modes and transmission solid state electronics

1. ECE-656: Fall 2011
Lecture 5:
Modes and Transmission
Professor Mark Lundstrom
Electrical and Computer Engineering
Purdue University, West Lafayette, IN USA
1
8/29/11

2. 2
key result from L4:
Lundstrom ECE-656 F11
I =
2q
h
γ E
( )π
D E
( )
2
f1
− f2
( )dE
∫
0 V
I
I device
( )
1
f E ( )
2
f E

3. Lundstrom ECE-656 F11 3
outline
1) Modes
2) Transmission
3) Discussion
4) Summary
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4. 4
modes or conducting channels
( )
1 2
2
2
q D
I f f dE
h
γ π
= −
∫ 0
U =
( ) ( ) 2 ?
E D E
γ π =
( )
( )
E
E
γ
τ
=

is the “broadening” and has units of energy.
( )
D E has units of 1/energy.
( ) ( ) ( )
2
E D E M E
γ π = is a number. We will show that M
is the number of conducting
channels at energy, E.
Lundstrom 2011

5. 5
a 2D ballistic channel
Let’s do an “experiment” to determine what γ (or τ) is.
Lundstrom 2011
EF1
( )
1
f E
1
F
E 2
F
E
( )
2
f E
x
y
θ
( ) ( ) 2 ?
M E D E
γ π
= =
I
I
L
W
υ

( ) ( )
2D
D E D E WL
= ( )
*
2 2
D
m
D E
π
=

(parabolic bands
with gV = 1)

6. 6
the “experiment”
( )
( )
( )
1 2
2
2
D E
q
I E f f
h
γ π −
( )
( )
( ) ( )
1 2
2
D E
N E f E f E
= +
 
 
(energy channels are independent)
( )
( )
1 2
1 2
f f
qN
I f f
γ
+
=
−

Apply V >> 0 to right contact, if f2 << f1 (injection from the
left contact only), then:
stored charge
current
qN
I
τ
γ
= = =

Lundstrom 2011

7. 7
transit time
stored charge
current
τ =
( ) ( )
( ) S x
I E qW n E E
υ+
=
S
N n WL
=
(ballistic transport)
x
( )
S
n x
L
( )
( )
( ) ( )
x
qN E L
E
I E E
τ
υ+
= =
“transit time”
Lundstrom 2011

8. 8
modes (conducting channels) in 2D
( )
1
f E
1
F
E 2
F
E
( )
2
f E
W x
y
υ

θ
cos
x
υ υ θ
+
=
/2
/2
cos 2
cos
d
π
π
θ θ
θ
π π
+
−
= =
∫
( )
*
2 C
E E
m
υ
−
=
I
I
L
Lundstrom 2011
2
x
υ υ
π
+
=

9. 9
a 2D ballistic channel
Lundstrom 2011
EF1
( )
1
f E
1
F
E 2
F
E
( )
2
f E
x
y
θ
( ) ( ) 2 ?
M E D E
γ π
= =
I
I
L
W
υ


10. 10
modes in 2D
( )
x
E
L
γ
τ υ+
= =
 
But how do we interpret
this result physically?
( ) ( ) ( ) 2
M E E D E
γ π
=
( )
( )
2
2
D
x
D E WL
M E
L
υ π
+
=

( ) ( )
2
4
x D
h
M E W D E
υ+
=
( ) ( )
2 2
4
D x D
h
M E W W D E
υ+
=
( ) ( )
1 1
4
D x D
h
M E D E
υ+
=
( ) ( )
3 3
4
D x D
h
M E A A D E
υ+
=
Lundstrom 2011

11. M E
( )= gV
W k
π
= gV
W
λB
E
( ) 2
11
interpretation
( )
*
2 2
D V
m
D E g
π
=

( )
( )
*
2 C
V
m E E
M E g W
π
−
=

( ) ( )
2
4
x D
h
M E W D E
υ+
=
( )
*
2 C
E E
m
υ
−
=
2
x
υ υ
π
+
=
( )
2 2
*
2
C
k
E k E
m
= +

Lundstrom 2011
M(E) is the number of electron half
wavelengths that fit into the width, W,
of the conductor.

