1. Chapter 6
Carrier Transport

2. DRIFT Definition-Visualization
Drift is charge-particle motion in response to an applied electric field.
The relaxation time ( 𝜏 𝑚 ) can be interpreted as mean free time ( 𝑡 ) between collisions, if a particle reaches equilibrium by the collision once.
= 𝑑𝑡 𝜏 𝑚
The probability that a particle experience collision during time dt:
If there are n(t) particles,
Collision becomes less with time.
𝑑𝑛(𝑡)=−𝑛(𝑡) 𝑑𝑡 𝜏 𝑚
𝑑𝑛(𝑡) 𝑑𝑡 =− 𝑛(𝑡) 𝜏 𝑚
𝑛(𝑡)=𝑛 0 𝑒 −𝑡/ 𝜏 𝑚
Number of particles experience a collision during dt.
𝑡 = 0 ∞ 𝑡𝑛(𝑡) 𝑑𝑡 0 ∞ 𝑛(𝑡) 𝑑𝑡 = 𝜏 𝑚
Average time,
𝑙 = 𝑡 ∙ 𝑣 𝑡ℎ
Mean free time
Mean free path

3. 𝑑 𝑝 𝑥 𝑑𝑡 | 𝜀−𝑓𝑖𝑒𝑙𝑑 + 𝑑 𝑝 𝑥 𝑑𝑡 | 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 =0
In steady-state, carriers drifts at a constant drift velocity by balancing between acceleration by electric field and deceleration by collision.
Total momentum at t
𝑑 𝑝 𝑥 𝑑𝑡 | 𝜀−𝑓𝑖𝑒𝑙𝑑 + 𝑑 𝑝 𝑥 𝑑𝑡 | 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 =0
𝑑 𝑝 𝑥 =− 𝑝 𝑥 𝑑𝑡 𝑡
𝑑 𝑝 𝑥 𝑑𝑡 | 𝜀−𝑓𝑖𝑒𝑙𝑑 =−𝑛𝑞 𝜀 𝑥
acceleration,
Momentum change due to collision during dt
deceleration,
𝑑 𝑝 𝑥 𝑑𝑡 | 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 =− 𝑝 𝑥 𝑡
< 𝑝 𝑥 >= 𝑝 𝑥 𝑛 =− 𝑑 𝑝 𝑥 𝑛𝑑𝑡 𝑡 =−𝑞 𝑡 𝜀 𝑥
Average momentum per electron,
< 𝑣 𝑥 >= < 𝑝 𝑥 > 𝑚 𝑛 ∗ =− 𝑞 𝑡 𝑚 𝑛 ∗ 𝜀 𝑥
1-D expression
In 3-D,
𝑣 𝑑 =− 𝑞 𝑡 𝑚 𝑛,𝑐 ∗ 𝜀 = 𝜇 𝑛 𝜀
for electron
𝑣 𝑑 = 𝑞 𝑡 𝑚 𝑝,𝑐 ∗ 𝜀 = 𝜇 𝑝 𝜀
for hole
Where are conductivity effective masses.
𝑚 𝑛,𝑐 ∗ , 𝑚 𝑝,𝑐 ∗
𝑚 𝑛,𝑐 ∗ = 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ 𝑚 𝑡 ∗ −1
𝑚 𝑝,𝑐 ∗ = 𝑚 𝑙ℎ ∗ 3/2 + 𝑚 ℎℎ ∗ 3/2 𝑚 𝑙ℎ ∗ 1/2 + 𝑚 ℎℎ ∗ 1/2

