What does Vf on a LED measure?

[Tessaiga]

What is the significance of a Vf on a LED and what are the pros and cons of high or low Vf?

Thanks

 

[Gunner12]

It means forward voltage.

Usually a lower Vf is desired. That would make a LED more efficient(less power consumed) and depending on the battery, could be easier on the driver.

 

[MrGman]

To add some detail to that of the previous poster.
It is the voltage drop that the chip itself has at any given current based on the dopant levels in the semiconductor and its junction turning on, which as the previous poster said goes to show the efficiency of the diode.

So pushing 1.000 amp through the die and one sample gives you a forward voltage of 3.731 Volts and another gives you 4.131V. The first is consuming exactly 3.731 watts of power (voltage times current) and the second is consuming more at 4.131 watts of power. If the second one is putting out the exact same amount of light in lumens as the first then its wasting that difference of power to get it to you. In this case its wasting 0.400 watts of power and not giving you anything for it. So if you were a manufacturer screening 10,000 chips, the best would be all those down in the 3.7 range and not the 4.1+ range, as an example.

Now if the unit that has a Vf of 4.131 is actually brighter at that level than the other chip at 3.731. Then you have to see if the lumens per watt ratios are actually the same or not. Chances are that the higher voltage drop unit is not given the same ratio of lumens per watt.

The lower the forward voltage of the chip, the lower the watts its dissipating on itself and also the cooler it runs. May not seem like a big deal but it all adds up when your heatsink is limited and you don't want the chip to "Die", so to speak. Hope that helps.

 

[Jarl]

I never figured out if it would do an equivalent level of light at 1A, regardless of Vf, or if it really was lumens/watt- so, more watts from higher Vf=more lumens.

 

[LukeA]

I never figured out if it would do an equivalent level of light at 1A, regardless of Vf, or if it really was lumens/watt- so, more watts from higher Vf=more lumens.

They're binned based on current, so a lower Vf spells higher efficiency.

 

[xcel730]

Nice explaination

To add some detail to that of the previous poster.
It is the voltage drop that the chip itself has at any given current based on the dopant levels in the semiconductor and its junction turning on, which as the previous poster said goes to show the efficiency of the diode.

So pushing 1.000 amp through the die and one sample gives you a forward voltage of 3.731 Volts and another gives you 4.131V. The first is consuming exactly 3.731 watts of power (voltage times current) and the second is consuming more at 4.131 watts of power. If the second one is putting out the exact same amount of light in lumens as the first then its wasting that difference of power to get it to you. In this case its wasting 0.400 watts of power and not giving you anything for it. So if you were a manufacturer screening 10,000 chips, the best would be all those down in the 3.7 range and not the 4.1+ range, as an example.

Now if the unit that has a Vf of 4.131 is actually brighter at that level than the other chip at 3.731. Then you have to see if the lumens per watt ratios are actually the same or not. Chances are that the higher voltage drop unit is not given the same ratio of lumens per watt.

The lower the forward voltage of the chip, the lower the watts its dissipating on itself and also the cooler it runs. May not seem like a big deal but it all adds up when your heatsink is limited and you don't want the chip to "Die", so to speak. Hope that helps.