Hi I am new here, I have made a simple infra-red LED array and put them in a circuit and installed them in a project box and it all works fine except I am a bit confused as to what value of resistor I need to put in as many on-line resistor calculators only calculate for LED's in series OR LED's all in parallel and not both at the same time.

Each LED's forward voltage is 1.6V and uses 140mA.
I have got two sets of 4 LED's in parallel with each other running of of a 9V battery and have got two 12ohm resistors, one at the start of each LED array. Now I do have dyscalculia which might be why I am getting comfused with what value of resistor I need.

I have attached a basic circuit diagram if that helps, Thanks.

 

You can add together the voltages of the LED's and treat it as a single one.
So, 4x1.6V = 6.4V

With a 9V battery and 140 mA, you get this:

9V - 6.4V = 140mA * R
R = 18.6 ohms

The current will drop as the battery discharges so you can play around with that a bit.

With a 12 ohm resistor you will have approximately:
I = (9V - 6.4V)/12 = 216 mA

 

Sorry, but I am having a hard time understanding what you are saying.

I have 8 1.6 volt 140mA LED's in total, I have 4 in series behind a 12ohm resistor and another 4 in series behind another 12ohm resistor in parallel with the other 4, What ohm resistor do I need to use as I don't think 12ohm's is enough.

Edit: I have just realised what you mean that with the on-line LED resistor calculators I can just put 2 LED's in the calculator but just 6.4v as the forward voltage of the LED's.

But I am still confused with the current, where it asks what is the current of each LED do I need to times that by 4 as-well? so instead of putting 140mA I put 560mA.

Thanks.

 

You can consider each string by itself since a battery has very low resistance compared to your components.
That means you can work with a single string of 4 LEDs and then just build it twice off the same battery.

Current in a series circuit is equal for all the components so you'd leave it as 140mA.

I showed the calculation, you would use two 18+ ohm resistors, one per string of 4 LEDs.

This calculator considers series/parallel and even suggests schematics for you:
http://led.linear1.org/led.wiz

That calculator gives 22 ohm resistors because you can't get an 18.6 ohm resistor.
http://www.elexp.com/t_eia.htm
With 10% resistors you go from either 18 ohms to 22 ohms. They pick the higher one because they're assuming you don't want to exceed the current you specified.

 

If you are using a 9v "transistor" battery, you will drain it very quickly. Your light output will diminish noticeably in the first few minutes of installing a fresh battery. 9v "transistor" batteries are rated at around 150mAH when discharged at a 20 hour rate (7.5mA); when you discharge them with a higher load, they go dead much more quickly.

Consider using an array of six AA alkaline batteries in a battery holder. They will be less expensive, and will last far, far longer.

 

Thankyou, that really made things easy to understand for a beginner like me!

I guess I am quite lucky that my LED's have not burned out yet using only 12ohm resistors and I will have a look through my big box of resistors tomorrow and swap them for some 22ohm ones.

 

If you are using a 9v "transistor" battery, you will drain it very quickly. Your light output will diminish noticeably in the first few minutes of installing a fresh battery. 9v "transistor" batteries are rated at around 150mAH when discharged at a 20 hour rate (7.5mA); when you discharge them with a higher load, they go dead much more quickly.

Consider using an array of six AA alkaline batteries in a battery holder. They will be less expensive, and will last far, far longer.

I am just using a normal 9V energizer battery.

Edit: Alkaline I belive so according to wikipedia I should get around 565mAh

 

OK, I was using rechargeable NiMH numbers instead of alkaline.

Still, you will have a very short battery life span even @ 565mAH. That's discharging at a 28.25mA constant rate for 20 hours, until the battery reaches around 7v. You want to discharge at a 140mA x 2 rate, or 260mA. That's almost 10x the 20 hour discharge rate.