Choosing The Resistor To Use With LEDs

This question gets asked every day in Answers and the Forums: What resistor do I use with my LEDs? So I've put together several different ways to figure it out.

Lets get right to it:
Each of the steps do the same thing. Step 1 is the simplest and we go downhill from there.

No mater what way you choose you must first know these three things:

• Supply voltage This is how much power you're putting into the circuit. Batteries and wall warts will have the output voltage printed on them somewhere. If you're using multiple batteries^*, add the voltage together.
• LED Voltage Sometimes "Forward Voltage" but usually just abbreviated "V".
• LED Current Sometimes "Forward Current". This is listed in milliamps or "mA".

Both of these last two can be found on the packaging for your LEDs or on your supplier's web site. If they list a range ("20-30mA") pick a value in the middle (25 in this case). Here are some typical values, but use your own values to be sure you don't burn out your LEDs!:

Red LED: 2V 15mA
Green LED: 2.1V 20mA
Blue LED: 3.2V 25mA
While LED: 3.2V 25mA

Okay, lets get started!

^* Batteries in series.

^{Introduction photo credits:
LED photo by Luisanto.
Resistor photo by oskay.}

Step 1: The Web Way

The easiest way is to use one of the online calculators provided below.

Just click on one and enter the info from the previous step and you're set! You only need to go to one.

The LED Center (For single LEDs)

The LED Center (For arrays of LEDs)

LED Calculator.net (For single or arrays of LEDs)

LED Calculator.com (For single or arrays of LEDs)

Comments

[jcook32]

Ok for the project I'm doing the LED calculator says i need a . I found LED's on eBay that come with resistor, but the resistors they have to pick from are 4.5V, 5V, 6V, 9V or 12V resistors can I use any of those? or do I still need to buy the 120 ohm resistors
http://www.ebay.com/itm/160878209257

[Grathio]

What they're saying in the listing is if you tell them what voltage you're running the LEDs, they'll calculate the correct resistor and send it along -- You don't need to calculate.

However I would verify that they send appropriate resistors once they arrive.

[justinmcg67]

Using math is often the fastest way. Simple algebra. It usually takes me longer to find a datasheet or some other information than it does to find & solve for R in Ohm's Law. R= V/I boom that simple.

[makerbuilderbaker]

This is just an application of ohms law.
V=IR (Voltage = Current times Resistance)
if the supply voltage is 5 the forward voltage is 2.5 and the forward current is 25 mA or .025 amps.
R=V/I so R= (5-2.5)/.025 so R=2.5/.025 which is 100 ohms. It is simple to use and more effective than an online calculator. There is a good site that explains this:
http://cuttingedgescienceclub.blogspot.com/2012/01/ohms-law-practice-problems.html
scroll down to the section labeled LED problems or you can look at the earlier sections to help.

[tutdude98]

thanks its help me :D

[Martythebest]

if i get it right...amps are enabling the charge to flow...so it doesnt matter how many amps are created by power supply?

[Grathio]

Right. As long as there are enough amps you're good. Too many amps is not a problem in any way.

[MACKattacksnipe]

so a 7.2 VDC 3000 mAh RC car battery all i need to worry about is that 7.2 VDC

[N3em]

Thank you so much its helps alot ,but i have qustion

how can i bulid led curcit like this
http://www.youtube.com/watch?v=E7Z6WqAP-SI

[Vick Jr]

So the supply current (like those listed on wall warts) makes absolutly no difference? It won't hurt to use a supply voltage with a very high current, as long as the resistor is calculated using the supply voltage, LED voltage, and LED current?

[Grathio]

Yes, that's correct.  You can perhaps think of Amps as "available power".  If you're not using it, it doesn't cause any trouble.

You need to make sure you have enough amps to supply everything you're powering, so if you're powering t12 LED's that pull 30 milliamps each you need a power supply with at least (12x30mA) 0.36 Amps. More than that will not cause any problems.

It's the same with the (somewhat) common task of buying a replacement computer power supply.  You can use pretty much any power supply as long as the voltage is the same and it has at least as many amps.

