目 錄
摘要 2
TRANSFER MATRIX 3
計算穿隧機率的方法 7
結果與討論 9
參考資料 22
摘 要
利用 TRANSFER MATRIX 方法,計算中心摻雜量子井之雙位 障共振穿隧二極體,入射粒子的能量和穿隧機率的關係。模擬程式 是用 LabVIEW 5.0 所寫成的,我們可以輸入的參數有:位能障的數 目、位能井的寬度、位能障的寬度、位能障的高度、偏壓的大小、
量子井中心摻雜正離子的濃度等。
模擬程式除了可將入射粒子能量對穿隧機率的關係圖直接顯示 於瑩幕上外,亦提供存檔的功能,可將計算得到的數值依指定的路 徑、檔名存成文字檔。
TRANSFER MATRIX
透過任何一個一維的位能障(或位能井),如圖 (1)所 示 ,來考慮 TRANSFER MATRIX 的一般形式。
V
x1 x2 x3 X 圖(1) 任何形式的一維位能障
將 Schrödinger equation 表示成如下的形式
where
是位能,E 是粒子的總量, ,且 x1是局部極小點,
。 ,選擇 為方程式(A2)的一個解
考慮 x1周圍的波函數
同理,考慮 x2周圍的波函數
(
12 1)
0" + − ψ =
ψ k V
) ( ) ( ) ( )
2 (
1 1
2 1 2
1 m E V x V x V x V x
k − = −
= h
) 2 (
2
x mV
h
0 ) ( 1
1 x =
V
) (x1 U E>
(A1)
(A2)
0 )
1(x ≡
V ψ1
ik x
x ) = 1 ( ) =
(
1 1' 11
ψ
ψ
(A3)) ( )
( 1 1
1
1 x B x
A ψ ψ
ψ = +
) ( )
( 2 2
2
2 x B x
A ψ ψ
ψ = +
=
=
22 21
12 11 2
2 1
1
θ θ
θ θ θ
θ B A B
A
(A4)
(A5)
(A6)
共軛複數 也會得到和 相同的 Schrödinger equation。因此 得到 。
比較方程式(A6)和方程式(A7)得到
Now equation (A8) lead to
我們也可用其他的表示方法來表示方程式(A9),cos(h)=sech(S) where 0≦h≦2π and sinh(S)=tan(h),因此
The transfer matrix from x1 to x2 is θl:
It is easily related to θ:
By equation (A12),the transmission coefficient of the current is the same from the left to the right and versa.
方程式(A1)程(A3)的解可表示成
ψ ψ
2 2 2
2ψ Aψ
B + B1ψ1+A1ψ1
=
2 2
1 1
A B A
B θ (A7)
=
11 12
12 11
θ θ
θ
θ θ (A8)
−
−
=
~) exp(
) cosh(
~) exp(
) sinh(
~) exp(
) sinh(
) exp(
) cosh(
2 1
2 1
α β
β θ α
i S
i S
i S
i S
k
k (A9)
− −
=
) cos(
) ) exp(
tan(
~ ) exp(
) tan(
~ ) ) exp(
cosh(
) exp(
2 1
1 2
h h i
i
h h i
i
k k
β α α β
θ (A10)
=
1 1
2 2
A B A
B
θl (A11)
= −
11 12
12 11
2 1
θ θ
θ θ θ
k k
l (A12)
If in some vicinity of x1 , then , By equation (A1) and (A3), is a plain wave. There
Elsewhere the difference between and this plain wave is, in virtue of equation (A13), of order of .Generally the best choice for x1,x2 are the points where V(x) has local minimum. In the vicinity of x1
If at x>x2 there is no barrier and thus there is only a transmitted wave B2=0,then
The current transmission t and reflection r coefficients equal
If a barrier is symmetric (and, in particular, ),then the Schrödinger equation allows for odd and even (with respect to the middle of the barrier) solutions. This leads to
in equation (A9). Thus characterizes the asymmetry of the barrier. To elucidate the physical meaning of the quantities S,α,β in equation (A9), consider several special cases.
