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Hi there - I want to make a device that blinks two LEDs both with 50% duty cycle and with a 180 degree offset to each other. They don't need to be super bright - 1-2ma I'd guess. I want it to run 24/7. It will be running inside and will be subjected to electric light during the day, and will be in the dark at night. I want the LEDs to run even when it isn't in the light. That means some sort of energy storage device.

I'm thinking a supercap for this: http://search.digikey.com/script...

So 100F at 2.7V gives me 364.5J (.5CV^2). So unless my maths are failing me - that means, if I'm pulling 1ma at 2V, I'd get about 50 hours run time ((364.5 W * seconds) / (2mW)) * (1 hour/3600 seconds), assuming perfect DC/DC efficiency. Of course I probably won't have have a DC/DC (unless somebody knows of one that would work well here - I'd just assume quiescent current would make one not practical). But still 50 hours is way more than I need. I really need only about 10 or so - so this seems reasonable.

Now for charging this capacitor - I was thinking I'd do something stupidly simple - diode in series with a solar panel (to prevent reverse current) connected across the leads of the capacitor. Zener diode across the leads of the capacitor to make sure it doesn't get overcharged. Or maybe is there a better way?

What do you all think?

Thanks!

1. Consider having a series resistor in the charging circuit to limit the current.

2. Use a 7555 or CD40106 to drive the LEDs. Both can handle a wide voltage range. Parallel the extra gates in the CD40106 to get more drive current.

3. Suggested LED circuit: connect the two LEDs and a series resistor for each across the power supply. V+ - Resistor - LED - resistor - LED - GND. Connect the drive to the midpoint.

4. Use high-efficiency LEDs.

If you think education is expensive, try ignorance.

If the solar panel is not totally oversized there is not need to limit the charging current. The solar panel will just not deliver much current, even in full sun light.

Take care of the leakage current of zener diodes, it can be real high with low voltage zeners. Sometimes a string of diodes/LEDs can be better than a low voltage zener.

Quote:
Take care of the leakage current of zener diodes, it can be real high with low voltage zeners. Sometimes a string of diodes/LEDs can be better than a low voltage zener.

I sugest to take off the zeners, they clamp the excess voltage dissipating on them...

Brunomusw

Regards,

Bruno Muswieck

First thing I notice is that your capacitor is not a constant voltage source. It will discharge as you draw current from it, lowering the voltage. A simple resistor in series won't do the trick. You'll need a boost type circuit to do the trick. There are LED drivers out there - look into those. They ought to do the trick.
Also, watch the voltage across the capacitor. I know supercaps are fairly low voltage but I've never worked with them so I don't know how hard the ratings are. Still, it's never a good idea to violate them, so check your solar panel output voltage vs. the capacitor rated voltage.

For optimum charging of the capacitor I would ideally use a constant-current source, maybe a LDO regulator in constant current mode, with a full-charge turn off sensor (probably a resistor sized so that if the capacitor is full, attempting to push more current though the regulator will force it off). Definitely use some sort of low-voltage diode to prevent reverse leakage.

Non-ideally I would use an LDO regulator with the output set to 2.5V - just below the rating on the capacitor. This will charge it a little more slowly but is easier. Again, a low-leakage diode to prevent reverse leakage is recommended.

I suggest using standard NiCd cells. Several reasons:
- The output voltage is (relatively) stable
-> No power regulator for the LED's necessary, resistor may be sufficient
- They tolerate overcharging (and convert excess energy in heat)
-> No charging circuit necessary as long as your solar cell is not charging more than 0.1C
- The excess capacity is useful for long weekends, etc. when nobody is in for a couple of days to turn on the light and it's been raining and there was not much natural daylight.

Markus

Markus

I would add to the list above
-> They are cheaper
-> They are easily replaced