AEL

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
====== LED Dropping Resistor Selection ======

An LED requires a resistor in series with either the anode or its cathode to reduce the current flowing through the LED to a safe level it can handle. Failure to add a series resistor (or choosing too small of a value) can cause the LED to be instantly destroyed after applying power.

{{:tools:led-series-resistor.png?400|}}

===== What you need to know =====

You need to know the forward voltage drop of the LED (**Vf**) that you're going to use; this can be measured with a simple multimeter in diode mode. However, having said that, some cheap multimeters don't have the ability to measure LED forward voltage at all. In this case, if you have access to the LED's datasheet, the information will be in there. If you don't (or know what datasheet applies), you can buy and use a cheap component tester like the ones shown below that are available all over the internet on shopping sites.

{{:tools:tester.png?200|}}
{{:tools:tester2.jpg?200|}}

You also need to know the supply voltage (**Vs**) and the desired current (**Iled**) the LED will be consuming. For most purposes, 20mA is about right. Depending on luminosity (milli-candelas - **mCd**) of the LED, a lower current can be used in the below calculation to get the required "brightness"; such as a lower brightness. This means the lower the current consumed by the LED, the higher the resistor value will be.

===== Simple calculation =====

For the purposes of this calculation, we'll choose an LED with a **Vf** of 2.2V, a supply voltage of 12V and a current of 20mA.

So:

**Rled = Vs - Vf / Iled**

**12 - 2.2 = 9.8 / 0.02 = 490**

So, 470 ohms should be fine (as it's the next closest value), but if you want to err on the side of caution - make the resistance to the next **higher** value; 560 ohm. It would be a good idea to calculate the amount of dissipation the resistor is going to experience so as to choose the correct wattage.

**P = V * I**; where **P** is in watts, **V** is in volts and **I** is in amps.

So (using the voltage drop across the resistor of 9.8V):

**9.8 * 0.02 = 0.196W (196mW).**

So, a 1/4W resistor should be fine, all though might get a bit "toasty" - so a 1/2W would be the better choice. It's always better to be over rated rather than under rated. This becomes more important the higher the **Vs** is. Let's take for example a **Vs** of 45V: **45 - 2.2 = 43.8 / 0.02 = 2,140 (2.2k)**; **43.8 * 0.02 = 0.856W!** So, at minimum, a 2W resistor would be required as even though a 1W would "work", it will get quite hot.

I hope you found this article useful ^_^