Mark Lundstrom SOLUTIONS: ECE 656 Homework (Week 10) Mark Lundstrom Purdue University 1)
Show that the following relation (eqn. (4.119) in Fundamentals of Carrier Transport) is true: e
−
π
R
ρ S MN ,OP
+e
−
π
R
ρ S NO ,PM
= 1 Solution: From slide 6 in Lecture 18, Fall 2011 (or eqns. (4.117) and (4.118) of FCT): RMN ,OP =
⎛ ( a + b) ( b + c ) ⎞
ρ S ⎛ ( a + b) ( b + c ) ⎞ π RMN ,OP
ln ⎜
= ln ⎜
⎟→
⎟ π ⎝ b( a + b + c ) ⎠
ρS
⎝ b( a + b + c ) ⎠
(i) RNO,PM =
⎛ ( a + b) ( b + c ) ⎞
ρ S ⎛ ( a + b) ( b + c ) ⎞ π RNO,PM
ln ⎜
= ln ⎜
⎟→
⎟ π ⎝
ac
ρS
ac
⎠
⎝
⎠
(ii) From (i) and (iii) e
−
e
π
R
ρ S MN ,OP
−
π
R
ρ S MN ,OP
+e
−
+e
π
R
ρ S NO ,PM
−
π
R
ρ S NO ,PM
=
b ( a + b + c) ) + ac
(a + b)(b + c)
=
ab + b2 + bc + ac
= 1 ab + ac + b2 + bc
= 1 2) For n-­‐type, bulk silicon doped at N D = 1017 cm-­‐3 the room temperature mobility is 800
cm2/V-­‐s. Answer the following questions. Some potentially useful information is: N C = 3.23 × 1019 cm-­‐3 NV = 1.83 × 1019 cm-­‐3 EG = 1.11 eV υT = 1.05 × 10 7 cm/s. Estimate the Seebeck coefficient. Make reasonable assumptions, but clearly state them. Solution: ⎤
⎛k ⎞⎡ Δ
S = − ⎜ B ⎟ ⎢ n − ηF ⎥ ⎝ q ⎠ ⎣ k B TL
⎦
n0 = N C eηF → ηF = ln ( n0 N C ) = −5.78 (non-­‐degenerate semiconductor) ECE-­‐656 1 Fall 2015 Mark Lundstrom ECE 656 Homework Solutions (Week 10) (continued) Assume Δ n = 2k BTL (non-­‐degenerate, constant mfp) S = − ( 86 )[ 2 + 5.78 ] = −669 µV/K S = −669 µV/K 3) A Hall effect experiment is performed on a n-­‐type semiconductor with a length of 2.65 cm, a width of 1.70 cm, and a thickness of 0.0520 cm, in a magnetic field of 0.5 T. The current in the sample along its length is 200 µA. The potential difference along the length of the sample is 195 mV and across the width is 21.4 mV. 3a) What is the carrier concentration of the sample? Solution: Recall that in 2D: 

 
E = ρ S J n + ( ρ S µn rH ) J n × B E y = ρ S J y − ( ρ S µn rH ) J x Bz = − ( ρ S µn rH ) J x Bz ⎛ 1
⎞
⎛ r ⎞
VH = −WE y = ( ρ S µ n rH ) I x Bz = ⎜
µ n rH ⎟ I x Bz = ⎜ H ⎟ I x Bz ⎝ nS qµ n
⎠
⎝ qnS ⎠
which is the same as eqn. (4.110) in FCT r
VH = H Bz I qnS
nS Bz I
0.5 × 200 × 10−6
=
=
= 2.92 × 1016 cm -2 −19
−3
rH qVH 1.6 × 10 21.4 × 10
(
)
nS rH
2.92 × 1016 cm -2
= nH =
= 5.62 × 1017 cm -3 t
0.0520 cm
nH =
n
= 5.62 × 1017 cm -3 rH
Not the carrier concentration, but the “Hall carrier concentration.” ECE-­‐656 2 Fall 2015 Mark Lundstrom ECE 656 Homework Solutions (Week 10) (continued) 3b) What is the mobility? Solution: 195 × 10−3
Rxx =
= 975 Ω 200 × 10−6
L
Rxx = ρ S = 975 Ω W
W
1.70
1
ρ S = 975 Ω =
× 975= 625 =
L
2.65
nS qµ n
1
1
µn =
=
16
nS q ρ S rH × 2.92 × 10 × 1.6 × 10−19 × 625
1
rH µ n = µ H =
= 0.3 cm 2 /V-s 16
−19
2.92 × 10 × 1.6 × 10 × 625
µ H = 0.3 cm 2 /V-s Not the mobility, but the “Hall mobility.” 3c) If the scattering time is 1ps, find the magnetic field for which this classical analysis of Hall effect is no longer valid? Solution: We require: ω cτ << 1 qBz
