I'm building a simple circuit that uses a couple solar cells to charge a (huge) capacitor. I'm using four solar cells, rated at 3V/20mA, in parallel to charge a 22F capacitor (I am aware the capacitor is huge, but this is a purely educational exercise). I use a 1N5817 Schottky diode in series with the solar cell array to prevent the capacitor from discharging. In addition, I've included a Zener diode in parallel with the capacitor. I'm not too sure what the point of the Zener diode is. I've looked at some schematics and they've included the Zener diode in parallel with the capacitor, so I did as well. My guess was that the Zener prevented damage to the solar cells if the capacitor discharged suddenly.

Here's my problem, I want to find out how long it takes (theoretically) for the capacitor to charge, but I'm confused about how to do it. Since there is no resistor in series with the capacitor in this circuit, is the maximum possible voltage across the capacitor (3V - Schottky diode voltage drop)? If this is the case, do I use the relation, I = C dV/dt, to find out the time it takes to reach the maximum theoretical voltage across the capacitor?

Edit for clarification: I couldn't find the correct symbols in EAGLE, so I labeled the components explicitly in the schematic.

Attached Images

solarcell.JPG (25.2 KB, 14 views)

 

Your solar battery (it is a photo-transistor in your schematic) produces its rated output only in summer, on the equator, at noon and when aimed directly at the sun. Its output is much less anywhere else or at any other time.
When the capacitor is nearly fully charged then its charging current is reduced so the voltage from the solar battery will be much higher but will be clamped by the zener diode.

Low voltage zener diodes are lousy. Your high current 3.3V one is the worst. It will leak so much current at 3V that it will quickly drain the capacitor at night.

Attached Images

Zener diode low voltage.PNG (55.5 KB, 7 views)

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Uncle $crooge

 

I can see no reason why the zener would be in that schematic in the first place, you don't exactly get light pulses of sufficient magnitude to warrant it. It could perhaps be to simulate the reverse breakdown characteristics of the solar cell itself?

From a 3V 150 ohm source (20ma max) it'll take about 8.5 hours to peak the capacitor to 3 volts. It only takes 50 minutes to get to 2 volts.

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The solar cell produces 3.0V with a load of 20mA. When the capacitor is nearly fully charged then its charging current becomes less and the voltage from the solar cell might reach 5V or more and blow up the low voltage capacitor. If a low voltage zener diode worked better without a high leakage current then it could be used as a 3.3V clamp. Five diodes in series would make a better 3.3V clamp.

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Uncle $crooge

 

Don't solar cells produce a constant current?

A 22F capacitor will take 22 seconds to charge to 1V given a current or 1A.

It will take 44 seconds to charge to 2V given a current of 1A.

It will take 22 seconds to charge to 2V if given a current of 2A.

It will take 220 seconds to charge to 2V if given a current of 200mA (0.2A).

Figure it out yourself, I should've given you enough information.

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