Question

Write the important considerations which are to be taken into account while fabricating a p-n junction diode to be used as a light emitting diode (LED). What should be the order of band gap of an LED, if it is required to emit light, in the visible range? Draw a circuit diagram and explain its action.

Answer 1

Just keep in mind that a light-emitting diode converts electrical energy into light energy. Also, A Light-emitting diode (LED) emits monochromatic light as well as white light. We can also say that light-emitting diode (LED) is a semiconductor that will glow when a voltage is applied to it.

Complete step by step solution:
Firstly, we will know about the light-emitting diode (LED).
A Light-emitting diode (LED) is defined as the special heavily doped p-n junction diode which emits spontaneous radiation when forward biased.

Now, if we want to fabricate a p-n junction, the important considerations to be taken are given as:
i) It should be a heavily doped p-n junction.
ii) The reverse breakdown of LEDs should be very low.
ii) The reverse breakdown of LED should be typically around $SV$.
Now, the semiconductor used for the LED to emit light in the visible range should have a bandgap of about $3eV$ to $1.8eV$.
Now, the circuit diagram for the light-emitting diode is given by

Now, we know that when a p-n junction is forward biased the electrons-injected to the p-side of the junction diode falls from the conduction band to the valence band and they will combine with the holes of the valence band. This process is equivalent to the jumping of electrons from the higher energy state to a lower energy state. Here, the energy will be released when the electron-hole recombines in the form of visible light.

Additional Information:
As we know, the energy of a photon of visible light is given by $h\nu = {E_g}$, where ${E_g}$ is the energy gap between the conduction band and valence band and $\nu$ is the frequency of emitted visible radiation. Now, the wavelength of visible light is given by
${E_g} = \dfrac{{hc}}{\lambda }$
$\Rightarrow \,\lambda = \dfrac{{hc}}{{{E_g}}}$
$\Rightarrow \,{E_g} \simeq \dfrac{{1.242}}{\lambda }eV$
Therefore, this is the expression. By knowing the value of energy released we can calculate the wavelength and vice-versa.

Note: Now, just remember that the V-I characteristics of LED will be the same as that of elemental semiconductor say silicon diode. Also, the knee voltage of the LED is much higher than that of the $Si$diode. But, LED’s get damaged at low reverse voltage, say $5V$.