# LED’s Resistor Calculation

Updated: Apr 28, 2020

Ohms law is a must in circuit design to calculate, Current, Resistance and Voltage in the different sections of a circuit. Ohms law simply represents the relationship between voltage, current and resistance. Let us see how Ohms law is applied to select the current limiting resistor for LED.

Ohms law can be expressed like this-

V (Voltage) = I X R

I (Current) = V / R – in Amps

R (Resistance) = V / I – in Ohms

Suppose the input voltage is 8 volts, current is 2 Amps and resistor is 4 Ohms. Then

V = I x R = 2 x 4 = 8 Volts

I = V / R = 8 / 4 = 2 Amps

R = V / I = 8 / 2 = 4 Ohms

This is the basic method to calculate voltage, current and resistance in circuit design.

**LED Series Resistor-**

A current limiting resistor is must for LED, otherwise it will burn instantly. The value of the resistor must be proper to get sufficient brightness and to increase the life of LED. Different LEDs have different forward voltage drop across them. These are the voltage drops of general type LEDs

Red – 1.7 V

Green – 1.8 V

Yellow – 1.7 V

Blue – 3.1 – 3.6 V

White 3- 3.6 V

But this may slightly vary depending on the type and make of LEDs

**LED RESISTOR-**

Current through the LED must be between 10 to 25 Milli-Ampere. Optimum current for good brightness is 20 mA. If the value of the resistor increases, current reduces followed by a reduction in the brightness. But this will increase the life of LED. If the value of resistor decreases, current increases followed by an increase in brightness.

Suppose the input voltage is 12 volts and the LED is standard diffuse type 5mm Red LED. It has a voltage drop of 1.7 volts. Fix the current through the LED as 20 mA (0.02 Amps)

Formula for selecting the resistor is

R = Vs – Vf / If

R is resistance in Ohms, Vs is input voltage, Vf is the Forward voltage drop of LED and If is the current through LED.

Out of the four values, we know three values

Vs = 12 V

Vf = 1.7 V

If = 0.02 A

So the value of resistor is-

R = Vs – Vf / If = 12 V – 1.7 V / 0.02 A = 10.3 V / 0.02 A = 515 Ohms

So when we select 515 Ohms resistor, exactly 0.02 Amps or 20 Milli Ampere current flows through the LED. But 515 Ohms resistor is hard to get, so we can select either 470 Ohms or 560 Ohms. Then the current will be

I = V / R = 10.3 / 470 = 0.021 Amps or 21 mA.