QBManager

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365592 Pure $S i$ at 300 $K$ has electron $(n_{e})$ concentrations of $1.5 \times 10^{16} m^{- 3}$ Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} .$ $n_{e}$ in the doped $S i$ is

1

5 \times 10^{9}

2

7 \times 10^{9}

3

9 \times 10^{9}

4

8 \times 10^{9}

Explanation:

We know that for a doped semiconductor in thermal equilibrium, we have
$n_{e} n_{h} = n_{i}^{2}$
As per given data, $n_{i} = 1.5 \times 10^{16} m^{- 3}$
$n_{h} = 4.5 \times 10^{22} m^{- 3}$
Thus $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2} m^{- 6}}{4.5 \times 10^{22} m^{- 3}} = 5.0 \times 10^{9} m^{- 3}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365593 Pick out the statement which is not correct.

1 At a low temperature, the resistance of a semiconductor is very high.

2 Movement of holes is restricted to the valence band only

3 Width of the depletion region increases as the forward bias voltage increases in case of a

N

-

P

junction diode

4 In a forward bias condition, the diode heavily conducts.

Explanation:

In case of $N$ - $P$ junction diode, width of the depletion region decreases as the forward bias voltage increases.

KCET - 2010

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365594 An intrinsic semiconductor has a resistivity of $0.50 Ω m$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} V^{- 1} S^{- 1}$ and $0.11 m^{2} V^{- 1} S^{- 1}$ respectively.

1

1.2 \times 10^{18} m^{- 3}

2

2.5 \times 10^{19} m^{- 3}

3

1.9 \times 10^{20} m^{- 3}

4

30 \times 10^{19} m^{- 3}

Explanation:

$ρ = 0.50 Ω m$
$μ_{e} = 0.39 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.11 m^{2} V^{- 1} s^{- 1}$
Resistivity of intrinsic semiconductor is
$\frac{1}{ρ} = e (μ_{e} n_{i} + n_{i} μ_{h})$
$\Rightarrow n_{i} = \frac{1}{ρ e (μ_{e} + μ_{h})}$
$n_{i} = \frac{1}{(0.5) (1.6 \times 10^{- 19}) (0.39 + 0.11)}$
$= 2.5 \times 10^{19} m^{- 3} .$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365595 In an intrinsic semiconductor, the number of conduction electron is $6 \times 10^{19}$ per cubic meter. How many holes are there in a sample of size $1 c m \times 1 c m \times 2 m m$ ?

1

5 \times 10^{2}

2

12 \times 10^{- 12}

3

12 \times 10^{12}

4

4 \times 10^{8}

Explanation:

Number of density of electron,
$n_{e} = 6 \times 10^{19} m^{- 3}$ .
Volume of sample, $V = 1 c m \times 1 c m \times 2 m m$
$= 10^{- 2} m \times 10^{- 2} m \times 2 \times 10^{- 3} m$
$= 2 \times 10^{- 7} m^{3}$
$∴$ No. of electrons in the given sample $= n_{e} \times v$ .
$\begin{aligned} = 6 \times 10^{19} \times 2 \times 10^{- 7} \\ = 12 \times 10^{12} . \end{aligned}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365596 In intrinsic semiconductor conductivity is

1 low

2 average

3 high

4 very low

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365592 Pure $S i$ at 300 $K$ has electron $(n_{e})$ concentrations of $1.5 \times 10^{16} m^{- 3}$ Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} .$ $n_{e}$ in the doped $S i$ is

1

5 \times 10^{9}

2

7 \times 10^{9}

3

9 \times 10^{9}

4

8 \times 10^{9}

Explanation:

We know that for a doped semiconductor in thermal equilibrium, we have
$n_{e} n_{h} = n_{i}^{2}$
As per given data, $n_{i} = 1.5 \times 10^{16} m^{- 3}$
$n_{h} = 4.5 \times 10^{22} m^{- 3}$
Thus $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2} m^{- 6}}{4.5 \times 10^{22} m^{- 3}} = 5.0 \times 10^{9} m^{- 3}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365593 Pick out the statement which is not correct.

