Semiconductor MCQ | Semiconductor Question and Answer with explanation

Ques.31. In the depletion region of a PN junction, there is a shortage of________

  1. Acceptor ions
  2. Donors ions
  3. Holes and electrons
  4. None of the above

Answer.3. Holes and electrons

Explanation:-

Consider the p-n junction at room temperature (say 300°K). The region to the left side of the junction is a p-type semiconductor and the region to the right is an n-type semiconductor. In an n-type semiconductor, the concentration of electrons is more than the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. Since particles have a tendency to diffuse from a region of high concentration to a region of low concentration, electrons and holes diffuse across the junction. Holes diffuse from p-side to n-side and electrons diffuse from n-side to p-side.

depletion region

When an electron diffuses to the p-side, it leaves behind a positively charged donor ion on the n-side. This donor ion cannot move as it is heavy and is bonded to the surrounding atoms. As the electrons continue to diffuse from the n-side to the p-side, a layer of positive charge is developed on the n-side of the junction. Similarly, when a hole diffuses to the n-side, it leaves behind a negatively charged acceptor ion on the p-side, which is immobile. As the holes continue to diffuse, a layer of negative charge is developed on the p-side of the junction. Thus, due to the diffusion of electrons and holes across the junction, a thin layer around the junction becomes devoid of free charge carriers. Consequently, a region which consists of the uncovered acceptor and donor ions is developed in the junction area. This region is called the depletion region. As the depletion region consists of immobile positive and negative ions, it is called a space-charge region. Its thickness is of the order of one-tenth of a micron. Thus in depletion region is a shortage of holes and electrons since it consists of donor and acceptor ions.

 

Ques.32.A reversed biased PN Junction has_______

  1. Very narrow depletion layer
  2. Almost no current
  3. Very low resistance
  4. Large current flow

Answer.2. Almost no current

Explanation:-

Reverse Biasing:- When the positive terminal of a dc source or battery is connected to the n-type and negative terminal is connected to the p-type semiconductor of a PN junction, as shown in Fig.. the junction is said to be in reverse bias.

When a PN junction is reverse biased, in the P-region the holes are attracted towards the negative terminal of the supply voltage. Similarly, the electrons are attracted to the positive terminal of the supply voltage in the N-region. Hence, the majority of carriers are drawn away from the junction. As the barrier potential is increased, it is very difficult for the majority carriers (holes in the P-type region and electrons in the N-type region) to diffuse across the junction. In the case, the applied reverse potential acts in such a way that it establishes an electric field which increases the field due to the potential barrier. Thus, the barrier potential at the junction is strengthened. The increased potential barrier prevents the flow of majority charge carriers across the junction. Thus, a high resistive path is established by the junction and practically no current flows through the circuit. However, minority carriers (electrons in the P-type region and holes in the N-type region) drift across the junction and the barrier potential helps the minority carriers to cross the junction.

reversed biased PN

The following points are important to note when a PN junction is reverse biased:

  • The junction potential barrier is strengthened.
  • The junction offers a high resistance (called reverse resistance, R..) to current flow.
  • It may be noted here that at reverse biasing, practically no current flows through the junction, but a little current (in µA) flows through the junction because of minority carriers available in the semiconductors even at room temperature. However, this current is neglected for all practical purposes.

 

Ques.33. A PN Junction, acts as a _________

  1. Unidirectional switch
  2. Bidirectional switch
  3. Controlled switch
  4. None of the above

Answer.1. Unidirectional switch

Explanation:-

  • A diode is a two-terminal device made of a p-n junction. It works like an electronic switch that allows current flow only in one direction.
  • It is a conductor in the positive direction and an insulator in the negative direction.
  • Once it starts conducting, the resistance of it is negligible. It is like a closed switch.
  • The voltage and current relationship across a diode have three regions: the forward bias region, reverse bias region, and reverse breakdown region.

unidirectional switch

  • When the voltage is larger than the forward bias voltage value (VF = 0.3 V for germanium diode, VF = 0.6 V to 0.7 V for a silicon diode),
  • VD > VF, the diode becomes a conductor and acts as a closed switch. Hence, the current increases very fast indicating very low resistance at this device. 
  • In this region, the diode can be considered as a closed switch with negligible resistance and a small voltage drop of VF.
  • When the diode is in the reverse bias region, VZ < VD < VF, then the diode acts as an insulator or open switch. In other words, it acts like a very large resistor.
  • When the VD < VZ, the diode operates in the breakdown region and acts as a closed switch. 

