The only info provided with them is that the voltage is 3.6-6 volts. Is there any way to test them with a meter for the actual voltage and the amperage they use?

 

ask on the electrical board. they should know. i would also like a specific answer to this question.

thanks
anthony

 

I think you'd have to burn one up to find the max voltage it will support.

It sounds like you can run 3 in series and be safe since that would be 13 volts across 3 for a total of 4.33 volts each.
2 in series might be too much if they are suppesed to be rated at 6 volts each. you could run 2 with an other diode in series to drop the volts down a bit just to be safe.

The amps they use should be very very small. run an ammeter in series with one of the LEDs with a 6 volt battery and see what the amperage reads.

 

Just for future reference, you don't "burn one" up to see the max voltage for an LED. Most LEDs need about two volts to come on. What you need to figure out is the amount of CURRENT going through the LEDs. For an LED that vary as low as 6 mA up to 100mA for a high brightness LED. Using ohm's law, you need to figure out the resistor value you can use to obtain the maximum current recommened by the manufacturer of the LED, knowing you have 13 or so Volts comming in.

 

Give me an example using Ohms law to show the max voltage of the LED before it burns up?

 

What Odyssey is saying is that the voltage is not what burns up the LED, one could have 100V on it, it's the current that runs thru it that damages it. A typical LED drops about 1.6 to 2 volts across it. If one knows the maximum current rating for the part, the resistor is easy to figure out. If we have a max system voltage of 16 volts (typical automotive spec), and a voltage drop of 2 volts for the LED, then the resistor will drop 14 volts. If the current rating for the LED is 20mA, then the current thru the resistor is also 20mA. Ohm's law states that R=V/I so the resistor value would be R=14V / 20ma = 700Ohm. If you run two LEDs in series, the current would stay the same, at 20mA, but the voltage drop would change. For two LEDS, the drop across the resistor would go from 14V to 12V, etc.

 

So, I oculd hook up the LED to a small 9volt battery and it shouldn't burn up since the 9v battery has very little current?

 

No... read RCR said. An LED will drop about 2 volts across it. A 9volt battery will for sure have more then 20 or so mA. Try more like a few hundreds. So yes, doing that WILL burn up the LED in short time. However if you had one very weak battery, then you could hook it up and it would be fine.

 

What would happen is the battery would cook itself.

Been there. Done that.

 

For anyone who is interested in calculating the proper value of resistor to use for a given LED, check this page http://linear1.org/ckts/led.php and input the source voltage (figure about 14.6 for automotive purposes), the forward voltage of the LED (usually between 2 and 6), and the desired current through the LED (from 20 to upwards of 50 for some types). Then hit 'Find R' to get the value needed.

Pete