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Topic: Forward voltage vs Current (Read 19230 times) previous topic - next topic

scswift

Apr 02, 2011, 01:27 pm Last Edit: Apr 02, 2011, 03:22 pm by scswift Reason: 1
Hey guys,

I was looking at the datasheet for this LED and I noticed it says the maximum forward voltage is 5.2v, which is more than the 5v I'd planned to supply the leds with.  The typical forward voltage is 4.3, so if I did use 5v, I could assume most of the LEDs I bought would work, but unless I don't understand forward voltage properly, some would not.

http://www.vishay.com/docs/81260/vlww9900.pdf

I did however notice a section of the datasheet where it graphs forward voltage vs current.  And I was wondering if someone could explain to me what that means.

I know that if I have a 5v source, and I use a 15ohm resistor, I'll put 50mA through the LED if the forward voltage is 4.3.  And I know how to calculate that with Ohm's law.  But that's the extent of my knowledge regarding that.

I guess what I'm wondering is... If I do get an LED with a forward voltage of greater than 5v, is that chart telling me how I should adjust the current so it will still function, or am I just screwed at that point and will have to test all my LEDS before I solder them into my circuit if I choose to go with this LED, and a 5v supply?  I'm guessing it's the latter.  But I would still like to understand what that graph is trying to tell me.





madworm

What ultimately kills the LED if it is overdriven is the heat it can't get rid of. As usual it is (forward voltage * current). As I is a function of Vf and vice versa, it is sufficient to make sure that the current won't exceed the maximum permissible value, in this case 50mA. Here this can be accomplished by using a suitable resistor (ignoring variations of temperature in the LED).

Depending upon your requirements, the variations in the "typical parameters" of this LED may or may not affect you (uniformity in brightness for a display...). Personally I have not had any serious issues with single color LEDs in that respect though, but with RGB matrices and color balance. The worst difference in Vf I had to deal with in a single batch of cheap LEDs was about 0.1V, but I use LED drivers, so it doesn't matter much.

I doubt you can actually buy LEDs that would have values such as: 4.3, 5.0, 4.1 ... in the same batch. These would come from a pretty shitty factory. I don't even know if it's possible to produce LEDs with such a deviation and call it a 'working process'.

scswift

Quote
What ultimately kills the LED if it is overdriven is the heat it can't get rid of. As usual it is (forward voltage * current). As I is a function of Vf and vice versa, it is sufficient to make sure that the current won't exceed the maximum permissible value, in this case 50mA. Here this can be accomplished by using a suitable resistor (ignoring variations of temperature in the LED).


I understand that heat kills an LED, and that the absolute maximum rating here is 50mA.  But I still don't understand the purpouse of the graph.

If the graph were showing me how much current I could put through the LED if it had different forward voltages, then wouldn't the current be inversely proportional to the forward voltage, rather than proportional to it?

In other words, let's say you had two leds, one with a Fv of 4.3v, and one with a Fv of 4.7v.  4.3*50mA is less than 4.7*50mA.  So one would think the led with the higher forward voltage would heat up faster.  But this graph seems to show the reverse relationship.  That an LED with a Fv of 4.3v is generating more heat than one with a Fv of 4.7v for the same mA.  How can that be?

tomm

Surely heat is directly related to current, so if you're putting the same current through each it doesn't matter?

The graph just shows how much current would flow through the LED if directly connected to that voltage, if you're using current limiting resistors, then the graph doesn't make much sense.

scswift

Actually I found this page which I think answers my question:
http://www.maxim-ic.com/app-notes/index.mvp/id/3070

"For a standard LED of 5mm diameter, Figure 1 shows the forward voltage (VF) vs. forward current (IF). Note that the voltage drop across an LED increases with forward current."

So I think what the graph is showing is how the forward voltage changes as I put more current through the LED.

With that in mind, I think that means if I reduce the current I'm trying to put through the LED, the forward voltage will drop.  So if these LEDs can have a max forward voltage of 5.3v at 50mA, then if I reduce that to say 40mA, I will have less of a chance of getting a batch of LEDs that simply won't work at 5v.

