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Fuse	#1 08/03/2013 - 18h25 Top
Class : Apprenti Posts : 5 Registered on : 08/03/2013 Off line	Sorry for what is probably going to be a pretty ignorant post, but I don't have any electronics background. I think what I'm asking is going to be pretty easy for people who have any experience with this. I'm looking to build a 3point clip IR system powered by an old phone charger. I have an old Xbox Live Vision camera that I can use with exposed film as a visible light filter, and would prefer not to have to remove the internal IR filter (but can if needed). I plan on using 3 SFH485P IR LEDs powered at 60 mA. It seems like wiring them in series is the way to go because it's simpler. I have two chargers to choose from, a 5v and a 3.7v. I'd prefer to use the 3.7v as it has the longer and lighter cord. I've tried using online calculators to figure out what resistor to use, but the output is a bit confusing. I think I need a 5.6 ohm resistor, or 1/4W? I think those are the same thing? How is my plan? Am I on the right track or have I horribly misunderstood something? Thanks for your time and any advice.
Steph	#2 08/03/2013 - 19h15 Top
Class : Moderator Posts : 656 Registered on : 16/11/2007 Off line	Hi, better take the 5V, it's easier to use in serial circuit with 10Ohm. Otherwise you need to go for a paralell circuit 3 x 35Ohm (for 60mA) or 3 x 27Ohm (for 80mA). 1/4W resistors are enough. Edited by Steph on 08/03/2013 at 19h17.
Fuse	#3 08/03/2013 - 22h56 Top
Class : Apprenti Posts : 5 Registered on : 08/03/2013 Off line	Thank you for your reply. Do I just need a single 1/4W resistor if using the 5v charger in a serial circuit? Or do I need 2? The Freetrack wizard says 1/2W when I use the 5v supply. Sorry. I'm sure that's a stupid question. Like I said, I have no electronics experience. Edited by Fuse on 08/03/2013 at 23h32.
Steph	#4 09/03/2013 - 12h43 Top
Class : Moderator Posts : 656 Registered on : 16/11/2007 Off line	Hi, Sorry. I'm sure that's a stupid question. There are no stupid questions, just stupid answers. (So I hope I will be good enough... ) There is confusion about resistance and dissipation. Resistors are rated according to their maximum power dissipation. Most discrete resistors in solid-state electronic systems absorb much less than a watt of electrical power and require no attention to their power rating. Such resistors in their discrete form, including most of the packages detailed below, are typically rated as 1/10, 1/8, or 1/4 watt. The power P dissipated by a resistor is calculated as: P = R x I² So for your serial 5V circuit with one 6.8 Ohm resistor you got: R = U / I R = (5Vpower supply - 3 x 1.5Vled) / 0.08A ( I go for 80mA with the sfh485) R = 0.5V / 0.08 = 6.25 ...next resistor to 6.25Ohm in E-Serie is 6.8Ohm. so I = U / R I = 0.5V / 6.8Ohm I = 0.074A Remember: P = R x I² P = 6.8Ohm x 0.074A P = 0.037W 1/4W = 0.250W So to answer your question with less than two lines: No, you just need one 1/4W 6.8Ohm resistor. But just another question: Do you have the reference of your power supplies, (both, 3.7V and 5V)? It's important to know, if they are regulated or not.
Fuse	#5 09/03/2013 - 14h46 Top
Class : Apprenti Posts : 5 Registered on : 08/03/2013 Off line	Thanks for providing both the detailed answer and "for dummies" version. I've been having some bad migraines and the meds have had my brain pretty scrambled. I admit I was trying to be lazy and let someone else do my work for me. My head is clearer today and I'm starting to understand some of this stuff. What do you mean by "reference" of the power supply? I assume you mean the actual manufacturer's part number, instead of the one Nokia or whoever slaps on after production? I don't think I see anything that looks like that. I do have an old voltmeter lying around. Can I use that to see if the output is being regulated? Once I confirm the output of the power supply I think I've got a handle on what I need to do. EDIT: Ok, so the leads on my multimeter are too big to test the 5V or 3.7V power supply, I'll need to find some wire to bridge it before I can test those. Will a paperclip or something work to bridge the connection without the resistance of the metal messing up the reading? Edited by Fuse on 09/03/2013 at 17h42.