1 At a low temperature, the resistance of a semiconductor is very high.

2 Movement of holes is restricted to the valence band only

3 Width of the depletion region increases as the forward bias voltage increases in case of a

N

-

P

junction diode

4 In a forward bias condition, the diode heavily conducts.

Explanation:

In case of $N$ - $P$ junction diode, width of the depletion region decreases as the forward bias voltage increases.

KCET - 2010

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365594 An intrinsic semiconductor has a resistivity of $0.50 Ω m$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} V^{- 1} S^{- 1}$ and $0.11 m^{2} V^{- 1} S^{- 1}$ respectively.

1

1.2 \times 10^{18} m^{- 3}

2

2.5 \times 10^{19} m^{- 3}

3

1.9 \times 10^{20} m^{- 3}

4

30 \times 10^{19} m^{- 3}

Explanation:

$ρ = 0.50 Ω m$
$μ_{e} = 0.39 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.11 m^{2} V^{- 1} s^{- 1}$
Resistivity of intrinsic semiconductor is
$\frac{1}{ρ} = e (μ_{e} n_{i} + n_{i} μ_{h})$
$\Rightarrow n_{i} = \frac{1}{ρ e (μ_{e} + μ_{h})}$
$n_{i} = \frac{1}{(0.5) (1.6 \times 10^{- 19}) (0.39 + 0.11)}$
$= 2.5 \times 10^{19} m^{- 3} .$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365595 In an intrinsic semiconductor, the number of conduction electron is $6 \times 10^{19}$ per cubic meter. How many holes are there in a sample of size $1 c m \times 1 c m \times 2 m m$ ?

1

5 \times 10^{2}

2

12 \times 10^{- 12}

3

12 \times 10^{12}

4

4 \times 10^{8}

Explanation:

Number of density of electron,
$n_{e} = 6 \times 10^{19} m^{- 3}$ .
Volume of sample, $V = 1 c m \times 1 c m \times 2 m m$
$= 10^{- 2} m \times 10^{- 2} m \times 2 \times 10^{- 3} m$
$= 2 \times 10^{- 7} m^{3}$
$∴$ No. of electrons in the given sample $= n_{e} \times v$ .
$\begin{aligned} = 6 \times 10^{19} \times 2 \times 10^{- 7} \\ = 12 \times 10^{12} . \end{aligned}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365596 In intrinsic semiconductor conductivity is

1 low

2 average

3 high

4 very low

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365592 Pure $S i$ at 300 $K$ has electron $(n_{e})$ concentrations of $1.5 \times 10^{16} m^{- 3}$ Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} .$ $n_{e}$ in the doped $S i$ is

1

5 \times 10^{9}

2

7 \times 10^{9}

3

9 \times 10^{9}

4

8 \times 10^{9}

Explanation:

We know that for a doped semiconductor in thermal equilibrium, we have
$n_{e} n_{h} = n_{i}^{2}$
As per given data, $n_{i} = 1.5 \times 10^{16} m^{- 3}$
$n_{h} = 4.5 \times 10^{22} m^{- 3}$
Thus $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2} m^{- 6}}{4.5 \times 10^{22} m^{- 3}} = 5.0 \times 10^{9} m^{- 3}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365593 Pick out the statement which is not correct.

1 At a low temperature, the resistance of a semiconductor is very high.

2 Movement of holes is restricted to the valence band only

3 Width of the depletion region increases as the forward bias voltage increases in case of a

N

-

P

junction diode

4 In a forward bias condition, the diode heavily conducts.

Explanation:

In case of $N$ - $P$ junction diode, width of the depletion region decreases as the forward bias voltage increases.

KCET - 2010

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365594 An intrinsic semiconductor has a resistivity of $0.50 Ω m$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} V^{- 1} S^{- 1}$ and $0.11 m^{2} V^{- 1} S^{- 1}$ respectively.

