# volt range for LED's

Discussion in 'General Electronics Chat' started by lotusmoon, Jun 29, 2013.

1. ### lotusmoon Thread Starter Member

Jun 14, 2013
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I am new to electronics and am building a 555 timer to run an LED cluster. If i use 15 volts there is enough power for the LED's. Is there a general volt range for LED's or is 15 volts at 200mA the same as 5 volts at 600mA ?

2. ### LDC3 Active Member

Apr 27, 2013
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The power available is the same, but LEDs require a minimum voltage in order to have the current flow through it. Let's say the LEDs require 2.5V. Three LEDs in series would require at least 7.5V. The 15V supply will turn on the LEDs, but the 5V supply will not.

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3. ### bertus Administrator

Apr 5, 2008
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Last edited: Jun 29, 2013
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4. ### lotusmoon Thread Starter Member

Jun 14, 2013
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Thank you for being so helpful.
I am going to have the LED's in parallel so I don't think there will be a voltage drop.
As long as there is not a maximum voltage I think I am OK.

Is there a minimum Amps to start an LED?

5. ### #12 Expert

Nov 30, 2010
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LEDs are all about current. They are not voltage operated devices. You must put a resistor in series with each LED or each string of LEDs in series to limit the current. The LEDs will decide their own voltage, you supply the current. Look here.

http://forum.allaboutcircuits.com/showthread.php?t=19075

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6. ### MrChips Moderator

Oct 2, 2009
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Putting LEDs in parallel is not a good thing because each LED has different current-voltage characteristics. The LEDs will not be the same brightness.

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7. ### WBahn Moderator

Mar 31, 2012
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Why would you think there wouldn't be a voltage drop if they are in parallel?

An LED, like most other diodes, has a characteristic voltage drop across it when it is conducting the amount of current that it is rated for. This voltage changes some with current and some with temperature and some with just being the third LED from the left of the die when it was fabbed.

The key thing you need to realize is that if you apply just a little bit less voltage than this that the current will drop sharply and, likewise, if you apply just a little bit more voltage then the current will rise sharply. For a normal diode, a change in voltage of just 60mV will result an a factor of ten change in current. LEDs will be similar.

So the circuit you use should be designed to limit the current to the desired value while allowing the LED to drop its characteristic voltage. The easiest way is to use a resistor. If you have 12V available and you have an LED that drops 2V at an intended current of 10mA, then putting a 1kΩ resistor in series of it will result in the needed 10V dropped across the resistor at 10mA.

The other thing you need to be aware of is that you should never just put diodes, including LEDs, in parallel. The reason is that remember when I said that the forward voltage depends on temperature? Well, the forward voltage decreases as the temperature increases. What happens as a device dissipates power? Its temperature increases. Also remember that I said that the forward voltage also depends on variations in fabrication (the third diode from the left as opposed to the seventh).

So here is what will happen if you were to take 10 LEDs, all nominally identical, and put them in a circuit in parallel with a single resistor. With ten LEDs, you would need 100mA, meaning that you would use a 100Ω resistor. So you hook it up and everything starts out just fine. All of the LEDs have the same voltage across them, but they don't have the same current because each one is slightly different. One of them will conduct more than the others at that voltage and so it will consume more current than the others. But that means that it is dissipating more power and, hence, heating up more. But as it heats, it will draw even more current at the same voltage and so more of the 100mA channels to that diode, which results in it heating up more and gobbling more of the current. This process is known as "thermal runaway". If the LED can't handle the full 100mA, it will burn out. As soon as it burns out, one of the other LEDs will take its place as the current hog and then it will burn out, too (only quicker). This chain reaction will continue until all of the LEDs are destroyed.

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8. ### lotusmoon Thread Starter Member

Jun 14, 2013
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Thank you for this. I am being saved from making a lot of mistakes.
So if I was having 50 LED's
the circuit was at 15 volts,
the LED's had a voltage drop of 2.5 v
this would be enough for 5 LED's in series.
and I would need 10 lines of 5 LED's each line with its own resistor.?

9. ### LDC3 Active Member

Apr 27, 2013
920
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It would work for most LEDs. Some LEDs have a forward voltage drop of 3V or more, in which case 15V will not turn on the LEDs. Look on the datasheet of the LEDs you wish to use to see what their forward voltage drop is. Remember, this is only a guide and the actually forward voltage drop varies by 10-20%.

