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

The correct option is 1

Concept:

• N-type semiconductors are created by doping a pure semiconductor with a material that has more valence electrons than the semiconductor. This creates an excess of free electrons (majority carriers) and the dopant atoms, after releasing their extra valence electrons, become positive ions. These positive ions are not free to move around; they are fixed in the crystal lattice, and therefore they are referred to as "immobile positive ions". 
• It's worth noting that while electrons are the majority carriers in N-type semiconductors, there are still holes (minority carriers) present due to inherent thermal generation in the semiconductor.

Additional Information:

• N-type semiconductors: An extrinsic semiconductor where the dopant atoms provide extra conduction electrons to the host material like Phosphorus P in Silicon Si.
  • This creates an excess of negative (n-type) electron charge carriers that are able to move freely.

Concept of Common-Base (CB) Configuration:

In a common-base BJT amplifier:

• Input signal is applied to the emitter

• Output is taken from the collector

• Base is common to both input and output (grounded for AC signals)

 Additional Information

1) Suitable for high-frequency applications

• Correct: The CB configuration has excellent high-frequency response due to:

  • Low input impedance

  • No Miller effect (capacitance multiplication)

  • Better bandwidth compared to CE configuration

2) Voltage gain is very high

• Correct: While current gain (α) is slightly less than 1, the CB amplifier provides high voltage gain (comparable to CE).

3) Operates as an amplifier in saturation region

• Incorrect (Answer):

  • BJT acts as an amplifier only in the active region (where IC = βIB holds).

  • In saturation, the transistor acts as a closed switch (not for amplification).

4) Works as an off switch when both junctions are reverse-biased

• Correct: This describes the cutoff region (transistor acts as an open switch).

Conclusion:

The incorrect statement is Option 3, because a CB amplifier must operate in the active region for proper amplification, not saturation.

Final Answer: 3) The CB configuration operates as an amplifier when the transistor is in the saturation region. 

Explanation:

Incorrect Statement Analysis for a Diode Operating in Forward Bias

In a forward-biased diode, the following processes occur that affect the depletion region and the potential barrier:

Correct Option:

Option 4: The reduction in the depletion region causes a heavy flow of minority carriers across the junction.

This statement is incorrect because, in a forward-biased diode, the reduction in the depletion region primarily facilitates the flow of majority carriers (electrons in the n-region and holes in the p-region) across the junction. The minority carriers (holes in the n-region and electrons in the p-region) do not play a significant role in the current flow in a forward-biased condition. The primary current flow in a forward-biased diode is due to the majority carriers overcoming the potential barrier and recombining at the junction.

Explanation of Correct Option:

When a diode is forward-biased, the external voltage applied across the diode reduces the potential barrier at the p-n junction. This reduction in the potential barrier is due to the narrowing of the depletion region, which occurs as the majority carriers are pushed towards the junction. The applied forward voltage causes the electrons in the n-region to move towards the p-region, and the holes in the p-region to move towards the n-region. As a result, the width of the depletion region decreases, allowing more majority carriers to cross the junction and recombine, leading to an increase in the current flow through the diode.

The key points to understand here are:

• The forward bias reduces the potential barrier, making it easier for majority carriers to cross the junction.
• The depletion region narrows, allowing for the increased flow of majority carriers.
• The current in a forward-biased diode is primarily due to the flow of majority carriers, not minority carriers.

Analysis of Other Options:

Option 1: The reduction in the width of the depletion region is due to the recombination of charge carriers and immobile ions near the junction.

This statement is correct. In forward bias, as the majority carriers move towards the junction, they recombine with the oppositely charged immobile ions (donors in the n-region and acceptors in the p-region), reducing the width of the depletion region.

Option 2: The reduction in the potential barrier occurs due to the narrowing of the depletion region.

This statement is correct. The applied forward voltage reduces the potential barrier at the junction by narrowing the depletion region, allowing more majority carriers to cross the junction.

Option 3: The reduction in the depletion region allows a majority carrier flow across the junction.

This statement is correct. The narrowing of the depletion region in forward bias facilitates the flow of majority carriers (electrons and holes) across the junction, leading to an increase in current.

In summary, the incorrect statement is option 4, as it incorrectly attributes the increased current flow in a forward-biased diode to the flow of minority carriers. The correct understanding is that the majority carriers are responsible for the increased current in a forward-biased diode.

Explanation:

A Zener diode has a breakdown voltage of Vz = 7 V at 300 K, with a temperature coefficient of 2.3 mV/°C. To find the new breakdown voltage Vz at 400 K, we need to consider the change in temperature and how it affects the breakdown voltage.

Calculation:

1. First, determine the temperature difference:

• The initial temperature (T1) = 300 K
• The final temperature (T2) = 400 K
• Temperature difference (ΔT) = T₂ - T₁ = 400 K - 300 K = 100 K

2. Convert the temperature difference from Kelvin to Celsius:

• Since the temperature coefficient is given in mV/°C, we need to use the equivalent Celsius temperature difference.
• Note: The temperature change is the same in Celsius and Kelvin, so ΔT = 100 °C.

3. Calculate the change in breakdown voltage using the temperature coefficient:

• Temperature coefficient = 2.3 mV/°C
• Change in breakdown voltage (ΔVz) = Temperature coefficient × Temperature difference
• ΔVz = 2.3 mV/°C × 100 °C = 230 mV

4. Convert the change in breakdown voltage to volts:

• ΔVz = 230 mV = 0.230 V

5. Determine the new breakdown voltage:

• Initial breakdown voltage (Vz) = 7 V
• New breakdown voltage (Vz_new) = Initial breakdown voltage + Change in breakdown voltage
• Vz_new = 7 V + 0.230 V = 7.23 V

Conclusion:

The new breakdown voltage at 400 K is 7.23 V, which corresponds to Option 2.

V-I characteristics of a PN junction diode

Forward Bias:

When the p-type side of the diode is connected to a higher potential than the n-type side, the diode is forward-biased. This reduces the width of the depletion region, making it easier for electrons to flow from the n-type side to the p-type side and holes to flow from the p-type side to the n-type side, resulting in a large forward current. 

Reverse Bias:

When the n-type side of the diode is connected to a higher potential than the p-type side, the diode is reverse-biased. This widens the depletion region, making it difficult for the majority carriers to cross the junction. The reverse saturation current, which flows due to the thermally generated minority carriers, is very small compared to the forward current. 

Explanation

• The forward current of a diode is always significantly greater than its reverse saturation current. When a diode is forward-bias, it allows a substantial amount of current to flow, while in reverse bias, only a very small amount of reverse saturation current flows.
• This is due to the depletion region being much narrower in forward bias, allowing majority carriers to cross the junction easily. At the same time, it's wider in reverse bias, hindering the flow of majority carriers and allowing only minority carriers to contribute to the current. 

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,

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