compass silva type 100

[dd85]

hello to all
i am looking for the reference ( part number ) of the red led or any solutions to improved her life ( wiring ,other led with reistor, dealer adress, ... )
thank you for your help

[BillyDoc]

dd85,

Most LED's have an almost infinite life if properly treated. They also generally come in "standardized" sizes, making replacement very easy (if you can get to it). I'm not familiar with that particular compass, but can you get to the LED to remove and replace it? I presume it is now dead. If so, remove the existing LED and measure it's dimensions. A typical size will be 5mm in diameter. Then go to a distributer such as digikey.com and shop for a replacement based on the qualities you prefer. Match the physical size first, color, diffused light output or not, etc. Watch that light output though, LED's are available now with 5 watts worth of output . . . and that would turn your compass into a searchlight! Oh, and LED's are polarized devices, so you do have a positive and a negative lead you have to keep straight (there's usually a flat area at the base of the LED to indicate the negative lead, but not always . . . check the data sheet).

Once you have picked an LED you like (or an assortment to try, as they are very cheap), then download the data sheet for it and check the "Forward Voltage Drop" and the typical operating current. For an LED with a brightness I would think you would want in a compass, not too much that is, the forward voltage will be something like 1.4 volts at a current of about 20 milliAmps.

Now you have to do a little simple math, using Ohm's law. You may know the voltage range of your system already, but if not we can make an educated guess. Assuming a 12 volt system, the maximum voltage is probably close to 14 volts when the motor is running and the battery is mostly charged, and at the other end of the cycle you might discharge the battery to as low as 10 volts (not a good idea, but it happens).

Your LED will use (say) 1.4 volts, so the question now becomes, what resistor can I put in series with it to drop the rest of the available volts to a current that is less than the current rating of the LED. If our highest voltage is 14, and the LED itself will use 1.4, then a resistor in series will be required to "drop" 14 minus 1.4 volts, or 12.6 volts.

This is where Ohm's law is useful. We want the value of a resistor that drops 12.6 volts at 20 milliamps, and we can find this by dividing the voltage by the current, 12.6 divided by 0.020, or 630 ohms for the smallest resistor we can use. A 630 ohm resistor might be difficult to find, however, so increase that to the next standard size, which is 680 ohms.

If we increase the resistance to 680 ohms we can then calculate the current at 12.6 volts, again using Ohm's very handy law. To do this just divide the voltage (12.6) by the resistance (680) to get the current, which in this case is 18.5 milliAmps . . . close enough.

For those of us who are mathematically "challenged," the relationships described by Ohm's little law are easy to remember with the following device: draw a large "T" and place an "E" over the top of it, with an "I" in the left armpit of the "T" and a "R" in the right armpit. "E" is the usual symbol for volts, "I" for current, and "R" for resistance. To use the device just divide I into E to get R, divide R into E to get I or multiply I times R to get E.

One last detail to pick your resistor. Multiply the voltage drop times the current to get the power that it will have to dissipate, and pick a resistor capable of dissipating at least that much power. In the present case with 0.0185 Amps and 12.6 volts that comes to a bit less than a quarter of a watt (0.2331 watts).

If it's convenient and you want to elaborate this a bit you can also calculate the resistance to use in a second variable resistor (potentiometer) and adjust the brightness of the LED as the need presents itself. Personally, I like the dimmest lights that still allow me to barely see the compass.