It's like using rope to climb down a cliff.  If you run short of rope (amps) you're in big trouble.  So you might as well bring at least as much as you need.

[Gaoh]

that's it!
just the answer for my problem.
the current of the power supply always confused me, cause i thought it was the source for LED to burnt.
thanks, this is helpful.

[Tommyhzy]

This post just cleared up everything that confuddlywizzled me for a long time. Thank you for using such easily interpreted analogies!

[Saprobic]

I have a question about over use of resistors. How much flexibility in teh value do you have? Say I calculate a required resistance of 120 Ohms. If I have a resistor of 470 Ohms will it still work?

[Grathio]

 It really depends on the circuit and the LEDs you're using.  While it's better to use larger (more Ohm) resistor than a smaller one, going too high will keep your LED from lighting much if at all.

As a rule don't go under by more than 10% or over by 25%.  If you're looking at 120 Ohm,  that would be a range of 108-150 Ohm.

However if you have several 470 ohm resistors you can connect them in parallel [wikipedia link] to divide their resistance.  In this case 4 in parallel would give you about 118 Ohm which would be just right for you.

[Saprobic]

Great thank you so much for that. So this leads me to wonder if I run resistors in series do they add?

[Grathio]

Yes!  So if you run two 68 Ohm resistors in serise you'll get 136 Ohms, which is more or less in the range for your needs.

[sinopro]

If I use 3 LED lamps in white, each is 3.2V and 25mAH, totoal is 3.2V 75mAH, and battery is 9V, not 12V, what the resistor should be?

[Grathio]

It entirely depends how you wire the LEDs. Are they in series or parallel?

I recommend that you go to Step 1, click on one of the links there and discover the answer yourself.

[phillyj]

What if you salvage LEDs. There's no sure way to know for sure their ratings, right?

[Grathio]

I don't know of any way to be 100% sure but you can get close:

I've done it this way. Get:

• Your LED
• Your DC power supply that you plan to use.
• A 10k potentiometer.
• A multimeter
• A prototyping breadboard (Not required but makes all of this a lot less messy.)
  

Connect pin 1 of the potentiometer (typically the leftmost pin) to +V on your power supply, connect pin 3 (the rightmost pin) to -V. Turn the potentiometer all the way to the left (off) and plug your LED between the middle pin on the pot and -V. (If your LED still has pins, the long pin goes to the pot, the short one to -V)

Slowly turn the potentiometer up. If you turn it most of the way and don't see a light try turning the LED around.

Turn the pot until you get a bright light without the LED warming up. Cut power to the circuit and use your multimeter to read the resistance between pins 1 and 2 on the pot. That value is roughly the resistor you need to use.

At least that's how I've done it. If someone else knows a better way, please post it.

This is probably a good subject for an Instructable...

[Jack A Lopez]

Look man, this isn't hard. Unless you think math is hard, and based on your title for step 3, maybe you do. I've heard similar sentiments from Teen Talk Barbie(tm).
;-)

But I digress.

Step 1: Start with a resistor and DC voltage source you know won't kill the LED. For example a 12V battery or wall-wart, and a 10K resistor.

You know the current drawn by this arrangement will be less than 1.2 mA, no matter what V_led is. How do you know this? According to your formula, the current through your resistor is: I = (V_supply - V_led)/R = V_supply/R - V_led/R
Maybe you don't know V_led exactly, but you expect it to be positive, thus: I < V_supply/R

Step 2: At this point you may be able to see a tiny amount of light comming out of the LED. If you do that means the LED is good, and biased in the forward direction.

Step 3: Using a mulimeter, measure the actual voltage drop across your 10K resistor. Suppose it happens to be 8.4 V
How much current is flowing? Because you used a power-of-10-sized resistor, the math is easy: 8.4V/10K = 0.84 mA

Step 4: Now you that you know the LED is working maybe want to get agressive, and swap in a smaller resistor to get more current (and more light output). Can you predict what the current will be with your new resistor, assuming the V_supply and V_led remain the same? Yes you can, since I = (V_supply - V_led)/R = V_constant/R.

