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A silicon bar is doped with donor impurities ND = 2.25 × 1015 cm-3
If the electron mobility μn = 1000 cm2/v-s then the approximate value of resistivity of silicon bar assuming partial ionization of 55% is __________ (Ω - cm)
- a)55 Ω – cm
- b)25 Ω – cm
- c)34 Ω – cm
- d)5 Ω – cm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A silicon bar is doped with donor impurities ND= 2.25 × 1015cm-3...
Given ND = 2.25 × 1015 cm-3
Partial ionization of 55%, number of electrons is
⇒ n = 0.55 × 2.25 × 1015 cm-3
= 1.2375 × 1015 cm-3
Electron mobility μn = 1000 cm2/v-s
Resistivity of n-type silicon = 
= 5.05 Ω – cm
= 5.05 Ω – cm
Most Upvoted Answer
A silicon bar is doped with donor impurities ND= 2.25 × 1015cm-3...
X 10^15 cm^-3. The intrinsic carrier concentration of silicon at room temperature is ni = 1.5 x 10^10 cm^-3. Determine the majority and minority carrier concentrations in the silicon bar.
To determine the majority and minority carrier concentrations in the silicon bar, we need to consider the donor impurities and the intrinsic carrier concentration.
The majority carriers in an n-type semiconductor (such as silicon doped with donor impurities) are the electrons provided by the donor impurities. In this case, the donor impurity concentration is given as ND = 2.25 x 10^15 cm^-3. Therefore, the majority carrier concentration is equal to ND.
Majority carrier concentration (n) = ND = 2.25 x 10^15 cm^-3
The minority carriers in an n-type semiconductor are the holes, which are generated due to thermal excitation. The intrinsic carrier concentration of silicon at room temperature is given as ni = 1.5 x 10^10 cm^-3. Since silicon is an n-type semiconductor, the majority carriers (electrons) are much greater in concentration than the minority carriers (holes).
Minority carrier concentration (p) = ni^2 / ND
Substituting the given values:
Minority carrier concentration (p) = (1.5 x 10^10 cm^-3)^2 / 2.25 x 10^15 cm^-3
Simplifying:
Minority carrier concentration (p) = 2.25 x 10^20 cm^-3 / 2.25 x 10^15 cm^-3
Minority carrier concentration (p) = 10^5 cm^-3
Therefore, the majority carrier concentration in the silicon bar is 2.25 x 10^15 cm^-3 (electrons), and the minority carrier concentration is 10^5 cm^-3 (holes).
To determine the majority and minority carrier concentrations in the silicon bar, we need to consider the donor impurities and the intrinsic carrier concentration.
The majority carriers in an n-type semiconductor (such as silicon doped with donor impurities) are the electrons provided by the donor impurities. In this case, the donor impurity concentration is given as ND = 2.25 x 10^15 cm^-3. Therefore, the majority carrier concentration is equal to ND.
Majority carrier concentration (n) = ND = 2.25 x 10^15 cm^-3
The minority carriers in an n-type semiconductor are the holes, which are generated due to thermal excitation. The intrinsic carrier concentration of silicon at room temperature is given as ni = 1.5 x 10^10 cm^-3. Since silicon is an n-type semiconductor, the majority carriers (electrons) are much greater in concentration than the minority carriers (holes).
Minority carrier concentration (p) = ni^2 / ND
Substituting the given values:
Minority carrier concentration (p) = (1.5 x 10^10 cm^-3)^2 / 2.25 x 10^15 cm^-3
Simplifying:
Minority carrier concentration (p) = 2.25 x 10^20 cm^-3 / 2.25 x 10^15 cm^-3
Minority carrier concentration (p) = 10^5 cm^-3
Therefore, the majority carrier concentration in the silicon bar is 2.25 x 10^15 cm^-3 (electrons), and the minority carrier concentration is 10^5 cm^-3 (holes).
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A silicon bar is doped with donor impurities ND= 2.25 × 1015cm-3If the electron mobility μn= 1000 cm2/v-s then the approximate value of resistivity of silicon bar assuming partial ionization of 55% is __________ (Ω - cm)a)55Ω – cmb)25Ω – cmc)34Ω – cmd)5Ω – cmCorrect answer is option 'D'. Can you explain this answer?
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