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TonyM	#1 09/10/2013 - 16h27 Top
Class : Apprenti Posts : 6 Registered on : 05/10/2013 Off line	So guys, I've finally decided to build my 3-point clip. The only thing I've managed to come by are 30° 840nm leds. Now, I want to use them in series, powered by an old USB cable that I found in a drawer. Question is : how many Ohms do I need? Here are the LED specs : Nominal Voltage : 1.3-1.5 V DC Nominal Current : 20mA ( 50mA max) Max Power : 100mW FOV : 30° WL : 840/850 nm Can you help me figuring out what resistence is needed for the build? I'd like if you could also explain a bit of theory behind it. Many thanks!
OverDriven	#2 12/10/2013 - 01h02 Top
Class : Apprenti Posts : 5 Registered on : 09/10/2013 Off line	Yea! Here is LED Assembly Wizard: http://www.free-track.net/english/hardware/calcled A little bit of theory is I=U/R For more: http://en.wikipedia.org/wiki/Ohm Edited by OverDriven on 12/10/2013 at 01h05.
TonyM	#3 12/10/2013 - 09h33 Top
Class : Apprenti Posts : 6 Registered on : 05/10/2013 Off line	I've already used that, and it gives me 68 ohm. What I want to know is if that is a correct value for my kind of setup.
OverDriven	#4 12/10/2013 - 11h43 Top
Class : Apprenti Posts : 5 Registered on : 09/10/2013 Off line	TonyM @ 12/10/2013 - 09h33 a dit: I've already used that, and it gives me 68 ohm. What I want to know is if that is a correct value for my kind of setup. Ok. Give me a kind of your led, i'll check a datasheet.
TonyM	#5 17/10/2013 - 23h19 Top
Class : Apprenti Posts : 6 Registered on : 05/10/2013 Off line	Hi, after having built it using a 15 ohm resistor as suggested, my resistor burnt up. I had no luck with Free track since now... Maybe the problem is that the resistor doesn't have enough ability to dissipate all that heat, I mean 1/8w might be too little, because after some calculation I got this : Total Power supplied by USB : 5V x 0.1A = 500mW (.5 W) Total power used by LEDs : 1.5V x 0.03A x 3 =135 mW Power needed to be dissipated : 500 - 135 = 365 mW Since resistor is placed before LEDs it uses the same current provided by USB, so R = (5V)x(5V)/(.365 W)~=68 Ohm. So I need a 68 Ohm resistor 1/2 W, right? Another solution is to use 2 x 176 Ohm 1/4W in parallel, to make them run cooler than a single one... Is all this right? Edited by TonyM on 18/10/2013 at 11h09.
Steph	#6 18/10/2013 - 19h31 Top
Class : Moderator Posts : 656 Registered on : 16/11/2007 Off line	Hi Tony, Your USB port can deliver up to 500mA. But your build only use up the current of one LED, (serial circuit). So with the specs of the LED in your first post I get: R = (5Vusb – 3 x Uled) / Iled R = 5Vusb - 4.5V / 0.04A R = 12.5 Ohm You do use 15 Ohm so: Iled = U / R Iled = 0.5V / 15 Ohm Iled = 0.033A Current flow in this circuit is 33mA. Now the power dissipated is: P = R x I² P = 15 Ohm x 0.033A² P = 0.016W (160mW) ...normaly no problem with a 1/4W resistor. The problem must be elsewhere. Are you sure about the specifications for your LEDs?
OverDriven	#7 18/10/2013 - 19h55 Top
Class : Apprenti Posts : 5 Registered on : 09/10/2013 Off line	TonyM! Please, if you know, give me a kind of your led, i'll check a datasheet and than we will talk and think P.S. Yes, Steph, I think about it too... Edited by OverDriven on 18/10/2013 at 19h57.
Galileo38	#8 21/11/2013 - 19h51 Top
Class : Apprenti Posts : 17 Registered on : 03/07/2013 Off line	be carefull when you are using ub from your mobo. You can destroy it if you accidentally make a short circuit.

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