Thursday, August 3, 2023

H C Verma solutions, SEMICONDUCTORS AND SEMICONDUCTOR DEVICES, Chapter-45, EXERCISES, Q21 to Q35, Concepts of Physics, Part-II

EXERCISES, Q21 - Q35


       21. The current-voltage characteristic of an ideal p-n junction diode is given by
i =iₒ (eeV/kT -1)
where the drift current iₒ equals 10 µA. Take the temperature T to be 300 K.
(a) Find the voltage Vₒ for which eeV/kT =100. One can neglect the term 1 for voltages greater than this value.
(b) Find an expression for the dynamic resistance of the diode as a function of V for V > Vₒ.
(c) Find the voltage for which the dynamic resistance is 0.2 Ω. 

ANSWER: (a) From the given condition we neglect the term 1 and equate, 
eeVₒ/kT =100
→eVₒ/kT =ln 100
→Vₒ =(kT*ln 100)/e
        =8.62x10⁻⁵*300*ln 100
        =(0.12 eV)/e
        =0.12 V.

(b)  The dynamic resistance is given as dV/di. We take the differential of both sides of the characteristic expression,
di =iₒ*(eeV/kT)*(edV/kT)
→(dV/di)*(iₒe/kT)*(eeV/kT) =1
→dV/di =kTe-eV/kT/eiₒ.

(c) Given the dynamic resistance dV/ di =0.2 Ω, iₒ =10 µA, T =300 K. Putting the values in the above expression,
0.2 =8.62x10⁻⁵*300*(e-eV/kT)/(e10x10⁻⁶)
→(e-eV/kT) =0.2e/2586
(Since the unit of k is eV/K, e from the denominator on the RHS will be eliminated for uniformity of units)
→eeV/kT =2586/0.2 =12930
→eV/kT =ln (12930) =9.5
→eV =8.62x10⁻⁵*300*9.5 =0.25 eV
→V =0.25 V.
 





 

       22. Consider a p-n junction diode having the characteristic i =iₒ (eeV/kT -1) where iₒ =20 µA. The diode is operated at T 300 K. 
(a) Find the current through the diode when a voltage of 300 mV is applied across it in forward bias. 
(b) At what voltage does the current double?

ANSWER: (a) V =300 mV =0.3 V, 
iₒ =20 µA =2x10⁻⁵ A, T =300 K.
The current through the diode is,
 i =2x10⁻⁵(e0.3e/8.62x10⁻⁵*300 -1)
  =2.1 A
  ≈2 A.

(b) Suppose at voltage V' the current doubles. So, i' =2i
→iₒ(eeV'/kT -1) =2iₒ(eeV/kT -1)
→eeV'/kT -1 =2eeV/kT -2
(Numerical terms -1 and -2 are negligible in comparison to the other terms on each side, so neglect.)
→e(eV'/kT -eV/kT) =2
→e(V'-V)e/kT =2
→(V' -V)e/kT =ln 2
→V' -V =(kT/e)*ln 2
→V' =V +(8.62x10⁻⁵*300)*ln 2
       =0.3 +0.018 volts
       =0.318 volts =318 mV.
 





       23. Calculate the current through the circuit and the potential difference across the diode shown in the figure (45-E1). The drift current for the diode is 20 µA.  
The figure for Q-23


ANSWER: (a) Since the diode is in reverse bias, it will not allow batteries to send current through it. But the drift current is independent of the bias and its direction is from n-type to p-type semiconductor which is in the direction of the potential difference in the given circuit. So the only current in the circuit will be the drift current =20 µA.  

(b) Suppose the potential difference across the diode is V. Then, 
iR +V =5 volts 
→V =5 -20x10⁻⁶*20 volts 
       =5 -0.0004 volts
       =4.9996 volts 
       ≈5 V.       





 
       24. Each of the resistances shown in the figure (45-E2) has a value of 20 Ω. Find the equivalent resistance between A and B. Does it depend on whether point A or B is at higher potential?    
The figure for Q-24


ANSWER: Since all four resistances have equal value =20 Ω, the given circuit forms a Wheatstone bridge. In a balanced Wheatstone bridge, the middle wire has zero current because the potential difference across it is zero. Thus we can remove the middle wire of the given circuit containing the diode. So the remaining circuit has two parallel resistances, each having a resistance of R =20 Ω +20 Ω =40 Ω. Let the equivalent resistance = R', Then, 
1/R' =1/40 +1/40 =2/40 =1/20 
→R' =20 Ω. 
 
The equivalent resistance R' will not depend on whether point A or B is at higher potential.   





 
-----------------------------
In problems, 25 to 30, assume that the resistance of each diode is zero in forward bias and is infinity in reverse bias.
  
