[time-nuts] Noise and non-linear behaviour of ferrite transformers

Alexander Pummer alexpcs at ieee.org
Sun Jul 20 21:50:11 EDT 2014

No, the current passing the outside f the shield  will not induce any 
voltage inside of the coax, but the voltage drop caused by the current 
on the ohmic resistance [!!!] of the shield will show upbetween the two 
ends of the cable -- and that will  show up as  it was added to to the 
voltage which is carried on the center conductor of the coax.

On 7/20/2014 6:10 PM, Chuck Harris wrote:
> I'm not sure what you are saying.
> skin depth = (2.6/sqrt(fhz))inches for copper.
> So, at 60Hz,   skin depth = 0.336 inches.
> and at 100KHz, skin depth = 0.008 inches.
> and at 1MHz,   skin depth = 0.0026 inches.
> Are you saying that at 60Hz, because the
> skin depth is deeper than the coax shield is
> thick, that current passing through the outside
> of the shield will induce voltage inside of
> the shield, and that at say, 100KHz where the
> skin depth is a little less than the shield
> thickness, or at 1 MHz, where the skin depth
> is only a small fraction of the thickness of
> the shield that it won't?
> Or something else?
> -Chuck Harris
> Bob Camp wrote:
>> Hi
>> The “coax is an antenna” problem comes in well before you get to DC. 
>> Even with no
>> transformer involved, the skin depth of the coax shield gives up well 
>> above 60 Hz
>> (and likely well above 100 KHz). If you want to do full isolation 
>> over a very wide
>> range you need some combination of shielding and balanced lines.
>> Bob
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