And Austin, I'd like to say that we've said that a dozen times already. Like I said before, this is much ado over nothing. I always drive an LED string (the number of lamps in the string may be just 1, or 3, or 6, or more) with a constant current source, or a voltage source plus an adequate series resistor.

In addition to computing the forward current in the string, I always compute the forward voltage as well. It's all important to know this. If I'm using a +12V source with series resistor, or current mirror/source/sink, to drive a string of 6 lamps, I need the Vf parameter. If the lamps are infrared w/ 1.50V maximum Vf, I can make it with a current source or current drive buck converter since 6 X 1.5V = 9V < 12V supply. But if the lamps are amber w/ a Vf of 2.2V, the string of 6 drops 13.2V, & forget it.

Nobody here is unaware of that. I'll bet every person in this thread that debated you is well aware of the importance of Vf & that it is intimately related to If. We live in an age where education is attainable by vast numbers of people from all backgrounds. This isn't 1840 when only kids from a well to do family could go to college. People dirt poor attended college with me back in the 1970's, some graduated & got good jobs.

Unless one is employed doing leading edge research, it is not likely that what that person knows is limited to just him. I worked for 11 years in currency validation. When you insert a $1 or $5 dollar bill into a vending machine, I developed the hardware that optically & magnetically scanned the currency, determined authenticity, & denomination, up to $100 bills, as well as 90 or so different countries.

My work put me in a position where I had nobody to lean on for help. My rivals who made this same kind of product did not & still do not publish their methods & neither did we. It's a trade secret. So I had to spend countless hours working in the lab studying money properties, scanning them optically & magnetically, & observing which parameters are worth exploring & which are not.

A lot of conventional wisdom assumptions are proven wrong. My point is that because I worked in an area that is trade secret protected, I acquired knowledge that few people in the world possess. When I left the company I was informed that if I go to work for a rival, the use of anything I learned there could result in a law suit. All departing personnel were issued this notice.

Anyway, info such as that is something I have that few others have in this world, & I am honor bound not to divulge it, as well as legally liable. But info such as how LED's work, how they best are driven, etc., is hardly a secret, just like Ohm's law, Shockley's diode equation, etc. I cannot understand how anyone would lecture people who have years of experience practicing the art, as well as lots of formal education in the same, thinking that they are telling us something new.

I'm sure you mean well, but believe me when I say that this is nothing new to us. Those who posted on this thread obviously know how to drive an LED(s) & they are certainly aware that If & Vf are inter-related in a strong sense. This stuff is taught at the undergrad level. The number of people w/ college degrees these days is staggering. I read that 90% of all scientists/engineers who ever lived, are alive today! That tells us something.

To get on a platform & lecture on electronics requires a Ph.D., or at the very least an MS plus lots of experience in the art. I've been doing this 34 yrs., in my dissertation stage of Ph.D., & most of what I post is well known already. When I graduated with the BE in the late 70's, I remember being told that holding a BE (or BS) just doesn't have the power it had a generation earlier, that an MS is the minimum if one wishes to stand out.

How true then, even more so now. Please reread your posting comments & you should see what I mean. Your points made are very well known to any practitioner with a BS (or BE) in full EE, & 5 or more years practicing actual hardware design. LED lamps are not a mystery, they weren't so even in the 70's when I studied EE. The diode properties & If/Vf relations, were old news in the 70's. I learned the way to drive a p-n junction around 1975 or so in electronics lab, 1st course. The Vf/If puzzled me during 1 week of the course, then I learned it, & have had no doubts since.

Sorry to run on, BR.

Claude

Very interesting post. I didn't mean to seem to make claims to know something new. If everyone here really is so familiar with diodes, which I'm sure they are, then why are they having such a hard time understanding the basic principles behind what I've been saying? Am I not being clear enough? Or are they just upset that I posted on a primarily educational forum, because I could potentially confuse noobies? We all agree on the same things, but there's still conflict, so obviously theres some sort of misconception, or something, keeping everyone from standing on common ground.

 

We are not having problems, we feel you are just so far off base the theory is in left field.

This basic assumption that you somehow are grasping that which we are not is the arrogance Cabraham is talking about. We understand what you are saying, and are rejecting it as futile. It does not follow, it has no real application. Or, like I said earlier, it is a distraction.

There are applications for the concept, but not with LEDs. Even then, the internal resistance of a diode is not a given parameter, because it is not the same between two virtually identical parts. Instead, we start from Vf and go from there. If it must be modeled, and I have never had to in almost 40 years of experience, it can be done on the fly.

The real problem is the nonlinear characteristics of the resistance. It tends to introduce distortion and harmonics. This is good for frequency multipliers, not so much for anything else. I have seen it used (rarely) for AGC circuits, but not often.

You really need to ask yourself why so many people are spending time on this thread arguing the point, instead of assuming we don't get it.

This is one area where electronics is actually very simple. LEDs are easy, it you let them be. We have had similar arguments with transistors, where there really was a case for the voltage model. There are even practical applications for the model, as in current mirrors. Irony or what?

