How To Power A LED / Role Of Resistor In LED Power Supply

A resistor is an electronic component that limits the flow of current and finally divides the voltage in a circuit. So, a resistor limits the current through LED (Light Emitting Diode). If a resistor is not used then LED will consume the current until it melts. The voltage drop across a LED depends on the colour of LED. For example, the voltage drop across 5mm blue LED is 3.4 volts. So if you want to lit up a 5mm blue LED with 5-volt power supply, you need a 320 Ω resistor. How?

Let's find it out -

At first, we need to find voltage drop across the resistor (Vr)?

We know, Voltage drop across resistor (Vr) = Specified supply voltage (Vs) - LED voltage drop (Vf)

Vr = Vs - Vf

So we have 5 V as specified supply voltage and the voltage drop across 5mm blue LED is 3.4 volts.
Therefore, voltage drop across resistor (Vr) = 5 - 3.4 = 1.6 volts.

We need to pass 5 mA (milliampere) current through 5mm blue LED to gives the good visibility of the LED. LED at 1 mA is easily visible.

i.e., Specified desired current (I) through LED = 5 mA or .005 ampere.

So, the resistance (R) = Voltage drop across resistor / Current through LED

R = V / I (Ohms law)
R= Vr / I R=Vs - Vf / I

     R = Vs - Vf / I
         = 5 - 3.4 / .005
         = 320 Ω.

The resistance of 320 Ω will give voltage drop across the resistor of 1.6 volts and the remaining 3.4 Volts drop on the blue LED.

If you know the forward voltage of a LED then you can easily find the voltage across the resistor by deducting the forward voltage with the supply voltage.

Now, if you know voltage drop across the resistor and the resistance capacity of a resistor than you can easily identify how much current the LED will consume.

If you take the example of 5mm blue LED,

Vr = 5- 3.4 = 1.6

Therefore, current I = V / R = 1.6 / 320 = .005 Ampere or 5 milli ampere.

Now, let us take the example of 5mm red LED,

Red LED has forward voltage drop = 1.8 to 2.2 volts.

Vr = 5 - 2 = 3 (Vf is taken as 2 volts as an average)
I = 3 / 320 = .009 Ampere or 9 milli ampere.

This indicates that red LED needs slightly higher current than the blue LED and the resistance should be more than 330 Ω.

But the maximum current through red LED should be below 20 mA otherwise it will melt.

So, in case of red LED, R = 5v-2v / .009amp = 333 Ω resistor is needed to lit up the 5 mm red LED with 5v power supply.

In case of green LED, the forward voltage drop of green LED is 2.2 volts. So, R = 5 - 2.2 / .005 = 560 Ω resistor is needed to lit up a 5mm green LED. 5 milli amp of current passes through it and the supply voltage is 5 volts.

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