 

Ques.34. The leakage current is least in_______

  1. Silicon
  2. Germanium
  3. Carbon
  4. Sulfur

Answer.1. Silicon

Explanation:-

When a PN-junction is reverse biased the holes in the P-region are attracted towards the negative terminal of the applied voltage source, whereas the electrons in the N-region are attracted to the positive terminal of the voltage source. Thus the majority carriers are drawn away from the junction. This widens the depletion layer and increases the barrier potential.

The increased barrier potential makes it very difficult for the majority carriers to diffuse across the junction. Thus there is no current due to majority carriers in a reverse bias PN junction. Or in other words, we can say that the junction offers very high resistance under reverse biased condition.

However, the barrier potential helps the minority carriers in crossing the junction. As a matter of fact, as soon as a minority carrier is generated, it is swept (i.e., drifted) across the junction because of the barrier potential. Hence a small amount of current (called reverse saturation current or leakage current or current due to minority charge carriers) does flow through the reverse biased PN junction. The amount of this current depends upon the generation of minority carriers diffusing across the junction.

It may be noted that the generation of minority carriers depends upon the temperature and is independent of the applied reverse voltage. Therefore the current due to the flow of minority carriers remains the same whether the applied voltage is increased or decreased. Because of this reason, the current is known as reverse saturation current.

The reverse saturation current is of the order of nanoamperes (nA) for silicon and microamperes (µA) for Germanium PN junctions. The reverse saturation current for, Si devices is smaller than that of Ge because silicon has a large energy gap (a 1.1 eV) between the conduction and valence bands compare to the energy band gap of (0.7 eV). As a result of this, fewer minority carriers are thermally generated at any temperature in silicon than that of germanium. This is the reason why Si is mostly preferred over Ge.

 

Ques.35. A reverse biased PN junction has a resistance________

  1. Of the order of Ω
  2. Of the order of KΩ
  3. Of the order of MΩ
  4. None of the above

Answer.3. Of the order of MΩ

Explanation:-

An ideal diode should offer zero resistance in forward bias and infinite resistance in reverse bias. But in practice no diode can act as an ideal diode, i.e. an actual diode does not behave as a perfect conductor when forward biased and as a perfect insulator, when reverse biased. 

Reverse resistance: It is the resistance offered by the PN junction diode under reverse bias condition. It is very large compared to the forward resistance, which is in the range of several MΩ.

 

Ques.36. The leakage current across a PN Junction is due to_______

  1. Majority carriers
  2. Minority carriers
  3. Junction capacitance
  4. None of the above

Answer.2. Minority carriers

Explanation:-

When a PN-junction is reverse biased the holes in the P-region are attracted towards the negative terminal of the applied voltage source, whereas the electrons in the N-region are attracted to the positive terminal of the voltage source. Thus the majority carriers are drawn away from the junction. This widens the depletion layer and increases the barrier potential.

The increased barrier potential makes it very difficult for the majority carriers to diffuse across the junction. Thus there is no current due to majority carriers in a reverse bias PN junction. Or in other words, we can say that the junction offers very high resistance under reverse biased condition.

However, the barrier potential helps the minority carriers in crossing the junction. As a matter of fact, as soon as a minority carrier is generated, it is swept (i.e., drifted) across the junction because of the barrier potential. Hence a small amount of current (called reverse saturation current or leakage current or current due to minority charge carriers) does flow through the reverse biased PN junction. The amount of this current depends upon the generation of minority carriers diffusing across the junction.

It may be noted that the generation of minority carriers depends upon the temperature and is independent of the applied reverse voltage. Therefore the current due to the flow of minority carriers remains the same whether the applied voltage is increased or decreased. Because of this reason, the current is known as reverse saturation current.

 

Ques.37. When the temperature of an extrinsic semi-conductor increase, the pronounced effect is on________

  1. Majority carriers
  2. Minority carriers
  3. Junction capacitance
  4. None of the above

Answer.2. Minority carriers

Explanation:-

Effect of Temperature: When the temperature of an n-type semiconductor is raised, the number of electron-hole pairs due to thermal excitation from the valence band to the conduction band will increase. The number of electrons coming from the donor level will remain constant as the donor atoms are already ionized. Thus at a very high temperature, the concentration of thermally generated free electrons from the valence band will be much larger than the concentration of free electrons contributed by the donors. At this situation, the hole and the electron concentrations will be nearly equal and the semiconductor will behave like an intrinsic one. The same argument shows that a p-type semiconductor will also behave like an intrinsic semiconductor at a very high temperature. The general result is that as the temperature of an extrinsic semiconductor increases, the semiconductor passes from an extrinsic to an intrinsic one.

