If you look at my previous posts, I pointed out that I was assuming constant temperature. I am aware that, in practice, it could be dangerous due to the heavy reliance of Vf on temperature, but beyond that everything I have said, I think you'll find, is correct.

You will also find it's useless. The Vf of a diode (from which resistance would be calculated):

1) varies widely from diode to diode
2) varies with current
3) varies with junction temp

Using it to make any kind of design calculation is pure lunacy. It's a useless parameter. And, as stated, driving from any voltage source will almost guarantee device failure due to thermal runaway. No upside there.

 

Nothing that has been stated here I really have a problem with. You guys can't seem to see the EXACT same thing in another light. Neither of us are wrong, but I can reach the same conclusion multiple ways. The voltage across a diode can be calculated given the current (and other parameters), meaning the opposite is also true. That is, current through a diode can be calculated given the voltage (and other parameters).

The only thing I've tried stating this entire time is that viewing LEDs as voltage-controlled resisters is a valid way to understand them, because it mimics how they work physically (the depletion region gets thinner as the voltage increases). Vf doesn't even truly exist, it's just a chosen point where current increases very quickly, that's all that I've been trying to explain. Not once have I made the point that LEDs are better off being voltage controlled, only that they, in the end, can be thought of as such. All components can be. Voltage and current are directly related, if you know one you can figure the other, period. If you had a precise temperature sensor and voltage source, you could run an LED off of a pure voltage source.

If I must, I'll describe what I've been explaining in more detail. It really is interesting and eye-opening, I just really want you guys to see that.

Besides, we're not TOTALLY off topic yet. Here's a question: LEDs are temperature-dependent, and at certain temperatures they draw more current with less voltage. Because their brightness seems to be based purely on current, are LEDs then more efficient at different temperatures?

 

Also, I don't think Vf changes with current, at least not directly. Vf is just the point when the diode begins to conduct more and more very rapidly with small increases in voltage. in other words, even with a large range of currents, the voltage drop across the diode will change little, so it will always stay pretty close to Vf.

 

Nothing that has been stated here I really have a problem with. You guys can't seem to see the EXACT same thing in another light. Neither of us are wrong, but I can reach the same conclusion multiple ways. The voltage across a diode can be calculated given the current (and other parameters), meaning the opposite is also true. That is, current through a diode can be calculated given the voltage (and other parameters).

The only thing I've tried stating this entire time is that viewing LEDs as voltage-controlled resisters is a valid way to understand them, because it mimics how they work physically (the depletion region gets thinner as the voltage increases). Vf doesn't even truly exist, it's just a chosen point where current increases very quickly, that's all that I've been trying to explain. Not once have I made the point that LEDs are better off being voltage controlled, only that they, in the end, can be thought of as such. All components can be. Voltage and current are directly related, if you know one you can figure the other, period. If you had a precise temperature sensor and voltage source, you could run an LED off of a pure voltage source.

If I must, I'll describe what I've been explaining in more detail. It really is interesting and eye-opening, I just really want you guys to see that.

Besides, we're not TOTALLY off topic yet. Here's a question: LEDs are temperature-dependent, and at certain temperatures they draw more current with less voltage. Because their brightness seems to be based purely on current, are LEDs then more efficient at different temperatures?

Ref bold quote - but if you adjusted the voltage source value based upon info from the temperature sensor, i.e. a closed loop system, you are attempting to achieve the same thing as using a current drive network, voltage source plus series resistor, or true current source.

An LED is not voltage driven for the thermal runaway reasons discussed by myself & others who are well informed of LED properties. If we monitor the temp, then adjust the drive voltage accordingly, that is not true voltage drive. With a current source, no temp sensing is needed. As the device heats up, thermal stability is already achieved through its natural modus operandi. A voltage driven network, however, must be modified by using additional info as a feedback mechanism, i.e. temp sensing.

Although a voltage source driving the LED might be feasible if the temp sensor & feedback were properly implemented, does it not make sense to use a current drive or its cousin, voltage drive with series resistor?

Of course I & V are inter-related, if you know 1 you can compute the other. I feel everybody here already knows that. Also, the light output of an LED varies with current in a linear sense approx., but it is true to say that it also varies w/ voltage as well in an exponential sense approx.

We say that LED lamps are current controlled because that is their natural modus operandi. To use a voltage source sans series resistance is driving them against their natural mode. Without additional feedback to mitigate thermal runaway, the device is doomed. You seem to expend a lot of effort conveying to us that voltage is involved with LED's & I say of course it is. It just happens that V is not the parameter we drive, I is. We control I, & allow V to be indirectly determined. The fact that for a given I there has to be a corresponding V is so trivial we just treat it as understood.

So when I say we must current drive an LED, I imply w/o explicitly stating such, that the current source or network driving the LED must have enough voltage output capability to meet the forward voltage drop of the devices driven. If we are driving a string of 5 LED's at 1.80V forward drop each, the current drive network must be able to output 9.0 volts.

