PHYSICS 02 SEMICONDUCTOR ELECTRONIC MATERISL DEVICE

 1: In an n-type silicon, which of the following statements is true:


(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.

Solution 1:
The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carries. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

3.Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to
(Eg  c    ,   (E g si   and (E g Ge . Which of the fo llowing statements is true?

 

a)  (E g si  <     (E g Ge    <   (Eg  

b) (Eg  c    <    (E g Ge       >    (E g si 

c) (Eg  c  >   (E g si  >   (E g Ge   

d) (Eg  c  =  (E g si    =   (E g Ge   

The correct answer is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that germanium is the least. The energy bang gap of the these elements are related as : c) (Eg  c  >   (E si  >   (E Ge   

4. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) Free electrons in the n-region attract them.
(b) They move across the junction by the potential difference.
(c) Hole concentration in p-region is more as compared to n-region.
(d) All the above

The correct statement is (c).
The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

5.When a forward bias is applied to a p-n junction, it


(a) Raises the potential barrier.
(b) Reduce the majority carrier current to zero.
(c) Lower the potential barrier.
(d) None of the above.

The correct statement is (c).
When forward bias is applied to a p-n junction, it lower the values of potential barrier. In the case of forward bias, the potential barrier is opposed by the applied voltage. Hence, the potential barrier across the junction gets reduced.

6.For transistor action, which of the following statements are correct.


(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.

The correct statement is (b), (c).
For a transistor action, the base region must be lightly doped so that the base region is very
thin. Also, the emitter junction must be forward-biased and collector junction should be
reversed-biased.

7.For a transistor amplifier, the voltage gain


(a) Remains constant for all frequencies.
(b) Is high at high and low frequencies and constant in the middle frequency range.
(c) Is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.

The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.

8.In half-wave rectifier, what is the output frequency if the input frequency is 50Hz. What is the output frequency of a full-wave rectifier for the same input frequency

 

i)20 Hz

ii)100Hz

iii) 60 Hz

iv)200Hz

Input frequency = 50Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴Output frequency = 2 × 50 = 100 Hz.

9.For a CE-transmitter amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ

 

i)the base current of the amplifier is
10μA.

ii)the base current of the amplifier is
7μA.

iii)the base current of the amplifier is
8μA.

iv)the base current of the amplifier is
9μA.

Collector resistance,Rc= 2kΩ= 2000Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor,β= 100
Base resistance,RB= 1kΩ= 1000Ω
Input signal voltage =Vi
Base current =IB

We have the amplification relation as :

Voltage amplification = V/Vi   = β  ( R / RB   )

Vi    = V RB / β  R

       = 2×1000 / 100×2000   = 0.01V

Therefore, the input signal voltage of the amplifier is 0.01V.
Base resistance is given by the relation:

RB   = Vi  / V
= 0.01/1000 = 10-6A


Therefore, the base current of the amplifier is
10μA.

10.Two amplifier are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.

 

i)the output AC signal of the given amplifier is 4V

ii)the output AC signal of the given amplifier is 6V

iii)the output AC signal of the given amplifier is 2V

iv)the output AC signal of the given amplifier is 8V

 

Voltage gain of the first amplifier, V= 10

Voltage gain of the second amplifier, V= 20

Input signal voltage, V= 0.01V

Output AC signal voltage =V0

The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,

V=V1  ×V 

=10× 20× 200
We have the relation:

V = V0   / V i   

    = 200 ×0.01 =2v 

Therefore, the output AC signal of the given amplifier is 2V

 

11. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? 

 

i)he photodiode cannot detect the signal.

ii) he photodiode can detect the signal.