 

We are not having problems, we feel you are just so far off base the theory is in left field.

This basic assumption that you somehow are grasping that which we are not is the arrogance Cabraham is talking about. We understand what you are saying, and are rejecting it as futile. It does not follow, it has no real application. Or, like I said earlier, it is a distraction.

There are applications for the concept, but not with LEDs. Even then, the internal resistance of a diode is not a given parameter, because it is not the same between two virtually identical parts. Instead, we start from Vf and go from there. If it must be modeled, and I have never had to in almost 40 years of experience, it can be done on the fly.

The real problem is the nonlinear characteristics of the resistance. It tends to introduce distortion and harmonics. This is good for frequency multipliers, not so much for anything else. I have seen it used (rarely) for AGC circuits, but not often.

You really need to ask yourself why so many people are spending time on this thread arguing the point, instead of assuming we don't get it.

This is one area where electronics is actually very simple. LEDs are easy, it you let them be. We have had similar arguments with transistors, where there really was a case for the voltage model. There are even practical applications for the model, as in current mirrors. Irony or what?

I gotcha now, I really do. You're not so much arguing against the concept, but "what's the point?". I'm fine if I look crazy caring about these types of curiosities, but not fine looking crazy like I don't have a clue, ha!

I've made an attachment, not because I don't think you guys know yet what I've been saying, but because maybe it'll peak some interest.

EDIT: The image shows two load lines, equal resistance but with different source voltage. It clearly shows how, in essence, the resistor drops a large portion of the increase in source voltage, leaving only a small bit for the LED, meaning a small rise in current even given a large change in source voltage, because the resistor drops a majority of the "excess" voltage increase.

 

And there is where we are truly breaking down. Gotcha?

As I have said, this is a teaching site. You started this as if you wanted to learn something new about LEDs. You didn't.

 

And there is where we are truly breaking down. Gotcha?

As I have said, this is a teaching site. You started this as if you wanted to learn something new about LEDs. You didn't.

Actually, it's all I wanted to do. I'm not the one who threw a fit about this whole irrelevant thing. I, literally, said something basically off-the-cuff, didn't expect any retaliation, I figured it was so obvious.

 

I'm not sure if this helps, but when I was looking at "brightness" and considering altering the brightness of LEDs, I knew that there had to be sufficient voltage across the LED to drive it. However, it was then up to current to affect intensity. For this I referred to the data sheet curve for the LEDs I bought (see below).



You'll see that between 1 and 10mA there is a pretty steep increase curve in luminous intensity. After that it tends to flatten out quick quickly. Hence in the end I opted to go to around 12 to 15mA maximum in my design as the intensity difference wasn't noticeable enough.

Oops. I read the first page of this thread and was answering it. Didn't notice the other 9 pages! I think I better put my glasses back on. Sorry chaps.

 

Oops. I read the first page of this thread and was answering it. Didn't notice the other 9 pages! I think I better put my glasses back on. Sorry chaps.

Non-sense! Very helpful post. Thank you very much
It gives me a solid bit of evidence to support some of my original "hunches".

 

Chris, Chris, Chris.
The last graph you posted shows Luminous Intensity vs current and shows the LED or your vision being overloaded above 2mA. I have never seen an LED so bad. Maybe it shows what happens when your iris is closing with bright continuous light.

Your graph does not show you that the perceived brightness is doubled when the current is 9 or 10 times higher.

The graph in the datasheet of an ordinary red 5mm LED I have used for years has its Luminous Intensity almost linear to changes in current. But it does not show how bright it looks.

 

Chris, Chris, Chris.
The last graph you posted shows Luminous Intensity vs current and shows the LED or your vision being overloaded above 2mA. I have never seen an LED so bad. Maybe it shows what happens when your iris is closing with bright continuous light.

Your graph does not show you that the perceived brightness is doubled when the current is 9 or 10 times higher.

The graph in the datasheet of an ordinary red 5mm LED I have used for years has its Luminous Intensity almost linear to changes in current. But it does not show how bright it looks.

Your graph is normalized, where 20mA = 1
It shows RELATIVE intensity with varying current.
Also, Chris's graph covers a larger range.

 

Here is a photo of some LEDs with the current increasing from 6mA to 24mA.
12mA is only slightly brighter than 6mA. 24mA is only slightly brighter than 12mA.
There is not much difference between 6mA and 24mA but the current is quadrupled.

 

Here is a photo of some LEDs with the current increasing from 6mA to 24mA.
12mA is only slightly brighter than 6mA. 24mA is only slightly brighter than 12mA.
There is not much difference between 6mA and 24mA but the current is quadrupled.

If the detector of light is the human eyeball, of course the difference between 6 & 24 mA is not that great. But occasionally the detector is a photodiode, CCD, etc. where the photocurrent induced due to the incident light would increase by 4 times since these sensors are not logarithmic, but linear. I just thought I'd mention it since there are uses for LEDs other than display. Of course for IR LEDs, the detector is not the eyeball. BR.

Claude