4. Two effective masses of carrier
Density of state effective mass, 𝑚 𝑑 ∗ , in the density of state function
Conductivity (or mobility) effective mass, 𝑚 𝑐 ∗ , in the expression for mobility
The density of states effective mass for electrons and holes is given by,
𝑚 𝑛 ∗ =𝑚 𝑑𝑒 ∗ = 𝑚 𝑥𝑥 𝑚 𝑥𝑥 𝑚 𝑥𝑥 1/3
= 𝑚 𝑥𝑥
for the 𝛤 -valley (= 𝑚 𝑥𝑥 = 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 )
= 𝑚 𝑙 𝑚 𝑡 2 1/3
for the X or L -valley
𝑚 𝑝 ∗ = 𝑚 ℎℎ ∗ 3/2 + 𝑚 𝑙ℎ ∗ 3/2 2/3
𝑚 𝑑ℎ ∗ 3/2 = 𝑚 ℎℎ ∗ 3/2 + 𝑚 𝑙ℎ ∗ 3/2
The conductivity (or mobility) effective mass for electrons and holes is given by,
1 𝑚 𝑛,𝑐 ∗ = 𝑚 𝑥𝑥 𝑚 𝑦𝑦 𝑚 𝑧𝑧 = 1 𝑚 𝑥𝑥
for the 𝛤 -valley (= 𝑚 𝑥𝑥 = 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 )
= 𝑚 𝑙 𝑚 𝑡
for the X or L -valley
1 𝑚 𝑝,𝑐 ∗ = 𝑚 𝑙ℎ ∗ 1/2 + 𝑚 ℎℎ ∗ 1/2 𝑚 𝑙ℎ ∗ 3/2 + 𝑚 ℎℎ ∗ 3/2
𝐽 𝑝 =𝑝𝑞 𝜇 𝑐𝑜𝑛 𝜀= 𝑝 ℎℎ 𝑞 𝜇 ℎℎ + 𝑝 𝑙ℎ 𝑞 𝜇 𝑙ℎ 𝜀∝ 𝑚 ℎℎ ∗ ∙ 1 𝑚 ℎℎ ∗ + 𝑚 𝑙ℎ ∗ 3/2 ∙ 1 𝑚 𝑙ℎ ∗
= 𝑚 ℎℎ ∗ 1/2 + 𝑚 𝑙ℎ ∗ 1/2 = 𝑚 𝑑ℎ ∗ 3/2 ∙ 1 𝑚 𝑝,𝑐 ∗ = 𝑚 ℎℎ ∗ 3/2 + 𝑚 𝑙ℎ ∗ 3/ 𝑚 𝑝,𝑐 ∗

5. Drift Current The formal definition of current, 𝐼 𝑃|𝑑𝑟𝑖𝑓𝑡 =𝑞𝑝 𝑣 𝑑 𝐴
In vector notation,
𝐽 𝑃|𝑑𝑟𝑖𝑓𝑡 =𝑞𝑝 𝑣 𝑑
Excluding situations involving large ℰ fields,
𝑣 𝑑 = 𝜇 𝑝 ℰ
where 𝜇 𝑝 ,the hole mobility, is the constant of proportionality constant
𝐽 𝑃|𝑑𝑟𝑖𝑓𝑡 =𝑞 𝜇 𝑝 𝑝 ℰ
similarly,
𝐽 𝑁|𝑑𝑟𝑖𝑓𝑡 =𝑞 𝜇 𝑛 𝑛 ℰ

6. Mobility Matthiessen’s Rule
The carrier mobility varies inversely with the amount of scattering taking place within the semiconductor.
To theoretically characterize mobility it is therefore necessary to consider the different types of scattering events that can take place inside a semiconductor.
Phonon (lattice) scattering
Ionized impurity scattering
Scattering by neutral impurity atoms and defects
Carrier-carrier scattering
Piezoelectric scattering
For the typically dominant phonon and ionized impurity scattering, single-component theories yield, respectively, to first order
𝜇 𝐿 ∝ 𝑇 −3/2
𝜇 𝐼 ∝ 𝑇 −3/2 / 𝑁 𝐼
where
𝑁 𝐼 = 𝑁 𝐷 𝑁 𝐴 −
Matthiessen’s Rule
Noting that each scattering mechanism gives rise to a “resistance-to-motion” which is inversely proportional to the component mobility, and taking the “resistance” to be simply additive (analogous to a series combination of resistors in an electrical circuit), one obtains
1 𝜇 𝑛 = 1 𝜇 𝐿𝑛 𝜇 𝐼𝑛 +…
1 𝜇 𝑝 = 1 𝜇 𝐿𝑝 𝜇 𝐼𝑝 +…

7. Doping/Temperature Dependence
The Si carrier mobility versus doping and temperature plots presented respectively in Figs 6.5 and 6.6 were constructed employing the empirical-fit relationship
𝜇= 𝜇 𝑚𝑖𝑛 + 𝜇 (𝑁/𝑁 𝑟𝑒𝑓 ) 𝛼
where
𝜇: carrier mobility
N: doping density(either NA or ND)
All other quantities are fit parameters that exhibit a temperature dependence of the form
𝛼= 𝐴 0 ( 𝑇 300 ) η
where
𝐴 0 : temperature-independent constant
T: temperature in Kelvin
η : temperature exponent for the given fit parameter

