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Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Electronic: Material, Devices And Simple Circuits4 Marks
Question
A photodiode is an optoelectronic device in which current carriers are generated by photons through photo-excitation i.e., photo conduction by light. It is a p-n junction fabricated from a photosensitive semiconductor and provided with a transparent window so as allow light to fall on its function. A photodiode can turn its current ON and OFF in nanoseconds. So, it can be used as a fastest photo-detector.
  1. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5eV. It can detect a signal of wavelength:
  1. 4000nm.
  2. 6000nm.
  3. 4000Â.
  4. 6000Â.
  1. Three photo diodes $D_1, D_2$ and $D_3$ are made of semiconductors having band gap of 2.5eV, 2eV, and 3 eV, respectively. Which one will be able to detect light of wavelength 6000Â?
  1. $D_1$
  2. $D_2$
  3. $D_3$
  4. $D_1$ and $D_2$ both.
  1. Photodiode is a device:
  1. Which is always operated in reverse bias.
  2. Which of always operated in forward bias.
  3. In which photo current is independent of intensity of incident radiation.
  4. Which may be operated in forward or reverse bias.
  1. To detect light of wavelength 500nm, the photodiode must be fabricated from a semiconductor of minimum bandwidth of:
  1. 1.24eV
  2. 0.62eV
  3. 2.48eV
  4. 3.2eV
  1. Photodiode can be used as a photodetector to detect:
  1. Optical signals.
  2. Electrical signals.
  3. Both (a) and (b).
  4. None of these.
✓

Answer

  1. (c) 4000Â.
Explanation:
$\lambda_\text{max}=\frac{\text{hc}}{\text{E}}=\frac{6.6\times10^{-34}\times3\times10^8}{2.5\times1.6\times10^{-19}}$
= 5000Â
$\therefore\lambda=4000\widehat{\text{A}}<\lambda_\text{max}$
  1. (b) $D_2$
Explanation:
Energy of incident photon, $\text{E}=\frac{\text{hc}}{\lambda}$
$=\frac{6.6\times10^{-34}\times3\times10^8}{6\times10^{-7}\times1.6\times10^{-19}}=2.06\text{eV}$
The incident radiation can be detected by a photodiode if energy of incident photon is greater than the band gap.
As $D_2 = 2eV,$ therefore $D_2$ will detect these radiations.
  1. (a) Which is always operated in reverse bias.
Explanation:
Photodiode is a device which is always operated in reverse bias.
  1. (c) 2.48eV
Explanation:
Let $E_g$ be the required bandwidth. Then
$\text{E}_\text{g}=\frac{\text{hc}}{\lambda}$
Here, he = 1240eV nm,
$\lambda=500\text{nm}$
$\therefore\text{E}_\text{g}=\frac{1240\text{eVnm}}{500\text{nm}}=2.48\text{eV}.$
  1. (a) Optical signals.
Explanation:
A photodiode is a device which is used to detect optical signals.

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