1

1.2 \times 10^{18} m^{- 3}

2

2.5 \times 10^{19} m^{- 3}

3

1.9 \times 10^{20} m^{- 3}

4

30 \times 10^{19} m^{- 3}

Explanation:

$ρ = 0.50 Ω m$
$μ_{e} = 0.39 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.11 m^{2} V^{- 1} s^{- 1}$
Resistivity of intrinsic semiconductor is
$\frac{1}{ρ} = e (μ_{e} n_{i} + n_{i} μ_{h})$
$\Rightarrow n_{i} = \frac{1}{ρ e (μ_{e} + μ_{h})}$
$n_{i} = \frac{1}{(0.5) (1.6 \times 10^{- 19}) (0.39 + 0.11)}$
$= 2.5 \times 10^{19} m^{- 3} .$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365595 In an intrinsic semiconductor, the number of conduction electron is $6 \times 10^{19}$ per cubic meter. How many holes are there in a sample of size $1 c m \times 1 c m \times 2 m m$ ?

1

5 \times 10^{2}

2

12 \times 10^{- 12}

3

12 \times 10^{12}

4

4 \times 10^{8}

Explanation:

Number of density of electron,
$n_{e} = 6 \times 10^{19} m^{- 3}$ .
Volume of sample, $V = 1 c m \times 1 c m \times 2 m m$
$= 10^{- 2} m \times 10^{- 2} m \times 2 \times 10^{- 3} m$
$= 2 \times 10^{- 7} m^{3}$
$∴$ No. of electrons in the given sample $= n_{e} \times v$ .
$\begin{aligned} = 6 \times 10^{19} \times 2 \times 10^{- 7} \\ = 12 \times 10^{12} . \end{aligned}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365596 In intrinsic semiconductor conductivity is

1 low

2 average

3 high

4 very low

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365592 Pure $S i$ at 300 $K$ has electron $(n_{e})$ concentrations of $1.5 \times 10^{16} m^{- 3}$ Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} .$ $n_{e}$ in the doped $S i$ is

1

5 \times 10^{9}

2

7 \times 10^{9}

3

9 \times 10^{9}

4

8 \times 10^{9}

Explanation:

We know that for a doped semiconductor in thermal equilibrium, we have
$n_{e} n_{h} = n_{i}^{2}$
As per given data, $n_{i} = 1.5 \times 10^{16} m^{- 3}$
$n_{h} = 4.5 \times 10^{22} m^{- 3}$
Thus $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2} m^{- 6}}{4.5 \times 10^{22} m^{- 3}} = 5.0 \times 10^{9} m^{- 3}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365593 Pick out the statement which is not correct.

1 At a low temperature, the resistance of a semiconductor is very high.

2 Movement of holes is restricted to the valence band only

3 Width of the depletion region increases as the forward bias voltage increases in case of a

N

-

P

junction diode

4 In a forward bias condition, the diode heavily conducts.

Explanation:

In case of $N$ - $P$ junction diode, width of the depletion region decreases as the forward bias voltage increases.

KCET - 2010

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365594 An intrinsic semiconductor has a resistivity of $0.50 Ω m$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} V^{- 1} S^{- 1}$ and $0.11 m^{2} V^{- 1} S^{- 1}$ respectively.

1

1.2 \times 10^{18} m^{- 3}

2

2.5 \times 10^{19} m^{- 3}

3

1.9 \times 10^{20} m^{- 3}

4

30 \times 10^{19} m^{- 3}

Explanation:

$ρ = 0.50 Ω m$
$μ_{e} = 0.39 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.11 m^{2} V^{- 1} s^{- 1}$
Resistivity of intrinsic semiconductor is
$\frac{1}{ρ} = e (μ_{e} n_{i} + n_{i} μ_{h})$
$\Rightarrow n_{i} = \frac{1}{ρ e (μ_{e} + μ_{h})}$
$n_{i} = \frac{1}{(0.5) (1.6 \times 10^{- 19}) (0.39 + 0.11)}$
$= 2.5 \times 10^{19} m^{- 3} .$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365595 In an intrinsic semiconductor, the number of conduction electron is $6 \times 10^{19}$ per cubic meter. How many holes are there in a sample of size $1 c m \times 1 c m \times 2 m m$ ?

