Semiconductor Electronics : Materials, Devices and Simple Circuits Easy Questions and Answers | Page - 18

Semiconductor Electronics : Materials, Devices and Simple Circuits

Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point?

1. metallic bonding
1. van der Waal’s bonding
1. ionic bonding
1. covalent bonding

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For a metal, conductivity decreases with increase in temperature. Also, metal has high melting point.

Correct Option: A

For a metal, conductivity decreases with increase in temperature. Also, metal has high melting point.

Which one of the following statement is FALSE ?

1. Pure Si doped with trivalent impurities gives a p-type semiconductor
1. Majority carriers in a n-type semiconductor are holes
1. Minority carriers in a p-type semiconductor are electrons
1. The resistance of intrinsic semiconductor decreases with increase of temperature

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Majority carriers in an n-type semiconductor are electrons.

Correct Option: B

Majority carriers in an n-type semiconductor are electrons.

A zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is

1. 10 mA
1. 15 mA
1. 20 mA
1. 5 mA

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Voltage across zener diode is constant.

Current in 1kΩ resistor,
(i)_1kΩ = 15 volt = 15 mA
1kΩ

Current in 250Ω resistor,
(i)_1kΩ = (20 - 15)V = 5V A
250 Ω 250 Ω

= 20 A = 20 mA
100

∴ (i)_{zener diode} = (20 - 15) = 5 mA

Correct Option: D

Voltage across zener diode is constant.

Current in 1kΩ resistor,
(i)_1kΩ = 15 volt = 15 mA
1kΩ

Current in 250Ω resistor,
(i)_1kΩ = (20 - 15)V = 5V A
250 Ω 250 Ω

= 20 A = 20 mA
100

∴ (i)_{zener diode} = (20 - 15) = 5 mA

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

1. n–type with electron concentration n_e = 5 × 10²² m^–3
1. p–type with electron concentration n_e = 2.5 ×10¹⁰ m^–3
1. n–type with electron concentration n_e = 2.5 × 10²³ m^–3
1. p–type having electron concentration n_e = 5 × 10⁹ m^–3

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n₁² = n_en_h
(1.5 × 10¹⁶ )² = n_e (4.5 × 10²² )
⇒ n_e = 0.5 × 10¹⁰
or n_e = 5 × 10⁹
Given n_h = 4.5 × 10²²
⇒ n_h >> n_e
∴ Semiconductor is p-type and
n_e = 5 × 10⁹ m^–3.

Correct Option: D

n₁² = n_en_h
(1.5 × 10¹⁶ )² = n_e (4.5 × 10²² )
⇒ n_e = 0.5 × 10¹⁰
or n_e = 5 × 10⁹
Given n_h = 4.5 × 10²²
⇒ n_h >> n_e
∴ Semiconductor is p-type and
n_e = 5 × 10⁹ m^–3.

In the following figure, the diodes which are forward biased, are

1. (C) only
1. (C) and (A)
1. (B) and (D)
1. (A), (B) and (D)

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Only in (A) and (C) diodes are forward biased as p-type should be at higher potential and n-type at lower potential.

Correct Option: B

Only in (A) and (C) diodes are forward biased as p-type should be at higher potential and n-type at lower potential.