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Calculating resistor value and power rating for LED driving
Infra Red LEDs - Voltage Advice

I need to make an array of IR LEDs, just want to check what size resistors I need.

I'll have 2 arrays of a 5 by 5 of IR LEDs. So essentially, I'm powering 50 IRs.

The power will come from 2 AA batteries (3V), these are the LEDs:

http://www.rapidonline.com/Electronic-Components/5-0-x-4-1mm-Oval-Infrared-55-30-deg-LED-200104

You can view the spec within that link. If I give every LED its own resister, what size will I need? I want to get the upmost power out of these LEDs.

Thanks

--- UPDATE ---

Also, whats the best way to wire these? Parallel, or Serial?

I'm now thinking about powering this from a power source, is this suitable:

http://www.rapidonline.com/Electrical-Power/Uniross-Power-Supply-600mA-2250mA-401024

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  • \$\begingroup\$ The fact your LEDs are in an array doesn't change the resistor value. If you have a certain driving voltage, an LED with a certain forward voltage, and a certain desired current, you've got your resistor value, regardless of array or not array. \$\endgroup\$
    – The Photon
    Commented Feb 1, 2013 at 19:45
  • \$\begingroup\$ also electronics.stackexchange.com/questions/25222/… This one seems to be the most straightforward one to answer your question. \$\endgroup\$
    – The Photon
    Commented Feb 1, 2013 at 19:51
  • \$\begingroup\$ Right, so with a 3V battery, a forward voltage of 1.8V, and a forward current of 80 (100Max), a resistor of 15Ω is good? \$\endgroup\$
    – Oliver Jones
    Commented Feb 1, 2013 at 20:02
  • \$\begingroup\$ Looks right to me. \$\endgroup\$
    – The Photon
    Commented Feb 1, 2013 at 20:04
  • \$\begingroup\$ Did you ever read How can I efficiently drive an LED? Did you perform any other basic research? You are about the 10000th person to ask some basic variation of "how do I drive an LED?" \$\endgroup\$
    – Phil Frost
    Commented Feb 1, 2013 at 21:19

1 Answer 1

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These diodes have a max \$V_f\$ of 1.8. With a supply voltage of 3V, you don't have the option to connect two or more diodes in series.

All diodes therefore, will require a current-limiting resistor.

Sample schematic

First four LED's shown.

$$R =\frac{E}{I}$$

$$R = \frac{(3 - 1.8)}{0.080} = 15\Omega$$

Total current will be 50 * 0.080 or 4A. Two AA batteries will not be able to supply this current, certainly not for long.

Edit:

Instead, as per my comments, I would use a higher voltage (12V), and connect LED's in series. This also requires fewer current-limiting resistors.

Series Schematic

This way you can connect 5 LED's in series. Each has a voltage drop of 1.8, so the series voltage drop is:

$$1.8 * 5 = 9V$$

This leaves 3V across the resistor. 80mA would still be the desired current flow through each series, so the resistor size should be:

$$R = \frac{3}{0.080} = 37.5\Omega$$

Now you only need one tenth as many resistors. (The above schematic only shows four series, but you'd continue the pattern for all 50 LED's.)

The total current would be 80mA * 10 (because there are 10 sets of diodes in series) for a total of 800 mA or 0.8 A.

My comment and link for a 12V 5A supply is incorrect: you don't need the same current capability at a higher voltage. For this design, I'd recommend a 12V 1A power supply.

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  • \$\begingroup\$ How long, is long? - What do you recommend as a power source then? Thanks \$\endgroup\$
    – Oliver Jones
    Commented Feb 1, 2013 at 20:46
  • \$\begingroup\$ Alkaline AA batteries might source about 2.5 amps for a short time, it depends on the manufacturer. Check the datasheet for them. I'd recommend a small, inexpensive power supply such as a wall wart, but given the option, I would use a higher voltage to allow several diodes to be connected in series. A higher voltage supply will allow you to connect more diodes in series, and reduce the number of current limiting resistors needed. You'll still need to source approximately 4A though. \$\endgroup\$
    – JYelton
    Commented Feb 1, 2013 at 21:14
  • \$\begingroup\$ One AA battery has at most a capacity of 3000 mAh. Under the most magical conditions, this will run a 5A load for \$ 3Ah / 5A = 0.6 h = 36 minutes \$. In reality, most batteries have a lower capacity, and by the time they have exhausted even half of their capacity, their internal resistance has increased significantly and their voltage dropped significantly. \$\endgroup\$
    – Phil Frost
    Commented Feb 1, 2013 at 21:14
  • \$\begingroup\$ Would something like this be ideal? rapidonline.com/Electrical-Power/… \$\endgroup\$
    – Oliver Jones
    Commented Feb 1, 2013 at 22:57
  • \$\begingroup\$ The maximum capacity of the one you linked is 2250 mA, or 2.25 A. I don't know good sources in Europe, but you need something capable of 4A or more, like this 12V 5A supply. \$\endgroup\$
    – JYelton
    Commented Feb 2, 2013 at 0:35

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