ECE606: Solid State Devices Lecture 3 - Purdue University

47
Klimeck – ECE606 Fall 2012 – notes adopted from Alam ECE606: Solid State Devices Lecture 3 Gerhard Klimeck [email protected] Klimeck – ECE606 Fall 2012 – notes adopted from Alam Periodic Structure E Motivation

Transcript of ECE606: Solid State Devices Lecture 3 - Purdue University

Page 1: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ECE606: Solid State DevicesLecture 3

Gerhard [email protected]

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

PeriodicStructure

E

Motivation

Page 2: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

2 2

20

( )2

d dU x i

m dx dt

Ψ Ψ− + Ψ =ℏℏ

2 2

20

( )2

dU x E

m dx

ψ ψ ψ− + =ℏ

2 2

20

( )( ) ( ) ( )

2

iEt iEt iEtd xU x x i x

m

iEe e e

dx

ψ ψ ψ− − −

− + = −ℏ ℏ ℏ

ℏℏ

Assume

20

2 2

2( ) 0

mdE U

dx

ψ ψ+ − =ℏ

/( , ) ( )

iEtx xt eΨ ψ −= ℏ

Time-independent Schrodinger Equation

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Time-independent Schrodinger Equation

( ) x xx De Eeα αψ − += +

22

20

d

dxk

ψ ψ+ =[ ]02m Uk

E −≡

22

20

d

dxαψ ψ− =[ ]02m U E

α−

≡ℏ

( ) ( ) ( )ikx ikx

x Asin kx B cos kx

A e A e

ψ−

+ −

= +

≡ +

20

2 2

2( ) 0

mdE U

dx

ψ ψ+ − =ℏ

If E >U, then ….

If U>E, then ….

Page 3: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

A Simple Differential Equation

0

* dp dx

i dxΨ Ψ

∞ =

∫ℏ

2 2

20

( )2

dU x E

m dx

ψ ψ ψ− + =ℏ

• Obtain U(x) and the boundary conditions for a given problem.

• Solve the 2nd order equation – pretty basic

• Interpret as the probability of finding an electron at x

• Compute anything else you need, e.g.,

2 *ψ ψ ψ=

0

* dE dx

i dtΨ Ψ

∞ = −

∫ℏ

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Additional notes»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Page 4: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

PeriodicStructure

E

Case 1:

Free electron

E >> U

Case 2:Electron in infinite wellE << U

Case 3:

Electron in finite well

E < U

Full Problem Difficult: Toy Problems First

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

d 2ψdx2

+ k2ψ = 0 [ ]02m E Uk

−≡

U(x) ~ 0

E

1) Solution ( ) ( ) ( )ikx ikx

x Asin kx B cos kx

A e A e

ψ−

+ −

= +

≡ +

2) Boundary condition ( ) positive going wave

= negative going wave

ikx

ikx

x A e

A e

ψ +

−−

=

Case 1: Solution for Particles with E>>U

Page 5: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

2 2 2* A or Aψ ψψ + −= =U(x) ~ 0

E

( ) ( ) ( )ikx ikx

x Asin kx B cos kx

A e A e

ψ−

+ −

= +

≡ +

Probability:

( ) positive going wave

= negative going wave

ikx

ikx

x A e

A e

ψ +

−−

=

0

or -* dp dx k k

i dxΨ Ψ

∞ = =

∫ℏ

ℏ ℏMomentum:

Free Particle …

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

PeriodicStructure

E

Case 1:

Free electron

E >> U

Case 2:Electron in infinite wellE << U

Case 3:

Electron in finite well

E < U

Full Problem Difficult: Toy Problems First

Page 6: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Case 2: Bound State Problems

• Mathematical interpretation of Quantum Mechanics(QM)

» Only a few number of problems have exact mathematical solutions» They involve specialized functions

