A silicon sample is uniformly doped with 1016 phosphorus atoms-cm3 and 2 - 1016 boron atoms -cm3- If all the dopants are fully ionized - the material is

Ans : (b) p type with carrier concentration of 10^16 /cm^3, nbsp; Given that, N_D = n = Phosphorus atoms = 10^16 /cm^3 N_A = p = Boron atoms = 2 × 10^16 /cm^3 ...

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Intrinsic and Extrinsic Semiconductor

A silicon sam...

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A silicon sample is uniformly doped with $10^{16}$ phosphorus $a t o m s / c m^{3}$ and $2 \times 10^{16}$ boron $a t o m s / c m^{3}$ . If all the dopants are fully ionized , the material is

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Solution

Ans : (b)
p-type with carrier concentration of $10^{16} / c m^{3}$ ,
Given that,
$N_{D} = n =$ Phosphorus atoms = $10^{16} / c m^{3}$
$N_{A} = p =$ Boron atoms = $2 \times 10^{16} / c m^{3}$
$∵ N_{A} >> N_{D}$
So, the resultant material will be p-type semi conductor carrier concentration
$= N_{A} - N_{D}$
$= 2 \times 10^{16} - 10^{16}$
$= 10^{16} / c m^{3}$

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A silicon sample is uniformly doped with 1016 phosphorus atoms/cm3 and 2×1016 boron atoms/cm3. If all the dopants are fully ionized , the material is

Ans : (b) p-type with carrier concentration of 1016/cm3, Given that, ND=n= Phosphorus atoms = 1016/cm3 NA=p= Boron atoms = 2×1016/cm3 ∵NA>>ND So, the resultant material will be p-type semi conductor carrier concentration =NA−ND =2×1016−1016 =1016/cm3

A silicon sample is uniformly doped with $10^{16}$ phosphorus $a t o m s / c m^{3}$ and $2 \times 10^{16}$ boron $a t o m s / c m^{3}$ . If all the dopants are fully ionized , the material is