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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Band Structure of an Open-Circuited PN Junction - Electrical Engineering (EE) MCQ

Test: Band Structure of an Open-Circuited PN Junction - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Band Structure of an Open-Circuited PN Junction

Test: Band Structure of an Open-Circuited PN Junction for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Band Structure of an Open-Circuited PN Junction questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Band Structure of an Open-Circuited PN Junction MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Band Structure of an Open-Circuited PN Junction below.
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Test: Band Structure of an Open-Circuited PN Junction - Question 1
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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?

Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 1

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.

Test: Band Structure of an Open-Circuited PN Junction - Question 2
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Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction?

Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 2

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.

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Test: Band Structure of an Open-Circuited PN Junction - Question 3
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 Calculate the Eo given that Nd=1.5*1010cm-3, Na=1.5*1010cm-3 at temperature 300K?

Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 3

 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.

Test: Band Structure of an Open-Circuited PN Junction - Question 4
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 In a p-n junction, the valence band edge of the p material is greater than which of the following band?
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 4

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.

Test: Band Structure of an Open-Circuited PN Junction - Question 5
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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)
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 5

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

Test: Band Structure of an Open-Circuited PN Junction - Question 6
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Calculate the value of Eo when pno=104cm-3 and ppo=1016cm-3 at T=300K.
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 6

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

Test: Band Structure of an Open-Circuited PN Junction - Question 7
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Calculate the value of Dp when µp=400cm/s and VT=25mV.
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 7

Dp= µp*VT
=400*10-2*25*10-3
=0.1.

Test: Band Structure of an Open-Circuited PN Junction - Question 8
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 What is the value of kT at room temperature?
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 8

 kT=1.38*10-23*300K
=4.14*10-21/ (1.6*10-19)
=0.0256eV.

Test: Band Structure of an Open-Circuited PN Junction - Question 9
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 Is Vo depends only on the equilibrium concentrations. Is it true or false?
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 9

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.

Test: Band Structure of an Open-Circuited PN Junction - Question 10
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Calculate Vo when ppo=1016cm-3, pno=104cm-3 and Vt=25mV.
Detailed Solution for Test: Band Structure of an Open-Circuited PN Junction - Question 10

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

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