequency equals to switching frequency,but 1.6k is a bit low from my opinion,how to eliminate your switch noise under such a low frenquecy?
lc filter's transfer function is a two order system,so u can achieve it's corner frequency which should be 1/7~1/10 of your spwm frequency,
here your spwm frequency equals to switching frequency,but 1.6k is a bit low from my opinion,how to eliminate your switch noise under such a low frenquecy?
0
回复
提示
@
letmegiveuahandlcfilter'stransferfunctionisatwoordersystem,soucanachieveit'scornerfrequencywhichshouldbe1/7~1/10ofyourspwmfrequency,hereyourspwmfrequencyequalstoswitchingfrequency,but1.6kisabitlowfrommyopinion,howtoeliminateyourswitchnoiseundersuchalowfrenquecy?
go on
how to achieve your L and C is often up to your experience.u can get L value first,or C first,use this value to get the other varialbe's value,here below is an example:
1) C value
accordding your inverter output current maxsimum value ,caculate the current value flowing the capacitor,then use the formula C=dU/dI,so C can be achieved.
2)L value
if C has been choosed,then u can use the corner frequecy to get your L value,
corner frequecy can be depicted as following:
f=1/[2*pi*sqr(L*C)],
usually f equal to 1/5~1/10 of fs(switching frequency).
now C and f has been calulated,so use the above formula,L can be drawn out.
by the way,much experience determines your choice for this two parameters.
contact me if still any questions
how to achieve your L and C is often up to your experience.u can get L value first,or C first,use this value to get the other varialbe's value,here below is an example:
1) C value
accordding your inverter output current maxsimum value ,caculate the current value flowing the capacitor,then use the formula C=dU/dI,so C can be achieved.
2)L value
if C has been choosed,then u can use the corner frequecy to get your L value,
corner frequecy can be depicted as following:
f=1/[2*pi*sqr(L*C)],
usually f equal to 1/5~1/10 of fs(switching frequency).
now C and f has been calulated,so use the above formula,L can be drawn out.
by the way,much experience determines your choice for this two parameters.
contact me if still any questions
0
回复
提示
@
goonhowtoachieveyourLandCisoftenuptoyourexperience.ucangetLvaluefirst,orCfirst,usethisvaluetogettheothervarialbe'svalue,herebelowisanexample:1)Cvalueaccorddingyourinverteroutputcurrentmaxsimumvalue,caculatethecurrentvalueflowingthecapacitor,thenusetheformulaC=dU/dI,soCcanbeachieved.2)LvalueifChasbeenchoosed,thenucanusethecornerfrequecytogetyourLvalue,cornerfrequecycanbedepictedasfollowing:f=1/[2*pi*sqr(L*C)],usuallyfequalto1/5~1/10offs(switchingfrequency).nowCandfhasbeencalulated,sousetheaboveformula,Lcanbedrawnout.bytheway,muchexperiencedeterminesyourchoiceforthistwoparameters.contactmeifstillanyquestions
can u explain about c=DU/DI,for example?
tks
tks
0
回复
提示
@morel
canuexplainaboutc=DU/DI,forexample?tks
sorry,made a mistake
i=Cdu/dt
u=Um*sin(w*t),
then,
Im=Um*C*w,
so C=Im/Um*w,
if u=Umsin(2*pi*50*t)
so,i=Um*C*2*pi*50*cos(2*pi*50*t)
Im=Um*C*2*pi*50
from the above,
C=Im/Um*2*pi*50
sorry,i can't remember clearly how to determine the value of the Im,maybe it's 1/n of the output current max value.I can tell u more if I find out where my material is
i=Cdu/dt
u=Um*sin(w*t),
then,
Im=Um*C*w,
so C=Im/Um*w,
if u=Umsin(2*pi*50*t)
so,i=Um*C*2*pi*50*cos(2*pi*50*t)
Im=Um*C*2*pi*50
from the above,
C=Im/Um*2*pi*50
sorry,i can't remember clearly how to determine the value of the Im,maybe it's 1/n of the output current max value.I can tell u more if I find out where my material is
0
回复
提示
@
letmegiveuahandlcfilter'stransferfunctionisatwoordersystem,soucanachieveit'scornerfrequencywhichshouldbe1/7~1/10ofyourspwmfrequency,hereyourspwmfrequencyequalstoswitchingfrequency,but1.6kisabitlowfrommyopinion,howtoeliminateyourswitchnoiseundersuchalowfrenquecy?
我用单片机控制
频率升不上去,定时中断受限,请问有什么好办法吗?
频率升不上去,定时中断受限,请问有什么好办法吗?
0
回复
提示
@
letmegiveuahandlcfilter'stransferfunctionisatwoordersystem,soucanachieveit'scornerfrequencywhichshouldbe1/7~1/10ofyourspwmfrequency,hereyourspwmfrequencyequalstoswitchingfrequency,but1.6kisabitlowfrommyopinion,howtoeliminateyourswitchnoiseundersuchalowfrenquecy?
请问至少多大频率可以消除开关噪音
.
.
0
回复
提示
@whualei
我用单片机控制频率升不上去,定时中断受限,请问有什么好办法吗?
oh
80c196 or tms320f240x DSPs.
when I stayed in college,f240DSP was very popular,20k fs was easily acchieved as well as PWM generator.but an obvious demerit was such that the A/D converter was not very good as expected anyway.
a guy realized pure sinusoidal output with one dsp and operation amplifier(even parallel operation was acchieved).the THD is below 0.5% even with diode rectifier load. and attracted much attention ,including lite-on corp. ,and so on
80c196 or tms320f240x DSPs.
when I stayed in college,f240DSP was very popular,20k fs was easily acchieved as well as PWM generator.but an obvious demerit was such that the A/D converter was not very good as expected anyway.
a guy realized pure sinusoidal output with one dsp and operation amplifier(even parallel operation was acchieved).the THD is below 0.5% even with diode rectifier load. and attracted much attention ,including lite-on corp. ,and so on
0
回复
提示
@
sorry,madeamistakei=Cdu/dtu=Um*sin(w*t),then,Im=Um*C*w,soC=Im/Um*w,ifu=Umsin(2*pi*50*t)so,i=Um*C*2*pi*50*cos(2*pi*50*t)Im=Um*C*2*pi*50fromtheabove,C=Im/Um*2*pi*50sorry,ican'trememberclearlyhowtodeterminethevalueoftheIm,maybeit's1/noftheoutputcurrentmaxvalue.IcantellumoreifIfindoutwheremymaterialis
我看到一本书上取的是1/2,不过我认为可以取到1/4.
The value of the Im is 1/n of the output current max value.The number n may set to 2 to 4.I think.
The value of the Im is 1/n of the output current max value.The number n may set to 2 to 4.I think.
0
回复
提示
