Question 1 |

In a semiconductor device, the Fermi-energy level is 0.35 \mathrm{eV} above the valence band energy. The effective density of states in the valence band at T=300 \mathrm{~K} is 1 \times 10^{19} \mathrm{~cm}^{-3}. The thermal equilibrium hole concentration in silicon at 400 \mathrm{~K} is ___ \_\times 10^{13} \mathrm{~cm}^{-3}. (rounded off to two decimal places).

Given \mathrm{kT} at 300 \mathrm{~K} is 0.026 \mathrm{eV}.

Given \mathrm{kT} at 300 \mathrm{~K} is 0.026 \mathrm{eV}.

63.36 | |

25.36 | |

45.25 | |

98.36 |

Question 1 Explanation:

Given,

E_{F}-E_{V}=0.35 \mathrm{eV} \quad \text { [Considering it is given at } 400 \mathrm{~K} \text { ] }

Also, V_{T_{1}}=K T_{1}=0.026 \mathrm{eV} at T_{1}=300 \mathrm{~K}

\therefore \frac{V_{T_{1}}}{V_{T_{2}}}=\frac{T_{1}}{T_{2}} \Rightarrow V_{T_{2}}=\frac{T_{2}}{T_{1}} \times V_{T_{1}}

\therefore V_{T_{2}}=\frac{400}{300} \times 0.026

V_{T_{2}}=0.03466 \mathrm{eV} at T_{2}=400 \mathrm{~K}

Now, N_{v}=1 \times 10^{19} / \mathrm{cm}^{3} at T_{1}=300 \mathrm{~K}

N_{v} \propto T^{3 / 2}

\frac{N_{V_{2}}}{N_{V_{1}}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}

N_{V_{2}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2} N V_{1}

\left(\because T_{2}=400 \mathrm{~K}\right) =\left(\frac{400}{300}\right)^{3 / 2} N_{V_{1}}

N_{V_{2}}=1.5396 \times 10^{19} / \mathrm{cm}^{3}

Now, hole concentration at 400 \mathrm{~K} is given as

\begin{aligned} & p=N_{V} e^{-\left(E_{F}-E_{V}\right) / k T_{2}}=1.5396 \times 10^{19} \times e^{-0.35 \mathrm{eV} / 0.03466 \mathrm{eV}} \\ & p=63.36 \times 10^{13} \mathrm{~cm}^{-3} \end{aligned}

E_{F}-E_{V}=0.35 \mathrm{eV} \quad \text { [Considering it is given at } 400 \mathrm{~K} \text { ] }

Also, V_{T_{1}}=K T_{1}=0.026 \mathrm{eV} at T_{1}=300 \mathrm{~K}

\therefore \frac{V_{T_{1}}}{V_{T_{2}}}=\frac{T_{1}}{T_{2}} \Rightarrow V_{T_{2}}=\frac{T_{2}}{T_{1}} \times V_{T_{1}}

\therefore V_{T_{2}}=\frac{400}{300} \times 0.026

V_{T_{2}}=0.03466 \mathrm{eV} at T_{2}=400 \mathrm{~K}

Now, N_{v}=1 \times 10^{19} / \mathrm{cm}^{3} at T_{1}=300 \mathrm{~K}

N_{v} \propto T^{3 / 2}

\frac{N_{V_{2}}}{N_{V_{1}}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}

N_{V_{2}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2} N V_{1}

\left(\because T_{2}=400 \mathrm{~K}\right) =\left(\frac{400}{300}\right)^{3 / 2} N_{V_{1}}

N_{V_{2}}=1.5396 \times 10^{19} / \mathrm{cm}^{3}

Now, hole concentration at 400 \mathrm{~K} is given as

\begin{aligned} & p=N_{V} e^{-\left(E_{F}-E_{V}\right) / k T_{2}}=1.5396 \times 10^{19} \times e^{-0.35 \mathrm{eV} / 0.03466 \mathrm{eV}} \\ & p=63.36 \times 10^{13} \mathrm{~cm}^{-3} \end{aligned}

Question 2 |

In an extrinsic semiconductor, the hole concentration is given to be 1.5 n_{i} where n_{i} is the intrinsic carrier concentration of 1 \times 10^{10} \mathrm{~cm}^{-3}. The ratio of electron to hole mobility for equal hole and electron drift current is given as ___

(rounded off to two decimal places).

