Chapter 4-Example and Quantum Dots

8/14/2019 Chapter 4-Example and Quantum Dots

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Example:

2,of 51018 cm-3 on the p-side and a donor concentration of 1016 cm-3 on the n-side. Thelife time of holes in n-region is 417 ns and the life time of electrons in the p-region is 5

ns. The drift mobilities of electron in p-regions and hole in n region are 120 and 440

cm2 V-1s-1, respectively.

a.Calculate the minority diffusion lengths at 300K.

b.What is the current when there is a forward bias of 0.6V across diode at 300 K?

a) DL =

026.0 eVkT

( ) 12cm10.3120. === s

eeD ee

cm.=== see hh

cm102.1105cm10.3 4912 === ssDL eee

34( )( ) cm108.2110417cm39.114912

=== ssDL hhh


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b) What is the current when there is a forward bias of 0.6V across diode at 300 K?

= 1expkT

eVJJ so

2

i

ae

e

dh

hso n

NLeD

NLeDJ

+=

I=AJ 2i

ae

e

dh

hso n

NL

eD

NL

eDAI

+=

For V>>kT/e, (V=0.6 V, and kT/e=0.026 V)

= kTI

kTII soso exp1exp

35


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eDeDeh 2

=

( )( ) ( )cmssC

NLNLi

aedh

so

310

3184-

12

3164-

12192 10

cm.13cm39.11106.1cm01.0

+=

A

cmcm

141036.8

cm.cm.

=

AV

V

AkT

eV

II so314

1096.0026.0

6.0

exp1036.8exp

=

=

=

36


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Consider a pnp Si BJT that has the following properties. The doping concentrations in

Example:

emitter, base and collector are 21018,1016, 1016 cm-3. The intrinsic concentration is1010 cm-3 The width of emitter and base are 2 m under the active operation. The holedrift mobility in the base is 400 cm2 V-1 s-1 and electron drift mobility in the emitter is

2 -1 -1 - 2

base is 400ns. Assume the emitter has 100% emitter efficiency. Calculate the CBcurrent transfer ratio and current gain . What is the emitter-based voltage if theemitter current is 1mA?

( )

12

cm36.10400

026.0

=== se

eV

e

kT

D hh

The minority carrier transit time t across the base.

- - -

t = B h= cm . cm s = . s

The base transport factor is

45

99517.010400

1093.111

9

9

=

===

s

s

h

t

T


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The current gain of the transistor is

2.20699517.01

.

1=

=

=

eVeVeAD

The emitter current is due to the hole diffusion since the emitter efficiency is 1,

=

=

kTkTWso

B

E expexp

2102 10 n

( )1610=== cm

Np

d

no

) ) ) )cmcmscmCpeADI nh

3412219

0 1036.101002.0106.1

==

A

cmWBso

141066.1

102

=

46

VI

I

e

kTV

so

EEB 64.0ln =

=


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Example

A solar cell under an illumination of 500Wm-2 has a short circuit current Isc of 150 mA

and an open circuit output voltage Voc of 0.530 V, what is the short circuit current and

open circuit voltage when the light intensity is doubled? Assume ideality factor=1.5.

The I-V characteristic under illumination is given

+= 1exp

ph kT

eVIIIo

For open circuit, I=0

e >>At RT,

hIkT eV

89

=o

ocIe

nexpph =

=kT

o


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IKI ==The short circuit current is the photocurrent, so at double the intensity this is

( )( ) mAmAIIII scsc 3002150

1

212 ==

=

At given temperature, the change in Voc is

=

=1

2

1

2

12 lnlnIeIe

VVph

ph

ococ

=1.5, the new open circuit is

2IkT

90

....1

12

Ieococ


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Chapter 5: nanoelectronics- The quantum dotso tical ro ert

Single electron device

91


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Principle of the quantum dots

One free electron is in three dimensional infinite potential box.

++=2

3

2

2

2

1

321 8 c

n

b

n

a

n

m

h

E nnn

( )2

22

2

2

3

2

2

2

1

2

321

88 ma

Nh

ma

nnnhE nnn =

++=a = b = c

2 )12( nh +

21

8mann+

Energy difference is smaller as a increases to macroscopic dimensions,

,


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Difference between nanoparticle and quantum dots:

Quantum dots: the size of quantum is small enough that the energy difference between

the consecutive energy levels can be distinguished, namely, the energy is discrete and

not continuous.

