An earlier post stated that a 150 microamp battery drain (when the FT is
OFF) would deplete the battery in 10 days.
I think the actual calculation should be:
Service Hours = (Battery Capacity in mA-hrs) / (Current drain in mA)
The typical FT-50 battery is 600 mA-hours capacity, and 150 uA = .15 mA,
Service Hours = 600 mA-hrs / .15 mA = 4000 hours
4000 hours = 23.8 weeks. A battery should not deplete in a mere 10 days
of non-use. Most likely, battery self-discharge would be a much more
significant effect in battery depletion.
My several FT-50s all pull 240 microamps (.24 mA) from the battery when
the radio is off, regardless of whether the battery is a 6, 7.2, or 9.6
volt unit.
Supposedly there is a Yaesu service bulletin out for some FT-50s that
have a much higher current draw when off. Up to 10 mA has been reported
by some!
Regards,
Mike / KK5F
--
FAQ on WWW: http://www.qsl.net/k7on/yaesu.html
Submissions: yaesu@contesting.com
Administrative requests: yaesu-REQUEST@contesting.com
Problems: owner-yaesu@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|