"Kris Krieger" <me@[EMAIL PROTECTED]
> wrote in message
news:kOGdnVPS95_Nuf7VnZ2dnUVZ_hSdnZ2d@[EMAIL PROTECTED]
> ehsjr <e.h.s.j.r.removethespampunctuation@[EMAIL PROTECTED]
> wrote in
> news:u3E8k.54$WJ.12@[EMAIL PROTECTED]
>
>> I think you've gone down a path that may be counter productive.
>> First: The amount of power you will get from the solar panel
>> will depend predominantly on how much surface area you can get
>> exposed to the sun. You can include instructions to the
>> consumer concerning installation for maximum exposure, but
>> the only factor _you_ can control is the surface area.
>>
>> What does that mean at this point? You know the size of what
>> you are building, so _you_ need to find solar cells that
>> will fit that area. *That* selection will dictate how much
>> power the design will have available, and it will be a range
>> from minimum (0 on a cloudy day) to maximum. The amount of
>> power available from the panel will vary throughout the day.
>>
>> Only when you know how much power will be available in a
>> typical *week* can you properly design the electronics.
>>
>> So, post again once you have found a solar panel (or
>> combination of panels) that will fit the device you are
>> making. How much power will be available to work with
>> under the "typical" conditions of your intended market?
>>
>> You want something simple and better than a typical solar
>> powered garden light. So do I - and I also want cheap
>> oil. Neither is generally available these days. You
>> can maximize what a typical solar powered garden light
>> (like cat# SPL-09 at http://www.allelectronics.com/
see
>> also cat# SPL-05) produces with increased complexity and
>> construction cost. If you can live with the performance
>> level of either of those, you job is done - just incor****ate
>> their panels & circuitry in your prroduct. Oterwise, you
>> have to put in the work to search for the best panels
>> you can get to fit your product, and obtain the best
>> performance possible from that power source.
>>
>> Without knowing how much power is available, it is impossible
>> to say how long it will take to charge a cell, nor can the
>> charging circuit be designed for best performance.
>>
>> IIRC, no one has told you to do that work - identifying
>> the panels - in the various threads, posts and replies since
>> you started looking for an answer. Of course, I may have
>> missed it, so if you have identified the panels, how much
>> charging power is available to work with?
>>
>> Ed
>>
>
> Actually, the initial consideration was finding how much brightness
would
> be enough, and which LEDs (and/or combination thereof) will provide it.
> From there, it goes backwards to find out how to run them off of a
> *maximum* of four 1.2V NiMH batteries. From there, working backwards
> again, is the charge controller IC - finding the one that will charge,
> and keep from overcharging, 3 to 4 of the batteries. THe solar cells
> (not panels, just cells) is actually the last thing. WHat I'm trying to
> figure out is whether I should have the solar cells producing the same
> total voltage as that which the batteries will have (3.6V for three of
> them, 4.8V for 4); since I've learned that NiMH are supposed to be fast-
> charged rather than trickle-charged, I'm tryign to figure whether solar
> cell amperage is more im****tant for doing that, with voltage being
> insignificant, or what.
>
> Only *after* I know all that, can I select the specific cells, and
> configuration therof...
>
> - Kris
The problem is total energy input vs. requirements. You can't get more
joules out of the system than you put in, so you need to figure out how
many
joules you'll use. How long will the lights be expected to stay on? How
much
power will the lights consume while they are on? The product of those
gives
the energy required.
Current solar technology gives you between 3% and 17% efficiency.
According
to Wikipedia, the irradiance of the earth averages out to 342W/m/m over a
24
hour period. So, you can get an average of 5MJ/Day from the most efficient
1
m^2 panel you can get. Not bad. However, the panel would need to track the
sun to get maximal efficiency. Without that, you get 1/5 of the energy
out.
Hmm. So, 1MJ/Day. Still not bad. Sadly, your panel probably isn't a square
meter. Lets say 1/10 of a meter on a side, right? Then that is 1/100 of a
square meter, so you now have 1e4J/Day. Still OK. However, suppose you
have
a bit of shade? My solar panels for my garden lights only get direct
sunlight about 30% of the day. So, lets say 1/2, so now 5e3J/Day. What
about
dirt? Another 50%, so now 2.5e3J/Day. This is with the best solar panels
that are available. You'll probably get the 3% variety, unfortunately. So,
adjust down to 18% of the prior, to 441J/Day. That seems closer to what I
expected, unless I got the math wrong.
Ok, now, how much energy do you want to expend? 5W for 4 hours? Hmm, that
is
5 * 3600 * 4 = 72kJ per day. Yikes! That is more than 100 times what
you'll
get with the little 4x4 solar panel above.
These are ballpark figures. However, they suggest that you are probably
only
going to be able to get, at most 1kJ/Day out of the panel, so your lamp
will
need to be much lower power. Lets go with something more realistic, like
two
20mA white LEDs. Each LED requires about 3.5V forward voltage, so the
power
requirement of the duo is 0.1W. So, given you have 1kJ/Day coming in, you
can run them for 10,000 seconds, which is 2.8 hours. Not too bad. That
seems
closer to the garden lights I have in my yard.
Regards,
Bob Monsen
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