LEDs. If you look at their specs, Vf, or voltage forward is mentioned. Like minimum 3.0v and maximum 3.8v. What does this spec mean? Safe operating range? What happens if you put more voltage through it? Like say 4.5 or 6.0 volts?
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MaxaBaker
Flashlight Enthusiast
To answer your quesation, I have no idea.
But, I just put as much voltage into them as I can.
But, I just put as much voltage into them as I can.
Negeltu
Enlightened
I think... that the non technical answer is... Forward voltage is the amount of voltage that is needed to cause current to flow through the led and light it up. It's a good idea not to exceed the max Vf of the led.
MaxaBaker
Flashlight Enthusiast
Oops...... /ubbthreads/images/graemlins/grin.gif /ubbthreads/images/graemlins/tongue.gif
Zelandeth
Flashlight Enthusiast
Okay, best thing I can think of here is an example.
Say you've got a 12V DC supply, and want to run a blue LED with a 3.7V Vf on it, with a maximum current of 25mA.
To work that out you'd use the following formula...as I remember anyway.
(VSupply - Vf)/I...in other words:
(12 - 3.8)/0.025 = 300 ohm.
Which actually sounds like the right number to me, so I'm guessing I'm right there with the calculation.
Electronically speaking, the Vf is the voltage which is required to break down the depletion layer within the P-N junction of the diode, and allow current to start flowing through the afforementioned P-N junction.
Say you've got a 12V DC supply, and want to run a blue LED with a 3.7V Vf on it, with a maximum current of 25mA.
To work that out you'd use the following formula...as I remember anyway.
(VSupply - Vf)/I...in other words:
(12 - 3.8)/0.025 = 300 ohm.
Which actually sounds like the right number to me, so I'm guessing I'm right there with the calculation.
Electronically speaking, the Vf is the voltage which is required to break down the depletion layer within the P-N junction of the diode, and allow current to start flowing through the afforementioned P-N junction.
asdalton
Flashlight Enthusiast
In practical terms, it's better to think of controlling the current and letting the voltage drop take care of itself. Small changes in applied voltage lead to very large changes in current, and it's the current that determines the light output and/or overheating.
It's all interrelated- if you provide more voltage then naturally more current will flow. This overdrives the LED and will shorten its life, perhaps by a lot.
You cannot provide more voltage without providing more current although what might happen is that the internal resistance of the batteries will lower their effective voltage when you try to apply that voltage to an object that doesn't have a lot of resistance. So lets say you connect that LED to a 9V smoke detector battery- the resistance of the battery itself will make the LED actually see much less than 9V.
Well, that's my guess anyway. I'm a CE, not an EE.
You cannot provide more voltage without providing more current although what might happen is that the internal resistance of the batteries will lower their effective voltage when you try to apply that voltage to an object that doesn't have a lot of resistance. So lets say you connect that LED to a 9V smoke detector battery- the resistance of the battery itself will make the LED actually see much less than 9V.
Well, that's my guess anyway. I'm a CE, not an EE.
Ken_McE
Flashlight Enthusiast
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Rule of thumb is that the more voltage you run through a led, the brighter it will be. The minimum is probably the least voltage that will make noticable light. You might be able to underdrive it and still get a little light. The max voltage is the hardest that you can safely run it. You can give it more than the max voltage and it'll glow real bright, but the lifespan will drop considerably. If you keep upping the voltage you eventually reach a point where it flashes brightly once and burns out instantly.
OK, thanks for the input. I've got one of those keychain LED lights that runs off 2 cr2016 lithium batteries. Which I guess put out about 6 volts. The LED is rated at 8,000 mcd, I believe. I'm getting a few of these
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=66952&item=3873608991&rd=1&ssPageName=WDVW
26,000 mcd bulbs from the group buy, and I was wondering if one of these new bulbs would work OK in the keychain flashlight. But I didn't know what the ramifications of the Vf 3.0-3.8 really implied. Do you think it would work?
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=66952&item=3873608991&rd=1&ssPageName=WDVW
26,000 mcd bulbs from the group buy, and I was wondering if one of these new bulbs would work OK in the keychain flashlight. But I didn't know what the ramifications of the Vf 3.0-3.8 really implied. Do you think it would work?
Negeltu
Enlightened
Most of the time the resistance in the lithium coin cells will protect the leds by limiting the amount of current that flows. Those cells have high resistance.
Leeoniya
Enlightened
yeah, current control is best.
