On Apr 7, 2:44*pm, wrote:
On Wed, 7 Apr 2010 14:19:12 -0700 (PDT), bpnjensen





wrote:
On Apr 7, 2:16*pm, wrote:
On Wed, 7 Apr 2010 13:28:18 -0700 (PDT), bpnjensen

wrote:
On Apr 7, 1:20*pm, dave wrote:
bpnjensen wrote:

Dale at PAR advertises that his matching transformer is wound on a
binocular core...but he does not go into precise detail. You'd have to
buy one and disassemble the unit casing to find out.

Bruce

Maybe we shouldn't have chased him away...

We've lost more smart people that way...

I think it was the part about buying one and disassembling it. Could
be using one of these. Probably type 43.http://www.surplussales.com/Inductor...FerMisc-5.html

Jim

Indeed! *I bet I know which one Dale uses ;-)

Maybe even cheaper with instructions and diagram:http://www.aytechnologies.com/TechData/9-to-1_XFMR.htm

Jim

Thanks Jim - I have two compound questions about this diagram that he
does not answer in the FAQ:

1 - In this wrapping technique, does the 50 ohm node at upper left
lead electrically to the 450 ohm node at upper right, or the feedline
ground? I assume the latter, but...this technically creates a half-
wrap somewhere, which would give either 2.5 or 3.5 turns. Same for
the 450 ohm node. It is not obvious from the diagram or the text.

2 - Is it automatically assumed that the feedline ground also goes to
a ground rod (somewhere?), or just the coax outer conductor? What
happens if there is no radio ground, just one at the antenna ground,
or vice-versa? What if both grounds are at the same rod?

Just wonderin'... :-)

Bruce