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FreeTrack Forum > FreeTrack : English Forum > Support : Tracking System > Double check me before I buy please?
Bulldogg629 | #1 22/07/2009 - 00h45 |
Class : Apprenti Off line |
Ok I think Ive figured out how to use the calculator but Id like to make sure I got it right. Using these LEDs:
http://cgi.ebay.com/20-pcs-5mm-850nm-Infrared-IR-LED-Free-Resistor_W0QQitemZ150359765688QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item230223f2b8&_trksid=p4634.c0.m14.l1262&_trkparms=|293%3A1|294%3A30 They have a forward voltage of 1.6v and max current of 60mah. So if I run them at 50mah with a 2 AA battery pack, rechargeables at 1.2v each, 2.4 total...then the calculator says I need 3, 16 Ohms 5% resistors, correct? Calculating this for 1.2v rechargeables, does that mean I should never use 1.5v alkalines in it? What does the Dissipated Power of 0.04watts mean? And assuming I read the numbers right but cant find 16 ohm resistors, which direction can I round to and still be ok? down to 15 or up to 20 for example? There should be no backlight to cause problems but Id like to have them as bright as I can to avoid removing the IR filter, I plan on using the microsoft 1000 cam. Thanks for any help. |
Jorge.PT | #2 22/07/2009 - 13h30 |
Class : Apprenti Off line |
For a construction with the leds in parallel:
I=[Voltage of Power supply - Led foward voltage]/Resistance Value Ex: I = [3-1,6]/16 = 87,5 mA (2 1,5 V batteries would burn your leds...) so you need a bigger resistance if want to use 1.5v batteries. R = [3-1.6]/0.05 = 28 ohms ; You need a 28 ohms resistance to feed 50mA with 2 1.5v batteries. With the 1.2v batteries and 28ohms: I = [2.4-1.6]/28 = 28.6 mA Keep in mind that the non-chargeable batteries have +/- 1200 mA.h and rechargeable ones +/- 1800 mA.h 1200mA.h / 3*50mA = 8 hours 1800mA.h / 3*28.6 mA = 21 hours Batteries don't give a constant voltage, so the rechargeable batteries will stop giving you the required 1.6v before they reach the 1800mA.h limit. The best solution is to have a non battery power supply like a USB, but I suggest a fuse for motherboard protection. EDIT: Dissipated Power is the power that they lose by heat and radiation.
Edited by Jorge.PT on 22/07/2009 at 13h31.
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Bulldogg629 | #3 23/07/2009 - 00h05 |
Class : Apprenti Off line |
Im more lost than when I started Was 16ohm for rechargable batteries correct? Are you saying that if I went with 28ohms resistors I would be able alkalines AND rechargables or just alkalines?
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benmeijer | #4 23/07/2009 - 01h06 |
Class : Habitué Off line |
If you want to use rechargeable batteries (2.4 v):
use 16 ohm resistors. If you want to use alkaline batteries (3.0 v): use 28 ohm resistors. you have to make a decision what batteries you use! 16 ohm / 3.0 v ---> your leds are to bright and will be destroyed. 28 ohm / 2.4 v ---> your leds are not bright enough. |
Bulldogg629 | #5 23/07/2009 - 03h41 |
Class : Apprenti Off line |
ah I see, thank you both. Last question...how do commercial products get away with being compatible with both types of batteries? tv remotes, cameras, etc.
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benmeijer | #6 23/07/2009 - 23h28 |
Class : Habitué Off line |
Commercial products have a complex layout that will assure that all the components have the right voltage and current.
You can built a setup that will work with both types of battery, but is more complex. you need (3 * 16 ohm) and (1 * 12 ohm) resistors ---------------[IIII]----- 16 ohm \ \ ---------------[IIII]----- --------------------[IIII]--------- + 3.0 v 16 ohm / | 12 ohm / |------------------- + 2.4 v ---------------[IIII]----- 16 ohm if you use 2.4 v (rechargeables) bypass the 12 ohm resistor (using extra switch). |
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