如何去算磁芯的气隙大小?

lg= μo*Ae*(N^2/L -1/AL)

thanks for your supports!

sorry;I can't understand full!
for example:E56/24/19core;Ae=337mm*mm;Le=107mm;Al=400nH/N2;Ue=159;then what is the air-gap ?
thanks!

Is your Al a gapped or a non-gapped value?

Approximately, lg =μo.le/μe or lg=μo.Ae/Al(gapped)

I saw the Ferroxcube's catalogue that said the small air-gaps the formula is Ue=Ui/(1+(Gap*Ui/Le));but I use the formula to calculate and the result is wrong!

Can you see the Ferroxcube catalogue?the E56/24/19 core:Ui=2700(3C81),Ue=157,Al=400nH/N2,and the air-gap is about 370um;but I don't know how to calculate?
now I have a project:E55/28/21;Ui=2300;air-gap is 100um;can you help me to calculate the Al?thanks very much!!

According to the Ferrocube catalog, if it is E56/24/19-E400,Al=400nH/N2,μe=101,lg=640μm (not 159 and 370 respectively as you have said). Using the formula for small gap, the calculated lg is 1020μm, a discrepancy exists. The one thing I am not so sure is whether the Ferroxcube gap is at the centre leg only or on all three legs.
The formula for small gap Al is Al(gapped)=μo.μr.Ae/(le+μr.lg). I think you can calculate it yourself.

But if you check the Philips catalog (Ferroxcube is now part of Philips),you will find that for E56/24/19-E400, μe is 101, while lg is 1240μm which is very close to the calculated value of 1020μm as in my messsage no.9.

thanks,
can you tell me what mean the Ur?and the Ae&Lg's units if is cm?

I think the air-gapis at the centre leg,because I used the ferroxcube core that the centre leg is grinded!if at three legs,I think the core don't grind and can use mylar or others to get the gap,do you think?
for my exprience:the centre leg gap thickness is twifold to the all three legs gap thickness.

yes,I check the philips catalogue,I don't know why there is discrepancy for the same campany?

The gap should be at the centre leg. I think the Philips data is more trustworthy.
μr is relative permeability, you can just use μi in place of it.
The units for Al is H/turn square, Ae metre square and lg metre. Or you can multiply the formula by 10^-3 and get Al in mH/T^2, Ae in cm^2 and lg in cm.

Yes,I agree the Philips data is more trustworthy!
I look for some datum and found the formula as below:
for EE,EF etc. square centre:Lg=Uo*N^2*Ae*10^4/L
for ER,EP etc. circle centre:Lg=Uo*N^2*Ae*10^4*(1+Lg/D)/L
note:L=inductance(uH)
     Ae:cm
     D:diametre(cm)
I calculate with this formula and the result is tight to the Philips data!

Uo=4*3.14*10^-7

D=the centre diametre
Uo=4*3.14*10^-7

For the circular centre formula, we've got Lg both on the left hand and right hand side, is this correct ? May I ask where do yo find this ? :)

<<开关电源中的磁性元件>>(赵修科老师)

谢谢,看到了.
1.矩形柱公式内的是Ag,不是Ae
2. 圆形柱公式内是(1+Lg/D)^2

好帖好贴!

yes,but when the air-gap is small,the Ag should be equaled Ae;the Ag should be considered at big air-gap;Do you think?

These 2 formulae take care of the fringing effect of the air gap by using the equivalent gap area Ag instead of Ae. Ag is taken as Ae plus a rim with a width of Lg/2 around Ae. Thus for square leg, Ag=(W+Lg).(B+Lg), for circular, Ag=(pi/4).(D+Lg)^2 =Ae.(1+Lg/D)^2. As you can see, the smaller Lg is, the closer Ag is to Ae. Yes, if Lg is very small, we can just use Ae; if Lg is big, we have to consider using Ag.

yes,I agree.thanks for your advice!!
if expediency,can contact me by E-mail:liucgood@163.com;

其实有一个笨方法,不知受不受用,一般说来呢,这个GAP可以不用提供给生产厂商的,只要生产厂商自己用磨床试做出您所需要的电感就可以了.我有一个朋友就是帮客户提供这个业务服务的.

其实有一个笨方法,不知受不受用,一般说来呢,这个GAP可以不用提供给生产厂商的,只要生产厂商自己用磨床试做出您所需要的电感就可以了.我有一个朋友就是帮客户提供这个业务服务的.

http://zhiqiang.blog.dianyuan.com/article.php?arti_id=32821

gap 也能计算吗?