Advertisements
Advertisements
Question
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true?
Options
(Eg)Si < (Eg)Ge < (Eg)C
(Eg)C < (Eg)Ge > (Eg)Si
(Eg)C > (Eg)Si > (Eg)Ge
(Eg)C = (Eg)Si = (Eg)Ge
Solution
(Eg)C > (Eg)Si > (Eg)Ge
Explanation:
Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.
The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge
APPEARS IN
RELATED QUESTIONS
If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is(ii) reveres biased?
With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason?
The width of depletion region of p-n junction diode is _______.
(A) 0.5 nm to 1 nm
(B) 5 nm to 10 nm
(C) 50 nm to 500 nm
(D) 500 nm to 1000 nm
Draw V − I characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.
Show the output waveforms (Y) for the following inputs A and B of (i) OR gate (ii) NAND gate ?
Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction ?
How does a light emitting diode (LED) work? Give two advantages of LED’s over the conventional incandescent lamps.
Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?
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 the band gap of an LED, if it is required to emit light in the visible range? Draw a circuit diagram and explain its action.
Explain photodiode.
What is the magnitude of the potential barrier across a Ge p-n junction?
Pressure P varies as P = `alpha/beta "exp" (- (alpha x)/"k"_"BT")`, where x denotes the distance, kB is the Boltzmann's constant, T is the absolute temperature and α and β are constant. The dimension of β is ______.
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. lt can detect a signal of wavelength ______.
For LED's to emit light in visible region of electromagnetic light, it should have energy band gap in the range of:
Read the following paragraph and answer the questions.
|
LED is a heavily doped P-N junction which under forward bias emits spontaneous radiation. When it is forward-biased, due to recombination of holes and electrons at the junction, energy is released in the form of photons. In the case of Si and Ge diode, the energy released in recombination lies in the infrared region. 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 of 1.8 eV. The compound semiconductor Gallium Arsenide – Phosphide is used for making LEDs of different colours.
|
- Why are LEDs made of compound semiconductor and not of elemental semiconductors?
- What should be the order of bandgap of an LED, if it is required to emit light in the visible range?
- A student connects the blue coloured LED as shown in the figure. The LED did not glow when switch S is closed. Explain why?
OR
iii. Draw V-I characteristic of a p-n junction diode in
(i) forward bias and (ii) reverse bias
Briefly explain how emf is generated in a solar cell.
Draw solar cells of I-V characteristics.
