1

UCONN ECE 4243/6243 Fall 2014 Nanoscience and Nanotechnology-I

L2 Energy levels in potential wells and density of states 1. Bonds and energy bands (page 38)

2

Bonds (page 43) Z

Y

X

S

Z

Y

X

P y

+-

one sp 3 orbitalformed due to an s

and a p y orbital

Z

4 sp 3 bonds

Tetrahedral primitive cell of Si lattice

3

L

V(x) = 0

0

0))x(VE(m2

dx

d22

2

2

h

Infinite Potential Well p.47

Solve Schrodinger Equation

In a region where V(x) = 0, using boundary condition that (x=0)=0, (x=L)=0. Plot the (x) in the 0-L region.

4

22

2 mE2

dx

d

0))x(VE(m2

dx

d22

2

In the region x = 0 to x = L, V(x) = 0

Let 2

2x

mE2k

xikxik xx BeAe The solution depends on boundary conditions.

Equation 2 can be written as D coskxx + C sinkxx.

At x=0, 00 D , so D= 0

(2)

xkC xsin*

(3)

Using the boundary condition (x=L) = 0, we get from Eq. 3 sin kxL = 0, L = Lx 2

2

mE

L

nk

nLk

x

xx

xxx

nx = 1,2,3… (4)

2

22

2

222

82 mL

nh

L

n

mE xx

n

5

22x

mE2k

xikxik xx BeAe The solution depends on boundary conditions. It satisfies boundary condition when Eq.6 is satisfied.

Page 52-53: Using periodic boundary condition (x+L) = (x), we get a different solution:

We need to have kx= x

x

L

n2 ---(6)

0))x(VE(m2

dx

d22

2

y

y

L

n2

Lz

nz2

If we write a three-dimensional potential well, the problem is not that much different ky =

n y= 1,2,3,…

kZ =

, n z = 1,2,3,….

)2

()2

()2

()(),,(Z

z

y

y

x

xnn L

znSin

L

ynSin

L

xnASinrzyx

The allowed kx, ky, kz values form a grid. The cell size for each allowed state in k-space is

ZYX

zyx

LLLV

VLLL

3)2(222

…(7)

…(8)

Density of States in 3D semiconductor film (pages 53, 54)

6

Density of states between k and k+dk including spin

N(k)dk =

dkk

3

2

)2(

42

…(10)

k

k+dk

Density of states between E and E+dE

N(E)dE=2 dE

E

kk

3

2

)2(

4

…11

E=Ec + cm

k

2

22 (12)

ee m

k

m

k

dk

dE 22

2

2 …(13)

N(E)dE = dEE

me 2/12/322

)2

(2

1

…14

VB

E

N(E)

dEEmn 2/12/3

22)

2(

2

1

N(E)dE = …15

Time dependent Schrodinger Equation p.68-69

7

8

9

Finite Potential Well p.70

10

In the barrier region

11

Wavefunction matching at well-barrier boundary

12

Wavefunction matching at well-barrier boundary

13

14

Electrons EC1 = 72.5 meV

Heavy Holes EHH1 = 22.3 meV

Light Holes ELH1 = 52.9 meV

15

αL/2

kL/2

Where, radius is

2

LV

2m02

eb

Electrons EC1 = 72.5 meVHeavy Holes EHH1 = 22.3 meVLight Holes ELH1 = 52.9 meV

Output: Equation (15) and (12), give k, α’ (or α). Once k is known, using Equation (4) we get: ew

22

2m

kE

(4)

EmL

eeff *

22

2

The effective width, Leff can be found by:

E1 is the first energy level.

16

2h

Summary of Schrodinger Equation in finite quantum well region.

In the well region where V(x) = 0, and in the barrier region V(x) = Vo = ∆Ec in conduction band, and ∆Ev in the valence band.

The boundary conditions are:

2L 2

L

0

AlxGa1-xAs AlxGa1-xAsGaAs

∆EV

∆EC

0

V(z)

z

-EG

∆EV

-EG+∆EV

z

∆Ec = 0.6∆Eg

∆Ev = 0.4∆Eg

2

Lz

2

Lz

2

Lz

dz

d

m

1

2

Lz

dz

d

m

1

eweb

(5) Continuity of the wavefunction

(6) Continuity of the slope

2ew2 E2m

k

20eb2 E)(V2m

α

kzsin Ckz cosC(z) '22

2Lzα

1eC(z)

2

Lz

Eigenvalue equation:

2

Lk

2

Lα

m

m

2

kLtan

eb

ew

(1)

2

02eb

2

eb

ew

2

2

LV

2m

2

αL

m

m

2

Lk

2

02eb

2'2

2

LV

2m

2

Lα

2

Lk

eb

ew'

m

mα

z)y,(x,E z)y,(x,V(z)2m

22

C. 2D density of state (quantum wells) page 84

17

)(2

EfdkdkV nz

yx

y

y

x

xyx L

dk

L

dk

Vdkdk

V /2/2

22

Carrier density number of states

(1)Without f(E) we get density of states

Quantization due to carrier confinement along the z-axis.

nz

nz

e EEHL

m)(

2

kt

dkt + kt

18

Si-SubstrateEc

Ev

EF

Vacuum

SiO

2

SiO

2

Si D

ot

Control Gate

qSi

qSiO2

qs

Ec

Eg

t1DNCt2

A

BC

DE

H

FG

Si Substrate

DrainSource

SiO2

Control Gate

Si Dots

Floating Gate

Fig .2. Schematic cross-section of a floating

Fig .4 energy band model to program a

Examples of Quantum dot lasers

19

20

Why quantum well, wire and dot lasers, modulators and solar cells?Quantum Dot Lasers:•Low threshold current density and improved modulation rate.•Temperature insensitive threshold current density in quantum dot lasers. Quantum Dot Modulators:•High field dependent Absorption coefficient (α ~160,000 cm-1) : Ultra-compact intensity modulator •Large electric field-dependent index of refraction change (Δn/n~ 0.1-0.2): Phase or Mach-Zhender ModulatorsRadiative lifetime τr ~ 14.5 fs (a significant reduction from 100-200fs). Quantum Dot Solar Cells: High absorption coefficent enables very thin films as absorbers. Excitonic effects require use of pseudomorphic cladded nanocrystals (quantum dots ZnCdSe-ZnMgSSe, InGaN-AlGaN)

Table I Computed threshold current density (Jth) as a function of dot size in for InGaN/AlGaN Quantum Dot Lasers

q and Jth

Quantum Dot Size 100100100Å 505050Å 353535Å

Defect Status

q Jth A/ cm2 =418nm

q Jth A/ cm2 =405nm

q Jth A/ cm2

=391nm

Negligible Dislocations

(ideal)

0.9 76

0.9 58

0.9 54

Traps N t=2.9x1017cm-

3 (Dislocations =1x1010 cm-2)

0.0068 10,118 0.0068 7,693 0.0068 7,147

Excitonic Enhancement (in presence of dislocations)

0.049 1,404 0.17 304 0.358 136

(Ref. F. Jan and W. Huang, J. Appl. Phys. 85, pp. 2706-2712, March 1999).

Nanophotonics

• Si nanophotonics

• Surface enhanced Raman effect via plasmon formation in thin metal films or gold nanoparticles.

• Plasmons are modified by functionalized nanoparticles enabling biosensing of proteins etc.

21

22

Semiconductor Background Review (ECE 4211 Chapters 1 and 2)

Energy bands in semiconductors: Direct and indirect energy gap

N- and p-type doping,

Carrier concentrations: np=ni2

Fermi-Dirac Statistics & Fermi level

Drift and diffusion currents

P-n junctions: Forward/Reverse biased Heterojunctions

23

Conductivity σ, Resistivity ρ= 1/ σ

Current density J in terms of conductivity and electric field E: J = E = (-V) = - V

I = J A = E (W d),

In n-type Si, nq n nno + q p pno

24

Carrier Transport: Drift and Diffusion

Drift Current: In = Jn A = - (q n n) E A

Diffusion Current density: Jn= +q Dn n, [Fick's Law]

Total current = Diffusion Current + Drift Current

Einstein’s Relationship: Dn/μn =kT/q

Pnonno=ni2 n-Si ND=Nn=nno

Pponpo=ni2 p-Si NA=Pp=Ppo

25

Drift and Diffusion of holes in p-Si

In p-type Si,

The conductivity is: nq p ppo + q n npo

Drift Current: Ip drift = Jp A = (q p p) E A

Diffusion Current density: Jp= - q Dp p, [Fick's Law]

Diffusion current: Ip diff = - q A Dp p

Einstein’s Relationship: Dn/μn =kT/q

Total hole current = Diffusion Current + Drift Current

Ip= - q A Dp p + (q p p) E A

26

Carrier concentrationWhen a semiconductor is pure and without impurities and defects,

the carrier concentration is called intrinsic concentration and it is denoted by ni. i.e.

n=p=ni.

ni as a function of Temperature, see Figure 17 (page 28) and Fig. 11 (page 69).

