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QUESTION: 1

The conduction band edge in the p material is not at the same level to that of conduction band edge in the n material. Is it true or false?

Solution:

In a p-n junction diode, the energy levels of the p material and n material will not be at same level. They will be different. So, the conduction band edge as well as the valence band edge of the p material will not be same to that of the n material.

QUESTION: 2

Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction?

Solution:

The shift in the energy of the energy level will be the difference of the conduction band edge of the p material and conduction band edge of n material. In the energy level diagram, the conduction band edge of p material is higher than that of the n material.

QUESTION: 3

Calculate the E_{o} given that N_{d}=1.5*10^{10}cm^{-3}, N_{a}=1.5*10^{10}cm^{-3} at temperature 300K?

Solution:

E_{o}=kTln((N_{d}*N_{a})/(n_{i})^{2})

Substituting k=1.38*10^{-23}/K, T=300k and the values ofN_{d},N_{a}and ni,

We get

E_{o}=0eV.

QUESTION: 4

In a p-n junction, the valence band edge of the p material is greater than which of the following band?

Solution:

When the p-n junction is formed, the energy levels of the p- material go higher than the n material. That’s why the valence band of the p material will be greater than that of the n material.

QUESTION: 5

Which of the following equations represent the correct expression for the band diagram of the p-n junction? (E1=difference between the fermi level of material and conduction band of n material and E2=difference between the conduction band of n material and fermi level of n material)

Solution:

From the energy band diagram of the p-n junction, the option ‘a’ satisfies that band diagram.

QUESTION: 6

Calculate the value of E_{o} when p_{no}=10^{4}cm^{-3} and p_{po}=10^{16}cm^{-3} at T=300K.

Solution:

E_{o}=kTln(p_{po}/p_{no})

Substituting the values, we get

E_{o}=0.7eV.

QUESTION: 7

Calculate the value of Dp when µp=400cm/s and V_{T}=25mV.

Solution:

Dp= µp*V_{T}

=400*10^{-2}*25*10^{-3}

=0.1.

QUESTION: 8

What is the value of kT at room temperature?

Solution:

kT=1.38*10^{-23}*300K

=4.14*10^{-21}/ (1.6*10^{-19})

=0.0256eV.

QUESTION: 9

Is Vo depends only on the equilibrium concentrations. Is it true or false?

Solution:

Vo is the contact potential of the junction when the junction is in equilibrium. If, the junction is not in the equilibrium, Vo can’t be calculated.

QUESTION: 10

Calculate Vo when p_{po}=10^{16}cm^{-3}, p_{no}=10^{4}cm^{-3} and Vt=25mV.

Solution:

V_{o}=V_{T} ln(p_{po}/p_{no} )

=25*10^{-3}*ln(10^{16}/10^{4})

=0.69V.

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