Semiconductor Devices Test

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Semiconductor Devices Test - 86

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Weekly Quiz Competition

Question 1
4 / -1

In a P–type semiconductor, germanium is dopped with

A
Gallium

B
Boron

C
Aluminum

D
All of these
Question 2
4 / -1

A piece of semiconductor is connected in series in an electric circuit. On increasing the temperature, thecurrent in the circuit will

A
Decrease

B
Remain unchanged

C
Increase

D
Stop flowing
Question 3
4 / -1

In extrinsic semiconductors

A
The conduction band and valence band overlap

B
The gap between conduction band and valence band is more than 16 eV

C
The gap between conduction band and valence band is near about 1 eV

D
The gap between conduction band and valence band will be 100 eV and more
Question 4
4 / -1

Resistivity of a semiconductor depends on

A
Shape of semiconductor

B
Atomic nature of semiconductor

C
Length of semiconductor

D
Shape and atomic nature of semiconductor
Question 5
4 / -1

Although carbon, silicon and germanium have same lattice structure and four valence electrons each, their band structure leads to the energy gaps as

A
E_g(Si) g(Ge) g(C)

B
E_g(SD> E_g(Ge)< E_g(C)

C
E_g(Si) < E_g(Ge)> E_g(C)

D
E_g(Si)> E_g(Ge)> E_g(C)
Question 6
4 / -1

Pure Si at 500 K has equal number of electron (n_e) and hole (n_h) concentrations of 1.5 × 10¹⁶ m^–3. Doping byindium increases n_h to 4.5 × 10²² m^–3. The doped semiconductor is of

A
n-type with electron concentration n_e = 2.5 × 10²³m^–3

B
p-type having electron concentration n_e = 5 × 10⁹m^–3

C
n-type with electron concentration n_e = 5 × 10²²m^–3

D
p-type with electron concentration n_e = 2.5 ×10¹⁰m^–3

Solution
Question 7
4 / -1

C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because

A
In case of C the valence band is not completely filled at absolute zero temperature

B
In case of C the conduction band is partly filled even at absolute zero temperature

C
The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third

D
The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit

Solution
Question 8
4 / -1

In the forward bias arrangement of a PN–junction diode

A
The N–end is connected to the positive terminal of the battery

B
The P–end is connected to the positive terminal of the battery

C
The direction of current is from N–end to P–end in the diode

D
The P–end is connected to the negative terminal of battery
Question 9
4 / -1

A p–n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength

A
6000 Å

B
4000 nm

C
6000 nm

D
4000 Å

Solution
Question 10
4 / -1

The reverse biasing in a PN junction diode

A
Decreases the potential barrier

B
Increases the potential barrier

C
Increases the number of minority charge carriers

D
Increases the number of majority charge carriers

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Semiconductor Devices Test - 86

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Semiconductor Devices Test - 86

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Question 1

Gallium

Boron

Aluminum

All of these

Question 2

Decrease

Remain unchanged

Increase

Stop flowing

Question 3

The conduction band and valence band overlap

The gap between conduction band and valence band is more than 16 eV

The gap between conduction band and valence band is near about 1 eV

The gap between conduction band and valence band will be 100 eV and more

Question 4

Shape of semiconductor

Atomic nature of semiconductor

Length of semiconductor

Shape and atomic nature of semiconductor

Question 5

Eg(Si) g(Ge) g(C)

Eg(SD> Eg(Ge)< Eg(C)

Eg(Si) < Eg(Ge)> Eg(C)

Eg(Si)> Eg(Ge)> Eg(C)

Question 6

n-type with electron concentration ne = 2.5 × 1023m–3

p-type having electron concentration ne = 5 × 109m–3

n-type with electron concentration ne = 5 × 1022m–3

p-type with electron concentration ne = 2.5 ×1010m–3

Solution

Question 7

In case of C the valence band is not completely filled at absolute zero temperature

In case of C the conduction band is partly filled even at absolute zero temperature

The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third

The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit

Solution

Question 8

The N–end is connected to the positive terminal of the battery

The P–end is connected to the positive terminal of the battery

The direction of current is from N–end to P–end in the diode

The P–end is connected to the negative terminal of battery

Question 9

6000 Å

4000 nm

6000 nm

4000 Å

Solution

Question 10

Decreases the potential barrier

Increases the potential barrier

Increases the number of minority charge carriers

Increases the number of majority charge carriers

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E_g(Si) g(Ge) g(C)

E_g(SD> E_g(Ge)< E_g(C)

E_g(Si) < E_g(Ge)> E_g(C)

E_g(Si)> E_g(Ge)> E_g(C)

n-type with electron concentration n_e = 2.5 × 10²³m^–3

p-type having electron concentration n_e = 5 × 10⁹m^–3

n-type with electron concentration n_e = 5 × 10²²m^–3

p-type with electron concentration n_e = 2.5 ×10¹⁰m^–3