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Chapter 2 Motion and Recombination of Electrons and Holes. 2.1 Thermal Motion. Average electron or hole kinetic energy. 2.1 Thermal Motion. Zigzag motion is due to collisions or scattering with imperfections in the crystal. Net thermal velocity is zero.
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Chapter 2 Motion and Recombination
of Electrons and Holes
2.1 Thermal Motion
Average electron or hole kinetic energy
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Hotpoint Probe can determine sample doing type
Hotpoint Probe distinguishes N and P type semiconductors.
Thermoelectric Generator (from heat to electricity ) and Cooler (from electricity to refrigeration)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
2.2.1 Electron and Hole Mobilities
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
t
m
v
q
E
mp
p
q
t
E
mp
=
v
m
p
=
m
=

m
v
E
v
E
p
n
2.2.1 Electron and Hole Mobilities
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
2.2.1 Electron and Hole Mobilities
v = E ; has the dimensions of v/E
Electron and hole mobilities of selected semiconductors
Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in highspeed devices?
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Drift Velocity, Mean Free Time, Mean Free Path
EXAMPLE: Given mp = 470 cm2/V·s,what is the hole drift velocity at E= 103 V/cm? What is tmp and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units.
Solution: n = mpE = 470 cm2/V·s 103 V/cm =4.7 105cm/s
tmp = mpmp/q =470 cm2/V ·s 0.39 9.11031 kg/1.61019 C
= 0.047 m2/V ·s 2.21012 kg/C = 11013s = 0.1 ps
mean free path = tmhnth ~ 1 1013 s 2.2107 cm/s
= 2.2106 cm = 220 Å = 22 nm
This is smaller than the typical dimensions of devices, but getting close.
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2.2.2 Mechanisms of Carrier Scattering
Phonon scatteringmobility decreases when temperature rises:
= q/m
T
vth T1/2
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
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.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Temperature Effect on Mobility
Question:
What Nd will make dmn/dT = 0 at room temperature?
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
J
p
unit
+
area
n
+
Jp = qpv A/cm2 or C/cm2·sec
If p = 1015cm3 and v = 104 cm/s, then
Jp= 1.61019C 1015cm3 104cm/s
=
Hole current density
EXAMPLE:
2.2.3 Drift Current and Conductivity
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
2.2.3 Drift Current and Conductivity
Jp,drift = qpv = qppE
Jn,drift = –qnv = qnnE
Jdrift = Jn,drift + Jp,drift = E=(qnn+qpp)E
conductivity (1/ohmcm) of a semiconductor is = qnn + qpp
1/ = is resistivity (ohmcm)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Relationship between Resistivity and Dopant Density
DOPANT DENSITY cm3
Ptype
Ntype
RESISTIVITY (cm)
= 1/
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EXAMPLE: Temperature Dependence of Resistance
(a) What is the resistivity () of silicon doped with 1017cm3 of arsenic?
(b) What is the resistance (R) of a piece of this silicon material 1m long and 0.1 m2 in crosssectional area?
Solution:
(a) Using the Ntype curve in the previous figure, we find that = 0.084 cm.
(b) R = L/A = 0.084 cm 1 m / 0.1 m2
= 0.084 cm 104 cm/ 1010 cm2
= 8.4 104
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
EXAMPLE: Temperature Dependence of Resistance
By what factor will R increase or decrease from T=300 K to T=400 K?
Solution: The temperature dependent factor in (and therefore ) is n. From the mobility vs. temperature curve for 1017cm3, we find that n decreases from 770 at 300K to 400 at 400K. As a result, R increases by
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Particles diffuse from a higherconcentration location to a lowerconcentration location.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
D is called the diffusion constant. Signs explained:
p
n
x
x
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Total Current – Review of Four Current Components
JTOTAL = Jn + Jp
Jn = Jn,drift + Jn,diffusion = qnnE+
Jp = Jp,drift + Jp,diffusion = qppE–
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
–
+
N type Si
N
E

