[Elecraft] how did I calculate the loss?

[drcuthbert@micron.com]

The lengths were for a line length of 180 degrees. BTW, even with 600 =
ohm line there is an awkward length around 180 feet. But if one is =
running let's say 100 feet things look good to me.

  Dave

-----Original Message-----
From: George, W5YR [mailto:w5yr@att.net]
Sent: Monday, December 15, 2003 4:11 PM
To: drcuthbert; elecraft@mailman.qth.net
Subject: Re: [Elecraft] how did I calculate the loss?


Dave, I missed the part where the ladderline and 600-ohm line have to be =
270
ft long and the coax only 180 ft.

Could you either repeat that part or attach a copy of your email?

I might mention that TLDetails is also an excellent program for handling
problems of this type. Google will find it for you.

73, George W5YR
w5yr@att.net


----- Original Message -----=20
From: <drcuthbert@micron.com>
To: <elecraft@mailman.qth.net>
Sent: Monday, December 15, 2003 4:05 PM
Subject: [Elecraft] how did I calculate the loss?


You might be wondering how I calculated the line loss with the =
mismatched
load. I used two methods. NEC (with a physical T-line) and with =
WinSmith.
But, I remembered a link to an online program that easily solves this =
type
of problem. Here is the link: http://fermi.la.asu.edu/w9cf/tran/

Let's solve the 30 -j500 ohm load with 180 feet of RG213:
Loss =3D 9.7 dB
Let's solve the 30 -j500 ohm load with 270 feet of 450 ohm ladder ohm =
line:
Loss =3D 2.8 dB
Let's solve the 30 -j500 ohm load with 270 feet of 600 ohm opn line:
Loss =3D 0.9 dB

Not that bad with 600 ohm line. Less line length means less loss (it's =
not
linear though). And with 600 ohm line we don't have to avoid any line
lengths to avoid an awkward impedance at the K2, as we might have to =
with 50
ohm line. For example, a 45 degree, 600 ohm line (68 feet) gives 10 +j50 =
at
the K2 and only 0.42 dB of line line. I am now of the opinion that =
feeding
the 8030 with open line, and tuning with the K2, is a viable option.

    Dave WX7G