Definition and explanation of forward voltage?

I asked this a while ago and several Flashaholics were kind enough to answer. Trouble is, they assumed elementary circuit knowledge that I simply don't have.

I think the forward voltage symbol is vf, as in voltage forward, or very frustrating? What's significant about that figure, what does it actually tell me in practical terms?

In the case of 5W luxeons I thought it was a measure of efficiency, but that appears to be covered by flux because flux tell you how many lumens per watt.

So if anyone can help me to understand this you will make this electronically challenged Flasholic very happy.

Thanks,
Brightnorm

From a similarly dumbfounded user, this is my take on Vf:

Vf is the voltage needed for an LS (assuming we're talkin' about LSs here) to light up and peform according to spec.

If you're AT Vf, your target current drain will be what is advertised. So when Lumileds says an LS is pulling 350ma at 3.5V, you will get 350ma draw from the batteries if your voltage is at 3.5V...

As far as the effects of increasing or decreasing voltage from that point (point of Vf), I'll leave that up to someone else. I'd venture to guess that if V=Ir, then the lesser you are from Vf results in an increased current flow, and vice versa. But then again, I might not have much of a clue as to what I'm talkin' about...
/ubbthreads/images/graemlins/crazy.gif

Well, that's a good start, but I don't know which is a good, better, best or worse forward voltage. What is it that we're looking for, and why is it good? Does it affect the brightness or burntime of the light?

Brightnorm

In simple terms (the only kind I have a hope of understanding) Vf is that voltage, specific to a LED, where a specified current will pass through the LED. If one asks what is the Vf of a LED, an assumed standard current level is understood or implied. For Lumiled 1 watt LED's this current is 350 mA. For the 5 W it is 700 mA. On a Nichia 5 mm white LED the assumed current is 20 mA. Other current levels can be considered but they should be stated as well. For instance if I want to over drive a 1W Luxeon at 500 mA I could ask what is the Vf of a 1W for 500 mA. In other words, what voltage would need to be applied to the LED for 500 mA to pass through it.

- Don

The Vf does give you an indication of efficiency, in the following manner (assuming my understanding of Lumileds data sheets is correct)

The Vf is the voltage which that particular luxeon needs to have supplied in order to give the luminous flux output which they specify, assuming it is also able to draw the recommended current (350mA for 1W or 700mA for 5W).

Lets say you have 2 Q-binned Luxeons, both of which will produce, say, 30 Lumens when driven to specification.

If the first luxeon has a rated Vf of 3.2V, that means it is rated to produce 30 lumens when supplied with 3.2V and 350mA - this means it requires 1.12W of power (power=voltage x current) to produce 30 lumens output.

If the second luxeon has a rated Vf of 3.6V, and produces the same 30 lumens when supplied with this 3.6V and 350mA, this means that it requires 1.26W of power to produce 30 lumens of output.

Therefore, the first luxeon is more efficient since it requires less power to produce the same output. The factor which tells you of this efficiency is Vf - the lower Vf meaning that the luxeon will operate at a lower voltage with the same amount of current, to produce its rated lumen output.

If you divide the lumen output by the power, then you can get a lumens per watt rating - in the first case, 25.8 lumens/watt, and in the second, 23.8 lumens/watt. This tells you that you are getting more light for the same power supplied to the luxeon.

In terms of effect on brightness and runtime, both of these are affected, depending on how you look at it - in the example above, it has no effect on brightness, since both luxeons are being driven to specification and both produce the same amount of light. However, since the first one requires less power to produce the light, it should have a (slightly) longer runtime, compared to the second one running from the same supply.

In addition, with the lower Vf, this means you could overdrive it by increasing the Vf you supply (and therefore the current) which will produce greater brightness. And since it started at a lower Vf, it means it will be brighter than the second one if driven at the same voltage as it.

Hope that made some sense..

Graham

So Luminous flux depends on vf to reach its stated spec.
I'm definitely starting to get it. It does seem a bit complicated, but the mists are beginning to clear.

Thanks all
Brightnorm

LEDs are diodes.
Diodes are non-linear current devices, the voltage drop across them is not I*R but a function approximating a log curve. For white LS LEDs this means something like this;
for I = 50ma Vf = 2.0V R = 40
100ma 2.8V 28
200ma 3.2V 16
300ma 3.4V 11
Spec 350ma 3.5v 10
450ma 3.6V 8

The Vf of an individual LED is determined by the variables in the manufacturing process and varies in a normal (hopefully) distribution from more than 3V to less than 4V, hence the different xxVx BIN codes.

Does this mean we can supply the LED with a tightly regulated 3.5V and everything is OK? No, we don't live in a perfect world, R is temperature dependent and drops as temp rises. As soon as the 350Ma rises (I squared R heating) we start into thermal runaway. If we limit (regulate) the current we reduce the effects of R's negative temp coefficient and the LED Vf stabilizes at a slightly lower Vf.

We can regulate the LED current by using various methods from as simple as using the battery's internal reistance (Direct Drive), the battery's internal resistance plus an external resistance (Resistored) or Boost voltage/current regulators (BadBoys, etc.) which automaticaly vary the Vf to get the desired current.

