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I need a charger and battery for a cell phone booster.

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Posted by philkryder on October 16, 2009, 7:11 am
 

Colleagues -

I would like to put a cell phone booster/repeater in a remote
location.

It is a Wilson 801245 model.

I believe it uses about 3 watts max.

The unit will be in an unsupervised location about 90% of the time.

It will be at about 2500 feet altitude.

Since it will not be supervised, I would like to have it be as un-
obtrusive as possible by having a solar panel lying flat on top of an
8x40 shipping container, so that it won't attract attention.

I would like to use some form of sealed battery to prevent gas
accumulation in the container.


I have several questions:

1) Is a SOLAR charger a good way to go?

2) If I go SOLAR, and it is lying flat rather that angled optimally,
what size would I need?

3) does this 45 watt unit (see link below) from harbor freight have a
large margin of excess capacity?
http://www.harborfreight.com/cpi/ctaf/displayitem.taf?Itemnumber599
?are there better options?

4) what size battery should I get to store power for nights?

5) where is a good place to purchase the battery and which style of
sealed batter should I get?

6) are there other critical considerations that I should consider?

Thanks much!
Phil





Posted by You on October 16, 2009, 6:09 pm
 
In article


If you lay it FLAT, and there is the possibility that the area gets snow
in the winter, plan on the system going offline when it snows.



Posted by Malcom \"Mal\" Reynolds on October 16, 2009, 6:40 pm
 In article
<8204478f-c75b-40b7-b413-bbff2033bb3b@u3
6g2000prn.googlegroups.com>,
 wrote:


How long between visits to the location.
It well might be cheaper to just change
charged batteries.

Posted by ghio on October 20, 2009, 5:11 am
 
TekSolar can provide a plug and play solution.

Posted by Ron Rosenfeld on October 20, 2009, 11:58 pm
 On Fri, 16 Oct 2009 00:11:45 -0700 (PDT), philkryder


That'll depend on the uplink power being generated.  According to the
specs, it requires 6v at 0.5A -- 1.5A.  That is 3watts -- 9 watts.

However, if you are going to supply DC to the 801245, you'll need to know
both the peak and and average power requirements.



It depends on the location, and the solar resource available at that
location.


It depends on the location and the optimal angle for that location.  But
it's probably close to 1/cos(optimal angle).  There may also be a problem,
with a flat panel, with keeping it clean and unobstructed.  So you may lose
50-100% of your power by keeping it flat.


It depends on your location and the solar resource at that location.


First you need to obtain information about the solar insolation available
at the location for the panels.  In particular, you will need to know the
solar insolation for the worst month of the year.


You not only need to store power for nights, but also for cloudy days.  And
if this is a critical system, you will want to design for worst case.


See above
--ron

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