L10. Density of Carriers - 3: EE511 Semiconductor Device Modeling
L10. Density of Carriers - 3: EE511 Semiconductor Device Modeling
L10. Density of Carriers - 3: EE511 Semiconductor Device Modeling
14/08/18
Equations for no and po at Equilibrium
∗ ∗ ∗ ∗
𝑚𝑐𝐷𝑜𝑆 = 𝑚𝑣𝐷𝑜𝑆 𝑚𝑐𝐷𝑜𝑆 ≠ 𝑚𝑣𝐷𝑜𝑆 ⇒ Ei not exactly at the center of the band gap
Semiconductor Si Ge GaAs
𝐸𝐺 𝑒𝑉 (𝑎𝑡 300𝐾) 1.11 0.66 1.42
𝑁𝐶 𝑐𝑚−3 (𝑎𝑡 300𝐾) 2.82 × 1019 1.05 × 1019 4.37 × 1017
𝑁𝑉 𝑐𝑚−3 (𝑎𝑡 300𝐾) 1.83 × 1019 3.92 × 1018 8.68 × 1018
𝑛𝑖 𝑐𝑚−3 (𝑎𝑡 300𝐾) 1.08 × 1010 1.83 × 1013 2.03 × 106
Temperature Dependence of Carrier Density
3/2
2𝜋𝑘𝑇𝑚𝑒∗
𝑁𝐶 = 2 1
ℎ2 4 3
3/2
3/2 2𝜋𝑘𝑇
2𝜋𝑘𝑇𝑚ℎ∗ ⇒ 𝑛𝑖 𝑇 = 2 𝑚𝑒∗ 𝑚ℎ∗ 3/4
𝑒 −𝐸𝐺 /2𝑘𝑇
𝑁𝑉 = 2 ℎ2
ℎ2
2
𝑛𝑖 = 𝑝𝑖 = 𝑁𝐶 𝑁𝑉 𝑒 −𝐸𝐺 /2𝑘𝑇
𝐸
• Asymmetric E-k due to crystal anisotropy
• E.g. mx different from my and mz
• From curvatures: mx > (my = mz) 𝐸𝐶
• Kinetic energy of electrons in CB can be written as along kx
2
𝑘𝑥 − 𝑘𝑥,𝑚𝑖𝑛 𝑘𝑦2 𝑘𝑧2
𝐸 − 𝐸𝐶 = ℏ2 + + 𝑘𝑥
2𝑚𝑥 2𝑚𝑦 2𝑚𝑧 𝑘𝑥,𝑚𝑖𝑛
Take the example of the volume of a box: height 2 cm, width 3 cm, length 4 cm.
Volume = 24 𝑐𝑚3
3
Geometric mean of its dimensions = 2 × 3 × 4 = 𝑥 (𝑠𝑎𝑦)
3 3
Volume of a box with x cm in all dimensions = x3 = 2×3×4 = 24 𝑐𝑚3
Conduction Band of Silicon
Two effective masses: 3/2
2 𝑚𝑒∗
1. Longitudinal effective mass ml (equivalent to mx) 𝜌 𝐸 = 2 𝐸 − 𝐸𝐶
𝐶𝐵 𝜋 ℏ2
2. Transverse effective mass mt (equivalent to my or mz)
The effective mass will be the geometric mean of x, y, and z values
Note: Six equivalent <100> directions
3
Thus, 𝑚𝑛∗ = 6𝑚𝑙 𝑚𝑡 𝑚𝑡
𝑚𝑙 = 0.98𝑚𝑜
𝑚𝑡 = 0.19𝑚𝑜
𝑚𝑛∗ = 1.08𝑚𝑜 (𝑚𝑛∗ = 0.067𝑚𝑜 for GaAs)
Valence Band of Silicon
3/2
2 𝑚ℎ∗
𝜌 𝐸 = 2 𝐸𝑉 − 𝐸
𝑉𝐵 𝜋 ℏ2
Three valence bands simultaneously co-exist
𝐸
Heavy hole band (low curvature)
𝐸𝑉 0
Light hole band (higher curvature) 𝑘𝑥
Split-off band (different energy, ES)
𝐸𝑆
∗ 3/2 ∗ 3/2
2 𝑚ℎℎ 2 𝑚𝑙ℎ
𝜌 𝐸 = 2 𝐸𝑉 − 𝐸 + 2 𝐸𝑉 − 𝐸
𝑉𝐵 𝜋 ℏ2 𝜋 ℏ2
3/2
2 𝑚ℎ∗ 2/3
𝜌 𝐸 = 2 𝐸𝑉 − 𝐸 ⇒ 𝑚ℎ∗ 3/2 = ∗ 3/2
𝑚ℎℎ + ∗ 3/2
𝑚𝑙ℎ ⇒ 𝑚ℎ∗ = ∗ 3/2
𝑚ℎℎ + ∗ 3/2
𝑚𝑙ℎ
𝑉𝐵 𝜋 ℏ2
∗ ∗
𝑚ℎℎ = 0.49𝑚𝑜 𝑚𝑙ℎ = 0.16𝑚𝑜
𝛼𝑇 2 Units for α and β ?
