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All questions of Semiconductor Electronics: Materials, Devices and Simple Circuits for JEE Exam
Arrange the materials in ascending order of energy band gaps- a)conductor, semiconductor, insulator
- b)insulator, semiconductor, conductor
- c)semiconductor, conductor, insulator
- d)insulator, conductor, semiconductor
Correct answer is option 'A'. Can you explain this answer?
Arrange the materials in ascending order of energy band gaps
a)
conductor, semiconductor, insulator
b)
insulator, semiconductor, conductor
c)
semiconductor, conductor, insulator
d)
insulator, conductor, semiconductor
|
Krishna Iyer answered |
- In the case of insulators, the energy gap is very large while in the case of conductors energy gap is very short.
- This energy gap refers to the energy difference between the valence band and conduction band whereas, in semiconductors, this band gap value is in between that of conductors and insulators.
Can you explain the answer of this question below:When a pure semiconductor is heated, its resistance- A:Goes down.
- B:Goes up.
- C:Remains the same.
- D:None of the above.
The answer is a.
When a pure semiconductor is heated, its resistance
A:
Goes down.
B:
Goes up.
C:
Remains the same.
D:
None of the above.
|
Nikita Singh answered |
As temperature increases electrons from the valence band go to the conduction band, so conductivity increases hence resistance decreases.
Can you explain the answer of this question below:What is the order of forbidden gap in the energy bands of silicon?
- A:
3.1 eV
- B:
0.1 eV
- C:
2.1 eV
- D:
1.1 eV
The answer is d.
What is the order of forbidden gap in the energy bands of silicon?
3.1 eV
0.1 eV
2.1 eV
1.1 eV
|
Anjali Iyer answered |
Eg : Silicon & Germanium. b) For silicon forbidden energy gap is 1.1 eV and for germanium 0.72 eV. c) At absolute zero, semiconductors behave as perfect insulators. d) Semiconductors are of two types.
The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015 cm-3and 5 x 1012 cm-3 respectively. The mobilities of electrons and holes are respectively 23 x 103 cm3 / Vs and 102 cm3 / Vs. Which type of semiconductor is it?- a)p-type only
- b)n-type only
- c)either n or p type
- d)both n and p type
Correct answer is option 'B'. Can you explain this answer?
The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015 cm-3and 5 x 1012 cm-3 respectively. The mobilities of electrons and holes are respectively 23 x 103 cm3 / Vs and 102 cm3 / Vs. Which type of semiconductor is it?
a)
p-type only
b)
n-type only
c)
either n or p type
d)
both n and p type
|
Maulik Dasgupta answered |
Given data:
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
A semiconductor is formed by- a)Covalent bonds
- b)Electrovalent bonds
- c)Coordinate bonds
- d)None of above
Correct answer is option 'A'. Can you explain this answer?
A semiconductor is formed by
a)
Covalent bonds
b)
Electrovalent bonds
c)
Coordinate bonds
d)
None of above
|
|
Wahid Khan answered |
The electrons surrounding each atom in a semiconductor are part of a covalent bond. A covalent bond consists of two atoms "sharing" a single electron. Each atom forms 4 covalent bonds with the 4 surrounding atoms. Therefore, between each atom and its 4 surrounding atoms, 8 electrons are being shared.##
In a semiconductor, the energy gap between valence band and conduction band is abouta)10 eVb)5 eVc)1 eVd)15 eVCorrect answer is option 'C'. Can you explain this answer?
|
Riya Banerjee answered |
In semiconductors, the forbidden gap between the valence band and conduction band is very small. It has a forbidden gap of about 1 electron volt (eV).
Mobilities of electrons and holes in a sample of intrinsic germanium semiconductor at room temperature are 0.36m2/volt-sec and 0.17 m2/volt-sec respectively. If the electron and hole densities are each equal to 2.5 X 1019/m3, the conductivity is ______.
- a)1.22 S/m
- b)3.12 S/m
- c)4.12 S/m
- d)2.12 S/m
Correct answer is option 'D'. Can you explain this answer?
Mobilities of electrons and holes in a sample of intrinsic germanium semiconductor at room temperature are 0.36m2/volt-sec and 0.17 m2/volt-sec respectively. If the electron and hole densities are each equal to 2.5 X 1019/m3, the conductivity is ______.
a)
1.22 S/m
b)
3.12 S/m
c)
4.12 S/m
d)
2.12 S/m
|
|
Krishna Iyer answered |
Given in question,
Mobilities of electrons μe = 0.36×m2V−1s−1
Mobilities of holes μh = 0.17×m2V−1s−1
densities of electron= densities of holes=2.5×1019m−3
As we know, conductivity,
σ=1/p=e(μene+μnnn)
=1.6×10−19[0.36×2.5×1019+0.17×2.5×1019)]
=2.12Sm−1
σ=1/p=e(μene+μnnn)
=1.6×10−19[0.36×2.5×1019+0.17×2.5×1019)]
=2.12Sm−1
So the electrical conductivity of germanium is 2.12Sm−1
The name of a diode that can be used to provide a variable capacitance is:- a)varactor diode
- b)varactor capacitor
- c)potential diode
- d)potential capacitor
Correct answer is option 'A'. Can you explain this answer?
The name of a diode that can be used to provide a variable capacitance is:
a)
varactor diode
b)
varactor capacitor
c)
potential diode
d)
potential capacitor
|
Dr Manju Sen answered |
The varactor diode is used in a place where the variable capacitance is required, and that capacitance is controlled with the help of the voltage. The Varactor diode is also known as the Varicap, Voltcap, Voltage variable capacitance or Tunning diode.
The knee voltage of a-n junction diode is 0.8 V and the with of the depletion layer is 2 μm. What is the electric field in the depletion layer?- a)0.4 KV/m
- b)0.4 MV/m
- c)4 MV/m
- d)4 KV/m
Correct answer is option 'B'. Can you explain this answer?
