Re: [SI-LIST] : Chassis hole opening and frequencies

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From: Chris Padilla (cpad@cisco.com)
Date: Thu Dec 30 1999 - 13:38:32 PST


Doug,

In antenna theory, a microstrip (for example) that is fed (by an ideal
infinite rise-time source) in one end and

(a) shorted at the other end will radiate at the frequency that corresponds
to a quarter wavelength

(b) open at the other end will radiate at the frequency that corresponds
to a half wavelength

So if our microstrip is 12 cm long and it is shorted at one end, we should
be able to radiate 625 MHz pretty well. If you imagine your ideal source
is in the hole of the chassis and connected across the longest dimension
of the hole, we have the same thing set up in (a) above. Our source is
referenced to ground and the short is, effectively, ground, too. Does this
make any sense???

So my long-winded answer is a "quarter wavelength of the longest dimension
that defines the hole."

This should get you very much in the ballpark.

WARNING::DEVIATION FROM QUESTION::WARNING---------------**************

Now the next question is: What if I have 2 or 3 or a whole matrix of holes
(air holes for cooling for instance) in my chassis. How will THAT radiate?

I can indirectly answer that question as follows:

We can get SE (shield effectiveness) from: 20*log(wavelength/2*max_length)

where max_length is the maximum dimension of the slot or aperture.

So the SE of a 100 mil opening at 5 GHz is 21.45 dB-->watch units!

Mulitple apertures reduce the shielding effectiveness. Lets call it MA_n.

The amount of reduction depends on (1) the spacing between the apertures,
(2) the frequency, and (3) the number of apertures. When apertures of equal
size are placed close together (less than a half wavelength), the reduction
in shielding effectiveness is approximately proportional to the square root
of the number of apertures n:

MA_n= -10*log(n)

So now our total shielding effectiveness, SE_tot = SE + MA_n

Example: 4 100 mils holes at 5 GHz provides a reduction in shielding of 6 dB
if the holes are less than a half wavelength or 1.18 inches apart.

So now our SE = 21.45 + (-6) = 15.45 dB at 5 GHz.

I know, I know, it is more than you wanted to know.

----->Chris

At 09:45 AM, you wrote:
>As I recall, there is a relationship between a hole in a chassis and the
>frequencies that can pass through that opening. I recall that the longest
>dimension of the hole defines the wavelength, or quarter wavelength, or
>something, of the lowest frequency than can conveniently enter or escape
>through the opening.
>
>Can anyone give me the correct relationship?
>
>Thanks.
>
>And Happy New Year to All...........
>
>Doug Brooks
>and all of us here at UltraCAD
>
>
>
>
>
>.
>************************************************************
>See our updated message re in-house seminars on our web page
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>Doug Brooks, President doug@eskimo.com
>UltraCAD Design, Inc. http://www.ultracad.com
>
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