I need help buying a resistor - - - - help me plz i don't know which one.

[#12]

That is as simple as it gets.

The red wire goes to a resistor. The resistor gets connected to the first LED. Only to one LED.
The first LED wil be connected to the resistor at one leg and another LED at the other leg.
The second LED is connected to the first LED at one leg and the black wire at the other leg of the LED.
The symbol marked "ground" is what you are calling "black".

 

[superflux]

I think that i understand (i think) now....

**broken link removed**
whats right?

There you go - that is correct.

 

[dominicanstan]

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[#12]

Your drawing is good. I only think you should put more safety on the LEDs by using a 62 or 68 ohm resistor. They last longer if you don't push them with all the current they can survive, and the brightness is not bothered by using a little bit more resistance.

 

[dominicanstan]

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[dominicanstan]

Thanks a lot #12....
You Helped me a lot with my questions!

 

[superflux]

wait if the resistor is higher the LEDs are going to receive less power? and the LEDs will have a longer "life" but will shine less right?

Slightly less power with a higher resistor. It will probably be annoyingly bright as it is.

 

[dominicanstan]

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[#12]

Yes. That is what I said. The LEDs will only be a little bit less bright. That is a good reason to put a little bit more resistance. You can't see the difference but the LEDs are happier.

You can not put more than 2 LEDs in series with a 5 volt supply because 1.9 + 1.9 + 1.9 = 5.7 volts.
If you try to use 3 LEDs in one series string with 5 volts, none of them will work.

 

[superflux]

because i thinks i can put the 7 LEDs in serial (in one wire) with a different resistor or not?

Not going to happen. Each LED is dropping 1.9 volts. 3 LEDs (3 * 1.9 volts) would be 5.7 volts and you only have 5 volts, so that is too many in SERIES.

The MAX you can do is 2 of them in SERIES for each chain. That is a 3.8 volt drop from your 5 volt supply which works well.

 

[dominicanstan]

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[#12]

If all the LEDs are perfectly the voltage the label says and you want to run them as bright as they will go, that will work.

 

[audioguru]

You can buy a 1.9V incandescent light bulb but you cannot buy a 1.9V LED.
LEDs have a range of voltage because some are 1.5V and others are are 2.4V even if they have the same part number.
So if they are connected in parallel then the one with the lowest voltage hogs all the current and burns out. Then the one with the next lowest voltage burns out. Then the current is too high for the remaining LEDs and they also burn out.

Cheap Chinese flashlights connect the LEDs in parallel because they buy thousands then test and sort them so the ones in one flashlight have exactly the same forward voltage.

Here are the spec's for an LED that I use:

 

[dominicanstan]

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[Reloadron]

OK, below is an image containing a few LED circuits. The circuit on the left is how I would do it. To explain what AG mentions look at the circuit on the right. If we parallel two LEDs and want each to draw .025 amp that is a .050 amp draw but... No two LEDs are exactly alike. That Vf of 1.9 could easily be 1.8 for one and 2.0 for the other. The 1.9 is a range and that is as good as it gets. This means that a pair of LEDs in parallel with a single series resistor will not get an even current split. That is why a parallel pair like drawn is unwise to make. If you follow the math you will see what will happen.

Personally if at all possible I would buy a pile of 130 Ohm 1/4 watt resistors and give each individual LED its own series resistance. That will ensure a long LED life. I worked with the numbers provided. In reality I would try a higher resistance and get the current down around 20 mA and see how the intensity looks.

Ron

Ron

 

[dominicanstan]

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[audioguru]

You didn't hear what I said about the range of voltages for LEDs and your stupid Calculator also doesn't know.

Your LEDs are not 1.9V. Some might be 1.5V and others might be 2.4V. Some might be 1.9V.
Simply do the simple math instead of using the stupid Calculator:
1) If the LEDs are actually 1.5V then six totals 9V. The 24 ohm series resistor will have 12V - 9V= 3V across it which results in a current of 3V/24 ohms= 125mA which will instantly fry the LEDs.
2) If the LEDs are actually 2.4V then six totals 14.4V and they will not light with only 12V.

Reduce the number of LEDs in series so that the current-limiting resistor has more voltage:
1) If the LEDs are actually 1.5V then four totals 6V. A 240 ohm resistor limits the current to 25mA which is bright.
2) If the LEDs are actually 2.4V then four totals 9.6V. The 240 ohm resistor will limit the current to 10mA which is not dim.

 

[dominicanstan]

audioguru thanks for your help but i dont need that kind of help from you!
Once again thanks! And i think i better look for help in another forum, because you aren't respectful... thanks for take your time!

bye bye and thanks everyone for help me.... especially #12.

 

[audioguru]

I was respectful to you but I was not respectful to your stupid Calculator. I am on all the other forums so I will see you there!
Ohm's Law is very simple to calculate yourself. Name-Brand LEDs have a detailed datasheet that you should look at.

You don't know if your LEDs are 1.5V, 1.9V or 2.4V unless you measure them. You might be lucky so that most of your LEDs are near 1.9V.

 

[Reloadron]

AG tried to give you good advice as to why placing LEDs in parallel is not a good or accepted practice. I posted a schematic to prove what he tried to convey. Then you say he wasn't respectful and blow away all your post? What you need to do is try and understand what was being told to you. OK in your little book AG wasn't respectful and you could use a lesson in growing up! I am sure another forum will tell you the same thing. Do the math!

Ron