The knee voltage of a-n junction diode is 0.8 V and the with of the depletion layer is 2 μm. What is the electric field in the depletion layer?
a)
0.4 KV/m
b)
0.4 MV/m
c)
4 MV/m
d)
4 KV/m
|
Gaurav Kumar answered |
Knee voltage- it is that forward voltage beyond which current start increasing rapidly, but below knee voltage variation of forward current and applied voltage is linear.
The electric field in a region is given by, E=V/l.
where V is the potential and l is the length or distance of the region in which it has to be measured.
now, E=V/l
E=0.8/2×10-6 m
E= 0.4 MV/m
The electric field in a region is given by, E=V/l.
where V is the potential and l is the length or distance of the region in which it has to be measured.
now, E=V/l
E=0.8/2×10-6 m
E= 0.4 MV/m
Can you explain the answer of this question below:Electrons moving like molecules of a gas exist in:
- A:
all solids
- B:
covalent solids
- C:
metallic solids
- D:
ionic solids
The answer is c.
Electrons moving like molecules of a gas exist in:
all solids
covalent solids
metallic solids
ionic solids
|
|
Rohan Singh answered |
Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements.
What is the value of dielectric constant of a air ?- a) Less than 1
- b)Zero
- c)Equal to 1
- d) Not determine
Correct answer is option 'C'. Can you explain this answer?
What is the value of dielectric constant of a air ?
a)
Less than 1
b)
Zero
c)
Equal to 1
d)
Not determine
|
Ambition Institute answered |
The dielectric constant of a medium K = ∈/∈0
where ∈ is the permittivity of the medium.
For air, ∈ = ∈0
⟹ K = 1
where ∈ is the permittivity of the medium.
For air, ∈ = ∈0
⟹ K = 1
In a common base transistor circuit, the current gain is 0.98. On changing emitter current by 5 mA, the change in collector current is- a)0.196 mA
- b)2.45 mA
- c)4.9 mA
- d)5.1 mA
Correct answer is option 'C'. Can you explain this answer?
In a common base transistor circuit, the current gain is 0.98. On changing emitter current by 5 mA, the change in collector current is
a)
0.196 mA
b)
2.45 mA
c)
4.9 mA
d)
5.1 mA
|
Lavanya Menon answered |
Given, That the current gain is 0.98
The change in emitter current is Ie=5mA
Therefore, the change in collector current is Ic=0.98×Ie=0.98×5=4.9mA
Therefore, the correct option is C.
The change in emitter current is Ie=5mA
Therefore, the change in collector current is Ic=0.98×Ie=0.98×5=4.9mA
Therefore, the correct option is C.
In a semiconductor, current conduction is due- a)Only to holes.
- b)Only to electrons
- c)To holes and electrons
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
In a semiconductor, current conduction is due
a)
Only to holes.
b)
Only to electrons
c)
To holes and electrons
d)
None of the above
|
|
Geetika Shah answered |
There are two types of semiconductors: p-type and n-type. Therefore both holes and electrons conduct current respectively.
A photodiode is used preferably in- a)to increase voltage
- b)in forward bias condition
- c)to convert electrical energy into light
- d)in reverse bias condition
Correct answer is option 'D'. Can you explain this answer?
A photodiode is used preferably in
a)
to increase voltage
b)
in forward bias condition
c)
to convert electrical energy into light
d)
in reverse bias condition
|
Rohit Shah answered |
In a photo-diode when light is incident, the fractional increase in the majority carriers is much less than the fractional increase in the minority carriers. Consequently, the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change due to the photo-effects on the majority carrier dominated forward bias current. Hence, photo-diodes are preferred to be used in the reverse bias condition to easily observe the variation of current with intensity.
The power amplification for a common emitter (CE) amplifier is 22110. Its current gain is 67 and internal resistance is 200 ohm. What is the value of the of the output resistance?- a)985 ohm
- b)66000 ohm
- c)41 ohm
- d)9125 ohm
Correct answer is option 'A'. Can you explain this answer?
The power amplification for a common emitter (CE) amplifier is 22110. Its current gain is 67 and internal resistance is 200 ohm. What is the value of the of the output resistance?
a)
985 ohm
b)
66000 ohm
c)
41 ohm
d)
9125 ohm
|
Prakhar Maheshwari answered |
The diode current depends on which of the following:- a)the diode voltage, temperature, and forward saturation current
- b)only diode voltage
- c)anode voltage
- d)the diode voltage, temperature, and reverse saturation current
Correct answer is option 'D'. Can you explain this answer?
The diode current depends on which of the following:
a)
the diode voltage, temperature, and forward saturation current
b)
only diode voltage
c)
anode voltage
d)
the diode voltage, temperature, and reverse saturation current
|
Nandita Ahuja answered |
The diode current depends on:
1. Diode Voltage:
The diode current is directly proportional to the diode voltage. When a diode is forward biased, meaning the positive terminal of the voltage source is connected to the anode and the negative terminal to the cathode, the diode allows current to flow through it. The magnitude of this current depends on the voltage applied across the diode. As the voltage increases, the diode current also increases.
2. Temperature:
The diode current is also affected by temperature. As the temperature increases, the diode current also increases. This is due to the fact that at higher temperatures, the mobility of charge carriers (electrons and holes) increases, leading to a higher current flow through the diode.
3. Reverse Saturation Current:
The reverse saturation current is the current that flows through a diode when it is reverse biased, meaning the positive terminal of the voltage source is connected to the cathode and the negative terminal to the anode. This current is typically very small, but it still has an impact on the overall diode current. The diode current is directly proportional to the reverse saturation current. As the reverse saturation current increases, the diode current also increases.
Conclusion:
In conclusion, the diode current depends on the diode voltage, temperature, and reverse saturation current. The diode voltage determines the magnitude of the current flow, the temperature affects the mobility of charge carriers, and the reverse saturation current influences the overall diode current. All three factors play a crucial role in determining the diode current and must be taken into consideration when analyzing or designing diode circuits.
