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Re: [Amps] placement of RF choke bypass cap.

To: garyschafer@comcast.net
Subject: Re: [Amps] placement of RF choke bypass cap.
From: Roger <sub1@rogerhalstead.com>
Date: Fri, 19 Mar 2010 21:36:03 -0400
List-post: <amps@contesting.com">mailto:amps@contesting.com>

Gary Schafer wrote:
>   
>> -----Original Message-----
>> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]
>> On Behalf Of Bill, W6WRT
>> Sent: Friday, March 19, 2010 6:28 PM
>> To: amps@contesting.com
>> Subject: Re: [Amps] placement of RF choke bypass cap.
>>
>> ORIGINAL MESSAGE:
>>
>> On Fri, 19 Mar 2010 14:41:00 -0400, "Gary Schafer"
>> <garyschafer@comcast.net> wrote:
>>
>>     
>>> There is no way for current to enter the center part of a tube or rod
>>>       
>> or
>>     
>>> flat sheet.
>>>       
>> REPLY:
>>
>> Imagine a large flat sheet of copper. You solder a wire in the center
>> of it, and another wire on the other side, directly opposite the first
>> one. Are you telling me that RF will not pass from one wire, through
>> the sheet into the other wire?  Or how about if the two wires were not
>> directly opposite each other, but spaced a few inches apart. Still no
>> current between them? Don't be silly.
>>
>> If what you say was true, all our radios and amps would not work.
>> There are dozens if not hundreds of places where RF passes through
>> sheet metal from one side to the other, such as through the shield
>> side of a coax connector.
>>
>> 73, Bill W6WRT
>>     
>
> Yes that's right. You can solder a wire to each side of a sheet of copper
> and there will be no current passed thru the copper sheet to the wire on the
> other side. You can even have a solid piece of wire going thru the copper
> sheet and if you solder it to the sheet on each side, nothing gets thru.
> Same for a screw going thru the sheet. If it is grounded all around the
> head, as it usually is, then nothing gets thru. The surface currents act on
> the head of the screw just like they do on the flat surface. They cancel
> currents deeper below the surface. This is "classic skin effect" at work.
>
> If it did get thru then the shields around IF cans would not work. Coax
> cable would not work as it does.
>
>
> Build a small oscillator and put it in a sealed metal box. If there are no
> open holes or open seams you will not be able to detect the oscillator
> inside.
>
> Roger gave the example of a screen room. If all joints are sealed and no
> holes in the screen a high power transmitter inside will not be detected
> outside. However, slip a small insulated wire thru the screen and it is like
> someone opened the door.
>   
Those doors used a "cam lock" and you could hear the transmitted signal 
until the cam was almost fully closed on the ramp. The last half inch 
was like a variable attenuator with absolutely no signal left when fully 
closed.
> Run a kw of RF thru some hardline and try to detect any current on the
> outside of the cable. There will be no signal there. 
> If what you are saying is true, that RF passes thru metal, then there would
> be signal on the outside of the cable as there is plenty of current on the
> inside of the shield.
>
> It is skin effect that keeps the current from passing thru the cable shield.
>
> There is a way for some leakage even with hard line that is due to the
> resistance of the shield and is known as "transfer impedance" but that is
> another story. Not what we are talking about here.
>
> Look at W8JIs experiment again. He has a bnc connector on one side of a
> sheet of copper. The center and shield are connected to the copper sheet an
> inch or so apart. He injects a signal into it basically forming a small loop
> for the signal to develop across. 
> He measures the signal level with a probe connected to the spectrum
> analyzer. On the side that the signal is injected there is signal in the
> sheet. Going to the other side with the probe there is no detectable signal
> at all. Even directly in line with the terminals on the other side of the
> sheet where the signal is injected. 
> Skin effect keeps the current from going thru the sheet.
>   
I'll admit it's counter intuitive, but it works.
If you perform the same experiment using DC you will be able to detect 
voltage on the other side. Not much, but it should be close to the same 
as the input side.  You might get to the point of needing a bridge to 
measure it though.  IIRC the lowest standards I worked with were 0.001 
ohm, but it might have been 0.0001 ohm. That was almost 30 years ago. 
It's interesting measuring the resistance of something when the 
resistance of the leads are at least an order or two of magnitude higher 
than the resistance of the object. 

73

Roger (K8RI)
> 73
> Gary  K4FMX
>
>
>
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