− ℏ2

2m∇2Ψ + VΨ = iℏ

∂∂t

Ψ

Ψn x( )= A1

2n n!Hn (Bx)e

−Cx 2

2Ψn x( )= Asin kn x( )Ψn r( )= ARn r( )Yn θ,ϕ( ) Ψn x( )= NAi ξn( )

V = ∞ V = ∞

V = 0

x = Lx = 0

V ∝ 1

r

V = ∞ V = ax

x = 0 x

V = 1

2kx2

0 xCoulomb Potential

by nucleus in an atom Particle in a box Harmonic Oscillator Triangular Potential Well

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1-D Particle in a Box – A Solution Guess

• (Step 1) Formulate time independent Schrödinger equation

− ℏ2

2m

d2

dx2 ψ x( )+ V x( )ψ x( )= Eψ x( ) V x( )=0 0< x < Lx

∞ elsewhere

where,

• (Step 2) Use your intuition that the particle will never exist outside the energy barriers to guess,

ψ x( )=0 0≤ x ≤ Lx

≠ 0 in the well

• (Step 3) Think of a solution in the well as:

ψn x( )= AsinnπLx

x

, n =1,2,3,…

V = ∞ V = ∞

V = 0x = Lxx = 0

Page 7: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1-D Particle in a Box – Visualization

• (Step 4) Plot first few solutions

ψn x( )= AsinnπLx

x

, n =1,2,3,…

ψn x( )2= A2 sin2 nπ

Lx

x

, n =1,2,3,…

x = Lxx = 0

n =1

x = Lxx = 0

n = 2

x = Lxx = 0

n = 3

x = Lxx = 0

n = 4

• (Step 5) Plot corresponding electron densities

The distribution of SINGLE particle

x = Lxx = 0

n = 2

x = Lxx = 0

n = 3

x = Lxx = 0

n = 4

ONE particle => density is normalized to ONE

Matches the condition we guessed at step 2!But what do the NEGATIVE numbers mean?

x = Lxx = 0

n =1

“s-type” “p-type” “d-type” “f-type”

ψ x( )=0 0≤ x ≤ Lx

≠ 0 in the well

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1-D Particle in a Box – Normalization to ONE particle

(Step 6) Normalization (determine the constant A)Method 1) Use symmetry property of sinusoidal function

ψn x( )2= A2 sin2 nπ

Lx

x

x = Lxx = 0 x = Lxx = 0

A2 Area( )=1= Lx

2× A2

∴ A = 2Lx

1= ψn x( )2dx

0

Lx∫ = A2 sin2 nπLx

x

dx

0

Lx∫ = A2

1− cos2nπx

Lx

20

Lx∫ dx = A2 Lx

2

∴ A = 2

Lx

∴ψn x( )= 2

Lx

sinnπLx

x

,

n =1,2,3,…

0 < x < Lx

ψn x( )2

0 ~ LxMethod 2) Integrate over

1= ψn x( )2dx

0

Lx∫ = A2 sin2 nπLx

x

dx

0

Lx∫ = A2

1− cos2nπx

Lx

20

Lx∫ dx = A2 Lx

2

Area( )=1= Lx

2× A2

∴ A = 2Lx

Page 8: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1-D Particle in a Box – The Solution

(Step 7) Plug the wave function back into the Schrödinger equation

ψn x( )= 2

Lx

sinnπLx

x

− ℏ2

2m

d2

dx2 ψ x( )+ V x( )ψ x( )= Eψ x( )

ℏ2

2m

n2π 2

Lx2

2

Lx

sinnπLx

+ 0× 2

Lx

sinnπLx

= En

2

Lx

sinnπLx

n =2

n = 3

n = 4

n = 1

ψn x( )= 2Lx

sinnπLx

x

En = ℏ2π 2

2mLx2 n2

n =1,2,3,…, 0< x < Lx

Discrete Energy Levels!

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1-D Particle in a Box – Quantum vs. Macroscopic

• Quantum world → Macroscopic world» What will happen with the discretized energy levels if we increase the length of the box?

