Electronic Devices and Circuits Questions and Answers for Experienced people on “Band Structure of an Open-Circuited p-n junction”.
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?
A. True
B. False
Answer: A
Clarification: 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.
2. Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction?
A. Eo = Ecn – Ecp
B. Eo = Ecp – Ecn
C. Eo = Ecp + Ecn
D. Eo = -Ecp – Ecn
Answer: B
Clarification: 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.
3. Calculate the Eo given that Nd=1.5*1010cm-3, Na=1.5*1010cm-3 at temperature 300K?
A. 1.5*1010eV
B. 0.256eV
C. 0eV
D. 4.14*10-21eV
Answer: C
Clarification: 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.
4. In a p-n junction, the valence band edge of the p material is greater than which of the following band?
A. Conduction band edge of n material
B. Valence band edge of n material
C. Conduction band edge of p material
D. Fermi level of p material
Answer: B
Clarification: 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.
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)
A. Ecn – E f = (1/2)*EG – E1
B. Ecn – E f = (1/2)*EG – E2
C. Ef – Ecp = (1/2)*EG – E1
D. Ecn – Ef = (1/2)*EG + E1
Answer: A
Clarification: From the energy band diagram of the p-n junction, the option ‘a’ satisfies that band diagram.
6. Calculate the value of Eo when pno=104cm-3 and ppo=1016cm-3 at T=300K.
A. 1meV
B. 0.7meV
C. 0.7eV
D. 0.1meV
Answer: C
Clarification: Eo=kTln(ppo/pno)
Substituting the values, we get
Eo=0.7eV.
7. Calculate the value of Dp when µp=400cm/s and VT=25mV.
A. 1
B. 0.01
C. 0.1
D. 10
Answer: C
Clarification: Dp= µp*VT
=400*10-2*25*10-3
=0.1.
8. What is the value of kT at room temperature?
A. 0.0256eV
B. 0.25eV
C. 25eV
D. 0.0025eV
Answer: A
Clarification: kT=1.38*10-23*300K
=4.14*10-21/ (1.6*10-19)
=0.0256eV.
9. Is Vo depends only on the equilibrium concentrations. Is it true or false?
A. True
B. False
Answer: A
Clarification: 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 calculateC.
10. Calculate Vo when ppo=1016cm-3, pno=104cm-3 and Vt=25mV.
A. 69V
B. 6.9V
C. 0.69V
D. 0.069V
Answer: C
Clarification: Vo=VT ln(ppo/pno )
=25*10-3*ln(1016/104)
=0.69V.
for Experienced people, electronic-devices-circuits-questions-answers/”>.