Also, ni can be obtained by multiplying n and p expressions (apge 68 of notes)

2kT

E3/4

pn

2/3

2i

kT

E3/2

4

2pn

22

g

g

e)mm(h

kT 22n

eh

(kT)mm π44

in

27

Extrinsic Semiconductors: Doped n- and p-type Si, GaAs, InGaAs, ZnMgSSe IIIrd or Vth group elements in Si and Ge are used to dope them to increase their hole and electron concentrations, respectively.

Vth group elements, such as Phosphorus, Arsenic, and Antimony, have one more electron in their outer shell, as a result when we replace one of the Si atoms by any one of the donor, we introduce an extra electron in Si.

These Vth group atoms are called as donors. Once a donor has given an electron to the Si semiconductor, it becomes positively charged and remains so. Whether a donor atom will donate its electron depends on its ionization energy ED. If there are ND donor atoms per unit cm3, the number of the

ionized donors per unit volume is given by

e

2

1+1

1-1 NN

kT

)E-E(D+D

fD

28

Fermi-Dirac Statistics

We have used a statistical distribution function, which tells the probability of finding an electron at a certain level E. This statistics is called Fermi-Dirac statistics, and it expresses the probability of finding an electron at E as

 

Ef is the energy at which the probability of finding an electron is ½ or 50%.

 In brief, donor doped semiconductors have more electrons than holes.

e+1

1 Ef

kT

)E-E( f )(

E

29

Acceptors and p-type semiconductors: We can add IIIrd group elements such as Boron, Indium and Gallium in Si. When they replace a Si atom, they cause a deficiency of electron, as they have three electrons in their outer shell (as compared to 4 for Si atom). These are called acceptor atoms as they accept electrons from the Si lattice which have energy near the valence band edge Ev. Eq. 12 expresses the concentration of ionized acceptor atoms (on page 71).

N-A is the concentration of the ionized acceptor atoms that have accepted

electrons. EA is the empty energy level in the acceptor atom.

Hole conduction in the valence band: The electron, which has been accepted by an acceptor atom, is taken out of the Si lattice, and it leaves an empty energy state behind. This energy state in turn is made available to other electrons. It is occupied by other electrons like an empty seat in the game of musical chairs. This constitutes hole conduction.  

e41

+1

1*N=N

kT

)E-E(A-A

fA

30

Donors and acceptors in compound semiconductors (see Problem set before chapter 1, p. 26)

Semiconductors such as GaAs and InGaAs or ZnMgSSe are binary, ternary, and quaternary, respectively. They represent III-V and II-VI group elements.

For example, the doping of GaAs needs addition of group II or VI elements if we replace Ga and As for p and n-type doping.

In addition, we can replace Ga by Si for n-type doping. Similarly, if As is replaced by Si, it will result in p-type doping. So Si can act as both n and p-type dopant depending which atom it replaces.

Whether Si is donor or acceptor depends on doping temperature.

31

Calculation of electron and hole concentrations in n-type and p-type semiconductors

Method #1: (simplest)Simple expressions for electron and hole concentrations in n-Si having ND

concentration of donors (all ionized). Electron concentration is n = nn or nno =ND, (here, the subscript n means on the n-side or in n-Si; additional subscript ‘o’ refers to equilibrium). Hole concentration is pno =ni

2/ND.

For p-Si having NA acceptor concentration (all ionized), we have p= pp =NA, and electron concentration npo= (ni

2)/NA,

 Method#2 (simpler)Here, we start with the charge neutrality condition. Applying charge neutrality, we get: total negative charge density = total positive charge densityi.e. qND

+ + qpno = qnno, here pno and nno are the hole concentrations in the

n-type Si at equilibrium. But pno or hole concentration = ni2/ND. Substituting pno

in the charge neutrality equation, we get electron concentration by solving a quadratic equation [Eq 8, page 71]. Its solution is: n4+NN

2

1=n 2

i2DDn

32

Method#3 (Precise but requires Ef calculations)

qND+ + qpno = qnno, Charge neutrality condition in n-type

semiconductor can be written as:[Eq. 6 on page 70]

eh

kTm22 = e

h

kTm22+

e+1

1-1N kT

E

2

n2

3

kT

)EE(-

2

p2

3

kT

)E-E(D

fgf

fD

Here, we have ignored the factor of ½ from the denominator of the first term. In this equation, we know all parameters except Ef. One

can write a short program and evaluate Ef or assume values of Ef

and see which values makes left hand side equal to the right hand side.