In a quasiclassical case, the comparison of equation (A16a) to the formulas 31(a1) for τ,ñ leads to
∫
− ′ ′ − ′ ′+ +
= ξ ξ ξ ξ ξ ψ ξ ξ
ξ
ψ 0 1
2 1 1
1
1( ) 1 ik ( )[V ( ) k ] ( )d (A13)
0 ) 0
1 (
1 =
−
= x x V
ξ (A14)
1 ≡0
V ψ1
)]
( exp[
)
( 1
1 x = ik x − x
ψ (A15a)
)
1( x ψ
∫ξ ξ′ ξ′
ξ 0V1( )d
4 1 1
1
1(x) − exp[ ik (x − x )] ~ (x − x )
ψ (A15b)
11 21 1
1 11
1
2 1
θ ρ θ
τ ≡ = θ ≡ =
A B A
A (A16a)
1 2 2 2
1 1
2 2 2
2 2
1 2 1
k k A
k A t k
A r B
τ ρ
=
=
=
=
(A16b)
k k k1= 2 ≡
2
~ π
β = (A17)
β π −β~≡
2
Thus,žS is the absolute value of the classically unavailable phase area (action).
The straightforward calculations in the case of a rectangular barrier, V(x)=0 at x<x1 , at x1<x<x2 and at x>x2 lead to θ with
A δ-function potential energy,
allows us to choose x1 = x0 - 0 , x2 = x0 + 0 , and leads to
2 1
2 1
)]
( 2 )
[(
) (
) ( )
2 (
) ( )
(
) (
1
1
2
1 1
1
2
1 1
1
x V m E
x k
dx x k dx
x k
dx x k dx
x k
dx x k S
S
x x
x x x
x
x x x
x
−
=
− +
≅
−
+
≅
−
=
≈
∫ ∫
∫ ∫
∫
′
′′
′
′′
″
′
h β π
α
(A18)
(A19)
(A20)
m x k
V 2
) ) (
(
2 2 2
1 +κ h
≡ V(x)≡(k22 +κ2)
d k
x x d k
k k k
k k
k d i k
k d d k
k i i
S
k d i k
k d d k
ik i
S
0
1 2 1
2 2
1 0
2 1 1
2
2 1 1
2 0
~
2
) sinh(
)]
( ) [(
) cosh(
) 1 ( 2 )
exp( 1 2 ) 1 exp( ~ ) sinh(
) sinh(
) ( ) [(
) cosh(
) 1 ( ) 2exp(
) 1 exp( ~ ) cosh(
+
=
−
=
−
′=
= +
− − +
′
−
=
+ − −
=
α α
κ κ κ κ
β
κ κ κ κ
α (A21)
(A22)
(A23)
(A24)
) ( )
(x v x x0
V = δ −
2
~ 1 2
) exp(
)
cosh( α = − β =π
k i iv
S
(A25)
(A26)
= 11∗
22 M
M M21=M12∗
k M k0 det =
計算穿隧機率的方法
V(z)
0 a a+b/2 a+b L z
double-barrier quantum well 的 potential 示意圖 z < 0
z > L
) ( )
) (
( )
( 0 0
L z ik L
z ik
z ik z
ik
e B e
B z
e A e
A z
−
−
−
− +
−
− +
+
= Ψ
+
= Ψ
=
− +
− +
A A M B
B
−
−
=
0 0
1 ) 1
0 , ( 1
1 2 1
ik ik
L S k
i k
i M
k M
k k
A
k T B
2 22
0
0 2
2
=
=
+ +
) (
)
( 11 0 22 21 0 12
22 k S
k i S k
S S k
M = + + −
21 2 12
0 12 2
0 11
0
) (
) (
4
k S S
k k
S S k
k k T
− +
+
=
In region Ⅰ
In region Ⅱ
In region Ⅲ
其中 V0:barrier height Va:bias voltage
Et:transverse energy in a GaAs layer
=
− −
=
−
=
a k a
k k
a k k
a k S
E E z
eV k
L V z V
z V
p p
p
p p
p
t p
a
cosh sinh
sinh cosh
) ( )
(
2
0
γ
γ γη γ
Ⅰ
{ }
= −
−
−
=
+ −
− +
=
−
cos 2 sin 2
sin 2 cos 2
) ( 0 2
) (
1 1
1
1 1
1 1
2 1
b b
b b
S
E z
eV E
L z V z z
V
p p
p
p p
p
t p
a
κ κ
κ
κ κ
κ η
κ
ε σ
Ⅱ
{ }
−
=
−
−
=
+
− +
=
−
cos 2 sin 2
sin 2 cos 2
) ( 0 2
) (
2 2
2
2 1
2 2
2 2
b b
b b
S
E z
eV E
L z V z z
V
p p
p
p p
p
t p
a
κ κ
κ
κ κ
κ η
κ
ε σ
Ⅲ
m*:effective mass S = SⅠSⅡSⅢSⅠ where
( )
h
GaAs GaAs
m m
z z m
*
*
*
2
) (
=
= η γ
( )
2
* 2
2
* 2
0
2 2
h h
eV a
E k m
E k m
= +
=
結 果 與 討 論
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25 0.30
