Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 (edited) The reason you use a resistor with LEDs isn't to limit the voltage across the LED, it's to limit the current. An 'on' LED has VERY low resistance so you could land up with a short circuit and melted wires by going without a resistor. By using a resistor, you can use Ohms law to limit the current in the circuit. You don't need a resistor on EVERY LED, so long as the resistor you use is rated high enough to handle the full current. Edit: OK, let's expand on this and do a little maths for you. Fill in your own numbers depending on the spec of the LEDs you use. As far as I can see, you're using 12 LEDs in a 2S6P setup. So, total voltage drop across this circuit will be that of two LEDs. Let's assume your LEDs have a voltage drop of 2.5V. Total drop is 2.5V x 2 = 5V. Now, you need enough current for all LEDs in the chain. Say they're rated at 50mA each. You want to limit the current to 12 x 50mA = 600mA. Assuming you're using a 6V supply, the voltage across the resistor would be 6V - 5V = 1V. You've now got everything you need to work out what resistor to use with Ohms law. V = I.R, so R = V/I. R = 1 / 0.6 R = 1.66666 Ohms - so you use the resistor you can find that's closest to that value. Seems to be a 1.6 Ohm resistor. http://uk.rs-online.com/web/search/searchB...t&R=0150616 Now we check that it wouldn't be overloaded. P = IV. P = 0.6 x 1. P = 0.6W. Perfect for that above resistor. If anyone has any trouble working out what resistor to use, give me a poke and tell me the specs of your LEDs and I can do the sums for you. Edited May 19, 2009 by Aterlatus Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 an english translation maybe lol Quote Link to comment Share on other sites More sharing options...
Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 OK, in English... If you don't want to run the risk of melting wires at best, and total failure at worst then you should ALWAYS use a resistor with LEDs. Tell me about your setup and the specs on your LEDs and I can offer suggestions for resistor values. Quote Link to comment Share on other sites More sharing options...
Hiks Posted May 19, 2009 Share Posted May 19, 2009 I've made that mistake before when I was wiring LED's for a different application, the black smoke poured out. Was quite scary I really thought I shagged my new computer as it was my first time at watercooling a PC. Quote Link to comment Share on other sites More sharing options...
Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 Sounds like a classic lack of resistor scenario - short circuit, lots of current, plenty of melting... Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 New 5mm dia. Ultra Bright White LED 14000 mcd Size: 5mm Forward Voltage (V):3.2~3.4 Forward Current (mA):20 View Angle: About 25 degree Luminous Intensity: 14000mcd Lens Color:Water Clear Emitted Colour:White this is the only info supplied with them Quote Link to comment Share on other sites More sharing options...
adolf hamster Posted May 19, 2009 Share Posted May 19, 2009 looks a nice guide there i'm not too shabby at soldering but i think i'd not bother with that, is it possible for you to get the pods to be bright enough for some functional lighting? would be pretty neat running in the dark perhaps a xenon bulb in place of all the leds? Quote Link to comment Share on other sites More sharing options...
Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 Forward Voltage (V):3.2~3.4 Forward Current (mA):20 this is the only info supplied with them That's all you need If you're setting up the same as the OP (2 series, 6 parallel) then total voltage drop would be 6.4-6.8V - unless you're running off a 7.2V pack you need to rethink your strategy. The voltage drop across an LED varies quite a bit, but generally the colour makes the biggest difference. White ones have quite a significant drop, along with blue. Green/yellow are somewhere in between, with red often being less than 2V. If you're trying to run off 9V though then we can do some numbers for you: Voltage drop would be 6.8V for 2s6p. That leaves 2.2V across the resistor. Max current would be 20mA x 12, so 240mA. R = V/I so 2.2/.24 = 9.16 Ohm. Let's go with 9.1Ohm then, limiting to 241mA (close enough). P = I.V so P = 2.2 x 0.241 = 0.53W. Let's find a resistor that'll do the job: http://uk.rs-online.com/web/search/searchB...t&R=0150795 Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 I'll be ruuning mine off a 6v hump pack with them all wired the same as the guide (2 series, 6 parallel) Quote Link to comment Share on other sites More sharing options...
Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 Then let there be darkness.... you'd be better wiring them all in parallel if you expect to run off a 6V pack. You'll be open-circuit if you try to wire them the same as the guide. 6V < (2 x 3.4V) LEDs aren't like regular bulbs in that they'll glow dimly if they don't get the right voltage. They're diodes. They're either open or closed and unless they get enough voltage over them then they're open (off). Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 so you reckon i am better wiring them all pos to neg together in one loop then ? Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 phewf seems i have opened a can of worms? I am wiring them in pairs so that i do not need a resistor when running off 6v (2 x 3.2v LED's) i'm not sure what you mean when you say they will not work like this (Then let there be darkness) i have done two sets of light pods and they do work? Open circuit? your electronics knowledge is obviously more advanced than my basic fumblings but i have tried it and it works so they will run off 6V. I see what you mean about a short causing wiring to overheat though so what resistor should i be installing then? let me know and i will modify my design & the guide..i use 12 leds per pod, 20mAh. So what resistor do you reckon for 6v supply and what for a 9v supply? Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 Hmm that could get messy Its actually easier to join the led's for 9v than it is for 6v have a looky at the ones i have started for keenan if you can make it out from the pic Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 just counted up and my lighting plans involve 6 pods @ 12 led's per pod , think i need a 12v lead acid in there (yes my new theme is based on a blackpool tram) lmao Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 just counted up and my lighting plans involve 6 pods @ 12 led's per pod , think i need a 12v lead acid in there (yes my new theme is based on a blackpool tram) lmao You will need a 13amp socket and a lot of extension cable Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 may have to cut it down to 2 pods up front plus the 2 reds on the back end Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 (edited) Just been doing some more research and i still dont think that i need a resistor (according to reliable sources). I will keep looking but i think that a small fuse could be used (if required) but i personally think the LED's will blow before they could draw enough current to melt the wiring. ??? will check with my electronics genius buddy at work tomorrow but he has okayed my work already. I do understand that a short could draw a lot of current but what will cause the short in a semi sealed light pod? Edited May 19, 2009 by perruzo Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 i was thinking of running a thin line of silicone around the outside edge to help seal it from water Quote Link to comment Share on other sites More sharing options...
Matt B Posted May 19, 2009 Share Posted May 19, 2009 Well voltage is forced, and current is drawn, so I'd have thought it would be ok provided it meets the voltage criteria, as if theres too many volts then it will kill it for sure. Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 i was thinking of running a thin line of silicone around the outside edge to help seal it from water Bang on mate, i think that the pod internals could also be filled with silicon for even more protection.... Quote Link to comment Share on other sites More sharing options...
paul2304 Posted May 19, 2009 Share Posted May 19, 2009 true mate Quote Link to comment Share on other sites More sharing options...
Topper Posted May 19, 2009 Share Posted May 19, 2009 Its actually easier to join the led's for 9v than it is for 6v have a looky at the ones i have started for keenan if you can make it out from the pic Cant really make it out,is it a case of soldering 3 together instead of 2 ? Also,you could seal them with liquid electrical tape, http://www.google.co.uk/products?hl=en&...=1&ct=title Quote Link to comment Share on other sites More sharing options...
perruzo Posted May 19, 2009 Author Share Posted May 19, 2009 (edited) Cant really make it out,is it a case of soldering 3 together instead of 2 ? Also,you could seal them with liquid electrical tape, http://www.google.co.uk/products?hl=en&...=1&ct=title Yes thats good stuff but i think i'm gonna use some silicon sealant as its cheap! AND just noticed - 100 posts woohoo And yeah you just wire them in threes for 9v. i'm doing some longevity tests on a 9v battery with Keenans pods. Start at the top of the pod and work towards the bottom.... Your LED's were posted today first class mate.. Edited May 19, 2009 by perruzo Quote Link to comment Share on other sites More sharing options...
Topper Posted May 19, 2009 Share Posted May 19, 2009 Yes thats good stuff but i think i'm gonna use some silicon sealant as its cheap! AND just noticed - 100 posts woohoo And yeah you just wire them in threes for 9v. i'm doing some longevity tests on a 9v battery with Keenans pods. Start at the top of the pod and work towards the bottom.... Your LED's were posted today first class mate.. Ok cool, How about updating the guide with the 9v option,that would be good Thanks,looking forward to build these now Quote Link to comment Share on other sites More sharing options...
Aterlatus Posted May 19, 2009 Share Posted May 19, 2009 Well voltage is forced, and current is drawn, so I'd have thought it would be ok provided it meets the voltage criteria, as if theres too many volts then it will kill it for sure. Not quite - a diode (including LEDs) that's allowing current to pass (IE, a lit LED) has a very low resistance, pretty much negligible. Given that I = V/R (Ohms law to the rescue) since the total resistance of the circuit is very low, the total current becomes very high. If you manage to get LEDs running without a resistor and it doesn't melt something, you're damn lucky. It's not gonna be the same case for everyone or every LED. It costs pennies to put an appropriate resistor in there to limit current, so why anyone wouldn't I don't know. Quote Link to comment Share on other sites More sharing options...
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