LEDs are rated for Current however. what that means is this: when you run the LED exactly at its rated current..say 350ma. then you can EXPECT to see the indicated Vf drop across it.
but there is a Vf variation always since not all LEDs are created equal, so putting the same rated Vf into identical Binned LEDs does not guarantee that both LEDs will run at their rated current. one might be slightly under, one might be slightly over.
current regulation actually regulates the voltage through a current sense resistor to make sure the LEDs get EXACTLY the rated current. so the only thing a lower Vf would indicate in a current regulated design is the fact that it will consume less power because it will require a lower voltage to drive the LED at its rated current. and since power = V*I. keeping the I constant leaves only one variable to change for efficiency. the Vf, which means a lower resistance of the LED itself.
and yes. since current is the same through all in-line components, you can use V=IR (total voltage drop = total current * total resistance).
if you know the difference in voltage between say 6V and your LED Vf of 3.44v, the remaining voltage you need to drop is 6-3.44 = 2.56V...and you already know the current you will need so I = 350ma
so 2.56V = 0.350*R
solve for R, you get 7.3 ohm resistor.
since you probably wont find a 7.3 ohm, you can drop the resistance to a common 6.8ohm and plug it back in to recalculate the current.
2.56V = I*6.8
solve for I you get 0.376A or 376ma, about a 26ma overdrive.
for resistor wattage rating all you need to do is multiply I*V so 2.56*0.376=0.96 Watts, this is a very good size resistor and a lot of energy wasted, and it will get quite warm, as you'd expect it to. the more voltage it drops and the higher the current at that voltage, the more head it needs to dissipate.
thats why a low Vf and high flux is preferable, because the LED will generate less heat and more power will go to producing light.
LEDs are rated for Current however. what that means is this: when you run the LED exactly at its rated current..say 350ma. then you can EXPECT to see the indicated Vf drop across it.
but there is a Vf variation always since not all LEDs are created equal, so putting the same rated Vf into identical Binned LEDs does not guarantee that both LEDs will run at their rated current. one might be slightly under, one might be slightly over.
current regulation actually regulates the voltage through a current sense resistor to make sure the LEDs get EXACTLY the rated current. so the only thing a lower Vf would indicate in a current regulated design is the fact that it will consume less power because it will require a lower voltage to drive the LED at its rated current. and since power = V*I. keeping the I constant leaves only one variable to change for efficiency. the Vf, which means a lower resistance of the LED itself.
and yes. since current is the same through all in-line components, you can use V=IR (total voltage drop = total current * total resistance).
if you know the difference in voltage between say 6V and your LED Vf of 3.44v, the remaining voltage you need to drop is 6-3.44 = 2.56V...and you already know the current you will need so I = 350ma
so 2.56V = 0.350*R
solve for R, you get 7.3 ohm resistor.
since you probably wont find a 7.3 ohm, you can drop the resistance to a common 6.8ohm and plug it back in to recalculate the current.
2.56V = I*6.8
solve for I you get 0.376A or 376ma, about a 26ma overdrive.
for resistor wattage rating all you need to do is multiply I*V so 2.56*0.376=0.96 Watts, this is a very good size resistor and a lot of energy wasted, and it will get quite warm, as you'd expect it to. the more voltage it drops and the higher the current at that voltage, the more head it needs to dissipate.
thats why a low Vf and high flux is preferable, because the LED will generate less heat and more power will go to producing light.
jtr1962
Flashaholic
[ QUOTE ]
lightmeup said:
LEDs. If you look at their specs, Vf, or voltage forward is mentioned. Like minimum 3.0v and maximum 3.8v. What does this spec mean? Safe operating range? What happens if you put more voltage through it? Like say 4.5 or 6.0 volts?
[/ QUOTE ]
The stated Vf is the voltage drop across the LED when running at the rated current, usually 20 mA. There is a range due to manufacturing variations, although 3.0V to 3.4V seems to cover most of the usual Vf range. White LEDs above 3.5V or below 3.0V at rated current are rare.
As for the second question, most LEDs will be destroyed running them at 4.5V and especially 6V. You drive an LED with a constant current circuit. This can be as simple as two transistors and two resistors or it can be a highly complex switching regulator. The idea is the same-drive the LED at the rated current and let Vf fall where it may.
lightmeup said:
LEDs. If you look at their specs, Vf, or voltage forward is mentioned. Like minimum 3.0v and maximum 3.8v. What does this spec mean? Safe operating range? What happens if you put more voltage through it? Like say 4.5 or 6.0 volts?
[/ QUOTE ]
The stated Vf is the voltage drop across the LED when running at the rated current, usually 20 mA. There is a range due to manufacturing variations, although 3.0V to 3.4V seems to cover most of the usual Vf range. White LEDs above 3.5V or below 3.0V at rated current are rare.
As for the second question, most LEDs will be destroyed running them at 4.5V and especially 6V. You drive an LED with a constant current circuit. This can be as simple as two transistors and two resistors or it can be a highly complex switching regulator. The idea is the same-drive the LED at the rated current and let Vf fall where it may.
