infrared led underpowered

[crazyking]

Hi, i'm having a little prob with making a homemade infrared camera. It really seemed much easier the way they shown it on youtube. 1.) I'd like the "IDIOTS GUIDE TO CHOOSING A CAPACITOR" 2.) I'd like help with my specific application. I've chosen 10 high-output infrared leds from radioshack, ( output = 16mw min , forward volt.=1.2v , forward curr.=100ma , wavelength= 940nm) 2 10ohm resistors & 2 9v batteries. I've arranged the leds in sequence, in 2 rows of 5 LEDs, with 1 10ohm resistor per row. I'm obliously doing something VERY wrong here because i can't get a flicker outta these things. I wan't at least get 30min of full intensity out of these things ( i'm sure this may be impossible but, PLEASE humor me).

 

[audioguru]

Most Instructables on YouTube are made by guys who know nothing about electronics.
Thanks for posting the link to the one you found.
1) Five 1.2V IR LEDs in series need 5V.
2) If your two 9V batteries are in series then they make 18V when they are brand new.
3) The 10 ohms resistor sets a current of (18V - 5V)/10 ohms= 1.3A which will quickly burn out the LEDs.
4) Two rows of LEDs results in a battery current of 2.6A so the batteries will last for about 1 minute or until the LEDs burn out.

 

[crazyking]

Most Instructables on YouTube are made by guys who know nothing about electronics.
Thanks for posting the link to the one you found.
1) Five 1.2V IR LEDs in series need 5V.
2) If your two 9V batteries are in series then they make 18V when they are brand new.
3) The 10 ohms resistor sets a current of (18V - 5V)/10 ohms= 1.3A which will quickly burn out the LEDs.
4) Two rows of LEDs results in a battery current of 2.6A so the batteries will last for about 1 minute or until the LEDs burn out.

I'm sorry, i can't seem to find the link again myself
When you said the LEDS would "burn out",did you mean run out of power?
Again, sorry for the stupid questions. I'm a novice aspiring to learn more.
What modifications would you make?

 

[blueroomelectronics]

No they'll overheat and eventually produce no output.

 

[crazyking]

No they'll overheat and eventually produce no output.

Obviously, i've put some money into this project. Is there any way to make this thing work or should i scrap it?

 

[arhi]

can you draw schematic of how did you connect led's and batteries so we do not have to check out the crystal ball ?

 

[crazyking]

can you draw schematic of how did you connect led's and batteries so we do not have to check out the crystal ball ?

arhi i'm sure you're smart but, there's no reason to get smart.
You could have just asked. I thought i explained how i connected them.


_____Ω
/.........................\
+..........................-
\........................./
_____Ω



l---(+9v- +9v-)---l


The switch is obviously connected (just not shown).
EXCLUDE THE PERIODS.

 

[arhi]

arhi i'm sure you're smart but, there's no reason to get smart.

That ment to be funny not to provoke you... in my business I have to ask client to do a copy paste of the shell output as in 99% cases they say they did one thing and they actually did another.

so you made a block of 5 leds + resistor in series and connected that block to 9V battery.

Vf = 1.2V
If = 100mA

5*1.2 = 11V so 5LED's need more V to operate with full intensity.

You can connect your 2 batteries in serial to get 18V. Use the 2 block in parallel (as in series you will need 22V meaning - more then you have).

I trough every block is (V-Vf)/R = 18-11/10 = 7/10 = 0.7A .. this is 7x more then those LED's are rated for. As If is rated 100mA (0.1A)
0.1A = 7/R -> R = 7/0.1 = 70Ω

so, connect the leds like in attached pic

hope this helps

 

[arhi]

Note that this will take some 200mA from your batteries. If we assume you are using some good duracell 9V batteries (Ah at 0.100A is ~0.3) we can calculate that you can theoretically get 1.5 hours of light. The theory and practice does not always agree as 9V batteries will drop below 9V much sooner so LED will be lit much longer but will start to loose brightness much sooner.

EDIT: source of the attachment: Discharge tests and capacity measurement of 9 volt transistor radio batteries

 

[blueroomelectronics]

Many small IR LEDs are rated for 100mA but that may be a pulsed rating not continuous. You have to check the datasheet.

 

[crazyking]

That ment to be funny not to provoke you... in my business I have to ask client to do a copy paste of the shell output as in 99% cases they say they did one thing and they actually did another.

so you made a block of 5 leds + resistor in series and connected that block to 9V battery.

Vf = 1.2V
If = 100mA

5*1.2 = 11V so 5LED's need more V to operate with full intensity.

You can connect your 2 batteries in serial to get 18V. Use the 2 block in parallel (as in series you will need 22V meaning - more then you have).

