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Tunneling Outline - Review: Barrier Reflection - Barrier Penetration (Tunneling) - Flash Memory 1
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### Transcript of 6.007 Lecture 42: Tunneling - MIT OpenCourseWare · Example: Barrier Tunneling . Question: What...

• Tunneling

Outline -  Review: Barrier Reflection -  Barrier Penetration (Tunneling) -  Flash Memory

1

• A Simple Potential Step

Region 1 Region 2

CASE I : Eo > V

In Region 1: In Region 2:

ψA = Ae−jk1x

ψB = Be−jk1x

ψC = Ce−jk1x

E = Eo

E = 0

x = 0x

Eoψ = − �2

2m

∂2ψ

∂x2

(Eo − V )ψ = − �2

2m

∂2ψ

∂x2

k21 =2mEo�2

k22 =2m (Eo − V )

�2

V

2

• A Simple Potential Step

CASE I : Eo > V

is continuous: is continuous:

ψ1 = Ae−jk1x +Bejk1x ψ2 = Ce−jk2x

ψ1(0) = ψ2(0) A+B = C

∂ ∂ψ(0) =

∂x

kψ2(0) A

∂x− 2B =

ψ

∂ψC

k1∂x

Region 1 Region 2

ψA = Ae−jk1x ψC = Ce−jk1x

ψB = Be−jk1x

E = 0

x = 0x

V

3

• A Simple Potential Step

CASE I : Eo > V

B

A=

1− k2/k11 + k2/k1

=k1 − k2k1 + k2

C

A=

2

1 + k2/k1

=2k1

k1 + k2

A+B = C

A−B = k2k1

C

Region 1 Region 2

ψA = Ae−jk1x ψC = Ce−jk1x

ψB = Be−jk1x

E = Eo

E = 0

x = 0x

V

4

• Quantum Electron Currents

Given an electron of mass that is located in space with charge density and moving with momentum corresponding to

… then the current density for a single electron is given by

m

ρ = q |ψ(x)|2

< p > < v >= �k/m

J = ρv = q |ψ|2 (�k/m)

6

• A Simple Potential Step

CASE I : Eo > V

JtransmittedTransmission = T =

Jincident=

JCJA

=|ψC |2(�k2/m)|ψA|2(�k1/m) =

∣∣∣∣CA∣∣∣∣2k2

JreflectedReflection = R =

k1

Jincident=

JBJA

=|ψB |2(�k1/m)|ψA|2(�k1/m) =

∣∣∣∣BA∣∣∣∣2

B

A=

1− k2/k11 + k2/k1

C

A=

2

1 + k2/k1

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

ψC = Ce−jk1x

E = Eo

E = 0

x = 0x

V

7

• A Simple Potential Step

CASE I : Eo > V

1

1

BReflection = R =

∣∣∣∣2

A

∣∣∣ =∣∣k1 − k2∣ ∣∣

2

k1 + k2

∣∣∣

Transmission = T = 1

∣−R

4k1k2=

T

R |k1 + k2|2T +R = 1

k2k1

=

√1− V

EoEo = V Eo = ∞

Region 1 Region 2

ψA = Ae−jk1x ψC = Ce−jk1x

ψB = Be−jk1x

E = 0

x = 0x

V

8

• IBM Almaden STM of Copper

9

• IBM Almaden Image originally created by the IBM Corporation.

10

• IBM Almaden Image originally created by the IBM Corporation.

11

• A Simple Potential Step

CASE II : Eo < V

In Region 1: In Region 2:

2�

Eoψ = − ∂2ψ

2m ∂x2

(Eo − V )ψ = − �2

2m

∂2ψ

∂x2

k21 =2mEo�2

κ2 =2m (Eo − V )

ψC = Ce−κx

�2

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = EoE = 0

x = 0x

V

12

• A Simple

ψ2 = Ce−κx

Potential Step

is continuous: is continuous:

CASE II : Eo < V

ψ1 = Ae−jk1x +Bejk1x

ψ1(0) = ψ2(0) A+B = C

∂ ∂ψ(0) =

∂x

ψ

∂ψψ2(0)

∂x

κA

∂x−B = −j

ψC = Ce−κx

Ck1

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = EoV

E = 0x

x = 0

13

• A Simple Potential Step

CASE II : Eo < V

B

Total reflection Transmission must be zero

A=

1 + jκ/k1 C

1− jκ/k1 A=

2

1− jκ/k1

R =

∣∣∣∣BA∣∣∣∣2

= 1 T = 0

A+B = C

A−B = −j κ

ψC = Ce−κx

Ck1

Region 1 Region 2

ψA = Ae−jk1x

ψB = Be−jk1x

E = EoE = 0

x = 0x

V

14

• 2a = L

T = 0

Quantum Tunneling Through a Thin Potential Barrier

Total Reflection at Boundary

Frustrated Total Reflection (Tunneling)

