Silicon is doped with boron to a concentration of 4 - 1017atoms-cm3- Assume the intrinsic carrier concentration of silicon to be 1-5 - 1010-cm3 and the value of kT-q to be 25 mV at 300 K-Compared to undoped silicon-the Fermi level of doped silicon

Ans : (c) Since, boron is p type impurity, therefore , fermi level goes down E_i E_f = kT lnN_An_i = 25 × 10^ 3ln4 × 10^171.5 × 10^10= 0.427 eV ...

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Question

Silicon is doped with boron to a concentration of $4 \times 10^{17} a t o m s / c m^{3}$ . Assume the intrinsic carrier concentration of silicon to be $1.5 \times 10^{10} / c m^{3}$ and the value of $k T / q$ to be $25 m V$ at $300 K$ .Compared to undoped silicon,the Fermi level of doped silicon

goes up by

0.13 e V

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goes down by

0.427 e V

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goes up by

0.427 e V

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goes down by

0.13 e V

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Solution

The correct option is B goes down by $0.427 e V$
Ans : (c)

Since, boron is p-type impurity, therefore , fermi level goes down
$E_{i} - E_{f} = k T ln \frac{N_{A}}{n_{i}}$
$= 25 \times 10^{- 3} ln \frac{4 \times 10^{17}}{1.5 \times 10^{10}} = 0.427 e V$

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Silicon is doped with boron to a concentration of 4×1017atoms/cm3. Assume the intrinsic carrier concentration of silicon to be 1.5×1010/cm3 and the value of kT/q to be 25 mV at 300 K.Compared to undoped silicon,the Fermi level of doped silicon

The correct option is B goes down by 0.427 eVAns : (c) Since, boron is p-type impurity, therefore , fermi level goes down Ei−Ef=kTlnNAni =25×10−3ln4×10171.5×1010=0.427 eV

Silicon is doped with boron to a concentration of $4 \times 10^{17} a t o m s / c m^{3}$ . Assume the intrinsic carrier concentration of silicon to be $1.5 \times 10^{10} / c m^{3}$ and the value of $k T / q$ to be $25 m V$ at $300 K$ .Compared to undoped silicon,the Fermi level of doped silicon

The correct option is B goes down by $0.427 e V$
Ans : (c)

Since, boron is p-type impurity, therefore , fermi level goes down
$E_{i} - E_{f} = k T ln \frac{N_{A}}{n_{i}}$
$= 25 \times 10^{- 3} ln \frac{4 \times 10^{17}}{1.5 \times 10^{10}} = 0.427 e V$