Module 6 : LIGHT EMITTING DIODE (LED)
Lecture : LED - I
  Example 2 :
  In the circuit shown, the forward biased LED has a voltage drop of 1.5 volts. If the battery voltage is 6 V, calculate the resistance to be connected to the circuit, if the current through the LED is 15 mA. How much power is dissipated in the resistor ?
  Solution :
 
If $ r$is the internal resistance of the LED he current through the resistors is $ V/(R+r) = 0.015 $which gives $ R+r= 6/0.015 = 400\ \Omega$. As the drop across LED is 1.6 V, the internal resistance $ r$is $ 1.5/0.015= 100\ \Omega$. The external resistance to be connected is $ R=400-100 = 300\ \Omega$. The power rating of the resistor should at least be $ RI^2= 300\times 2.25\times 10^{-4} = 77.5$mW.
  The peak inverse voltage of an LED is rather low. This requires care to be taken when operating an LED with an alternating supply. A rectifying diode is added to the circuit to protect the diode from damage every other half cycle.
  \includegraphics{led9c.eps}
  Exercise :
  The working voltage of an LED is 1.8 volts. If the desired current flow is 15 mA, how much power is dissipated in a resistor that must be connected to an LED circuit operated on a d.c. voltage of 12 V.(Ans. 0.153 W)
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