EDIT: Okay so I'm changing the LED I'm using. Now I'm using a 6 Watt LED. There's more info on the link, its a EcoSmart MR16 6-Watt (20W Equivalent) LED Flood Light Bulb (16 WW FL). I know I'm going to have to redo my circuit a bit. But can you help me now? Thanks.

I'm trying to power a 6 Watt LED using 3 NiCad batteries (3.6 volts). But the LED spec sheet says its a 12V LED. Should I use a booster? I don't really have much experience in electronics.

EDIT: (I deleted some of the old stuff)

I also designed a circuit: circuit

Now the problem seems to be that there's not enough voltage from the battery or enough current in the circuit to power the battery for a long period of time. I'm powering the circuit through this generator. The generator only provides 8.2 - 9 V and 17 mA. I'm not really experienced in this so can someone help me figure out if a better generator (such as this one) will help me? - btw when I tested how much the output from the motor was, it came to a max of 16V when I was cranking it really quickly.

Another problem with using 3 AA NiCad Batteries seems to be that they won't power the LED for a long enough period of time. Will using capacitors eradicate this problem? And if so which capacitors?

To clarify: The generator is hand cranked and through some testing its max output is about 16 V (cranking it really fast). This charges up the batteries and the batteries then power the 6W LED. I know I need diodes, resistors and capacitors in the circuit, but I've never really worked with this before so any help is much appreciated.

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    \$\begingroup\$ Can you link to the LED in question? \$\endgroup\$ – Oli Glaser Jul 23 '12 at 20:05
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    \$\begingroup\$ The more information you give in your question the better. A schematic is always welcomed. \$\endgroup\$ – m.Alin Jul 23 '12 at 20:09
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    \$\begingroup\$ Damn! An LED that requires 7W? Is this thing going in a lighthouse? \$\endgroup\$ – Polynomial Jul 23 '12 at 21:04
  • \$\begingroup\$ What made you decide the LED requires 20 V? You can't just pick any combination of volts and amps that comes out to 7 W. The LED unit will most likely want a specific voltage. You have to find out what that voltage is. \$\endgroup\$ – Olin Lathrop Jul 24 '12 at 21:24
  • \$\begingroup\$ The new 6W LED requires 12V not 20V- I was completely wrong about that. Thanks for pointing it out. \$\endgroup\$ – Om23 Jul 26 '12 at 15:37

Since your lamp is equipped with a standard connector for halogen lamps, I would suppose that it needs 12V. (It would be rather unusual for a lamp with this socket to have a higher voltage, and for a LED lamp double so).

So you would need a step-up converter, which can deliver about 0.6 ampere at 12V, when powered by 3.6V. As Olin already calculated, your batteries would then need to provide about 2A (even more since the converter doesn't operate at 100% efficiency).

Given that your hand crank can create 16V with 17mA (which is about 0.27W), this would mean that 24h working on the crank gives you 1h of lamp usage (assuming 100% efficiency - in reality this would be much worse).

Sorry - but with a hand crank delivering 0.3W electrical power in the best case won't power a 7W lamp for long.

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No bare LED requires 20 volts. Either you have a string of LEDs, you have something with a power supply built in, or you got confused in looking at the spec sheet. You need to find out what you really have before anyone can offer concrete advice. Give us a link to the spec sheet and we can see what you really have and demystify it for you.

However, you still need to consider the battery current consumption. No matter what voltage your LED assembly ultimately needs, you know it requires 7 W. With 3.6 V in, that means the battery will be drained at a rate of 7 W / 3.6 V = 1.9 A, and that's the bare minimum without any conversion losses. In reality it will be at least 2 A, probably a little more.

2A is certainly within range of many NiCd cells, including AA. However, you should consider how long they can keep that up. If these are AA, then 1.5 A-h capacity is probably not too far off. 1.5 A-h / 2 A = 45 minutes, which is the theoretical run time of fully charged cells. Reality is probably closer to 30 minutes, especially if you want the cells to have a reasonable lifetime.

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  • \$\begingroup\$ @Om23: That link is mostly a marketing blurb. It doesn't contain any electrical information other than it consumes 7 W. Shows us the real datasheet. \$\endgroup\$ – Olin Lathrop Jul 24 '12 at 15:53
  • \$\begingroup\$ That's all I have to work with. I'll try getting more specs but can't promise anything. \$\endgroup\$ – Om23 Jul 24 '12 at 16:04
  • \$\begingroup\$ So I was doing some recalculations on my circuit. Since I only have the 7 Watts to work with I added an additional battery to make my battery supply 4.8 Volts. I also changed my capacitor value to 600uF and my last resistor value to 10e-3 Ohms. With these changes the circuit lab says that there is 4.138 V and 1.315 A right before the LED giving it 5.44 Watts of power. Is my reasoning wrong since I want the LED to be powered for lets say 4 hours after cranking the lever for 10 minutes- for which the motor's output is about says 17mA and when I tested it, it said 16 Volts for the output... \$\endgroup\$ – Om23 Jul 24 '12 at 16:36
  • \$\begingroup\$ @Om23: We can't tell what is needed without electrical specs for the LED. \$\endgroup\$ – Olin Lathrop Jul 24 '12 at 17:00

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