Electronic Devices MCQ Quiz - Objective Question with Answer for Electronic Devices - Download Free PDF

The relationship between carrier concentrations in a semiconductor is given by the mass action law, which states:

n × p = n_i²

Where:

• n = concentration of electrons (minority carriers in p-type material, in cm^-3).
• p = concentration of holes (majority carriers in p-type material, in cm^-3).
• n_i = intrinsic carrier concentration of silicon (in cm^-3).

Given data:

• Intrinsic carrier concentration, n_i = 1.5 × 10¹⁰ cm^-3.
• Acceptor concentration (boron doping), N_A = 1.5 × 10¹⁴ cm^-3.
• In a p-type semiconductor, the majority carrier concentration (holes) is approximately equal to the acceptor concentration: p ≈ N_A.

Step 1: Apply the mass action law

Using the mass action law:

n × p = n_i²

Substitute p ≈ N_A:

n × N_A = n_i²

Solve for the electron concentration, n:

n = n_i² / N_A

Step 2: Substitute the given values

Intrinsic carrier concentration: n_i = 1.5 × 10¹⁰ cm^-3

Acceptor concentration: N_A = 1.5 × 10¹⁴ cm^-3

Substitute these values into the equation:

n = (1.5 × 10¹⁰)² / (1.5 × 10¹⁴)

First, calculate n_i²:

n_i² = (1.5 × 10¹⁰) × (1.5 × 10¹⁰) = 2.25 × 10²⁰

Now, divide by N_A:

n = (2.25 × 10²⁰) / (1.5 × 10¹⁴)

n = 1.5 × 10⁶ cm^-3

The concentration of electrons in the silicon sample is 1.5 × 10⁶ cm^-3.

The correct option is 2

Explanation:

In an n-MOSFET, the gate polysilicon needs to be doped to make it conductive. For an n-MOSFET, the channel is n-type, and the gate polysilicon is typically doped with an n-type impurity to lower its resistance and to help set the threshold voltage.

Common n-type impurities used for doping silicon are Group V elements. Let's look at the options:

1. Boron: Boron is a Group III element and is a p-type dopant. It creates holes in silicon.
2. Antimony: Antimony (Sb) is a Group V element and is an n-type dopant. It contributes free electrons to silicon.
3. Gallium: Gallium (Ga) is a Group III element and is a p-type dopant.
4. Carbon: Carbon (C) is a Group IV element, like silicon. It's typically used to modify silicon's lattice properties, not as a primary dopant to create n-type or p-type conductivity.

Therefore, for doping gate polysilicon in an n-MOSFET to make it n-type, Antimony would be a suitable impurity. Other common n-type dopants for silicon include Phosphorus and Arsenic.

Explanation:

Operation of BJT in Linear Region

Definition: A Bipolar Junction Transistor (BJT) operates in the linear region (also known as the active region) when it functions as an amplifier. In this region, the transistor is neither fully "on" (saturation) nor fully "off" (cutoff). The active region is characterized by specific biasing conditions of the junctions within the BJT.

Biasing Conditions:

• Base-Emitter (B-E) Junction: The B-E junction is forward biased, meaning the voltage at the base is higher than the voltage at the emitter by at least the threshold voltage (approximately 0.7V for silicon BJTs).
• Base-Collector (B-C) Junction: The B-C junction is reverse biased, meaning the voltage at the base is lower than the voltage at the collector.

These biasing conditions allow the transistor to act as an amplifier by controlling the flow of current between the collector and emitter terminals based on the input signal applied to the base terminal.

Correct Option Analysis:

The correct option is:

Option 1: B-E junction is forward biased, and B-C junction is reverse biased.

This option correctly describes the biasing conditions required for a BJT to operate in the linear region. The forward biasing of the B-E junction ensures that the transistor is "on," while the reverse biasing of the B-C junction prevents the transistor from entering saturation. Together, these conditions enable the transistor to function as an amplifier

• The drain current (I_D) of a MOSFET in the saturation region is given by the equation:
  
  I_D = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²,
  
  where:
  • μ_n: Electron mobility
  • C_ox: Gate-oxide capacitance per unit area
  • W: Width of the MOSFET channel
  • L: Length of the MOSFET channel
  • V_GS: Gate-to-source voltage
  • V_TH: Threshold voltage
• All other parameters (μ_n, C_ox, and L) are constant in this problem.
• Therefore, the drain current is directly proportional to both the width of the MOSFET (W) and the square of the overdrive voltage (V_GS - V_TH).

Solution:

Let the parameters of the second MOSFET (MOSFET 2) be:

• Width: W
• Overdrive voltage: (V_GS - V_TH)
• Drain current: I_D2

For the first MOSFET (MOSFET 1), the parameters are:

• Width: 2W (double the width of MOSFET 2)
• Overdrive voltage: 2(V_GS - V_TH) (double the overdrive voltage of MOSFET 2)
• Drain current: I_D1

The drain current for MOSFET 2 is:

I_D2 = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

The drain current for MOSFET 1 is:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × [2(V_GS - V_TH)]²

Simplifying I_D1:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × 4(V_GS - V_TH)²

I_D1 = 4 × (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

I_D1 = 8 × I_D2

Conclusion:

The ratio of the drain currents of the two MOSFETs is:

I_D1 : I_D2 = 8 : 1

The correct answer is Option 3.

