Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A sun jar is an object that stores solar energy in a battery and then releases it during dark hours through a led.


  • a $65cm^2$ solar panel
  • a 12h/12h light/dark cycle
  • insolation of $2.61kWh/m^2/day$
  • perfectly efficient components (i.e. without violating Entropy laws or other theoretical limits)
  • light is emitted by an ideal "white" source

How bright can we make the jar so that it shines for 12 hours?

Useful links:

share|cite|improve this question
I'm curious. Why $65cm²$ ? – Malabarba Jan 10 '11 at 1:34
Because the top of my sun jar is 9cm in diameter... :-) – Sklivvz Jan 10 '11 at 8:15
up vote 2 down vote accepted

Using that ideal white source on the wiki link, that states 251 lm/W.

The insolation level is $2.61kWh/m^2d = \frac{2610}{24}W/m^2$

which across $65cm^2 (= 0.0065m^2)$ gives 0.7W. If everything were 100% efficient, then you'd have $0.7 \times 251 lm = 178 lm$

Now we derate on some maximum theoretical efficiencies. The biggest theoretical derating will be on the PV, and that will completely dwarf any loss on the maximum theoretical efficiency of a round trip into storage and back.

For a single-junction n-p PV cell in unconcentrated sunlight, the maximum efficiency (Shockley–Queisser, DOI:10.1063/1.1736034) is 30%, giving you about 53 lm. By layering multiple junctions, you could theoretically get 42% (2 junctions, 74 lm), 49% (3 junctions, 87 lm), tending to 68% ($n\to\inf$, 121 lm) (doi: 10.1088/0022-3727/13/5/018)

Now, if you are allowed to put a concentrating lens onto the $65 cm^2$ cell, so that the sun jar harvests light from a much larger area, then we can really go to town. From the second link, the maximum efficiencies for concentrating PV, as the number of junctions tends to infinity, is 86.8%. So then you've got to find out what the maximum concentration could be without the whole lot bursting into flames ... that 86.8% is based on a concentration factor of 45 900 , so it's a PV cell made of pure unobtainium, and I'll stop right there.

share|cite|improve this answer

Well, basic answer is that it can be about twice "brighter" than sun (i.e. it will emit twice the amount of light absorbed), if LED emits only visible light, while battery absorbs everything from IR to UV.

Reality is of course much much worse.

share|cite|improve this answer

2.61kWh/m^2/yr * 65cm^2 = 0.002 watts; since the time to charge/uncharge is the same, this is both the power in and the power out.

Conversion to light at 300 lumens/watt: 0.6 lumens

If it shines in all directions, about 0.05 candela - so about as powerful as one-twentieth of a candle.

So. Not very brightly, then.

share|cite|improve this answer
Note the comments on the question---there was a units error in the initial text, so this should be in $\text{W}\text{h}/\text{m}^2/\text{day}$, or about 365 brighter...18ish candela. – dmckee Aug 4 '11 at 15:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.