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Semiconductor Electronics Material Devices and Simple Circuits

150663 The forbidden energy gap for ' ${Ge}^{'}$ crystal at ' 0 ' $K$ is

1

2.57 eV

2

6.57 eV

3

0.071 eV

4

0.71 eV

Explanation:

D The energy Band gap of germanium crystal is $E_{g} = 0.71 eV$
Semiconductor Band gap (0 K) Band gap (300K)
$Si$
$1.21 eV$
$1.1 eV$
$Ge$
GaAs
$0.785 eV$
$0.72 eV$
$1.52 eV$
$1.47 eV$

Karnataka CET-2022

Semiconductor Electronics Material Devices and Simple Circuits

150673 The width of forbidden gap in silicon crystal is $1.1 eV$ . When the crystal is converted into $n$ type semiconductor the distance of Fermi level from conduction band is:

1 greater than

0.55 eV

2 equal to

0.55 eV

3 lesser than

0.55 eV

4 equal to

1.1 eV

Explanation:

C Given that, $E_{g} = 1.1 eV$
For $n$ type semiconductors
$E_{f} > E_{i}$
Where,
$E_{f} =$ Fermi energy level for silicon
$E_{i} =$ Fermi energy level for intrinsic semiconductor
$E_{g} =$ Energy band gap for silicon
$E_{c} =$ Energy are conduction band.
Now,
$E_{c} - E_{f} = \frac{E_{g}}{2} - (E_{f} - E_{i})$
$E_{c} - E_{f} = \frac{1.1}{2} - (E_{f} - E_{i}) (∵ E_{g} = 1.1 e v)$
$E_{c} - E_{f} = 0.55 - (E_{f} - E_{i})$ So, the distance of Fermi level from conduction band is lesser than $0.55 eV$ .

AP EAMCET(Medical)-1999

Semiconductor Electronics Material Devices and Simple Circuits

150695 A p-n junction has acceptor impurity concentration of $10^{17} {cm}^{- 3}$ in the $p$ -side and donor impurity concentration of $10^{16} {cm}^{- 3}$ in the $n$ -side. What is the contact potential at the junction ( $k T =$ thermal energy, intrinsic carrier concentration $n_{i} = 1.4 \times 10^{10} {cm}^{- 3})$ ?

1

(kT / e) \ln (4 \times 10^{12})

2

(kT / e) \ln (2.5 \times 10^{23})

3

(kT / e) \ln (10^{23})

4

(kT / e) \ln (10^{9})

Explanation:

B Given that, Acceptor impurity concentration in the p-side,
$n_{a} = 10^{17} {cm}^{- 3}$
Donor impurity concentration in the n-side,
$n_{d} = 10^{16} {cm}^{- 3}$
intrinsic carrier concentration $(n_{i}) = 1.4 \times 10^{10} {cm}^{- 3}$
Constant potential at the junction-
$V_{constant} = \frac{kT}{e} \ln (\frac{n_{a} n_{d}}{n_{i}^{2}})$
$= \frac{kT}{e} \ln (\frac{10^{17} \times 10^{16}}{{(1.4 \times 10^{10})}^{2}})$
$= \frac{kT}{e} \ln (5.1 \times 10^{12})$
$V_{constant} \approx \frac{kT}{e} \ln (4 \times 10^{12})$

VITEEE-2008

Semiconductor Electronics Material Devices and Simple Circuits

150696 The energy gap of silicon is $1.14 eV$ . At what wavelength the silicon will stop to absorb the photon?

1

10877 \AA

2

9888 \AA

3

1087.7 \AA

4

1000 \AA

Explanation:

B Energy gap,
$E = 1.14 eV = 1.14 \times 1.6 \times 10^{- 19}$
$∵ E = \frac{hc}{λ} \Rightarrow λ = \frac{hc}{E}$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.14 \times 1.6 \times 10^{- 19}}$
$= 0.00000108882$
$= 10888 \times 10^{- 10}$
$λ = 10888 \AA$
$λ = 10877 \AA$

