Which IR-Led should i buy ?

[protocol217]

1. SFH 4715S IR OSLON Black 3.4 W , half angle ±45°, typ. 1030 mW & 440mW/sr at 1 A dc
http://www.osram-os.com/osram_os/en...500mw/emitters-with-850nm/sfh-4715s/index.jsp
datasheet: http://www.osram-os.com/Graphics/XPic8/00100755_0.pdf/SFH 4715S Lead (Pb) Free Product - RoHS Compliant.pdf

2. SFH 4783 Dragon Dome 2.1 W , half angle ±12°, typ. 470mW & 2300 mW/sr at 1 A
**broken link removed**
datasheet: http://www.osram-os.com/Graphics/XPic0/00119243_0.pdf/SFH 4783, Lead (Pb) Free Product - RoHS Compliant.pdf

which of this two leds is more powerful, brighter or can light up more - further
and explain to me whu is there two radiant intensitys (1030 mW & 440mW/sr) what does 1030 mean and what does 440 ?

 

[KeepItSimpleStupid]

Ok, you can look up steradian. mw/sr is an intensity. The area is a steradian.

The +-12 and +-45 is probably more important to you. This is the the "cone of emmission" so to speak. So, you can calculate how much circular area it will illuminate. +-12 deg will see further.

 

[protocol217]

i am planing on installing the diode in housing that has a magnifying glass and can be focus-able.
so does the +-12 and +-45 mater in my case or should i buy the stronger one ?

and i don't understand your explanation about radiant intensitys, my English is bad so i don't understand the word "steradian"

what led from these two will light up more when it is installed in an housing that has zooming- focus with one magnifying glass? like this one:

 

[KeepItSimpleStupid]

You changed the game when you add a lens.

Anyway, the sun's intensity at AM 1.5 G spectrum is 100 mW/sqcm, so 100 mW falls on every sqcm.

A Steradian is a curved surface. Use the pics, here: https://en.wikipedia.org/wiki/Steradian

My optics experience is weak, but in general, you want to fill the lens with a collimated beam, so all the light is used.

See: https://en.wikipedia.org/wiki/Lens_(optics)

So, now the distance of the LED to the lens and the size of the lens matters.

 

[protocol217]

my lens is 2 cm ( 0.787 inch ) away from diode when its in focus ( i see the square when locking throe night vision camera )
an in diameter my lens is also 2 cm
so what led of these two is better in my case ?

 

[KeepItSimpleStupid]

If I did this right, the optimal distance from the LED to the lens is about 1.5 cm: =COS(1/1.5)*360/(2*3.14159)

You would need a 50 degree 1/2 angle spread: COS(1/2)*360/(2*3.14159)

The numbers:
360/2*PI = the conversion from radians to degrees.

You have a right triangle that is 1/2 the lens diameter and it;s the opposite side.
The 1 in 1/2 is 1/2 of 2 cm.
The distance of 2 is the adjacent side.
Opposite/adjacent is COS.

The *2 is twice the length of the right triangle.

Note: big time edit - half angle removed the *2

Also see this https://www.electro-tech-online.com/threads/led-viewing-angle.142316/#post-1194079 post.

 

[protocol217]

i am sorry i don't understand any of your calculations, am i asking you which one of these two leds would you use with a lens like mine ?

 

[KeepItSimpleStupid]

The 45 degree one, moved to 1.5 cm from the lens if possible.

 

[protocol217]

then i will buy the 45 degree one. Thank you very much.

 

[protocol217]

I have a driver for that led (820 ma) with LM317, but it gets very hot an i cant fit it in my housing, so i want to buy one of those 17mm in diameter driver Bord like this one:
**broken link removed**

but that one is 1000ma which is max for my led (and i don't want to fry my led) , can someone find me a suitable 17mm driver for my led ( SFH 4715S IR OSLON Black 3.4 W)
Forward Current: 1 A
Forward Voltage: 2.9 V
i think the ideal current would be 900ma am i right?

so here i found a driver and a 5w led, would this driver work for my LED ?
**broken link removed**
in the specs below it seas Working Current 900mA is that for the driver or for the led that is there?

 

[KeepItSimpleStupid]

They are both cheap enough, but you can definately play with the first one: See: https://www.google.com/url?sa=t&rct...=2neXIHbeMUE11y2f9rWXMA&bvm=bv.76247554,d.cWc

and set the current wherever you like. It might take a little reverse engineering.

1A seems too high, so I would agree with your 900 mA assessment.

If you wanted to do some testing, just add a few diodes in series (0.6V drop each) and a resistor. Measure the voltage across the resistor under varying input voltages and see if it stays constant. Remember to check the power with the expected drop or the full supply voltage.

 

[protocol217]

my soldering skills are not that good to solder small components like those on the board, so i need a original 17mm driver.
And what do you think about the second driver, that one would work right?
it is 20mm but i think i can get it to be 17mm.
here is the second driver again with little better specs:
**broken link removed**

 

[KeepItSimpleStupid]

The other driver should work.

Buy two or three of the one you want and don't want. They are small and cheap. You might only have to change one component. There is also a good chance that the same IC is used in both drivers.

 

[protocol217]

i found one more: https://www.dx.com/p/3-6v-9v-800ma-...affiliate&utm_medium=dealextreme#.VCm8WVeLVhU

i think that one is the best for my led am i right ?
and does this driver give 925ma with one Li-ion 3.7 battery ? or do i need two batterys ?

 

[KeepItSimpleStupid]

Vf = 2.9 (not sure if that is min or max) Vf needs to be 0.5 to 0.9 V higher, so 3.4-3.8

They say and I say: (not suitable for 1 white LED from 1 Li-ion cell)

You also have to look at the usable discharge curve for the Li-Ion battery as well.

 

[protocol217]

and with two 3.7v battery's, this ones: **broken link removed**
they are half of the size of a normal li-ion cell. it should work right?

 

[KeepItSimpleStupid]

Bear in mind that you'll get a little less than an hour of use out of the light. ie. the battery theoretically can deliver 850 mA for an hour. There is also an efficiency issue.

 

[protocol217]

a half of hour is good for me, i will order this driver: https://www.dx.com/p/3-6v-9v-800ma-...affiliate&utm_medium=dealextreme#.VCm8WVeLVhU