Calculate the intrinsic carrier concentration in gallium arsenide and germanium at T = 300 K . (Ans. GaAs, n i = 1.80 × 10 6 cm − 3 ; Ge, n i = 2.40 × 10 13 cm − 3 )

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Answer 1

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Chapter 1, Problem 1.1EP

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at $T = 300 K$ . (Ans. GaAs, $n i = 1.80 × 10 6 cm − 3$ ; Ge, $n i = 2.40 × 10 13 cm − 3$ )

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

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Answer 2

Textbook Question

100%

Chapter 1, Problem 1.1EP

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at $T = 300 K$ . (Ans. GaAs, $n i = 1.80 × 10 6 cm − 3$ ; Ge, $n i = 2.40 × 10 13 cm − 3$ )

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

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Answer 3

Textbook Question

100%

Chapter 1, Problem 1.1EP

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at $T = 300 K$ . (Ans. GaAs, $n i = 1.80 × 10 6 cm − 3$ ; Ge, $n i = 2.40 × 10 13 cm − 3$ )

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

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Answer 4

Textbook Question

100%

Chapter 1, Problem 1.1EP

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at $T = 300 K$ . (Ans. GaAs, $n i = 1.80 × 10 6 cm − 3$ ; Ge, $n i = 2.40 × 10 13 cm − 3$ )

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

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Answer 5

Textbook Question

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Answer 6

Textbook Question

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Answer 7

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

Answer 8

Expert Solution & Answer

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer 12

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $niGaAs=1.80×106cm−3$ and in Germanium is $niGe=2.40×1013 cm−3$ .

Answer 13

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $Eg=1.4eV$ .

Specific semiconductor coefficient is $B=2.10×1014cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

For Germanium,

Band gap energy is $Eg=0.66eV$ .

Specific semiconductor coefficient is $B=1.66×1015 cm−3K−3/2$ .

Boltzmann’s constant is $k=86×10−6eV/K$ .

Concept Used:

The intrinsic carrier concentration formula is given as

$ni=BT3/2e(−Eg/2kT)$ (1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

$ni=(2.10× 10 14)(300)32e( −1.4 2( 86× 10 −6 )( 300 ) )=1.79770×106cm−3≃1.80×106cm−3$

For Germanium, substitute the given values in equation (1).

$ni=(1.66× 10 15)(300)32e( −0.66 2( 86× 10 −6 )( 300 ) )=2.4036×1013cm−3≃2.40×1013cm−3$

Answer 14

Ch. 1 - Calculate the intrinsic carrier concentration in...Ch. 1 - (a) Calculate the majority and minority carrier...Ch. 1 - Consider ntype GaAs at T=300K doped to a...Ch. 1 - Consider silicon at T=300K . Assume the hole...Ch. 1 - Determine the intrinsic carrier concentration in...Ch. 1 - (a) Consider silicon at T=300K . Assume that...Ch. 1 - Using the results of TYU1.2, determine the drift...Ch. 1 - The electron and hole diffusion coefficients in...Ch. 1 - A sample of silicon at T=300K is doped to...Ch. 1 - (a) Calculate Vbi for a GaAs pn junction at T=300K...

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at T = 300 K . (Ans. GaAs, n i = 1.80 × 10 6 cm − 3 ; Ge, n i = 2.40 × 10 13 cm − 3 )

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