Electronics Hot Resistors

i built a LED array for my keyboard to make it glow, it all works fine, but the resistor i am using (27 Ohms, 2 watt) seems to get overly hot, as in within 3 seconds of plugging the ciruit up, it gets so hot that i cant even touch it. Is this ok, or does something need to be changed ?

 

your led's i take it they are wired to the keyboard power supply. the power disapted by the resistor should be about .9W which is with in's range. Resistors do get hot, so just make sure its not near to anything that can be easily damaged.
Personaly i would put a higher wattage resistor in.

what are your LED wired into?

 

no, its connected to my pc 12v rail

the resistor is a 2 watt resistor



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12v    Gnd

each led is 1.8v and 20mA

so 3x 1.8 =5.4
20x13 =260

12-5.4 = 6.6

R = v / i
r = 6.6 / 0.26
r = 25 Ohms


5.4 x .26 = watts
1.404 = watts


resistor i got, 27 ohms, 2 watt

so why is this resistor getting so hot?, and should i put a ramsink on it to cool it down

 

1.5 watts on a small object = hot.

You can knock the temperature down by using say four 110R (or 120R) in parallel, or lower the led current to 10-15mA, or use 6 strings of 4 leds, leaving less voltage to lose.

I'd be tempted to split your array into several sections, each with it's own dropper resistor - avoids "current-stealing" by any of the strings.

<bi-centeniall post

 

Congratulations

 

Actually thats the power used by the LED array (1.8v * 3=5.4v), the voltage dropped by the resistor is 6.6v, so the power used by the resistor is 6.6 * 0.26=1.716W. This is getting a bit too close to the power rating of the resistor.

 

what if i put a heatsink on the resistor using a little bit of Arctic silver epoxy? cuse i got a small heatsink (its actually a ramsink) and i got the epoxy already from previous projects

would this do?

 

dont know it might work the 50 watt ones from rapid go upto 100 watt if u put a specially made sink on them so might be worth giving it a go

does arctic silver epoxy conduct?

 

If you don't want to spread the load using several resistors then just replace it with a 5Watt version and have done.

 

its connected to a 12 v power supply, the resistor is going to have a potential difference of about 12v. which means the power disapated is going to be 5.3W.

power dissapted in a resistor = v squared x r

 

Actually the PD is 6.6V

 

yeah fair enough, didn't ead that far originaly. opps

 

Try V^2/R

 

typing error.

god i'm being a total spaz i know its v^2 /R and I^2 * R, i just typed it wrong. otherwise i would have said the wattage was like 1176.12 W using v^2 * R where v = 6.6 and r = 27. and not many people have 1000W Psu's (yet)