# 9V battery to power up a 6 pcs & 12 pcs of 20mA 4-pin 130Ãï¿½ÃÂ° Power LED (White)

Hello, I'm new to this and I have a question (again) about 9V battery with power switch to powering up a 6 pcs of 20mA 4-pin 130° Power LED (White); I would like to arrange it in series and a 12 pcs(more or less) of the same LED with the same battery voltage (9V).

After I read some references, (I hope I'm not mistaken...) it is all about the **resistors**?

So what would I like to know is, between a 270 ohm and 120 ohm resistors which one more suitable for the 6 pcs modules and the 12 pcs if I wanted each modules gives suitable brightness(on a scale 1 for dim and 5 for very bright, 3(maybe will work) - 4(if I'm not asking to much :D), and how do I arrange it ? I've been reading the Instructables Q&A around 9V battery and LED lately.

I'm afraid I haven't catch the solution, and I hope my question will helping anyone whose have the same specific problem, and be useful for everyone else generally.

Thanks in advance everyone.

## Comments

9 years ago

Well, when you wire it in series, each LED is going to drop the voltage a bit. When it's lit, this drop is 3.5V, as listed by the datasheet. Since you want 6 LED's in series, they will require 21V. This is much more than your battery can supply, so the LED's will not light up. (or be so dim you can't see them.)

If you were to bump it down to 2 LED's then the voltage drop of the two at your specified current would be 7V. Meaning the other 2 volts have to be dropped in the resistor. Since the current is 20mA and the voltage is 2V, you can find the resistor value using Ohm's Law:

V=IR

R=V/I

=100 Ohms

If you want six LED's, do this setup three times in parallel. This means your battery will have to supply 60mA, which will really drain the battery.

If this isn't clear, I can draw a simple diagram that will help.

Answer 9 years ago

Hi sshuggi,

So clearly, I have to make a module contains 2 LED w/ 1 resistor place it in series, then put each (series)module as parallel?

i.e for 6 LED :

(2LED+1R) -

(2LED+1R) --- power switch --- battery

(2LED+1R) -

(*pardon me if I'm mistaken)

It will be very helpful for the diagrams :D

So it 's still possible to be made, with the components I had

Now the next problem is the power source, I think ....

(please correct me if I'm wrong...)

Thanks a lot :D

Answer 9 years ago

You have it completely right, although you can save yourself two resistors by putting the resistor

beforethe branch point. You just need to divide it by three, because the current into a node has to equal the current out. Ergo, the new resistor will have to give 60mA for the three modules of LED's, dropping it value down to 33 Ohms.As for your 9V source, it is just a regular 9V battery. (You know the kind that looks like a block and has two terminals at the top that you can feel if you touch your tongue to it.)

If you still want to use only what you have, you run into a problem. Those resistors are way too high, they'll give your LED's almost no current. Solution: Wire the resistors in parallel. The equation for total resistance in parallel is as follows:

1/R

_{T}=1/R_{1}+1/R_{2}+...So, if you have the two resistors you mentioned in you question in parallel, your total resistance will be about 83 Ohms. Your LED's will light up, but with less than half the brightness. My suggestion, just go to radioshack and get a ≈33 Ohm resistor. It'll probably come in a 5 pack for like a dollar. They'll also have switches and batteries there if you need them.