Wiring up your LED's:
(Data taken from Kingbright web site, http://www.us.kingbright.com/. Specific data sheet in this case is for the W2523SURC 5mm LED - http://www.us.kingbright.com/images/catalog/SPEC/W2523SURC.pdf)
Volt drop across LED (from data sheet): 1.9V at 20 mA. This is the Vf spec, or forward voltage. Notice they specify a maximum rating of 2.5V - this should not be used in normal operation. The current rating (20mA) is specified because the voltage drop will vary with the current.
20mA is the current generally used for your bog-standard LEDs. Some superbright LEDS demand higher currents - I've seen some that are heatsinked and can take 1A - you can't look at the damn things they are so bright.
Anyway, the forward voltage drop determines your minimum battery spec. In your case you will be driving 2 LEDs, so you will need a minimum of 2 x 1.9 V, or 3.8V. Note this is only if you are using this LED. You can get other types with a lower forward voltage drop. To be assured of the correct voltage drop, jack up the voltage a bit to take device variance into account. In your case you say you are going to use a 9V battery - that's fine - your major concern here is the current through the devices.
OK, so now you have two LEDs in series - you want 20mA through them both (which will automatically give you a forward volt drop on each of them of 1.9V, no matter what the supply voltage, as long as it is above 3.8V)
Use Ohm's law: R = V/I.
I = 0.020 A
=> R = 9/0.02 for a 9V battery = 450 ohms. Closest standard value is 470 ohms, which will give a slightly reduced current of 19mA (from I = V/R = 9/470 = 0.019).
Note that you can use any voltage you like, as long as you choose the correct resistor. For example, if you use 3 penlight (AA) batteries, which will give you 4.5V, you then have:
R = 4.5/0.02 = 225 ohms. You've dropped the voltage, so the resistance drops accordingly. In this case the closest standard value is 240 ohms. You could probably use 220 ohms as well, as the maximum current this device can handle is 30mA, but let's not go there for this explanation.
Just a quick note on wattage of the resistor: For your 9V battery, you have a resistor of 470 ohms. You will have 19mA running through it, so the power it will have to dissapate is I squared x R = 0.019^2 * 470 = 0.169 Watts, so choose a 1/4 Watt resistor.
Setup of the circuit:
(+) ------/\/\/\/\---LED----LED----- (-) Make sure your LED orientation is correct. Longer leg is generally the anode (+), shorter leg is the cathode (-). Check the data sheet!
Sorry if you found it all a bit boring
(Data taken from Kingbright web site, http://www.us.kingbright.com/. Specific data sheet in this case is for the W2523SURC 5mm LED - http://www.us.kingbright.com/images/catalog/SPEC/W2523SURC.pdf)
Volt drop across LED (from data sheet): 1.9V at 20 mA. This is the Vf spec, or forward voltage. Notice they specify a maximum rating of 2.5V - this should not be used in normal operation. The current rating (20mA) is specified because the voltage drop will vary with the current.
20mA is the current generally used for your bog-standard LEDs. Some superbright LEDS demand higher currents - I've seen some that are heatsinked and can take 1A - you can't look at the damn things they are so bright.
Anyway, the forward voltage drop determines your minimum battery spec. In your case you will be driving 2 LEDs, so you will need a minimum of 2 x 1.9 V, or 3.8V. Note this is only if you are using this LED. You can get other types with a lower forward voltage drop. To be assured of the correct voltage drop, jack up the voltage a bit to take device variance into account. In your case you say you are going to use a 9V battery - that's fine - your major concern here is the current through the devices.
OK, so now you have two LEDs in series - you want 20mA through them both (which will automatically give you a forward volt drop on each of them of 1.9V, no matter what the supply voltage, as long as it is above 3.8V)
Use Ohm's law: R = V/I.
I = 0.020 A
=> R = 9/0.02 for a 9V battery = 450 ohms. Closest standard value is 470 ohms, which will give a slightly reduced current of 19mA (from I = V/R = 9/470 = 0.019).
Note that you can use any voltage you like, as long as you choose the correct resistor. For example, if you use 3 penlight (AA) batteries, which will give you 4.5V, you then have:
R = 4.5/0.02 = 225 ohms. You've dropped the voltage, so the resistance drops accordingly. In this case the closest standard value is 240 ohms. You could probably use 220 ohms as well, as the maximum current this device can handle is 30mA, but let's not go there for this explanation.
Just a quick note on wattage of the resistor: For your 9V battery, you have a resistor of 470 ohms. You will have 19mA running through it, so the power it will have to dissapate is I squared x R = 0.019^2 * 470 = 0.169 Watts, so choose a 1/4 Watt resistor.
Setup of the circuit:
(+) ------/\/\/\/\---LED----LED----- (-) Make sure your LED orientation is correct. Longer leg is generally the anode (+), shorter leg is the cathode (-). Check the data sheet!
Sorry if you found it all a bit boring