Tuesday, July 25, 2023

H C Verma solutions, SEMICONDUCTORS AND SEMICONDUCTOR DEVICES, Chapter-45, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

EXERCISES, Q11 - Q20


       11. The conductivity of a pure semiconductor is roughly proportional to T3/2 e-ΔE/2kT where ΔE is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4 K to 300 K? 

ANSWER: Given, T =4 K, T' =300 K;  
and ΔE =0.74 eV, ΔE' =0.67 eV.

From the given proportionality of the conductivity, the factor by which the conductivity increases is  
=(T'3/2/T3/2)(e-ΔE'/2kT'/e-ΔE/2kT)
=(T'/T)3/2 e(ΔE/T -ΔE'/T')/2k
=(300/4)3/2 e(0.74/4 -0.67/300)/2k
=650*e¹⁰⁶⁰ =650*e⁵⁶⁰e⁵⁰⁰
{after putting the value of k}
=650*1.6x10²⁴³*1.4x10²¹⁷
=1.5x10⁴⁶³.
            





 

       12. Estimate the proportion of the boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7x10¹⁵ holes per cubic meter. The density of silicon is 5x10²⁸ atoms per cubic meter.   

ANSWER: In pure silicon, the number of holes and the number of conduction electrons are equal. Hence the total number of charge carriers in silicon per cubic meter is, 
=2*7x10¹⁵
=14x10¹⁵.

The conductivity will increase 100 times if the number of charge carriers increases 100 times. So it will be equal to 100x14x10¹⁵ =14x10¹⁷. Though a boron atom adds one hole in the semiconductor, it will also replace a silicon atom in the material. Suppose X number of boron atoms are added to one cubic meter of silicon replacing X number of silicon atoms. 
         Given that number of silicon atoms in a cubic meter N =5x10²⁸.
       Now the number of conduction electrons per cubic meter of the doped silicon (provided by silicon atoms only)
={(N-X)/N}*7x10¹⁵ 
    This is also the number of holes provided by silicon in the doped semiconductor.
     Contribution of holes by boron =X. 

   Equating the total number of charge carriers in view of the increased conductivity. We get, 
X+{(N-X)/N}*7x10¹⁵+{(N-X)/N}*7x10¹⁵ =14x10¹⁷
→XN+2(N-X)*7x10¹⁵ =N*14x10¹⁷
→XN +N*14x10¹⁵ +X*14x10¹⁵=N*14x10¹⁷ 
→X(N -14x10¹⁵) =N*14x10¹⁵*(100 -1)
Putting the value for N, we get
→X =5x10²⁸*14x10¹⁵*99/(5x10²⁸ -14x10¹⁵)
       ≈1.4x10¹⁸ 

Hence the proportion of boron impurity is
    =X/N
    =1.4x10¹⁸/(5x10²⁸-1.4x10¹⁸)
    ≈1.4x10¹⁸/5x10²⁸
    ≈1/3.5x10¹⁰  
i.e. 1 in about 3.5x10¹⁰.         





 
       13. The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of the conduction electrons in germanium is 6x10¹⁹ per cubic meter. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to 2x10²³ per cubic meter. Find the concentration of the holes in the doped germanium.   

ANSWER:  Before doping the germanium, the concentration of conduction electrons
 =concentration of holes =6x10¹⁹ per m³ 
After doping the concentration of conduction electrons =2x10²³ per m³, 
Let the concentration of holes per m³ after doping =X. 

  Since the product of the concentration of holes and conduction electrons are the same before and after doping, we can write 
X*2x10²³ =6x10¹⁹*6x10¹⁹
→X =18x10¹⁵
       =1.8x10¹⁶ per cubic meter.           




 

       14. The conductivity of an intrinsic semiconductor depends on the temperature as σ =σₒe-ΔE/2kT where σₒ is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double its value at T =300 K. Assume that the gap for germanium is 0.630 eV and remains constant as the temperature is increased.   

ANSWER: Given that band gap ΔE =0.630 eV
T =300 K. 
Suppose at temperature T', the conductivity is double its value at T. So,
σ' =2σ
→σₒ e-ΔE/2kT' =2σₒ e-ΔE/2kT
→e(-ΔE/2kT' +ΔE/2kT) =2
→(ΔE/2k)*(1/T -1/T') =ln 2 
→(0.630/2k)(1/300 -1/T') =ln 2 
→3654.3*(1/300 -1/T') =ln 2
→1/300 -1/T' =1.9x10⁻⁴  
→1/T' =3.14x10⁻³ 
→T' =318 K.         





       15.  A semiconducting material has a band gap of 1.0 eV. Acceptor impurities are doped into it which creates acceptor levels 1 meV above the valence band. Assume that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also, if kT is more than twice the gap, the upper levels have the maximum population. The temperature of the semiconductor is increased from 0 K. The concentration of the holes increases with temperature and after a certain temperature, it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.    