R = 1 T = 0

15

• CASE II : Eo < V

Region 1 Region 2 Region 3

In Regions 1 and 3: In Region 2:

A Rectangular Potential Step

for Eo < V :

κxψA = Ae

−jk1x ψC = Ce−

ψB = Bejk1x

ψF = Fe−jk1x

ψD = Deκx

VE = Eo

E = 0 −a a0

Eoψ = − �2

2m

∂2ψ

∂x2

(Eo − V )ψ = − �2

2m

∂2ψ

∂x2

k21 =2mEo�2

κ2 =2m(V − Eo)

�2

T =

∣∣∣∣FA∣∣∣∣2

=1

1 + 14V 2

Eo(V−Eo) sinh2(2κa)

16

• A Rectangular Potential Step

x=0 x=L

2a = L

Real part of Ψ for Eo < V, shows hyperbolic

(exponential) decay in the barrier domain and decrease

in amplitude of the transmitted wave.

for Eo < V :

Transmission Coefficient versus Eo/V for barrier with

T =

∣∣∣∣FA∣∣∣∣2

=1

1 + 14V 2

Eo(V−Eo) sinh2(2κa)

T =

∣∣∣∣FA∣∣∣∣2

≈ 11 + 14

V 2

Eo(V−Eo)e−4κa

sinh2(2κa) =[e2κa − e−2κa]2 ≈ e−4κa

2m(2a)2V/� = 16

17

• Flash Memory

Electrons tunnel preferentially when a voltage is applied

Erased 1

Stored Electrons

Programmed 0

Image is in the public domain

Insulating Dielectric

CONTROL GATETunnel Oxide

FLOATING GATE

SOURCE CHANNEL

Substrate

Channel

Floating Gate

DRAIN

18

• MOSFET: Transistor in a Nutshell

Conducting Channel

Tunneling causes thin insulating layers to become leaky !

Image is in the public domain

Conduction electron flow

Control Gate

Semiconductor

19

UNPROGRAMMED PROGRAMMED

To obtain the same channel charge, the programmed gate needs a higher control-gate voltage than the unprogrammed gate

How do we WRITE Flash Memory ?

CONTROL GATE

FLOATING GATE

SILICON

CONTROL GATE

FLOATING GATE

20

• 0 L

V0

x

Eo metal metal

air gap

Example: Barrier Tunneling

Question: What will T be if we double the width of the gap?

•  Let s consider a tunneling problem:

An electron with a total energy of Eo= 6 eV approaches a potential barrier with a height of V0 = 12 eV. If the width of the barrier is L = 0.18 nm, what is the probability that the electron will tunnel through the barrier?

L = 2a

T =

∣∣∣∣FA∣∣∣∣2

≈ 16Eo(V − Eo)V 2

e−2κL

κ =

√2me�2

(V − Eo) = 2π√

2meh2

(V − Eo) = 2π√

6eV

1.505eV-nm2≈ 12.6 nm−1

T = 4e−2(12.6 nm−1)(0.18 nm) = 4(0.011) = 4.4%

21

• Multiple Choice Questions

Consider a particle tunneling through a barrier:

1. Which of the following will increase the likelihood of tunneling?

a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the particle

2. What is the energy of the particles that have successfully escaped ?

a. < initial energy b. = initial energy c. > initial energy

0 L

V

x

Eo

Although the amplitude of the wave is smaller after the barrier, no energy is lost in the tunneling process

22

• Application of Tunneling: Scanning Tunneling Microscopy (STM)

Due to the quantum effect of barrier penetration, the electron density of a material extends beyond its surface:

material STM tip One can exploit this to measure the ~ 1 nm electron density on a material s surface:

Sodium atoms on metal:

STM images

Single walled carbon nanotube:

V E0

STM tip material

Image originally created by IBM Corporation

23

• Reflection of EM Waves and QM Waves

Then for optical material when μ=μ0:

= probability of a particular = probability of a particular photon being reflected electron being reflected

photonsP = �ω ×

s cm2

P =|E|2η

PreflectedR =

E=

Pincident

∣r∣o∣∣

2

Eio

∣∣∣∣

R =

∣∣B∣∣A∣∣∣∣2

=

∣∣∣∣k1 − k2k1 + k2∣∣∣∣2

=

∣∣∣∣n1 + n2

electronsJ = q ×

2

n1 + n2

∣∣∣∣

s cm2

J = ρv = q |ψ|2 (�k/m)

JreflectedR =

=| B |

Jincident |ψA|2

R =

∣∣B∣∣2

A

∣∣∣ =∣∣k1 − k2∣ ∣∣

2

k1 + k2

∣∣∣∣

24

• MIT OpenCourseWarehttp://ocw.mit.edu

6.007 Electromagnetic Energy: From Motors to LasersSpring 2011