A silicon bar is doped with donor impurities ND- 2-25 - 1015atoms-cm3- Given the intrinsic carrier concentration of silicon at T-300 K is ni - 1-5 - 1010 cm-3- Assuming complete impurity ionization- the equilibrium electron and hole concentrations are

Ans : (d) Since N_D gt; gt; n_i therefore equilibrium electron concentration is N≃ N_D= 2.25 × 10^5 cm^ 3 And equilibrium hole concentration is given by m ...

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Intrinsic and Extrinsic Semiconductor

A silicon bar...

Question

A silicon bar is doped with donor impurities $N_{D} = 2.25 \times 10^{15} a t o m s / c m^{3}$ . Given the intrinsic carrier concentration of silicon at $T = 300 K$ is $n_{i} = 1.5 \times 10^{10} c m^{- 3}$ . Assuming complete impurity ionization, the equilibrium electron and hole concentrations are

n_{0} = 1.5 \times 10^{10} c m^{- 3}, p_{0} = 1.5 \times 10^{15} c m^{- 3}

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n_{0} = 2.25 \times 10^{15} c m^{- 3}, p_{0} = 1.5 \times 10^{10} c m^{- 3}

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n_{0} = 1.5 \times 10^{16} c m^{- 3}, p_{0} = 1.5 \times 10^{5} c m^{- 3}

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n_{0} = 2.25 \times 10^{15} c m^{- 3}, p_{0} = 1.5 \times 10^{5} c m^{- 3}

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Solution

The correct option is D $n_{0} = 2.25 \times 10^{15} c m^{- 3}, p_{0} = 1.5 \times 10^{5} c m^{- 3}$
Ans : (d)

Since $N_{D} >> n_{i}$ therefore equilibrium electron concentration is
$N ≃ N_{D} = 2.25 \times 10^{5} c m^{- 3}$
And equilibrium hole concentration is given by mass action law
$p = \frac{n_{i}^{2}}{N_{D}} = \frac{(1.5 \times 10^{10})^{2}}{2.25 \times 10^{15}} = 1 \times 10^{5} c m^{- 3}$

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A silicon bar is doped with donor impurities ND=2.25×1015atoms/cm3. Given the intrinsic carrier concentration of silicon at T=300 K is ni=1.5×1010cm−3. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are

The correct option is D n0=2.25×1015cm−3, p0=1.5×105cm−3Ans : (d) Since ND>>ni therefore equilibrium electron concentration is N≃ND=2.25×105cm−3 And equilibrium hole concentration is given by mass action law p=n2iND=(1.5×1010)22.25×1015=1×105cm−3