The correct answer is: 3) both (1) and (2)

Explanation:

• Junction Parasitics (like capacitance and resistance at the PN junctions) slow down the switching speed of the device, especially at high frequencies.

• Electron Mobility affects how quickly charge carriers can move through the semiconductor. Lower mobility results in slower device response, thus limiting high-frequency operation.

Both factors contribute to the frequency limitation of bipolar semiconductor devices.

Mistake Points

The question is asking about the number of valence electrons, and not about the valency of the atom. Since a valence electron is the number of outer shell electrons that is associated with an atom, Phosphorous will have 5, and Silicon will have 4 valence electrons.

The correct answer is 5 and 4.

Explanation:

• A valence electron is an outer shell electron that is associated with an atom.
• These electrons can participate in the formation of a chemical bond.

Key Points

• Silicon has two electrons in its first shell, eight electrons in the second shell, and four (4) electrons in the third shell.
• Since the electrons in the third shell are the outermost electrons, silicon has four valence electrons.
• Phosphorus having an atomic number 15 is a pentavalent element, which means it has 5 valence electrons in its outermost shell.

Property of Semiconductors:

• Semiconductors are the materials that have a conductivity between conductors (generally metals) and non-conductors or insulators (such as ceramics).
• Semiconductors can be compounds such as gallium arsenide or pure elements, such as germanium or silicon.
• Semiconductors have an almost empty conduction band and an almost filled valence band.
• In a semiconductor, the mobility of electrons is higher than that of the holes.
• Its Resistivity lies between 10^-5 to 10⁶ Ωm
• Conductivity lies between 10⁵ to 10^-6 mho/m
• The temperature coefficient of resistance for semiconductors is Negative.
• The current Flow in the semiconductor is mainly due to both electrons and holes.

• The avalanche breakdown is a phenomenon in which there is an increase in the number of free electrons beyond the rated capacity of the diode; This results in the flow of heavy current through the diode in reverse biased condition
• Avalanche breakdown occurs in lightly doped diode
• Zener breakdown mainly occurs because of a high electric field; When the high electric field is applied across the PN junction diode, then the electrons start flowing across the PN-junction; Consequently, develops little current in the reverse bias
• The Zener breakdown occurs in heavily doped diodes

Important difference between Avalanche and Zener breakdown:

Transistor can be acted as

1) Resistor in current mirror

2) Capacitor in level shifter

3) Closed or ON switch in saturation region

4) Inverter in cutoff and saturation region

5) Amplifier in active region

Nickel-Iron cell:

• Nickel-iron cell is in the charged condition, the active material on positive plates is Ni(OH)₄ and that on the negative plates is iron (Fe)
• The positive and negative plates are held in a nickel-plated steel container; the plates being insulated from each other by hard rubber strips
• The container contains 21 per cent solution of KOH (electrolyte) to which is added a small amount of lithium hydrate (LiOH) for increasing the capacity of the cell
• It has lesser weight and longer life than that of a lead-acid cell
• The emf of this cell is about 1.36 V
• These cells are very suitable for portable work

Effect of temperature on resistance of different material:

Conductor: When the temperature of conducting material increases, the resistance of that particular material increases.

Insulator: When the temperature of conducting material increases, the resistance of that particular material decreases.

Semiconductor: When the temperature of semiconducting material increases, the resistance of that particular material decreases. 

A negative coefficient for a material means that its resistance decreases with an increase in temperature. Hence, pure semiconductor materials (silicon and germanium) typically have negative temperature coefficients of resistance.

Concept:

• The combining capacity of an atom of an element to form a chemical bond is called its valency.
• The outermost electron shell of an atom is called the valence shell.
• The electrons present in the outermost shell of an atom are called valence electrons.
• The valence electron of an atom takes part in a chemical reaction because they have more energy than all the inner electrons.
• The valency of an element is
  • Equal to the number of valence electrons
  • Equal to the number of electrons required to complete eight electrons in the valence shell.
• Valency of a metal=No. of Valence electrons
• Valency Of a non-metal=8-No. of valence electrons
• For Ex:
  • Sodium(Z=11) Electronic Configuration=2,8,1
    • Valency=1
  • Magnesium(Z=2) Electronic Configuration=2,8,2
    • Valency=2
  • Chlorine(Z=17) Electronic Configuration=2,8,7
    • Valency=8-7=1
  • Oxygen= 8 Electronic Configuration=2,6
    • Valency=8-6=2

In a pn junction, if the doping concentration increases, the recombination of electrons and holes increases, thereby increases the voltage across the barrier.

\(V = \frac{{KT}}{q}{\rm{ln}}\left( {\frac{{{N_a}{N_d}}}{{n_i^2}}} \right)\)

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.

Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

\(\alpha = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_E}}}\)

Where, I_E = I_C + I_B

Calculation:

Given,

I_E = 2 mA

I_B = 20 μA = 0.02 mA

From above concept,

I_C = 2 mA - 0.02 mA = 1.98 mA

Current amplification factor is given as,

\(\alpha=\frac{I_C}{I_E}=\frac{1.98}{2}=0.99\)