On 6/27/18 3:38 PM, Zivney, Terry wrote:
Jim Lux said:
Actually not - the center conductor in an N is 0.120" - you might be thinking of the pointy
locator pin in the middle (0.063"), but in a "properly mated" N connector, that's
not carrying any power.
In a BNC plug, the center pin is 0.053" in diameter (slightly smaller than the
locator on an N) although i confess, having just looked through a variety of drawings on
google (BNC dimension image), there is a HUGE variation. Just like in an N connector,
you're counting on the mating surface at the shoulder of that pin to carry the RF current
(and set the impedance) (typically 0.081 to 0.087") and that's about 70% of the size
of the N connector's center pin
********
So, if the connection on the N is just the flat donut contact, that is even
less surface area than the BNC pin! So, e N have a lower power rating than the
BNC?
Not really... it *is* more complex than that. And skin effect is what
it's all about -
It's really only the outer few thousandths of an inch of the annulus
that carry the current in an N, but that's still a reasonable area. the
circumference is about 3/4".
In a BNC the annulus is even smaller in diameter (the pin is smaller)
if you look at the drawing here:
http://www.rfacsolutions.com/rf-connectors/bnc/
on the plug side it's dimension F that's important (the OD of the pin,
in the dielectric) (0.084 +/- 0.03)
On the jack side it's dimension J (also the OD of the jack, in the
dielectric.
The shoulder of the center pin (between diameter G and diameter F) is
the actual mating surface.
So the annulus is 0.081-0.087 inches in diameter. 0.084 is the nominal..
In the N connector, that mating surface is 0.120" in diameter. So the N
has about 50% more current carrying area
Now, if you're talking about DC or low frequencies, where skin depth is
significantly greater than (F-G)/2 (about 0.015"), then the current
starts to flow through the "sliding" contact in the BNC (and in the same
on the N).
The pins are probably gold plated brass or beryllium copper or something
like that. Let's hope it's a high copper content brass.
remember, at 7MHz, skin depth in copper is a mil (0.001) - and the
mating surface in the BNC is 15 mils wide - 15 skin depths..
To get to where *significant* current is flowing in the tapered sliding
contact, you need to be down in the few hundred kHz range (skin depth
goes as Sqrt(f))
At 100 kHz, skin depth in copper is 8 mils.. so the contact area is
still almost 2 skin depths wide - 63% of the current is carried in the
first skin depth, and 86% of the current is carried in the outer 2 skin
depths together.
BTW, the real problem with BNC is the bayonet outer connection. TNC has
the same pin dimensions, and is a much better connector. With the BNC,
the shield connection is that step (referring to the figure at the
website above) between dimension B and A on the jack. The critical
dimension is the ID of the shield, which actually isn't specified on
that drawing. it's the surface next to the dielectric.
THe interface between the OD of the dielectric post on the Jack
(dimension A) has to match the ID of the dielectric on the plug (also
Dimension A).
Hoewver, if you look at them, the Post has a max OD of 0.186, and the
socket (on the plug) as a min ID of 0.190, so there's 4 mils of slop there.
If the two shields don't line up, then the RF current doesn't flow from
one to the other cleanly, and with the funky spring loading of a BNC,
not lining up (as well as bent pins, etc.) is surprisingly common.
BTW, this is why connectors like SMA have a torque spec you want just
the right amount of crush - basic rule - if it has wrench flats, it's
supposed to be torqued. If it has knurls, it's supposed to be hand
tightened.
For all this, though, the impedance humps and mismatch on ANY of these
connectors is negligible at HF. PL-256/SO-239, N, whatever..
Unless you want to use lapped APC-7s (with guide pins, of course) on
your HF gear - for the ultimate in repeatability and low VSWR.
Reflection coefficient uncertainty is in the 0.001 range or better.
1000Vrms working voltage to 18GHz
Loss specified as <0.03 * GHz dB - so for 40 meters, .00021 dB - perhaps
we should call that 12 microBel?
return loss is specified as >= 32dB to 18 GHz, so with that kilowatt,
you'll see a maximum of a half watt reflected back to you.
Serious bragging rights there when you say that's why you busted the
pileup by using those $250 connectors.
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