Electronic Devices

Carrier transport
 Transport Processes

Primary carrier action inside a semiconductor:

Drift: charged particle motion in response to an applied electric
field.
 Diffusion: charged particle motion due to concentration gradient
or temperature gradient.
 Recombination-Generation: a process where charge carriers
(electrons and holes) are annihilated (destroyed) and created.
Thermal Motion

 Electrons and atoms in the Si lattice are always in random

thermal motion.
 Electrons make frequent collisions with the vibrating atoms.
 “Lattice scattering” or “phonon scattering” increases with
increasing temperature.
3 1 2
 Average electron or hole kinetic energy  kT  mvth
2 2

 At 300 K, vth  ? Take m = 0.25 m0

vth  2.3 *10 cm / s

7
Thermal Motion

Random scattering The field gives a net

events (R-G centers) drift superposed on top

• Due to scattering, electrons in a semiconductor do not

achieve constant velocity nor acceleration.
• However, they can be viewed as particles moving at a
constant average drift velocity vd
Mobility

 c :mean free time between the collisions, typically ~ 1 ps

𝐹 = −𝑞ℇ; 𝑞ℇ
𝑎=−
𝑚𝑛
𝑞ℇ
𝑣𝑛 = − 𝜏
𝑚𝑛 𝑐
𝑞𝜏𝑐
= − ℇ
𝑚𝑛

Similarly, 𝑣𝑝 = 𝜇𝑝 ℇ
Mobility

mu_psd.m
m has the dimensions of v/E:  cm/s cm2 
  
 V/cm V  s 

Electron and hole mobility of selected

intrinsic semiconductors (T = 300 K)

Si Ge GaAs InAs
m n (cm /V·s)
2
1400 3900 8500 30000
m p (cm2/V·s) 470 1900 400 500

Based on the above table alone, which semiconductor and which

carriers (electrons or holes) are attractive for applications in high-
speed devices?
EXAMPLE: Given mp = 470 cm2/V·s,

what is the hole drift velocity at E = 103 V/cm?

What is p and what is the distance traveled between collisions

(called the mean free path)?

Take mp = 0.39 m0

Solution: n = mpE = 470 cm2/V·s  103 V/cm = 4.7 105 cm/s

p = mpmp/q =470 cm2/V ·s  0.39  9.110-31 kg/1.610-19 C
= 0.047 m2/V ·s  2.210-12 kg/C = 110-13s = 0.1 ps
mean free path = pnth ~ 1 10-13 s  2.2107 cm/s
= 2.210-6 cm = 220 Å = 22 nm
This is smaller than the typical dimensions of devices, but getting
close.
 Mechanisms of carrier scattering

There are two main causes of carrier scattering:

1. Phonon Scattering
2. Ionized-Impurity (Coulombic) Scattering

Phonon scattering mobility decreases when temperature rises:

1 1
m phonon   phonon    T 3 / 2

phonon density  carrier thermal velocity T  T 1 / 2

m = q/m T
vth  T1/2
 Impurity (Dopant)-Ion Scattering or Coulombic Scattering

Boron Ion Electron

_
- -
Electron +
Arsenic
Ion

There is less change in the direction of travel if the electron zips by

the ion at a higher speed.

vth3 T 3/ 2
mimpurity  
Na  Nd Na  Nd
 1 1
mi
1 1
Effective Mobility = 
m
i
mi ml
Total mobility vs. impurity concentration

Mobilities and diffusivities

in Si and GaAs at 300 K
as a function of impurity
concentration.
Drift Velocity vs. Electric Field: Velocity saturation

• Linear relation holds at low field intensity, ~5103 V/cm

• Velocity saturation is observed at higher Electric field
• This dependence of a Ϭ upon E is an example of a hot
carrier effect, which implies that the carrier drift velocity vd
is comparable to the thermal velocity vth.
•This limit occurs near the mean thermal velocity (≈ 10^7 cm/s)
and represents the point at which added energy imparted by
the field is transferred to the lattice rather than increasing the
carrier velocity. The result of this scattering limited velocity is a
fairly constant current at high field.
•The result of this scattering limited velocity is a fairly constant current
at high field.
Drift Current and Conductivity