9. Experimental values for lightly doped Si,
𝜇 𝐿 ∝ 𝑇 −3/2
from first order theory
Experimental values for lightly doped Si,
for electron
𝜇≈ 𝜇 𝐿 ∝ 𝑇 −2.3±0.1
∝ 𝑇 −2.2±0.1
for hole

10. For GaAs,

11. High-Field/Narrow-Dimensional Effects
Under low electric field
Carrier gains energy from the electric field and loses the energy through collisions with low energy acoustic phonons or impurities.
The averages energy of the electrons ≈ 3 2 𝑘𝑇at thermal equilibrium ( 𝑻 𝒆 ≈ 𝑻 𝒍𝒂𝒕𝒕𝒊𝒄𝒆 )
Drift velocity 𝑣 𝑑 ∝ 𝜀 and current density 𝐽 ∝ 𝜀 .
Velocity Saturation
under high electric field
Electrons gain energy from the field faster than they can lose it to the lattice.
The electron distribution can be characterized by effective temperature, 𝑇 𝑒 .
( 𝑻 𝒆 > 𝑻 𝒍𝒂𝒕𝒕𝒊𝒄𝒆 : hot electron effect)
Drift velocity 𝑣 𝑑 and current density 𝐽 are no longer linear with 𝜀 . (nonohmic)
Electrons can transfer energy to the lattice by the generation of high energy optical phonons. This causes saturated drift velocity (𝒗 𝒅𝒔𝒂𝒕 ).
In Si at 300 K, 𝑣 𝑑𝑠𝑎𝑡 ≈ 10 7 cm/sec for both electrons and holes at 𝜀 ≈ 10 7 V/cm.
Temperature dependence of 𝑣 𝑑𝑠𝑎𝑡 for electrons in Si can be modeled by the empirical-fit expression.
𝑣 𝑑𝑠𝑎𝑡 = 𝑣 𝑑𝑠𝑎𝑡 0 1+𝐴 𝑒 𝑇/ 𝑇 𝑑
𝑣 𝑑𝑠𝑎𝑡 0 =2.4× 10 7 𝑐𝑚/𝑠𝑒𝑐
A = 0.8
Td = 600 K

12. Intervalley Carrier Transfer
For GaAs
ellipsoidal constant
energy surface
spherical constant
energy surface
∆ Γ𝐿
= 0.29 eV
𝑣 𝑑 𝜀
Electrons in Γ; 𝑚 𝑛Γ,𝑐 ∗ =𝑚 𝑛 ∗ = 𝑚 0
Electrons in L; 𝑚 𝑙 ∗ =1.9 𝑚 0 , 𝑚 𝑡 ∗ =0.075 𝑚 0
𝑚 𝑛𝐿,𝑐 ∗ =0.55 𝑚 0
𝑚 𝑛𝐿,𝑐 ∗ ≈10 𝑚 𝑛Γ,𝑐 ∗
𝜀 𝑐 =3.3× 10 3 𝑉/𝑐𝑚
Under normal circumstances, the Γ−valley is the only one occupied, but for an applied field of ~ 3.5 KV electrons begin to be transferred to the L-valley. The resulting negative differential conductance occurs when the carriers are transferred from low mass, high velocity states to high mass, low velocity states is referred to as the “Gunn Effect”.

13. 𝑛 Γ = 1 2 𝜋 2 2 𝑚 Γ ∗ ℏ 2 3/2 0 ∞ 𝐸 1 2 𝑒𝑥𝑝 − 𝐸− 𝐸 𝐹 𝑘 𝑇 𝑒 𝑑𝐸
for nondegenerated semiconductor
= 𝑚 Γ ∗ 𝑘 𝑇 𝑒 𝜋ℏ /2 𝑒𝑥𝑝 𝐸 𝐹 𝑘 𝑇 𝑒
where Te is an electron temperature
𝑛 𝐿 =4∙ 1 2 𝜋 𝑚 𝐿 ∗ ℏ /2 ∆Γ𝐿 ∞ 𝐸 𝑒𝑥𝑝 − 𝐸− 𝐸 𝐹 𝑘 𝑇 𝑒𝐿 𝑑𝐸
= 2 𝑚 𝐿 ∗ 𝑘 𝑇 𝑒 𝜋ℏ /2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒𝐿 𝑒𝑥𝑝 𝐸 𝐹 𝑘 𝑇 𝑒𝐿
n 𝐿 n Γ =4 𝑚 𝐿 ∗ 𝑇 𝑒𝐿 𝑚 Γ ∗ 𝑇 𝑒 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒𝐿 𝑒𝑥𝑝 𝐸 𝐹 𝑘 1 𝑇 𝑒𝐿 − 1 𝑇 𝑒
If TeL = Te is an electron temperature,
𝑛 𝐿 ≈ 𝑛 Γ at Te = 950 K.
For temperature higher than this, the upper valley has a higher density of states occupied.
Thus when an electron initially in the Γ-valley at energy of E = ∆ 𝛤𝐿 is scattered, it is more likely to undergo an intervalley scattering to L-valley.
The total conductivity for carriers in the two set of valleys,
𝜎= n Γ 𝑞 𝜇 Γ + n 𝐿 𝑞 𝜇 𝐿
where n = n Γ + n 𝐿
The change in the conductivity with electric field, assuming 𝜇 is only a very weak function.
𝑑𝜎 𝑑𝜀 ≈𝑞 𝜇 Γ 𝑑 n Γ 𝑑𝜀 +𝑞 𝜇 𝐿 𝑑 n 𝐿 𝑑𝜀 = 𝜇 Γ − 𝜇 𝐿 𝑑 n Γ 𝑑𝜀
𝑑 n Γ 𝑑𝜀 =− n 𝐿 𝑑𝜀