Let's not make too much ado over nothing. BR.

Claude

 

"We control I, & allow V to be indirectly determined."

YES! Yeah, Finally. This is what I was trying to say. You're wanting to keep I at a certain level, and, knowingly or not, you ARE controlling the input voltage to the LED to do so when you use a series resistor.

Here's an intuitive understanding of why/how:

http://upload.wikimedia.org/wikipedia/en/thumb/e/ed/Load_line_diode.png/300px-Load_line_diode.png

Where the two lines intersect is the operating point. The horizontal axis is the voltage across the diode. Notice, that if you change the input voltage, you're essentially pulling the point on the y-axis, where the resistor line intersects, up and down, which has an effect on the operating point and the voltage across the diode, but not as much as it would be without the resistor (even if you don't put in your own resistor, this rule applies, but because there's VERY low resistance, the load line of the resistor does little to affect the operating point). When temperature changes, you're essentially shifting the diode line left and right, and, notice, when you do so, because of the resistor, the current isn't allowed to change as much as it would otherwise, AND, the voltage across the diode changes to help keep the current stable.
You can also use this to see how the resistor essentially regulates how much voltage it drops to help keep I stable.

Hopefully this makes a lot more sense now!

 

"We control I, & allow V to be indirectly determined."

YES! Yeah, Finally. This is what I was trying to say. You're wanting to keep I at a certain level, and, knowingly or not, you ARE controlling the input voltage to the LED to do so when you use a series resistor.

Here's an intuitive understanding of why/how:

http://upload.wikimedia.org/wikipedia/en/thumb/e/ed/Load_line_diode.png/300px-Load_line_diode.png

Where the two lines intersect is the operating point. The horizontal axis is the voltage across the diode. Notice, that if you change the input voltage, you're essentially pulling the point on the y-axis, where the resistor line intersects, up and down, which has an effect on the operating point and the voltage across the diode, but not as much as it would be without the resistor (even if you don't put in your own resistor, this rule applies, but because there's VERY low resistance, the load line of the resistor does little to affect the operating point). When temperature changes, you're essentially shifting the diode line left and right, and, notice, when you do so, because of the resistor, the current isn't allowed to change as much as it would otherwise, AND, the voltage across the diode changes to help keep the current stable.
You can also use this to see how the resistor essentially regulates how much voltage it drops to help keep I stable.

Hopefully this makes a lot more sense now!

Once again Austin, when I say "current controlled" or "current driven" it is understood that voltage is eventually adjusted to maintain the target current. The textbook definition of a current source is one that outputs whatever voltage is needed in order to maintain a fixed current. With an LED current drive buck converter, I maintain steady current in the LED lamp string. If the temp should suddenly increase, let's say an internal part got warm, i.e. a motor was turned on, the junction temp of the LED's increases. The Vf will momentarily drop. But the inductor will maintain steady current through the LED. When the switch cycle ends the inductor will have more energy left than it did before the temp increase. This results in a slight increase in current on the ensuing power cycles. The feedback etects this current increase & reduces duty factor accordingly. The lower Vf means that the duty factor will decrease to maintain equilibrium.

Again, it is trivial to point this out. When we say "current driven" all of the other activity that goes along with it are understood. We don't need to say that the voltage is important & that the control network influences and/or is influenced by said voltage. We know that.

We are not, however, "controlling the LED voltage to be a certain level". We are controlling LED current, but allowing the voltage to be indirectly determined, there is a difference. In controlling LED If, the Vf assumes whatever value corresponds to If per the diode characteristic curve. We do not control Vf, rather physical conditions do. We control If.

Make sense? Cheers.

Claude

 

How will you hold the temperature of the self-heating LED constant? Very difficult.
Current causes it to heat up which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more....

Then when connected to a constant voltage it gets hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and....

 

How will you hold the temperature of the self-heating LED constant? Very difficult.
Current causes it to heat up which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more....

Then when connected to a constant voltage it gets hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and....

Good point. Although external temp can be sensed, it is the internal temp of the junction that determines the Is value. Although what he proposes might be feasible, it is grossly involved. With current drive on the other hand, it is ultra simple & ultra stable thermally. Why even mess with trying to implement a servo control of temp? Current drive is the natural modus operandi of any forward biased p-n junction. So LED's, SCR's, bjt's, etc., are always current driven. It sure makes sense. Thanks for your input, you post good info.

Claude

 

How will you hold the temperature of the self-heating LED constant? Very difficult.
Current causes it to heat up which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more current which cause a higher temperature which causes more....

Then when connected to a constant voltage it gets hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and hotter and....

Yes, I understand this, but you're missing what my point was, which was that the diode doesn't know the difference between a pure voltage source and a voltage coming in through a resistor after it's dropped some of the voltage.

