Semiconductor EquilibriumEquilibrium
No external forces (voltages, electric fields, temp.gradients)
FirstConsider pure crystal
ThenConsider addition of dopants
Semiconductor EquilibriumCharge carriers
Electrons in conductancen(E) = gc(E)fF(E)n(E) - prob. dens. of electronsgc(E) - conductance densityfF(E) - Fermi-Dirac prob. function
Holes in valencep(E) = gV(E)(1 - fF(E))p(E) - prob. dens. of holesgv(E) - valence densityfF(E) - Fermi-Dirac prob. function
Semiconductor EquilibriumCharge carriers(cont.)
€
n0 = gc∫ (E) fF (E)dE
€
n0 = Nc exp−(EC − EF )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥
€
NC = 22πmn
*kT
h2
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
€
p0 = gv∫ (E)(1− fF (E))dE
€
Nv = 22πmp
*kT
h2
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
€
p0 = Nv exp−(EF − Ev )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥
Semiconductor EquilibriumCharge carriers(cont.)Example
Find the probability that a state in the conduction band is occupiedand calculate the electron concentration in silicon at T = 300K. Assume Fermi energy is .25 eV below the conductance band
Note low probability per state but large number of states implies reasonable concentration of electrons
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fF (E) ≈ exp−(Ec − EF )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= exp(−0.25 /.0259) = 6.43⋅10−5
€
n0 = Nc (E)exp−(Ec − EF )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= (2.8 ⋅1019)(6.43⋅10−5) =1.8 ⋅1015cm−3
Semiconductor EquilibriumCharge carriers(cont.)
For intrinsic semiconductor, concentration of electrons in conductance band is equal to holes in the valence band. Thus,
€
ni2 = nipi = NcNv exp
−(Ec − Ev )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= NcNv exp
−EgkT
⎡
⎣ ⎢
⎤
⎦ ⎥
Semiconductor EquilibriumDopant Atoms (n-type semiconductor)
Phosphorous has 5 valence electrons
Energy-band diagram
Semiconductor EquilibriumDopant Atoms (p-type semiconductor)
Boron has 3 valence electrons
Energy-band diagram
Semiconductor EquilibriumThe Extrinsic Semiconductor
n-type p-type
Semiconductor EquilibriumThe Extrinsic Semiconductor
ExampleConsider doped silicon at 300K. Assume that the Fermi enery is .25 eV below the conduction band and .87 eV above the valence band. Calculate the thermal equilibrium concentration of e’s and holes
€
n0 = Nc (E)exp−(Ec − EF )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= (2.8 ⋅1019)exp(−0.025 /0.0259) =1.8 ⋅1015cm−3
€
p0 = Nv (E)exp−(EF − Ev )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= (1.04 ⋅1019)exp(−0.087 /0.0259) = 2.7 ⋅104cm−3
Semiconductor EquilibriumThe Extrinsic Semiconductor
The n0p0 product
That is, the product of n0 and p0 is a constant for a given semiconductor at a given temperature.€
n0p0 = NcNv exp−(Ec − EF )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥exp
−(EF − EV )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥= NcNv exp
−EgkT
⎡
⎣ ⎢
⎤
⎦ ⎥= ni
2
Semiconductor EquilibriumStatistics of donors and acceptors
Ratio of electrons in donor state total electrons
ExampleConsider phosporous doped silicon at T = 300K and at a concentration of Nd = 1016 cm-3. Find the fraction of electrons in the donor state.
€
ndno + nd
=1
1+Nc
2Ndexp
−(Ec − Ed )
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥
€
ndno + nd
=1
1+2.8 ⋅1019
2 ⋅1016exp
−0.045
0.0259
⎡ ⎣ ⎢
⎤ ⎦ ⎥
= .41%
Semiconductor EquilibriumCompensated semiconductors
Formed by adding both donor and acceptor impuritiesin the same region
Energy-band diagram
Semiconductor EquilibriumCompensated semiconductors (cont.)
With the assumption of charge neutrality, we can derive
ExampleConsider a silicon semiconductor at T = 300K in which Na = 1016 cm-3 and Nd = 3 1015 cm-3. Assume ni = 1.5 1010 cm-3 and find p0 and n0.
€
n0 =(Nd −Na )
2+
Nd −Na2
⎛
⎝ ⎜
⎞
⎠ ⎟2
+ ni2
€
p0 =(Na −Nd )
2+
Na −Nd2
⎛
⎝ ⎜
⎞
⎠ ⎟2
+ ni2
€
p0 =(1016 − 3⋅1015)
2+
1016 − 3⋅1015
2
⎛
⎝ ⎜
⎞
⎠ ⎟
2
+ (1.5 ⋅1010)2 ≈ 7 ⋅1015cm−3
€
n0 =ni
2
p0
= 3.21⋅104
Semiconductor EquilibriumPosition of Fermi energy level
As a function of doping levels As a function of temperature for a given doping level