33

Effect of Temperature on Carrier Concentration

The intrinsic and extrinsic concentrations depend on the temperature.

For example, in Si the intrinsic concentrations at room temperature (~300K) is ni =1.5x1010 cm-3. If you raise the temperature, its value

increases exponentially (see relation for ni).

34

Another way of looking at the carrier concentration expression [pages 52-59]

The electron concentration in conduction band between E and E+dE energy states is given by

dn = f(E) N(E) dE.To find all the electrons occupying the conduction band, we need to integrate the dn expression from the bottom of the conduction band to the highest lying level or energy width of the conduction band. That is, [see page 56 notes]

This leads to electron concentration (see page 57 and page 68): This equation assumes that the bottom of the conduction band Ec = 0.

0

)()( dEEfENn

e h

kTm22=n kT

E

2

n

3/2f

An alternate expression results, if Ec is not assumed to be zero.

e h

kTm22=n kT

)E-E(

2

n

3/2fc

35

 Direct and Indirect Energy Gap Semiconductors

k vector

Energy

E-K diagram of an indirectsemiconductor

Energy Gap Eg

k vector

Energy

E-K diagram of an directsemiconductor

Energy Gap Eg

Fig. 10b. Energy-wavevector (E-k) diagrams for indirect and direct semiconductors(page 22). Here, wavevector k represnts momentum of the particle (electron in the conduction band and holes in the valence band). Actually momentum is = (h/2)k = k

36

Electrons & Holes Photons Phonons

Statistics F-D & M-B Bose-Einstein Bose-Einstein

Velocity 

vth ,vn

1/2 mvth2 =3/2 kT

Light c or v = c/nr

nr= index of refraction

Soundvs = 2,865 meters/s in GaAs

Effective Mass 

mn , mp

(material dependent)

No mass No mass

 Energy 

E-k diagramEelec=25meV to 1.5eV

ω-k diagram (E=hω)ω~1015 /s at E~1eVEphotons = 1-3eV

ω-k diagram (E= ω)ω~5x1013/s at E~30meVEphonons = 20-200 meV

 Momentum 

P= kk=2π/λλ=2πvelec/ω

momentum: 1000 timessmaller than phonons and electrons

P= kk=2π/λλ=2πvs/ω

37

P-n Junctions (See Overview, Pp. 7-11] Before Going to Chapter 2)

n

NNq

kT=

p

p

q

kT=voltage in-Built=V 2

i

da

no

pobi lnln

)NN(2

NNqp

p

qkT

=)NN(2

NNqV

=W

DAor

DA

no

po2

1

DAor

DA

bi2

1

ln

Use (Vbi – Vf) for forward-biased junctions, and (Vbi + Vr )for reverse-biased

junctions.

38

Shockley's equation

1)-e(J=J kTVq

s

f

1)-e(I=I=A J kTVq

s

f

Here, the reverse saturation current Is = A Js, and reverse saturation current

density Js is

 

Here, Dp is the diffusion coefficient of holes, pno is the hole concentration

under equilibrium on n-Si side, Lp is the diffusion length of holes (L2p = Dp x

p). p is the average lifetime of injected holes.

Similarly, Ln is the diffusion length of electrons injected from n-side into p-

side and npo is the minority electrons on p-side at equilibrium. 

L

nDq+

L

pDq=J

n

pon

p

nops

39

Charge distribution, Field, Built-in Voltage

xNq

-=xNq

-=E noDoAlGaAs

pooGaAs

A

max

Poisson's Equation ·D =

Or, 2V= -r,= q(ND+ + p- NA

- -n)]

(W)E2

1-=)x+x(E

2

1-=V=- mponombipn

WxNq

2

1=V or,

WxNq

2

1=V

or

noAbi

or

poAbi

=XX nopo

40

Vbi

W

0-Xp0 Xn0

P N

X

q Voltage=q

Vbi

W

0-Xp0Xn0

P N

X

Voltage

Edx-=d

Edx-=-=d xx-pnno

po

n

p

(a) Voltageas a function of distance (Since field E is negative, the negative sign makes itpositive. As a result the potential increases parabolically as we go from p-side to the n-side).

p

n

(b) Electron energy as a function of distance (Multiply the voltage with electroncharge which is negative q. The negative q makes q to change sign with respect topart (a). That is, the energy of electron is higher in p-Si than in n-Si).