There are power limited voltage boosters that limit current by designing them to provide safe current in a range of Vf more or less and adjustable regulated voltage boosters you can "tune" a wider range of Vfs the desired current (MadMax).

If you want to learn from the basic , try here :

Lessons in Electric Circuit - Vol III (Semiconductors) Chapter 3: Diodes and Rectifiers

Perhaps an analogy? Hmmmm, now what?

Perhaps a see-saw (teeter-totter). If the distance walking up the end on the ground is volts and the distance the other end drops is current, than it could work?

As you increase the voltage from nothing (beging walking up the see-saw) nothing happens for a while (ie the volts from 0V seem to do nothing). At a given point as you climb however, the other end begins to drop (the current begins to increase and the LED lights). When the end drops a certain measured distance then you have the 'rated' current' (dictated by the manufacturer). The distance you have walked up at this point is the Vf.

Now what if you take another step? The other end comes crashing down for such a small step! A massive increase in current for small increase in voltage, perhaps damaging the LED. Under these conditions, such as direct driving, the current is only limited by the internal resistance of the power source - perhaps see this as your weight. Some batteries will destroy the LED, others with high internal resistance wont.

Is that imaginative enough?

Chris

Hello there,

Im not sure if this is what you were looking for, but...

Voltage is the driving force that makes the flow of
charge carrying particles (current) begin. Once
this process begins, things begin to happen, such as
something lighting up (bulb or LED).
The current is what actually does stuff, while the
voltage makes sure the current flows.

Without the current, nothing would happen, but without
the voltage, current couldnt flow, so you need both
in most environments.

In any device (LED, bulb, etc.), you need to meet the
voltage spec so that current can flow through it.

With some devices, you can reduce the voltage quite a
bit and still get some reasonable current flow. With
other devices, this isnt true because the relationship
between the voltage across the device and the current
through the device is very unusual (such as an LED or
diode).

In the case of the LED, there is a spec that tells you
what the voltage should be (approximately) in order
to get the kind of current level you would like to achieve
in order to get maximum usage out of the device.

In trying to find out what the relationship between
current and voltage is for a particular device (like
an LED) it's usually a good idea to start by pumping a
small current though the device and then measuring the
voltage across it, and writing these two numbers down.
Then, the current is increased a little and the voltage
is measured again and these two numbers again are written
down. Continually increasing the current and making
measurements and logging the results until the max current
is reached forms a profile of the devices' static
voltage/current relationship.


Any questions? ;-)

Good luck with your LED circuits,
Al

[ QUOTE ]
vicbin said:
If you want to learn from the basic , try here :

Lessons in Electric Circuit - Vol III (Semiconductors) Chapter 3: Diodes and Rectifiers



[/ QUOTE ]

BEST. LINK. EVER.

Back to lurk mode...

Keep in mind that LEDs are *horrible* diodes as diodes go. They have really bad curves and very bad part-to-part variability. This is one of the reasons that 'direct drive' designs suffer from a lot of unpredictability and unstability.

Oh, and the I-V characteristic of a diode in forward bias is an exponential curve, no log. Same thing, flipped about the X-Y axis, so no blood no foul. /ubbthreads/images/graemlins/smile.gif

Sorry for the repost to the very old thread, but the link above is no longer valid.

Here's the current one as of today.

http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/SEMI_3.html

Willmore said:

Keep in mind that LEDs are *horrible* diodes as diodes go. They have really bad curves and very bad part-to-part variability. This is one of the reasons that 'direct drive' designs suffer from a lot of unpredictability and unstability.

Oh, and the I-V characteristic of a diode in forward bias is an exponential curve, no log. Same thing, flipped about the X-Y axis, so no blood no foul. /ubbthreads/images/graemlins/smile.gif

Click to expand...

That was my impression.

Geez, they sound as bad as the old electron beam pentode tubes. Is there anyone else here who even remembers those?

magellan said:

That was my impression.

Geez, they sound as bad as the old electron beam pentode tubes. Is there anyone else here who even remembers those?

Click to expand...

I don't think that I have any of the ones you reference, but I have some nice valves.

Chris

LOL

Nice. I'm impressed.

And then there's the good ol' silicon-controlled rectum finder for those who can't find their *** with both hands.

marklein said:

[ QUOTE ]
vicbin said:
If you want to learn from the basic , try here :

Lessons in Electric Circuit - Vol III (Semiconductors) Chapter 3: Diodes and Rectifiers



[/ QUOTE ]

BEST. LINK. EVER.

Back to lurk mode...

Click to expand...

It says that page is no more.

Although several dozen lines down it says "Defcon goons dog balls niner."

Well, all I can say is 10-4, Defcon goons, this is cat scrotum six-niner, over and out!

We got ourselves a convoy!

I'm dating myself again.

magellan said:

It says that page is no more.

Although several dozen lines down it says "Defcon goons dog balls niner."

Well, all I can say is 10-4, Defcon goons, this is cat scrotum six-niner, over and out!

We got ourselves a convoy!

I'm dating myself again.

Click to expand...

Here's the current link:

http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/SEMI_3.html

Will we ever see reverse tunnel effect LED?
Such one would not require driver, just couple of passives