(c) holes are the minority carriers and pentavalent atoms

Semiconductor Electronic Materials Devices and

Simple Circuits

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

(a) Electrons are the majority carriers and trivalent atoms are the dopants.

(b) Electrons are the minority carriers and pentavalent atoms are the dopants.

(c) Holes are the minority carriers and pentavalent atoms are the dopants.

(d) Holes are the majority carriers and trivalent atoms are the dopants.

Solution:

An n-type semiconductor is obtained by doping the Ge or Si with pentavalent

atoms. In an n-type semiconductor, electrons are the majority carriers and holes are

the minority carriers, and hence, option (c) is true.

2) Which of the statements given in Exercise 14.1 is true for p-type

semiconductors?

Solution:

An n-type semiconductor is obtained by doping the Ge or Si with pentavalent

atoms. In an n-type semiconductor, electrons are the majority carriers and holes are

the minority carriers, and hence, option (c) is true.

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3) Carbon, silicon, germanium have four valence electrons each. These are

characterized by valence and conduction bands separated by energy band

gaps.

Equal to (Eg)c, (Eg)si and (Eg)Ge. Which of the following statements is true?

(a) (Eg)si< (Eg)Ge< (Eg)c (b) (Eg)c< (Eg)Ge > (Eg)si

(c) (Eg)c > (Eg)si> (Eg)Ge (d) (Eg)c = (Eg)si = (Eg)Ge

Solution:

The energy band gap is maximum for carbon and minimum for germanium, out of

the three given elements. Hence, answer (c) is correct.

4) In an unbiased p-n junction, holes diffuse from the p-region to n-egion

because:

(a) Free electrons in the n-region attract them

(b) They move across the junction due to potential difference

(c) Hole concentration in p-region is more as compared to n-region

(d) All the above

Solution:

In an unbiased p-n junction, the diffusion of charge carriers across the junction

takes place from higher concentration to lower concentration.

Thus, (c) is the correct answer.

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

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(a) Raises the potential barrier

(b) Reduces the majority Carrier current to zero

(c) Lowers the potential barrier

(d) None of the above

Solution:

When a forward bias is applied to a p-n junction, the applied voltage opposes the

barrier voltage. Due to this, the potential barrier across the junction is lowered.

Hence, answer (c) is correct.

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

(a) The 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) The emitter junction as well as the collector junction is forward biased.

Solution:

When a transistor is biased, the charge carriers move from the emitter region to the

base. Hence, to minimize the recombination of charge carriers, the base region is

doped lightly.

For a transistor action, the emitter junction is forward biased and the collector

junction is reverse biased. Hence, options (b) and (c) are correct.

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7) For a transistor amplifier, the voltage gain:

(a) Remains constant for all frequencies

(b) Is 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

Solution:

Option (c)

The voltage gain is low at high and low frequencies, and constant at mid

frequencies

8) In half wave rectification, what is the output frequency is 50 Hz?

What is the output frequency of a full wave rectifier for the same input

frequency?

Solution:

In half wave rectification, only one half of the wave is rectified, but in a full wave

rectification, both the waves are rectified. Hence, the output ripple frequency is 50

Hz for a half wave rectifier and 100 Hz for a full wave rectifier.

9) For a CE transistor amplifier, the audio signal voltage across the collected

resistance of 2 kΩ is 2V. 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.

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Solution:

According to the problem,

Output resistance, R0 = 2000Ω

Output voltage, V0 = 2 volts

Current gain, β = 100

Base resistance, Ri = 1000Ω

As Av =

= β

Vi =

Vi =

(

) = 0.01 V

Ib =

=

.

Ω = 10µA

10) Two amplifiers are connected one after another in series (cascaded). The

first amplifier has a voltage gain 10 and the second has a voltage gain of 20.

If the input signal is 0.01 volt, calculate the output ac signal.

Solution:

Total voltage gain,

Av =

= Av1 x Av2

Or ∆V0 = ∆Vi x Av1 X Av2

∆V0 = 0.01 x 10 x 20 = 2 V

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11) A p-n photodiode is fabricated from a semiconductor with a band gap of

2.8 e V.

Can it detect a wavelength of 6000 nm?