• Energy level spacing goes smaller and smaller as physical dimension increases.

• In macroscopic world, where the energy spacing is too small to resolve, we see continuum of energy values.

• Therefore, the quantum phenomena is only observed in nanoscale environment.

En = ℏ2π 2

2mLx2 n2

Lx = 5nm Lx = 50nm Lx = 50cm

Page 9: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Additional notes»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

PeriodicStructure

E

Case 1:

Free electron

E >> U

Case 2:Electron in infinite wellE << U

Case 3:

Electron in finite well

E < U

Full Problem Difficult: Toy Problems First

Page 10: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

( ) 0

( ) 0

x

x

ψψ

= −∞ == +∞ =

d 2ψdx2

+ k 2ψ = 0

B B

B B

x x x x

x x x x

d d

dx dx

ψ ψ

ψ ψ

− +

− +

= =

= =

=

=

1)

Solution Ansatz

2N unknowns

for N regions

2) Boundary Conditions at the edge

Reduces 2 unknowns

3) Boundary Condition at each interface:

Set 2N-2 equations for

2N-2 unknowns (for continuous U)

Det (coefficient matrix)=0And find E by graphical or numerical solution

4) 2( , ) 1x E dxψ

−∞=∫5)

Normalization of unity probability

for wave function

( ) x xx De Eeα αψ − += +( ) ikx ikxx A e A eψ −

+ −= +

Five Steps for Closed System Analytical Solution

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

( ) 0

( ) 0

x

x

ψψ

= −∞ == +∞ =

U(x)E

2) Boundary

conditions

at the edge

0 a

sin cosA kx B kxψ = +

x xMe Ceα αψ − ++=x xDe Neα αψ − ++=

1)

E

Case 2: Bound-levels in Finite Well (steps 1,2)

Page 11: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

0 a

sin cosA kx B kxψ = +

xCeαψ = xDe αψ −=

B B

B B

x x x x

x x x x

d d

dx dx

ψ ψ

ψ ψ

− +

− +

= =

= =

=

=

C B

C kAα=

= −

sin( ) cos( )

cos( ) sin( )

a

a

A ka B ka De

kA ka kB ka De

α

αα

+ =− = −

sin( ) co

0 0 00 0 0

0 0

00

s( )

cos( ) sin(

1

)

1

/

a

a

A

B

Cka ka e

ka ka e k D

αα

α−

=

−−

3) Boundary at each interface

Case 2: Bound-levels in Finite Well (steps3)

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

0 a

sin cosA kx B kxψ = +

xCeαψ = xDe αψ −=

sin( ) co

0 0 00 0 0

0 0

00

s( )

cos( ) sin(

1

)

1

/

a

a

A

B

Cka ka e

ka ka e k D

αα

α−

=

−−

2 (1 )tan( )

2 1a

ξ ξα ξξ

−=

det (Matrix)=0

02

0

2 U

U

E mξ α≡ ≡ℏ

Only unknown is E

(i) Use Matlab function (ii) Use graphical method

Case 2: Bound-levels in Finite Well (step 4)

Page 12: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

2 5x x= +

2 5y x= +21y x=

y

x

0

2 (1 )

2a ( )

1t n a

ξ ξξ

α ξ −=

ζ=E/U

E1x0

Case 2: Bound-levels in Finite Well (steps 4 graphical)

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ζ=E/U

0 a

E1

0

2 (1 )

2a ( )

1t n a

ξ ξξ

α ξ −=

ζ=E/U

E1

Obtained the eigenvalues=> could stop here in many cases

Case 2: Bound-levels in Finite Well (steps 4 graphical)

Page 13: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

sin( ) co

0 0 00 0 0

0 0

00

s( )

cos( ) sin(

1

)

1

/

a

a

A

B

ka ka e

ka

C

DDe k

k

ka

α

α

α

α

=

− − xCeαψ = xDe αψ −=

sin cosA kx B kxψ = +

2 (1 )tan( )

2 1a

ξ ξα ξξ

−=

det (Matrix)=0

02

0

2 U

U

E mξ α≡ ≡ℏ

=> Linear dependent system=> Only 3 variables are unique=> One variable is undeterminedLet’s assume A can be freely Chosen => can get B,C,D

Need another boundary condition!