𝐸𝐺 𝑇 = 𝐸𝐺𝑜 − An empirical model
𝛽+𝑇
Energy gap at 0K
E
ro
Material 𝐸𝐺𝑜 (𝑒𝑉) 𝛼 (𝑒𝑉 𝐾) 𝛽(𝐾) T ↑
r
Ge 0.74 4.77 x 10-4 235
Si 1.17 4.73 x 10-4 636
GaAs 1.52 5.41 x 10-4 204 ro Useful to roughly predict
a few material properties
Bond formation
Distance between atoms
𝛼𝑇 2 2𝜋𝑘𝑇
3/2
𝐸𝐺 𝑇 = 𝐸𝐺𝑜 − 𝑛𝑖 𝑇 = 2 𝑚𝑒∗ 𝑚ℎ∗ 3/4 𝑒 −𝐸𝐺 /2𝑘𝑇
𝛽+𝑇 ℎ2
For silicon
4.73 × 10−4 × 𝑇 2
𝐸𝐺 𝑇 = 1.17 − 𝑒𝑉
636 + 𝑇
1.22
1.20 − 𝐸𝐺 2𝑘
1.18
Band Gap (eV)
1.16
1.14
1.12
1.10
1.08
1.06
1.04
0 100 200 300 400 500
T (K)
3/2
2𝜋𝑘𝑇
𝑛𝑖 𝑇 = 2 𝑚𝑒∗ 𝑚ℎ∗ 3/4
𝑒 −𝐸𝐺 /2𝑘𝑇
ℎ2
• For any given temperature, intrinsic carrier concentration exponentially decreases with EG
• These carriers are always present and, hence, can produce current if field is applied
• This current, termed ‘leakage current’, is detrimental – Power dissipation / Heat
• From EG values: Ge has higher ni and, thus, higher leakage current than Si
⇒ One of the reasons why Si is preferred to Ge
With Doping
no
(log scale) Not controlled by doping; Equal e-s and h+s
Intrinsic
Useful region for device operation
Extrinsic
ND Smaller range for Ge than for Si
Ionization/
Carrier Freeze-out
− 𝐸𝐺 2𝑘
1/T
T
Compensation
𝑁𝐷+ = 𝑁𝐷 − 𝑛𝑑
Assume uniformly doped semiconductor at equilibrium, with all dopants ionized 𝑁𝐷+ = 𝑁𝐷 𝑁𝐴− = 𝑁𝐴
For the semiconductor to be electrostatically neutral, No. of +ve charges = No. of -ve charges
𝑝𝑜 + 𝑁𝐷 = 𝑛𝑜 + 𝑁𝐴 𝑝𝑜 − 𝑛𝑜 + 𝑁𝐷 − 𝑁𝐴 = 0
𝑛𝑜 𝑝𝑜 = 𝑛𝑖2
𝑛𝑖2 𝑁𝐷 − 𝑁𝐴 𝑁𝐷 − 𝑁𝐴 2 1/2
𝑛𝑖2
− 𝑛𝑜 + 𝑁𝐷 − 𝑁𝐴 = 0 𝑛𝑜 = + + 𝑛𝑖2
𝑛𝑜 𝑝𝑜 =
2 2 𝑛𝑜
For 𝑁𝐷 − 𝑁𝐴 ≫ 𝑛𝑖 , 𝑛𝑜 ≈ 𝑁𝐷 − 𝑁𝐴