-20 -15 -10 -5 0 5
lnT
Energy(eV)
number of barrier : 3 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25 0.30
-30 -25 -20 -15 -10 -5 0 5
lnT
Energy(eV)
number of barrier : 4 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25 0.30
-40 -30 -20 -10 0
lnT
Energy(eV)
number of barrier : 2 well width : 50Å
barrier width : 20Å barrier height : 0.5eV bias voltage : 0V
doped consistence : 0 #/cm2
0.0 0.1 0.2 0.3 0.4 0.5
-14 -12 -10 -8 -6 -4 -2 0 2
lnT
Energy(eV)
number of barrier : 3 well width : 50Å
barrier width : 20Å barrier height : 0.5eV bias voltage : 0V
doped consistence : 0 #/cm2
0.0 0.1 0.2 0.3 0.4 0.5
-20 -15 -10 -5 0 5
lnT
Energy(eV)
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0.02V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25 0.30
-15 -10 -5 0
lnT
Energy(eV)
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0.04V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25 0.30
-14 -12 -10 -8 -6 -4 -2 0 2
lnT
Energy(eV)
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0.06V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25
-12 -10 -8 -6 -4 -2 0 2
lnT
Energy(eV)
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0.08V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25
-14 -12 -10 -8 -6 -4 -2 0
lnT
Energy(eV)
number of barrier : 2 well width : 100Å barrier width : 40Å barrier height : 0.3eV bias voltage : 0.10V
doped consistence : 0 #/cm2
0.00 0.05 0.10 0.15 0.20 0.25
-14 -12 -10 -8 -6 -4 -2 0
lnT
Energy(eV)
圖(2)是在 barrier wdith = 40Å、well width = 100Å、barrier height
= 300meV、Nd = 0cm-2(量子井中心摻雜正離子的濃度)、bias voltage = 0.00V(black)、0.02V(read)、0.04V(green)、0.06V(blue)、0.08V(gyan)、
0.10V(magenta)的條件下,穿隧機率跟入射粒子能量的關係圖。從圖 中可以發現 bias voltage 越大 peak 數目會減少且向左移動。
圖(2) 在不同 bias voltage 的情況下,入射粒子的能量和穿隧機 率(transmission probability) 的關係圖。
圖(3)是在 barrier width = 40Å、well width = 100Å、barrier height = 300meV、bias voltage = 0.02V、Nd = 0cm-2(black)、2E11 cm-2(blue)的
0.0 0.1 0.2 0.3
- 2 0 - 1 5 - 1 0 -5 0 5
lnT
Energy(eV)
條件下,穿隧機率跟入射粒子能量的關係圖。從圖中我們發現,量 子井中心有無摻雜正離子對 spectrum 的影響並不大,正離子的作用 甚至可以不考慮。
圖(3) 量子井中心摻雜正離子的濃度對穿隧機率的影響
0 . 0 0 0 . 0 5 0 . 1 0 0 . 1 5 0 . 2 0 0 . 2 5 0 . 3 0
- 1 5 - 1 0 -5 0
lnT
Energy(eV)
參 考 資 料
1. R. Tsu and L. Eskai, Appl. Phys. Lett. 22, 562 (1973)
2. L. D. Landau and E. M. Lifshitz, Quantum Mechanics (Pergamon, New York, 1958), p. 63.
3. B. Jogai and K. L. Wang, Appl. Phys. Lett. 46 (2) (1985) 4. M. Ya. Azbel*, Physical Review B, 22, 8 (1983)
5. M. O. Vassel, Johnson Lee, and H. F. Lockwood, J. Appl. Phys. 54, 5026 (1983)
6. E. O. Kane, in Tunneling Phenomena in Solids, edited by E. Burstein and S. Lundqvist (Plenum, New York, 1969), Chap. 1.