What do the LED spec mean?

Forward voltage is the voltage drop across the LED when you apply the drive current. You can use this to determine the load resistor, for example, or to determine how many LEDs in a string you can get before there isn't enough voltage.

Current is what you *provide*, voltage is the *result*.

Case #1. Let's say you had a 5v supply and a green LED. If is 20mA, Vf is 2.2v. What size resistor do you need? How do you compute it? Well, you know the current you want... can you figure out the voltage across the resistor? It's 5v - Vf, or 5-2.2 = 2.8v. So the resistor has 2.8v across it, and needs to have 0.02 amps through it, R = V/I =

Case #2. My exposure box provides 39 VDC. Each LED has a Vf of 3.8v. How many can I string together? 39 / 3.8 = 10.2, or 10 LEDs. In practice, I strung 9 together with an LM317, which needs some volts to work - 9*3.8 = 34.2v, leaving about 4.8 volts for the LM317.

Well, zero current will let it last for years. The spec should have expected lifetime, but it probably depends somewhat on heat - if you have a lot of LEDs in an enclosed space, they won't last as long.

Thanks Delorie, So this means that the Voltage rating only serves one single purpose: To calculate. It has no other function. There is no other reason for needing to know it.

The forward voltage is usually a range - that is the voltage you measure across the led with a specified amount of current - and varies from led to led and from batch to batch. It is really only of use when you want to know how many you can put in series and get them to light with a specific voltage across the string.

There's also usually a de rating curve (chart) to show life expectancy or maximum current or power dissipation at a range of ambient temperatures the led might encounter. Hotter ambient, less allowed dissipation just like other semiconductors.

The safest way to drive leds is with a constant current source. A resistor is cheap and easy. If the led is expensive or needs to last, it pays to take the time and effort to use constant current. Use an IC driver intended for leds or roll your own (takes two transistors and a couple of resistors is all).

Life expectancy is not often specified and there seems to be no standardization when life is specified. Leds "never" burn out but some cheaper ones do fail. "Life" is usually specified as the point where the LED is down to 50% or 70% of the original light output in years.

To test the life they run them at higher currents - and try to predict how long they'd last at normal current.

One this groups regular respondents built a led tester to wean out the good suppliers on Ebay from the bad ones. He runs 20 ma leds with a constant current source set at 30 milliamps then measures the light output days or months later to see how fast it is dropping - but the technique, will useful to separate good ones from bad ones, doesn't predict the absolute life but the relative life between different suppliers.

Buy a good brand from a reputable source, if you need longevity and pay for quality.

If you run them at lower current they last longer and dim more slowly. So a high efficiency led rated at 20 milliamps, can often be run at much less current with the same light output as a less efficient one.

Also look at the beam spread - you can get Leds with impressive mcd numbers that only have very narrow beams (usually measured to the 1/2 power points in degrees). Buy with the beam spread and application in mind. If you want to see a pilot light from anywhere in the room - it needs to be a wide beam, if you need to light an area some feet from the source a narrow beam may do a better job.

Usually given as typical and maximum, to be read as "at the specified forward current (If), we expect that this part will typically show a voltage drop of 1.7 V. We further expect that the +3*sigma voltage drop will be no more than 2.2 V at that current." (Using Vf-typ 1.7 and Vf-max 2.2 and you hope that they're using 3 sigma and not one).

If your voltage source minimum spec is, say, 4.5 V and you wanted to guarantee at least 10 mA through the LED then you'd need to drop 2.3 V across your limiting resistor (since the LED might drop as much as 2.2 V). Thus, if you're using common 5% resistors, use a 220 ohm.

If your supply is typically 5 V and you have a typical LED, then you'll be pushing about 16 mA (if the resistor is at the other end of its own tolerance band).

You'll generally want the typical If to run at half the rated max or less. Check for a derating curve of If vs temperature, also.

One can not know, Grasshopper, one can only estimate the probability. ;-)

Isn't one enough? ;-)

Cheers! Rich

Apologies if I appear a bit dense. I almost got it...

So if an LED says forward currents 350mA (as my Luxeon LED says) and the forward voltage is 3.4V that means that if I hook up a battery with an exact voltage of 3.4V directly to the LED then there will be a current of 350mA flowing through it (no resistor needed), correct?

