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A sample of GaAs doped with NA = 1017 cm-3. For GaAs intrinsic concentration is n = 2.2 × 106 cm-3, mobility of electron is μn = 5300 cm2/V-sec, and mobility of hole is μp = 230 cm2/V sec.
If the sample is illuminated such that the excess electron concentration is 1016 cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?
Correct answer is between '12.48,12.60'. Can you explain this answer?
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Most Upvoted Answer
A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentra...
× 10^6 cm^-3 and the effective density of states in the conduction band is Nc = 2.8 × 10^19 cm^-3.
To find the electron concentration in the doped GaAs sample, we can use the equation:
n = ni^2/NA * exp(Ef/kT)
where ni is the intrinsic carrier concentration, NA is the acceptor concentration (doping level), Ef is the Fermi level, k is the Boltzmann constant, and T is the temperature.
Plugging in the values, we get:
n = (2.2 × 10^6)^2 / (10^17) * exp(Ef/kT)
Simplifying and taking the natural logarithm of both sides:
ln(n) = ln[(2.2 × 10^6)^2 / (10^17)] + Ef/kT
ln(n) = 16.412 + Ef/kT
To solve for Ef, we need to know the temperature. Let's assume a temperature of 300 K (room temperature). The value of kT at 300 K is 25.85 meV.
Substituting this value and the given doping concentration into the equation, we get:
ln(n) = 16.412 + Ef/25.85
Ef = (ln(n) - 16.412) * 25.85
Ef = (ln(NA/Nc) - 16.412) * 25.85
Ef = (-3.18 - 16.412) * 25.85
Ef = -506.6 meV
Therefore, the Fermi level in the doped GaAs sample is -506.6 meV below the conduction band edge. The electron concentration can be found by plugging this value of Ef into the equation we used earlier:
n = ni^2/NA * exp(Ef/kT)
n = (2.2 × 10^6)^2 / (10^17) * exp(-506.6/25.85)
n = 3.8 × 10^16 cm^-3
So the electron concentration in the doped GaAs sample is approximately 3.8 × 10^16 cm^-3.
To find the electron concentration in the doped GaAs sample, we can use the equation:
n = ni^2/NA * exp(Ef/kT)
where ni is the intrinsic carrier concentration, NA is the acceptor concentration (doping level), Ef is the Fermi level, k is the Boltzmann constant, and T is the temperature.
Plugging in the values, we get:
n = (2.2 × 10^6)^2 / (10^17) * exp(Ef/kT)
Simplifying and taking the natural logarithm of both sides:
ln(n) = ln[(2.2 × 10^6)^2 / (10^17)] + Ef/kT
ln(n) = 16.412 + Ef/kT
To solve for Ef, we need to know the temperature. Let's assume a temperature of 300 K (room temperature). The value of kT at 300 K is 25.85 meV.
Substituting this value and the given doping concentration into the equation, we get:
ln(n) = 16.412 + Ef/25.85
Ef = (ln(n) - 16.412) * 25.85
Ef = (ln(NA/Nc) - 16.412) * 25.85
Ef = (-3.18 - 16.412) * 25.85
Ef = -506.6 meV
Therefore, the Fermi level in the doped GaAs sample is -506.6 meV below the conduction band edge. The electron concentration can be found by plugging this value of Ef into the equation we used earlier:
n = ni^2/NA * exp(Ef/kT)
n = (2.2 × 10^6)^2 / (10^17) * exp(-506.6/25.85)
n = 3.8 × 10^16 cm^-3
So the electron concentration in the doped GaAs sample is approximately 3.8 × 10^16 cm^-3.
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Community Answer
A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentra...
Concept:
The conductivity for a doped semiconductor is obtained as:
σ = q n0 μn = q p0 μp
where,
ni = Intrinsic carrier concentration.
μn = electron mobility
μp = Hole mobility
When a semiconductor crystal is optically excited, one hole is created for every electron, i.e. electron-hole pass are created.
Application:
When the light is ON, excess electron and holes are generated in pairs, i.e.
Δp = Δn = 1016 cm3
Before Illumination:
NA = 1017 cm-3 and ni = 2.2 × 106 cm-3
Since NA ≫ ni, the excess hole concentration at thermal equilibrium will be:
p0 = NA = 1017 cm-3
The electron concentration using mass-action law is obtained as:
The electron concentration before illumination is negligible.
After illumination:
When an electron is generated optically, one hole is also created, i.e. they always occur in pairs.
∴ The excess hole concentration and excess electron concentration will be the same, i.e.
Δp = Δn = 1016 cm-3
Now, the electron and hole concentration after illumination will be:
n = n0 + Δn
n = 4.8 × 10-5 + 1016
n = 1016
Similarly,
p = p0 + Δp
p = 1017 + 1016 cm-3
p = 11 × 1016 cm-3
∴ the conductivity of the given sample when the light is ON will be:
σ = q(n0 + Δn)μn + q(p0 + Δp)μp
σ = q[(1016)μn + (11 × 1016)μp]
= (1.6 × 10-19) (1016 × 5300 + (11 × 1016) (230))
= 1.6 × 10-19 (5.3 × 1019 + 2.53 × 1019)
= 1.6 × (5.3 + 2.53)
σ = 12.528 (Ω-cm)-1
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A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. Can you explain this answer?
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A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. 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 A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. 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 A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. Can you explain this answer?.
A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. 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 A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. 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 A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. Can you explain this answer?.
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ample number of questions to practice A sample of GaAs doped with NA= 1017cm-3. For GaAs intrinsic concentration is n = 2.2 × 106cm-3, mobility of electron is μn= 5300 cm2/V-sec, and mobility of hole is μp= 230 cm2/V sec.If the sample is illuminated such that the excess electron concentration is 1016cm-3. What will the conductivity [in (Ω-cm)-1] of this sample, when the light is ON?Correct answer is between '12.48,12.60'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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