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I made a little windmill which is turning a small generator to light up an LED, but the voltage is too low even with high wind speed for the LED to glow. I thought maybe the electricity could be stored somehow maybe by a capacitor and released when the voltage is high enough, making the LED blink according to the wind speed.

I know little about electronics so I'm asking for help. Thank You.

Edit: What I want to know is how to make a circuit that charges up electricity from a low power source and discharge when the voltage is high enough. The frequency of discharges should increase when the power (wind speed) is increased. About the generator. It's actually a small electric motor. I tested it with another motor, connected their rotating heads together, applied 5V to one of them and measured about 4V coming from the other, so I thought it's not a bad idea to use it as a generator. This is a really small project and it shouldn't be taken too seriously.

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The point is to blink the LED using only the generated electricity. By "making the LED blink according to the wind speed." i meant that with higher wind speed electricity would be stored faster making it discharge more frequently. –  hellwisp Apr 30 '11 at 18:11
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4 Answers

Technical Explanation: The open circuit voltage produced by your generator is too low to drive any current through the LED.

Straightforward Explanation: Your problem is probably that there aren't enough volts produced by the generator, even when the LED isn't connected. You can measure this with a simple multimeter when spinning the generator. Set the multimeter to 20V AC or DC and see what voltage you measure. To get the LED working you should see at least 1.4 volts for a RED LED, 2.5 volts for blue, or 3 volts for a white LED.

Others have suggested a resistor is necessary to prevent damage to the LED. Normally, this would be true, but in this case as the LED starts to light up, it will slow down the wind turbine, protecting itself in all but the strongest winds. Additionally, most generators have the equivalent of a resistor inside them.

One suggestion is to use a microcontroller/capacitor. The problem with this solution is you will not be able to get a high enough voltage to power up the microcontroller.

My solution:

  • Use a Red rather than white/blue LED. Use no resistor (or if you do, choose something below 300 ohms)

  • Spin the generator faster - use gears to up the speed and lower the torque if necessary

  • Get a different type of generator designed for slow speed use.

  • If you have multiple LED's, connect them all in parallel (ie. every LED connects direct to the generator)

  • Make sure you're spinning the generator the right way - test spinning it backwards, or connect the LED the other way round.

Note: Human eyes are very sensitive - if the output voltage of the generator is high enough, even a tiny current flow will cause the LED to visibly light up in the dark.

Additionally, there are many types of generator, AC/DC, brushed and brushless etc. In this situation, it doesn't really matter which you choose.

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With energy harvesting, plenty of power will be available for an MCU and LED. –  Leon Heller Apr 30 '11 at 16:33
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Without thinking about it much I would have the generator charge a CAP and regulate that voltage to a low-power micro (perhaps a PIC12F615 or similar). The micro would mostly sleep but wake on a edge transition that is created every rotation. I would use the TIMER1 count between rotation interrupts to calculate the speed. I would turn on the LED proportional to the speed and go to SLEEP. The slower the speed the more the uP would be asleep and take less power.

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Charge an electrolytic capacitor in your generator's output an use a Joule Thief to drive the LED. You can light up any LED as long as the input voltage is at least 0.6 volts I think.

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What about flashing the LED to indicate the wind speed? –  Leon Heller Apr 30 '11 at 16:59
    
Theoretically the capacitor will charge until the voltage for the Joule Thief to lit the LED is enough. Then when the LED lights up, the capacitor will discharge so things will start again from the start. –  LaLeX Apr 30 '11 at 17:04
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Once it gets to a certain speed, which could be quite low, won't the LED be on continuously? –  Leon Heller Apr 30 '11 at 17:14
    
In this case, he could use a divider to charge the capacitor slower. Or use a smaller capacitor. Or both. It's trial and error. –  LaLeX Apr 30 '11 at 17:22
    
He said he wanted the LED to blink according to the wind speed. That looks like flashing to me! –  Leon Heller Apr 30 '11 at 17:51
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To store small amounts of generated electrical energy and get it to do something useful you need to use "energy harvesting" (EH) techniques. Suitable hardware is available from several sources and this company's EH300 kit might be suitable for your application. Farnell stocks the EH300; I bought one some time ago but haven't got round to trying it. The EH300 can deliver 25 mA for 68 ms at 1.8V - 3.6V, which is more than enough for an MCU and LED. Here is another company making EH hardware. Their technology is used with wireless transceivers and MCUs, taking a lot more power than an LED, powered by small solar cells.

You could power a suitable low-power MCU such as an XLP PIC or an MSP430 from the EH device, measure the voltage provided by the EH unit, and flash the LED at a rate proportional to the wind speed when there was sufficient stored power. The RPM of the windmill could be ascertained by measuring the voltage generated using an ADC input on the ADC. XLP PICs typically take under 20 nA in deep sleep mode, and can operate down to 1.8V.

I just tested the EH300 with input from my bench supply, varying it from 1.1V to 2V to represent the varying output from a generator. After a few minutes, it was delivering a stable 1.8V from a 1.1V input, so the EH technique will definitely work. Once the capacitors are charged, it should function indefinitely, provided that there is enough wind to turn the windmill at a reasonable speed from time to time. A battery could also be charged with the EH300 output.

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