 

Here's a question: LEDs are temperature-dependent, and at certain temperatures they draw more current with less voltage. Because their brightness seems to be based purely on current, are LEDs then more efficient at different temperatures?

Because our vision's response to brightness is logarithmic so we can see a wide range from moonlight to sunlight (disregarding the extra range from our iris) an LED with double or half the current looks almost the same brightness.
The temperature affect is small unless there is thermal runaway.

 

Once again Austin, when I say "current controlled" or "current driven" it is understood that voltage is eventually adjusted to maintain the target current. The textbook definition of a current source is one that outputs whatever voltage is needed in order to maintain a fixed current. With an LED current drive buck converter, I maintain steady current in the LED lamp string. If the temp should suddenly increase, let's say an internal part got warm, i.e. a motor was turned on, the junction temp of the LED's increases. The Vf will momentarily drop. But the inductor will maintain steady current through the LED. When the switch cycle ends the inductor will have more energy left than it did before the temp increase. This results in a slight increase in current on the ensuing power cycles. The feedback etects this current increase & reduces duty factor accordingly. The lower Vf means that the duty factor will decrease to maintain equilibrium.

Again, it is trivial to point this out. When we say "current driven" all of the other activity that goes along with it are understood. We don't need to say that the voltage is important & that the control network influences and/or is influenced by said voltage. We know that.

We are not, however, "controlling the LED voltage to be a certain level". We are controlling LED current, but allowing the voltage to be indirectly determined, there is a difference. In controlling LED If, the Vf assumes whatever value corresponds to If per the diode characteristic curve. We do not control Vf, rather physical conditions do. We control If.

Make sense? Cheers.

Claude

It is definitely assumed to be known, but when people started throwing a hissy when I mentioned that, in the end, you're still in some sense controlling voltage, I had to assume otherwise.

In the end, we're controlling current in a relatively simple way and controlling voltage in a relatively complex way when we use a series resistor. Like I said before, it's semantics, because both are directly related, you can look at the problem either way and get identical results in the end. I just like the voltage-controlled model in theory because it fits the physical model better, and is interesting mathematically. Varistors, which act like diodes, are considered voltage-controlled resistors, so it's not like I'm alone with this idea. http://en.wikipedia.org/wiki/Varistor

We all agree, we just haven't figured that out yet

EDIT: "You can not force the voltage across a LED to change, except with things like temperature." - Bill Marsden
Well, almost. XD
With more current across an LED, the voltage across the LED changes, though slightly, and not only because of temperature. It's obvious when you look at the current VS voltage graph of an LED to see that.

 

Because our vision's response to brightness is logarithmic so we can see a wide range from moonlight to sunlight (disregarding the extra range from our iris) an LED with double or half the current looks almost the same brightness.
The temperature affect is small unless there is thermal runaway.

Didn't really answer my question, but thanks. I already knew about the logarithmic nature of our eyes, I had suspected that to be the case in my original post

If an LED was, say, really cold and conducted a lot more current at a lower voltage, even though the overall power would be reduced, would the light output be the same? That is, would you be converting more of the actual power into light?

EDIT: This is similar to one of my original questions, which was if LEDs with naturally lower Vf (Like red LEDs), were more efficient than others with higher Vf.

 

It is definitely assumed to be known, but when people started throwing a hissy when I mentioned that, in the end, you're still in some sense controlling voltage, I had to assume otherwise.

In the end, we're controlling current in a relatively simple way and controlling voltage in a relatively complex way when we use a series resistor. Like I said before, it's semantics, because both are directly related, you can look at the problem either way and get identical results in the end. I just like the voltage-controlled model in theory because it fits the physical model better, and is interesting mathematically. Varistors, which act like diodes, are considered voltage-controlled resistors, so it's not like I'm alone with this idea. http://en.wikipedia.org/wiki/Varistor

We all agree, we just haven't figured that out yet

Ref bold quote - a small but very vocal minority has this view. We know that I & V cannot exist independently except static cases, & that both are involved & inter-related. But some devices due to their inherent nature, are much more amenable to being current driven, while others are best suited to be voltage driven. With either type of device, the variable not being directly driven is still involved, & it is controlled but in an indirect sense.

For some reason, you & others just have a natural propensity to focus on the voltage as primary, & current as secondary, despite the fact that current is the directly driven variable. As far as the physical model goes, the mere fact that voltage is present does not make it the main player & likewise for current. For LED emission, the physical model is that of electrons colliding with lattice ions, resulting in electrons transitioning from the conduction band (high energy) to the valence band (lower energy).

The conservation of energy results in energy emission in the form of photons & hence light is generated. Increasing the current is basically increasing the number of charge carriers per second & hence increasing the number of collisions, which increases the number of emitted photons. There is a strong physical relationship between forward current & photon emission.

