On Mon, 28 Jul 2008 12:53:15 -0700 (PDT), terry
put finger to keyboard and composed:

The unit has two sensor, that look like diodes mounted in cavities in
the brass silver plated transmission line of the unit. The ouputs of
the diodes are connected to a 6 position switch. Three FWD. and three
REV. power positions. The output of the switch is just two unshielde
short leads to the meter iteself. There is also a tip/ring headphone
style jack on the side of the unit which disconnects the internal
meter and extends the circuit so it can be read by a remote or
external meter. So at that stage believe everything 'after' the sensor
diodes is low current DC. So it would appear necessary to amplify the
DC micro-voltage/current by about 3 to operate the replacement meter.
BTW tempoarily arranged the existing (damaged) meter (disconnected
from the unit) and the replacement in series with a low voltage source
and a varaible resitor. With the old meter at (as far as coule be
judged due to the damage) at full scale the replacement was about 0ne
third. Haven't measured the DC resistance of each yet because that
might have an effect on whatever circuit is used to 'step uop' the DC
current to operate the meter.

Excuse my ignorance, but wouldn't the 6-position switch be inserting
three different series resistances to vary the meter's sensitivity for
three different power ranges? If so, then why not just replace these
resistors with lower values, assuming the diodes are not affected by
the additional current?

FWIW, I found this home brew design that uses a 100uA meter:
http://www.qsl.net/vu2msy/homebrew/S...newsletter.pdf

- Franc Zabkar
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