(rounded off to two decimal places).

1.15 | |

2.25 | |

2.85 | |

3.36 |

Question 2 Explanation:

Given, intrinsic carrier concentration n_{i}=1 \times 10^{10} \mathrm{~cm}^{-3}

Hole concentration,

\quad p=1.5 \times n_{i}

p=1.5 \times 10^{10} \mathrm{~cm}^{-3}

Given, electron and hole current are equal

\begin{aligned} I_{p \text { dritt }} & =I_{n \text { dritt }} \\ p q \mu_{p} E A & =n q \mu_{n} E A \\ 1.5 \times 10^{10} \mu_{p} & =n \mu_{n} \quad ...(i) \end{aligned}

But according to mass action law,

\begin{aligned} n p & =n_{i}^{2} \\ n & =\frac{n_{i}}{1.5}=\frac{10^{10}}{1.5} \mathrm{~cm}^{-3} \end{aligned}

Put in equation (i)

\therefore \quad 1.5 \times 10^{10} \mu_{p}=\frac{10^{10}}{1.5} \times \mu_{n}

\frac{\mu_{n}}{\mu_{p}}=2.25

Hole concentration,

\quad p=1.5 \times n_{i}

p=1.5 \times 10^{10} \mathrm{~cm}^{-3}

Given, electron and hole current are equal

\begin{aligned} I_{p \text { dritt }} & =I_{n \text { dritt }} \\ p q \mu_{p} E A & =n q \mu_{n} E A \\ 1.5 \times 10^{10} \mu_{p} & =n \mu_{n} \quad ...(i) \end{aligned}

But according to mass action law,

\begin{aligned} n p & =n_{i}^{2} \\ n & =\frac{n_{i}}{1.5}=\frac{10^{10}}{1.5} \mathrm{~cm}^{-3} \end{aligned}

Put in equation (i)

\therefore \quad 1.5 \times 10^{10} \mu_{p}=\frac{10^{10}}{1.5} \times \mu_{n}

\frac{\mu_{n}}{\mu_{p}}=2.25

Question 3 |

For an intrinsic semiconductor at temperature T=0 \mathrm{~K}, which of the following statement is true?

All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons. | |

All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons. | |

All energy states in the valence and conduction band are filled with holes. | |

All energy states in the valence and conduction band are filled with electrons. |

Question 3 Explanation:

Intrinsic semiconductor at \mathrm{T}=0 \mathrm{~K} behaves as an insulator.

Hence, valence band is completely filled with electron and conduction band is completely empty.

Hence, valence band is completely filled with electron and conduction band is completely empty.

Question 4 |

In a semiconductor, if the Fermi energy level lies in the conduction band, then the semiconductor is known as

degenerate n-type. | |

degenerate p-type. | |

non-degenerate n-type. | |

non-degenerate p-type. |

Question 4 Explanation:

As the Fermi lies inside the conduction band hence it is degenerate n-type semiconductor.

Question 5 |

Select the CORRECT statement(s) regarding semiconductor devices.

Electrons and holes are of equal density in an intrinsic semiconductor at equilibrium. | |

Collector region is generally more heavily doped than Base region in a BJT. | |

Total current is spatially constant in a two terminal electronic device in dark under
steady state condition. | |

Mobility of electrons always increases with temperature in Silicon beyond 300 K. |

Question 5 Explanation:

At equilibrium n = p = n_i
for intrinsic
semiconductor

Collector region is generally lightly doped then base region in BJT. Hence option B is wrong.

By increasing temperature above 300K, mobality of electrons decreases hence option (D) is also wrong

Collector region is generally lightly doped then base region in BJT. Hence option B is wrong.

By increasing temperature above 300K, mobality of electrons decreases hence option (D) is also wrong

There are 5 questions to complete.