J and K CET- 2005

Semiconductor Electronics Material Devices and Simple Circuits

150663 The forbidden energy gap for ' ${Ge}^{'}$ crystal at ' 0 ' $K$ is

1

2.57 eV

2

6.57 eV

3

0.071 eV

4

0.71 eV

Explanation:

D The energy Band gap of germanium crystal is $E_{g} = 0.71 eV$
Semiconductor Band gap (0 K) Band gap (300K)
$Si$
$1.21 eV$
$1.1 eV$
$Ge$
GaAs
$0.785 eV$
$0.72 eV$
$1.52 eV$
$1.47 eV$

Karnataka CET-2022

Semiconductor Electronics Material Devices and Simple Circuits

150673 The width of forbidden gap in silicon crystal is $1.1 eV$ . When the crystal is converted into $n$ type semiconductor the distance of Fermi level from conduction band is:

1 greater than

0.55 eV

2 equal to

0.55 eV

3 lesser than

0.55 eV

4 equal to

1.1 eV

Explanation:

C Given that, $E_{g} = 1.1 eV$
For $n$ type semiconductors
$E_{f} > E_{i}$
Where,
$E_{f} =$ Fermi energy level for silicon
$E_{i} =$ Fermi energy level for intrinsic semiconductor
$E_{g} =$ Energy band gap for silicon
$E_{c} =$ Energy are conduction band.
Now,
$E_{c} - E_{f} = \frac{E_{g}}{2} - (E_{f} - E_{i})$
$E_{c} - E_{f} = \frac{1.1}{2} - (E_{f} - E_{i}) (∵ E_{g} = 1.1 e v)$
$E_{c} - E_{f} = 0.55 - (E_{f} - E_{i})$ So, the distance of Fermi level from conduction band is lesser than $0.55 eV$ .

AP EAMCET(Medical)-1999

Semiconductor Electronics Material Devices and Simple Circuits

150695 A p-n junction has acceptor impurity concentration of $10^{17} {cm}^{- 3}$ in the $p$ -side and donor impurity concentration of $10^{16} {cm}^{- 3}$ in the $n$ -side. What is the contact potential at the junction ( $k T =$ thermal energy, intrinsic carrier concentration $n_{i} = 1.4 \times 10^{10} {cm}^{- 3})$ ?

1

(kT / e) \ln (4 \times 10^{12})

2

(kT / e) \ln (2.5 \times 10^{23})

3

(kT / e) \ln (10^{23})

4

(kT / e) \ln (10^{9})

Explanation:

B Given that, Acceptor impurity concentration in the p-side,
$n_{a} = 10^{17} {cm}^{- 3}$
Donor impurity concentration in the n-side,
$n_{d} = 10^{16} {cm}^{- 3}$
intrinsic carrier concentration $(n_{i}) = 1.4 \times 10^{10} {cm}^{- 3}$
Constant potential at the junction-
$V_{constant} = \frac{kT}{e} \ln (\frac{n_{a} n_{d}}{n_{i}^{2}})$
$= \frac{kT}{e} \ln (\frac{10^{17} \times 10^{16}}{{(1.4 \times 10^{10})}^{2}})$
$= \frac{kT}{e} \ln (5.1 \times 10^{12})$
$V_{constant} \approx \frac{kT}{e} \ln (4 \times 10^{12})$

VITEEE-2008

Semiconductor Electronics Material Devices and Simple Circuits

150696 The energy gap of silicon is $1.14 eV$ . At what wavelength the silicon will stop to absorb the photon?

1

10877 \AA

2

9888 \AA

3

1087.7 \AA

4

1000 \AA

Explanation:

B Energy gap,
$E = 1.14 eV = 1.14 \times 1.6 \times 10^{- 19}$
$∵ E = \frac{hc}{λ} \Rightarrow λ = \frac{hc}{E}$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.14 \times 1.6 \times 10^{- 19}}$
$= 0.00000108882$
$= 10888 \times 10^{- 10}$
$λ = 10888 \AA$
$λ = 10877 \AA$