1. Diode Voltage:
The diode current is directly proportional to the diode voltage. When a diode is forward biased, meaning the positive terminal of the voltage source is connected to the anode and the negative terminal to the cathode, the diode allows current to flow through it. The magnitude of this current depends on the voltage applied across the diode. As the voltage increases, the diode current also increases.
2. Temperature:
The diode current is also affected by temperature. As the temperature increases, the diode current also increases. This is due to the fact that at higher temperatures, the mobility of charge carriers (electrons and holes) increases, leading to a higher current flow through the diode.
3. Reverse Saturation Current:
The reverse saturation current is the current that flows through a diode when it is reverse biased, meaning the positive terminal of the voltage source is connected to the cathode and the negative terminal to the anode. This current is typically very small, but it still has an impact on the overall diode current. The diode current is directly proportional to the reverse saturation current. As the reverse saturation current increases, the diode current also increases.
Conclusion:
In conclusion, the diode current depends on the diode voltage, temperature, and reverse saturation current. The diode voltage determines the magnitude of the current flow, the temperature affects the mobility of charge carriers, and the reverse saturation current influences the overall diode current. All three factors play a crucial role in determining the diode current and must be taken into consideration when analyzing or designing diode circuits.
In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called- a)donor
- b)acceptor
- c)extrinsic semiconductor
- d)intrinsic semiconductor
Correct answer is option 'D'. Can you explain this answer?
In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called
a)
donor
b)
acceptor
c)
extrinsic semiconductor
d)
intrinsic semiconductor
|
Preeti Iyer answered |
In the case of an intrinsic semiconductor (say Si) where each Si is having 4 outermost electrons, its crystal structure consists of making 4 covalent bonds with 4 neighbouring Si atoms. Each bond consists of two electrons.
Now if one of the bonds gets broken due to some reason (collisions or high temperature) then one electron gets free and it will be having sufficient energy to cross the band gap and be ready for conduction- So in intrinsic semiconductors, current flows due to breakage of crystal bonds.
Now if one of the bonds gets broken due to some reason (collisions or high temperature) then one electron gets free and it will be having sufficient energy to cross the band gap and be ready for conduction- So in intrinsic semiconductors, current flows due to breakage of crystal bonds.
A common emitter (CE) amplifier has a current gain of 49. The voltage drop across the collector terminal is 1.8 V. The collector resistance is 2.2 kilo-ohm. What is the base current in the CE amplifier?- a)1.41 x 10-5 A
- b)1.81 x 10-5 A
- c)1.27 x 10-5 A
- d)1.67 x10-5 A
Correct answer is option 'D'. Can you explain this answer?
A common emitter (CE) amplifier has a current gain of 49. The voltage drop across the collector terminal is 1.8 V. The collector resistance is 2.2 kilo-ohm. What is the base current in the CE amplifier?
a)
1.41 x 10-5 A
b)
1.81 x 10-5 A
c)
1.27 x 10-5 A
d)
1.67 x10-5 A
|
Kiran Sanodiya answered |
The conductivity of P – type semiconductor is due to- a)both electrons and holes
- b)holes
- c)electrons
- d)none of the above.
Correct answer is option 'B'. Can you explain this answer?
The conductivity of P – type semiconductor is due to
a)
both electrons and holes
b)
holes
c)
electrons
d)
none of the above.
|
|
Lavanya Menon answered |
A P-type semiconductor is formed when a trivalent electron deficient impurities such as boron group elements are doped with intrinsic semiconductor. As the impurities are electron deficient, they take electrons from the valence band creating a number of holes. Due this reason conductivity in P-type semiconductor is mainly due to holes rather than electrons.
Which of the following is true for a forward-biased diode?- a)The anode is more positive with respect to the cathode.
- b)The anode is negative with respect to the cathode.
- c)The anode potential is equal to the cathode potential.
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Which of the following is true for a forward-biased diode?
a)
The anode is more positive with respect to the cathode.
b)
The anode is negative with respect to the cathode.
c)
The anode potential is equal to the cathode potential.
d)
None of the above
|
Vîshåkhâ Sïñgh answered |
There are two types of ions- anions are the - ve ones & cations are the +ve ones.As you know current flow from +ve to - ve, so During electrolysis or forward biasing anions move to anodes that are +ve in nature whereas cations move to cathodes which are -ve in nature.
you can remember it by a goes to a & c goes to c.
you can remember it by a goes to a & c goes to c.
An N-type Ge is obtained on doping the Ge- crystal with- a)boron
- b)gold
- c)aluminum
- d)phosphorus
Correct answer is option 'D'. Can you explain this answer?
An N-type Ge is obtained on doping the Ge- crystal with
a)
boron
b)
gold
c)
aluminum
d)
phosphorus
|
Ishani Patel answered |
Explanation:The addition of pentavalent impurities such as antimony, arsenic or phosphorous contributes free electrons, greatly increasing the conductivity of the intrinsic semiconductor.
A diode as a rectifier converts:- a)a.c. into d.c.
- b)varying d.c. current into constant direct current
- c)high voltage into low voltage and vice-versa
- d)d.c. into a.c.
Correct answer is option 'A'. Can you explain this answer?
A diode as a rectifier converts:
a)
a.c. into d.c.
b)
varying d.c. current into constant direct current
c)
high voltage into low voltage and vice-versa
d)
d.c. into a.c.
|
Sanjeevini Angadi answered |
Rectifier is a device which converts A.C into D.C.....therefore diode as a rectifier converts ac into dc
In an unbiased p-n junction, holes diffuse from the p-region to n-region because- a)Free electrons in the n-region attract them
- b)They move across the junction by the potential difference
- c)Hole concentrate in p-region is more as compared to n-region
- d)All the above
Correct answer is option 'C'. Can you explain this answer?