14. From the current density equation, 𝐽= 𝜎𝜀
𝑑𝐽 𝑑𝜀
The differential conductivity,
𝑑𝐽 𝑑𝜀 =𝜎+𝜀 𝑑𝜎 𝑑𝜀 = n Γ 𝑞 𝜇 Γ + n 𝐿 𝑞 𝜇 𝐿 +𝑞𝜀 𝜇 Γ − 𝜇 𝐿 𝑑 n Γ 𝑑𝜀
𝑑𝐽 𝑑𝜀 <0 If
(-)function
n Γ 𝑞 𝜇 Γ + n 𝐿 𝑞 𝜇 𝐿 <−𝑞𝜀 𝜇 Γ − 𝜇 𝐿 𝑑 n Γ 𝑑𝜀
− 𝜇 Γ − 𝜇 𝐿 𝜇 Γ + n 𝐿 n Γ 𝜇 𝐿 𝜀 n Γ 𝑑 n Γ 𝑑𝜀 >1
𝝁 𝚪 > 𝝁 𝑳
𝜇 Γ ≈7000 𝑐 𝑚 2 /𝑉∙𝑠𝑒𝑐
for GaAs.
𝜇 L ≈100 𝑐 𝑚 2 /𝑉∙𝑠𝑒𝑐
− 𝑑 n Γ 𝑑𝜀 > n Γ 𝜀
also,
− 4 𝑚 𝐿 ∗ 𝑚 Γ ∗ 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒 𝑑 n Γ 𝑑𝜀 + n Γ 𝑑 𝑑𝜀 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒
where
𝑑 n Γ 𝑑𝜀 =− 𝑑 n 𝐿 𝑑𝜀 = If
𝑇 𝑒𝐿 ≈ 𝑇 𝑒
n 𝐿 n Γ =4 𝑚 𝐿 ∗ 𝑇 𝑒𝐿 𝑚 Γ ∗ 𝑇 𝑒 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒𝐿 𝑒𝑥𝑝 𝐸 𝐹 𝑘 1 𝑇 𝑒𝐿 − 1 𝑇 𝑒
≈4 𝑚 𝐿 ∗ 𝑚 Γ ∗ 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒

