In this lesson, you will apply the Ohm’s law and learn how to use LED safely.

## Forward voltage, or Vf

LEDs drops voltage, which is called “Forward voltage”, or $V_{f}$. Typical $V_{f}$ are shown below.

Color $V_{f}$ in V
Red 1.6 - 2
Blue 2.5 - 3
Yellow 2.1 - 2.2
Amber 2 - 2.1
Green 1.9 - 4

When the applied voltage is less than $V_{f}$, the LED acts like an open switch, and no current flows. To turn on an LED, you need to apply voltage more than $V_{f}$. When the voltage is more than $V_{f}$, the voltage across an LED is always $V_{f}$.

What if an LED is connected to a 5 V battery? It will break because an LED is not a resistor, and its resistance is (almost) zero. It just drops voltage. The Ohm’s law says:

$I = { V \over R } = { { V_{cc} - V_{f} } \over 0 } = { { 5 - V_{f} } \over 0 } = \infty$

In theory, infinite current will flow thorough the LED.

In reality, infinite current will not flow because the battery has its own resistance, and cannot source infinite current. However, it will be enough current to break the LED.

## Current limiting resistor

To protect the LED, a resistor is used very often. The resistor limits the current that flows thorough the LED. To calculate the value of the resistor, you need $I_{f}$, or forward current, which is the current flows into the LED.The data sheet, or the product page, of the LED shows typical $I_{f}$ for the LED. A typical value is between 10 mA and 20 mA. As recent LEDs are very bright, even a few mA would work. Then you need the $V_{cc}$, the voltage of the power source. Here, two AAA batteries, or 3 V, are used as an example. $V_{f}$ is 1.8 V, and $I_{f}$ is 10 mA. The voltage across the current limiting resistor is:

$V_{cc} - V_{f} = 3 - 1.8 = 2.2 (V)$

From the Ohm’s law:

$R = { V \over I } = { { V_{cc} - V_{f} } \over I_{f} } = { { 3 - 1.8 } \over { 0.01 } } = 120 (ohm)$

## Power rating

Another important factor in choosing current limiting resistor is power rating. The power rating of small resistors you typically find are 1/8 W, or 125 mW. The power used by the current limiting resistor is:

$P = VI = 2.2 \times 0.01 = 0.022 (W) = 22 (mW)$

As it is much lower than 125 mW, a typical 1/8 W resistor can be used.

Use two times higher wattage rating than you actually need. If $P$ is 100 mW, do not use 1/8 W resistor. Use 1/4 W instead.

## Other lessons

Other lessons in Electronics Basic Course: