# Do LEDs use their max voltage? End LEDs dimmer

Discussion in 'The Projects Forum' started by Imdsm, Feb 24, 2011.

1. ### Imdsm Thread Starter Member

Feb 11, 2011
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0
Just a quickie, I've noticed that the end two LEDs in one of my projects (a blue and a UV led) are dimmer than the others. The LEDs came with a specsheet that said for example, 3.0 to 3.6 volts.

I calculated the resistance needed by choosing the middle, 3.3, thinking they'd be powered not fully but enough. No need to put the throttle to full etc.. but the end two are dimmer than the others. With the red, you can't really notice, but the others you can.

It has just occurred to me that, the voltage is not going to equal itself out among the three LEDs, they are not going to say, here have some of my volts.. am I right in assuming that the first LED's will have taken 3.6 each, leaving the last one with only 2.7 v rather than 3.3?

Cheers

2. ### Wendy Moderator

Mar 24, 2008
20,765
2,536
As I explained in my blog, LEDs are first and foremost current devices. Once you exceed their Vf, their dropping voltage (which is a constant), then the current is what does the trick. Nothing else. You need the Vf to make sure the power supply can drive the LED properly, and the power supply should be way over the total Vf.

Many people new to electronics and LEDs have a bit of problem with the concept, but it is both true and important.

Most DVMs have a current meter, you can use it for this. Or you can measure the voltage across the current limiting resistors (absolutely required) and calculate the current through the LED via ohm's law. If you aren't sure how to do do this... ASK!!!

Are you trying to share the resistor between two LEDs? This is another common mistake. Resistors cost around 3¢ each, it is not economical if you try to save such a fundamental component on a simple circuit.

Chapters 1 and 2 both cover LEDs in detail for beginners. It's the core reason I wrote it, the rest kinda just flowed out.

LEDs, 555s, Flashers, and Light Chasers

How's the project coming along?

BTW, LEDs are very long life devices, but they are not immortal or invulnerable. Exceed their rated current and they will show colors you never thought you'd see, just before they go dark. If the current is weak they will be dim.

Imdsm likes this.
3. ### Imdsm Thread Starter Member

Feb 11, 2011
39
0
Hi Bill,

What I have is "chains" of LEDs, like so:

And for the Blue LED for example, the spec sheet says:

Forward Voltage VF (Min.) 3.0 (Typ.)--- (Max.)3.6 V
IF=20mA

I wasn't sure if it would drop the max Vf or not, so when finding the resistance for the CLR I chose the middle value, 3.3, then did Vs - (V1 + V2 + V3) / IF, which gave me 12 - 9.9 / 0.02 or 2.1 / 0.02 which gives me 105 Ohm, so I used a 100 Ohm resistor.

However, the LED furthest from the resistor seems dimmer, for both the blue and UV, I was wondering if this is because it goes:

Power supply: 12v
100 Ohm Resistor: -2 to 10v
1st LED: -3.6 to 6.4v
2nd LED: -3.6 to 2.8v
3rd LED: only has 2.8v left?

However what confuses is me is that if the electrons flow from - to +, surely they should hit the dimmest LED first, so wouldn't that one get the first voltage drop?

The resistor does make the current go to 20 mA (ish), so I know the LED is getting enough current (as the others on the same chain are working OK), but could my idea about the voltage dropping be right?

(I will just say, there is a chance that I have got it totally wrong, so if so, I do apologise!)

With regards to the project, it's working fine at the moment apart from the dimmed LEDs, as you can see here, the blue can't be seen as well on the left:

I am slightly confused, but I am also sure it's probably extremely obvious..

Thanks again, and sorry for the bad image quality.. I have thousand pound nikon sitting here but opted to take the picture with my iphone...

4. ### Wendy Moderator

Mar 24, 2008
20,765
2,536
There is no "hitting" a component first. Everything in electronics is pretty close to speed of light, and the choices of where you put a resistor is a bit arbitrary (it doesn't matter). Physical considerations count to plan for Murphy's Law (what happens if I short out here? type problems), but the electrons, they don't care.

You want 20ma (or 0.02A) through those LEDs. You have 2.8V left over for the resistor.

Ohm's Law is simple and algebraic.

V=IR, I=V/R

You have 2.8V ÷ 1000Ω, you have 2.8ma, about 1/10 what you want. The LEDs will be dim.

Instead, match the resistor to the LEDs.

You want 20ma. You have 2.8V, then

I=V/R=2.8V/0.02=140Ω ≈ 150Ω

Ohm's Law is your friend, it is usually the first major math you learn in electronics, and is needed for LEDs.