Energy band gap of the given photodiode, Eg= 2.8 eV

 

Wavelength, λ = 6000nm= 6000×10-9

 

The energy of a signal is given by the relation:

E = hc/λ

where,

 

h=planks’s constant  

  = 6.626×10-34   js

C= speed of light

= 3×108  m/s 

E = 6.626 × 10-34    ×  3 ×  108   /    6000 × 10 -9

  = 3.313 × 10 -20   J

 

 but 1.6 × 10 -20  J= 1e V 

∴ E =3.313 × 10 -20    J

=  3.313 × 10 -20     

   ____________       =   0.207eV

    1.6 ×  10-19 

 

The energy of signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8

eV- the energy band gap of a photodiode. Hence, the photodiode cannot detect

the signal.

Additional exercises

 12: The number of silicon atoms per mis 5 ×10 28   .  .This is doped simultaneously with  5 ×10 22  .  atoms per m3   of Arsenic and 5 ×10 20  per matoms of indium. Calculate the number of electrons and holes. Given that ni = 1.5 ×10 16 m -3   Is the material n-type or p-type?

Number of silicon atoms,  N =5×10 28 atoms/m3

Number of arsenic atoms, nas = 5×10 22atoms/m3

Number of indium atoms, nin = 5×10 20atoms/m3

Number of thermally generated electrons,ni  =1.5×10 16electrons/m3

Number of electrons, nc =5×10 22 – 1.5×10 16 ≈ 4.99× 10 22

Number of holes =nh

In thermal equilibrium, the concentration of electrons and holes in a semiconductor are related as 

ncnh   = ni 2   

∴  nh  = ni 2 / ne

= (1.5×10 16 ) 4.99×10 22

≈ 4.51× 10 9

Therefore, the number of electrons is approximately
4.99×10 22 and the number of holes is about  4.51 ×10 9   . Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

 13:In an intrinsic semiconductor the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300k? Assume that the temperature dependence of intrinsic carrier concentration n i   is  given by n i  =n 0    exp ( – E / 2KT ) 

Where  n is constant 

Energy gap of the given intrinsic semiconductor, E  = 1.2 eV

the temperature dependence of the intrinsic carrier concentration is written as ; 

n = n0  exp  ( – E / 2KT ) 

where ,

K= Boltzmann constant = 8.62 × 10 -5  eV/K 

T = temperature

n0   =  constant 

Initial  temperature ,  T1  =300k

The intrinsic carrier-concentration at this temperature can be written as:

ni1   = n0  exp  ( – E / 2K×300  )  ………..(1)

Final temperature, T 2 = 600K

The intrinsic carrier-concentration at this temperature can be written as:

ni2  = n0  exp  ( – E / 2K×600  ) …………(2) 

The ratio between the conductivities at 600K and at 300K is equal to the ratio between the respective intrinsic carrier concentrations at these temperatures.

   ni2                    n0  exp  ( Eg / 2KB  ×600  )

_____    =       ______________________________

   ni1                 n0  exp  ( – E / 2K×300  )

= exp E / 2K [ (1/300)   –  ( 1/600) ]

= exp [11.6]   = 1.09 × 105

Therefore, the ratio between the conductivities is 1.09 × 105

14 .In a p-n junction diode, the current I can be expressed as

                  eV 

I = I 0 exp      ___    –  1 

                           2K

 Where  I is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative bias, and I  is the current through the diode, kB is the Boltzmann constant  ( 8.6 ×  10-5  eV /K)  and T is the absolute temperature. If for a given diode  I 0 =   5 × 10 -12 A and T = 300 K,then

 

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is

increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1V to 2V?

In a p-n junction diode, the expression for current is given as:

 

                  eV 

I = I 0 exp      ___    –  1 

                           2K

Where,

= Reverse saturation current  = 5 × 10 -12

 

T = Absolute temperature = 300K

 

k B= Boltzmann constant  = 8.6 × 10 -5     eV / K = 1.376 × 10  23   JK-1

 

V = Voltage across the diode

 

(a) Forward voltage, V = 0.6 V

 

=5× 10-1  exp1.6 × 10 -19  × 0.6 / 1.376 × 10 -23 × 300 ) – 1 ]

 


∴Current, I

 

= 5× 10-12   × exp [ 22.36 ] =0.063A

 

there fore , the forward current is about 0.063A

 