Case 2: Bound-levels in Finite Well (steps 5 wavefunction)

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

=> Linear dependent system=> Only 3 variables are unique=> One variable is undeterminedLet’s assume A can be freely Chosen => can get B,C,D

Non-linear => no simple linear algebra expression

sin( ) co

0 0 00 0 0

0 0

00

s( )

cos( ) sin(

1

)

1

/

a

a

A

B

ka ka e

ka

C

DDe k

k

ka

α

α

α

α

=

− −

1

cos( )

0 0

0 0

sin(

1

)

1

0 aka

B

C kA

D A kae α

α−

− = − −

cos(

0 0

0 0

sin

1

)) 0

1

(a

B

C kA

D kae Aka α

α−

= − −

−xCeαψ = xDe αψ −=

sin cosA kx B kxψ = +

2

1dxψ∞

−∞= ⇒∫

( ) ( )20 2 2 2 2

01

ax x

ae dx A sin kx B sin kx dx eDC dxα α∞ −

−∞ + + + = ∫ ∫ ∫

Another boundary condition!

Step 5: Wave-functions

Get “A”

Get “B,C,D”

Page 14: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

xCeαψ = xDe αψ −=

sin cosA kx B kxψ = +

Key Summary of a Finite Quantum Well

• Problem is analytically solvable • Electron energy is quantized and wavefunction is lo calized• In the classical world:

» Particles are not allowed inside the barriers / wal ls => C=D=0

• In the quantum world:» C and D have a non-zero value!» Electrons can tunnel inside a barrier

SiO2 SiSi

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Additional notes»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Page 15: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Transmission through a single barrierScattering Matrix approach

Incident : A

Reflected : B Incident : F

Transmitted : E

Define our system : Single barrier

One matrix each for each interface: 2 S-matrices

No particles lost! Typically A=1 and F=0.

D

C

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Tunneling through a single barrier

Wave-function each region,

Incident : AReflected : B Incident : F

Transmitted : ED

C

Page 16: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Single barrier case

Applying boundary conditions at each interface (x=0 and x=L) gives,

Which in matrix can be written as,

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Generalization to Transfer Matrix Method

• The complete transfer matrix

• In general for any intermediate set of layers, the TMM is expressed as:

• For multiple layers the overall transfer matrix will be

• Looks conceptually very simple and analytically pleasing• Use it for your homework assignment for a double barrier structure!

An−1+

An−1−

=M11 M12

M21 M22

An+

An−

AB F

EDC

Page 17: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Single barrier case

Transmission can be found using the relations between unknown constants,

Case: E<Vo

Case(γL<<1): Weak barrier

Case(γL large): Strong barrier

Case: E>Vo

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Single barrier : Concepts

•Transmission is finite under the barrier – tunneling!•Transmission above the barrier is not perfect unity!•Quasi-bound state above the barrier. Transmission goes to one.

•Computed with – http://nanohub.org/tools/pcpbt

Case: E>Vo

Page 18: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Effect of barrier thickness below the barrier

•Increased barrier width reduces tunneling probability•Thicker barrier increase the reflection probability below the barrier height.

•Quasi-bound states occur for the thicker barrier too.

•Computed with – http://nanohub.org/tools/pcpbt

Case: E>Vo

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Single Barrier – Key Summary

• Quantum wavefunctions can tunnel through barriers• Tunneling is reduced with increasing barrier height and

width

• Transmission above the barrier is not unity»2 interfaces cause constructive and destructive

interference »Quasi bound states are formed that result in unity

transmission

Page 19: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Additional notes»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double Barrier Transmission:Scattering Matrix approach

Left Incident

Reflected Right Incident

Transmitted

Define our system : Double barrier

One matrix each for each interface: 4 S-matrices

No particles lost! Typically Left Incident wave is normalized to one.Right incident is assumed to be zero.

