From: Tony Williams
Subject: Re: Core gapping techniques and general SMPS magnetics stuff
Date: Tue, 17 Dec 2002 12:36:42 +0000 (GMT)
NNTP-Posting-Date: Tue, 17 Dec 2002 12:36:55 +0000 (UTC)
User-Agent: Pluto/1.14i (RISC-OS/3.60)
In article ,
Christopher R. Carlen wrote:
> This brings up an interesting question. I have determined that H=NI/Le
> for a closed magnetic path, where N is number of turns, I is current,
> and Le is the effective path length in meters. Thus, B may be
> determined from the permeability (which is of course non-linear, but a
> reasonable approximation of peak flux density may be made from the
> effective permeability, ue, of the core.)
> Of course that leads to way to much flux without a gap in my case. I
> will tell you how I dealt with this in my example calculations, and you
> can tell me if this is neglecting some coupling between the relations
> that I have overlooked:
> Take the E19/8/5 core in 3C90, with ue=1650 ungapped and Al=1170
> (nH/turn^2). I want to make for example a 100uH inductor, to carry a
> peak current of 1.2A without exceeding 160 mT.
> To make 100uH, I need 9.245 turns. The effective path length Le=0.04m,
> so H=9.245*(1.2A)/(0.04m)=277 A/m. That gives a B=ue*u0*H=574 mT.
> Egad! That's well past the saturation point.
> A gap of 0.35mm is shown to reduce Al and ue so that Al=100 and ue=140.
> Now I'd need N=31.6 turns to get 100uH.
My quick doodles (in oersteds and gauss, so probably wrong)
suggest that, with a 0.35mm gap, ue drops to about 107.
But that is theory, disregarding the effect of fringing,
which will be large with such a small gap.
> With the new number of turns I get H=948 A/m.
H = 4.pi.n.I/10.le for an ungapped core.
H = 4.pi.n.I/10(le + u.d) for a core with gap= d.
(le and d are in cm in this old system.)
Which I think brings your H down to about 71 A/m.
Which is more consistent with the fact that the resultant
flux density, B, has come down from 540mT to 167mT.