1

5 \times 10^{2}

2

12 \times 10^{- 12}

3

12 \times 10^{12}

4

4 \times 10^{8}

Explanation:

Number of density of electron,
$n_{e} = 6 \times 10^{19} m^{- 3}$ .
Volume of sample, $V = 1 c m \times 1 c m \times 2 m m$
$= 10^{- 2} m \times 10^{- 2} m \times 2 \times 10^{- 3} m$
$= 2 \times 10^{- 7} m^{3}$
$∴$ No. of electrons in the given sample $= n_{e} \times v$ .
$\begin{aligned} = 6 \times 10^{19} \times 2 \times 10^{- 7} \\ = 12 \times 10^{12} . \end{aligned}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365596 In intrinsic semiconductor conductivity is

1 low

2 average

3 high

4 very low

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365592 Pure $S i$ at 300 $K$ has electron $(n_{e})$ concentrations of $1.5 \times 10^{16} m^{- 3}$ Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} .$ $n_{e}$ in the doped $S i$ is

1

5 \times 10^{9}

2

7 \times 10^{9}

3

9 \times 10^{9}

4

8 \times 10^{9}

Explanation:

We know that for a doped semiconductor in thermal equilibrium, we have
$n_{e} n_{h} = n_{i}^{2}$
As per given data, $n_{i} = 1.5 \times 10^{16} m^{- 3}$
$n_{h} = 4.5 \times 10^{22} m^{- 3}$
Thus $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2} m^{- 6}}{4.5 \times 10^{22} m^{- 3}} = 5.0 \times 10^{9} m^{- 3}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365593 Pick out the statement which is not correct.

1 At a low temperature, the resistance of a semiconductor is very high.

2 Movement of holes is restricted to the valence band only

3 Width of the depletion region increases as the forward bias voltage increases in case of a

N

-

P

junction diode

4 In a forward bias condition, the diode heavily conducts.

Explanation:

In case of $N$ - $P$ junction diode, width of the depletion region decreases as the forward bias voltage increases.

KCET - 2010

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365594 An intrinsic semiconductor has a resistivity of $0.50 Ω m$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} V^{- 1} S^{- 1}$ and $0.11 m^{2} V^{- 1} S^{- 1}$ respectively.

1

1.2 \times 10^{18} m^{- 3}

2

2.5 \times 10^{19} m^{- 3}

3

1.9 \times 10^{20} m^{- 3}

4

30 \times 10^{19} m^{- 3}

Explanation:

$ρ = 0.50 Ω m$
$μ_{e} = 0.39 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.11 m^{2} V^{- 1} s^{- 1}$
Resistivity of intrinsic semiconductor is
$\frac{1}{ρ} = e (μ_{e} n_{i} + n_{i} μ_{h})$
$\Rightarrow n_{i} = \frac{1}{ρ e (μ_{e} + μ_{h})}$
$n_{i} = \frac{1}{(0.5) (1.6 \times 10^{- 19}) (0.39 + 0.11)}$
$= 2.5 \times 10^{19} m^{- 3} .$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365595 In an intrinsic semiconductor, the number of conduction electron is $6 \times 10^{19}$ per cubic meter. How many holes are there in a sample of size $1 c m \times 1 c m \times 2 m m$ ?

1

5 \times 10^{2}

2

12 \times 10^{- 12}

3

12 \times 10^{12}

4

4 \times 10^{8}

Explanation:

Number of density of electron,
$n_{e} = 6 \times 10^{19} m^{- 3}$ .
Volume of sample, $V = 1 c m \times 1 c m \times 2 m m$
$= 10^{- 2} m \times 10^{- 2} m \times 2 \times 10^{- 3} m$
$= 2 \times 10^{- 7} m^{3}$
$∴$ No. of electrons in the given sample $= n_{e} \times v$ .
$\begin{aligned} = 6 \times 10^{19} \times 2 \times 10^{- 7} \\ = 12 \times 10^{12} . \end{aligned}$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365596 In intrinsic semiconductor conductivity is

1 low

2 average

3 high

4 very low

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: