680 Ohm Resistor Getting Hot.

[Lost_in_the_HTTP]

I have a 12VDC supply feeding a standard discrete LED about 40 feet away. Wire is 24G copper (telephone type cable). Resistor packaging calls for a 680 ohm resistor which I had on hand and installed. But it's getting hot to the touch, almost too hot. LED is working, but may be a bit bright.

Do I change the resistor to higher value or lower? I have some 470s and 330s as well as 1Ks and higher.

All are 1/4 watt.

 

[[DHT]Osiris]

I have a 12VDC supply feeding a standard discrete LED about 40 feet away. Wire is 24G copper (telephone type cable). Resistor packaging calls for a 680 ohm resistor which I had on hand and installed. But it's getting hot to the touch, almost too hot. LED is working, but may be a bit bright.

Do I change the resistor to higher value or lower? I have some 470s and 330s as well as 1Ks and higher.

All are 1/4 watt.

Going about this all wrong, make a little liquid cooling loop for the resistor!

 

[NutBucket]

Simplest thing to do, put two 330 in series and spread the heat dissipation.

 

[Red Squirrel]

You will just need a higher wattage resistor, maybe 5w.

Another option is to slightly reduce the voltage of the power supply as that will reduce the amount of current the resistor needs to bleed off. Basically LEDs need a constant current source and not a constant voltage source, so a cheap way to do this is to introduce a resistor which bleeds off some current. Its basically a cheap way to do it without needing a constant current power supply.

 

[pcgeek11]

Use a heat sink.

 

[NutBucket]

Pretty sure you don't need a 25W resistor

 

[bba-tcg]

I have a 12VDC supply feeding a standard discrete LED about 40 feet away. Wire is 24G copper (telephone type cable). Resistor packaging calls for a 680 ohm resistor which I had on hand and installed. But it's getting hot to the touch, almost too hot. LED is working, but may be a bit bright.

Do I change the resistor to higher value or lower? I have some 470s and 330s as well as 1Ks and higher.

All are 1/4 watt.

The best way would be to use a higher wattage resistor. 12 volts with a 680 watt resistor dissipates slightly over .21 watts. Your 1/4 resistor will do it, but be hot to the touch. A one watt resistor would run cooler.

P=VI or (V^2)/R =144/680 = .211765 watts.

Two 330 ohm resistors in series will do the job too, but look less elegant, depending upon the way you have to mount the resistors.

 

[VashHT]

If you want it to be dimmer up the resistance, you're pulling close to 1/4 watt like others said so it'll be a bit hot. I'd agree with Sick Willie just go for a higher wattage resistor, they're like less than a cent on digikey.

 

[shortylickens]

If you wanna use the same resistance, they do make larger bodies that can dissipate 1/2 watt or even 1 watt.
OR you can do some kind of active or passive cooling.
OR you can use something with HIGHER resistance, which lowers current flow. Which reduces heat. But also fucks up your circuit, which was designed and build by somebody much smarter than you.

 

[Lost_in_the_HTTP]

Switched to a 1K. Still warm to the touch, but not as hot. May go up a bit more yet. It's just a pilot indicator. It doesn't need to be bright, just enough to know it's on.

 

[pcgeek11]

OR you can use something with HIGHER resistance, which lowers current flow. Which reduces heat.

Higher resistance increases heat.

HEATING EFFECT OF AN ELECTRIC CURRENT

Elinikäinen ja toimintavarma kotimainen oppimisen ympäristö

peda.net

 

[bba-tcg]

Higher resistance increases heat.

HEATING EFFECT OF AN ELECTRIC CURRENT

Elinikäinen ja toimintavarma kotimainen oppimisen ympäristö

peda.net

You'd think so, but consider this:

Using the same formula from above: P=VI or (V^2)/R, if you had a voltage of 12v and a resistance of 12 ohms, you would have: (12*12)/12 = 12 watts.

Now if you had 12v and 144 ohms: (12*12)/144 = 1 watt.

Which is going to produce more heat?

And that formula is right there on the page you linked. It's obvious that the higher the resistance, the lower the power consumption.