After a bit of digging, I found that forward voltage is considered to be the actual voltage drop across a diode. I wish they'd just call it voltage drop, but oh well. The way it was explained to me is that the forward voltage was the point on a current VS Voltage graph where the current reached/exceeded 1 Amp, or at least where current became swinging up fairly quickly (at the knee). This has probably been a point of confusion in our communication thus far.

 

For led arrays current drive is the best way to drive them ,not only from the self destruct view but from evenness of lighting .Each led can have quite a large variation in forward voltage ,especially high power white leds ,current driving them will give the same light intensity to the array , i realise i am disregarding variations in colour temperature with that statement .The other problem to take into account is junction temperature , if this becomes too high it will destroy the led just as easily as current or voltage .So a temperature monitoring current supply is a good way to go ,where the rise in temperature above nominal reduces the current until an equilibrium is achieved.
Another misleading statement by led manufacturers is the vast difference there is in the life of an led . Its usefull life is vastly different to its actual life.For example manufacturer quotes 100,000 hrs life (which is true ) ,but the light intensity after 7 to 10,000 hrs has dropped by half ,for lighting purposes that would be the end of its usefull life not the 100,000 quoted.

 

After a bit of digging, I found that forward voltage is considered to be the actual voltage drop across a diode. I wish they'd just call it voltage drop, but oh well. The way it was explained to me is that the forward voltage was the point on a current VS Voltage graph where the current reached/exceeded 1 Amp, or at least where current became swinging up fairly quickly (at the knee). This has probably been a point of confusion in our communication thus far.

Most LEDs are 5mm and have their range of forward voltage spec'd at 20mA. 1A is a very high current and will blow up most LEDs.
Here are the spec's for the red LED I have and the graph of forward voltage is only for a "typical" one:

 

Most small leds will blow up with 1A current ,but there are a lot of white leds out there that 1A is no where near the current they consume. and a difference between 1.5 v and 2.4 is an enormous forward voltage difference,which proves the case for current drive .

 

I do realize 1A is a hell of a lot of current for most LEDs, but because of the V I characteristic of an LED, the difference in voltage between 30mA and 1A probably isn't going to be more then half a volt.
I'm realizing how much more accurate the "rule-of-thumb" measurement likely is. I'll have to look at it in a bit more depth to make sure I'm getting it right, but it seems like it automatically will give you the "balance" point, but it won't give you a GENERAL solution, you'd still need to know the Vf for the different currents you'd want to use.

The problem is, to find the voltage across the LED (which you need to know to calculate the current through the resistor) is based on the current going through the LED. So the question needs the solution to answer itself. It's the same issue from a different perspective that I had before. I'm thinking that, maybe, if you know the current you want, and you know the Vf of the LED you're using at that current, if you ASSUME the current through the LED is what you're aiming for at first, and you use that assumption to calculate what the resistor value should be to get that current, then REALIZE that current, it'll actually be at the balance point. Or something like that. Hopefully that makes sense. I'm liking this discussion a lot so far, learning a lot and GAINING perspective.

 

Most small leds will blow up with 1A current ,but there are a lot of white leds out there that 1A is no where near the current they consume. and a difference between 1.5 v and 2.4 is an enormous forward voltage difference,which proves the case for current drive .

Actually the difference between 1A and 20ma is simple design, not Vf. They are actually very close. You will need a heat sink with a 1A, as the wattage a 0.7A LED is 3W, while a 1.4A LED is 6W. These are real world values.

A white LED is a blue LED with a phosphor, unless you are talking RGB, which is three LEDs and a totally different mechanism.

Typical ranges are the same for power LEDs as they are 20ma LEDs, which I describe in the article I presented earlier.

LEDs, 555s, Flashers, and Light Chasers

Somehow I suspect the OP didn't bother to read it, which is a pity.

Here is the approximate Vf of most LEDs:

......... Older Generation ... Newer Generation
Current ....... 10ma ................... 20ma
Red ............ 1.5V .................... 2.5V
Yellow ........ 2.0V .................... 3.0V
Green ......... 2.0V .................... 3.0V
Blue ....................................... 3.5V
White ..................................... 3.5V

Some internal physics mechanisms have changed a little between the older and new generation, as well as the base efficiencies. Nowdays you have watch out, even a 20ma LED can damage your eyes with brightness. Base voltage Vf can vary as much as ±0.5V, though a ±0.3V is more typical.

 

Oh i didnt realise we were going to get into a discussion about white leds not being white ,but various shades of blue depending die and colour temperature . I thought we were talking about voltage vs current .so lets be up front the BEST way to drive leds is with a temperature monitoring current drive,because the two most important criteria are current and temperature ,the voltage ,so long as it is high enough is irrelevant

 

Actually I responded to your contention that a high wattage LED has a different Vf than a low wattage, which is false. The fact is, Vf is a highly individual characteristic dependent on color and manufacturing process, it can be all over the place, but centered on core values (and wattage isn't part of it).

A blue and white LED have the same Vf because they are actually the same kind of LED.

And yes, Vf was part of this conversation. If you go by the data sheet (always a worthy goal) the current is a target value, not the Vf.

 

First off i wish to publish an apology. to Bill for being a bit sharp ,i think we all agree the importance of current drive for leds and the tread is going off the original intent.

 

Agreed.