Plus I've been told I should stay 80% below absolute maximum ratings anyway, so 40mA would not be unreasonable.

scswift

#5
Apr 02, 2011, 07:43 pm Last Edit: Apr 02, 2011, 07:45 pm by scswift Reason: 1
Quote
Surely heat is directly related to current, so if you're putting the same current through each it doesn't matter?


I'm relatively new to this whole circuit design thing, but I'm pretty sure that current isn't the only thing that matters.  What matters, I believe, is the power which you can see here is calculated with voltage x current:
http://www.the12volt.com/ohm/ohmslawcalculators.asp

To use a water analogy, let's say you have two pipes going to a pool, and the diameter of the pipe is the current, and the pressure which the water is under is the voltage.  If both pipes are 3" in diameter, but one has twice the pressure, then you can see there's a lot more energy there and it will fill your pool twice as fast.

So to go back to the LED example, the led with the lower forward voltage is using up more of the voltage inside it to generate light and heat, so it should get a little hotter than the one with the higher forward voltage.

floresta

#6
Apr 02, 2011, 09:21 pm Last Edit: Apr 02, 2011, 09:33 pm by floresta Reason: 1
Quote
I'm relatively new to this whole circuit design thing, but I'm pretty sure that current isn't the only thing that matters.
As far as LEDs are concerned current is the only thing that matters.  The LED 'forward voltage' is not something that you apply, it is a result of the LED characteristics and the current.  As long as the current is within the LED ratings the forward voltage and the resultant power dissipation will also be within the LED ratings.  You can't use an Ohm's law calculator for an LED because an LED is not a resistor (it is non-linear) and Ohm's law does not apply.  That's why you need a graph to get the relationship between current and forward voltage.

Quote
"For a standard LED of 5mm diameter, Figure 1 shows the forward voltage (VF) vs. forward current (IF). Note that the voltage drop across an LED increases with forward current."

So I think what the graph is showing is how the forward voltage changes as I put more current through the LED.

With that in mind, I think that means if I reduce the current I'm trying to put through the LED, the forward voltage will drop.  So if these LEDs can have a max forward voltage of 5.3v at 50mA, then if I reduce that to say 40mA, I will have less of a chance of getting a batch of LEDs that simply won't work at 5v.
You interpreted this part correctly.

Quote
To use a water analogy, let's say you have two pipes going to a pool, and the diameter of the pipe is the current, and the pressure which the water is under is the voltage.  If both pipes are 3" in diameter, but one has twice the pressure, then you can see there's a lot more energy there and it will fill your pool twice as fast.

So to go back to the LED example, the led with the lower forward voltage is using up more of the voltage inside it to generate light and heat, so it should get a little hotter than the one with the higher forward voltage.
AARRGGHH.

Don

scswift

#7
Apr 03, 2011, 01:48 am Last Edit: Apr 03, 2011, 01:54 am by scswift Reason: 1
I don't understand how the LED can lower the voltage without actually using up power.  Power = voltage x current, and energy cannot be created or destroyed, so where does the voltage go once it hits the LED if not into generating light and heat?

Also, if you put two leds in series, you can get away with using a smaller resistor and generating less waste heat because there's less power that the resistor needs to dissipate.  But if one led uses 30mA, and two leds in series also use 30mA, how are they generating twice the light, and twice the heat, if not by converting the voltage part of the equation to light and heat?

madworm

#8
Apr 03, 2011, 02:36 am Last Edit: Apr 03, 2011, 02:39 am by madworm Reason: 1
The LED just says: "You can't make me!"

And as it is not alone in the circuit, the dumb resistor gets to do the dirty job and gets a little hotter ;-)

Quote
so where does the voltage go once it hits the LED if not into generating light and heat?


It never gets there. If the LED 'decides' to take 2.7V at a given current, the remainder is dropped 'elsewhere' (series resistor, driver chip...).


floresta

Quote
where does the voltage go
The voltage doesn't 'go' anywhere and it doesn't 'hit' anything.  You apply voltage to a series circuit and current flows.  The amount of current is determined by the components in the circuit.  When everything settles out there is a certain amount of current flowing and a certain amount of voltage appearing across the various components.  Did you ever wonder why it takes so many years to earn an engineering degree?