Jp

Hole current density Jp = qpv A/cm2 or C/cm2·sec

EXAMPLE: If p = 1015cm-3 and v = 104 cm/s, then

Jp= 1.610-19C  1015cm-3  104cm/s
= 1.6 C/s  cm 2  1.6 A/cm 2
 Jp|drift = qpvd = qmppE
Jdrift = Jn|drift + Jp|drift =q(mnn+mpp)E = E
E E

 Conductivity (1/-cm) of a semiconductor:  = q(mnn+mpp)

 Resistivity (-cm) of a semiconductor: =1/
 For n-type material:  For p-type material:

1 1
 
qmn N D qm p N A
 Example

Consider a Si sample at 300 K doped with 1016/cm3 Boron. What is its

resistivity?
Take mp = 437 cm2/V-s
NA = 1016 cm–3 , ND = 0 (NA >> ND  p-type)

p  1016 cm–3, n  104 cm–3

1

q mn n  q mp p
1

qmp p
 [(1.6 1019 )(437)(1016 )]1
 1.430   cm
 Band diagram with Electric field: Band bending

At equilibrium ( with no external field )

EC
All these
energies Eİ Pure/undoped semiconductor
are
horizontal
EV

How these energies will change with an applied field ?

+ - EC
qV EfF
e-
Eİ
n – type

Electric field hole

EV
Electron movement
Hole flow rseb
Electric field applied to semiconductor
𝑑𝐸𝑐
𝐹𝑜𝑟𝑐𝑒 = −𝑞ℇ = −∇ 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = −
𝑑𝑥

Since we are only interested in grad(PE),

we can use Ec, EF, Ei or Ev.
It is convenient to use Ei

1 𝑑𝐸𝑖
ℇ=
𝑞 𝑑𝑥

Define Electrostatic Potential using the relation

𝐸𝑖
𝒱=−
𝑞
• Since electrons drift in a direction opposite to the field, we
expect the potential energy for electrons to increase in the
direction of the field.
• we know the slope in the bands must be such that electrons
drift "downhill" in the field. Therefore, E points "uphill" in the
band diagram.
 Hall Effect

 Consequence of forces exerted on moving charges by

Electric and Magnetic fields
 Determines majority carrier type (n or p),carrier concentration
and mobility
z
Bz
d
x
y
VH w
vx vx
Eyp Eyn

L
Va
Ix
 Consider p-type semiconductor

Lorentz Force
𝐹 = 𝑞(𝑣 × 𝐵)
In upward direction
= 𝑞𝑣𝑥 𝐵𝑧

Holes accumulate at the top  Produce Downward electric

field Ey
𝑞ℇ𝑦 = 𝑞𝑣𝑥 𝐵𝑧 Hall field
In steady state
Using 𝐽𝑝 = 𝑞𝑝𝑣𝑝
𝐽𝑝
ℇ𝑦 = 𝐵𝑧
𝑞𝑝
1 Hall Coefficient
= 𝑅𝐻 𝐽𝑝 𝐵𝑧 ; 𝑅𝐻 =
𝑞𝑝
Carrier concentration

Hall Voltage is given by

𝐽𝑝
𝑉𝐻 = ℇ𝑦 𝑊 ℇ𝑦 = 𝐵
𝑞𝑝 𝑧
𝐽𝑝
= 𝐵𝑧 𝑊
𝑞𝑝
𝐼
= 𝐵𝑧 𝑊
𝑊𝑑 (𝑞𝑝 )

𝐼𝐵𝑧 𝐼𝐵𝑧
p-type 𝑉𝐻 = 𝑝=
𝑞𝑝𝑑 𝑞𝑑𝑉𝐻

𝐼𝐵𝑧 𝐼𝐵𝑧
n-type 𝑉𝐻 = − 𝑛=−
𝑞𝑛𝑑 𝑞𝑑𝑉𝐻
Example

A sample of Si is doped with 1016 phosphorus atoms/cm3. Find RH

and VH in a sample with W = 500 μm, cross section A = 2.5 x 10-3
cm2,I = 1 mA, and Bz = 1 Tesla.
Note: 1 Tesla = 1 Wb/m2 = 104 G.

