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In order to create a photovoltaic cell all I need is a photodiode connected parallel to a capacitor right?

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  • \$\begingroup\$ This is a photovoltaic cell. They are used to generate power. A photodiode and a capacitor would be some kind if photodetector. \$\endgroup\$ – JRE Nov 25 '19 at 19:45
  • \$\begingroup\$ What are you trying to build? \$\endgroup\$ – JRE Nov 25 '19 at 19:45
  • \$\begingroup\$ JRE we have to store somewhere current dont we? \$\endgroup\$ – Mrs Chemistry Nov 25 '19 at 19:50
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    \$\begingroup\$ Then a diode and a capacitor will not work. You need a solar cell and a rechargeable battery. The solar cell charges then battery in the sunlight, and the battery powers the LEDs in the dark. \$\endgroup\$ – JRE Nov 25 '19 at 20:37
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    \$\begingroup\$ The amount of energy you can collect depends on the area illuminated by the sun and the length of time it is lit. The area of a photodiode is too small to be useful. \$\endgroup\$ – JRE Nov 25 '19 at 20:38
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For what you want to do, you should use a photovoltaic cell (also known as a solar cell.)

The one in the link looks like this:

enter image description here

That's an AM5412. According to its datasheet it can produce 93 milliwatts. That's 36 milliamperes at 2.6 volts. It is a solar cell.

enter image description here

This is a photodiode.

It looks like this:

enter image description here

According to its datasheet it can produce upto 35 microamperes of current at a maximum voltage of 0.39 volts.

enter image description here

If you could get maximum current and maximum voltage at the same time, that would be about 14 microwatts.

Lets say you want to drive a 3.3 volt LED at 20 milliamperes for 1 hour. Energy:

$$3.3V \times 0.02 amperes \times 3600 seconds = 237.6 joules $$

The solar cell will need \$\frac{237.6}{0.093} = 2555 seconds\$ to collect the required energy.

The photo diode would need something on the order of 7000 times longer.

This is why you use photovoltaic cells instead of photodiodes.

A 1F super capacitor charged to 3.3 volts contains 5.4 joules of energy.

To hold the required 237 joules of energy to light your LED for one hour, you would need 44 1 farad supercapacitors.

They look like this:

enter image description here

They cost about four dollars each.

Or, you buy a couple of rechargeable NiMh cells that have far more capacity than all your supercapacitors together. That might cost you 2 dollars.

That's why you use batteries instead of capacitors.

All of that ignores losses that will occur in charging your storage system, in regulating the current, and preventing your battery from backfeeding the solar cell when it's dark.

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