Well, if I understand what you guys are saying, then running this new led at 6v is a no-no. And since it is a pocket keychain light, there is no place to add resistors or change the circuit. Right? So, could I try to replace my 2 CR2016's which are putting out 6v together with 2 CR2032's which only will be putting out about 3v together? Will that make Faraday and Maxwell smile from their graves?
jtr1962
Flashaholic
Actually, I think you meant replacing two CR2016s with one CR2032. Yes, that would work, and would likely run the LED under rated current, or in cases where Vf is 3V it would run exactly at 20 mA. I don't think there is any chance of damaging a white LED off 1 cell. The lowest Vf I've seen is about 2.9V (and about 3V at 30 mA). This is well within the safe zone. In some cases you might get by running an LED off two CR2016s because the internal resistance of the cells would cause the LED to see way less than 6V. I'd still put in a resistor just to play it safe with 2 cells.
I've seen a lot of talk around here about 'over-driving' LEDs to get more brightness out of them. Does this mean at more than the specified Vf range (3.0-3.4 in my case)? How much 'over-drive' is tolerable before it becomes very damaging? Does this cause the bulb to burn out faster? What would be the likely effect of me running one of my new bulbs at 6v? Shorter lifespan, quick burnout, or what? Would this have a big effect in terms of more output or less longevity? I'm just wondering what the tradeoffs are like in real world terms?
gadget_lover
Flashaholic
[ QUOTE ]
jtr1962 said:
[ QUOTE ]
lightmeup said:
LEDs. If you look at their specs, Vf, or voltage forward is mentioned. Like minimum 3.0v and maximum 3.8v. What does this spec mean? Safe operating range? What happens if you put more voltage through it? Like say 4.5 or 6.0 volts?
[/ QUOTE ]
The stated Vf is the voltage drop across the LED when running at the rated current, usually 20 mA. There is a range due to manufacturing variations, although 3.0V to 3.4V seems to cover most of the usual Vf range. White LEDs above 3.5V or below 3.0V at rated current are rare.
As for the second question, most LEDs will be destroyed running them at 4.5V and especially 6V. You drive an LED with a constant current circuit. This can be as simple as two transistors and two resistors or it can be a highly complex switching regulator. The idea is the same-drive the LED at the rated current and let Vf fall where it may.
[/ QUOTE ]
To expand on this just a touch....
Each individual led has only one voltage at which it draws the current it is designed for. That is known as the Vf. The manufacturer is too lazy (or cheap or some other reason) to test and label each one individually, so they simply tell you the range within which the Vf should fall. You'd think they could dump all the 3.26 to 3.29 Vf leds in one bin.
To answer the second question; You will burn out the LED (sometimes slowly) if you use significantly higher voltage than it's rated for. By significantly higher I mean 20% or more. Watch batteries can not produce enough energy to burn out the LED at 6 volts, but AA batteries or bigger can.
Daniel
jtr1962 said:
[ QUOTE ]
lightmeup said:
LEDs. If you look at their specs, Vf, or voltage forward is mentioned. Like minimum 3.0v and maximum 3.8v. What does this spec mean? Safe operating range? What happens if you put more voltage through it? Like say 4.5 or 6.0 volts?
[/ QUOTE ]
The stated Vf is the voltage drop across the LED when running at the rated current, usually 20 mA. There is a range due to manufacturing variations, although 3.0V to 3.4V seems to cover most of the usual Vf range. White LEDs above 3.5V or below 3.0V at rated current are rare.
As for the second question, most LEDs will be destroyed running them at 4.5V and especially 6V. You drive an LED with a constant current circuit. This can be as simple as two transistors and two resistors or it can be a highly complex switching regulator. The idea is the same-drive the LED at the rated current and let Vf fall where it may.
[/ QUOTE ]
To expand on this just a touch....
Each individual led has only one voltage at which it draws the current it is designed for. That is known as the Vf. The manufacturer is too lazy (or cheap or some other reason) to test and label each one individually, so they simply tell you the range within which the Vf should fall. You'd think they could dump all the 3.26 to 3.29 Vf leds in one bin.
To answer the second question; You will burn out the LED (sometimes slowly) if you use significantly higher voltage than it's rated for. By significantly higher I mean 20% or more. Watch batteries can not produce enough energy to burn out the LED at 6 volts, but AA batteries or bigger can.
Daniel
jtr1962
Flashaholic
[ QUOTE ]
lightmeup said:
I've seen a lot of talk around here about 'over-driving' LEDs to get more brightness out of them. Does this mean at more than the specified Vf range (3.0-3.4 in my case)?