Also this problem is analytically solvable! => Homework assignment

Page 20: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Reminder: Single barrier

•Transmission is finite under the barrier – tunneling!•Transmission above the barrier is not perfect unity!•Quasi-bound state above the barrier. Transmission goes to one.

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double barrier: Concepts• Double barriers allow a transmission probability of one / unity for discrete energies• (reflection probability of zero) for some energies below the barrier height.• This is in sharp contrast to the single barrier case • Cannot be predicted by classical physics.

Page 21: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double barrier: Quasi-bound states

• In addition to states inside the well, there could be states above the barrier height.• States above the barrier height are quasi-bound or weakly bound.• How strongly bound a state is can be seen by the width of the transmission peak.• The transmission peak of the quasi-bound state is much broader than the peak for the state inside the well.

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Effect of barrier height

•Increasing the barrier height makes the resonance sharper.•By increasing the barrier height, the confinement in the well is made stronger, increasing the lifetime of the resonance.

•A longer lifetime corresponds to a sharper resonance.

Page 22: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Effect of barrier thickness

•Increasing the barrier thickness makes the resonance sharper.

•By increasing the barrier thickness, the confinement in the well is made stronger, increasing the lifetime of the resonance.

•A longer lifetime corresponds to a sharper resonance.

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double barrier energy levels Vs Closed system

The well region in the double barrier case can be thought of as a particle in a box.

Page 23: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Particle in a box

• The time independent Schrödinger equation is

− ℏ2

2m

d2

dx2 ψ x( )+ V x( )ψ x( )= Eψ x( ) V x( )=0 0< x < Lx

∞ elsewhere

where,

• The solution in the well is:

ψn x( )= AsinnπLx

x

, n =1,2,3,…

ψn x( )= 2Lx

sinnπLx

x

En = h2π 2

2mLx2 n2

n =1,2,3,K , 0< x < Lx

• Plugging the normalized wave-functions back into the Schrödinger equation we find that energy levels are quantized.

n = 2

n = 3

n = 4

n = 1

V = ∞ V = ∞

V = 0x = Lxx = 0

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double barrier & particle in a box

• Double barrier: Thick Barriers(10nm), Tall Barriers(1eV), Well(20nm).• First few resonance energies match well with the particle in a box

energies.• The well region resembles the particle in a box setup.

• Green: Particle in a box energies.

• Red: Double barrier energies

Page 24: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Open systems Vs closed systems

• Green: Particle in a box energies.

• Red: Double barrier energies

• Double barrier: Thinner Barriers(8nm), Shorter Barriers(0.25eV), Well(10nm).• Even the first resonance energy does not match with the particle in a box energy.• The well region does not resemble a particle in a box. • A double barrier structure is an OPEN system, particle in a box is a CLOSED

system.

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Reason for deviation?

• Wave-function penetrates into the barrier region.

• The effective length of the well region is modified.

• The effective length of the well is crucial in determining the energy levels in the closed system.

Potential profile and resonance energies using tight-binding.

First excited state wave-function amplitude using tight binding.

Ground state wave-function amplitude using tight binding.

En = h2π 2

2mLwell2 n2

n =1,2,3,K , 0< x < Lwell

Page 25: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Double Barrier Structures - Key Summary

• Double barrier structures can show unity transmission for energies BELOW the barrier height»Resonant Tunneling

• Resonance can be associated with a quasi bound state»Can relate the bound state to a particle in a box

»State has a finite lifetime / resonance width

• Increasing barrier heights and widths:»Increases resonance lifetime / electron residence time

»Sharpens the resonance width

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Additional notes»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Page 26: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1 Well => 1 Transmission Peak

• Vb=110meV, W=6nm, B=2nm

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

2 Wells => 2 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm Bonding/Anti-bonding State

Page 27: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

3 Wells => 3 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

4 Wells => 4 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Page 28: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