E
E
E
(x)
(x)
(x)
f
c
v
0.7V
+
x
2.4 Relation Between the Energy Diagram and V, E
Ec and Ev vary in the opposite
direction from the voltage. That
is, Ec and Ev are higher where the voltage is lower.
E(x)=
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
2.5 Einstein Relationship between D and
Consider a piece of nonuniformly doped semiconductor.
Ec(x)
Ef
n
=

q
E
kT
Ev(x)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
=
m
D
p
p
q
2.5 Einstein Relationship between D and
dn
n
=

q
E
dx
kT
dn
at equilibrium.
=
+
=
m
J
qn
qD
0
E
n
n
n
dx
qD
=

m
n
E
0
qn
qn
E
n
kT
Similarly,
These are known as the Einstein relationship.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
What is the hole diffusion constant in a piece of silicon with p = 410 cm2 V1s1 ?
Solution:
Remember: kT/q = 26 mV at room temperature.
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2.6 ElectronHole Recombination
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Ec
Direct Recombination
is unfavorable in silicon
Recombination
centers
Ev
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Trap
Indirect band gap
Example: Si
Direct recombination is rare as k conservation is not satisfied
Direct band gap
Example: GaAs
Direct recombination is efficient as k conservation is satisfied.
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Rate of recombination (s1cm3)
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
A bar of Si is doped with boron at 1015cm3. It is exposed to light such that electronhole pairs are generated throughout the volume of the bar at the rate of 1020/s·cm3. The recombination lifetime is 10s. What are (a) p0, (b) n0 , (c) p’, (d) n’, (e) p , (f) n, and (g) the np product?
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Solution:
(a) What is p0?
p0= Na = 1015 cm3
(b) What is n0 ?
n0 = ni2/p0= 105 cm3
(c) What is p’?
In steadystate, the rate of generation is equal to the rate of recombination.
1020/scm3 = p’/
p’= 1020/scm3 · 105s = 1015 cm3
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
(d) What is n’?
n’= p’= 1015 cm3
(e) What is p?
p= p0 + p’= 1015cm3 + 1015cm3 = 2×1015cm3
(f) What is n?
n = n0 + n’= 105cm3 + 1015cm3 ~ 1015cm3 since n0 << n’
(g) What is np?
np ~ 21015cm3 ·1015cm3 = 21030 cm6>> ni2 = 1020 cm6. The np product can be very different from ni2.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
If n’ is negative, there are fewer electrons than the equilibrium value.
As a result, there is a net rate of thermal generation at the rate of n/t .
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Whenever n’ = p’ 0, np ni2.We would like to preserve and use the simple relations:
2.8 Quasiequilibrium and QuasiFermi Levels
Even when electrons and holes are not at equilibrium, within each group the carriers can be at equilibrium. Electrons are closely linked to other electrons but only loosely to holes.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
EXAMPLE: QuasiFermi Levels and LowLevel Injection
Consider a Si sample with Nd=1017cm3 and n’=p’=1015cm3.
(a) Find Ef .
n = Nd = 1017 cm3 = Ncexp[–(Ec– Ef)/kT]
Ec– Ef = 0.15 eV. (Ef is below Ec by 0.15 eV.)
Note: n’ and p’ are much less than the majority carrier
concentration. This condition is called lowlevel
injection.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
EXAMPLE: QuasiFermi Levels and LowLevel Injection
Now assume n = p = 1015 cm3.
(b) Find Efn and Efp .
n = 1.011017cm3 =
Ec–Efn = kT ln(Nc/1.011017cm3)
= 26 meV ln(2.81019cm3/1.011017cm3)
= 0.15 eV
Efn is nearly identical to Ef because nn0 .
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
p = 1015 cm3 =
Efp–Ev = kT ln(Nv/1015cm3)
= 26 meV ln(1.041019cm3/1015cm3)
= 0.24 eV
Ec
Ef
Efn
Efp
Ev
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
=
m
v
E
p
p

=
m
v
E
n
n
=
m
J
qp
E
,
p
drift
p
=
m
J
qn
E
,
n
drift
n
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
is the recombination lifetime.
n’ and p’ are the excess carrier concentrations.
n = n0+ n’
p = p0+ p’
Charge neutrality requires n’= p’.
rate of recombination = n’/ = p’/
Efnand Efpare the quasiFermi levels of electrons and holes.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)