J and K CET- 2005

Semiconductor Electronics Material Devices and Simple Circuits

150663 The forbidden energy gap for ' ${Ge}^{'}$ crystal at ' 0 ' $K$ is

1

2.57 eV

2

6.57 eV

3

0.071 eV

4

0.71 eV

Explanation:

D The energy Band gap of germanium crystal is $E_{g} = 0.71 eV$
Semiconductor Band gap (0 K) Band gap (300K)
$Si$
$1.21 eV$
$1.1 eV$
$Ge$
GaAs
$0.785 eV$
$0.72 eV$
$1.52 eV$
$1.47 eV$

Karnataka CET-2022

Semiconductor Electronics Material Devices and Simple Circuits

150673 The width of forbidden gap in silicon crystal is $1.1 eV$ . When the crystal is converted into $n$ type semiconductor the distance of Fermi level from conduction band is:

1 greater than

0.55 eV

2 equal to

0.55 eV

3 lesser than

0.55 eV

4 equal to

1.1 eV

Explanation:

C Given that, $E_{g} = 1.1 eV$
For $n$ type semiconductors
$E_{f} > E_{i}$
Where,
$E_{f} =$ Fermi energy level for silicon
$E_{i} =$ Fermi energy level for intrinsic semiconductor
$E_{g} =$ Energy band gap for silicon
$E_{c} =$ Energy are conduction band.
Now,
$E_{c} - E_{f} = \frac{E_{g}}{2} - (E_{f} - E_{i})$
$E_{c} - E_{f} = \frac{1.1}{2} - (E_{f} - E_{i}) (∵ E_{g} = 1.1 e v)$
$E_{c} - E_{f} = 0.55 - (E_{f} - E_{i})$ So, the distance of Fermi level from conduction band is lesser than $0.55 eV$ .

AP EAMCET(Medical)-1999

Semiconductor Electronics Material Devices and Simple Circuits

150695 A p-n junction has acceptor impurity concentration of $10^{17} {cm}^{- 3}$ in the $p$ -side and donor impurity concentration of $10^{16} {cm}^{- 3}$ in the $n$ -side. What is the contact potential at the junction ( $k T =$ thermal energy, intrinsic carrier concentration $n_{i} = 1.4 \times 10^{10} {cm}^{- 3})$ ?

1

(kT / e) \ln (4 \times 10^{12})

2

(kT / e) \ln (2.5 \times 10^{23})

3

(kT / e) \ln (10^{23})

4

(kT / e) \ln (10^{9})

Explanation:

B Given that, Acceptor impurity concentration in the p-side,
$n_{a} = 10^{17} {cm}^{- 3}$
Donor impurity concentration in the n-side,
$n_{d} = 10^{16} {cm}^{- 3}$
intrinsic carrier concentration $(n_{i}) = 1.4 \times 10^{10} {cm}^{- 3}$
Constant potential at the junction-
$V_{constant} = \frac{kT}{e} \ln (\frac{n_{a} n_{d}}{n_{i}^{2}})$
$= \frac{kT}{e} \ln (\frac{10^{17} \times 10^{16}}{{(1.4 \times 10^{10})}^{2}})$
$= \frac{kT}{e} \ln (5.1 \times 10^{12})$
$V_{constant} \approx \frac{kT}{e} \ln (4 \times 10^{12})$

VITEEE-2008

Semiconductor Electronics Material Devices and Simple Circuits

150696 The energy gap of silicon is $1.14 eV$ . At what wavelength the silicon will stop to absorb the photon?

1

10877 \AA

2

9888 \AA

3

1087.7 \AA

4

1000 \AA

Explanation:

B Energy gap,
$E = 1.14 eV = 1.14 \times 1.6 \times 10^{- 19}$
$∵ E = \frac{hc}{λ} \Rightarrow λ = \frac{hc}{E}$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.14 \times 1.6 \times 10^{- 19}}$
$= 0.00000108882$
$= 10888 \times 10^{- 10}$
$λ = 10888 \AA$
$λ = 10877 \AA$

