I am trying to power a high power LED from a 9V battery. I have 3 10k 1/2 W resistors in series to throttle the current. I most likely will run it at 200-300mA. Is there a better way to do this instead of using resistors? I am new and tinkering around, but could I use a n mosfet and voltage divider with the same 9v battery to throttle the current that way? How does would a non-newb approach this to not waste the battery with heat.
9V batteries are meant for low current. They have high internal resistance and low capacity. They will not power your LED for long at all.
It is much better to use either:
- A single 18650 LiIon cell (but you will have to purchase a suitable charger)
- Or several AAs in series. This is much safer for a beginner (no Lithium fire) and I'm pretty sure you got NiMH rechargeables in the house already, so it's cheap.
A constant current DC-DC converter will have much higher efficiency than a resistor. You can get those at places like dealextreme, aliexpress or ebay... but it would be simpler to just buy a flashlight.
The best voltage to use is one only a little higher (~1 volt) than that required to light the LED. If it is a white LED, about 3 Volts is fine. Other LEDs work with even lower voltages. Once you have a good voltage level, because a LED is a DIODE, you need a way to limit the current. A dropping resistor works fine, but really only works properly with the one voltage it is designed for. It is possible to construct a current source which will provides a fairly constant current for a wider range of voltages. A fairly good one can be constructed with two transistors and three resistors. See 'Currrent-Limiting' wikipedia topic. Remember, the one transistors will be dissipating the excess power from your voltage supply so with a lot of excess voltage, they can drop quite a bit of power and may need heat sinks.