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
a)
Free electrons in the n-region attract them
b)
They move across the junction by the potential difference
c)
Hole concentrate in p-region is more as compared to n-region
d)
All the above
|
|
Vîshåkhâ Sïñgh answered |
There are two types of extrinsic semiconductor
p type - when intrinsic(pure form)semiconductor is doped with a trivalent (grp 13)element like B, Al,Ga,In, then its 3 valence electrons form covalent bonds with 3 Ge or Si atoms while the 4th valence electron of Ge or Si is not able to form the bond,thus there remain a hole.
n type - when Ge or Si crystal is doped with pentavalent (grp 15) like P,As,Sb,then it form covalent bond with all 4 electrons of Ge or Si atom and one valence electron is left,thus there is a free electron.
Since p type(+ve) has larger no. of holes (represent +ve)than no. of electrons and anything moves from its higher concentration to its lower concentration if no ext. force is applied, holes move from p to n region.
p type - when intrinsic(pure form)semiconductor is doped with a trivalent (grp 13)element like B, Al,Ga,In, then its 3 valence electrons form covalent bonds with 3 Ge or Si atoms while the 4th valence electron of Ge or Si is not able to form the bond,thus there remain a hole.
n type - when Ge or Si crystal is doped with pentavalent (grp 15) like P,As,Sb,then it form covalent bond with all 4 electrons of Ge or Si atom and one valence electron is left,thus there is a free electron.
Since p type(+ve) has larger no. of holes (represent +ve)than no. of electrons and anything moves from its higher concentration to its lower concentration if no ext. force is applied, holes move from p to n region.
Hole is- a)An anti-particle of electron
- b)An artifically created particle
- c)Absence of free electrons
- d)A vacancy created when an electron leaves a covalent bond
Correct answer is option 'D'. Can you explain this answer?
Hole is
a)
An anti-particle of electron
b)
An artifically created particle
c)
Absence of free electrons
d)
A vacancy created when an electron leaves a covalent bond
|
|
Zeal Gosai answered |
Actually hole is not anything we just asume that when an electron moves it creates a vacancy which we call a hole.
Transistors are useful in equipments such as hearing aids and small radio receivers because transistors :- a)do not distort sound
- b)Can stand rough use
- c)Consume little power
- d)Are inexpensive
Correct answer is option 'A'. Can you explain this answer?
Transistors are useful in equipments such as hearing aids and small radio receivers because transistors :
a)
do not distort sound
b)
Can stand rough use
c)
Consume little power
d)
Are inexpensive
|
Om Desai answered |
Transistors consume very less energy and are therefore used in these equipment because these equipment don't have space for big power source.
Why does current flow in the forward-bias direction but not flow in the reverse-bias direction in a pn junction?- a)The anode is negative.
- b)Minority carriers are pushed to the junction.
- c)Majority carriers are pushed to the junction.
- d)The intrinsic carriers are removed.
Correct answer is option 'C'. Can you explain this answer?
Why does current flow in the forward-bias direction but not flow in the reverse-bias direction in a pn junction?
a)
The anode is negative.
b)
Minority carriers are pushed to the junction.
c)
Majority carriers are pushed to the junction.
d)
The intrinsic carriers are removed.
|
Pooja Mehta answered |
The PN junction has the very useful property that electrons are only able to flow in one direction. As current consists of a flow of electrons, this means that current is allowed to flow only in one direction across the structure, but it is stopped from flowing in the other direction across the junction.
A solid having uppermost energy – band partially filled with electrons is called- a)none of the above
- b)a conductor
- c)a semi – conductor
- d)an insulator
Correct answer is option 'B'. Can you explain this answer?
A solid having uppermost energy – band partially filled with electrons is called
a)
none of the above
b)
a conductor
c)
a semi – conductor
d)
an insulator
|
|
Riya Banerjee answered |
Conductor is an object or type of material that allows the flow of an electrical current in one more directions. A metal wire is a common electrical conductor. In metals such as copper or aluminium, the mobile charged particles are welcome.
The efficiency of a full wave rectifier is- a)Double as that of a half wave rectifier
- b)same as that of a half wave rectifier
- c)half as that of a half wave rectifier
- d)One third as that of a half wave rectifier
Correct answer is option 'A'. Can you explain this answer?
The efficiency of a full wave rectifier is
a)
Double as that of a half wave rectifier
b)
same as that of a half wave rectifier
c)
half as that of a half wave rectifier
d)
One third as that of a half wave rectifier
|
|
Kiran Khanna answered |
Efficiency of a Full Wave Rectifier
The efficiency of a rectifier refers to how effectively it converts alternating current (AC) into direct current (DC). In the case of a full wave rectifier, it is known to have a higher efficiency compared to a half wave rectifier.
1. Half Wave Rectifier
- A half wave rectifier is a simple circuit that uses a single diode to convert AC to DC.
- It works by allowing only one half of the input AC waveform to pass through, while blocking the other half.
- The output waveform produced by a half wave rectifier is characterized by a series of positive half cycles, with the negative half cycles being eliminated.
- The rectified output waveform has a large amount of ripple and contains only half of the input power.
- Therefore, the efficiency of a half wave rectifier is relatively low.
2. Full Wave Rectifier
- A full wave rectifier is a more complex circuit that uses four diodes arranged in a bridge configuration.
- It works by allowing both halves of the input AC waveform to be rectified, resulting in a full wave rectified output waveform.
- The rectified output waveform has a smaller amount of ripple compared to a half wave rectifier, as it includes both positive and negative half cycles.
- The full wave rectifier utilizes the entire input power, resulting in a higher efficiency compared to a half wave rectifier.
3. Comparison of Efficiencies
- The efficiency of a rectifier can be defined as the ratio of the DC power output to the AC power input.
- For a half wave rectifier, the maximum efficiency is around 40.6%.
- In contrast, a full wave rectifier has a maximum efficiency of around 81.2%.
- Therefore, the efficiency of a full wave rectifier is double that of a half wave rectifier.
Conclusion
The efficiency of a full wave rectifier is double that of a half wave rectifier. This is because the full wave rectifier utilizes both halves of the input AC waveform, resulting in a higher power conversion efficiency.