Steph	#6 09/03/2013 - 17h49 Top
Class : Moderator Posts : 656 Registered on : 16/11/2007 Off line	Newer power supplies (USB-charger, mobile phone charger) are mostly regulated switch mode power supplies with voltage that varies little with load. Older ones had large transformers, (therefore they are easily identifiable by their weight.) They are unregulated with high voltage drop on load.( 10-30%) I do have an old voltmeter lying around. Can I use that to see if the output is being regulated? Yes you (we) can. (Heard that already anywhere... ) Just measure the output without any load. Above 10% higher than the given voltage it's very probably not regulated. What do you mean by "reference" of the power supply? I assume you mean the actual manufacturer's part number, instead of the one Nokia or whoever slaps on after production? I don't think I see anything that looks like that. I mean the model name of it (Nokia AC...) so that I can identify it in the net. So that I can tell you what type it is. I can give you both circuit (3.7V and 5V) later. Will a paperclip or something work to bridge the connection without the resistance of the metal messing up the reading? It's ok. Edited by Steph on 09/03/2013 at 17h50.
Fuse	#7 09/03/2013 - 22h38 Top
Class : Apprenti Posts : 5 Registered on : 08/03/2013 Off line	I've actually found a couple more power supplies I can use, I'll go ahead and list them all. Sorry if I'm throwing too much at you, just let me know if you're sick of me. haha Also, now that I realize I'm going to be buying resistors in a pack I'm really not committed to series. I guess let me know if there's a particular reason I should use one of these power supplies over the other, either in a series or parallel circuit. All are AC to DC wall adapters. First bit is the name; second is what it is labeled as outputting; third is hopefully the model number, fourth is the voltage reading using the multimeter. "Hon-Kwang ITE Power Supply"; 12V 800mA; Model HK-C110-A12; Reads 11.9V "Nokia Direct Plug-In Transformer Unit"; 3.7V 340mA; Type ACP-7U, Model 15.1312; Reads 8.8V "Creative Switching Adapter"; 5V 1.5A; Model TESA3-051500; I am not able to get a steady reading from the voltmeter. It may be because the plug (on the output end) is damaged, but I'm not sure. It looks like I should still be making good contact. I can strip the end off if needed, but would rather not if you believe I should use one of the other available power supplies. (It's a back-up for a back-up charger.) "Now & Zen CAT No. 35-4.5-350"; 4.5V 350mA; Model EIA172-H (I'm not sure if that's actually the model. The label on this one is clearly not from the original manufacturer.); Reads 7V Should I use the 12V? I'd need a 100Ohm resistor if my math is right (agrees with the wizard), but it's appears to be regulated which I assume is always preferred. Thanks again for helping me out. Edited by Fuse on 09/03/2013 at 22h40.
Steph	#8 10/03/2013 - 00h07 Top
Class : Moderator Posts : 656 Registered on : 16/11/2007 Off line	Creative Switching Adapter This one looks good for me. Also this will be the easiest way to go as I always give you the right calculation above for a serial circuit. You can go further with this topic. You do not need the PTC-fuse as you do not use your PC USB. If you do not have a 6.8 Ohm resistor you can use a 10Ohm either.
Fuse	#9 10/03/2013 - 05h01 Top
Class : Apprenti Posts : 5 Registered on : 08/03/2013 Off line	Cool. Thanks again for helping me out. Tomorrow I'll strip the plug off the Creative adapter and double-check its output, then order the LEDs and probably resistors (Not sure there's an electronics store near by anymore). I'll post back in a week or two when I have all the parts and let you know how it turns out.

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