Advertisements
Advertisements
प्रश्न
Explain how a potential barrier is developed in a p-n junction diode.
Advertisements
उत्तर १
During the formation of p-n junction, and due to the concentration gradient across p-, and n- sides, holes
diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carries gives rise to diffusion current across the junction. When an electron diffuses from n → p, it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n → p, a layer of positive charge (or positive space-charge region) on n-side of the junction is developed.
उत्तर २
In a p-n junction diode, holes are the majority carriers on the p side whereas electrons are the majority carriers on the n-side of the semiconductor. Due to the diffusion of majority carriers from p-region to n-region, the p-region becomes less positive and n-region becomes less negative. An imaginary voltage is developed across the junction which prevents further movement of majority carriers across the junction. The voltage so developed is known as the potential barrier.
संबंधित प्रश्न
With the help of a neat circuit diagram, explain the working of a photodiode.
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?
Write briefly the important processes that occur during the formation of p−n junction. With the help of necessary diagrams, explain the term 'barrier potential'.
Briefly explain its working. Draw its V - I characteristics for two different intensities of illumination ?
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~μA). What is the reason, then, to operate the photodiode in reverse bias?
Draw the circuit diagram of a full wave rectifier using p-n junction diode.
Explain its working and show the output, input waveforms.
Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?
An ideal diode should pass a current freely in one direction and should stop it completely in the opposite direction. Which is closer to ideal-vacuum diode or a p-njunction diode?
Explain the three processes involved in the working of a solar cell.
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
