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

€

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