I trough every block is (V-Vf)/R = 18-11/10 = 7/10 = 0.7A .. this is 7x more then those LED's are rated for. As If is rated 100mA (0.1A)
0.1A = 7/R -> R = 7/0.1 = 70Ω

so, connect the leds like in attached pic

hope this helps

You've clearly proven i haven't been to school in a long time

1.)The picture that you uploaded seems similar to what i have, is there any difference that i'm not seeing here?

2.)The battery requirements are 2 or 3 9v?

3.)The 70ohm you figured is the resistor i should go with?

4.)This diagram doesn't seem to have a capacitor, does my project need one?

 

[arhi]

As Bill told ya, re-check the 100mA forward current rating as I personally never saw the IRED that can work continuously with 100mA, only pulsed.

1. I'm not the ascii art expert so dunno .. I saw your ascii art differently then what I have drawn and I understood your explanation differently from what I have drawn.

2. if you connect them the way I have drawn, with 2 75R resistors, 18V (2 batteries) will do the job (will provide 100mA per chain). In order to find out how long that will last - you have to try. In order to double the time add another 2 batteries in parallel to these two, or use 2 batteries to power one chain and another 2 to power second chain.

3. 70R is the "ideal" size (you need one per chain) if you power the chain with 18V (70R does not exist iirc, you need to use "first bigger" that is 75R). And I did not "fiured" it, I just calculated it ... the formula is simple I = (V-Vf)/R where Vf is sum of all Vf's of the led's you have in chain.

4. no need for it, you are powering the LED's from batteries, they are "clean" power source so no need to "smooth" it

NOTE that if you misread the datasheet and these IRED's can only be pulsed at 100mA (and max constant current is ~30mA) they will burn out after few sec's (maybe even faster).

btw, how are you making camera with led's ?? do you just want to make IR light source so your ordinary camera can "see better" or ?

 

[blueroomelectronics]

I guess he pulled the IR filter out. I heard they can be used on the beach as X-Ray cameras

 

[audioguru]

The calculation of 5*1.2V=11V is wrong. It is only 6V. Before I was wrong and said it was 5V.
With an 18V battery and a 70 ohm current-limiting resistor the current is 171mA.
The LEDs will quickly burn out.

I think the new lenses in my eyes will perform on the beach like the IR camera.
Hee, hee. Look what I see!

 

[arhi]

I would really like to know how the hack did I multiply 5*1.2 and got 11 ?!?!?!?!?! REALLY !!

the formula stands, only, someone with basic multiplication skills needs to implement it

18V batt, 6V Vf, (18-6)/0.1 = 120R

 

[audioguru]

I would really like to know how the hack did I multiply 5*1.2 and got 11 ?!?!?!?!?! REALLY !!

the formula stands, only, someone with basic multiplication skills needs to implement it

18V batt, 6V Vf, (18-6)/0.1 = 120R

The original Instructable had the moron designer using only 10 ohms (1.2A). Two strings were straining the battery at 2.4A but the LEDs would almost instantly burn out.

 

[arhi]

I saw a circuit (CNR) that uses 9V zinc battery and 5 LED's (3red, 3green, 2yellow IIRC) and NO RESISTOR ... and, it was working for days .. they the battery died and the kid changed the battery (with duracell alkaline ) and LED's died in a sec, one actually "exploded", 3 just died and 1 still works (that was in some plastic kid's gun, press button and the led's turn on)

so maybe the original moro^H^H^H^Hdesigner tought that 9V batt cannot output more then 100mA ... or he calculated 5*1.2 = 17

 

[audioguru]

The original designer who knows nothing about electronics used two dead 9V batteries.

 

[crazyking]

Many small IR LEDs are rated for 100mA but that may be a pulsed rating not continuous. You have to check the datasheet.

I would really like to know how the hack did I multiply 5*1.2 and got 11 ?!?!?!?!?! REALLY !!

the formula stands, only, someone with basic multiplication skills needs to implement it

18V batt, 6V Vf, (18-6)/0.1 = 120R

I guess he pulled the IR filter out. I heard they can be used on the beach as X-Ray cameras

I've checked the information provided by RADIOSHACK and i cannot find any information indicating that it needs to be pulsed.
1.) If it does need to be pulsed what would i need to do to get it to work?
2.) 2-120R resistors will sustain the bat/LEDs how long?
3.) I'm actually trying to make my own IR lens and see how that works
(that CNN vid on youtube about x-ray cameras is pretty cool ).

 

[audioguru]

RadioCrap does not know anything about the rejected parts that they sell at extremely high prices.
You need to know the manufacturer's name so you can look at the detailed datasheet. RadioCrap doesn't even know who made the IR LEDs.

Most IR LEDs have a max allowed continuous current of about 30mA. The peak allowed current (for very short duration pulses) is from 100mA to 1A depending on which company made them.
Your LEDs might burn out if they use an 18V battery and a 120 ohm current-limiting resistor that limits the current to 100mA continuously or until the battery voltage runs down.