Answer:
RH = -625 cm3/C
VH = -1.25 mV
Mobility

Using the expression for p in

𝐽𝑝 = 𝑞𝑝𝑣𝑝
= 𝑞𝑝𝜇𝑝 ℇ𝑥
we can calculate hole mobility

𝐼𝐿
𝜇𝑝 =
𝑞𝑝𝑉𝑥 𝑊𝑑
 Carrier Diffusion
 Particles diffuse from regions of higher concentration to
regions of lower concentration region, due to random thermal
motion (Brownian Motion).

Diff_psd
Diffusion currents

p n
J p  qD p J n  qDn
x x
p(x) Current flow
n(x) Current flow
+ -
Hole flux Electron flux

• Dn, Dp  Diffusion constants or diffusivity for

electrons and holes
• Unit of diffusivity: cm2/s
 Current density equations

In presence of Electric field and concentration gradient

Total current density = (drift + diffusion) current density
𝜕𝑛
𝐽𝑛 = 𝑞𝜇𝑛 𝑛ℇ + 𝑞𝐷𝑛
𝜕𝑥
𝜕𝑝
𝐽𝑝 = 𝑞𝜇𝑝 𝑝ℇ − 𝑞𝐷𝑝
𝜕𝑥
𝐽 = 𝐽𝑛 + 𝐽𝑝
 Drift current flows when an electric field is applied.
 Diffusion current flows when a gradient of carrier concentration
exist.
 Current Flow Under Equilibrium Conditions

 In equilibrium, there is no net flow of electrons or holes :

𝐽𝑛 = 0; 𝐽𝑝 = 0
The drift and diffusion current components must balance each
other exactly.
A built-in electric field of ionized atoms exists, such that the
drift current exactly cancels out the diffusion current due to the
concentration gradient.
𝑑𝑛
𝐽𝑛 = 𝑞𝜇𝑛 𝑛ℇ + 𝑞𝐷𝑛 =0
𝑑𝑥
Consider a piece of non-uniformly doped n-type semiconductor:
 EF  Ec  kT
n  NCe
dn N C  EF  Ec  kT dEc
 e
dx kT dx
n dEc

kT dx
dn q
 nE
dx kT
 Electrons will diffuse from left to right (diffusion current right to left)
 This moves negative charge (electrons) to RHS
 Creates electric field from left to right which induces drift current that
draws electrons back to LHS
 Electrons diffuse until drift current balances diffusion current
Einstein relation

But, under equilibrium conditions, Jn = 0 and Jp = 0

𝑑𝑛
𝐽𝑛 = 𝑞𝜇𝑛 𝑛ℇ + 𝑞𝐷𝑛 =0
𝑑𝑥
dn q
 nE
dx kT
𝑘𝑇
𝐷𝑛 = 𝜇𝑛
𝑞
Very useful formula
Einstein relation
Relates mobility and diffusivity
 Example
What is the hole diffusion coefficient in a sample of silicon at 300
K with mp = 410 cm2 / V.s ?

 kT 
D p    m p  (26 mV )  410 cm 2 V 1s 1  11 cm 2 /s
 q 

Remember: kT/q = 26 mV at room

temperature.
 Generation and Recombination Processes

In thermal equilibrium
n.p=ni2
External perturbation may lead to non-equillibrium
n.p>ni2 ; carrier excess
n.p<ni2 ; carrier deficit
Equilibrium is restored via Recombination-Generation (R-G)
Recombination: a process by which conduction electrons
and holes are annihilated.
Generation: a process by which conduction electrons and
holes are created.
Generation Processes

Band-to-Band R–G Center Impact Ionization

1 dEc
E
q dx

• Due to lattice defects or

unintentional impurities
• Also called indirect
generation

• Only occurs in the

presence of large E
 Recombination Processes

Band-to-Band R–G Center Auger

Collision

• Prominent in • Rate is limited by • Occurs in heavily

direct gap minority carrier trapping doped material
semiconductors • Primary recombination
way for Si
 Direct and Indirect Gap Semiconductors

E-k Diagrams
Ec
Ec

Phonon
Photon
Photon

Ev Ev
GaAs, GaN Si, Ge
(direct semiconductors) (indirect semiconductors)

• Little change in momentum is • Large change in momentum is

required for recombination required for recombination
• Momentum is conserved by • Momentum is conserved by
photon (light) emission mainly phonon (vibration)
emission + photon emission
Direct Recombination

Consider thermal equilibrium

Gth: Carrier generation rate
(No. of e-h pairs generated/cm3/s
Rth: Carrier Recombination rate

At thermal equilibrium
Gth = Rth and p.n = ni2
Rate of direct recombination
R = np
For n-type semiconductor at thermal equilibrium
Gth = Rth = nn0pn0
thermal equilibrium
n-type semiconductor
 G = GL + Gth
R =  nn pn = (nn0 +n)(pn0 +p)
Shine light n & p are excess carrier conc.
n = p; Charge neutrality cond.