ANSWER:  There are two acceptor levels, one due to impurities which is 1 meV, and the other is the original which is 1.0 eV.
    For the temperature at which hole concentration will be constant 
kT =2*1 meV
→T =(2/1000)/8.62x10⁻⁵ =23 K. 
Next band gap is =1.0 -1.0x10⁻³ eV 

The transition in this band gap will be possible only when kT is more than 1/50 of this gap. So
  kT =(1.0 -1.0x10⁻³)/50   
→T =232 K. 
Between temperatures 23 K to 232 K, the hole concentration is constant. Thus the order of the temperature range in which the hole concentration is approximately constant is 20 K to 230 K.       





 
       16. In a p-n junction, the depletion region is 400 nm wide. and the electric field of 5x10⁵ V/m exists in it. (a) Find the height of the potential barrier. (b) What should be the minimum kinetic energy of a conduction electron that can diffuse from the n-side to the p-side?   

ANSWER: (a) Width of the depletion region 
d =400 nm =4x10⁻⁷ m
The existing electric field in this region,
E =5x10⁵ V/m

The height of the potential barrier =potential difference across the depletion region
=d*E
=4x10⁻⁷*5x10⁵ volts 
=20x10⁻² volts
=0.2 volts.   

(b) For diffusion, the minimum kinetic energy of a conduction electron should be equal to the work done in resisting it by the electric field. i.e. 
=eE*d 
= e*(0.2 volts) 
=0.2 eV.        





 
       17. The potential barrier existing across an unbiased p-n junction is 0.2 volts. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if (a) the junction is unbiased. (b) the junction is forward biased at 0.1 volts and (c) the junction is reverse biased at 0.1 volts?   

ANSWER: (a) The hole is assumed to have a positive charge equal to e. Suppose there is an electric field of E volts/meter across the depletion region of the unbiased p-n junction. For the diffusion of a hole from the p-side to the n-side, the minimum kinetic energy required by it is equal to the work done by the electric field in resisting the diffusion. i.e.
=eE*d, {d is the width of the depletion region}
=e(Ed) 
=e*V,   {Ed =V=potential barrier =0.2 volts} 
=e*(0.2 volts)  
=0.2 eV.   

(b) When the junction is forward-biased at 0.1 volts, the potential barrier is reduced by 0.1 volts because the field provided by the battery is opposite to the field at the depletion region. Now the reduced potential barrier, 
V =0.2 -0.1 =0.1 volts. 
Hence the minimum kinetic energy possessed by a hole for diffusion now is, 
=e*(0.1 volts) 
=0.1 eV.  

(c) In the reversed bias, the direction of the field at the depletion region and the direction of the field provided by the battery are the same, and hence they add up. The new potential barrier now is 
=0.2 +0.1 =0.3 volts. 
So the minimum kinetic energy possessed by a hole for diffusion is  
=e*(0.3 volts) 
=0.3 eV.        




 

       18. In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side.    

ANSWER: (a) When the hole approached the junction from the p-side, it moves across the junction against the electric field in the depletion region. The force exerted by the field is opposite to the movement. So it will lose 250 meV of its kinetic energy in crossing the junction. So the remaining kinetic energy of the hole when it crosses the junction =300 meV -250 meV 
=50 meV.   

(b) When the hole approached the junction from the n-side, it moves along the existing electric field in the depletion region. So the force exerted by the field is along the movement that increases its kinetic energy by the amount of the potential barrier. So the kinetic energy of the hole when it crosses the junction  
=300 meV +250m meV 
=550 meV.         





 
       19. When a p-n junction is reverse-biased, the current becomes almost constant at 25 µA. When it is forward-biased at 200 mV, a current of 75 µA is obtained. Find the magnitude of the diffusion current when the diode is (a) unbiased (b) reverse-biased at 200 mV and (c) forward-biased at 200 mV.  

ANSWER:  When a p-n junction is reverse-biased, the diffusion stops. So the diffusion current is zero. The only current is the drift current which is independent of the bias and remains constant. The 25 µA current in the reverse bias is the drift current. 

(a) In the unbiased condition, the diffusion current and the drift current are equal and opposite in direction. Hence the magnitude of the diffusion current when the diode is unbiased =25 µA.  

(b) Since the diffusion stops in the reverse bias, the diffusion current is zero.  

(c) When the p-n junction is forward-biased, the diffusion current from the p-side to the n-side is more than the drift current from the n-side to the p-side. So 
net current =diffusion current -drift current  
→diffusion current =net current +drift current 
→diffusion current =75 µA +25 µA 
                              =100 µA.      





 
       20. The drift current in a p-n junction is 20.0 µA. Evaluate the number of electrons crossing a cross-section per second in the depletion region.   

ANSWER: In a drift current in the depletion region, holes from newly created hole-electron pairs move toward the p-side, and the electrons toward the n-side. If in a drift current, the number of holes and electrons crossing a cross-section per second are n each then the total charge crossing per second =2ne. Hence the drift current is 
i =2ne =20.0 µA 
→2ne =20.0x10⁻⁶   
 
Charge on an electron e =1.602x10⁻¹⁹ C
So the number of electrons crossing per second in the depletion region
n =20.0x10⁻⁶/2e
=20.0x10⁻⁶/(2*1.602x10⁻¹⁹)  
=6.2x10¹³.        

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Links to the Chapters


CHAPTER- 45- Semiconductors and Semiconductor Devices




CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-II

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


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Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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