15. 𝑑 n Γ 𝑑𝜀 =− 4 𝑚 𝐿 ∗ 𝑚 Γ ∗ 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒 𝑑 n Γ 𝑑𝜀 + n Γ ∆ Γ𝐿 𝑘 𝑇 𝑒 2 𝑑 𝑇 𝑒 𝑑𝜀
=− n 𝐿 n Γ 𝑑 n Γ 𝑑𝜀 + n Γ ∆ Γ𝐿 𝑘 𝑇 𝑒 2 𝑑 𝑇 𝑒 𝑑𝜀
=− n 𝐿 n Γ 𝑑 n Γ 𝑑𝜀 − n 𝐿 ∆ Γ𝐿 𝑘 𝑇 𝑒 2 𝑑 𝑇 𝑒 𝑑𝜀
n Γ +n 𝐿 n Γ 𝑑 n Γ 𝑑𝜀 =− n 𝐿 ∆ Γ𝐿 𝑘 𝑇 𝑒 2 𝑑 𝑇 𝑒 𝑑𝜀
− 𝑑 n Γ 𝑑𝜀 = ∆ Γ𝐿 𝑘 𝑇 𝑒 n Γ 𝑇 𝑒 n 𝐿 n Γ +n 𝐿 𝑑 𝑇 𝑒 𝑑𝜀
− 𝑑 n Γ 𝑑𝜀 > n Γ 𝜀
> n Γ 𝜀
n 𝐿 n Γ +n 𝐿 ∆ Γ𝐿 𝑘 𝑇 𝑒 𝜀 𝑇 𝑒 𝑑 𝑇 𝑒 𝑑𝜀
>1
Assuming that the electron temperature increases linearly with electric field,
𝜀 𝑇 𝑒 𝑑 𝑇 𝑒 𝑑𝜀 ≈1
𝑚 𝐿 ∗ 𝑚 Γ ∗ −3/2 𝑒𝑥𝑝 ∆ Γ𝐿 𝑘 𝑇 𝑒
n 𝐿 n Γ =4 𝑚 𝐿 ∗ 𝑚 Γ ∗ 3/2 𝑒𝑥𝑝 − ∆ Γ𝐿 𝑘 𝑇 𝑒
∆ Γ𝐿 𝑘 𝑇 𝑒 >1+ n Γ 𝑛 𝐿 =1+ 1 4
simple transcendental equation

16. 𝑣 𝜀 Read “Ballistic transport/velocity overshoot”.
𝑚 𝐿 ∗ 𝑚 Γ ∗ −3/2 𝑒𝑥𝑝 ∆ Γ𝐿 𝑘 𝑇 𝑒
∆ Γ𝐿 𝑘 𝑇 𝑒 >1+ n Γ 𝑛 𝐿 =1+ 1 4
∆ Γ𝐿 𝑘 𝑇 𝑒
There are two regions of where this inequality is not satisfied:
i) At very high electron temperature( not of interest)
ii) At low electron temperature of interest
∆ Γ𝐿 𝑘 𝑇 𝑒 ≈5.8
Upper limit of
Lower limit of electron temperature, 𝑇 𝑒 ≈600 𝐾
𝑛 𝐿 n Γ ≈0.15
So that negative differential conductivity sets when as little as 15 % of the electrons transferred to the upper valleys. 𝑣
increasing ∆ Γ𝐿 𝜀
Read “Ballistic transport/velocity overshoot”.

17. Related Topics Resistivity/Conductivity 𝜺 =ρ 𝑱 𝑱 =𝜎 𝜺 = 1 𝜌 𝜺 or
𝑱 =𝜎 𝜺 = 1 𝜌 𝜺
In a homogeneous material,
𝑱 = 𝑱 𝑑𝑟𝑖𝑓𝑡 = 𝑱 𝑁|𝑑𝑟𝑖𝑓𝑡 + 𝑱 𝑃|𝑑𝑟𝑖𝑓𝑡 =q( 𝜇 𝑛 n + 𝜇 𝑝 p) 𝜺
∴ , 𝜌= 1 q( 𝜇 𝑛 n + 𝜇 𝑝 p)
resistivity
[Ω∙𝑐𝑚]
conductivity,
𝜎=q( 𝜇 𝑛 n + 𝜇 𝑝 p)
𝜌= 1 q 𝜇 𝑛 𝑁 𝐷
for n-type semiconductor
𝜌= 1 q 𝜇 𝑝 𝑁 𝐴
for p-type semiconductor

18. Sheet Resistance 𝑅 𝑠 = 𝜌 𝑡 𝑅=𝜌 𝐿 A =𝜌 𝐿 W∙t = 𝑅 𝑠 𝐿 W [Ω/⎕]
𝑅 𝑠 = 𝜌 𝑡
[Ω/⎕]
𝑅=𝜌 𝐿 A =𝜌 𝐿 W∙t = 𝑅 𝑠 𝐿 W
Four-point probe technique
1) For thick sample (s << t)
At probe 1,
𝜀 𝑟 =𝜌 𝐽 𝑟 =− 𝑟 𝜕𝑉( 𝑟 ) 𝜕𝑟 1 2 3 4 D t
𝐽 𝑟 = 𝑟 𝐼 2𝜋 𝑟 2
where
𝑉 𝑟 =− 𝜌𝐼 2𝜋 𝑟 2 𝑑𝑟+ 𝐶 1
= 𝐼𝜌 2𝜋𝑟 + 𝐶 1 I
In spherical coordinate system
Due to symmetry of current path, V is not function of 𝜃 and ∅ 𝑟 𝑟
𝑉 21 = 𝐼𝜌 2𝜋𝑠 + 𝐶 1 𝑟
𝑉 31 = 𝐼𝜌 2𝜋(2𝑠) + 𝐶 1

