D
Dae
Guest
sorry for the silly question but I really wonder that. thanks a lot!!
Dae.
Dae.
Mark2 said:Power kills the light. (power = voltage * current)
Voltage and current are linked, i.e. to get more current flowing, you need more voltage, and if you apply more voltage, more current will flow. Note: Both LEDs and Incandescents are *not* linear "resistors", but for both, the following rule is valid: more voltage = more current = more power (= more heat).
evan9162 said:Here's a primer on how voltage and current are related in LEDs:
http://candlepowerforums.com/vb/showthread.php?t=77221
and another:
http://candlepowerforums.com/vb/showthread.php?t=72528
Geogecko said:Well, actually, an LED is a diode, which means it has a fixed forward voltage (may vary slightly with current level, but generally, it's fixed, look at a datasheet for Vf).
With that respect, it depends on how you are driving the LED. If it's a passive circuit, which means, basically just a series resistor with the LED, the current will depend on what value of resistor you choose, and what the Vf of the LED is. Here is an example.
Vf of a particular LED is 4V. If you have a battery that had a nominal voltage of 6V, and you wanted to drive the LED at 1A (amp), then this is how you would select the resistor.
V=I * R (ohm's law, I = current in A, R = resistance in ohms, V = voltage in V)
R=V / I
R=(Vp - Vf) / I [Vp is the voltage provided by the batteries, since the voltage across the LED is fixed, you can assume at this point, that the voltage across the resistor is now fixed (as long as the battery stays constant)]
R=(6 - 4) / 1
R=2 ohms.
Notice the VERY low resistance value. Most high power LED's don't use this type of configuration, for one, the power lost in the resistor is wasted at heat, and secondly, using an active circuit can provide regulation of the light output of the LED, even when the battery voltage is declining.
Active circuits usually consist of either a PWM (pulse width modulation) circuit, which basically looks like a square-wave of pulses at a high frequency, which in the end, average out to be an average voltage, or a constant current source.
This is probably over the scope of this thread, but thought I'd point some things out.
So, based on the above example though, over driving an LED can be accomplished several ways, leaving everything else constant, you could do the following:
1. Decrease the resistance; more current flow, with the same voltage.
2. Increase the battery voltage; more current flow, with the same resistor.
Gwaihir said:Yes, the needed resistance is only 2 Ohm, BUT : what batteries you use ?
Every battery have internal resistance. With above configuration, if you use 4AAA or 4AA batteries (alkaline or manganese - no matter)- you don't need ballast resistor at all. The internal resistance of small-size batteries is big enouth to limit the current. And if you want to use lithium or rechargeables - better to use resistor or regulator.
Gwaihir said:Yes, the needed resistance is only 2 Ohm, BUT : what batteries you use ?
Every battery have internal resistance. With above configuration, if you use 4AAA or 4AA batteries (alkaline or manganese - no matter)- you don't need ballast resistor at all. The internal resistance of small-size batteries is big enouth to limit the current. And if you want to use lithium or rechargeables - better to use resistor or regulator.
evan9162 said:Go ahead and direct drive a Luxeon III off 4AA Alkaline batteries and see what happens. You're going to easily be pushing 2A, which is quite unsafe, to say the least.
evan9162 said:A ticket will cost you $12, or one likely-to-be-killed luxeon
My experience shows that 3 fresh AA alkalines will drive a typical luxeon I/III at 1.5A - 4 have more than enough "kick" to kill a luxeon in short order.
Dae said:thanks a lot, chesterqw & Mark!
the reason why I raised this thread is that, 18650 is 3.6V as 16340 but is twice in length and the current is almost three times as much as 16340's.