5. ### John P AAC Fanatic!

Oct 14, 2008
1,634
224
The relevant issue is that all three LEDs and the resistor are in series. If you did have one LED that had an unusually high voltage drop, it would indeed reduce the current through that leg of the circuit, but all the devices in that leg (3 LEDs and a resistor) would all be getting the same current. The order in which the current reaches the components is totally unimportant.

Anyway, the electrons that really make up electrical current have a negative charge. What we call "electric current" flowing from positive to negative, is actually electrons going the other way. Too much information, possibly.

Last edited: Feb 24, 2011
6. ### Wendy Moderator

Mar 24, 2008
20,765
2,536
Just curious, have you read the material I linked? If not you really, really need to, otherwise we're both wasting time.

I went off half cocked last post, and used your numbers for the last post, but going through it from scratch (the remanents of this flu have still got me down, it was especially nasty this time).

This is a redraw similar to what is in Chapter 1.

Note how the Vf of the LEDs add up. you have 1.2V across a 1KΩ resistor.

I = 1.2V / 1KΩ = 1.2ma

You want 20 ma. So,

R = V/I = 1.2V / 0.02A = 60Ω ≈ 62Ω (a standard value).

Remember, current is the important number. Voltage only sets it up.

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7. ### Imdsm Thread Starter Member

Feb 11, 2011
39
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Thanks for the replies, I apologise for the confusing I did actually put under the picture at the time of posting to ignore the resistor values as I just quickly drew up the diagram to show you how they are laid out, but it must have got deleted whilst I was writing the post.

I have read your fantastic articles, all of them under DC. The reason I didn't put the resistor values in is 1) I was just showing how I have them laid out, and 2) I don't have my scribbling book with the LED specs with me (it's at home).

The resistors I am actually using are a result of adding up the voltages of all the LEDs, taking that away from the supply voltage, and then using Ohms law to of course divide by the current required.

R = (Vs - Vf) / I
R = (12 - (3.3 + 3.3 + 3.3)) / 0.02
R = (12 - 9.9) / 0.02
R = 2.1 / 0.02
R = 105 Ohms

Again sorry for not mentioning that I'm not just using 1k resistors for every set of LED's.

I think my questions are pretty much answered, but one final question would be:

If I have three LEDs which all require 2v each, but I only supply 5v, do they all get 1.6v each or do two get 2.0v and one get 1v?

This is the part I don't seem to be able to get a concise answer for. The way I see it, and read it, is each "component" has a voltage drop, but in what order does this occur?

Or is does happen almost all at once? The components with the more resistance, the higher the voltage drop on them? If there are two LEDs with the same resistance lets say, and a resistor which takes 6v from a 12v supply, would both the LEDs share the remaining 6v between them and take 3v each?

P.S, if so, why would one LED in a series of 3 be dimmer than the other 2? Shouldn't they all be equally dim?

8. ### Wendy Moderator

Mar 24, 2008
20,765
2,536
LEDs are like people, they are individuals.

Now for the bad news. They (blue) are ESD sensitive. You may simply have an LED that is not up to snuff.

<sigh> Please . read . the . links. They way it is going I might as well copy and paste it here, I am repeating each and everything I have already stated. Several times. Chapters 1 and 2 explain all of this and more, in detail. Especially in Chapter 1. LEDs, 555s, Flashers, and Light Chasers

The voltage drop of LEDs is a constant. If it drops 3.6V, it will not work with less. If there is less than 3.6 it may glow dimly, but it will not light properly. You can not get the LED to drop more. If you feed the LED direct battery voltage it will try to force the battery to it's dropping voltage while it burns up.

If you do not met the minimum voltage requirements you will not be able to predict what will happen. It is out of tolerance, it will not work.

Voltage, once the minimums are met, is a distraction. Focus on current instead. It is also the reason you can't ignore the resistors value, it is important.

9. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
One item that hasn't been mentioned is LED aging. If used with their typical rated current, LEDs might last 50,000 to 100,000 hours. However, they darken (emit less light) as they age.

If you subject LEDs to more current than they are rated for, they age much more rapidly.

It could be that during testing of your LEDs, that you subjected the darker one to excess current for several moments.

10. ### Markd77 Senior Member

Sep 7, 2009
2,803
594
Even in a fresh batch there will be some brightness variation. If it's really critical to have them the same then you should get more than you need and find sets that match in brightness.
Also unless you test the forward voltages with the current you will be using, you should assume Vf is the minimum, because if they all happen to have the minimum Vf then your calculation will give you more current than they can handle.