Pick a reistor that's half as large (5K), and you expect the current to double (1.68 mA)

Pick a resistor that's 10 times smaller (1K), and you expect the current to get 10 times bigger (8.4 mA)

Pick R_new= (1/a)*R_start, and expect the current to get a times bigger.

Etc.

[phillyj]

And you would stop this when, visually, the LED doesn't increase in brightness?

[Jack A Lopez]

This is a good question. You stop increasing the current before you get scared that it might be too much.

For a random plastic LED in a T1 or T1+3/4 case, I think 5 mA is a good conservative limit. Or set the limit at 10 mA, if you like to live on the "edge". Recall the LED from my previous example that drew 0.84 mA in series with a 10K resistor. If I want to the current to be 5 mA, that's about 5 or 6 times more current, so I want a resistor thats 5 or 6 times smaller than 10K. Suppose the resistors I happen to have in my collection that are close to this are:
2.2 K = 10K/4.5454 ==> I = 4.5454*0.84 mA = 3.81mA
1.5 K= 10K/6.6666 ==> I = 6.6666*0.84 mA = 5.60 mA

So if I were being strict about my 5mA limit, I guess I'd have to choose the 2.2K resistor.

BTW you can probably afford to burn up (over current) some LEDs, just to see what it does. They can turn interesting colors, like incandescent orange, the actual color of something burning, but only briefly, and only once.

You said these LEDs were "salvage", right? To me that means found, via dumpster diving, junk given to you by a friend, basically free, like manna from heaven. Thus burning up a few of them is not a big deal anyway.

BTW throughout this example, I have assumed that these "salvage" LEDs of yours are the cheap kind. LEDs designed to handle more current (e.g.100s of mA or more) are physically different in appearance. They are physically larger, the case is water-clear, and they have large heat sinks, or a die made for mounting to a large heat sink.

As an example of one of these "beefier" LEDs, the attached pic shows a 3 watt (800 mA @ 3.6 V) white, LED mounted inside an aluminum enclosure that used to be a 5oz can of "Vienna Sausages" in its former life.

[rickharris]

A nice description.

http://led.linear1.org/1led.wiz will do the calculations for you for a single LED

http://led.linear1.org/led.wiz will calculate the array for many LEDs drive from the same power source (much harder for many to work out.

[snoopy2]

Go to besthongkong.com to get resistor info on 12vdc values, but do not ever
use 1/4 watt resistors as they tend to heat up, always use @ least 1/2 watt.
Don't need to go to site, as here are the values:
First, LED'S need to be in series - + - + etc, longer lead is +, Can put resistor on end of either -/+ but electronically speaking - is where it should be placed 'cause
electrons actually go there 1st.
LED COLORS - -W,B,G,Cyan/Aqua, B/G color changing:
LED#s 1= 470ohm, 2=270, 3=100 USE NO MORE THAN 3 IN CIRCUIT!
COLORS-- R,Y,Amber, Orange; #'s 1=560, 2=390, 3=330, 4=220, 5=100
USE NO MORE IN CIRCUIT..... Problem with household led bulbs - they're using even smaller than 1/4 watt resistors, and resistors are burning up, appearing as burnt out leds, while it's the circuit actually burnt up. If you want to use the led's that are inside the bulb, being extremely safe, break lightbulb, unsolder all leds
now u have a bunch of leds! Manufacturers have finally found out about not using resistors in circuit, and soon we will have the EXTREMELY CHEAPEST way of lighting a house, Flourescent is cheaper than incandescent, led lighting is BY FAR Cheaper than any known lights, as they use negligable amounts of elec. 'DA , 'DA', THAT'S ALL FOLKS.... NOT REALLY
As Christ gave all Grace, I pray Peace and Love be unto ALL - - -
God's slave, Servant to ALL Ron Spears a.k.a. [e-mail]
Official Jesus freak

[Uncle Kudzu]

thanks! this looks like good information for those of us trying to fathom all the great LED instructables here. i'm motivated to try out some of the bike light projects and your 'ible will doubtless come in very handy!

[Bongmaster]

definitely useful :)

[LuminousObject]

I've always wondered how people did this. Thanks!