[amb]

Your link doesn't point to a specific switch.
Anyway, if it says it's for 12V, then it means it has a built-in resistor that is optimized for 12V without any additional resistors. Using less voltage than 12V would make it not illuminate or very dim. Using more voltage would require additional resistance. Depending on the color of the LED you could deduce what the internal LED forward voltage is, and therefore calculate what the internal resistance might be.

 

[Navyblue]

Oops, link fixed.

Ok, assuming the forward voltage is 12V (taking into account of whatever is inside the switch), with the assumption of 20 ma current needed. The LED calculator says I need a 1k ohms 1W resistor. Does the values sound about right?

 

[rembrant]

Hmm, I have 12v LEDs. There is one in my head amp now. It has been run on 10.3v or so for about two years with no resistor. The point is, they do exist.

 

[amb]

Quote:

Originally Posted by Navyblue /img/forum/go_quote.gif Ok, assuming the forward voltage is 12V (taking into account of whatever is inside the switch), with the assumption of 20 ma current needed. The LED calculator says I need a 1k ohms 1W resistor. Does the values sound about right?

First of all, you'll almost never need to run 20mA through a small LED. Many LEDs get plenty bright anywhere from 2mA to 10mA.

Second, as I said, no small LED has a forward voltage of 12V. The fact that the switch LED is rated that way, is because it already has a resistor inside and its value is optimized for 12V. Without knowing what LED current that would make, you'll have to apply 12V and measure the current with an ammeter to know. Once you have that (let's just say it's 5mA), and since it's got a blue LED inside and most blue LEDs have ~4V forward voltage, we can calculate the internal resistor's value.

R = V / I = (12V - 4V) / 5mA = 1.6K

If you want to run the LED at 30V, then you need this much total resistance:

R = V / I = (30V - 4V) / 5mA = 5.2K

Since there is already a 1.6K resistor inside, you only need another 3.6K resistor. For the wattage, since you have 5mA flowing through the resistors,
we can calculate the dissipation:

P = I² * R = 5mA² * 3.6K = 90mW

Thus, a 1/4W resistor will be more than adequate.

Quote:

Originally Posted by rembrant Hmm, I have 12v LEDs. There is one in my head amp now. It has been run on 10.3v or so for about two years with no resistor. The point is, they do exist.

The resistor is built-in to that type of LED.

 

[Navyblue]

Thanks again.

3.6k ohms 1/4W it shall be.