       25. Find the currents through the resistances in the circuits shown in Figure (45-E3).    
The Figure for Q-25


ANSWER: (a) E =2 V, R =2 Ω, Both the diodes are in forward bias and hence will have zero resistance. So from Ohm's law, the current through the resistance will be 
i =E/R 
  =2/2 A 
  =1 A.  

(b) Here one diode is in forward bias giving zero resistance while another is in reverse bias giving infinite resistance. Since they are connected in series, the total resistance in the circuit will be infinite and no current will flow. 
i =E/∞ =zero.  

(c) Both of the parallel connected diodes are in forward bias offering no resistance. Hence the circuit will be similar to the first circuit. Current through the resistance 
i =2/2 =1 A. 
 
(d) In this circuit one of the two parallel connected diodes is in forward bias resulting in zero resistance. Another one is in reverse bias giving infinite resistance. So all the current will pass through the first diode without resistance. The equivalent resistance is zero. We also can see it by calculating two resistances zero and ∞ in parallel.  
1/R =1/0 +1/∞ =∞ +0 =∞ 
→R =1/∞ =0. 
  So only resistance in the circuit is 2 Ω. The current through the resistance is, 
i =2/2 =1 A.     





 
       26. What are the readings of the ammeters A₁ and A₂ shown in figure (45-E4). Neglect the resistances of the meters.    
The figure for Q-26


ANSWER: The diode is in reverse bias giving infinite resistance, so the current in the upper loop will be zero. The reading of ammeter A₁ will be zero.
          Since the resistance of the diode is infinite, the effect will be as if no upper loop but only a lower loop. In this loop, E =2 V, R =10 Ω. So the current in the lower loop 
i =V/R =2/10 =0.2 A.
   Hence the reading of ammeter A₂ =0.2 A.
 





 
       27. Find the current through the battery in each of the circuits shown in Figure (45-E5).    
The figure for Q-27


ANSWER: In the first figure, both of the diodes are in forward bias and hence no resistance by them. The equivalent resistance of the circuit will be for two 10 Ω parallel resistances. Hence, 
1/R =1/10 +1/10 =1/5 
→R =5 Ω. 
E =5 V. 
Hence the current through the battery 
  i =E/R
    =5/5 =1 A.  

In the second figure, the lower diode is in reverse bias, thus having infinite resistance. This diode will act as a break in the wire. Since the upper diode is in the forward bias with zero resistance, the effective simple circuit will have a 5 V battery and a 10 Ω resistance. The current through the resistance and the battery will be 
i =5/10 =0.5 A.     





 
       28. Find the current through the resistance R in figure (45-E6) if (a) R =12 Ω (b) R =48 Ω.  
The figure for Q-28


ANSWER: (a) When R =12 Ω, the net potential difference between the two ends will be applying a reverse bias on the diode. Thus no current through the diode and it will act as a break in the wire. The remaining circuit has emf =4+6 =10 V and resistance =12+12 =24 Ω. Hence current through R is  
i =10/24 =0.42 A.  

(b) When R =48 Ω. Suppose the current through 4 V battery = i, through 6 V battery =i' and through 1 Ω resistance =i". From Kirchoff's loop law in the upper loop, 
12i -4 -6 +48i' =0 
→12i +48i' =10  ---- (i)   
     For the lower loop, 
-48i' +6 +1i" =0
→-48i' +i" =-6  
→i" =48i' -6   ----- (ii)
At the left junction, 
i =i' +i"
From (i)  
12i' +12i" +48i' =10 
→60i' +12i" =10     
→30i' +6i" =5
→30i' +6(48i' -6) =5
→318i'  -36 =5
→i' =41/318 =0.13 A. 
  




 
     29. Draw the current-voltage characteristics for the device shown in Figure (45-E7) between terminals A and B.   
The figure for Q-29

ANSWER: From problems 25 to 30, given that the resistance of each diode is zero in forward bias and infinity in reverse bias. So the current will be zero if B is at a higher potential than A. When A is at a higher potential than B, then diode resistance is zero due to forward bias and the current will be i =V/R =V/10. The current-voltage characteristic will be as follows.
Figure - 1


In the second figure, when A is at a higher potential than B, the lower branch is ineffective due to reverse bias. The upper diode is in forward bias, so it has zero resistance. The current will be through 10 Ω resistance and its value, 
i =V/10. 
We plot this potential difference and current on the positive X-axis and Y-axis respectively. 
  When B is at a higher potential than A, the upper diode is reverse biased and the lower one is forward biased. So the current will flow only through the lower diode and resistance. The current through the device is, 
i =V/20.  
     We plot this potential difference and the current on the negative X-axis and Y-axis respectively. The current-voltage characteristic drawn for the device is as below. 
Figure - 2

 

  



       30. Find the equivalent resistance of the network shown in Figure (45-E8) between points A and B.    
The figure for Q-30


ANSWER: We have to take the resistance of a forward-biased diode as zero and reverse-biased as infinite.