5 Wells => 5 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

6 Wells => 6 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Page 29: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

7 Wells => 7 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

8 Wells => 8 Transmission Peaks

• Vb=110meV, W=6nm, B=2nm

Page 30: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

9 Wells => 9 Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

19 Wells => 19 Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

Page 31: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

29 Wells => 29 Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

39 Wells => 39 Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

Page 32: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

49 Wells => 49 Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

N Wells => N Transmission Peaks

• Bandpass filter formed• Band transmission not symmetric

• Bandpass sharpens withincreasing number of wells

Page 33: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

1 Well => 1 Transmission Peak => 1 State

• Bandpass filter formed• Band transmission not symmetric

• Bandpass sharpens withincreasing number of wells

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

2 Wells => 2 Transmission Peaks => 2 States

• Bandpass filter formed• Band transmission not symmetric

Page 34: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

3 Wells => 3 Transmission Peaks => 3 States

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

4 Wells => 4 Transmission Peaks => 4 States

• Bandpass filter formed• Band transmission not symmetric

Page 35: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

5 Wells => 5 Transmission Peaks => 5 States

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

6 Wells => 6 Transmission Peaks => 6 States

• Bandpass filter formed• Band transmission not symmetric

Page 36: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

7 Wells => 7 Transmission Peaks => 7 States

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

8 Wells => 8 Transmission Peaks => 8 States

• Bandpass filter formed• Band transmission not symmetric

Page 37: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

9 Wells => 9 Transmission Peaks => 9 States

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

19 Wells => 19 Transmission Peaks => 19 States

• Bandpass filter formed• Band transmission not symmetric

Page 38: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

29 Wells => 29 Transmission Peaks => 29 States

• Bandpass filter formed• Band transmission not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

39 Wells => 39 Transmission Peaks => 39 States

• Bandpass filter formed• Band transmission not symmetric

Page 39: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

49 Wells => 49 Transmission Peaks => 49 States

• Bandpass filter formed• Band transmission not symmetric

• Cosine-like band formed• Band is not symmetric

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

N Wells => N Transmission Peaks => N States

• Bandpass filter formed• Band transmission not symmetric

• Cosine-like band formed• Band is not symmetric

Page 40: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

N Wells => N States => 1 Band

• Vb=110meV, W=6nm, B=2nm => ground state in each well=> what if there were excited states in each well => Vb=400meV

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

N Wells => 2N States => 2 Bands

Vb=110meV, W=6nm, B=2nm

Vb=400meV, W=6nm, B=2nm

Page 41: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

N Wells => 2N States => 2 Bands

Vb=110meV, W=6nm, B=2nm

Vb=400meV, W=6nm, B=2nm

1 state/well => 1band

2 states/well => 2bands

Can we get more states/well?

=> Increase well width

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

X States/Well => X Bands

Vb=110meV,W=6nm, B=2nm

Vb=400meVW=6nm, B=2nm

1 state/well=> 1 band

2 states/well => 2 bands

3 states/well => 3 bands

Vb=400meVW=10nm, B=2nm

Page 42: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

X States/Well => X Bands

Vb=110meV,W=6nm, B=2nm

Vb=400meVW=6nm, B=2nm

1 state/well=> 1 band

2 states/well => 2 bands

3 states/well => 3 bands

Vb=400meVW=10nm, B=2nm

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Formation of energy bands

•Each quasi-bond state will give rise to a resonance in a well. (No. of barriers -1)

•Degeneracy is lifted because of interaction between these states.•Cosine-like bands are formed as the number of wells/barriers is increased

•Each state per well forms a band•Lower bands have smaller slope = > heavier mass

Page 43: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Procedure Summary• Additional notes

»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

( ) 0

( ) 0

x

x

ψψ

= −∞ == +∞ =

d 2ψdx2

+ k 2ψ = 0

B B

B B

x x x x

x x x x

d d

dx dx

ψ ψ

ψ ψ

− +

− +

= =

= =

=

=

1)

