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

 

http://johnbokma.com/pet/scorpion/detection-using-uv-leds.html

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
around 6 x 20 mA = 120 mA.

Note that I did this calculation, later checked it with a voltage meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

 

:

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.

 

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

 

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

 

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

 

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
http://johnbokma.com/pet/scorpion/detection-using-uv-leds.html

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

 

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.

 

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

 

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

 

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

 

Batteries in toy racing cars?

 

"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.

 

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.

 

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....

 

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:
http://www.epitex.com/Catalog_PDF/08_Point_source_LED/L590CE-34F.PDF
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.
;-)

 

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

 

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."