In an N-type semiconductor, the number of free electrons does not change appreciably with the increase In temperature, while the number of holes increases. In a P-type semiconductor, the number of free electrons increases with the increase in temperature, while the number of holes remains constant.

From the figure the following points can be deduced. (Tc= Curie temperature.)

IV graph semiconductor

1. At T= 0 Kelvin: The carrier concentration is zero hence, conductivity is zero and extrinsic semiconductor will be working as an insulator.

2. At O Kelvin < T < 300K: As temperature increases, majority and minority carriers are created and thus conductivity of extrinsic semiconductor increases with temperature.

3. At T= 300 Kelvin: The conductivity of extrinsic semiconductor will become maximum.

4. At 300K < T < Tc

(i) As temperature increases, the mobility of the charge carrier and, therefore, conductivity increases.

(ii) As temperature increases, majority carrier concentration will remain almost a constant.

(iii) As temperature increases, minority carrier concentration increases with the temperature.

5. At T= TC (Curie temperature): The concentration of minority carriers becomes equal to majority carrier concentration and therefore, extrinsic semiconductor will become intrinsic and the conductivity will be minimum.

6. At T > Tc The semiconductor is now intrinsic and conductivity increases with the temperature.

7. At very high temperature, extrinsic semiconductor will behave as intrinsic semiconductor (at Curie temperature).

8. At low temperature, the conductivity of extrinsic semiconductor increases with temperature (below 300 K).

9. At Curie temperature, extrinsic semiconductor becomes intrinsic.

10. In an extrinsic semiconductor, as the temperature is increased (above 300K), its conductivity decreases.

Note: Majority carrier concentration remains constant as temperature increases while minority carrier concentration increases with temperature Majority carrier concentration is almost independent of temperature.

 

Ques.38. With forward bias to a PN junction the width of the depletion layer__________

  1. Decreases
  2. Increases
  3. Remains the same
  4. None of the above

Answer.1. Decreases

Explanation:-

Forward bias: In this case, the positive terminal of the battery is connected to the P-side and negative terminal to the N-side. A large amount of current flows through the junction under this condition. Whenever a PN junction is forward biased, the holes are repelled by the positive terminal of the voltage source. Then holes are forcefully moved towards the junction. Correspondingly, the electrons are repelled by the negative terminal of the voltage source and these electrons are also forcefully moved towards the junction. As carriers (holes and electrons) get energy from the external voltage source, some of the holes and electrons are able to enter into the depletion layer and recombination takes place. Consequently, the width of the depletion layer decreases as well as the potential barriers (VB at the depletion region is decreased ). Therefore, more majority carriers diffuse across the junction and a large amount of current will flow through the PN junction. An ideal PN junction has a voltage drop across the crystal.

forward biased PN

The applied forward voltage (VF) establishes an electric field which acts against the field due to the potential barrier. Therefore, the resultant field is weakened and the barrier height is reduced at the junction. As potential barrier voltage is very small (0.3 to 0.7V), therefore, a small forward voltage is sufficient to completely eliminate the barrier. Once the potential barrier is eliminated by the forward voltage, junction resistance becomes almost zero and a low resistance path is established for the entire circuit. Therefore, current flows in the circuit. This is called forward current.

 

Ques.39. The leakage current in a PN junction is of the order of______

  1. A
  2. mA
  3. kA
  4. μA

Answer.4. µA

Explanation:-

The reverse saturation current is of the order of nanoamperes (nA) for silicon and microamperes (µA) for Germanium PN junctions. The reverse saturation current for, Si devices is smaller than that of Ge because silicon has a large energy gap (a 1.1 eV) between the conduction and valence bands compare to the energy band gap of (0.7 eV). As a result of this, fewer minority carriers are thermally generated at any temperature in silicon than that of germanium. This is the reason why Si is mostly preferred over Ge.

 

Ques.40. The valence shell in a silicon atom has the letter designation of

  1. A
  2. K
  3. L
  4. M

Answer.4. M

Explanation:-

Silicon was discovered in 1823. It is found extensively in the earth’s crust as a white c sometimes colorless compound, silicon dioxide.

silicon structure

Semiconductors are atoms that contain four valence electrons. Silicon (Si) belongs to group IV of the periodic table. Its atomic number is 14. The figure shows a shell structure of the silicon atom. The silicon atom contains 14 electrons (2 + 8 + 4 = 14), which occupy the first three shells. Two electrons occupy the first K shell, eight electrons occupy the second L shell, and four valence electrons occupy the third M shell. Ten out of 14 electrons are tightly bound to the nucleus. The outermost shell is called the valence shell, and the electrons in the valence shell are called the valence electrons.

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