[/ QUOTE ]
Overdrive basically means running the LED above its maximum continous current (this is 30 mA for most 5mm LEDs). Again, Vf is irrelevant. You can severely overdrive an LED at 3.5V or you can underdrive it. It depends upon the characteristics.
[ QUOTE ]
How much 'over-drive' is tolerable before it becomes very damaging? Does this cause the bulb to burn out faster? What would be the likely effect of me running one of my new bulbs at 6v? Shorter lifespan, quick burnout, or what? Would this have a big effect in terms of more output or less longevity? I'm just wondering what the tradeoffs are like in real world terms?
[/ QUOTE ]
For 5mm LEDs at anything above 100 mA you're in danger of melting the bond wires. Under that you're OK but heat will take its toll in the form of greatly shortened life. Based on my testing, there is not much point to going beyond about 50 or 60 mA as there is little extra light to be found and greaty shortened life. Here are two graphs for my eBay 26,000 mcd LEDs:
The lifetime graph is based on empirical equations from one of the white LED lifetime studies I found online. Most 5mm LEDs will have similar characteristics. In this case "lifetime" is defined as the time to 50% brightness.
lightmeup said:
I've seen a lot of talk around here about 'over-driving' LEDs to get more brightness out of them. Does this mean at more than the specified Vf range (3.0-3.4 in my case)?
[/ QUOTE ]
Overdrive basically means running the LED above its maximum continous current (this is 30 mA for most 5mm LEDs). Again, Vf is irrelevant. You can severely overdrive an LED at 3.5V or you can underdrive it. It depends upon the characteristics.
[ QUOTE ]
How much 'over-drive' is tolerable before it becomes very damaging? Does this cause the bulb to burn out faster? What would be the likely effect of me running one of my new bulbs at 6v? Shorter lifespan, quick burnout, or what? Would this have a big effect in terms of more output or less longevity? I'm just wondering what the tradeoffs are like in real world terms?
[/ QUOTE ]
For 5mm LEDs at anything above 100 mA you're in danger of melting the bond wires. Under that you're OK but heat will take its toll in the form of greatly shortened life. Based on my testing, there is not much point to going beyond about 50 or 60 mA as there is little extra light to be found and greaty shortened life. Here are two graphs for my eBay 26,000 mcd LEDs:
The lifetime graph is based on empirical equations from one of the white LED lifetime studies I found online. Most 5mm LEDs will have similar characteristics. In this case "lifetime" is defined as the time to 50% brightness.
Thanks, great answer. I'm starting to think the old LED in my keychain light is probably supposed to be driven by a 3v rather than 6v system. With these new 26,000 mcd LEDs, what would you guess the performance difference would be between driving it with 2 CR2016 versus one CR2032? Which setup would be brighter and what would the likely difference in bulb life be like? Your recommendation or comments? Thanks....
jtr1962
Flashaholic
On one CR2032 you might get over 20,000 mcd if the LED has a fairly low Vf (3.0 or 3.1V). If not, it'll be somewhat dimmer. On 2 CR2016 it would certainly be very bright but you might be in danger of exceeding 100 mA and having the LED fail prematurely. It depends upon the internal resistance of the cells. You can short a cell using a multimeter on the 10A scale, and calculate the internal resistance approximately using the equation R=3/I. This will give you some idea if you'll be safe. My best guess is that a CR2016 has an internal resistance above 30 ohms, so two in series would put less than 50 mA into a white LED.
My UV keychain light has 2 2016s and the LED is still working fine some I'm guessing it's safe.
I need to mention here that one common mistake made is that people will run an LED on two lithium button cells and mistakeningly think they're running the LED at 6V. They're not-more than likely the LED sees 3.5V or less due to the cells' internal resistance. However, they'll think it survived 6V and then put it on a 6V power supply or on 5 NiMH in series, the LED will burn out instantly, and they'll wonder why. Running almost any white LED on even 3 NiMH cells is probably pushing it, and I'm not a proponent of directly drive anyway except when the power source is something like a button cell with very high internal resistance.
My UV keychain light has 2 2016s and the LED is still working fine some I'm guessing it's safe.
I need to mention here that one common mistake made is that people will run an LED on two lithium button cells and mistakeningly think they're running the LED at 6V. They're not-more than likely the LED sees 3.5V or less due to the cells' internal resistance. However, they'll think it survived 6V and then put it on a 6V power supply or on 5 NiMH in series, the LED will burn out instantly, and they'll wonder why. Running almost any white LED on even 3 NiMH cells is probably pushing it, and I'm not a proponent of directly drive anyway except when the power source is something like a button cell with very high internal resistance.
Is there any polarity requirement with those two leads? Does it matter which wire gets + or - side of battery?Similar threads
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