12 LED resistance circuit help

Hi, I want to make a simple LED circuit with 12 LEDs running off a nine volt battery. I've managed to dig up enough information about most things, so I know I'll have to wire them in parallel. But the resistance I should be using still confuses me. Should I have one (or more) resistors at the beginning of the circuit? Or one before each LED in the circuit?

The LEDs I'm using have a 3.6v voltage drop, and they are supposed to get 20mA I beleive. So that's 12 LEDs off a 9v battery.

I have a bunch of 27 Ohm resistors because that's what I (probably mistakenly) calculated I should have before each LED. But I have no idea what i'm doing. So if anyone can help me out on what resistors I should use and where I should put them that would be greatly appreciated!

Thanks, Steve

Reply to
Brilla
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9V, 3.6V drop ->

9V - ( 2 x 3.6 V ) = 1.8 V

so you want to have 1.8V over your resistor.

the 2 LEDs in series use 20mA ->

1.8V
--
20 mA  = 90 Ohm


With 12 LEDs, you need 6 resistors. And the total current should be 
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Reply to
John Bokma

meter,

I came up with the same calculation as you. If LEDs are dieing, then

20mA may be a bit too much current. Since there are two LEDs in series, one may hog more current than the other resulting in its demise. You might consider not driving them so hard. There is probably a relatively insignificant brightness difference between 10mA and 20mA anyway.
Reply to
Anthony Fremont

Sounds like you want the leds to be powered individually. You need a 270 ohm resister to limit the current to 20 mA. Each led should get its own resister and then wire the led/resister units in parallel.

1/4 watt resisters will be fine. When you put leds in series they will pass the same current but one my have a larger voltage drop, consequently dissipate more power, because of a variance of their characteristics. Regards, Tom
Reply to
Tom Biasi

--
ciao Ban
Bordighera, Italy
Reply to
Ban

--
So, you\'re saying that because the LEDs may not be identical one may
be drawing more current than the other?
Reply to
John Fields

"Anthony Fremont" wrote

In a series circuit the current is equal through each component.

Reply to
dB

"Brilla" schreef in bericht news: snipped-for-privacy@posting.google.com...

Well,

The resistor voltage will be 9-3.6=5.4V. Using Ohms law the resistor should be R=5.4/20=270 Ohm. So 27 Ohm is way too low. Using the right resistors will require your battery to provide 240mA. You will need a pretty big battery to light your LEDs for even some minutes. Maybe to 4,5V batteries in series will do.

Most of the energy from the battery is dissipated in the resistors. Really a waste. You can try to use two LEDs in series. The resistor will be reduced to (9-2*3.6)/20=90 Ohm. Power efficiency is much better this way. However, practical 9V batteries tend to loose some voltage pretty fast when in use. So the current through your LEDs will decrease accordingly.

To make a real good battery powered LED-light you need to go electronic. But that's a different story.

petrus bitbyter

Reply to
petrus bitbyter

"John Fields" wrote

series,

Perhaps "dissipate more power" would have been more appropriate than "hog more current".

Reply to
Anthony Fremont

And where does it store that extra current?

In my case: 30 mA is max, so feeding them 22 mA shouldn't be that bad. Moreover, the LED died when the circuit at

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was connected to 9V, or maybe even 7V (less then 12V anyway).

--
John                               MexIT: http://johnbokma.com/mexit/
                           personal page:       http://johnbokma.com/
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Reply to
John Bokma

Since they are in series, yes. And this is possible if the voltage *over* the LED differs (which it very likely does). However the current should stay 20 mA, and due to the resistor, one LED can only have a higher voltage if another LED has less. The current will stay the same though.

It's like you can hog water in an open tube, it has to go somewhere.

--
John                               MexIT: http://johnbokma.com/mexit/
                           personal page:       http://johnbokma.com/
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Reply to
John Bokma

series,

Heh. Some advice handed out in the NG is really atrocious, ain't it?

relatively

Reply to
Watson A.Name - "Watt Sun, th

John Bokma wrote

"Due to the resistor"? What a strange thing to say.

The resistor limits the current, it has no direct effect on the voltage developed across each l.e.d. The actual value across each l.e.d. varies from device to device at any current. The data sheets give a "typical" Vf and sometimes a max figure.

Reply to
dB

In this case we should not trust the value, 3.6V, which was given earlier. That should be checked with a voltmeter in reality.

If these LEDs have a lower voltage that would explain why some suddenly die.

Send 10mA through a LED and measure the voltage over it, and the voltage over the resistor. Do the math and find out how much current is passing the resistor, and the LED.

--
 Roger J.
Reply to
Roger Johansson

I found an author of an article published in Electronics Now commenting about the voltage through the circuit...

Reply to
Lord Garth

Batteries in toy racing cars?

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John                               MexIT: http://johnbokma.com/mexit/
                           personal page:       http://johnbokma.com/
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Reply to
John Bokma

"Anthony Fremont" wrote

I neither said nor implied that it wasn't.

Yes, the curve is in an article on my site. (I can suck eggs.)

You seem to have misundersood my post which was regarding the ambiguous sentence "However the current should stay 20 mA, and due to the resistor, one LED can only have a higher voltage if another LED has less."

Reply to
dB

I guess the restatement I made is not good enough. Should I start another thread and offer a formal apology to the world for making such a heinous mis-statement about current vs. dissipation? Maybe I could help make amends by belittling others, nit-picking posts and posting a bunch of OT crap?

Lets see if we can't get on to the road to recovery now. Speaking of good advice, why are you trying to get a poster to use non-rechargeable alkaline batteries when he clearly expressed a preference for rechargeable? Hey Fields, are you ever going to acknowledge/correct your mistake in S.E.D about max collector current on the 2N4401? sheez....

Reply to
Anthony Fremont

Vf _is_ dependant upon current. At extremely low currents, Vf will be significantly lower than the nominal value. As current increases, so will Vf. The curve is steep, but it is not vertical.

If it is a detailed datasheet it will also specify a test condition clause giving the current associated with the stated Vf. Like this one for example:

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The datasheet sometimes specifies a minimum Vf as well. The max Vf can be as much as double the min Vf. Like Ripley says, believe it or not. ;-)

Reply to
Anthony Fremont

the other

Yeah, I know. I messed up, I should have said power, so sue me. ;-)

Reply to
Anthony Fremont

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