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Explain the design, working, uses and benefits of light emitting diode. |
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Answer» Solution :It is a heavily doped p-n junction which under forward bias emits spontaneous radiation, hence DIODE have to provide forward bias to operate. The diode is encapsulated with transparent COVER so that emitted light can come out. The outline of the LED is shown in figure.  When the diode is forwardbiased, electrons are sent from `n to p` where they are minority carriers and hole are sent from `p to n` where they are minority carriers. At the junction boundary the concentration of minority crriers increases when there no bias. Thus at the junction boundaryon either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction as a result, the energy is RELEASED in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. When the forward current of the diode is small, the intensity of light emitted is small. As the forward current increases, intensity of light increases and reaches a maximum. Further increase in the forward current results in DECREASE of light intensity. LED are biased such that the light emitting efficiency is maximum. The V-I characteristics of a LED is similar to that of a Si junction diode. But the threshold voltages are much HIGHER and slightly different for each colour. The reverse breakdown voltages of LED are very low, typically around 5V. So care should be taken that high reverse voltages do not appear across them. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap `E_(g)` of 1.8eV (spectral range of visible light is from about 0.4 `mu`m to 0.7`mu`m, that is from about 3 eV to 1.8eV). Hence `E_(g_(1))=(hc)/(lambda_(1)) and E_(g_(2)) = (hc)/(lambda_(2))` `=(6.625xx10^(-34)xx3xx10^(8))/(4xx10^(-7)xx1.6xx10^(-19))eV` `=3.10eV` `=3eV` and `E_(g_(2))=(hc)/(lambda_(2))=(6.625xx10^(-34)xx3xx10^(8))/(7xx10^(-7)xx 1.6xx10^(-19))eV` `=1.774 eV` `=1.8eV` The compound semiconductor Gallium Arsenidephosphide `[GaAs_(1-x)P_(x)]`is used for making LEDs of different colours. `GaAs_(0.6)P_(0.4)[E_(g)=1.9eV]` is used for red LED. GaAs `[E_(g)=1.4eV]` is used for making infrared LED. The uses of LED are in remote controls, burglar alarm systems (For prevent stealing) optical communication. Extensiveresearch is being done for developingwhite LEDs which can replace incandecent lamps. LEDs have following advantagesover conventionalincandescent low powerlamps. (1) Low operational voltageand less power. (2) Fast action and no warm-up time required. (3) The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly monochromatic. (4) Long life and ruggedness. (5) Fast ON/OFF switching capability. |
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