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.
73
Alex
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
_______________________________________________
time-nuts mailing list -- time-nuts@febo.com
To unsubscribe, go to
https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
_______________________________________________
time-nuts mailing list -- time-nuts@febo.com
To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