As current increases, recombination requires a finite time to happen so that more carriers accumulate along each edge of the depletion zone & forward voltage drop increases since the barrier has been raised due to additional charge accumulation. But this raised barrier does not produce increased photon emission, the increased current does.

Frankly I cannot fathom how anyone can question the fact that LED's are current controlled light sources. Not only must they be current driven for thermal stability reasons, they also output photons in proportion to current value. That sure is current controlled to me.

Current conforms to the LED physical model MUCH better than that of voltage. Voltage is incidental, a result of charge accumulation along depletion zone edge due to collisions w/ lattice structure. Finite recombination time results in voltage changing with current changes. But photonic emission is controlled exclusively by current. Semiconductor literature detail this. BR.

Claude

 

Ref bold quote - a small but very vocal minority has this view. We know that I & V cannot exist independently except static cases, & that both are involved & inter-related. But some devices due to their inherent nature, are much more amenable to being current driven, while others are best suited to be voltage driven. With either type of device, the variable not being directly driven is still involved, & it is controlled but in an indirect sense.

For some reason, you & others just have a natural propensity to focus on the voltage as primary, & current as secondary, despite the fact that current is the directly driven variable. As far as the physical model goes, the mere fact that voltage is present does not make it the main player & likewise for current. For LED emission, the physical model is that of electrons colliding with lattice ions, resulting in electrons transitioning from the conduction band (high energy) to the valence band (lower energy).

The conservation of energy results in energy emission in the form of photons & hence light is generated. Increasing the current is basically increasing the number of charge carriers per second & hence increasing the number of collisions, which increases the number of emitted photons. There is a strong physical relationship between forward current & photon emission.

As current increases, recombination requires a finite time to happen so that more carriers accumulate along each edge of the depletion zone & forward voltage drop increases since the barrier has been raised due to additional charge accumulation. But this raised barrier does not produce increased photon emission, the increased current does.

Frankly I cannot fathom how anyone can question the fact that LED's are current controlled light sources. Not only must they be current driven for thermal stability reasons, they also output photons in proportion to current value. That sure is current controlled to me.

Current conforms to the LED physical model MUCH better than that of voltage. Voltage is incidental, a result of charge accumulation along depletion zone edge due to collisions w/ lattice structure. Finite recombination time results in voltage changing with current changes. But photonic emission is controlled exclusively by current. Semiconductor literature detail this. BR.

Claude

There's no difference in either model. Neither is better than the other. They're just inverses. I can understand the model you're describing better now, and I'm glad you gave me such good info, but it's still really irrelevant. When designing, you usually have a voltage source, not a current source, so it's easier for me to think in those terms.
In other words, I'm usually given voltage, and have to find the current, not the other way around.

 

Just out of curiosity though, how would YOU calculate/describe the current going through an LED circuit with a series resistor given a specific source voltage? Not using the approximation, either. It's, of course, approximately going to be (Vs-Vf)/R, but that's not going to be 100% accurate, because that assumes an ideal diode. In my mind, I think of it's "effective resistance" decreasing with voltage, and I use that idea to understand it and calculate it, but how would you?

 

Also, I don't think Vf changes with current, at least not directly.

Wrong again. VF (the forward voltage drop of a diode) does indeed change with current as given by:

VF change = kT/q ln (I1/I2)

 

Good point. Although external temp can be sensed, it is the internal temp of the junction that determines the Is value.

Claude

It is impossible to get an accurate die temp unless there is some kind of calibrated diode designed onto the die for the purpose.

 

Just out of curiosity though, how would YOU calculate/describe the current going through an LED circuit with a series resistor given a specific source voltage? Not using the approximation, either. It's, of course, approximately going to be (Vs-Vf)/R, but that's not going to be 100% accurate, because that assumes an ideal diode. In my mind, I think of it's "effective resistance" decreasing with voltage, and I use that idea to understand it and calculate it, but how would you?

Assuming the drop across the resistor is several V or more, the diode voltage changes will be less than 1% error induced into resistor current.

But to answer your question, you would include the delta VF term I posted above to account for the microscopic change in VF due to current change caused by the voltage change across the resistor:

delta VF = kT/q ln (I1/I2) usually assumed to be = .026 ln (I1/I2) in Volts

You could also include a term for current change if you know the power dissipation, thermal resistance, etc as:

delta VF approximately -2mV/C

which is another vanishingly small term compared to the drop across the resistor.

 

Basically, when an LED is fed a current from a series resistor it works perfectly.
But if it is fed a voltage then either it doesn't light or it blows up.

Why not do it correctly?

 

Basically, when an LED is fed a current from a series resistor it works perfectly.
But if it is fed a voltage then either it doesn't light or it blows up.

Why not do it correctly?

I do do it correctly, I've mentioned that several times. Right now we're just talking about the validity of each models.