The semiconductor used for the fabrication of visible LEDs must at least have a band gap of:

CONCEPT:

Light Emitting Diode:

• It is a heavily doped p-n junction that under forward bias emits spontaneous radiation.
• The diode is encapsulated with a transparent cover so that emitted light can come out.
• When the diode is forward biased, electrons are sent from n to p (where they are minority carriers) and holes are sent from p to n (where they are minority carriers).
• At the junction boundary, the concentration of minority carriers increases compared to the equilibrium concentration (i.e., when there is no bias).
• Thus at the junction boundary on either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction.
• On recombination, the energy is released in the form of photons. Photons with energy equal to or slightly less than the bandgap are emitted.
• When the forward current of the diode is small, the intensity of light emitted is small.
• As the forward current increases, the intensity of light increases and reaches a maximum. Further, an increase in the forward current results in a decrease in light intensity.
• LEDs are biased such that the light-emitting efficiency is maximum.
• The V-I characteristics of a LED are similar to that of a-Si junction diode.
• But the threshold voltages are much higher and slightly different for each color.
• The reverse breakdown voltages of LEDs are very low, typically around 5V.
• So care should be taken that high reverse voltages do not appear across them.
• LEDs can emit red, yellow, orange, green, and blue light that are commercially available.
• The semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV (spectral range of visible light is from about 0.4 μm to 0.7 μm, i.e., from about 3 eV to 1.8 eV).
• The compound semiconductor Gallium Arsenide – Phosphide is used for making LEDs of different colors.
• Gallium Arsenide is used for making infrared LED.

EXPLANATION:

• We know that the semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV (spectral range of visible light is from about 0.4 μm to 0.7 μm, i.e., from about 3 eV to 1.8 eV). Hence, option 3 is correct.