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Re: [Amps] Shielding Effectiveness

To: amps@contesting.com
Subject: Re: [Amps] Shielding Effectiveness
From: "Roger (K8RI)" <sub1@rogerhalstead.com>
Date: Wed, 06 Oct 2010 14:23:37 -0400
List-post: <amps@contesting.com">mailto:amps@contesting.com>

On 10/6/2010 1:03 PM, Sam Carpenter wrote:
> So...The screen on this retrofit might not be the best:
>
> http://www.hamradiomarket.com/Articles/Ten%20Tec%20Centurion.htm
>
> Looks a little wide and not to well snugged down.
What I'd do in that case is take a strips of 1/8" Aluminum about 1/2" 
wide and sandwich the expanded SS between them and the cabinet.
I'd probably use #8 or #10 pan head screws (plated Brass or SS) about 
one per inch or two. What ever it'd take to make good, continuous contact.
The size of the strips, or even material is not important as long as 
it's strong enough to do the job and not magnetic like steel.  
Appearance may or may not play a part in the selection.

OTOH  I've seen commercial amps that only used a blade protector to keep 
fingers out of the fan in that location.

73

Roger (K8RI)

> N9FUT
>
> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
> Behalf Of Roger (K8RI)
> Sent: Wednesday, October 06, 2010 12:48 PM
> To: amps@contesting.com
> Subject: Re: [Amps] Shielding Effectiveness
>
>
>
>
> It's important to remember that hardware cloth which comes in many
> sizes, or screen does not normally give a good contact around the edge.
> If fender washers are used they basically need to be edge to edge.
> Commercial equipment uses a bracket, or frame to hold the screen in
> tight contact with the cover/chassis around the entire periphery.
> Although screen can be a much better shield at VHF, that advantage is
> lost if that contact is not continuous around the edge.
>
> 73
>
> Roger (K8RI)
>
> On 10/6/2010 11:13 AM, Ward Silver wrote:
>> Per Ott's "Electromagnetic Compatibility Engineering", section 6.10 -
> shielding effectiveness in dB is:
>> S = 20 log (wavelength / 2xL) where L is the maximum linear dimension of
> the aperture.
>> If you have multiple apertures, such as an array of holes, the general
> procedure is to calculate S for one hole and then reduce S as 10 log (n)
> where n is the maximum number of holes in a straight line in the array.
>> For example, if S for one hole is 52 dB (a 2.5 mm hole at 2 meters) and
> there are at most 20 holes in a line, S would be reduced by 10 log (20) = 13
> dB to S = 52 - 13 = 39 dB.
>> This is an approximation, of course, plus there are seams and other things
> to worry about.  But it's a good way to at least compare types of apertures
> in shields.
>> 73, Ward N0AX
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