 Often, the disturbance from equilibrium is small, such that the majority
carrier concentration is not affected significantly:

 For an n-type material Δ𝑝 ≪ 𝑛0 , 𝑛 ≈ 𝑛0 p  p0

 For a p-type material Δn ≪ 𝑝0 , 𝑝 ≈ 𝑝0 n  n0

• Low-level injection condition

However, the minority carrier concentration can be significantly

affected.
Example: low level injection

Consider Si dope with ND =1014 cm-3 at room temperature subject to

a perturbation p = n = 109 cm-3.Find out n and p. How do they
compare with p, n ?
A perturbation is given and then the system is left to relax
𝜕𝑝𝑛 0
= 𝐺𝐿 + 𝐺𝑡ℎ − 𝑅
𝜕𝑡
= 𝛽𝑛𝑛0 𝑝𝑛0 − 𝛽 𝑛𝑛0 + Δ𝑛 𝑝𝑛0 + Δp
𝑝𝑛 − 𝑝𝑛0 𝑝𝑛 − 𝑝𝑛0
=− =−
1/𝛽𝑛𝑛0 𝜏𝑝
p is the minority carrier lifetime: the average time an excess
minority carrier will survive in the sea of majority carriers

𝜕𝑝𝑛 𝑝𝑛 − 𝑝𝑛0
=−
𝜕𝑡 𝜏𝑝

The above expression is obtained for direct recombination, but

holds good for indirect recombination (R-G centers) as well.
Continuity Equations:
Overall effect of drift, diffusion and recombination
Consider the slice at x, thickness dx
Rate of change of electrons in the slice

𝜕𝑛 𝐽𝑛 𝑥 𝐴 𝐽𝑛 𝑥 + 𝑑𝑥 𝐴
𝐴𝑑𝑥 = − + ( 𝐺𝑛 − 𝑅𝑛 )𝐴𝑑𝑥
𝜕𝑡 −𝑞 −𝑞

No. of electrons entering No. of electrons leaving Recombination

Generation
Taking Taylor expansion
𝜕𝐽𝑛
𝐽𝑛 𝑥 + 𝑑𝑥 = 𝐽𝑛 𝑥 + 𝑑𝑥
𝜕𝑥
𝜕𝑛 1 𝜕𝐽𝑛
= + (𝐺𝑛 − 𝑅𝑛 ) for electrons
𝜕𝑡 𝑞 𝜕𝑥
Similarly
𝜕𝑝 1 𝜕𝐽𝑝
=− + (𝐺𝑝 − 𝑅𝑝 ) for holes
𝜕𝑡 𝑞 𝜕𝑥
For minority carriers under low injection

𝑑𝑛
Using the current density expression 𝐽𝑛 = 𝑞𝜇𝑛 𝑛ℇ + 𝑞𝐷𝑛
𝑑𝑥

𝜕𝑛𝑝 𝜕ℇ 𝜕𝑛𝑝 𝜕 2 𝑛𝑝 𝑛
𝑛𝑝 − 𝑛𝑝0 p-type
= 𝑛𝑝 𝜇𝑛 + 𝜇𝑛 ℇ + 𝐷𝑛 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜏𝑛
𝜕𝑝𝑛 𝜕ℇ 𝜕𝑝𝑛 𝜕 2 𝑝𝑛 𝑝 𝑝𝑛 − 𝑝𝑛0 n-type
= −𝑝𝑛 𝜇𝑝 − 𝜇𝑝 ℇ + 𝐷𝑝 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜏𝑝

Poisson Equation 𝑑ℇ 𝜌𝑠 𝜖𝑠 ∶ s/c dielectric permittivity

= ;
𝑑𝑥 𝜖𝑠 𝜌𝑠 ∶ space charge density
= q(p − n + ND+ − NA−)

