Advertisements
Advertisements
Question
Explain how a potential barrier is developed in a p-n junction diode.
Advertisements
Solution 1
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.
Solution 2
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.
RELATED QUESTIONS
State its any ‘two’ uses of photodiode.
How is a photodiode fabricated?
Show the output waveforms (Y) for the following inputs A and B of (i) OR gate (ii) NAND gate ?

The plate resistance of a triode is 8 kΩ and the transconductance is 2.5 millimho. (a) If the plate voltage is increased by 48 V and the grid voltage is kept constant, what will be the increase in the plate current? (b) With plate voltage kept constant at this increased value, by how much should the grid voltage be decreased in order to bring the plate current back to its initial value?
The wavelength and intensity of light emitted by a LED depend upon ______.
What is the magnitude of the potential barrier across a Ge p-n junction?
Name the device which converts the change in intensity of illumination to change in electric current flowing through it. Plot I-V characteristics of this device for different intensities. State any two applications of this device.
For LED's to emit light in visible region of electromagnetic light, it should have energy band gap in the range of:
Which one of the following is not the advantage of LED?
What energy conversion takes place in a solar cell?
