Hi,
I would appreciate your feedback on my calculations for a solar PV + battery +
LED system.
I am assuming NiCd battery can be discharged to 1.0 V and charged to 1.41 V.
A red LED needs about 1.8 V, and a blue LED needs up to 2.4 V, to get the
rated light. Therefore, I can use two batteries to power the red LED, but
I need three batteries to power the blue LED (two discharged batteries
have 2.0 V and the blue LED needs 2.4 V). So, building a toy with red and
blue LEDs I need three batteries. I don't want to use voltage up-converters /
charge-pumps.
I want to power four bright LEDs with 20 mA each.
Option 1.
Four LEDs in parallel, requiring 80 mA total.
Assuming a winter night at 12 hours (at least), I need 12 * 80 = 960 mAh or ~1 Ah.
Therefore my battery set needs three batteries at 1 Ah each, as batteries are
connected in series.
Batteries will be charged up to 3 * 1.41 V = 4.23 V. Assuming a solar cell
giving 0.49 V with a load (sunny day), I need 9 cells (9 * 0.49 V = 4.41 V).
And here comes a problem, I measured recently voltage and current from
a cell in a direct sun, and then with clouds. Cloudy day (Summer-time) gives
about 75 % less current (1/4)! That was what I measured! A cloudy winter day
will give even less. Open circuit voltage dropped to 0.48 V. Loaded cells
output about 12 % less voltage, giving about 0.48 V * 0.88 = 0.42 V. All of
that means, I need significantly overdesigned solar panels to work with heavy
winter clouds for a week or more - at least where I live. I need 10 cells
("almost" delivering voltage), and those cells need to be rated with four
times higher current - 4 times bigger 
?
Lets assume I will charge batteries with 200 mA for five hours. That will
give me 1 Ah. When NiCd batteries are charged at at C/5 (one fifth of capacity;
200 mAh is 1/5 of 1 Ah) efficiency factor is about 63 % (0.63), so I need
to increase current, 200 mA / 0.63 = 317 mA. Now, with winter clouds, I will
need a factor of four, and cells need to be rated at
4 * 317 mA = 1268 mA (~1.3 A).
Is that right? That's a huge difference in size and costs. Something
may not be right in my cloudy assumptions.
What do you think about my calculations, so far?
Current saving can be done having a full light for 5 hours, and then dropping
by 1/2 to 10 mA per LED, and finally turning off, if batteries are discharged.
Note: turned off LEDs at night might be viewed as a malfunction of the system.
Total capacity of a battery system is now at 5 h * 80 mA + 7 h * 40 mA = 680 mAh.
The charging current drops relatively (680/960) * 317 mA = 225 mA or the
oversized one at (680/960) * 1268 mA = ~900 mA.
What is the size of a cell delivering 900 mA? It looks like 2 x 2.5 inch
Multi-Crystalline, 15 % Conversion Efficiency.
Still, lots of current. And more, self discharge of about 2 % per day
(5 % at hot days) adds the charging current.
Option 2. Lets see how it looks for having two diodes in series? (Two
branches with two diodes each)
I need 40 mA of current. Two blue LEDs need 4.8 V, so I need five batteries in
series. Capacity calculates as 12 h * 40 mA = 480 mAh (~0.5 Ah), allowing
for AAA NiMH battery.
I have five batteries now. Charging voltage is 5 * 1.41 V = 7.05 V.
How many PV cells needed? Assuming cloudy day 7.05 / 0.42 V => 17 cells !!!
That's painful :-( :-( :-(
With current saving option I need capacity of 5 h * 40 mA + 7 h * 20 mA = 340 mAh
Five hour charging at C/5 and 0.63 efficiency requires (340/5)/0.63 = 108 mA.
That's as a major difference in current ... but ... I need to increase by that
factor of ?four to compensate for a winter cloudy day, so, cells rated at
about 400 mA are needed.
The size of a PV cell delivering 400 mA is 0.9" x 2.5".
Option 1: surface is 10 * 2" * 2.5" = 50 inch^2
Option 2: surface is 17 * 0.9" * 2.5" = 38 inch^2 (19" * 2 ")
If I would calculate the cell surface without the cloudy day factor, I would need
for Option 2 - 8.5 " * 1 " of cells - not bad, but ...
Conclusion: going with two branches (two LEDs each) requires less PV cells.
Also, voltage is higher, allowing easier building of a control system.
Your feedback is appreciated!
Elektronix
