Question

Silicon is doped with which of the following to obtain p-type semiconductor?

• A. Phosphorus
• B. Arsenic
• C. Boron
• D. Antimony
• A. \[5 \times 10^{22}\,\text{m}^{-3}\]
• B. \[1.5 \times 10^{6}\,\text{m}^{-3}\]
• C. \[9 \times 10^{8}\,\text{m}^{-3}\]
• D. \[4.5 \times 10^{9}\,\text{m}^{-3}\]
• A. current is mainly due to drifting of majority carriers.
• B. current is mainly due to drifting of minority carriers.
• C. diffusion and drift currents are equal.
• D. current is of the order of 1 A.
• A. 0.2 V
• B. 0.5 V
• C. 0.7 V
• D. 1.5 V
• A. 0.01 eV
• B. 0.05 eV
• C. 0.1 eV
• D. 0.15 eV

Answer 1

The Correct Option is C

Concept: Pure silicon is known as an intrinsic semiconductor. Its electrical conductivity can be increased significantly by adding a small amount of suitable impurity atoms. This process is called doping. When a trivalent impurity is added to silicon, one covalent bond remains incomplete because the impurity atom possesses only three valence electrons instead of four. This creates a hole, and the semiconductor becomes a p-type semiconductor. Trivalent impurities are called acceptor impurities because they accept electrons and create holes as majority charge carriers.

Step 1: Identify the type of impurity required. To obtain a p-type semiconductor, a trivalent impurity must be added. Common trivalent impurities are: \[ \text{Boron (B), Aluminium (Al), Gallium (Ga), Indium (In)} \]

Step 2: Examine the given options. \[ \text{Phosphorus} \rightarrow \text{Pentavalent} \] \[ \text{Arsenic} \rightarrow \text{Pentavalent} \] \[ \text{Boron} \rightarrow \text{Trivalent} \] \[ \text{Antimony} \rightarrow \text{Pentavalent} \] Among the given options, only Boron is a trivalent impurity.

Step 3: Conclude the answer. Since Boron is a trivalent impurity, it produces holes as majority charge carriers and converts silicon into a p-type semiconductor. \[ \boxed{\text{Boron}} \] Hence, the correct answer is \[ \boxed{\text{(C)}} \]

Answer 2

Concept: For a semiconductor in thermal equilibrium, the product of electron concentration and hole concentration remains constant and is given by \[ np=n_i^2 \] where \[ n=\text{electron concentration} \] \[ p=\text{hole concentration} \] \[ n_i=\text{intrinsic carrier concentration} \] This relation is known as the mass action law.

Step 1: Write the given quantities. Electron concentration: \[ n=5\times10^{22}\,\text{m}^{-3} \] Intrinsic carrier concentration: \[ n_i=1.5\times10^{16}\,\text{m}^{-3} \]

Step 2: Apply the mass action law. Using \[ np=n_i^2 \] we obtain \[ p=\frac{n_i^2}{n} \] Substituting the values, \[ p= \frac{(1.5\times10^{16})^2} {5\times10^{22}} \] \[ = \frac{2.25\times10^{32}} {5\times10^{22}} \] \[ = 0.45\times10^{10} \] \[ = 4.5\times10^{9}\,\text{m}^{-3} \]

Step 3: Verify the result. Since the electron concentration is extremely large, the hole concentration must be very small so that the product \(np\) remains constant. The calculated value satisfies this condition. Final Answer: \[ \boxed{ p=4.5\times10^{9}\,\text{m}^{-3} } \] Therefore, the correct option is \[ \boxed{\text{(D)}} \]

Answer 3

Concept: A p-n junction diode consists of a p-type semiconductor and an n-type semiconductor joined together. Due to the concentration difference of charge carriers on the two sides, majority carriers diffuse across the junction and form a depletion region. This depletion region establishes a built-in potential barrier. In an unbiased p-n junction: \[ \text{Diffusion Current} = \text{Drift Current} \] and the net current through the junction is zero. When the diode is forward biased, the external voltage opposes the barrier potential, reducing the width of the depletion layer and allowing majority charge carriers to cross the junction easily.