Solution Ansatz

2N unknowns

for N regions

2) Boundary Conditions at the edge

Reduces 2 unknowns

3) Boundary Condition at each interface:

Set 2N-2 equations for

2N-2 unknowns (for continuous U)

Det (coefficient matrix)=0And find E by graphical or numerical solution

4) 2( , ) 1x E dxψ

−∞=∫5)

Normalization of unity probability

for wave function

( ) x xx De Eeα αψ − += +( ) ikx ikxx A e A eψ −

+ −= +

Five Steps for Closed System Analytical Solution

Page 44: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Open System: Generalization to Transfer Matrix Method

• The complete transfer matrix

• In general for any intermediate set of layers, the TMM is expressed as:

• For multiple layers the overall transfer matrix will be

• Looks conceptually very simple and analytically pleasing• Use it for your homework assignment for a double barrier structure!

An−1+

An−1−

=M11 M12

M21 M22

An+

An−

AB F

EDC

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

Presentation Outline

• Time Independent Schroedinger Equation• Analytical solutions of Toy Problems

»(Almost) Free Electrons»Tightly bound electrons – infinite potential well»Electrons in a finite potential well»Tunneling through a single barrier

• Numerical Solutions to Toy Problems»Tunneling through a double barrier structure»Tunneling through N barriers

• Procedure Summary• Additional notes

»Discretizing Schroedinger’s equation for numerical implementations

Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool

http://nanohub.org/tools/pcpbt

Page 45: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

d 2ψdx2

+ k2ψ = 0

x

U0(x)

k > 0α = ik

[ ]02k m E U( x ) /≡ − ℏ

α = ik

Numerical solution of Schrodinger Equation

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

U(x)

x0 1 2 3 N+1 N

a

2 2

20

(2

)d

Em dx

U xψ ψ ψ− + =ℏ

U1

ψ3

(1) Define a grid …

Page 46: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

( ) ( )0 0

2 2

00 22x a x a

d a da ...

dx dx

xx aψ ψψ ψ

= =

= + + ++

( ) ( )0 0

2 2

00 22x a x a

d a da ...

dx dx

xx aψ ψψ ψ

= =

= − + −−

( ) ( ) ( )0

22

20 00 2x a

xd

ax a adx

xψψ ψψ=

+ −+ − =

d 2ψdx2

i

= ψ i−1 − 2ψ i +ψ i+1

a2

Second Derivative on a Finite Mesh

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

(2) Express equation in Finite Difference Form

( )2

20 2

( )Ud

t xa Edx

ψ ψ ψ− + =

0 x1 2 N+1

d 2ψdx2

i

= ψ i−1 − 2ψ i +ψ i+1

a2

( )0 1 0 0 12i ii i iEUt t tψ ψ ψ ψ− + − + + − =

2

0 202

tm a

≡ ℏ

ψ(0) = 0 ψ(L) = 0N unknowns

i-1 i i+1

Page 47: ECE606: Solid State Devices Lecture 3 - Purdue University

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

(3) Define the matrix …

−t0ψ i−1 + 2t0 + ECi( )ψ i − t0ψ i+1[ ]= Eψ i

Hψψψψ = Eψψψψ

N x N N x 1

(i = 2, 3…N-1)

−t0 (2t0 + EC1) −t0

−t0 (2t0 +ECi) −t0 = E

( )0 0 0 1 0 22 Ci it t E t Eψ ψ ψ ψ− + + − = (i = 1)

( )0 1 0 0 12N Ci N N it t E t Eψ ψ ψ ψ− +− + + − = (i = N)

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

(4) Solve the Eigen-value Problem

U(x)

1 x32 4 N (N-1)

a

Hψψψψ = Eψψψψ

ε1

ε2

ε3

ε4Eigenvalueproblem; easily solved with MATLAB& nanohubtools

wrong