The efficiency of a rectifier refers to how effectively it converts alternating current (AC) into direct current (DC). In the case of a full wave rectifier, it is known to have a higher efficiency compared to a half wave rectifier.
1. Half Wave Rectifier
- A half wave rectifier is a simple circuit that uses a single diode to convert AC to DC.
- It works by allowing only one half of the input AC waveform to pass through, while blocking the other half.
- The output waveform produced by a half wave rectifier is characterized by a series of positive half cycles, with the negative half cycles being eliminated.
- The rectified output waveform has a large amount of ripple and contains only half of the input power.
- Therefore, the efficiency of a half wave rectifier is relatively low.
2. Full Wave Rectifier
- A full wave rectifier is a more complex circuit that uses four diodes arranged in a bridge configuration.
- It works by allowing both halves of the input AC waveform to be rectified, resulting in a full wave rectified output waveform.
- The rectified output waveform has a smaller amount of ripple compared to a half wave rectifier, as it includes both positive and negative half cycles.
- The full wave rectifier utilizes the entire input power, resulting in a higher efficiency compared to a half wave rectifier.
3. Comparison of Efficiencies
- The efficiency of a rectifier can be defined as the ratio of the DC power output to the AC power input.
- For a half wave rectifier, the maximum efficiency is around 40.6%.
- In contrast, a full wave rectifier has a maximum efficiency of around 81.2%.
- Therefore, the efficiency of a full wave rectifier is double that of a half wave rectifier.
Conclusion
The efficiency of a full wave rectifier is double that of a half wave rectifier. This is because the full wave rectifier utilizes both halves of the input AC waveform, resulting in a higher power conversion efficiency.
Which of the statements is true for p-type semiconductors?- a)Electrons are minority carriers and pentavalent atoms are the dopants
- b)Electrons are majority carriers and trivalent atoms are the dopants
- c)Holes are minority carriers and pentavalent atoms are the dopants.
- d)Holes are majority carriers and trivalent atoms are the dopants
Correct answer is option 'D'. Can you explain this answer?
Which of the statements is true for p-type semiconductors?
a)
Electrons are minority carriers and pentavalent atoms are the dopants
b)
Electrons are majority carriers and trivalent atoms are the dopants
c)
Holes are minority carriers and pentavalent atoms are the dopants.
d)
Holes are majority carriers and trivalent atoms are the dopants
|
Sushil Kumar answered |
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.
In a p-n junction, as the diffusion process continues the width of the depletion zone- a)decreases
- b)increases
- c)remains the same
- d)oscillates
Correct answer is option 'B'. Can you explain this answer?
In a p-n junction, as the diffusion process continues the width of the depletion zone
a)
decreases
b)
increases
c)
remains the same
d)
oscillates
|
|
Preeti Iyer answered |
In reverse biasing, the positive terminal of the battery is connected to the n-type whereas the negative terminal is connected to the p-type junction. So the positive terminal tends to pull the electrons (near to the depletion layer) in n-type towards itself whereas the negative terminal pulls the holes towards itself which results in an increase in the width of the depletion layer.
An oscillator is nothing but an amplifier with:- a)positive feedback
- b)No feedback
- c)Negative feedback
- d)Reverse feedback
Correct answer is option 'A'. Can you explain this answer?
An oscillator is nothing but an amplifier with:
a)
positive feedback
b)
No feedback
c)
Negative feedback
d)
Reverse feedback
|
|
Rajat Kapoor answered |
An oscillator is an electronic circuit, which generates alternating voltage. In an oscillator the output power is returned back to the input, in phase with the starting power (i.e., as a positive feedback). Oscillator works on the principle of positive feedback.
What is the voltage gain in a common emitter, where input resistance is 3Ω and the load resistance is 24Ω, β = 0.6- a)2.4
- b)8.4
- c)4.8
- d)480
Correct answer is option 'C'. Can you explain this answer?
What is the voltage gain in a common emitter, where input resistance is 3Ω and the load resistance is 24Ω, β = 0.6
a)
2.4
b)
8.4
c)
4.8
d)
480
|
|
Lavanya Menon answered |
Voltage gain =β× (Rc/Rb)
Or, Voltage gain =0.6×(24/3)=4.8
Or, Voltage gain =0.6×(24/3)=4.8
The input voltage applied to a cascaded amplifier (consisting of two amplifiers) is 0.02 V and the output voltage is 6 V. The voltage gain of one of the two amplifiers is 10. What is the voltage gain of the other amplifier?
- a)1/30
- b)30
- c)3
- d)1/3
Correct answer is option 'B'. Can you explain this answer?
The input voltage applied to a cascaded amplifier (consisting of two amplifiers) is 0.02 V and the output voltage is 6 V. The voltage gain of one of the two amplifiers is 10. What is the voltage gain of the other amplifier?
a)
1/30
b)
30
c)
3
d)
1/3
|
|
Roshni Desai answered |
**Given Information:**
- Input voltage (Vin) = 0.02 V
- Output voltage (Vout) = 6 V
- Voltage gain of one amplifier (A1) = 10
**Calculating the Voltage Gain of the Cascaded Amplifier:**
1. The voltage gain of a cascaded amplifier is the product of the voltage gains of the individual amplifiers.
2. Let's assume the voltage gain of the other amplifier (A2) is 'x'.
3. The voltage gain of the cascaded amplifier (A_cascade) can be calculated as follows:
A_cascade = A1 * A2
A_cascade = 10 * x
4. We can calculate the voltage gain of the cascaded amplifier by using the input and output voltages:
A_cascade = Vout / Vin
A_cascade = 6 / 0.02
A_cascade = 300
5. Now we can equate the two expressions for the voltage gain of the cascaded amplifier:
10 * x = 300
6. Solving for 'x', we get:
x = 300 / 10
x = 30
**Conclusion:**
The voltage gain of the other amplifier (A2) is 30. Therefore, the correct answer is option 'B'.