We often get people who wish to emphasis the parameters that don't matter, and are unpredictable. Odd part is, if you were to ask me where diode resistance does matter I could answer that, I have worked on many RF circuits. It is not theoretical to me, I am a tech with almost 40 years experience. Even then however, the actual resistance is neither speced nor important, usually feed back is heavily involved.

For LEDs Vf and current rating is everything, the other stuff isn't important. Even when calculating wattage it is a matter of real Vf and current, resistance never enters into it.

I have also worked with many LEDs, including power units. I love do nothing circuits (flashy stuff).

 

Just to clear things up, I never meant to imply that you'd ever really need to use effective resistance, just that it was a valid idea, and that I thought it was a cool way to look at the operation of LEDs.

I should draw up a graph and show how the (Vs-Vf)/R = I IF Vf is based on I (Because Vf changes with I) REALLY works. Vf and I are related, and therefor you can't solve this algebraically, BUT, you can still use it. Yeah, that'll be interesting

 

Hello,

The following page from Maxim might interest you:
http://www.maximintegrated.com/app-notes/index.mvp/id/1883
It has some graphs on Vf VS current and brightness VS current.

This is a link from the page on leds from the EDUCYPEDIA:
http://educypedia.karadimov.info/electronics/composemiconductorsled.htm

Bertus

 

Actually the difference between 1A and 20ma is simple design, not Vf.

In effect, a high current LED is a whole lot of small LEDs in parallel fabricated over a larger surface area of the die.

 

Most small leds will blow up with 1A current ,but there are a lot of white leds out there that 1A is no where near the current they consume..

They are designing white LEDs for car headlights that handle a lot more than 1A. Getting the power dissipation out is the problem that has always stopped the use of LEDs for headlights but supposedly they have some that work. Seen them for motorcycle headlight units.... the back is covered with aluminum fins for heat sinking.

 

The problem is, to find the voltage across the LED (which you need to know to calculate the current through the resistor) is based on the current going through the LED. So the question needs the solution to answer itself.

No, it's actually easy. Done it many times when calculating resistors and currents to be used in current mirrors. You have a small resistor in series with each "diode" of the transistor and different currents through each side. You account for temp rise on the higher current side, incresed VBE due to higher current, and the current used for base drive (which is beta dependent and therefore, temp dependent). Just because calculations have more than one variable does not mean they can't be solved.

 

Do you want me to show how easy it is to calculate a current-limiting resistor for an LED?

My red LED has a spec'd forward voltage range of 1.5V to 2.4V at 20mA.
The lowest voltage LED is 1.5V so calculate a resistor to supply it with 27mA.
(9V - 1.5V)/27mA= 277 ohms. 270 ohms is the nearest available value.

What happens when my LED actually has a forward voltage of 2.4V?
Then (9V - 2.4V)/270 ohms= 24mA which looks like the same brightness.

 

The purpose of the op is NOT to get a circuit to work. I think it's about studying the minutia that never becomes significant in a real circuit.

 

To a technician, this is unnatural idea. Many engineers too. A parameter that has no real use, is not stable, is anathema to designing.

 

No, it's actually easy. Done it many times when calculating resistors and currents to be used in current mirrors. You have a small resistor in series with each "diode" of the transistor and different currents through each side. You account for temp rise on the higher current side, incresed VBE due to higher current, and the current used for base drive (which is beta dependent and therefore, temp dependent). Just because calculations have more than one variable does not mean they can't be solved.

I'm just bringing to attention the actual mathematics behind it all, which is fascinating. Given a series resistor, it's actually impossible to solve algebraically to find the Vf across the diode, it's transcendental. Ron H, I believe, is the one who taught me this originally

Let me show you... set I to it's TRUE value, based on characteristics of a diode and Vf.
a*e^(Vf*b)=(Vs-Vf)/R Where a and b are based on characteristics of a diode. e is Eulers constant. Vf is the voltage drop across the diode, Vs is the source voltage, and R is the resistance of the series resistor. Try and solve for Vf. Just try it.

Not saying you couldn't do it more easily with a different method, though. I'm seeing now that you could, and without having to resort to approximations, which is what I'm most happy about at this point.

 

Do you want me to show how easy it is to calculate a current-limiting resistor for an LED?

My red LED has a spec'd forward voltage range of 1.5V to 2.4V at 20mA.
The lowest voltage LED is 1.5V so calculate a resistor to supply it with 27mA.
(9V - 1.5V)/27mA= 277 ohms. 270 ohms is the nearest available value.

What happens when my LED actually has a forward voltage of 2.4V?
Then (9V - 2.4V)/270 ohms= 24mA which looks like the same brightness.

I have a feeling you haven't read the whole discussion in detail, which is fine, because theres a lot to read, but I am well aware of that method, and it's precisely what I use in practice.

I've found now that this is a much better approximation than I previously thought, however, because you can just adjust Vf for the current you're aiming for. So, if you're aiming for 20 mA with a 9V source, and the Vf at that point is 2V, you can do the whole (9-2)/Ω=20mA thing and be good. AND, if you wanted to aim for 5mA instead, you'd find the Vf at that point, say it's 1.9V, you can do the math differently (9-1.9)/Ω=5mA, and you'll be able to calculate the resistor more accurately than you would be able to assuming Vf is constant. It's a small change, but I'm very excited that the rule can be made more "general", because not all diodes are LEDs, and some require a good range of currents and accuracy.