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

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From: LATOURRETTE,JEFF (HP-SanJose,ex1) (jeff_latourrette@agilent.com)
Date: Mon Jan 10 2000 - 12:14:23 PST


Doug & Chris:

I agree with Ray's 1/10th wavelength and have even heard 1/20th recommended.
Remember that on your 1/4 wave example that the 1/4 wave frequency is the
most efficient which will radiate, but certainly other frequencies both
higher and lower can and will radiate from that line. The same is true for
apertures sizes in a chassis. It is true, however that you can achieve
cut-off by having sigificant depth in your aperture so that it resembles a
waveguide. The amount of attenuation to signals below the cutoff frequency
is directly proportional to the length of the waveguide structure. I've
captured some of this in an Application note, AN 1166: [18pp/195KB (Pub.
3/99)] "Minimizing Radiated Emissions of High-Speed Data Communications
Systems" which can be downloaded at:

http://www.semiconductor.agilent.com/fiber/fiberapps.html

Also, Make sure meshes used actually make conductive contact at its own
cross points, as well as on its edges to the chassis. Some cheaper screens
are weaved wires without electrical contact. Some use two mesh screens and
for extra high performance use an extruded mesh structure with depth (like a
honeycomb) in place of a screen.

Regards,

Jeff La Tourrette
  SJ Applications Manager
  Agilent Technologies Fiber-Optics Components
  370 West Trimble Rd. M/S 90TH
  San Jose, CA 95131
Voice: 408 435-4083 FAX: 408 435-4915

-----Original Message-----
From: Ray Waugh [mailto:ray_waugh@agilent.com]
Sent: Monday, January 10, 2000 11:09 AM
To: si-list@silab.eng.sun.com
Subject: RE: [SI-LIST] : Chassis hole opening and frequencies

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: ray_waugh@agilent.com

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

-----Original Message-----
From: Chris Padilla [mailto:cpad@cisco.com]
Sent: Thursday, December 30, 1999 1:39 PM
To: si-list@silab.eng.sun.com
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
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
>.
>Doug Brooks, President doug@eskimo.com
>UltraCAD Design, Inc. http://www.ultracad.com
>
>
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