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For a particular semiconductor material following parameters are observed:
μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3
These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)
Correct answer is '5.86'. Can you explain this answer?
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For a particular semiconductor material following parameters are obse...
The conductivity of a semiconductor material depends on various factors, including the carrier mobility and the carrier concentration. In this question, we are given the values of carrier mobilities and the carrier concentration for a particular semiconductor material.
Given parameters:
- Electron mobility (μn) = 1000 cm2/V-s
- Hole mobility (μp) = 600 cm2/V-s
- Effective density of states in the conduction band (Nc) = 1019 cm-3
- Effective density of states in the valence band (Nv) = 1019 cm-3
We are also given the measured conductivity of the intrinsic material at T = 300 K, which is 10-6 Ω cm-1.
To find the conductivity at T = 500 K, we need to consider the effect of temperature on the carrier concentration.
The carrier concentration in an intrinsic semiconductor can be calculated using the following equation:
ni^2 = Nc * Nv * exp(-Eg / (k * T))
Where:
- ni is the intrinsic carrier concentration
- Eg is the energy bandgap of the semiconductor
- k is the Boltzmann constant
- T is the temperature in Kelvin
Since the semiconductor material is intrinsic, the electron and hole concentrations are equal, i.e., ni = np.
Now, let's calculate the intrinsic carrier concentration at T = 300 K:
ni^2 = Nc * Nv * exp(-Eg / (k * T))
ni^2 = (1019 cm-3)^2 * exp(-Eg / (8.617333262 * 10-5 eV/K * 300 K))
Given that Nc = Nv = 1019 cm-3, we can simplify the equation:
ni^2 = (1019 cm-3)^2 * exp(-Eg / (8.617333262 * 10-5 eV/K * 300 K))
ni^2 = (1019 cm-3)^2 * exp(-Eg / 25.85)
At T = 300 K, the material is intrinsic, so ni = np.
Now, to calculate the conductivity at T = 500 K, we need to consider the change in carrier concentration due to temperature.
The conductivity of a semiconductor can be calculated using the following equation:
σ = q * (μn * n + μp * p)
Where:
- σ is the conductivity
- q is the elementary charge
- μn is the electron mobility
- μp is the hole mobility
- n is the electron concentration
- p is the hole concentration
As the material is intrinsic, n = p = ni.
Let's calculate the conductivity at T = 300 K using the given parameters:
σ_300K = q * (μn * ni + μp * ni)
σ_300K = q * (1000 cm2/V-s * ni + 600 cm2/V-s * ni)
σ_300K = q * (1600 cm2/V-s * ni)
Given that the measured conductivity at T = 300 K is 10-6 Ω cm-1, we can equate it to σ_300K and solve for ni:
10-6 Ω cm-1 = q * (1600 cm2/V-s * ni)
ni = (10-6 Ω cm-1) /
Given parameters:
- Electron mobility (μn) = 1000 cm2/V-s
- Hole mobility (μp) = 600 cm2/V-s
- Effective density of states in the conduction band (Nc) = 1019 cm-3
- Effective density of states in the valence band (Nv) = 1019 cm-3
We are also given the measured conductivity of the intrinsic material at T = 300 K, which is 10-6 Ω cm-1.
To find the conductivity at T = 500 K, we need to consider the effect of temperature on the carrier concentration.
The carrier concentration in an intrinsic semiconductor can be calculated using the following equation:
ni^2 = Nc * Nv * exp(-Eg / (k * T))
Where:
- ni is the intrinsic carrier concentration
- Eg is the energy bandgap of the semiconductor
- k is the Boltzmann constant
- T is the temperature in Kelvin
Since the semiconductor material is intrinsic, the electron and hole concentrations are equal, i.e., ni = np.
Now, let's calculate the intrinsic carrier concentration at T = 300 K:
ni^2 = Nc * Nv * exp(-Eg / (k * T))
ni^2 = (1019 cm-3)^2 * exp(-Eg / (8.617333262 * 10-5 eV/K * 300 K))
Given that Nc = Nv = 1019 cm-3, we can simplify the equation:
ni^2 = (1019 cm-3)^2 * exp(-Eg / (8.617333262 * 10-5 eV/K * 300 K))
ni^2 = (1019 cm-3)^2 * exp(-Eg / 25.85)
At T = 300 K, the material is intrinsic, so ni = np.
Now, to calculate the conductivity at T = 500 K, we need to consider the change in carrier concentration due to temperature.
The conductivity of a semiconductor can be calculated using the following equation:
σ = q * (μn * n + μp * p)
Where:
- σ is the conductivity
- q is the elementary charge
- μn is the electron mobility
- μp is the hole mobility
- n is the electron concentration
- p is the hole concentration
As the material is intrinsic, n = p = ni.
Let's calculate the conductivity at T = 300 K using the given parameters:
σ_300K = q * (μn * ni + μp * ni)
σ_300K = q * (1000 cm2/V-s * ni + 600 cm2/V-s * ni)
σ_300K = q * (1600 cm2/V-s * ni)
Given that the measured conductivity at T = 300 K is 10-6 Ω cm-1, we can equate it to σ_300K and solve for ni:
10-6 Ω cm-1 = q * (1600 cm2/V-s * ni)
ni = (10-6 Ω cm-1) /
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Community Answer
For a particular semiconductor material following parameters are obse...
Σ1 = eni(μn + μp)
10-6 = (1.6 x 10-19)(1000 + 600)ni
At T = 300 K, ni = 3.91 x 109 cm-3
Ni = 2.29 x 1013 cm-3
= (1.6 x 10-19)(2.29 x 1013)(1000 + 600)
= 5.86 x 10-3(Ω-cm)-1
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For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer?
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For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer?.
For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a particular semiconductor material following parameters are observed:μn = 1000 cm2/V - s, μp = 600 cm2/V - s, Nc = Nv = 1019 cm-3These parameters are independent of temperature. The measured conductivity of the intrinsic material = 10-6 (Ω cm)-1 at T = 300 K. The conductivity (Ω cm)-1 at T = 500 K is ___ × 10-3 (Answer up to two decimal places)Correct answer is '5.86'. Can you explain this answer?.
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