Thanks again

Don't try.

The problem here is that the "resistance" of an LED is exponential, so if your battery offered 3.5V instead of 3.4V, the current could go up a *lot*. Likewise, if you give it 3.3V you might not see any light at all.

If you operate the LED at the current for which the voltage is stated as a range, then you will get that range of voltages most of the time. Not always, even then. It's a guidance, that's all.

More fully, the voltage is at least a curve over reasonable ranges of driven current (unreasonable currents make the curve still more interesting), if not a multidimensional surface relating even more parameters, such as ambient temperature, aging, heat sinking, the actual distribution of manufactured LED behavior, and maybe even incident light from other sources, to name some.

What saves you from too much detail is that the voltage near to the specification range often goes by a factor that is a simple, linear function of current (equivalent to a resistor for short excursions beyond the range given), so that you can estimate the voltage by something like this:

V_fwd = V_cal + I * R_cal

With only one point specified in the data sheet, you really can't figure out R_cal. But if you make a two point measurement on some actual devices, you can average the results and compute a reasonable value for Rcal that will "work mostly" for a wider-ranging design.

To go much beyond the local area near the specification, you need to incorporate an LED voltage that varies with the emission coefficient (N) times 60mV per 10-fold current change plus a little more also for the change in current times its ohmic resistance. (In LEDs, N is often higher than 2.)

In other words, you need to go from a linear (resistor) interpretation to a more 'curvy' view that follows an exponential/logarithmic (same thing, just depends on your parameter plotting) curve:

V(I) = N*(k*T/q)*ln(I/I_s)+I*R_s

Where k*T/q at room temps is about 25.9mV and I_s and R_s and N are model parameters for your LED. This is the more sophisticated model (though it develops from even deeper understandings of the physical layout of the diode, dopant concentrations and so on.)

Small signal variation wise, rates of change in voltage relative to rates of change in current will be:

d[V(I)]/dI = n*k*T/q*(1/I) + R_s

That becomes the R_cal mentioned above, if you already know the approximate (I) for which you want to estimate its value. The V_cal can be had from the V(I) equation, using the same value of (I).

Jon

There's no such thing as "exact" in the real world.

No.

Vf=3.4V means that if you apply 3.3V you'll get 1uA flowing through it, while if you apply 3.5V you'll get 10A. Well; that's not strictly accurate, but it's a reasonable model.

Vf is the point at which the V-I curve transitions from near-horizontal to near-vertical.

This is why you need a resistor or a regulator.

Suppose that you have a 3.4V LED which you want to run at around 15-20mA.

If you have a 5V supply, a 3.4V LED, and a 100-Ohm resistor, then the current will be (5V-3.4V)/100R = 1.6V/100R = 16mA.

If the supply voltage is actually 5.2V, you get 18mA, if it's 4.8V you get

OTOH, if you have a 4V supply, a 3.4V LED and a 33-Ohm resistor, then the current will be (4V-3.4V)/33R = 0.6V/33R = 18mA.

If the supply voltage is actually 4.2V, you get 24mA, if it's 3.8V you get

For a string of seven Cree LEDs that I'm driving with a boost regulator, I get the following voltages and currents (and power):

I(string) V(string) V(LED) P(LED)

Thus the effective series resistance of each LED will be R = (4.46-3.36)/(0.88-0.28) = 1.83 ohms, with a zener voltage of V = 4.46 -

So, you can drive one of these LEDs with three 1.2 VDC NiCd cells for 3.6 volts, and you could use three 1.5 VDC cells for 4.5 volts and limit with a small resistor or even a silicon diode. But best results are with a current regulator which can be linear or PWM and can approach 98% efficiency with the addition of an inductor.

The forward voltage of the LED is fairly constant with temperature but I don't have any specs, so again a current regulator is your best bet. I have made a variation of the "Joule Thief" which provides current regulation and runs on as little as 2.5 VDC. I posted an LTSpice schematic previously but I can provide it again. I also made one that uses 6-15 VDC and a power MOSFET which is what I used for the tests shown above with 7 white LEDs at up to 27 watts at 77% efficiency. Not bad for a simple circuit with no ICs.

Paul