🚀 go-pugleaf

In article <1992Oct8.120225.6247@fys.ruu.nl> vreeburg@fys.ruu.nl (Jurriaan Vreeburg) writes:
>In <bvr5pd.14o@wang.com> garyf@wang.com (Gary Field) writes:
>
>>dhina@dxcern.cern.ch (Mourad Dhina) writes:
>
>>>i am trying to realize a ... coax cable of 25 ohms impedance.

>>Put two pieces of 50 Ohm cable in parallel. It should be exactly the same.

>After reading Gary's answer I had some doubt of its correctness. It implies
>that by putting two 50 Ohm coax cables in series, one is able to create a 100 Ohm
>coax cable, which is certainly not true.

	No, that is NOT implied, unless you are making an analogy
to simple resistors (which is not exactly what Gary was doing).

>I think that putting two cables in parallel doesn't change the impedance, since
>the impedance is defined as:
>
>                   L
>Z = square_root ( --- )
>                   C
>L: inductance per unit length
>C: capacitance per unit length
>
>What happens if we put two cables in parallel: both the inductance and the capacitance
>increase by a factor of two -> Z doesn't change!!!

	Huh?  The capacitance goes UP by a factor of two, and the
inductance goes DOWN by a factor of two, so Z is halved.

	And, in the case of two cables in series, the inductance is
doubled, BUT the length is as well (so inductance per unit length
is not changed).  Same argument for capacitance per unit length.

	The original argument, two 50 ohm cables in parallel, is correct.
It isn't a good idea, though, unless you're SURE you can get the 
signals of interest through the 50 ohm cables with comparable losses
and shielding efficiency to the semirigid cable.  Around here,
we use semirigid only for exotic, high-performance applications,
and there is no good substitute.

	John Whitmore