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Problem Set # 3 Solution
ECEN 3320 Fall 2013
Semiconductor Devices
September 9, 2013 โ Due September 18, 2013
- A semiconductor has Eg = 1.40 eV and mโ h = 0. 5 m 0 ; mโ e = 0.1m 0 at T = 300โฆK.
(a) Calculate the position of the intrinsic Fermi level, Ef , with respect to the middle of the band gap, Ei โ Emidgap = Ei โ Eg/2. Solution: The intrinsic Fermi level is defined in terms of the bandgap, temperature and effective carried masses as
Ei โ Ec โ Ev 2
kT ln
( mโ h mโ e
) .
Using the given values, we find that the distance of the Fermi level from the center of the gap is
Ei โ Ec โ Ev 2
0 .0258 ln(5) = 0. 750. 02581. 61 โ 0 .0313eV
(b) Calculate the intrinsic carrier concentration, ni. Solution: The intrinsic carrier concentration in terms of the effective densities of states, Nc and Nv and the bandgap energy as
ni =
โ NcNv exp [โEg/kT ].
Recalling that the effecive densities of states are given by
Nc = 2
( mโ ekT 2 ฯยฏh^2
) 3 / 2
Nv = 2
( mโ hkT 2 ฯยฏh^2
) 3 / 2
we find that for a bare electron
N 0 = 2
( m 0 kT 2 ฯยฏh^2
) 3 / 2 = 2. 61 ร 1019 cmโ^3.
Using the above, we find that
ni โ (0.05)^3 /^2 2. 61 ร 1019 ร exp(โ 0. 70 / 0 .0258) โ 4. 8 ร 106 cmโ^3.
(c) Impurity atoms are added so that the Fermi level, Ef , is 0.35 eV above the middle of the band gap. i. Are the impurities donors or acceptors? Solution: The impurities are donors as donors raise the Fermi level. The intrinsic Fermi level is raised by only 0.03 eV relative to the center of gap. The โmeasuredโ 0.35 eV is much larger than this indicating the semiconductor is an extrinsic n- doped semiconductor. ii. What was the concentration of impurities added? Solution: We need to make an additional assumption here, the assumption that the donor impurities are shallow. It is only the shallow impurities (within a few kT of the band edge) that will contribute to the repositioning of the Fermi level relative to the intrinsic Fermi level.
ND = n 0 = ni exp
[ Ef โ Ei kT
]
that evaluates to
ND โ 1. 07 ร 1012 cmโ^3.
- Determine the equilibrium electron and hole concentrations in silicon for the following con- ditions.
Solution: Using the fact that there is no free charge density in an equilibrium material, we can write that
n 0 โ p 0 + N (^) aโ โ N (^) d+ = 0.
The law of mass action n 0 p 0 = n^2 i combined with the above gives that
n^20 + (N (^) a+ โ N (^) dโ )n 0 โ n^2 i = 0 p^20 + (N (^) dโ โ N (^) a+ )p 0 โ n^2 i = 0
with solutions (omitting the spurious one generated by multiplying through by n 0 (p 0 )), we find
n 0 =
( N (^) dโ โ N (^) a+ 2
)
โโ โโ ( N (^) dโ โ N (^) a+ 2
) 2
p 0 =
( N (^) a+ โ N (^) dโ 2
)
โโ โโ ( N (^) a+ โ N (^) dโ 2
) 2
These relations can generally be simplified significantly. For an intrinsic semiconductor, N (^) d+ ni and N (^) aโ ni and, therefore, n 0 โ ni + (N (^) d+ โ N (^) aโ ) and p 0 โ ni โ (N (^) d+ โ N (^) aโ ). For an extrinsic semiconductor, for example, an n-type, ni N (^) d+ and n 0 โ N (^) d+ with p 0 = n 0 /n^2 i. In our present case, the intrinsic carrier concentration is given by
ni =
โ NcNv exp
[ โ ฮตg 2 kT
] โ 1 ร 1010 cmโ^3.
- (a) If Ec โ Ef = 0.2eV in GaAs at T = 500โฆK, calculate the values of the equilibrium carrier concentrations, n and p. Solution: The equilibrium carrier concentrations are given by
n = Nc(T = 500โฆC) exp
[ โ
Ec โ Ef kT |T =500โฆC
] = 9. 05 ร 1015 cmโ^3.
The band gap of GaAs at room temperature is roughly 1.424 eV. As EcโEf = 0.2eV, then Ef โ Ev = 1.224eV and we can evaluate
p = Nv exp
[ โ Ec โ Ef kT
] = 7. 56 ร 106 cmโ^3.
(b) Assuming that the value of n you obtained in (a) remains constant, calculate Ec โ Ef and p at T = 300โฆK. Solution: We have
Ec โ Ef = โkT ln
( n Nc
) .
Plugging in
Ec โ Ef = โkT ln
( n = 9. 05 ร 1015 Nc(500โฆC)
) โ 0 .1eV.
The value of Ef โ Ev should now be circa 1.324 eV. Evaluating
p = Nv exp
( โ
Ef โ Ev kT
) โ 4. 40 ร 10 โ^4 cmโ^3.
- A certain semiconductor is doped with Nd = 2 ร 1013 cmโ^3 and NA = 0. The intrinisic carrier concentration for this semiconductor is ni = 2 ร 1013 cmโ^3.
(a) Determine the majority and minority carrier concentrations at thermal equilibrium. Solution: Here, we have that N (^) d+ โ ni so we must use the general solution for concen- tration in terms of doping to find the effective concentrations. Writing
n 0 =
N (^) d+ โ N (^) a+ 2
โโ โโ(^ N + d โ^ N^ a+ 2
) 2
- n^2 i โ 3 ร 1013 cmโ^3.
(b) Determine the position of the energy level relative to the intrinsic Fermi level, Ef โ Ei. Solution: We have that
Ef โ Ei = kT ln
( n ni
) โ 0 .01eV