Don

floresta

#10
Apr 03, 2011, 02:59 am Last Edit: Apr 03, 2011, 03:22 am by floresta Reason: 1
Lets go back to some earlier posts:
Quote
But I would still like to understand what that graph is trying to tell me.
The graph is telling you the relationship between current and voltage for the LED.  You need a graph because the device is non-linear and Ohm's law does not apply.

Quote
If the graph were showing me how much current I could put through the LED if it had different forward voltages, then wouldn't the current be inversely proportional to the forward voltage, rather than proportional to it?
You've got the independent and dependent variables reversed.  The graph is telling you how much forward voltage you will get for different forward currents.  You said this yourself here:
Quote
So I think what the graph is showing is how the forward voltage changes as I put more current through the LED.


Quote
I don't understand how the LED can lower the voltage without actually using up power.
The LED does not lower the voltage.  The voltage is determined by the current flow and the current flow is determined by the forward voltage drop of the LED and the resistance.  If you see a chicken/egg situation here you are correct.  That's where the time is used up in getting the engineering degree.  

Basically you decide what current you want and you look at the graph to find out the forward voltage at that current.  You subtract that voltage from the supply voltage to determine the voltage across the resistor.  You apply Ohm's law to determine the required resistance.  You choose a resistor near that value, put it in the circuit, and measure the current.  If the current is too high you raise the resistance and if it is too low you lower the resistance.   Remember you are dealing with more than one approximation.  The graph is for a 'typical' LED, yours may be different.  The marked resistance on a resistor is a 'nominal' value, yours may be different.


Don

scswift

But if one LED "decides" to take 4.3v, and other "decides" to take 4.7v, are you telling me the one that uses 4.7v at 50mA isn't gonna get hotter than the one which takes 4.3v at 50mA?

scswift

Also, where does this leave me regarding the issue with some of the leds potnetially haivng a forward voltage of 5.3v?  If I lower the current I'm tying to put through it, will I also lower that forward voltage?

GaryP

This must be the first time I heard about term forward voltage, and I just don't get the point? Is this some sort of special LED that requires so much more... something ... than normal LED does?
It must be my narrow capasity in english...

Only thing I see here, is that voltage potential is divided in to two pieces with resistor, which also limits the max. current. As it was said, voltage don't disappear, nobodys eating it, current just turns to heat at some point. http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Is there any other point? Again, I would like to learn if there's something I am missing.
Thanks!

Cheers,
Kari
The only law for me; Ohms Law: U=R*I       P=U*I
Note to self: "Damn! Why don't you just fix it!!!"

scswift

#14
Apr 03, 2011, 12:06 pm Last Edit: Apr 03, 2011, 12:09 pm by scswift Reason: 1
Forward voltage is the voltage drop across an LED.  In other words, if you put an LED in a circuit with a 5v source, and its forward voltage is 2.4v, then you will read 5v-2.4v, ie 2.6v at the anode of the LED.  (I think!)

You would then do your Ohm's law calculation on the remaining voltage, and the amount of current you want to put through the LED, and you get:
R = E/I
R = 2.6v / 0.020a (20mA)
R = 130ohms

And you would then find the next largest standard resistor size which is closest to that.

This is why you can put only a few LEDS in series.  In the above example, you could put two LEDs in series before the voltage you have left is too low to light another.  Add a third LED, and no power will flow in the circuit and all the LED will remain dark.

And that's why I'm asking about this particular LED.  What makes this LED special is that depending on the batch of LEDs I get, the forward voltage for a single LED can range from 4.3v to 5.2v.  But my supply is only 5v.  So if I get a bunch of LEDs with a forward voltage of 5.2v, they won't work.

At least that's what I assume.  What I was asking is if the forward voltage and the current you put through the LED are related, if I could lower the current I put through them from the 50mA used in the datasheet, and then always be certain these LEDs will work in a 5v circuit.

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