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Led/Switch/Resistor Circuit Query ;)

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November 6, 2012 8:09:49 AM

Hey Everyone,

I'm doing a build atm and have decided to do my own wiring for some of the leds/cathodes/fans in the build. I only want to do simple on/off switches..

For the Led's I have these:
http://www.ebay.com.au/itm/180668636315?ssPageName=STRK...


For the Switch I intend on purchasing something like this :
http://www.ebay.com.au/itm/1pcs-B203-Panel-Mount-Wiring...


And my question is, what size resistor will i need for the setup I explained below?
(I already have about 15 of these, and if i can somehow use these instead of buying more then thats good!)



My plan is to intercept the 12v (lights) and 5v (fans) wires with a switch on both 5v and 12v lines.Led's will be wired in parallel with the line, because obviously they need their own resistor, and i dont want it intefearing with the main lines.. For my led's I need to know the resistance i will need for each (5v and 12v)?? If i can match the resistance i need with the resistors i have even if its some weird setup then i shall do that :) 

i know its sorta basic in the electricity field but im just not sure..

Heres the layout of a molex connector if you need it:
http://pinouts.ru/Power/BigPower_pinout.shtml

Heres a schematic of what im trying to explain:





Any answers or tips to this plan would be greatly appreciated!!
btw, i plan on placing the switches through the front bay covers

Thanks

Best solution

November 6, 2012 8:56:02 AM

your LEd's have ~ 3.4V voltage drop

With a 330ohms resistor you will have (using Ohms Law V=IR, I = V/R)

V = 5v-3.4V
= 1.6v

Current through the LED (I) is then

I = 1.6v / 389 Ohms

= 0.004A

or 4mA

this will light your LED's, which have a continuous forward current of 20mA, just not that brightly. To run at 20mA you would need a lower resistance, here

R = V/I

V is known - 1.6V, I is known 20mA, 0.02A, therefore R is

R = 1.6V / 0.02Ohms

R = 80 Ohms

this assumes you are using the 5V line



Using the 12V line

V = 12V-3.4V
V = 8.6V


R = 8.6V / 0.02Ohms
R = 430 Ohms


You could always run a few of the resistors in parallel. Two 389 Ohms resistors in parallel will give a resistance of ~ 195 Ohms, on the 5V line this would give a current of

I = 1.6V / 195 Ohms

I = 0.004A

or 8 mA

four in parallel would give a resistance of ~ 98 Ohms and a current of 16mA

That should light your LED's quite brightly, personally I would just get some more resistors - they are quite inexpensive.
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November 6, 2012 9:21:37 AM

Best answer selected by prodigydoo.
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November 6, 2012 9:26:15 AM

stuart72 said:

your LEd's have ~ 3.4V voltage drop


Thank You so much Stuart, does the voltage drop on the leds mean the voltage to the rest of the circuit will be effected? or because its in parallel does this mean it wont be?
I'm asking this because i dont want it to slow my fans or dim my internal lighting..

I knew resistors were cheap but after a quick search on ebay returning 100 430ohm resistors for 1 dollar, i think i will invest in some ;) 


Also, will operating the led's at their max current be a bad thing? like will they get really hot or eventually burn out?


Thanks again Stuart, the depth in your answer will really help me out in the future aswell :D 
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November 6, 2012 12:17:14 PM

"does the voltage drop on the leds mean the voltage to the rest of the circuit will be effected? or because its in parallel does this mean it wont be? "

Skipping-over-the-physics, the answer here is No

and from meory, your LED data sheet reckoned the safe peak current was 75mA, running at 20mA will be fine, they shouldn't get that hot, wattge per LED at that current will be ~ 0.07 Watts
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