19. : thickness correction factor
At probe 4, I
𝑉 𝑟 = 𝜌𝐼 2𝜋 𝑟 2 𝑑𝑟+ 𝐶 2 𝑟 𝑟 𝑟
=− 𝐼𝜌 2𝜋𝑟 + 𝐶 2
𝑉 24 =− 𝐼𝜌 2𝜋(2𝑠) + 𝐶 2
𝑉 34 =− 𝐼𝜌 2𝜋𝑠 + 𝐶 2
𝑉 2 = 𝑉 21 + 𝑉 24 = 𝐼𝜌 2𝜋𝑠 + 𝐶 1 − 𝐼𝜌 2𝜋 2𝑠 + 𝐶 2 = 𝐼𝜌 2𝜋 2𝑠 + 𝐶 1 + 𝐶 2
𝑉 3 = 𝑉 31 + 𝑉 34 = 𝐼𝜌 2𝜋 2𝑠 + 𝐶 1 − 𝐼𝜌 2𝜋𝑠 + 𝐶 2 =− 𝐼𝜌 2𝜋 2𝑠 + 𝐶 1 + 𝐶 2
𝑉= 𝑉 2 − 𝑉 3 = 𝐼𝜌 2𝜋𝑠
∴𝜌=2𝜋𝑠 𝑉 𝐼 𝐹 1
where
𝐹 1
: thickness correction factor
𝐹 1 = 𝑡/𝑠 2ln sinh 𝑡 𝑠 sinh 𝑡 2𝑠 ⁡
𝜌=2𝜋𝑠 𝑉 𝐼
If t >> s, 𝐹 1 →1,
𝐹 1 → 𝑡/𝑠 2𝑙𝑛2 ,
If t << s,
𝜌= 𝜋𝑡 𝑙𝑛2 𝑉 𝐼

20. 2) For thin sample (s >> t)
At probe 1,
𝜀 𝑟 =𝜌 𝐽 𝑟 =− 𝑟 𝜕𝑉( 𝑟 ) 𝜕𝑟
𝐽 𝑟 = 𝑟 𝐼 2𝜋𝑟𝑡
where
𝑉 𝑟 =− 𝜌𝐼 2𝜋𝑟𝑡 𝑑𝑟+ 𝐶 1
=− 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑟+ 𝐶 1 1 2 3 4 D t
𝑉 21 =− 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑠+ 𝐶 1
𝑉 31 =− 𝐼𝜌 2𝜋𝑡 𝑙𝑛2𝑠+ 𝐶 1 I
At probe 4, 𝑟
𝑉 𝑟 = 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑟+ 𝐶 2
In cylinderical coordinate system
𝑉 24 = 𝐼𝜌 2𝜋𝑡 𝑙𝑛2𝑠+ 𝐶 2
𝑉 34 = 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑠+ 𝐶 2
𝑉 2 = 𝑉 21 + 𝑉 24 =− 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑠+ 𝐶 1 + 𝐼𝜌 2𝜋𝑡 𝑙𝑛2𝑠+ 𝐶 2 = 𝐼𝜌 2𝜋𝑡 𝑙𝑛2 + 𝐶 1 + 𝐶 2
𝑉 3 = 𝑉 31 + 𝑉 34 =− 𝐼𝜌 2𝜋𝑡 𝑙𝑛2𝑠 + 𝐶 1 + 𝐼𝜌 2𝜋𝑡 𝑙𝑛𝑠+ 𝐶 2 =− 𝐼𝜌 2𝜋𝑡 𝑙𝑛2+ 𝐶 1 + 𝐶 2
𝑉= 𝑉 2 − 𝑉 3 = 𝐼𝜌 𝜋𝑡 𝑙𝑛2

