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Consider an n-type semiconductor in which the dopant concentration ND is 1014/cm3. If the intrinsic concentration of semiconductor 1010/cm3 then the concentration of electrons (n) and holes (P) at equillibrium are
- a)n = 1014/cm3, P = 106/cm3
- b)n < 1014/cm3, P > 106/cm3
- c)n > 1014/cm3, P < 106/cm3
- d)n < 1014/cm3, P < 106/cm3
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Consider an n-type semiconductorin which the dopant concentrationND is...
= 1014/cm3, P = 0
c)n = 1010/cm3, P = 1014/cm3
d)n = 0, P = 1014/cm3
To determine the concentration of electrons (n) and holes (P) at equilibrium, we need to consider the balance between the dopant concentration (ND) and the intrinsic concentration (ni).
In an n-type semiconductor, the dopant concentration (ND) is higher than the intrinsic concentration (ni). This means that there are more donor electrons (n) than holes (P) at equilibrium.
Since the question states that the dopant concentration (ND) is 1014/cm3 and the intrinsic concentration (ni) is 1010/cm3, we can conclude that the concentration of electrons (n) is 1014/cm3 and the concentration of holes (P) is 0.
Therefore, the correct answer is:
b) n = 1014/cm3, P = 0
c)n = 1010/cm3, P = 1014/cm3
d)n = 0, P = 1014/cm3
To determine the concentration of electrons (n) and holes (P) at equilibrium, we need to consider the balance between the dopant concentration (ND) and the intrinsic concentration (ni).
In an n-type semiconductor, the dopant concentration (ND) is higher than the intrinsic concentration (ni). This means that there are more donor electrons (n) than holes (P) at equilibrium.
Since the question states that the dopant concentration (ND) is 1014/cm3 and the intrinsic concentration (ni) is 1010/cm3, we can conclude that the concentration of electrons (n) is 1014/cm3 and the concentration of holes (P) is 0.
Therefore, the correct answer is:
b) n = 1014/cm3, P = 0
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Community Answer
Consider an n-type semiconductorin which the dopant concentrationND is...
For N-type semiconductor, n=Nd=10^14/cm3
and p=ni²/Nd = 10^20 / 10^14 = 10^6/cm3
and p=ni²/Nd = 10^20 / 10^14 = 10^6/cm3
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Consider an n-type semiconductorin which the dopant concentrationND is 1014/cm3. If the intrinsicconcentration of semiconductor1010/cm3 then the concentration ofelectrons (n) and holes (P) at equillibrium area)n = 1014/cm3, P = 106/cm3b)n < 1014/cm3, P > 106/cm3c)n > 1014/cm3, P < 106/cm3d)n < 1014/cm3, P < 106/cm3Correct answer is option 'A'. Can you explain this answer?
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