Re: Another Novice Q. - recharging - Volts and Amps

"Kris Krieger" <me@xxxxxxxxxx> wrote in message news:kOGdnVPS95_Nuf7VnZ2dnUVZ_hSdnZ2d@xxxxxxxxxxxxxxxx
ehsjr <e.h.s.j.r.removethespampunctuation@xxxxxxxxxxxxxxxx> wrote in
u3E8k.54$WJ.12@trnddc04:">news:u3E8k.54$WJ.12@trnddc04:

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 incorporate
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 important 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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