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Hy,
I posted some question about the HPLPowerManagement
component for micaz and I found some answer in the archive of the
forum.
I implemented the command Enable() in my code like
in the example apps/CountRadio. I tried to run the
application on the mote and to test the current
consumption and I checked that the current in that state (with the
SleepRadio application) goes down to (about) 50
microA.
I red the post of Joe Polastre about the condition
with them the node goes to the sleep mode: just call "Enable()" in
the init() part of the beginning code and when the
node check the condition (radio off, interrupt off, ecc...) the node
goes to sleep.
Now I'm working with an application that use the
MDA300 and I would increment the lifetime of the node. If I plug the micaz on
the sensorboard,
the current goes up
to (about) 37 mA. I tried to call the command "SamplerControl.stop()" for stop the use of the
channel, but without
effect. My questions are:
1. How can I reduce the current when
I'm using the MDA300?
2. Is possible reduce the current only
when the node is "unplugged"?
3. For stop the interrupt, I only have to
call the relative command .stop() of the component?
4. Why in the example CountRadio/RadioSleepM use 2
timer for arrive at the sleep state?
Thanks in advance for any suggestion or link to
some document about this topic.
Alessandro Balvis
Centro Ricerche ENEL
Pisa- ITALY
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