10. ### #12 Expert

Nov 30, 2010
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Also, each color of LED has a different range of voltages. Red is the lowest, blue and white are in the highest range.

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12. ### WBahn Moderator

Mar 31, 2012
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This is probably not a workable. 5 LEDs at 2.5V apiece would be 12.5V meaning that you need to drop the other 2.5V across the current limiting resistor. But if the voltage of the LEDs is high or low by 10%, then you would have a change of about 50% in your LED current, which is probably not acceptable. The more voltage you drop across the current limiting resistor (as a fraction of the voltage dropped across the LEDs) the more accurate and stable your LED current will be. On the other hand, you will be dumping a higher fraction of your power in the current limiting resistor. That's the basic tradeoff -- stability for efficiency.

13. ### Wendy Moderator

Mar 24, 2008
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If the OP is willing to test select the resistor (tweak the value for the current required) it could work, but this is usually more trouble than people are willing to go to.

14. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
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brilliant thank you this is such a great way of learning so much good information I will spend time looking through this.
There is one thing I do not understand I put out 7 red LED's in series on 13 volts they all light ok. I was expecting them to be different brightness as they would be the same amps, but the 1st would have 12 volts second 10 volts the 3rd 8 volts ect. But they were all the same brightness so is it the volt drop x amps = watts ?

15. ### Metalmann Active Member

Dec 8, 2012
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I'm one of those guys willing to waste the time, trying different LED/Resistors out.

One thing I have noticed, no two LEDs act exactly alike. Just like resistors. Doing it on paper is great, but I want to see it with my own eyes.

I'm so slow at it, I'm glad I don't have to do this for a living.

16. ### Wendy Moderator

Mar 24, 2008
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If you did it for a living reliability in design becomes much more critical, you don't have time to waste doing it. That is why it is worth learning how to do it reliably. Thing is, LEDs are about the easiest component in electronics to learn, transistors have similar wide variations you have to plan around when designing.

If you don't feel the need to push them to the edge of performance specs it is really easy. When I design concept circuits I just through a resistor value I know won't stress the LED and am done with it.

Apr 5, 2008
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18. ### WBahn Moderator

Mar 31, 2012
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For the most part, LEDs of the same part number will put out very close to the same brightness for the same amount of current through them. By putting them in series, you force them to have the same current.

The notion that the first has 12V and the second has 10V and so on is a common misconception. All any component cares about is the voltage DIFFERENCES between its terminals and the current THROUGH its terminals. It neither knows nor cares about voltages and currents elsewhere (unless there is a coupling mechanism, e.g., such as in a transformer or optocoupler).

So if the first LED has 12V on one side and 10V on the other, then it has 2V across it. If the fourth has 6V on one side and 4V on the other, then it has 2V across it.

Think of it like a cooler (the kind you put ice water or lemonade it) that has a spigot at the bottom. You know that the higher the top of the liquid is above the spigot the faster the liquid will come out when you open it. For instance, if the level of the liquid is 1cm above the spigot you expect an extremely weak flow but if it is 10cm you expect a decent flow and if it were 1m (assuming the cooler is even that tall), that you will get an extremely strong flow. But what about a cooler that is on a table at the entrance to a 30 story building versus a cooler that is on a table on the roof of that building? The top of the liquid in the second is somewhere around 100m higher than the top of the liquid in the first. What about those compared to ones at sea level or ones at 1000m? The answer, of course, is that you wouldn't expect any discernable difference. What matters is the height DIFFERENCE between the top of the liquid and the spigot on each individual cooler.

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19. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
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I did read it but I will give it another go as i see new things each time and there were many articles

Last edited: Jul 1, 2013
20. ### Metalmann Active Member

Dec 8, 2012
700
223
"By putting them in series, you force them to have the same current.

The notion that the first has 12V and the second has 10V and so on is a common misconception. All any component cares about is the voltage DIFFERENCES between its terminals and the current THROUGH its terminals. It neither knows nor cares about voltages and currents elsewhere (unless there is a coupling mechanism, e.g., such as in a transformer or optocoupler).

So if the first LED has 12V on one side and 10V on the other, then it has 2V across it. If the fourth has 6V on one side and 4V on the other, then it has 2V across it."

Very good way to describe it, Bahn.

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