B
Bob Monsen
- Jan 1, 1970
- 0
Dave.H said:According to the LED's datasheet (which is a 16,000 mcd red unit) the
power dissipation is 80 mW, forward current is 20 uA, Peak pulse
forward current is 150 mA. The LED draws about 20 mA at 2.0-2.3
volts.
His point is that at 6V, the resistor is dissipating more power (power is
voltage times current) than the LED.
There are special power supplies for driving big LEDs which use a technology
called "switch mode power supply". It uses a trick with inductors to
minimize this power loss in the resistor. However, for 20mA, it is probably
overkill. If you have a big lantern battery, the thing will last for days
pulling 20mA. The incandescent light bulb is probably pulling 5x or more of
that current. If you used a switch mode power supply, it might double the
time before the battery fails to 10x the endurance of the lamp with an
incandescent bulb. However, it is somewhat complicated, and will require
special chips or a purchased power supply to do it.
Regards,
Bob Monsen