I have a 1.2v, 100mA IR LED, powered by 5v of voltage and 40mA of current. What resistor should I use.
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Yes, get a couple of NPN transistors at Radio Shack similar to the 2N3904; you will also need a 68 ohm (1 watt) resistor, a 1K ohm (1/4 W) resistor and a 9 V battery. Hook it up as follows:
1) Connect the transistor emitter to ground. Be sure to look carefully at the EBC connection diagram.
2) Connect one end of the 1K resistor to the base of the transistor and connect the other end to the output of your micro-controller switch.
3) Connect the collector of the transistor to one end of the led. Connet the other end of the led to one end of the 68 ohm resistor. Connect the other end of the 68 ohm resistor to the + side of the 9V battery.
4) Connect the (-) side of the battery to ground.
Now, when the micro-controller output voltage goes positive to about 5 volts, it will turn on the NPN transistor. The NPN will "saturate" and current will flow through the transistor and the LED. If the voltage across the LED is about 1.2 volts, the current through the transistor and LED will be about 110 mA. The power dissipated in the resistor will be about 0.85 watts so the resistor must be able to handle this power (it will get very hot anyway).
The 9V battery will discharge fairly quickly (a few hours) if you leave the LED "on", so you may want to buy or scrounge up an AC adapter that plugs into the wall outlet an puts out 9 volts. Use that 9V source instead of the battery.
1) Connect the transistor emitter to ground. Be sure to look carefully at the EBC connection diagram.
2) Connect one end of the 1K resistor to the base of the transistor and connect the other end to the output of your micro-controller switch.
3) Connect the collector of the transistor to one end of the led. Connet the other end of the led to one end of the 68 ohm resistor. Connect the other end of the 68 ohm resistor to the + side of the 9V battery.
4) Connect the (-) side of the battery to ground.
Now, when the micro-controller output voltage goes positive to about 5 volts, it will turn on the NPN transistor. The NPN will "saturate" and current will flow through the transistor and the LED. If the voltage across the LED is about 1.2 volts, the current through the transistor and LED will be about 110 mA. The power dissipated in the resistor will be about 0.85 watts so the resistor must be able to handle this power (it will get very hot anyway).
The 9V battery will discharge fairly quickly (a few hours) if you leave the LED "on", so you may want to buy or scrounge up an AC adapter that plugs into the wall outlet an puts out 9 volts. Use that 9V source instead of the battery.
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How could you expect to light up a 100mA device with 40mA power source ?
How could your eye can see infra-red ?
Are you sure you IR LED rated 1.2V ? So far no LED can be lighted up under 1.5V .
How could your eye can see infra-red ?
Are you sure you IR LED rated 1.2V ? So far no LED can be lighted up under 1.5V .
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(5 - 1.2)v/(40/1000)A = 3.8*1000/40 ohms...approx 100 ohms <<<