The intrinsic electrical conductivity of GaAs at 75 ° C .

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Answer 1

Question

Chapter 14.12, Problem 83AAP

(a)

To determine

The intrinsic electrical conductivity of GaAs at 75°C.

(a)

Expert Solution

Answer to Problem 83AAP

The intrinsic electrical conductivity of GaAs at 75°C is 8.305×10−6 Ω−1⋅m−1.

Explanation of Solution

Write the expression for the conductivity of GaAs at 300 K.

σ300 K=niq(μn+μp) (I)

Here, the intrinsic carrier is ni, the charge is q, the mobility of the electron is μn and the mobility of the hole is μp.

Convert the temperature from degree Celsius to Kelvin.

(0°C+273) K=273 K

(75°C+273) K=348 K

Write the expression for the conductivity of GaAs at 348 K.

σ348 K=σ300 Ke(−Eg2kT300 K)e(−Eg2kT348 K) (II)

Here, the Boltzmann constant is k and the band energy gap is −Eg.

Conclusion:

Refer to table 14-6 “ electrical properties of intrinsic semiconducting compounds at room temperature 300 K” to obtain the mobility of electron and hole for GaAs as 0.720 m2/V⋅s and 0.020 m2/V⋅s respectively.

Refer to table 14-6 “ electrical properties of intrinsic semiconducting compounds at room temperature 300 K” to obtain the intrinsic carrier and energy band gap as 1.4×1012 m−3 and 1.47 eV respectively.

Here, the charge is 1.6×10−19 C and the Boltzmann constant is 8.62×10−5 eV/K.

Substitute 1.6×10−19 C for q, 1.4×1012 m−3 for ni, 0.720 m2/V⋅s for μn and 0.020 m2/V⋅s for μp in Equation (II).

σ300 K=(1.4×1012 m−3)(1.6×10−19 C)(0.720 m2/V⋅s+0.020 m2/V⋅s)=(2.24×10−7 C⋅m−3)(0.74 m2/V⋅s)(1 A⋅s1 C)(1 V/A1 Ω)=1.65×10−7 Ω−1⋅m−1

Substitute 1.65×10−7 Ω−1⋅m−1 for σ300 K, 1.47 eV for Eg, 8.62×10−5 eV/K for k, 348 K for T348 K and 300 K for T300 K in Equation (II).

σ348 K=(1.65×10−7 Ω−1⋅m−1)e(−1.47 eV2(8.62×10−5 eV/K)300 K)e(−1.47 eV2(8.62×10−5 eV/K)348 K)=(1.65×10−7 Ω−1⋅m−1)e(−28.42)e(−24.50)=(1.65×10−7 Ω−1⋅m−1)4.54×10−132.285×10−11=8.305×10−6 Ω−1⋅m−1

Thus, the intrinsic electrical conductivity of GaAs at 75°C is 8.305×10−6 Ω−1⋅m−1.

(b)

To determine

The intrinsic electrical conductivity of InSb at 75°C.

(b)

Expert Solution

Answer to Problem 83AAP

The intrinsic electrical conductivity of InSb at 75°C is 2.7386×104 Ω−1⋅m−1.

Explanation of Solution

Write the expression for the conductivity of InSb at 300 K.

σ300 K=niq(μn+μp) (III)

Here, the intrinsic carrier is ni, the charge is q, the mobility of the electron is μn and the mobility of the hole is μp.

Convert the temperature from degree Celsius to Kelvin.

0°C+273=273 K75°C+273=348 K

Write the expression for the conductivity of InSb at 348 K.

σ348 K=σ300 Ke(−Eg2kT300 K)e(−Eg2kT348 K) (IV)

Here, the Boltzmann constant is k and the band energy gap is −Eg.

Conclusion:

Here, the charge is 1.6×10−19 C and the Boltzmann constant is 8.62×10−5 eV/K.

Refer to table 14-6 “ electrical properties of intrinsic semiconducting compounds at room temperature 300 K” to obtain the mobility of electron and hole for InSb as 8.00 m2/V⋅s and 0.045 m2/V⋅s respectively.

Refer to table 14-6 “ electrical properties of intrinsic semiconducting compounds at room temperature 300 K” to obtain the intrinsic carrier and energy band gap as 1.35×1022 m−3 and 0.17 eV respectively.

Substitute 1.6×10−19 C for q, 1.35×1022 m−3 for ni, 8.00 m2/V⋅s for μn and 0.045 m2/V⋅s for μp in Equation (II).

σ300 K=(1.35×1022 m−3)(1.6×10−19 C)(8.00 m2/V⋅s+0.045 m2/V⋅s)=(2160 C⋅m−3)(8.045 m2/V⋅s)(1 A⋅s1 C)(1 V/A1 Ω)=1.7377×104 Ω−1⋅m−1≈1.74×104 Ω−1⋅m−1

Substitute 1.74×104 Ω−1⋅m−1 for σ300 K, 0.17 eV for Eg, 8.62×10−5 eV/K for k, 348 K for T348 K and 300 K for T300 K in Equation (II).

σ348 K=(1.74×104 Ω−1⋅m−1)e(−0.17 eV2(8.62×10−5 eV/K)300 K)e(−0.17 eV2(8.62×10−5 eV/K)348 K)=(1.74×104 Ω−1⋅m−1)e(−3.2869)e(−2.8335)=(1.74×104 Ω−1⋅m−1)0.037360.05880=2.7386×104 Ω−1⋅m−1

Thus, the intrinsic electrical conductivity of InSb at 75°C is 2.7386×104 Ω−1⋅m−1.

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