τ << 1 m*
µn =
qτ
m*
τ=
(i) m*µ n
q
(ii) With (ii), (i) becomes qBz m*µ n
<< 1→ Bz µ n << 1 m* q
ECE-­‐656 3 Fall 2015 Mark Lundstrom ECE 656 Homework Solutions (Week 10) (continued) Using the Hall mobility as an estimate of the real mobility Bz µ n = 0.5 × 0.3 = 0.15 << 1 so this experiment is in the low B-­‐field regime. The largest the B-­‐field can be to be in the low field regime is Bz =
1
1
=
= 3.33 T µ n 0.3
Bz = 3.33 T 4) Contact resistances are important. They can complicate measurements of semiconductor transport parameters, and they can degrade device performance. The constant resistance is specified by the interfacial contact resistivity, ρC , in Ω-cm 2 . A very good value is ρC ≈ 10−8 Ω-cm 2 . Consider n+ Si at room temperature and doped to N D = 1020 cm -3 . What is the lower limit to ρC ? (Assume a fully degenerate semiconductor and use appropriate effective masses for the conduction band of Si.) Solution: The lower limit resistance must be the ballistic contact resistance: RB =
1
1
=
2
GB
2q h T
(
)
M
Assuming that one-­‐half of the ballistic resistance is associated with each of the two contacts: ρC =
RB A
h
= 2
2
4q T
1
M A
Assume a strongly degenerate semiconductor: ρC =
ECE-­‐656 h
1
2
4q T ( E F ) M ( E F ) A
4 Fall 2015 Mark Lundstrom The lower limit occurs when the transmission is one ρCmin =
h
1
2
4q M ( E F ) A
Need to find the Fermi level. Recall that at 0 K, n0 =
EF
∫ D ( E ) dE 3D
EC
D3D
n0 =
(m )
( E) =
*
DOS
3/2
π 23
EF
EF
∫ D ( E ) dE = ∫
3D
EC
n0 =
2 ( E − EC )
(m )
(
3π 2  3
) (E
π 
2 3
dE =
(
2 m*DOS
π 
)
3/2
EF
∫ (E − E )
1/2
C
2 3
dE EC
3/2
− EC ) 3/2
F
⎛ 2 3⎞
( EF − EC ) = m*1 ⎜⎝ 3π  ⎟⎠
2 2
DOS
M 3D ( E F ) =
2 ( E − EC )
3/2
*
DOS
EC
2 2 m*DOS
2/3
(n )
2/3
0
m*DOM
( E − EC ) 2π  2 F
m*DOM 1 ⎛ 3π 2  3 ⎞
M 3D ( E F ) = *
mDOS 2π  2 ⎜⎝ 2 2 ⎟⎠
2/3
(n )
2/3
0
m*DOM
= *
mDOS
⎛3 π ⎞
⎜
⎟
⎝ 8 ⎠
2/3
n02/3 For Si, we have to consider the ellipsoidal bandstructure: m*DOS = ( 6 )
2/3
(m m )
2
t
1/3

= 1.06m0 m*DOM = 2mt* + 4 mt*m* = 2.04m0 (See: Jeong, Changwook; Kim, Raseong; Luisier, Mathieu; Datta, Supriyo; and Lundstrom, Mark S., "On Landauer versus Boltzmann and full band versus effective mass evaluation of thermoelectric transport coefficients,” J. Appl. Phys., 107, 023707, 2010.) ECE-­‐656 5 Fall 2015 Mark Lundstrom ECE 656 Homework Solutions (Week 10) (continued) m*
M 3D ( E F ) = DOM
m*DOS
⎛3 π ⎞
⎜
⎟
⎝ 8 ⎠
2/3
n02/3 =
( )
2.04
(0.762) 1020
1.06
2/3
= 3.16 × 1013 cm -2
and finally: ρCmin =
h
1
1
= 6.48 × 103
= 2.1× 10−10 Ω-cm 2 2
4q M ( E F ) A
3.16 × 1013
ρCmin = 2.1× 10−10 Ω-cm 2
So even for a very good interfacial contact resistivity, T ( E F ) ≈ 0.01 For more on this topic, see: ECE-­‐656 J. Maassen, C. Jeong, A. Baraskar, M. Rodwell, and M. Lundstrom, “Full band calculations of the intrinsic lower limit of contact resistivity,” Appl. Phys. Lett., Vol. 102, 2013. DOI: 10.1063/1.4798238 6 Fall 2015