Test: Band Structure Of An Open-Circuited PN Junction


10 Questions MCQ Test Electronic Devices | Test: Band Structure Of An Open-Circuited PN Junction


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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 Eo given that Nd=1.5*1010cm-3, Na=1.5*1010cm-3 at temperature 300K?

Solution:

 Eo=kTln((Nd*Na)/(ni)2)
Substituting k=1.38*10-23/K, T=300k and the values ofNd,Naand ni,
We get
Eo=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 Eo when pno=104cm-3 and ppo=1016cm-3 at T=300K.

Solution:

Eo=kTln(ppo/pno)
Substituting the values, we get
Eo=0.7eV.

QUESTION: 7

Calculate the value of Dp when µp=400cm/s and VT=25mV.

Solution:

Dp= µp*VT
=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 ppo=1016cm-3, pno=104cm-3 and Vt=25mV.

Solution:

 Vo=VT ln⁡(ppo/pno )
=25*10-3*ln(1016/104)
=0.69V.

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