- Input voltage (Vin) = 0.02 V
- Output voltage (Vout) = 6 V
- Voltage gain of one amplifier (A1) = 10
**Calculating the Voltage Gain of the Cascaded Amplifier:**
1. The voltage gain of a cascaded amplifier is the product of the voltage gains of the individual amplifiers.
2. Let's assume the voltage gain of the other amplifier (A2) is 'x'.
3. The voltage gain of the cascaded amplifier (A_cascade) can be calculated as follows:
A_cascade = A1 * A2
A_cascade = 10 * x
4. We can calculate the voltage gain of the cascaded amplifier by using the input and output voltages:
A_cascade = Vout / Vin
A_cascade = 6 / 0.02
A_cascade = 300
5. Now we can equate the two expressions for the voltage gain of the cascaded amplifier:
10 * x = 300
6. Solving for 'x', we get:
x = 300 / 10
x = 30
**Conclusion:**
The voltage gain of the other amplifier (A2) is 30. Therefore, the correct answer is option 'B'.
In an n- type silicondiode, which of the following statement is true :- a)Electrons are majority carriers and trivalent atoms are the dopants.
- b)Holes are majority carriers and trivalent atoms are the dopants
- c)Electrons are minority carriers and pentavalent atoms are the dopants.
- d)Holes are minority carriers and pentavalent atoms are the dopants.
Correct answer is option 'D'. Can you explain this answer?
In an n- type silicondiode, which of the following statement is true :
a)
Electrons are majority carriers and trivalent atoms are the dopants.
b)
Holes are majority carriers and trivalent atoms are the dopants
c)
Electrons are minority carriers and pentavalent atoms are the dopants.
d)
Holes are minority carriers and pentavalent atoms are the dopants.
|
|
Kulsoom Fatima answered |
Ye ese kaise ho skta hai.. pta nhi kyu ye hai ans
A p- type semiconductor can be obtained by adding- a)gallium to pure silicon
- b)phosphorus to pure germanium
- c)arsenic to pure silicon
- d)antimony to pure germanium
Correct answer is option 'A'. Can you explain this answer?
A p- type semiconductor can be obtained by adding
a)
gallium to pure silicon
b)
phosphorus to pure germanium
c)
arsenic to pure silicon
d)
antimony to pure germanium
|
Amrutha Chaudhary answered |
Understanding P-Type Semiconductors
P-type semiconductors are created by introducing certain impurities into intrinsic (pure) semiconductors. The type of impurity added determines the conductivity type of the semiconductor.
What is P-Type Semiconductor?
- A p-type semiconductor is characterized by an abundance of "holes" or positive charge carriers.
- These holes are created when trivalent impurities (elements with three valence electrons) are introduced into a pure semiconductor.
Why Gallium in Silicon?
- Gallium (Ga) is a trivalent element. When added to pure silicon, which has four valence electrons, gallium forms bonds with silicon atoms.
- Hole Creation: Gallium contributes only three electrons for bonding, leaving one silicon atom without a complete bond, creating a "hole."
- Majority Carriers: These holes act as positive charge carriers, making the semiconductor p-type.
Comparison with Other Options
- Phosphorus in Germanium: Phosphorus is a pentavalent element that adds extra electrons, leading to n-type conductivity, not p-type.
- Arsenic in Silicon: Similar to phosphorus, arsenic is also pentavalent, resulting in n-type conductivity.
- Antimony in Germanium: Antimony is another pentavalent element, which also leads to n-type conductivity.
Conclusion
- By understanding the role of impurities, we see that adding gallium to pure silicon effectively creates a p-type semiconductor. This process is crucial for various electronic applications, including diodes and transistors.
P-type semiconductors are created by introducing certain impurities into intrinsic (pure) semiconductors. The type of impurity added determines the conductivity type of the semiconductor.
What is P-Type Semiconductor?
- A p-type semiconductor is characterized by an abundance of "holes" or positive charge carriers.
- These holes are created when trivalent impurities (elements with three valence electrons) are introduced into a pure semiconductor.
Why Gallium in Silicon?
- Gallium (Ga) is a trivalent element. When added to pure silicon, which has four valence electrons, gallium forms bonds with silicon atoms.
- Hole Creation: Gallium contributes only three electrons for bonding, leaving one silicon atom without a complete bond, creating a "hole."
- Majority Carriers: These holes act as positive charge carriers, making the semiconductor p-type.
Comparison with Other Options
- Phosphorus in Germanium: Phosphorus is a pentavalent element that adds extra electrons, leading to n-type conductivity, not p-type.
- Arsenic in Silicon: Similar to phosphorus, arsenic is also pentavalent, resulting in n-type conductivity.
- Antimony in Germanium: Antimony is another pentavalent element, which also leads to n-type conductivity.
Conclusion
- By understanding the role of impurities, we see that adding gallium to pure silicon effectively creates a p-type semiconductor. This process is crucial for various electronic applications, including diodes and transistors.
How does the dynamic resistance of diode vary with temperature?- a)Directly proportional
- b)Inversely proportional
- c)Independent
- d)Directly to the square of temperature
Correct answer is option 'A'. Can you explain this answer?
How does the dynamic resistance of diode vary with temperature?
a)
Directly proportional
b)
Inversely proportional
c)
Independent
d)
Directly to the square of temperature
|
EduRev NEET answered |
The dynamic resistance can be defined from the I-V characteristic of a diode in forward bias. It is defined as the ratio of a small change to voltage to a small change in current, i.e.
VT = Thermal voltage
I = Bias current
∴ The dynamic resistance of the diode is directly proportional to the temperature.
The dynamic resistance is given by the inverse of the slope of i-v characteristics as shown:
VT = Thermal voltage
I = Bias current
∴ The dynamic resistance of the diode is directly proportional to the temperature.
The dynamic resistance is given by the inverse of the slope of i-v characteristics as shown:
A diode whose terminal characteristics are related as Id = Is(eV/ηVT - 1) is biased at Id = 2 mA. Its dynamic resistance is:
(Given η = 2 and VT = 25 mV)- a)25 Ω
- b)12.5 Ω
- c)50 Ω
- d)22.5 Ω
Correct answer is option 'A'. Can you explain this answer?
