Power sources connected in **series** result in their voltage being added together. For example, AA batteries are 1.5V each. So three in series are 4.5V (1.5+1.5+1.5).

LEDs use a certain amount of voltage when turned on:

Color |
Typical Forward Voltage (V_{F}) |

Red, orange | 2.0 |

Yellow | 2.1 |

Green (older yellow-green variety) |
2.2 |

Blue, white, ultraviolet and newer “true” green | 3.3 |

The total forward voltage used by LEDs in series is also added up:

Your power source must be at least as high as the LED’s total forward voltage or they won’t turn on. So with a battery, in math terms *V _{BAT}* >=

*V*

_{LED. }Typical small LEDs use around **20mA** of current to produce full brightness. (Ultra-bright LEDs can use 100mA or 1000mA or even more.) If you give them less current, they will be dimmer. If you give them too much current, they will burn out quicker. To calculate the resistance needed to allow a certain amount of current to flow, use Ohm’s Law.

We need to calculate the current flowing through the resistor and the rest of the circuit, so we subtract the voltage used by the LED(s) from the power source voltage. So out of the total voltage in the circuit (V_{BAT}_{ }), some of the voltage is dropped through the LED (V_{LED}) and the rest (V_{BAT} – V_{LED})** **is dropped across the remainder of the circuit (the resistor and the wires).

**R = V ÷ I**

**R = (**

*V*–_{BAT}*V*) ÷ I_{LED}By the way, the wires do have some resistance since nothing is a *perfect* conductor, but it’s so small we can ignore it for this example. (If you have dozens of feet of wire, then you should factor it in.)

In Ohm’s Law, **I** is current, expressed in Amperes, or Amps, or A. Remember that the LED spec is 20mA, or 20 milliAmps, or 20/1000A, or 0.02A.

Let’s say we have one red LED, a 9V battery, and we want the LED to look fully bright but get just a bit lower than 20mA of current. This is a common trick to extend the life of a battery because the LED is just a tiny bit dimmer and the naked eye can’t see the difference. So let’s give it **18mA**. 18 milliAmps can also be written as **0.018A**.

**R = (9.0 – 2.0) ÷ 0.018 = 388.9 Ohms**

But **resistors generally come in a limited set of values,** so we round up to the next common size which is** 390 Ohms**, but in practice if necessary anything from 300-500 ohms will work.

If we have **two** red LEDs, use the same equation but now with V_{LED} = 2V+2V = 4V.

**R = (9.0 – 4.0) ÷ 0.018 = 277.8 Ohm**

The closest standard resistor value is 300mA or sometimes 330mA.

If you use one of the calculators, you will see **the next highest standard value instead of the precise value**. (Ideally the calculator would show both!) Here’s the above scenario entered into http://ledcalculator.net (note that the LED voltage drop entered is *per LED*, the calculator will multiply it by number of LEDs for you):

Note that the calculator, and circuit schematics in general, usually show ground in a different place than power. Although usually in the physical world they will be right next each other in a single power source. Schematics omit that to keep the drawing as minimal and clear as possible.

Also note that in the above circuit with two LEDs, a total forward voltage of 4V would be too much if you were using a 3V coin cell battery. You would not be able to light up both LEDs no matter how low the resistance was. Does that mean you can only light up one LED with a coin cell battery? No!

You can also hook up LEDs in **parallel**. In this case each branch of the circuit gets the same voltage, so you calculate a resistor for every branch *individually*. (Although the voltage is the same from the battery in each branch, the total current the battery can supply must still be divided between them. In other words, there is always a fixed limit on the total power that any battery or power source can provide. In math terms, Power = Voltage x Current.) So here’s the same formula executed with a 3V battery, for each LED in parallel:

**R = (3.0 – 2.0) ÷ 0.018 = 55.6 Ohm (for LED 1)**

**R = (3.0 – 2.0) ÷ 0.018 = 55.6 Ohm (for LED 2)**

Here’s a complex example to illustrate the point:

Because of Ohm’s Law, *raising the voltage* with the same resistance must result in higher current. Likewise, using the same voltage but *lowering the resistance* must also result in higher current. And obviously, *lowering the voltage* with the same resistance would result in *lower* current. Since you can change either voltage or resistance to give the LEDs more or less current, which is best to adjust? Read this for more info – https://learn.adafruit.com/all-about-leds/which-to-adjust.

Three LEDs in series vs parallel on a breadboard. Notice the green highlights reminding you of what is internally connected. (Also note that the location of the black ground wire could be anywhere, it just so happens to be on the top on the series circuit for simplicity and taking up less space, and the bottom on the parallel circuit for clarity.)

## Leave a Reply