## Introduction: Multiple LEDs on a Single Resistor

This Instructable will show you how to calculate the required values to connect multiple LEDs with a single resistor vice an individual resistor per LED.

## Step 1: Using Ohm's Law

Ohm's Law dictates the relationship between voltage, current, and resistance. It is defined as the following equation:

Current (I) = Voltage (E) / Resistance (R)

In a circuit, all voltages gets used up, and each component offers some resistance that lowers the voltage. Knowing this, the above equation comes in handy for things like figuring out what resistor value to match up with an LED.

## Step 2: Understanding Parameters of LEDs

LEDS have predefined voltage drops across them and are designed to operate at a particular current value.

**Typical Maximum Current Value of an LED : 20mA (0.02 A)**

__Predefined Voltages for LEDs:__

**Red and Yellow: 2 Volts**

**Green and Blue: 2.9 Volts**

**White: 3.3 Volts**

## Step 3: Determining Values to Create Circuit

Remember Ohm's Law? We'll be using it to help determine how many LEDs we can get connect in **Series (CONNECTING LEDS IN PARALLEL WILL NOT WORK)**. This is dependent on the voltage source. Determining resistance helps us ensure we do not exceed 20mA.

__Example__:

*You have a 9 V Source and a handful of **RED** LEDs. Determine how many LEDs you can pair in series and determine the resistance value required.*

__So here's the information we do know:__

*The maximum current allowed is 20mA.*

*Red LEDs have a voltage drop of 2V.*

*Our supply voltage is 9V.*

First let us determine the resistance required.

**- USE OHM'S LAW-**

*I = V / R => 20 mA = 9 V / (R) => R = 9 / (0.02) = **450 Ohms*

Therefore you need a minimum resistance of 450 Ohms to prevent damage to the LEDs. **You should however, pick a slightly higher resistance value to ensure you do not exceed 20mA.**

**You should also note that the higher the current the brighter the LED will be.**

Second, let us determine the number of LEDs we can connect in Series. This is dependent on the voltage source.

The first rule we must apply is that the voltage drop across the LEDs combined should not exceed 80% of the voltage source. Therefore since we have a 9V source, we can only have 7.2V combined voltage drop of the LEDs in series.

We do know that each red LED has a 2V drop. Therefore, we divided are maximum allowable voltage drop by the voltage drop of a single LED.

*# of red LEDs = 7.2V / 2V = 3.6*. **YOU WILL ALWAYS ROUND DOWN!!!!**

**Therefore, you can connect (3) red LEDs in Series.**

## Step 4: Build Your Circuit

Now, you determined the resistance value of the resistor and how many LEDs you can connect.

Build your circuit in accordance with circuit diagram above.

## Step 5: CHEAT SHEET!!!!

If you want to skip the reading, here's a cheat sheet.

You can also combined different color LEDS as well.Ensure the calculations are made prior to building in accordance with Step 3.

**WARNING: FAILING TO PROPER CALCULATE VALUES MAY POTENTIALLY DESTROY LEDS AND OTHER COMPONENTS.**

Thanks again for reading!!!!!

## 7 Comments

3 months ago on Step 5

Well done, and a handy quick reference too. Thanks.

Another calculator for this is here along with a plethora of other calculators: https://www.amplifiedparts.com/tech-articles/led-parallel-series-calculator

4 years ago

A helpful calculator:

http://ledcalc.com/

Note the links for Single led-Series leds-Parallel leds

4 years ago

"The first rule we must apply is that the voltage drop across the LEDs combined should not exceed 80% of the voltage source."Where does this come from ? That's wrong. You can apply a kind of security margin but 20% is too much. I frequently put 4 red Leds on 9V source with a single resistor of 47ohm (for ~20mA). In the manufacturer's data, the Led forward voltage is given for the max. admissible intensity, this voltage increases if the current decreases.

4 years ago

Hi. Actually I use use in my apps. them LED's to be limited to 10~15mA thus NOT giving the full brightnes, bu to raise from 15 => to 20mA wan't make the brightnes to increase noticebly. The wattage of the resistor is of a less matter, e.g. if needed voltage drop from the source to LED's (RED=2,0V), then with 9V:

9-2=7V ; 7V*0,02A = 0,35W thus a 1/2 Watt resistor will do the job.

4 years ago

However, where have I gone wrong. If each led drops say 2 volts, then 3 in series drops a total of 6 volts. For a 9 volt supply, 6 volts are dropped in total by the 3 LEDs in series we now need to drop 3 volts across the resistor. For a current of 0.02 Ampers and using V/I =R , 3/.02 = 150 ohms. The 3 LEDs in series each dropping 2 volts being the same as a mythical 1 led which drops 6 volts?

Reply 4 years ago

I had the same issue when I was building this Instructable. I would recommend buying these LEDs. It seems I can only get this to work with these type.

https://www.amazon.com/gp/product/B00UWBJM0Q/ref=oh_aui_detailpage_o06_s01?ie=UTF8&psc=1

I still haven't discovered why your situation exist with other LEDs but I will share as soon as I find out.

4 years ago

One other issue when calculating the proper resistance, you also need to make sure you use the proper wattage resistor as well. And that is the current times the voltage to arrive at wattage. I like to at least double the wattage capability to make sure the dropping resister doesn't get hot. But you posted a good tutorial about LEDs and dropping reisitors.