  In the given figure, the diode is in the forward bias if point A is at a higher potential than B. So it will offer no resistance and the network will act as two 10 Ω resistances connected in parallel. Its equivalent resistance R is given as,  
 1/R =1/10 +1/10 =1/5   
→R =5 Ω. 

   When point B is at a higher potential than A, the diode is in the reverse bias offering infinite resistance and no current will flow through that branch. So only remaining resistance is 10 Ω in the lower wire. So the equivalent resistance of the network in this condition is R =10 Ω.   



 

       31. When the base current in a transistor is changed from 30 µA to 80 µA, the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain ß.    

ANSWER: The current gain is given as, 
ß =ΔIC/ΔIB

Given that change in the base current 
ΔIB =80 -30 =50 µA =50x10⁻³ mA
Change in collector-current
ΔIC =3.5 -1.0 =2.5 mA.

Hence,
ß =(2.5 mA)/(50x10⁻³ mA)
    =50.
  




 

       32. A load resistor of 2 kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain ß =50. The input resistance of the transistor is 0.50 kΩ. If the input current is changed by 50 µA, (a) by what amount does the output voltage change, (b) by what amount does the input voltage change and (c) what is the power gain?    

ANSWER: The input current is the base current. Here change in input current
ΔIB =50 µA. Current gain ß =50. Hence the change in collector current,
ΔIC =ßΔIB
       =50*50µA
       =2500 µA

The connected load resistance in the collector branch =2 kΩ.

(a) The change in output voltage,
 ΔVC =R*ΔIC
      =2x10³*2500x10⁻⁶ V
      =5.0 V.

(b) The input resistance of the transistor
r =0.5 kΩ =500 Ω
The change in input current ΔIB =50 µA
Hence the change in input voltage,
ΔVB =r*ΔIB
        =500*50x10⁻⁶ V
        =0.025 V
        =25 mV.

(c) Voltage gain =ΔVC/ΔVB
               =5.0/(25x10⁻³) =200.
Given that current gain ß =50.
Hence, power gain =voltage gain*current gain
                =200*50
                =10⁴.
 




 

       33. Let X =AB̅C̅ +BC̅A̅ +CA̅B̅. Evaluate X for 
(a) A =1, B =0, C =1,
(b) A =B =C =1 and
(c) A =B =C =0.         

ANSWER: The given function may be written as,
X =A AND NOT(B AND C) +B AND NOT(C AND A) +C AND NOT(A AND B)

(a) If A =1, B =0,C =1 then,
B AND C =0, NOT(B AND C) =1,
So, A AND NOT(B AND C) =1

for the next term,
C AND A =1, NOT (C AND A) =0,
So, B AND NOT(C AND A) =0

For the last term,
A AND B =0, NOT (A AND B) =1,
So, C AND NOT(A AND B) =1

The value of X =1 +0 +1 =1
Because the + symbol represents OR. If any of the variables on RHS is 1 in the OR function, the result is 1.

(b) A =B =C =1 
For the first term,
B AND C =1, NOT(B AND C) =0,
A AND NOT(B AND C) =0
 Similarly, the other two terms will also be zero. Now the function X is 
X =0 +0 +0
   =0 OR 0 OR 0
   =0. 

(c) A =B =C =0
For the first term, 
B AND C =0, NOT(B AND C) =1 
A AND NOT(B AND C) =0 AND 1 =0. 
    Similarly, the other two terms will also be 0.
Hence, X =0 +0 +0 
→X =0 OR 0 OR 0 
       =0.        



 
 

       34. Design a logical circuit using AND, OR, and NOT gates to evaluate AB̅C̅ +BC̅A̅.    

ANSWER: The given expression may be written as, 
A AND NOT(B AND C) +B AND NOT(C AND A)
The logical circuit to evaluate it may be understood through the following figure.   
Design of logical circuit for Q-34






                                                                                                                 
       35. Show that AB +A̅B̅ is always 1.    

ANSWER: AB +A̅B̅ means, 
(A AND B) OR NOT(A AND B). 

A or B may take a value of either 1 or 0. Depending upon the values of A and B, the value of AB will be either 1 or 0. So the value of "NOT(AB)" will be 0 and 1 respectively. Thus the given expression is either, 
1 OR 0 =1
or
0 OR 1 =1.      

So the given expression is always zero.
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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

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OBJECTIVE-I

OBJECTIVE-II


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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






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CHAPTER- 8 - Work and Energy

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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