Electronic – powering a 6 Watt LED

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.

Firstly, you need to go back to school to learn the basics. What you are trying has no hope of working.

An LED is a current driven device, and it has a fixed forward voltage drop. Quite frankly, you are lucky the LEDs are even working at all any more - you must have LEDs that have a forward voltage at (or around) the 3V the batteries are supplying.

You have to run the LEDs with some form of current limiting device. At the moment that current limiting device is the batteries themselves being in the non-linear portion of the LEDs IV curve, which is very fortunate for you, or they would have died.

So your LEDs have a forward voltage of (around) 3V. Putting two in series makes that forward voltage 6V, so you will need at least 6V to drive them in series. You also need to limit that current to the right value to run the LEDs (no idea what that should be for your specific LEDs, but 20mA is common for normal LEDs).

So running from 9V would be sensible in this situation. Assuming 20mA and 3V forward voltage, you will need a resistor, of:

$$ V_{LED}=9V - 6V = 3V $$ $$ R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{3}{0.02} = 150\Omega $$

^{simulate this circuit – Schematic created using CircuitLab}

You could also run from 4.5V (3xAA) and have as many LEDs in parallel as you wish. Each LED must have its own resistor in this case:

^{simulate this circuit}

The resistor of 75Ω is worked out in the same way:

$$ V_{LED}=4.5V - 3V = 1.5V $$ $$ R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{1.5}{0.02} = 75\Omega $$

As a side note, it is more common to put the switch in the + side of the circuit, not the - side. It doesn't really make any difference, it's just that's where others would expect it to be who don't know the circuit you have designed.