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### Topic: LED Gurus? (Read 1 time)previous topic - next topic

#### brhyaa729

##### Nov 20, 2012, 11:40 pm
Hey guys - how is everyone doing? Was hoping I could get some direction in setting up an array of 940nm 8mm LEDs. I watched a few YouTube Videos and found a guy that used 850nms and gave the specs to wire them in series but I want to be sure I understand the specs on the LEDs I am purchasing. Here is are the specs of the units.

Wavelength : 940nm
Vf: 1.4~1.5V
Vr: 5V
If: 140mA
Ipusle: 700mA
Power Dissipation:200mW
Tjun: 125'C
Tsol: 5second @ 260'C
MCD=50@ 350mA, MCD=70@400mA
Angle: 30'

I also found an LED Calculator online here: http://led.linear1.org/led.wiz

Am I correct in assuming that diode forward voltage is 1.4 (or 1.5 which should I use?)V and diode forward current is 140ma? I would like to use 5 of these in series with a 9 volt battery and I am trying to determine which size resistor I would use to prevent destroying the LEDs. If anyone could tell me what the abbreviations on the specs mean I would appreciate it. Thanks guys!

#### Tom Carpenter

#1
##### Nov 20, 2012, 11:51 pm
Yup, Forward Current (aka If) = 140 [mA] = 0.14 [A]. Forward Voltage (aka Vf) is somewhere in the region of 1.4 [V] to 1.5 [V], so I would work on the theory that it is 1.45 [V] (the actual voltage will depend on the LED, temperature, current), but this approximate figure should be good enough.

If we do the math, 5 diodes in series x 1.45 [V] each = 7.25 [V] dropped across the diodes. For a nominal 9 [V] supply, this leaves 9-7.25=1.75 [V] across your resistor. V=IR, so R = V/I = 1.75/0.14 = 12.5 [Ohm]. You will need an minimum P=IV = 0.14*1.75 = 0.245 [W] rated resistor.

However... A 9V battery supplying 140mA will run flat very very very quickly. You would be lucky if you got half an hour out of it. The other thing is that you would be wasting a lot of energy in that resistor.
~Tom~

#### betomax

#2
##### Nov 20, 2012, 11:56 pm
i'm not a LED guru (in fact anything else at all !!), but the spec as in most datasheet are Maximum absolute, so you must use more conservative vals, in your link i choose a 10mA (as usual) to get started, and the result are:

Solution 0: 5 x 1 array uses 5 LEDs exactly
+9V   R = 150 ohms
The wizard says: In solution 0:
each 150 ohm resistor dissipates 15 mW
the wizard says the color code for 150 is brown green brown
the wizard thinks 1/4W resistors are fine for your application
together, all resistors dissipate 15 mW
together, the diodes dissipate 75 mW
total power dissipated by the array is 90 mW
the array draws current of 10 mA from the source.

if  (a.k.a. foward current) is directly linked with 'brightness' but keep it as low as you can to give long life to your LEDS !

#### brhyaa729

#3
##### Nov 21, 2012, 12:01 am
If I would only get 30 mins or so out of the 9 volt battery are there any longer life battery alternatives that would still be feasible to carry around attached to a small 3 inch x 2 inch project box?

#### brhyaa729

#4
##### Nov 21, 2012, 12:52 am
I think I have the answer to my own question, I could use two 9 volt batteries in parallel to double the life and remain at 9 volts without adding much bulk to the project.

#### retrolefty

#5
##### Nov 21, 2012, 01:12 am

I think I have the answer to my own question, I could use two 9 volt batteries in parallel to double the life and remain at 9 volts without adding much bulk to the project.

Have you priced what 9 volt batteries retail for? They are a very poor choice both from a capacity and cost point of view. Either convert to series alkaline AA batteries for use a DC power module. 9 volt batteries are for smoke alarms.

Lefty

#### Tom Carpenter

#6
##### Nov 21, 2012, 01:23 am
For maximum power density, you would need to start looking at rechargeable Li-Po batteries like those used for RC planes and helicopters.

As a gauge for how long they last, a battery rated for 1100mAH would supply 1100mA for an hour, or in your case 140mA for 1100mAH/140mA = 7.9Hours.
~Tom~

#### brhyaa729

#7
##### Nov 21, 2012, 01:52 am
I have several 9 volt rechargeables and a charger.

#### Simpson_Jr

#8
##### Nov 21, 2012, 02:36 am
Question, what are you going to use those high power IR-leds for ?

Continuous illumination will indeed require quite some power, you can use 'm, but you'll have to recharge those batteries a lot. In a "TV-remote/TVBgonekit on steroids"-project though, the leds will hardly be on and  a 9v battery could be a perfect choice.

#### brhyaa729

#9
##### Nov 21, 2012, 03:33 am
An IR Illuminator to connect to a Sony Handycam with nightvision. Probably won't ever be turned on for more than a few mins at a time.

#### brhyaa729

#10
##### Nov 21, 2012, 04:06 am
I also have a ton of AA Eneloops - plus chargers for those, was thinking a caddy with 8 of those would give me 9.6 volts. This would probably be better than the 9 volts correct?

#### Chagrin

#11
##### Nov 21, 2012, 05:19 am
Read the battery; it should state a "mah" (milli amp hour) rating.

For alkaline batteries, a 9V battery has around 550mah meaning it can supply 500 milliamps (.5 amps) of current for an hour (or 250 milliamps for 2 hours, etc.). An AA battery is around 2000mah.

#### brhyaa729

#12
##### Nov 21, 2012, 12:02 pm
So would this battery give me 28 hrs of run time?
http://www.amazon.com/Venom-Group-International-4000mAh-Universal/dp/B000W7WWFW/ref=sr_1_4?ie=UTF8&qid=1353495538&sr=8-4&keywords=li-po+battery+11.1v+and+charge

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