21. summary ∴𝜌= 𝜋𝑡 𝑙𝑛2 𝑉 𝐼 𝐹 2 where where 𝐹 2 𝐹 2
∴𝜌= 𝜋𝑡 𝑙𝑛2 𝑉 𝐼 𝐹 2
where
where
𝐹 2
𝐹 2
: size correction factor
𝐹 2 = 𝑙𝑛2 𝑙𝑛2+𝑙𝑛 ( 𝐷 𝑠 ) ( 𝐷 𝑠 ) 3 −3 ⁡
𝜌= 𝜋𝑡 𝑙𝑛2 𝑉 𝐼
If D >> s, 𝐹 2 →1,
: same as before
summary
i) If t >> s and D >> s,
𝜌=2𝜋𝑠 𝑉 𝐼
ii) If the condition for t >> s and D >> s is not satisfied,
𝜌=2𝜋𝑠 𝑉 𝐼 𝐹 1 𝐹 2
𝐹 1 → 𝑡/𝑠 2𝑙𝑛2
iii) In most cases, t << s and D >> s,
, 𝐹 2 →1,
𝜌= 𝜋𝑡 𝑙𝑛2 𝑉 𝐼
𝑅 𝑠 = 𝜋 𝑙𝑛2 𝑉 𝐼 =4.53 𝑉 𝐼

22. Hall Effect Lorenz force in the sample assuming p-type,
𝐹 =𝑞 𝑣 𝑑 × 𝐵 +𝑞 𝜀
𝑥 𝐹 𝑥 𝑦 𝐹 𝑦 𝑧 𝐹 𝑧 =𝑞 𝑥 𝑦 𝑧 𝑣 𝑑𝑥 𝑣 𝑑𝑦 𝑣 𝑑𝑧 𝐵 𝑧 +𝑞 𝑥 𝜀 𝑥 𝑦 𝜀 𝑦 𝑧 𝜀 𝑧
𝐹 𝑥 =𝑞 𝑣 𝑑𝑦 𝐵 𝑧 +𝑞 𝜀 𝑥
𝐹 𝑦 =−𝑞 𝑣 𝑑𝑥 𝐵 𝑧 +𝑞 𝜀 𝑦
Lorenz force in y-direction must be balanced under steady state.
𝐹 𝑦 =−𝑞 𝑣 𝑑𝑥 𝐵 𝑧 +𝑞 𝜀 𝑦 = 0
𝑣 𝑑𝑥 = 𝐽 𝑥 𝑞𝑝
Moreover,
∴− 𝐽 𝑥 𝐵 𝑧 𝑞𝑝 + 𝜀 𝑦 =0
𝐽 𝑥 =𝑞 𝑝𝑣 𝑑𝑥 →
𝑅 𝐻 = 𝜀 𝑦 𝐽 𝑥 𝐵 𝑧 = 1 𝑞𝑝
Hall coefficient,
for p-type
𝑅 𝐻 =− 1 𝑞𝑛
for n-type

23. With measurable quantity,
𝑅 𝐻 = 1 𝑞𝑝 = 𝜀 𝑦 𝐽 𝑥 𝐵 𝑧 = 𝑉 𝐻 /𝑑 𝐼 𝑤𝑑 𝐵 = 𝑉 𝐻 𝑤 𝐵𝐼 = 𝑉 𝐻 𝑤 𝐵𝐼
If VH is given in volts, w in cm, B in gauss,
I in amps, and RH in cm3/coul.
The resistance of the bar is just VA/I.
𝑅= 𝑉 𝐴 𝐼 =𝜌 𝑙 𝑤𝑑
𝜌= 1 𝑞𝑝 𝜇 𝐻 = 𝑉 𝐴 𝐼 𝑤𝑑 𝑙 →
The Hall mobility,
𝜇 𝐻 = 1 𝑞𝑝 ∙ 1 𝜌 = 𝑅 𝐻 𝜌
More exact analysis gives,
𝑅 𝐻 = 𝑟 𝐻 𝑞𝑝
for p-type
Hall factor
𝑟 𝐻 ≈1
𝑅 𝐻 =− 𝑟 𝐻 𝑞𝑛
for n-type
The relationship between hall mobility and drift mobility
𝜇 𝐻 = 𝑟 𝐻 𝜇 𝑑𝑟𝑖𝑓𝑡
For GaAs, 𝑟 𝐻 > 1 → 𝜇 𝑑𝑟𝑖𝑓𝑡 < 𝜇 𝐻

