batteriesled
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.
You wanted a calculation. Here is the basic form of the calculation.
A typical red LED has a forward voltage drop of 1.8 V, and a maximum continuous current of around 20 mA.
Now what's our voltage? Lets say we want to use a 3 V source.
So we will have a voltage drop of 3.0 V - 1.8 V = 1.2 V over our resistor. The current through the resistor will be 20 mA, so our power is 1.2 V * 20 mA = 24 mW. That is not really a lot of power, although it is a significant fraction of power consumption of the LED. The LED itself uses 1.8V * 20mA = 36 mW.
If you power off 9V then consider using some of the 3W LEDs in series: -
Typically red leds need about 2.5V across them so wiring three in series will need at least 7.5V then there should be some current limiting circuit that might take 0.5V.
Typically for blue or green leds they need 3.4V across them so maybe only series together two LEDs.
You might just about get enough brightness with the NCP5623 chip this way but if not you'll need a current limit circuit and mosfets. The NTD80N02 looks OK but you need to limit current into the LEDs.
Best Answer
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 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.