12. 12
waveguide modes
Assume that there is one subband associated with confinement in the z-
direction. How many subbands (channels) are there associated with
confinement in the y-direction?
x
y
z
t
W
( , ) sin
x
ik x
y
x y e k y
ψ ∝
1,2,...
y
k m W m
π
= =
Lundstrom 2011
( )
( ) 2
B
W
M E
E
λ
= When ( ) ( )
1 2
B
M E E W
λ
=
→ =
2
B
λ
largest ky
2 2 *
2
y
E k m
= 

13. 13
waveguide modes
x
y
z
t
( , ) sin
x
ik x
y
x y e k y
ψ ∝
1,2,...
y
k m W m
π
= =
M = # of electron half
wavelengths that fit into W.
Lundstrom 2011
W
( )
( ) 2
B
W
M E
E
λ
= When ( ) ( )
2 B
M E E W
λ
=
→ =
largest ky
smaller ky

14. 14
density of states (for parabolic energy bands)
Lundstrom 2011
( ) ( ) ( )
*
2D 1
2
m
D E AD E A E ε
π
= = Θ −

( ) ( )
( )
( )
* *
3 2 3
2 C
D C
m m E E
D E D E E E
π
−
=
Ω =
Ω Θ −

( ) ( )
( )
( )
*
1 1
1
2
D
L m
D E L D E E
E
ε
π ε
= = Θ −
−

D3D
E
D2D
E
D1D
E
( )
( )
2 2 *
2
C
E k E k m
= + 

15. M2D
E
15
number of modes (for parabolic energy bands)
Lundstrom 2011
( ) ( )
( )
( )
*
1
2D
2
C
m E
M E W M E W H E E
ε
π
−
= −

( )
( )
2 2 *
2
C
E k E k m
= + 
( ) ( ) ( ) ( )
*
3D 2
2
C C
m
M E AM E A E E H E E
π
= = − −

M3D
E
( ) ( ) ( )
1D C
M E M E H E E
= = −
M1D
E
1

16. 16
summary
Lundstrom 2011
1) The density of states is used to compute carrier densities.
2) The number of modes (channels) is used to compute the
current.
3) The number of modes at energy, E, is proportional to the
average velocity (in the direction of transport) at energy, E
times the DOS(E).
4) M(E) depends on the bandstructure and on dimensionality.

17. 17
outline
Lundstrom 2011
1) Modes
2) Transmission
3) Discussion
4) Summary

18. 18
ballistic transport in 2D
EF1
I
L
2
F
E
( )
2
f E
W
x
y
υ

θ
( )
1
f E
1
F
E
(scattering from boundaries assumed
to be negligible – or specular)
I
Lundstrom 2011

19. I
W
19
diffusive transport in 2D
EF1
I
L
( )
1
f E
1
F
E 2
F
E
( )
2
f E
x
y
θ
X
X
X
X
X
X
X
• Electrons undergo a random walk as they go from left to right contact.
• Some terminate at contact 1, and some at contact 2.
• The average distance between collisions is the mfp, 
• “Diffusive” transport means Λ >> 
• The diffusive transit time will be much longer than the ballistic transit
time. Lundstrom 2011

20. 20
ballistic vs. diffusive transport
Lundstrom 2011
1) Ballistic:
Electrons travel without scattering from the injecting
contact to the absorbing contact.
( )
2
x E
υ υ
π
+
=
2) Diffusive:
Injected electrons undergo a random walk and leave
the device either through the contact from which they
were injected or from the other one.
?
x
υ+
=

21. 21
diffusive transport
γ
τ
=

?
τ =
Lundstrom 2011
Assume a channel that is much longer than the
mean-free-path for backscattering,
then, injected carriers diffuse to the other contact.
Fick’s Law of diffusion should apply.
A cm (2D)
S
n
dn
L J qD
dx
λ
>> = −

22. 22
transit time
L
x
( )
S
n x
∆
( )
0
S
n
∆
( ) 0
S
n L
∆ ≈
( )
0
S
n
n
I WqD
L
∆
=
( )
( )
2
0 2
0 2
S
n S n
Wq n L
qN L
I WqD n L D
τ
∆
= = =
∆
L λ
>>
Lundstrom 2011
Inject from the left contact, and collect at the right contact.

23. 23
diffusive transport
D
D
γ
τ
=

B
x
L
τ
υ+
=
Lundstrom 2011
( )
( )
( )
( )
( )
( ) ( ) ( )
B B
D B B
B D D
E E
E E T E E
E E
τ τ
γ γ γ
τ τ τ
= × = × ≡ ×

B
B
γ
τ
=
 2
2
D
n
L
D
τ =
γ D
E
( )π
D E
( )
2
= T E
( ) γ B
E
( )π
D E
( )
2










= T E
( )M E
( )
(ballistic) (diffusive)
T E
( )=
τB
E
( )
τD
E
( )
<< 1 (diffusive) T E
( )→ 1 (ballistic)

24. 24
diffusive transport
D
D
γ
τ
=

Lundstrom 2011
B
B
γ
τ
=

(ballistic) (diffusive)
B
x
L
τ
υ+
=
2
2
D
n
L
D
τ =
2
x
n
D
υ λ
+
=
T E
( )=
τB
E
( )
τD
E
( )
<< 1
T E
( )=
L υx
+
L2
2Dn
=
2Dn
υx
+
×
1
L
T E
( )=
λ E
( )
L
λ(E) = “mean-free-path for
backscattering”