24. Anisotropic Conductivity
The equations, so far, for the mobility, conductivity, and Hall constant are applicable for electrons in spherical band minima.
The situation is somewhat more complicated, when the carrier transport in an ellipsoidal minima.
For nonspherical energy surface with one ellipsoidal conduction band minimum at 𝛤 (𝑘=0)
𝑘 𝑦
transverse
𝐸( 𝑘 )− 𝐸 𝐶 ≅ ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 𝑘 𝑥 2 𝑚 𝑥𝑥 + 𝑘 𝑦 2 𝑚 𝑦𝑦 + 𝑘 𝑧 2 𝑚 𝑧𝑧
𝑘 𝑥
𝑘 𝑧
longitudinal
𝐽 𝑥 = 𝑞 2 𝑛 𝑡 𝑚 𝑥𝑥 𝜀 𝑥 =𝑞𝑛 𝜇 𝑥 𝜀 𝑥
The total current density is therefore,
𝐽 𝑦 = 𝑞 2 𝑛 𝑡 𝑚 𝑦𝑦 𝜀 𝑦 =𝑞𝑛 𝜇 𝑦 𝜀 𝑦
𝐽 = 𝑞 2 𝑛 𝑡 𝑚 ∗ ∙ 𝜀
𝐽 𝑧 = 𝑞 2 𝑛 𝑡 𝑚 𝑧𝑧 𝜀 𝑧 =𝑞𝑛 𝜇 𝑧 𝜀 𝑧
where 𝑚 ∗ is the effective mass tensor.

25. 1 𝑚 ∗ = 1 𝑚 𝑥𝑥 𝑚 𝑦𝑦 𝑚 𝑧𝑧
This can also be put in the form,
𝐽 = 𝜎 ∙ 𝜀
where 𝜎 is the conductivity tensor.
𝜎 = 𝑛𝑞 𝜇 𝑥 𝑛𝑞 𝜇 𝑦 𝑛𝑞 𝜇 𝑧
𝐽 𝑥 𝐽 𝑦 𝐽 𝑧 = 𝑛𝑞 𝜇 𝑥 𝑛𝑞 𝜇 𝑦 𝑛𝑞 𝜇 𝑧 ∙ 𝜀 𝑥 𝜀 𝑦 𝜀 𝑧
If 𝜀 is off axis and three diagonal terms are not equal, the current is not in same direction as 𝜀 .
“anisotropic conductivity” y
𝜀 is off axis with ellipsoidal conduction band minima. 𝜀
(𝑚 𝑥𝑥 ≠ 𝑚 𝑧𝑧 ≠ 𝑚 𝑦𝑦 , 𝑞𝑛 𝜇 𝑥 ≠𝑞𝑛 𝜇 𝑦 ≠𝑞𝑛 𝜇 𝑧 )
𝜀 is on axis. 𝐽
𝜀 is off axis but spherical conduction band minima.
(𝑚 𝑥𝑥 = 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 , 𝑞𝑛 𝜇 𝑥 =𝑞𝑛 𝜇 𝑦 =𝑞𝑛 𝜇 𝑧 )
𝜀 is on axis. x 𝐽 𝜀

26. 6 equivalent conduction band minima in the directions
For multiple equivalent ellipsoidal conduction band minimum at X or L (𝑘≠0) ky
𝐸( 𝑘 ) = ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 𝑘 𝑥 2 𝑚 𝑥𝑥 + 𝑘 𝑦 2 𝑚 𝑦𝑦 + 𝑘 𝑧 2 𝑚 𝑧𝑧
(𝑚 𝑥𝑥 ≠ 𝑚 𝑦𝑦 ≠ 𝑚 𝑦𝑦 ) kx
For example, Si:
6 equivalent conduction band minima in the directions
of 𝑘 = 𝜋 𝑎 (1, 0, 0)
𝜀 𝑥
Concentration of electron in each minima is n/6. kz
When the electric field in x-direction, the total current in the x-direction
𝐽 𝑥 = 𝑞 2 𝑛 𝑡 𝑚 𝑥𝑥 𝑚 𝑦𝑦 𝑚 𝑧𝑧 𝜀 𝑥 = 𝑛𝑞 𝜀 𝑥 3 𝜇 𝑥 + 𝜇 𝑦 + 𝜇 𝑧 =𝜎 𝜀 𝑥
Similar expressions can be obtained for y- and z-directions and for any electric field.
𝐽= 𝑞 2 𝑛 𝑡 𝑚 𝑐𝑜𝑛 ∗ 𝜀=𝑞𝑛 𝜇 𝑐𝑜𝑛 𝜀=𝜎𝜀
Compare with the expression for the conductivity,
𝜇 𝑐𝑜𝑛 = 𝜇 𝑥 + 𝜇 𝑦 + 𝜇 𝑧 3
: conductivity mobility
1 𝑚 𝑐𝑜𝑛 ∗ = 𝑚 𝑥𝑥 𝑚 𝑦𝑦 𝑚 𝑧𝑧
Isotropic conductivity
: conductivity
effective mass
The current and the electric field are always in the same direction.
scalar quantity