Hello!
I am attempting to make night vision goggles, but I am confused about the IR LED component
I've been reading many posts on this site for a few weeks and this is what ive gathered (Please correct me if it's wrong):
At first I tried putting my friends 9 V battery to my 6 IR LEDs all in parallel with no resistors, and they lit up perfectly through my camera and illuminated well (across my room), but then later i put a new 9 V to a single LED and it fried. This doesnt make sense to me because in the parallel circuit wasnt 9 volts being applied to each LED? why didnt they all break? -- I came to the conclusion that my friends battery must have been pretty old making the voltage much less?
After this happened i researched and now Im using a 9V battery (typical one from wal mart), and will connect 5 LEDs in series with a resistor, then repeat but attach it in parrallel so i have a total of 10 LEDs. It sounds like the common Vf for IR LEDs is about 1.5 V, so I assumed that my IR LEDs (which i took from remote controls) have a Vf of 1.5 V. Also, SgtWookie has said a number of times to be safe apply a max current of 25 mA, but i need these LEDs to be as bright as possible so is 100 mA okay? My understanding is that with a Vf of 1.5 V and 5 LEDs with 100 mA would give a resistor of 15 ohms:
R = E/I
R = [9 V - (1.5x5)]/0.1 A
R= 15 Ω
So if i have 9 V going to 5 remote control IR LEDs in series with a 15 Ω resistor, will I have bright IR LEDs and no malfunctions? also, will this make it so that there is 1.8 V (9/5) and 100 mA going to each LED ?
A lot of that is probably wrong, so the important question is: how can I make these IR LED's as bright as possible?
Thanks for your time!
I am attempting to make night vision goggles, but I am confused about the IR LED component
I've been reading many posts on this site for a few weeks and this is what ive gathered (Please correct me if it's wrong):
At first I tried putting my friends 9 V battery to my 6 IR LEDs all in parallel with no resistors, and they lit up perfectly through my camera and illuminated well (across my room), but then later i put a new 9 V to a single LED and it fried. This doesnt make sense to me because in the parallel circuit wasnt 9 volts being applied to each LED? why didnt they all break? -- I came to the conclusion that my friends battery must have been pretty old making the voltage much less?
After this happened i researched and now Im using a 9V battery (typical one from wal mart), and will connect 5 LEDs in series with a resistor, then repeat but attach it in parrallel so i have a total of 10 LEDs. It sounds like the common Vf for IR LEDs is about 1.5 V, so I assumed that my IR LEDs (which i took from remote controls) have a Vf of 1.5 V. Also, SgtWookie has said a number of times to be safe apply a max current of 25 mA, but i need these LEDs to be as bright as possible so is 100 mA okay? My understanding is that with a Vf of 1.5 V and 5 LEDs with 100 mA would give a resistor of 15 ohms:
R = E/I
R = [9 V - (1.5x5)]/0.1 A
R= 15 Ω
So if i have 9 V going to 5 remote control IR LEDs in series with a 15 Ω resistor, will I have bright IR LEDs and no malfunctions? also, will this make it so that there is 1.8 V (9/5) and 100 mA going to each LED ?
A lot of that is probably wrong, so the important question is: how can I make these IR LED's as bright as possible?
Thanks for your time!