Jim,
You might want to look at "Electromagnetic Shielding Material and
Performance" by Donald R.J. White. All of chapter contains references
related to shielding.
Best Regards,
William D'Orazio
CAE Electronics Ltd.
Electrical System Designer
Phone: (514) 341-2000 (X4555)
Fax: (514)340-5552
Email: [email protected]
-----Original Message-----
From: [email protected] [mailto:[email protected]]
Sent: Friday, October 20, 2000 10:31 AM
To: Paolo Roncone; [email protected]
Subject: Re:Holes, waveguides and honeycombs
forwarding for [email protected]
____________________Reply Separator____________________
Subject: Holes, waveguides and honeycombs
Author: Paolo Roncone <[email protected]>
List-Post: [email protected]
Date: 10/20/00 12:03 PM
Group,
We are working on a couple of designs of telecom gear contained in metal
s.u.b.-racks and we have to meet emission limits up to 40 GHz.
I need some advice on the workings of waveguides below cut-off and
honeycombs, because we have to include ventilation openings without
possibly degrading the shielding effectiveness.
First, I made an inquiry on textbook formulas for circular and rectangular
waveguide cut-off frequencies. I was happy to find consistency among three
different sources (I found the same formulas although rearranged in
different fashions).
From Ott's "Noise reduction techniques in electronic systems" I found:
fc = [6.9/d] GHz for circular waveguides
fc = [5.9/l] GHz for rectangular waveguides
where fc = cutoff frequency
d = diameter of circular section (inches)
l = longer side of rectangular section (inches)
Now my question is: what about honeycomb panels ?
Can I use the same formulas for honeycombs ? Here the single cells are
neither circular nor rectangular. Can I still apply these formulas with
good accuracy ? If not, anyone knows of other formulas that apply in this
case?
As for attenuation (shielding effectivenes) of one single waveguide
opening, if the frequency is well below cutoff , this is proportional to
the ratio of length/diameter of the waveguide. The recommended ratio is 2:1
to 4:1 in order to get good attenuation.
Now, I just found a formula for attenuation of honeycomb panels as function
of frequency, length-to-width ratio of each cell and also number of cells:
S [dB] = 20log(fc/f) + 27.3(t/W) - 10log(n) (f < fc/10)
where:
S [dB] = Shielding Effectiveness in dB
fc = cutoff frequency of waveguide
f = frequency
t = cell length (or thickness)
W = cell section width
n = number of cells in honeycomb panel
I have no problems with the first two terms in the above equation. As for
the third term, that means that increasing the number of cells (n) in the
honeycomb panel degrades the shielding effectiveness of the panel (ex.
1000 cells means 30 dBs lost).
Before finding this formula I had a feeling that due to the skin-effect
each honeycomb cell could be treated as a single cell.
So far I wasn't able to find other formulas for honeycomb panels. So I'd
like to have some feedback on this.
I hope to get some useful directions.
Thank you in advance,
Paolo Roncone
Cisco Photonics, Italy
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