Continuity Eqs. +Focus

Poisson Eq.course
of this + Boundary Cond.
Provideunder
To obtain these solutions unique solutions
various simplifying approximations
 Minority Carrier Diffusion Equations

 The minority carrier diffusion equations are derived from the general
continuity equations, and are applicable only for minority carriers.
 Simplifying assumptions:
 The electric field is small (neutral region, i. e. ), such
that:
𝜕𝑛 𝜕𝑛
𝐽𝑛 = 𝑞𝜇𝑛 𝑛ℇ + 𝑞𝐷𝑛 ≈ 𝑞𝐷𝑛 • For n-type material
𝜕𝑥 𝜕𝑥
𝑑𝑝 𝜕𝑝
𝐽𝑝 = 𝑞𝜇𝑝 𝑝ℇ − 𝑞𝐷𝑝 ≈= −𝑞𝐷𝑝 • For p-type material
𝑑𝑥 𝜕𝑥
 Equilibrium minority carrier concentration n0 and p0 are independent
of x (uniform doping).
 Low-level injection conditions prevail.
 Indirect thermal recombination-generation is the dominant thermal
R-G mechanism.
 Particular system under analysis is one-dimensional.
 Starting with the continuity equation for holes in n-type s/c:
0
0
𝜕𝑝𝑛 𝜕ℇ 𝜕𝑝𝑛 𝜕 2 𝑝𝑛 𝑝 𝑝𝑛 − 𝑝𝑛0
= −𝑝𝑛 𝜇𝑝 − 𝜇𝑝 ℇ + 𝐷𝑝 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜏𝑝
Small electric field assumption
With pn = pn0 + pn , we get
𝜕(𝑝𝑛0 + Δpn ) 𝜕 2 (𝑝𝑛0 + Δpn ) 𝑝 Δpn
= 𝐷𝑝 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜏𝑝
Assuming constant pn0

𝜕Δpn 𝜕 2 Δpn 𝑝 Δpn

= 𝐷𝑝 2
+ 𝐺𝐿 − For holes in n-type s/c
𝜕𝑡 𝜕𝑥 𝜏𝑝
𝜕Δnp 𝜕 2 Δnp 𝑛
Δnp For electrons in p-type s/c
= 𝐷𝑛 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜏𝑛

Minority Carrier Diffusion Equations

 Simplifications: Special Cases

np pn
 Steady state:  0, 0
t t

 2 np  2 pn
 No diffusion current: DN  0, DP 0
x 2 x 2

np pn
 No thermal R–G:  0, 0
n p

 No other processes: GL  0
Detailed Balance, Steady State
No net clockwise flow Steady clockwise flow
… and Transients
Unsteady flow
 Uniform Illumination: Transient analysis

t=0

ND = 1015 /cm3
n-type semiconductor p = 10-6 s
GL = 1017 /cm3-s
T = 300 K
 t < 0 : Equilibrium, pn = 0
 t = 0 : Illumination is switched on
 t > 0 : pn, Recombination 
 t   : Gp = Rp  Steady state
What happens in the steady state ?
0 0 0 0
2
𝜕𝑝𝑛 𝜕ℇ 𝜕𝑝𝑛 𝜕 𝑝𝑛 𝑝 𝑝𝑛 − 𝑝𝑛0
= −𝑝𝑛 𝜇𝑝 − 𝜇𝑝 ℇ + 𝐷𝑝 2
+ 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜏𝑝
Steady state Negligible field Uniform doping Δ𝑝𝑛 = 𝐺𝐿 𝜏𝑝
 Δpn (𝑡 → ∞) = 𝐺𝐿 𝜏𝑝