Solution:

According to question,

Incident wave length, λ = 6000 nm = 6000

Band gap of the semiconductor, E = 2.8 e V

Energy of the incident wavelength, E’ = hc/λ

E’ = 12400/600 e V

E’ = 2.667

As E’ <E, the p-n junction cannot detect the radiation of the given wavelength.

12) The number of silicon atoms per m3 is 5 x 1028

. This is doped

simultaneously with 5 x 1022

atoms per m3 of arsenic and 5 x 1020

per m3

atoms of indium. Calculate the number of electrons and holes. Given that ni =

1.5 x 1016

m-3

. Is the material n-type or p-type?

Solution:

ne = 5 x 1022

– 5 x 1020

= (5 – 0.05) x 1022

= 4.95 x 1022

m-3

nh =

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= (.×)

.×

= 4.54 x 109 m

-3

As ne>nh, the material is an n-type semiconductor.

13) In an intrinsic semiconductor, the energy gap, Eg, is 1.2 e V. its hole

mobility is much smaller than its electron mobility, and is independent of

temperature. What is the ratio between the conductivity at 600 K and that at

300 K? Assume that the temperature dependence of the intrinsic carrier

concentration ni is given by

ni= n0 exp (

), where n0 is a constant

Solution:

According to problem:

Energy gap, Eg = 1.2 e V

Boltzmann constant, kB = 8.62 x 10-5

e V/k

Intrinsic carrier concentration at 600 K (T1) is (ni)1

And intrinsic carrier concentration at 300 K (T2) is (ni)2

ni = n0 exp (

)

ni/n0 = !"

#$

%&

%& =

'

!" #$

'

!" #$

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%&

%& =

!"

#%

$

$ &

Here,

%

−

] =

.

×).*×!+[

*−

,] = 11.6

%&

%& = .* = (2.718)

11.6 = 1.09 x 10

5

The conductivity of the semiconductor is directly proportional to the number of

electrons.

-

-. =

%&

%& = 1.09 x 10

5

14) In a p-n junction diode, the current, I, can be expressed as I= I0 exp exp

(

- 1]

Where I0 is called the reverse saturation current; V is the voltage across the

diode, and is positive for forward bias and negative for reverse bias; I is the

current through the diode; kB is the Boltzmann constant (8.6 x 10-5

e V/K);

and T is the absolute temperature. If, for a given diode, I0 = 5 x 10-12

A and T

= 300 K, then:

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

Solution:


Reverse saturation current, I0 = 5 x 10-12

A

Absolute temperature, T = 300 K

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e V K-1

KB = 8.6 x 10-5

x 1.6 x 10-19

J K-1

(a) Forward voltage, V = 0.6 V

Then,

'

=

.*×!/×.*

).*×!+×.*×!/×, = 23.26

I = I0 [ exp('

- 1)]

I = 5 x 10-12

x e22.26

I = 5 x 10-12

x 4.64 x 109

I = 0.023 A

15) In a p-n junction diode, the current, I, can be expressed as I = I0 [exp('

- 1)]




e V/K);


A and T

= 300 K, then:

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

increased to 0.7 V?

Solution:


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A



e V K-1

KB = 8.6 x 10-5

x 1.6 x 10-19

J K-1

(b) If V = 0.7 V, then

'

=

.*×!/×.0

).*×!+×.*×!/×, = 27.13

I = I0 [exp ('

- 1)]

I = 5 x 10-12

x e26.14

I = 5 x 10-12

x 2.22 x 1011

I = 5 x 10-12

x e26.14

I = 5 x 10-12

x 2.22 x 1011

I = 1.11 A

Increase in current, ∆I = (1.11-0.023) = 1.087 A

16) In a p-n junction diode, the current, I, can be expressed as I = I0 [ exp('

- 1)]




e V/K);


A and T

= 300 K, then:

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(c) What is the dynamic resistance?

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

V?

Solution:



A



e V K-1

KB = 8.6 x 10-5

x 1.6 x 10-19

J K-1

(c) Since ∆I = 1.087 A

∆V = 0.7 – 0.6 = 0.1 A

Dynamic resistance = 12

13

= 0.1/1.087

= 0.092Ω

(d) When reverse bias voltage changes from 1 V to 2 V,

V = -1

'

=

.*×!/×

).*×!+×.*×!/×, = -38.8

e-38.8

= 0 (nearly)

I0 = 0

This shows that the diode possesses practically infinite dynamic resistance in the

reverse bias.

