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Test: Equilibrium Electrons & Holes Concentration - Electronics and Communication Engineering (ECE) MCQ


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6 Questions MCQ Test Electronic Devices - Test: Equilibrium Electrons & Holes Concentration

Test: Equilibrium Electrons & Holes Concentration for Electronics and Communication Engineering (ECE) 2024 is part of Electronic Devices preparation. The Test: Equilibrium Electrons & Holes Concentration questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Equilibrium Electrons & Holes Concentration MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Equilibrium Electrons & Holes Concentration below.
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Test: Equilibrium Electrons & Holes Concentration - Question 1

Consider the following uniformly doped n-type Si sample of length 100 μm, maintained at T = 300K,  

Light incident on the surface is absorbed at x = 0, resulting in pn (0) - p0 = 108 / cm3 excess holes at x = 0. The generation rate for x > 0 is zero)
The expression for minority charge carrier (holes) as function of x is

Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 1

Initial (without light) hole concentration under thermal equilibrium is

After light incident the excess hole concentration is


Test: Equilibrium Electrons & Holes Concentration - Question 2

A sample of silicon at T = 300 K is doped with boron at a concentration of 1.5 × 1015 cm-3 and with arsenic at a concentration of 8 × 1014 cm-3. The intrinsic carrier concentration of Si at T = 300 K is 1010 cm-3. The material is

Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 2

Boron is an acceptor atom (i.e a p-type impurity)
Na = 1.5 × 1015 cm-3
Arsenic is a donor atom (i.e. an n-type impurity)
Nd = 8 × 1014 cm-3
Since, Na > Nd, the given semiconductor is a p-type semiconductor.
Also, since Na - Nd ≫ ni the majority carrier hole concentration will be:
p0 = Na - Nd
p0 = 1.5 × 1015 – 8 × 1014 cm-3
p0 = (15 - 8) × 1014 cm-3
p0 = 7 × 1014 cm-3
Using mass action law

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*Answer can only contain numeric values
Test: Equilibrium Electrons & Holes Concentration - Question 3

Consider a silicon sample doped with ND = 1×1015/cm3 donor atoms. Assume that the intrinsic carrier concentration ni = 1.5×1010/cm3. If the sample is additionally doped with NA = 1×1018/cm3 acceptor atoms, the approximate number of electrons/cm3 in the sample, at T = 300 K, will be ______.


Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 3

According to law of mass action 
Application:
The newly created semiconductor is a compensated semiconductor with doping profile, 
The hole concentration due to NA is very large than Nd 
here, NA >> ND
So, we can neglect ND and the electron concentration can be found as, using law of mass action 

*Answer can only contain numeric values
Test: Equilibrium Electrons & Holes Concentration - Question 4

A pure silicon crystal with recombination constant αT has rate of recombination of 4.3 × 105 (cm3-sec)-1. The  intrinsic carrier concentration is  1.5 × 1010 cm-3. A dopant with concentration 4.7 × 1016 cm-3 is added then the new constant of recombination at equilibrium is ________ × 10-15 s-1


Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 4

At equilibrium recombination rate R = carrier generation rate G
i.e. R = G = αTnp = αTni2
Even in extrinsic semiconductors, the equilibrium concentration of product np is the same hence αT will remain constant.

Test: Equilibrium Electrons & Holes Concentration - Question 5

A sample of silicon at T = 300 K is doped with boron at a concentration of 1.5 × 1015 cm-3 and with arsenic at a concentration of 8 × 1014 cm-3. The intrinsic carrier concentration of Si at T = 300 K is 1010 cm-3.
Which of the following statements about the semiconductor material is/are false?

Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 5




Test: Equilibrium Electrons & Holes Concentration - Question 6

Holes are injected into n-type Ge so that the at the surface of the semiconductor hole concentration is 1014/cm3. If diffusion constant of hole in Ge is 49cm2/sec and minority carrier life time is τp = 10-3 sec. Then the hole concentration Δp at a distance of 4mm from the surface is ______1014/cm3.

Detailed Solution for Test: Equilibrium Electrons & Holes Concentration - Question 6


= 1.6 × 1013/cm3
= 0.16 × 1014/cm3

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