A diode whose terminal characteristics are related as Id = Is(eV/ηVT - 1) is biased at Id = 2 mA. Its dynamic resistance is:
(Given η = 2 and VT = 25 mV)
(Given η = 2 and VT = 25 mV)
a)
25 Ω
b)
12.5 Ω
c)
50 Ω
d)
22.5 Ω
|
|
Ambition Institute answered |
Concept:
- The dynamic resistance can be defined from the I-V characteristic of a diode in forward biased
- It is defined as the ratio of small change to voltage to a small change in current,
- It is the inverse of the slope of I-V characteristics curve
The dynamic resistance is given by the inverse of the slope of i-v characteristics
Calculation:
Given that, current (I) = 2 mA
We know that voltage (VT) = 25 mV
Dynamic resistance
Given that, current (I) = 2 mA
We know that voltage (VT) = 25 mV
Dynamic resistance
The output from a full wave rectifier is- a)a pulsating unidirectional voltage
- b)a pulsating dc voltage
- c)zero
- d)none
Correct answer is option 'A'. Can you explain this answer?
The output from a full wave rectifier is
a)
a pulsating unidirectional voltage
b)
a pulsating dc voltage
c)
zero
d)
none
|
Ciel Knowledge answered |
The output from a full wave rectifier is a pulsating unidirectional voltage. This is because while the rectifier converts AC to DC, the output is not a pure DC voltage; it contains fluctuations due to the rectification process. These fluctuations make the output a pulsating DC, which is unidirectional. To achieve a steady DC output, additional filtering components like capacitors are often used to smooth out the voltage.
In the case of metals the valence and conduction bands have- a)no overlap, energy gap is large
- b)no overlap, energy gap is small
- c)overlap, energy gap =0
- d)no overlap, energy gap =0
Correct answer is option 'C'. Can you explain this answer?
In the case of metals the valence and conduction bands have
a)
no overlap, energy gap is large
b)
no overlap, energy gap is small
c)
overlap, energy gap =0
d)
no overlap, energy gap =0
|
Anoushka Chopra answered |
Explanation:The materials can be classified by the energy gap between their valence band and the conduction band. The valence band is the band consisting of the valence electron, and the conduction band remains empty. Conduction takes place when an electron jumps from valence band to conduction band and the gap between these two bands is forbidden energy gap. Wider the gap between the valence and conduction bands, higher the energy it requires for shifting an electron from valence band to the conduction band.In the case of conductors, this energy gap is absent or in other words conduction band, and valence band overlaps each other. Thus, electron requires minimum energy to jump from valence band. The typical examples of conductors are Silver, Copper, and Aluminium.In insulators, this gap is vast. Therefore, it requires a significant amount of energy to shift an electron from valence to conduction band. Thus, insulators are poor conductors of electricity. Mica and Ceramic are the well-known examples of insulation material.Semiconductors, on the other hand, have an energy gap which is in between that of conductors and insulators. This gap is typically more or less 1 eV, and thus, one electron requires energy more than conductors but less than insulators for shifting valence band to conduction band.
Electrons are forbidden in a band (in a crystal) called the- a)valence band
- b)forbidden band
- c)conduction band
- d)memory band
Correct answer is option 'B'. Can you explain this answer?
Electrons are forbidden in a band (in a crystal) called the
a)
valence band
b)
forbidden band
c)
conduction band
d)
memory band
|
Sinjini Tiwari answered |
Explanation:A region of values of energy that electrons in an ideal crystal (without defects) cannot have. In semiconductors the forbidden band separating the valence band and the conduction band is usually considered. In this case the energy difference between the lower level (bottom) of the conduction band and the upper level (ceiling) of the valence band is called the width of the forbidden band.
The conductivity of a photosensitive semiconductor- a)increases as the number of electrons decreases
- b)does not depend on light and number of electrons
- c)increases with light
- d)decreases with light
Correct answer is option 'C'. Can you explain this answer?
The conductivity of a photosensitive semiconductor
a)
increases as the number of electrons decreases
b)
does not depend on light and number of electrons
c)
increases with light
d)
decreases with light
|
|
Priya Patel answered |
Photoconductivity, as a well-known optical and electrical phenomenon in semiconductor, is an effect that the electrical conductivity increases due to the absorption of light radiation (Bube, 1960, 1992; Rose, 1963).
The main difference between conductors, semiconductors and insulators is because of- a)Work function
- b)Mobility of electrons
- c)Energy of electrons
- d)Width of forbidden energy gap
Correct answer is option 'D'. Can you explain this answer?
The main difference between conductors, semiconductors and insulators is because of
a)
Work function
b)
Mobility of electrons
c)
Energy of electrons
d)
Width of forbidden energy gap
|
Niti Saha answered |
Explanation:Forbidden gap plays a major role for determining the electrical conductivity of material. Based on the forbidden gap materials are classified in to three types, they are : Insulators : The forbidden gap between the valence band and conduction band is very large in insulators. The energy gap of insulator is approximately equal to 15 electron volts (eV).Conductors: In a conductor, valence band and conduction band overlap each other. Therefore, there is no forbidden gap in a conductor.Semiconductors: In semiconductors, the forbidden gap between valence band and conduction band is very small. It has a forbidden gap of about 1 electron volt (eV).
The depletion layer in the p-n junction is caused- a)drift of electrons
- b)drift of holes
- c)migration of impurity ions
- d)diffusion of carrier ions
Correct answer is option 'D'. Can you explain this answer?
The depletion layer in the p-n junction is caused
a)
drift of electrons
b)
drift of holes
c)
migration of impurity ions
d)
diffusion of carrier ions
|
Aaditya Ghoshal answered |
Depletion region or depletion layer is a region in a P-N junction diode where no mobile charge carriers are present. Depletion layer acts like a barrier that opposes the flow of electrons from n-side and holes from p-side.
