# RE: [SI-LIST] : Chassis hole opening and frequencies

From: Ray Waugh ([email protected])
Date: Mon Jan 10 2000 - 11:08:35 PST

Doug...

Have you ever seen a parabolic dish antenna which uses a mesh (instead
of a solid metal surface)? It lets the wind and rain through. As a
rule of thumb, keep the holes in the mesh under a tenth wavelength in
dimension.

I'd suggest doing as the RF instrument people do -- cut a hole in your
chassis large enough for your fan, and mount a mesh over it. To avoid
having the mesh act as a RF radiator, make certain that it is
continuously attached to the rim of the chassis.

You could probably build and test a dummy chassis quicker than you could
do a proper analysis.

Ray
------------------------------------------------------
Raymond W. Waugh - WSD Diode Applications
E-mail: [email protected]

USPS : Agilent Technologies
Wireless Semiconductor Division
39201 Cherry Street, MS NK20
Newark, California 94560
------------------------------------------------------

-----Original Message-----
Sent: Thursday, December 30, 1999 1:39 PM
To: [email protected]
Subject: Re: [SI-LIST] : Chassis hole opening and frequencies

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

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
>
>
>
>
>
>.
>************************************************************
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>.
>Doug Brooks, President [email protected]
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