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@Prajakta Kelapure • 21 Jul, 2014 • 1 like
Numericals

1. Determine the concentration of electrons and holes in a sample for germanium at 300°K in which concentration of donor atoms is 2*10(14) [raise to power 14] atoms/cm(3) and that of acceptor atoms = 3* 10(14) atoms/cm3 . Is this p-type or n-type?

a. Repeat for donor and acceptor concentration = 1*10(15) atoms/cm3

Formula used: NA+ n = ND + p , n*p = ni(2)

Note: After getting the answer put the comment stating why p or n-type

2. Find the concentration of hole and electron in p-type Ge if conductivity is 100(Ωcm)-1

Repeat the process for n-type Si if conductivity is 0.1(Ωcm)-1

Formula used: conductivity σ= σn+ σp = nqµn + pqµp , n*p = ni(2)

Note : if only p and n to be calculatd then long method but for dependent problem go for short method.

Short method: State that since bar is p-type, thus concentration of holes is more and conductivity is because of holes and then modify the formula as σ = σp

3. Show that resistivity of intrinsic Ge at 300°K is 45 Ωcm.

Formula used: for intrinsic conductivity, σi= niq(µn + µp) and resistivity= 1/ σi

a.If donor impurity is added to the extent of one atom per 10(8) Ge atoms. Then prove that resistivity droops to 37 Ωcm
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