Pooja kulhade last edited by
answer is d. please explain
Vivaan gandhi last edited by
In reverse biased p-n junction diode, the free electrons begin their journey at the negative terminal whereas holes begin their journey at the positive terminal. Free electrons, which begin their journey at the negative terminal, find large number of holes at the p-type semiconductor and fill them with electrons.
The atom, which gains an extra electron, becomes a charged atom or negative ion or motionless charge. These negative ions at p-n junction (p-side) oppose the flow of free electrons from n-side.
On the other hand, holes or positive charges, which begin their journey at the positive terminal, find large of free electrons at the n-type semiconductor and replace the electrons position with holes.
The atom, which loses an electron, becomes a charged atom or positive ion. These positive ions at p-n junction (n-side) oppose the flow of positive charge carriers (holes) from p-side.
If the reverse biased voltage applied on the p-n junction diode is further increased, then even more number of free electrons and holes are pulled away from the p-n junction. This increases the width of depletion region. Hence, the width of the depletion region increases with increase in voltage. The wide depletion region of the p-n junction diode completely blocks the majority charge carriers. Hence, majority charge carriers cannot carry the electric current.