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A semiconductor sample at room temperature has an intrinsic concentration of 2.5 X 1017 /m3. After doping what will be the minority carrier concentration if the majority carrier concentration is given as 5.5 X 1021 /m3.
  • a)
    1.136 X 1013 /m3
  • b)
    2.2 X 104 /m3
  • c)
    3.33 X 1010 /m3
  • d)
    4.54 X 10-5 /m3
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A semiconductor sample at room temperature has an intrinsic concentra...
In a pure Semiconductor (Intrinsic Semiconductor), the electron and hole concentrations are n1p1 respectively. By doping impurity atoms the SC becomes extrinsic then the electrons and hole concentrations n2p2 respectively, then the following equations are acceptable
n1p1 = n2p2 = ni2
For Intrinsic Semiconductor, n=p=ni2 and as per questions before doping n1p1=ni2
Therefore,
p2=ni2n2
Free Test
Community Answer
A semiconductor sample at room temperature has an intrinsic concentra...
Given:
Intrinsic concentration (ni) = 2.5 x 10^17 /m^3
Majority carrier concentration (N) = 5.5 x 10^21 /m^3

To find:
Minority carrier concentration after doping

Solution:

1. Calculation of Intrinsic Carrier Concentration:
The intrinsic carrier concentration (ni) of a semiconductor can be calculated using the equation:

ni^2 = Nc * Nv * exp(-Eg / (2 * k * T))

Where:
Nc = Effective density of states in the conduction band
Nv = Effective density of states in the valence band
Eg = Band gap energy
k = Boltzmann's constant
T = Temperature

For silicon, the values are:
Nc = 2.8 x 10^25 /m^3
Nv = 1.04 x 10^19 /m^3
Eg = 1.12 eV
k = 1.38 x 10^-23 J/K
T = Room temperature (assumed to be 300K)

Substituting these values into the equation, we can solve for ni^2:

ni^2 = (2.8 x 10^25 /m^3) * (1.04 x 10^19 /m^3) * exp(-1.12 eV / (2 * (1.38 x 10^-23 J/K) * 300K))

ni^2 ≈ 1.45 x 10^20 /m^6

Taking the square root of ni^2, we get:
ni ≈ 3.81 x 10^10 /m^3

2. Calculation of Minority Carrier Concentration:
The minority carrier concentration (p or n) can be calculated using the equation:

p = (ni^2) / N

For the given majority carrier concentration (N = 5.5 x 10^21 /m^3), we can substitute the value of ni to find the minority carrier concentration (p):

p = ( (3.81 x 10^10 /m^3)^2) / (5.5 x 10^21 /m^3)
p ≈ 1.136 x 10^13 /m^3

Therefore, the minority carrier concentration after doping is approximately 1.136 x 10^13 /m^3, which corresponds to option A.
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A semiconductor sample at room temperature has an intrinsic concentration of 2.5 X 1017 /m3. After doping what will be the minority carrier concentration if the majority carrier concentration is given as 5.5 X 1021 /m3.a)1.136 X 1013 /m3b)2.2 X 104 /m3c)3.33 X 1010 /m3d)4.54 X 10-5 /m3Correct answer is option 'A'. Can you explain this answer?
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