WOODS, RICHARD wrote:
> I am about to give a short seminar to our Mechanical Engineers and
> Designers
> on enclosure design for EMC compliance. There is only one problem - I
> have
> no faith in the theory I have for the attenuation through openings.
> The
> following formula is from the "EMC Handbook", Vol 3, by Don White.
> Assuming
> the frequencies of interest are below the waveguide cutoff frequency,
> the
> formula is
>
> A(dB) = KL/G - 20 log N where,
>
> K = 32 for round holes or 27 for square holes
> L = thickness of panel
> G = hole diameter
> N = number of holes.
>
> According to this formula, one 1/4 inch hole in a 0.090 inch panel
> would
> have an attenuation of 11 dB, and ten holes would have no attenuation
>
> whatsoever. This does not match my experience in typical ITE. Does
> anyone
> have any usable "rules of thumb" for Mechicanical types?
>
> Richard Woods
> Sensormatic Electronics
> [email protected]
> Views expressed by the author do not necessarily represent those of
> Sensormatic.
Hi Rich,
You can find the formulae in the german mil standard VG95376 Part 4.
As far as I know, the formula is OK under following restrictions
1.High frequency
The frequency is much higher than the cut-off frequency of the wave
guide.
The cut off frequency for cylindrical guides is
f=c0/(1.71*D)
whereas c0=3e8m/s, D= diameter
The cut off frequency for rectangular waweguides is
fc=c0/(2a)
whereas a= length of the long side of the wave guide
The attenuation of one cylindrical waweguide is
A=31,9*(L/D)*sqrt(1-(f/fc)^2)
For rectangular or square wave guides
A=27,3 *(L/a*)sqrt(1-(f/fc)^2)
If the frequency is much higher than fc, than
A=31,9*L/D
2 High Attenuation
If one can assume that you can compute the field behind the perforated
wall
by adding the absolute values of the fields coming through each hole or
wave guide.
Than
A=31,9*L/D -20*lg(N)
3. Plane wave, far field
Of course only plane wave, far field and infinite wall conditions were
taken into consideration.
Reflections and resonances inside an enclosure were not taken into
consideration.
The above formula should therefore regarded as a pretty good guess.
The same german military standard also gives some formulae for
attenuation
of an infinite and very thin perforated wall.
Case 1: one small hole in an infinite wall, where
r0<lambda/10
r= distance of point P to the hole:
r0 < r < 0.15*lambda
outside
IIIIIIIIIIIIIIIIIIIII<--------2r0--------->
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIwall
inside \
\
r
\
\
+P
Attenuation of en external field
A[dB]= 13,5 + 60*lg(r/r0)
Case 2: one small hole in a spherical shield
A[dB]= 10 + 60*lg(r/r0)
For N holes, add -20*lg(N) to the formulae above. Also see restrictions
above.
A[dB]= 13,5 + 60*lg(r/r0) -20*lg(N)
A[dB]= 10 + 60*lg(r/r0) -20*lg(N)
Hope this helps
Best regards,
George
--
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* [email protected] * Manager EMC Lab *
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