J and K CET- 2005

Semiconductor Electronics Material Devices and Simple Circuits

150663 The forbidden energy gap for ' ${Ge}^{'}$ crystal at ' 0 ' $K$ is

1

2.57 eV

2

6.57 eV

3

0.071 eV

4

0.71 eV

Explanation:

D The energy Band gap of germanium crystal is $E_{g} = 0.71 eV$
Semiconductor Band gap (0 K) Band gap (300K)
$Si$
$1.21 eV$
$1.1 eV$
$Ge$
GaAs
$0.785 eV$
$0.72 eV$
$1.52 eV$
$1.47 eV$

Karnataka CET-2022

Semiconductor Electronics Material Devices and Simple Circuits

150673 The width of forbidden gap in silicon crystal is $1.1 eV$ . When the crystal is converted into $n$ type semiconductor the distance of Fermi level from conduction band is:

1 greater than

0.55 eV

2 equal to

0.55 eV

3 lesser than

0.55 eV

4 equal to

1.1 eV

Explanation:

C Given that, $E_{g} = 1.1 eV$
For $n$ type semiconductors
$E_{f} > E_{i}$
Where,
$E_{f} =$ Fermi energy level for silicon
$E_{i} =$ Fermi energy level for intrinsic semiconductor
$E_{g} =$ Energy band gap for silicon
$E_{c} =$ Energy are conduction band.
Now,
$E_{c} - E_{f} = \frac{E_{g}}{2} - (E_{f} - E_{i})$
$E_{c} - E_{f} = \frac{1.1}{2} - (E_{f} - E_{i}) (∵ E_{g} = 1.1 e v)$
$E_{c} - E_{f} = 0.55 - (E_{f} - E_{i})$ So, the distance of Fermi level from conduction band is lesser than $0.55 eV$ .

AP EAMCET(Medical)-1999

Semiconductor Electronics Material Devices and Simple Circuits

150695 A p-n junction has acceptor impurity concentration of $10^{17} {cm}^{- 3}$ in the $p$ -side and donor impurity concentration of $10^{16} {cm}^{- 3}$ in the $n$ -side. What is the contact potential at the junction ( $k T =$ thermal energy, intrinsic carrier concentration $n_{i} = 1.4 \times 10^{10} {cm}^{- 3})$ ?

1

(kT / e) \ln (4 \times 10^{12})

2

(kT / e) \ln (2.5 \times 10^{23})

3

(kT / e) \ln (10^{23})

4

(kT / e) \ln (10^{9})

Explanation:

B Given that, Acceptor impurity concentration in the p-side,
$n_{a} = 10^{17} {cm}^{- 3}$
Donor impurity concentration in the n-side,
$n_{d} = 10^{16} {cm}^{- 3}$
intrinsic carrier concentration $(n_{i}) = 1.4 \times 10^{10} {cm}^{- 3}$
Constant potential at the junction-
$V_{constant} = \frac{kT}{e} \ln (\frac{n_{a} n_{d}}{n_{i}^{2}})$
$= \frac{kT}{e} \ln (\frac{10^{17} \times 10^{16}}{{(1.4 \times 10^{10})}^{2}})$
$= \frac{kT}{e} \ln (5.1 \times 10^{12})$
$V_{constant} \approx \frac{kT}{e} \ln (4 \times 10^{12})$

VITEEE-2008

Semiconductor Electronics Material Devices and Simple Circuits

150696 The energy gap of silicon is $1.14 eV$ . At what wavelength the silicon will stop to absorb the photon?

1

10877 \AA

2

9888 \AA

3

1087.7 \AA

4

1000 \AA

Explanation:

B Energy gap,
$E = 1.14 eV = 1.14 \times 1.6 \times 10^{- 19}$
$∵ E = \frac{hc}{λ} \Rightarrow λ = \frac{hc}{E}$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.14 \times 1.6 \times 10^{- 19}}$
$= 0.00000108882$
$= 10888 \times 10^{- 10}$
$λ = 10888 \AA$
$λ = 10877 \AA$

J and K CET- 2005