Nanopartilce: the size of nanoparticles is small (nanoscale) and the energy differencebetween the consecutive energy levels cannot be distinguished, namely, the energy is

.

Characteristic size:

Methods for calculating the critical size:

1.De Bro lie wavelen th at Fermi ener metal

2.Exciton radius (semiconductor)


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a should be comparable to the De Broglie wavelength of the electron

Emph

p e*2 2, ==

Fe

FEm

h

*2=

3/2

*

2 3

8

=

n

m

hE

e

F

3/12

2F =

For Copper, n=8.851028 m-3, F=0.46 nm, if the particle size issmaller or comparable to 0.46nm, it is called quantum dots.


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Si bulk single crystal, the number of atoms is

very large, therefore the conduction band and

For nanoscale-size, the number of

the atoms is significantly reduced

The energies in conduction band and

valence band become discrete

1

m

Direct-gap interband transitions in quantum

dots from nth level valence band to mthg

n

1level conduction band can occur forincident photon energies.

h ++= mng


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mng EEEh ++=

2222

2*

22

8 Rm

nhE

e

n =

2*2*

88 RmRm

Ehhe

g ++=

112h =

8

8

*2

2

**2

g

he

g

m

hE

mm

+=

,**

***

he

her

mm

mmm

+=

Reduced mass

96


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Exciton:

Ec

CBInterband transition in semiconductor normally assumed that

the process of absorption (of a photon, thermal energy,

etc.)create free electron and a free hole. each of which can

Ev

EFi

con r u e o con uc on.

Another kind of transition: after the electron transition, the

VB

their mutual Coulomb attraction, forming a quasi particle

(electron and hole as a whole) known as an exciton.

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Exciton is very similar to the two-particle hydrogen system.

.Therefore the exciton can move around the crystal.

The mass now is reduced mass:

**

*** her

mm

mmm

+=

The permittivity is 0rinstead of0,

**4*

Substituting the 0 with 0rand me with mr , the bind energy of electronto hole and radius of ground state are given by

eVmRmnhE re

rY

re

r

r

r

6.138 2*2*22220 ===

**2

98r

er

r

er

r

rex A

ma

mema )53.0(

*0*4*

0 ===


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In quantum dots, exciton effects often play a dominant role in determining optical

properties since the hole and electron are very close in quantum dots.

The optical transition in quantum dots are usually associated with exciton and an

approximation expression known as the Brus equation model, the transition energy

in spherical dots:

8.1 22 ehh +=

,

48

**

***

0

he

her

rr

g

mm

mmm

m

+=

99


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In bulk semiconductors, the exciton radius is given by,

o

r

er

r

er

r

rex A

m

ma

m

m

em

ha )53.0(

4*

*

0*

*

4*

2

0 ===

For quantum dots, the actual separation between electron and hole is influenced

by the size of the dot. In this case, we will consider aex to be the excitation Bohr

radius, which is often taken as the measure of quantum confinement in quantum

dots.

ex,

R>aex, then the confinement is weak;

R


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Example:

For cadmium selenide (CdSe) particle with R=2.9nm. Given Eg (bulk)=1.74 eV,

me*=0.13me and mh*=0.45 me and r=9.4, for transition from the conduction tovalence bandedge.

Amma

mm

emha o

r

er

r

er

r

rex )53.0(4 *

*

0*

*

4*

2

0 === Strong confinementeffect

Without considering the exciton effect,

nm.=

,8 *2 r

gmR

hEh +=

nm

e

593

.

=

=

**

,**he

e

r mmm += Yellow light

101


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,

8.12 22 eh==

,580

48 0*2

nm

RmR rrg

=

Well consistent with the experimental results

102


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Bull's-eye. Red quantum dots injected into a live mouse mark the location of a

.

X.H.Gao et.al. Science, 300 (2003) 80.

103

Chapter 4-Example and Quantum Dots

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