= 1011 /cm3 Low level injection holds

The transient solution can be obtained by solving the equation

𝜕Δ𝑝 𝑛 Δ𝑝 𝑛 𝜕Δ𝑝 𝑛 Δ𝑝 𝑛 𝑑𝑦
= 𝐺𝐿 − 𝑜𝑟 + = 𝐺𝐿 + 𝑃𝑦 = 𝑄
𝜕𝑡 𝜏𝑝 𝜕𝑡 𝜏𝑝 𝑑𝑥

I. F.=𝑒 𝑃𝑑𝑥
and solution is y(I.F)= 𝑄 𝐼. 𝐹. 𝑑𝑥 + 𝑐.
Subject to the boundary condition
Δ𝑝𝑛 𝑡 𝑡=0 =0
General Solution
Δ𝑝𝑛 𝑡 = 𝐺𝐿 𝜏𝑝 + 𝐶e−t/τ p , where C= -GL τp
Putting in the boundary condition
Δ𝑝𝑛 𝑡 = 𝐺𝐿 𝜏𝑝 (1 − 𝑒 −𝑡/𝜏 𝑝 )
 GLp 1

0.8

0.6
 pn (t)
0.4

0.2

0
0 2 4 6
t/ p
 Steady state injection from one side

Consider a sample under constant illumination by a light source.

Calculate pn(x).

ND = 1015 /cm3
n-type sample

x=0 x
CONDITIONS
 Semi-infinite bar, uniformly doped
 Illumination creates pn (0) = 1010 /cm3 excess holes at x=0
 No light penetrates into the interior (x>0)
 pn  0 as x  
 0 0
0
𝜕𝑝𝑛 𝜕ℇ 𝜕𝑝𝑛 𝜕 2 𝑝𝑛 𝑝
0 𝑝 −𝑝
𝑛 𝑛0
= −𝑝𝑛 𝜇𝑝 − 𝜇𝑝 ℇ + 𝐷𝑝 + 𝐺𝐿 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 2 𝜏𝑝
Steady state
Electric field = 0

d 2 pn pn  pn0 d 2 pn pn

Dp  0  2  0, L p   p D p
dx2 p dx 2
Lp
 x / Lp x / Lp
pn ( x )  Ae  Be Diffusion length
• Boundary conditions for a long sample:
pn ()  0  B  0
pn (0)  pn 0
• Final solution:
pn ( x)  pn0 e  x LP
 pn (x )
pn (0)

pn 0
x
Lp

pn ( x)  pn0   pn (0)  pn0 e

 x / Lp
 Quasi-equilibrium and Quasi-Fermi Levels

 Whenever Δn = Δp ≠ 0 then np ≠ ni2 and we are at non-

equilibrium conditions.
 In this situation, now we would like to preserve and use the
relations:
( EF  Ei ) kT ( Ei  EF ) kT
n  ni e , p  ni e
 Above equations imply np = ni2, which does not apply anymore.
 Fermi level is defined only at equilibrium
 Even when electrons and holes are not in equilibrium, within
each group the carriers can be in quasi-equilibrium
 –electrons and holes loosely coupled via G-R (~ 1μs)
 –electrons (or holes) tightly coupled via scattering
(0.1ps)
 The solution is to introduce quasi-Fermi levels EFn and EFp
such that:
𝑛 = 𝑛𝑖 𝑒 (𝐹𝑛 −𝐸𝑖 )/𝑘𝑇

𝑝 = 𝑛𝑖 𝑒 (𝐸𝑖 −𝐹𝑝 )/𝑘𝑇

𝑛
𝐹𝑛 = 𝐸𝑖 + 𝑘𝑇 ln
𝑛𝑖
𝑝
𝐹𝑝 = 𝐸𝑖 − 𝑘𝑇 ln
𝑛𝑖

The quasi-Fermi levels is useful to describe the carrier

concentrations under non-equilibrium conditions
 Equations for Current with Quasi-Fermi levels
𝑑𝑝
Recall the current equation 𝐽𝑝 = 𝑞𝜇𝑝 𝑝ℇ − 𝑞𝐷𝑝
𝑑𝑥
Using the defining relation
𝑝 = 𝑛𝑖 𝑒 (𝐸𝑖 −𝐹𝑝 )/𝑘𝑇
for EFp
𝑑𝑝 𝑛𝑖 (𝐸 −𝐹 )/𝑘𝑇 𝑑𝐸𝑖 𝑑𝐹𝑝
Differentiate wrt x = 𝑒 𝑖 𝑝 −
𝑑𝑥 𝑘𝑇 𝑑𝑥 𝑑𝑥
𝑞𝑝 𝑝 𝜕𝐹𝑝
= ℇ−
Putting in the equation for Jp
𝑘𝑇 𝑘𝑇 𝜕𝑥
𝜕𝐹𝑝
𝐽𝑝 = 𝜇𝑝 𝑝 Flow of current
𝜕𝑥
Similarly for n Gradient of
𝜕𝐹𝑛
𝐽𝑛 = 𝜇𝑛 𝑛 Quasi Fermi Levels
𝜕𝑥
 Consider a Si sample at 300 K with ND = 1017 cm–3 and
Δn = Δp = 1014 cm–3.

