LED ON AC

semiconductor devices work on DC and it is very difficult to glow LED directly on AC. In this tutorial i am going to tell you how to glow LED directly on AC. To make the circuit we need following components

• 10K 10W Resistor
• IN5408
• Red LED

the circuit diagram of the project is shown below

WORKING

LED requires 1.5 V supply and 20 milli amperes so to make circuit working we have two challenges

• drop 218V across resistor
• to protect LED from getting reverse biased 

To complete the above challenge we are using 10W RESISTOR. it is very important to use this resistor. now the question arises why only this value?

the resistance value is given by the relation

                         R = {(220-1.5)/20} X 1000

                             =  10.95 K ohms

The power dissipation across resistor is given by

                         P = V x I

                             = 220V x .02( 20 milli amperes)

                             = 4,4W

thats why we use 10 W resistor

What happen if we use quarter watt resistor?

the resistor would burst like a small bomb. this is because the large voltage drop across resistor would cause enough heating to make it burst.

ALSO we are using 1N5408 diode which have large surface area to dissipate the heat caused by such large voltage.

CAUTIONS

• circuit may burst if you use small resistor
• never touch the circuit while circuit is in operation.
• any change in the circuit may cause short circuiting.
• don't use Chinese digital multimeter as its wire may burst