Majority current carriers in N – types are- a)holes
- b)negative ions
- c)positive ions
- d)electrons
Correct answer is option 'D'. Can you explain this answer?
Majority current carriers in N – types are
a)
holes
b)
negative ions
c)
positive ions
d)
electrons
|
Yash Kumar answered |
The majority current carriers in an N-type semiconductor are electrons.
Explanation:
N-type semiconductor:
An N-type semiconductor is formed by doping a pure semiconductor material (such as silicon or germanium) with a pentavalent impurity, which introduces extra electrons into the crystal lattice. The most commonly used impurity for N-type doping is phosphorus (P), which has five valence electrons.
Electron behavior in an N-type semiconductor:
1. Extra electrons: When the pentavalent impurity is added to the pure semiconductor material, the impurity atoms replace some of the original semiconductor atoms in the crystal lattice. Each impurity atom contributes an extra electron to the crystal structure, as there is one more valence electron than required for covalent bonding.
2. Majority carriers: These extra electrons become the majority current carriers in the N-type semiconductor. They are free to move within the crystal lattice, contributing to the flow of electric current.
3. Energy levels: The extra electrons occupy energy levels closer to the conduction band, which is the energy band where electrons can move freely. This is because the energy levels of impurity atoms are closer to the conduction band than the valence band.
4. Minority carriers: The majority carriers (extra electrons) are accompanied by a small number of holes, which are the minority carriers. Holes are created when electrons from the valence band move to occupy the vacancies left by the extra electrons in the impurity atoms.
5. Electron mobility: Electrons in the N-type semiconductor have higher mobility compared to holes. This is because electrons are lighter and have a negative charge, allowing them to move more easily in response to an electric field.
Conclusion:
In an N-type semiconductor, the majority current carriers are electrons. They are introduced into the crystal lattice by doping the semiconductor material with a pentavalent impurity. These extra electrons become free to move and contribute to the flow of electric current.
Explanation:
N-type semiconductor:
An N-type semiconductor is formed by doping a pure semiconductor material (such as silicon or germanium) with a pentavalent impurity, which introduces extra electrons into the crystal lattice. The most commonly used impurity for N-type doping is phosphorus (P), which has five valence electrons.
Electron behavior in an N-type semiconductor:
1. Extra electrons: When the pentavalent impurity is added to the pure semiconductor material, the impurity atoms replace some of the original semiconductor atoms in the crystal lattice. Each impurity atom contributes an extra electron to the crystal structure, as there is one more valence electron than required for covalent bonding.
2. Majority carriers: These extra electrons become the majority current carriers in the N-type semiconductor. They are free to move within the crystal lattice, contributing to the flow of electric current.
3. Energy levels: The extra electrons occupy energy levels closer to the conduction band, which is the energy band where electrons can move freely. This is because the energy levels of impurity atoms are closer to the conduction band than the valence band.
4. Minority carriers: The majority carriers (extra electrons) are accompanied by a small number of holes, which are the minority carriers. Holes are created when electrons from the valence band move to occupy the vacancies left by the extra electrons in the impurity atoms.
5. Electron mobility: Electrons in the N-type semiconductor have higher mobility compared to holes. This is because electrons are lighter and have a negative charge, allowing them to move more easily in response to an electric field.
Conclusion:
In an N-type semiconductor, the majority current carriers are electrons. They are introduced into the crystal lattice by doping the semiconductor material with a pentavalent impurity. These extra electrons become free to move and contribute to the flow of electric current.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of- a)copper decreases and that of germanium increases
- b)each of them increases
- c)each of them decreases
- d)copper increases and that of germanium decreases
Correct answer is option 'A'. Can you explain this answer?
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
a)
copper decreases and that of germanium increases
b)
each of them increases
c)
each of them decreases
d)
copper increases and that of germanium decreases
|
Geetika Tiwari answered |
We know, a piece of copper is a metal while that of germanium is a semiconducting material. For metals resistance increases with increase in temperature. The semiconductor has a negative temperature coefficient of resistance. Hence, when it is cooled its resistance increases.
In an N-P-N transistor, P-type crystal is- a)collector
- b)base
- c)grid
- d)emitter
Correct answer is option 'B'. Can you explain this answer?
In an N-P-N transistor, P-type crystal is
a)
collector
b)
base
c)
grid
d)
emitter
|
Gauri Khanna answered |
Explanation:When the p-type crystal is grown between relatively wide sections of n-type crystals then the transistor is called NPN transistor.
In a pure, or intrinsic, semiconductor, valence band holes and conduction-band electrons are always present- a)such that number of holes is greater than the number of electrons
- b)in equal numbers
- c)such that number of electrons is greater than the number of holes
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
In a pure, or intrinsic, semiconductor, valence band holes and conduction-band electrons are always present
a)
such that number of holes is greater than the number of electrons
b)
in equal numbers
c)
such that number of electrons is greater than the number of holes
d)
none of these
|
|
Sinjini Tiwari answered |
Explanation:An intrinsic semiconductor, also called an undoped semiconductor or i-type semiconductor, is a pure semiconductor without any significant dopant species present. The number of charge carriers is therefore determined by the properties of the material itself instead of the amount of impurities. In intrinsic semiconductors the number of excited electrons and the number of holes are equal: n = p.
The number of valence electrons in a good conductor is generally- a)six or more than six
- b)four
- c)five
- d)three or less than three
Correct answer is option 'D'. Can you explain this answer?
The number of valence electrons in a good conductor is generally
a)
six or more than six
b)
four
c)
five
d)
three or less than three
|
Rishika Chauhan answered |
Explanation:The electron theory states that all matter is composed of atoms and the atoms are composed of smaller particles called protons, electrons, and neutrons. The electrons orbit the nucleus which contains the protons and neutrons. It is the valence electrons that we are most concerned with in electricity. These are the electrons which are easiest to break loose from their parent atom. Normally, conductors have three or less valence electrons; insulators have five or more valence electrons; and semiconductors usually have four valence electrons.
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