The ionised concentration for donors is given by :
\(\dfrac{N_D}{1+2\exp\left[(E_F-E_D)/KT\right]}\)
What is being asked. A donor atom sitting at energy ED is ionised when it has given its electron to the conduction band. The question wants the density of such ionised donors, \(N_D^{+}\), out of a total donor density ND.
Step 1 — the occupied donors follow a modified Fermi function. The fraction of donor sites still holding their electron is not the plain Fermi–Dirac function, because a donor level can accept an electron of either spin but cannot hold two. That gives a degeneracy factor g = 2:
\(\dfrac{n_D}{N_D}=\dfrac{1}{1+\dfrac{1}{g}\exp\left[\dfrac{E_D-E_F}{kT}\right]}=\dfrac{1}{1+\dfrac{1}{2}\exp\left[\dfrac{E_D-E_F}{kT}\right]}\)
Step 2 — ionised means unoccupied. The donors that have lost the electron are the rest of them:
\(N_D^{+}=N_D-n_D=N_D\left(1-\dfrac{1}{1+\tfrac{1}{2}e^{(E_D-E_F)/kT}}\right)\)
Step 3 — simplify. Combining over a common denominator and inverting the exponent's sign,
\(N_D^{+}=\dfrac{N_D}{1+2\exp\left[\dfrac{E_F-E_D}{kT}\right]}\)
which is exactly option 1.
Two checks that pin down the answer without the algebra.
The sign in the exponent. When the Fermi level lies far below the donor level, EF − ED is large and negative, the exponential vanishes, and \(N_D^{+}\to N_D\) — complete ionisation, which is the normal room-temperature situation for silicon. When EF rises above ED (heavy doping or low temperature) the exponential blows up and \(N_D^{+}\to 0\), the freeze-out regime. Both limits behave correctly.
The denominator must never vanish. Options 2 and 4 carry a minus sign, so for some value of EF the denominator would pass through zero and the ionised density would become infinite and then negative — physically impossible. That alone eliminates half the choices.
The factor of 2. Option 3 forgets the spin degeneracy. The factor 2 is the standard result for a donor state; for acceptors the corresponding degeneracy factor is 4.
Hence, the ionised donor concentration is \(\dfrac{N_D}{1+2\exp\left[(E_F-E_D)/KT\right]}\).
Similar Questions
Consider the following statements :
If an electric field is applied to an n-type semiconductor bar, the electrons and holes move in opposite directions due to their opposite charges. The net current is
1. due to both electrons and holes with electrons as majority carriers.
2. the sum of electron and hole currents.
3. the difference between electron and hole current.Which of these statements is/are correct ?
Consider the following :
1. In semiconductors, the mobility of electrons is more than that of holes.
2. In semiconductors, the resistivity increases with the increase in temperature.
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4. In metals, thermal conductivity is inversely proportional to electronic conductivity at a particular temperature.Which of the following statements are correct ?
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(A) N = 1015 cm–3
(B) N = 1017 cm–3
(C) N = 3 × 1015 cm–3
(D) N = 1016 cm–3
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Match the following :
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A conducting line on an IC chip is 2.8 mm long and has a rectangular cross-section of 1 μm x 4 μm.. A current of 5 mA produces a voltage drop of 100 mV across the line. If the electron mobility is 500 cm2/V-s, the electron concentration is
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(a) \(ni ^ {2}\) (b) \(N_cN_v\exp\left(\dfrac{Eg}{kT}\right)\) (c) \(N_cN_v\exp\left(-\dfrac{Eg}{kT}\right)\) (d) \(N_v\exp\left(-\dfrac{Eg}{kT}\right)\)
Which of the above are correct ?
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Mobility and conductivity are related by which of the following equations?
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