Step 1: Understand the effect of forward biasing. In forward biasing: \[ p\text{-side} \rightarrow \text{Connected to positive terminal} \] \[ n\text{-side} \rightarrow \text{Connected to negative terminal} \] The applied voltage reduces the barrier potential. As a result, \[ \text{Depletion layer width decreases} \] and majority carriers can move across the junction more easily.

Step 2: Identify the dominant current. Since a large number of majority carriers cross the junction due to concentration difference, the current produced is called diffusion current. The diffusion current becomes much larger than the drift current under forward bias. Therefore, the current in a forward-biased diode is mainly due to the motion of majority carriers.

Step 3: Analyse the given options. Option (A): Majority carriers dominate the current flow. \checkmark Option (B): Drift current is due to minority carriers and dominates in reverse bias. \(\times\) Option (C): Diffusion current and drift current are equal only in equilibrium (unbiased condition). \(\times\) Option (D): Current depends on circuit conditions and cannot always be taken as \(1\,A\). \(\times\) Final Answer: The current in a forward-biased p-n junction is mainly due to majority charge carriers crossing the junction. \[ \boxed{\text{(A) current is mainly due to drifting of majority carriers}} \]

Answer 4

Concept: The threshold voltage, also called the cut-in voltage or knee voltage, is the minimum forward bias voltage that must be applied across a diode before it starts conducting appreciable current. Initially, the applied voltage is used in overcoming the barrier potential of the depletion region. Once this barrier is sufficiently reduced, the diode begins to conduct rapidly. The threshold voltage depends on the semiconductor material used.

Step 1: Recall standard threshold voltages. For commonly used semiconductor diodes: \[ \text{Germanium Diode} \approx 0.3\,V \] \[ \text{Silicon Diode} \approx 0.7\,V \] These values are standard and frequently used in electronic circuit analysis.

Step 2: Understand why silicon has a larger threshold voltage. Silicon has a larger energy band gap than germanium. \[ E_g(\text{Si}) \approx 1.1\,eV \] \[ E_g(\text{Ge}) \approx 0.7\,eV \] Therefore, a larger forward voltage is required before significant conduction begins in a silicon diode.

Step 3: Compare with the options. Among the given options, \[ 0.7\,V \] is the accepted threshold voltage of a silicon diode. Final Answer: \[ \boxed{0.7\,V} \] Hence, the correct option is \[ \boxed{\text{(C)}} \]

Answer 5

Concept: Germanium is a group 14 semiconductor having four valence electrons. When it is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony, each impurity atom contributes five valence electrons. Four of these electrons participate in covalent bond formation with neighbouring germanium atoms, while the fifth electron remains only weakly bound to the impurity atom. This extra electron can be easily detached even at room temperature and contributes to electrical conduction. Such impurities are called donor impurities because they donate free electrons to the semiconductor crystal.

Step 1: Understand donor doping in germanium. When a pentavalent atom is introduced into a germanium crystal: \[ \text{Number of valence electrons of Ge} = 4 \] \[ \text{Number of valence electrons of donor atom} = 5 \] Out of these five electrons:
• Four electrons form covalent bonds with neighbouring Ge atoms.
• One electron remains weakly attached to the donor atom. Therefore, only a very small amount of energy is required to free this electron.

Step 2: Recall the donor ionisation energy. The donor energy level lies very close to the conduction band. For germanium, the ionisation energy of the donor electron is approximately \[ 0.01\,\text{eV} \] This value is much smaller than the energy gap of germanium, which explains why donor electrons can easily become free charge carriers.

Step 3: Compare with the given options. The available options are: \[ 0.01\,\text{eV} \] \[ 0.05\,\text{eV} \] \[ 0.10\,\text{eV} \] \[ 0.15\,\text{eV} \] The accepted value for the donor ionisation energy in germanium is \[ 0.01\,\text{eV} \]

Step 4: Conclude the answer. Hence, the ionisation energy of the fifth weakly bound electron is \[ \boxed{0.01\,\text{eV}} \] Therefore, the correct option is \[ \boxed{\text{(A)}} \]