 

(b) For forward voltage, V=0.7 V, we can write:

 

I = 5 × 10 -12   [  exp 1.6  × 10 -19 ×  0.7 /   1.376 × 10 -23  × 300 )  × (-1)  ]

 

= 5 × 10 -12  × exp  [ 26.25 ] = 2.972A 

 

Hence, the increase in current, ΔI=I=I

 

=1.257 – 0.0256 =1.23A

 

 

(c)

Dynamic resistance  = Changeinvoltage  Changeincurrent

 

=( 0.7 – 0.6A)/ 2.972 = 0.0336/Ω

 

(d) If the reverse bias voltage changes from 1V to 2V, then the current(I) will

almost remain equal to I B   in both cases. Therefore, the dynamic

resistance in the reverse bias will be infinite

 

15.You are given he two circuits as shown in figure. Show that current (a) acts as OR gate while  the circuit (b) acts as AND gate

(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate.

This is shown in the following figure 

   

                                                                 

                                                                      ______

Hence, the output of the NOR Gate =  A+B 

                                                                                                             _____

                                                                                                             _____

This will be the input for the NOT Gate. Its output will be    A+B   =   A+B 

∴Y=  A+B 

Hence, this circuit fuctions as on OR Gate.

(b) A and B are the inputs and Y is the output of the given circuit (b) A and B are

the Inputs and Y is the output of the given circuit. It can be observed from the following figure that the Inputs of the right half NOR Gate are the outputs of the two note Gates.

Hence, the ouput of the given circuit can be written as:

       ____     _  _

       _    _     _  _

Y=  A+B  = A.B =A.B

Hence, this circuit functions as an AND Gate.

16.Write the truth table for a NAND Gate connected as given in figure.

A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure. 

Hence, the output can be written as :  

      ___     _     _     _ 

Y= A.A =A + A = A  …….. (1) 

The truth table for equation (i) can be drawn as :

 

This circuit fuctions as a NOT gate. The symbol for this logic circuit is shown as:

17.You are given two circuits as shown In figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

 

In both the given circuits, A and B are the inputs and Y is the output.

(a) The output of the left NAND gate will be AB. , as shown in the following figure.

Hence, the ouput of the combination of the two NAND gate is given as:

       ______________  

           ___          ___         __        __

 Y= ( A.B ) . (A.B) = AB +AB =AB

Hence, this circuit functions as an AND gate.

(b)

__

A is the output of the upper left of the NAND gate and

__

B is the output of the lower half of the NAND gate, as shown

in the following figure.

 

Hence, the ouput of the combination of the NAND gates will be given as:

                   _      _

      _   _      _      _

Y= A. B = A +B  =A + B

Hence, this circuit functions as an OR gate.

 

18.Write the truth table for circuit given in figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = O, B = 1 then A and B inputs of second NOR  gate will be O and hence  Y=1.Similarly work out the   values of Y for other combinations of A and B.  Compare with the truth table of OR, AND, NOT gates  and find the correct one )

A and B are the inputs of the given circuit. The output of the first NOR gate is

____

A+B   it  can be observed from the following figure that the inputs of the

second NOR gate become the output of the first one.

 

 

Hence, the output of the combination is given as:

 

 

 

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

 

 

19.Write the truth table for the circuits given in figure consisting of OR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

A acts as the two input of the NOR gate and Y is the output, as shown in the following figure. Hence, the output of the circuit

___

A+B.

                      ___       _ 

Output, Y = A+A = A

The truth table for the same is given as:

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate. (b) A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a) we can infer that the outputs of the

                                              _         _

first two NOR gates are Aand B, as shown in the following figure.

_           _ 

A and B are the inputs for the last NOR gate. Hence,

the output for the circuit can be written As:

       ____

       _   _    =  =

Y = A+B = A.B= A.B

The truth table for the same can be written as:

This is the truth table for an AND gate. Hence, this circuit function as AND gate

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