a) What are p and n? • The sample is an n-type

2
n
n0  N D  1017 cm 3 , p0  i  103 cm3
n0
n  n0  n  1017 +1014  1017 cm 3
p  p0  p  103 +1014  1014 cm 3

b) What is the np product?

np  1017 1014 =1031cm3

 Consider a Si sample at 300 K with ND = 1017 cm–3 and
Δn = Δp = 1014 cm–3.
0.417 eV
Ec
FN
c) Find FN and FP?
Ei
FN  Ei  kT ln  n ni  FP

 
5 Ev
FN  Ei  8.62 10  300  ln 10 10 17 10
0.238 eV
 0.417 eV

FP  Ei  kT ln  p ni  np  ni e
FN  Ei  kT
 ni e Ei  FP  kT
0.417 0.238
 1010 e 0.02586
1010 e 0.02586

 0.238 eV  1.000257  1031

 1031cm 3
Haynes – Shockley Experiment

Measurement of minority carrier mobility (mp) & diffusivity (Dp)

Carrier injection
Carrier collection
DSO

n-type

V1

 Carrier injection: light pulse or electrical pulse to forward

biased p-n junction
 Carrier collection: Reverse biased pn junction
 V1 establishes electric field
 𝜕ℇ
After pulse is applied 𝐺𝑝 = 0; =0
𝜕𝑥
𝜕𝑝𝑛 𝜕𝑝𝑛 𝜕 2 𝑝𝑛 𝑝𝑛 − 𝑝𝑛0
= 𝜇𝑝 ℇ + 𝐷𝑝 2
−
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜏𝑝

2
𝑁 𝑥 − 𝜇𝑝 ℇ𝑡 𝑡
𝑝𝑛 𝑥, 𝑡 = exp − − + 𝑝𝑛0
4𝜋𝐷𝑝 𝑡 4𝐷𝑝 𝑡 𝜏𝑝
N: No of e-h pairs generated per unit area
 𝑁 (𝑥 − 𝜇𝑝 ℇ𝑡)2 𝑡
exp − −
4𝜋𝐷𝑝 𝑡 4𝐷𝑝 𝑡 𝜏𝑝

1/e points:

𝑥 2 = 4𝐷𝑝 𝑡 ⟹ 𝑥 = 2 𝐷𝑝 𝑡

Δ𝑥 = 4 𝐷𝑝 𝑡

x
 𝑁 (𝑥 − 𝜇𝑝 ℇ𝑡)2 𝑡
exp − −
4𝜋𝐷𝑝 𝑡 4𝐷𝑝 𝑡 𝜏𝑝

Hse_psd.m
Calculation of mobility, diffusivity

In Haynes-Shockley experiment on n-type Ge semiconductor,

given the bar is 1 cm long, L = 0.95 cm, V1 = 2 V, and time for
pulse arrival = 0.25 ns. Find μp and Dp.

Vd 0.95 / ( 0.25  10 3 )
mp    1900 cm 2 / V  s
d 2

(x) 2 (t.vd) 2
Dp    49.4 cm 2 / s
16t 16t

Dp kT
 0.026 
mp q
Minority carrier lifetime

In a Haynes-Shockley experiment, the maximum amplitudes of

the minority carriers at t1 = 100 μs and t2 = 200 μs differ by a
factor of 5. Calculate the minority carrier lifetime.
𝑁
Peak amplitude = exp(-t/𝜏𝑝 )
4𝜋𝐷𝑝 𝑡

Δ𝑝 𝑡1 𝑡2 𝑒 −𝑡 1 /𝜏 𝑝 200 (200−100)/𝜏 μs
= = 𝑒 𝑝 =5
Δ𝑝 𝑡2 𝑡1 𝑒 −𝑡 2 /𝜏 𝑝 100

𝜏𝑝 = 79 μs