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17) You are given the two circuits as shown in Figure. Show that circuit (a)

acts as an OR gate, while circuit (b) acts as an AND gate.

Solution:

In Fig. (a), the output of the NOR gate is made the input for thr NOT gate. The

output of the NOR gate is

Y’ = 4 + 6

Then the Boolen expression is

Y = 4 + 6 = A+B

Hence, an OR gate is formed.

In fig. (b), the outputs of the two NOR gate are made the input for the NOT gate.

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The Boolean expression is

Y = 4 + 6 = 4.6 = A.B

Hence, an AND gate is formed.

18) For a circular coil of radius R and N turns carrying current I, the

magnitude of the magnetic field at a point on its axis at a distance x from its

centre is given by,

B = 789

(: ; ),/

(a) Show that this reduces to the familiar result for field at the centre of the

coil.

Solution:

(a) Magnetic field at a point on the axis of a circular coil, B = 789

(: ; ),/

((=9> )

(> ;: ),/)

Where x = distance of point on the axis at a distance from centre.

If x = 0 then B = 798

Which is the standard result for the magnetic field at the centre of the coil.

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19) (b) Consider two parallel co axial coils of equal radius R, and number of

turns N, carrying equal currents in the same direction, and separated by a

distance R. show that the field on the axis around the midpoint between the

coils is uniform over a distance that is small as compared to R, and is given

by, B = 0.7279?8

, approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a

small region is known as Helmholtz coils].

Solution:

(b) In a small region of length 2d about the midpoint between the coils, magnetic

field due to one of the coils, say B1

B1 = μABC

2[R2 + (

+d)

2]

3/2

B1 = = μABC

2R3[1 +

+D

+

D

]

3/2

B1 = 798

[1 +

+D

+

D

]

3/2

Since, d<< R, D

term is neglected.

B1 = 798

(5/4)

-3/2 (1+4d/5R)

-3/2

According to binomial theorem,

(1+4d/5R)-3/2

= 1+,

(D

) +

)(D

)

2 +…..

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Neglecting higher point,

(1+D

)

-3/2 = 1+

*D

B1 = 798

(5/4)

-3/2 (1+

*D

)

Similarly,

Magnetic field at the same point due to other coil, say B2.

B2 =798

(5/4)

-3/2 (1-

*D

)

The net magnetic= field at that point.

B = B1 + B2

798

(5/4)-3/2 [1+

*D

+1 -

*D

]

= (5/4) -3/2

798

Neglecting the term containing d2/R

2 and higher power of d

2/R

2

Since, d/R<<<1

B= 0.72798

Which is independent of ‘d’.

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20) You are given two circuits as shown in Figure, which consist of NAND

gates. Identify the logic operation carried out by the two circuits.

Solution:

In Fig. (a), the Boolean expression is y = 4. 6 = A.B i.e. an AND gate is created.

In fig (b) ,the Boolean expression is 4. 6 = 4 + 6G

= A + B i.e an OR gate is created.

21) Write the truth table for the circuit given in the figure below consisting of

NOR gates, and identify the logic operation (OR, AND, NOT) that this

circuit is performing.

(Hint: A = 0, B = 1, then input A and B of the second NOR gate will be 0,

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.)

Solution:

The Boolean expression for the given circuit is

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Y1 = 4 + 6 = 4. 6 and y = H = is 4. 6 = 4 + 6G = A + B

It means that an OR gate is created. The truth table for this gate is shown in fig.

22) Write the truth table for the circuit given in the figure below consisting of

NOR gates, and identify the logic operation (OR, AND, NOT) that this

circuit is performing.

Solution:

For the gate shown in

figure (a), the Boolean

expression is Y= A, i.e. a NOT gate is created. Its

truth table is given in

figure (c).

For the gate shown in

figure (b), the Boolean

expression is y = 4. 6 = 4 + 6G = A.B

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i.e an AND gate is produced. Its truth table is shown in figure (d).

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(c) holes are the minority carriers and pentavalent atoms

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