25. 25
transmission
I =
2q
h
γ E
( )π
D E
( )
2
f1 − f2
( )dE
∫ ⇔ I =
2q
h
T E
( )M E
( ) f1 − f2
( )dE
∫
T E
( )=
λ E
( )
λ E
( )+ L
λ is the “mean-free-path
for backscattering”
This expression can be
derived with relatively few
assumptions.
L >> λ T =
λ
L
<< 1
1) Diffusive:
2) Ballistic: L << λ T = 1
3) Quasi-ballistic: L ≈ λ T < 1
Lundstrom 2011

26. 26
outline
Lundstrom 2011
1) Modes
2) Transmission
3) Discussion
4) Summary

27. 27
summary
Lundstrom 2011
( )
( )
( )
1 2
2
2
D E
q
I E f f dE
h
γ π
= −
∫
( ) ( )( )
1 2
2q
I T E M E f f dE
h
= −
∫
0 V
I
I device
( )
1
f E ( )
2
f E

28. 28
key parameters
Lundstrom 2011
( ) ( )( )
1 2
2q
I T E M E f f dE
h
= −
∫
( ) ( )
2 2
4
D x D
h
M E W W D E
υ+
=
( ) ( )
1 1
4
D x D
h
M E D E
υ+
=
( ) ( )
3 3
4
D x D
h
M E A A D E
υ+
=
modes transmission
T E
( )=
λ E
( )
λ E
( )+ L

29. 29
M(E) in 1D
Lundstrom 2011
( ) ( )
1 1
4
D x D
h
M E D E
υ+
=
1
2 1
( )
D
D E
π υ
=

( )
1
2
4
D
h
M E υ
π υ
=

M1D
E
( )= 1

30. 30
M(E) in 1D: physical picture
E
BW
E
π a
−π a
E
ε1
ε1
top
E
M E
( )
BW
ε1
ε1
top
1

31. 31
multiple subbands
E
M E
( )
ε1
ε2
1
2 2
*
2
i
i
k
E
m
ε
= +

( )
E k
k
2
ε
1
ε
2 2
*
2 i
k
m

2
T E
( )

32. 32
M(E) in 3D (parabolic bands)
Exercise: Determine M(E) for a parabolic band
semiconductor in 3D.

33. 33
M(E) vs. DOS (parabolic bands)
M3D
E
M2D
E
M1D
E
1
D3D
E
D2D
E
D1D
E
E
E
E
1 E

34. 34
DOS and conductivity effective mass
n = NCF1/2 ηF
( )
ηF =
EF − EC
kbTL
( )
3/2
*
3/2 3
2
4
D B
C
m k T
N
π
=

mD
*
= 62/3
ml
*
mt
*
( )
1/3
“density of states effective
mass” (silicon)
µ =
qτ
mc
*
1
mC
*
=
1
3
1
ml
*
+
2
mt
*








“conductivity effective
mass” (silicon)

35. 35
M(E) for graphene
Recall:
( ) ( )
4
x
h
M E W D E
υ+
=
for graphene:
2 2
( ) 2 F
D E E π υ
= 
2
x F
υ υ
π
+
=
2
( )
F
E
M E W
π υ
=


36. 36
M(E) for graphene (ii)
2
( )
F
E
M E W
π υ
=

F
E k
υ
= 
2
( ) F
F
k
M E W
υ
π υ
=


M(E) = 2
Wk
π
M(E) = gV
W
λB
2
( )
How do we interpret
this physically?

37. 37
M(E) vs. D(E) for graphene
E
( )
D E
( )
D E E
∝
2 2
( ) 2 F
D E E π υ
= 
F
E
E
( )
M E
( )
M E E
∝
F
E
( ) 2 F
M E W E π υ
= 

38. Lundstrom ECE-656 F11 38
outline
1) Modes
2) Transmission
3) Discussion
4) Summary

39. Lundstrom ECE-656 F11
39
summary
( ) ( )( )
1 2
2q
I T E M E f f dE
h
= −
∫
( ) ( )
2 2
4
D x D
h
M E W W D E
υ+
=
( ) ( )
1 1
4
D x D
h
M E D E
υ+
=
( ) ( )
3 3
4
D x D
h
M E A A D E
υ+
=
modes transmission
T E
( )=
λ E
( )
λ E
( )+ L

40. Lundstrom ECE-656 F11
40
additional information
Transmission is related to scattering, which we will
discuss later in this course.
To see how to compute M(E) for an arbitrary E(k), see:
Changwook Jeong, Raseong Kim, Mathieu Luisier, Supriyo Datta, and
Mark Lundstrom, “On Landauer vs. Boltzmann and Full Band vs.
Effective Mass Evaluation of Thermoelectric Transport Coefficients,” J.
Appl. Phys., 107, 023707, 2010.

41. Lundstrom ECE-656 F11 41
questions
1) Modes
2) Transmission
3) Discussion
4) Summary