Well - yes - I think you are missing something - let's say - since I
don't have the EXACT value - that the radio stops working at 8.5
volts - and you have .4 volt drop between the battery and the load
(the radio) - so when the battery is discharged to 8.9 volts - the
radio quits on TX. Where if you could reduce the voltage drop between
the battery and the load to say .15 volts, then the battery could go
down to 8.65 before the radio sops working. By having less voltage
drop between the source and the load, you are able to use more of the
available battery capacity.
--- In FT817@y..., bbadger@y... wrote:
--- In FT817@y..., KG4CHX@A... wrote:
While experimenting I found the following voltage drops under
full
load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock
wiring.
This is all lost in heat energy. 2amp X .5 volts = 1 watt lost to
voltage drop!!
Does it really matter? According to K6XX analysis of current drain
during transmit, the current drawn by the power amplifier is largely
unaffected by voltage. As a result, that 1 watt that we could
conceivable recover by replacing the power connectors will just
result
in 1 watt more power dissipation in the MOSFETs.
Or am I missing something?
--
Brian N0KZ