P N

qVbi

(c) Energy band diagram of a p-n junction. Electron energy in the conduction bandof p-Si is higher than of electron in the conduction band of n-Si.

-Xp0Xn0

0

Fig. 4. Schematic representation of energy band diagram for a p-n junction showing electron energy.

41

Energy band diagrams: Homojunction & Heterojunction

Vbi

W

0-Xp0 Xn0

Ec

Ev

Ef

P N

p-AlGaAs n-GaAs

Ec

Ev

W

0-Xp0 Xn0

Ec

Ev

Ef

Eg ~ 1.9 eV

Eg = 1.424 eV

Homojunction

Heterojunction

42

Equilibrium

(VA)

-xp xn0 lnlp

p n

NA = 1019 cm-3

NA = 1019 cm-3

VA

I

Non-equilibrium

x

nepe

-xpo xno0 lnlp

p n

o

NA = 1019 cm-3

ND = 1016 cm-3

n(x)o

p(x)o

E(x)

NA

NDFig. 5.Carrier distribution in a p-n junction under equilibrium and under non-equilibrium (under forward biasing).

43

Energy band diagrams in Heterojunctions

N-AlGaAsp-GaAs

Eg1

Eg2

EfN2

Efp1

Ec2

Ec1

Ev2

Ev1

q2 q1

Ec

Reference vacuum level

q2

q1

Ei Ei

Ec-Efp

Ei-Efn

44

Energy band diagrams in Heterojunctions

By definition, the built in voltage is the difference between the two Fermi levels (Efp1 and Efn2 for p-GaAs and n-AlGaAs, respectively. Here, we have not used

the Ei we use the difference between the Fermi level and the band edge (i.e.

either Ec2-EfN2 or Efp1-Ev1).

qVbi = - q(q[(c2 - Efn2) –{gp1 – (Efp1 –Ev1)}]

= - q[(gp1 + (Ec2 - Efn2) + (Efp1 –Ev1)]

qVbi = Ec + gp1 -(Ec2 - Efn2) - (Efp1 –Ev1)

Vbi =1/q[Ec + gp1 -(Ec2 - Efn2) - (Efp1 –Ev1)]

Here, (Ec2 - Efn2) = (kT) ln(NC/n)

 and, (Efp1 –Ev1) = + (kT) ln(Nv/p)

45

p-AlGaAs n-GaAs

Ec

Ev

W

0-Xp0 Xn0

Ec

Ev

Ef

Eg ~ 1.9 eV

Eg = 1.424 eV

Energy band diagram in a heterojunction P-AlGaAs/n-GaAs.

46

L8, 2/10/05 Heterojunctions and Junction Fabrication Techniques

Heterojunctions?General: • Higher injection efficiency with lower doping levels in the wider energy gap

semiconductor

• Laser Diodes: Carrier confinement in a narrow layer, if needed (useful in lasers to generate photons in a narrow layer (smaller d); minority carriers are not as readily injected from a narrower gap material into wider gap material.

• Laser Diodes:Photon confinement in a three layer sandwich of low-high-low index of refraction (e.g. AlGaAs-GaAs-AlGaAs); nrAlGaAs)<nr(GaAs). See homework #1.

• Quantum Well Lasers: thin low energy gap active layer permits confinement of carriers in very narrow layer (~50-70Angstroms) forming quantum wells and providing lower threshold operation.

• Quantum wire and quantum dot lasers: Lower threshold and temperature insensitivity

L

)pD(+

L

)nD(L

)nD(

=

1)-e(L

pDqA+

L

nDqA

1)-e(L

nDqA

=)x(-I+)x(I

)x(I

p

nop

n

pon

n

pon

kT

Vq

p

nop

n

pon

kT

Vq

n

pon

nppn

pninj

f

f

47

• Solar cells: wider gap semiconductor acts as the window where photons enter the device and are absorbed in the lower energy gap material

• Solar cells: Provides higher operating voltage for a given current; and minimize recombination of carriers at the surface.

• Heterojunction Bipolar transistors (HBTs): High injection efficiency even with lower emitter concentrations permits the use of a very highly doped base. This in turn reduces the base transit time 9one of the main factors limiting the unit gain cutoff frequency fT.

• Flexibility in designing higher current gain, reducing resistance (e.g. sub-collector).

Why heterojunctions?