what burns(over-drive) a light, high voltage or high current?

D

Dae

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sorry for the silly question but I really wonder that. thanks a lot!!

Dae.
 

chesterqw

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i think both.

well high voltage as you know, is the level of power and if the source is 3v it will burst at 4v

for high current , as example for luxeon leds, if you drive a luxeon at 3.6v and the current at 500amh above the bin code, it will fly to the moon too.

that is why regulators rocks.
 

Mark2

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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).
 
D

Dae

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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).

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.
 

Geogecko

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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.
 

chimo

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Over current on a LED can damage the phosphor due to excess heat and reduce lumen output.

However, if you are wondering what causes the sudden flash and no light, it is often one of the small bond wires on the die overheating and failing (burning out - like a fuse) due to excess current.

You can run a small experiment with a 5mm LED if you slowly ramp up the current. It's a cheap and "fun" experiment.
 

AW

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Oh yes. I have seen too many times of those blue smokes coming out.....
 

Lynx_Arc

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LEDs *act* like a resistance but not a simple resistor in a circuit as they draw more current at higher voltages but only draw current related to voltage supplied to them.
If you are well below the Vf of the LED then you could have a 10,000 amp source and never hurt it but if your source is higher than the Vf you will overdrive it, also if the LED Vf drops for some reason during operation then your overvoltage will harm it.

So essentially voltage is the agent of frying LEDs with current being the weapon doing the harm.
 

The_LED_Museum

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LEDs are current-driven devices - a small increase in forward voltage usually results in a significant increase in forward current.
That's why you can blow them up with just a slight increase in voltage.
 

Gwaihir

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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.

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.
 

Geogecko

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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.

I guess my main point was that high power LED's should really be current controlled via an active circuit. The fact that battery voltage varies a lot is one reason not to direct drive them.

Low power LED's would probably need dropping resistors in order to have a long life, unless you get an LED that has the exact same Vf of the voltage of the battery (at new capacity).

The benifits of using a current controlled source out weigh the simplicity of almost zero extra components.

Active Control
Pro

  • Regulated Light Output
  • Better Use Of Battery Life
  • Longer Lasting LED
Con

  • Requires More Components
  • More Expensive
  • May Have Cut-off Point
Passive Control
Pro

  • Simple Design
  • Possible Higher Output W/ New Batteries
  • Some Light Available W/ Battery Almost Dead
Con

  • No Constant Light Output W/ Life Of Battery
  • Shorter LED Life (Maybe)
  • No "Smart" Controls/Features
  • Could Damage LED W/ Different Battery Types
  • Battery Wasted Due To Heat Lost In Dropping Resistor
Basically, my example was to prove out what effects voltage and current have on the LED. I'd only recommend using a dropping resistor when using low output LED's that did not need to be digitally controlled (for features or something). It has a better use for indicators and what not.

But that is a good point, when the battery voltage is close to the Vf, and the desired current flow requires a very small resistance, there probably is no need for the resistor, at least, when using batteries.
 

evan9162

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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.

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.
 

Geogecko

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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.

Yes, because 4 AA Alkaline batteries put out 6V nominally, and the Vf of your typical white Luxeon III is well below 6V.

I'd at least buy a ticket (if they weren't too much) to watch the demonstration myself...
 

evan9162

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A ticket will cost you $12, or one likely-to-be-killed luxeon :p

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.
 

Geogecko

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evan9162 said:
A ticket will cost you $12, or one likely-to-be-killed luxeon :p

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.

And even 3 is pushing it.

$12 is too much. I'd rather donate that to the Katrina Relief Raffle.:poke:
 

attowatt

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Does any one know if the dorcy aaa circuit is current or voltage regulated

I am thinking of using this board in a few setups that will be driven by 2aa or even 123's.
 

jwillson

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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.

Actually, the physically larger batteries have a much higher "capacity" (expressed in miliwatt hours), but will not provide any additional current in a given circuit. The 18650's will simply be able to run the light for a longer time. If a single 16340 presents no risk to the LED, then neither does a single 18650.

- Jared
 
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