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A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsic
carrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming complete
impurity ionization, the equilibrium electron and hole concentrations are
carrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming complete
impurity ionization, the equilibrium electron and hole concentrations are
- a)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3
- b)n0 = 1.5 x 1010 cm-3, p0 = 1.5 x 1015 cm-3
- c)n0 = 2.25 x 1015 cm-3, p0 = 1.5 x 1010 cm-3
- d)n0 = 2.25 x 1015 cm-3, p0 = 1.5 x 105 cm-3
Correct answer is option 'D'. Can you explain this answer?
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A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / ...
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A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / ...
Given parameters:
ND = 2.25 x 1015 atoms / cm3
ni = 1.5 x 1010 cm-3
T = 300 K
Equilibrium electron and hole concentrations can be calculated using the following formulae:
n0 = ND
p0 = ni^2 / ND
Calculation:
n0 = 2.25 x 1015 cm-3
p0 = (1.5 x 1010 cm-3)^2 / 2.25 x 1015 cm-3 = 1.5 x 105 cm-3
Therefore, the correct option is 'D' - n0 = 2.25 x 1015 cm-3, p0 = 1.5 x 105 cm-3.
Explanation:
1. Donor impurities increase the electron concentration in the silicon bar.
2. At equilibrium, the electron concentration is equal to the donor concentration as all the donors are ionized.
3. The hole concentration can be calculated using the law of mass action, which states that the product of electron and hole concentrations at equilibrium is a constant, equal to the square of the intrinsic carrier concentration.
4. In this case, the hole concentration is much smaller than the electron concentration as the donor concentration is much larger than the intrinsic carrier concentration.
5. Therefore, the correct option is 'D', where the electron concentration is equal to the donor concentration and the hole concentration is calculated using the law of mass action.
ND = 2.25 x 1015 atoms / cm3
ni = 1.5 x 1010 cm-3
T = 300 K
Equilibrium electron and hole concentrations can be calculated using the following formulae:
n0 = ND
p0 = ni^2 / ND
Calculation:
n0 = 2.25 x 1015 cm-3
p0 = (1.5 x 1010 cm-3)^2 / 2.25 x 1015 cm-3 = 1.5 x 105 cm-3
Therefore, the correct option is 'D' - n0 = 2.25 x 1015 cm-3, p0 = 1.5 x 105 cm-3.
Explanation:
1. Donor impurities increase the electron concentration in the silicon bar.
2. At equilibrium, the electron concentration is equal to the donor concentration as all the donors are ionized.
3. The hole concentration can be calculated using the law of mass action, which states that the product of electron and hole concentrations at equilibrium is a constant, equal to the square of the intrinsic carrier concentration.
4. In this case, the hole concentration is much smaller than the electron concentration as the donor concentration is much larger than the intrinsic carrier concentration.
5. Therefore, the correct option is 'D', where the electron concentration is equal to the donor concentration and the hole concentration is calculated using the law of mass action.
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A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer?
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A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer?.
A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer?.
Solutions for A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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Here you can find the meaning of A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A silicon bar is doped with donor impurities ND = 2.25 x 1015 atoms / cm3. Given the intrinsiccarrier concentration of silicon at T = 300 K is ni = 1.5 x 1010 cm-3. Assuming completeimpurity ionization, the equilibrium electron and hole concentrations area)n0 = 1.5 x 1016 cm-3, p0 = 1.5 x 105 cm-3b)n0 = 1.5 x 1010cm-3, p0 = 1.5 x 1015 cm-3c)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 1010 cm-3d)n0 = 2.25 x 1015cm-3, p0 = 1.5 x 105 cm-3Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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