# Wiring 5mm IR LEDs in parallel [duplicate]

Im trying to wire 9 5mm IR leds in parallel using a 9V battery. the forward voltage is about 1.5 and Im looking for about 100mA instead of the ussual 20mA with a standard LED. When I punch these numbers into a resistor calculator I get 10ohms BUT when i linked this up on a breadboard I blew out all my IR leds!! Does anyone have any insight, Id like to be sure before trying again, these things are expensive!

## marked as duplicate by Autistic, PeterJ, JRE, Daniel Grillo, Olin LathropMar 2 '16 at 11:58

• Quick check: 10V / 10 ohm = 1000mA – Mario Mar 2 '16 at 6:23
• The LEDs have to match well. – Mario Mar 2 '16 at 6:24
• Before you 'light up' an LED check the datasheet first, get the I\V curve and find out if you need to current limit it. They will usually give you the recommended current. – laptop2d Mar 2 '16 at 7:04
• Welcome to SE.EE, make sure you read the forum rules. You didn't actually ask a question. And you didn't do research. You need to make sure you do both of these before you write a question or it will get downvoted. Please provide all necessary information, like the type of component and datasheet. Please use the circuit editor when possible. – laptop2d Mar 2 '16 at 7:06
• @Mario: $$\dfrac{ \dfrac {10V}{10\Omega}}{9LED} \approx \text {110 mA per LED,}$$ but since LED's have a negative temperature coefficient of resistance and there's no such thing as a perfect match, it's possible for one LED to heat up, hog the current and fail open, whereupon it'll cause the next weakest LED to heat up and fail open and... The solution is to either use a separate ballast resistor for each LED, or - if the source voltage is high enough - to run a a series-connected string of LEDs with a single ballast resistor supporting the string. – EM Fields Mar 2 '16 at 10:08

I don't know what kind of calculator you were using, but chuck it out and get a new one:

$$\frac{9V-1.5V}{100mA} = 75\Omega$$

But you've also got another problem: the 9V battery.

You are trying to draw 900mA from a 9V battery. First off, take a look at 9V capacity vs. discharge curves: if it can even supply the current, its actual capacity will likely be less than 50% of its rated capacity. Discharge tests show that at 1000mA, you should expect a capacity of 100mAh, ~300mAh if you're lucky. Additionally, let's say it's got a $1.5\Omega$ internal resistance: you'll be losing $1.5V \cdot 900mA = 1.35V$ in the battery alone.

If your 9 LEDs are all in parallel and you want to run 100 milliamperes through each one, that's 900 mA total, and the ballast resistance to make that happen will be:

$$R = \frac{Vsupply - Vled}{I_{LED} \times n LED},$$

where, in your case, Vsupply is 9 volts, Vled is 1.2 volts (since they're all in parallel), Iled is the current through a single LED and nLED is the number of LEDs in the array

So,

$$R = \frac{9V - 1.2V}{0.1A\times 9} \approx 8.7\text { ohms}$$

However, running LEDs in parallel with a single ballast is a Bad Thing because (since LEDs have a negative temperature coefficient of resistance) it's possible for the LED with the lowest forward voltage to hog the current, heat up, hog more and more current as it heats up more and more, and eventually fail open, whereupon it'll cause the next weakest LED to heat up and fail open and... POOF! One by one, all of your LEDS lose their magic smoke.

Plus, as @uint128_t mentioned, you're going to wind up with serious battery life problems.

A solution to both of these problems is to run the LEDs in series with a single ballast resistor for the entire string.

For example, if you have a 9 volt supply and your LEDs have a Vf of 1.2 volts, the Vf's add, so the number of LEDs you can run off of 9 volts is:

$$nLED =\frac{Vsupply}{V_F(_{LED})} = \frac{9V}{1.2V} = 7.5 \text { LEDs}$$

You obviously can't run half an LED, so if you need nine LEDs in your array you could run a string of 7 with its ballast resistor in parallel with a string of 2 and its ballast resistor.

Or a string of 4 in parallel with a string of 5, or...

The nice part about doing it this way is that even though you'll be drawing 200 milliamperes out of the battery (which still won't make it very happy) you'll be getting the same amount of light out, while wasting less power and increasing the battery's life than if you were sucking 1 ampere out of it.

Finally, you might want to re-evaluate the need for a 9 LED load, since if you go to 7 you'll only be taking 100 mA from the battery, making it even happier. :)

• Wow, this really made things more coherent . Thank you for the help. I'll try with 7 LEDs :) – Dana Williams Mar 4 '16 at 0:29