Semiconductor Tutorial 2

8/6/2019 Semiconductor Tutorial 2

http://slidepdf.com/reader/full/semiconductor-tutorial-2 1/20

1Professor N Cheung, U.C. Berkeley

Semiconductor Tutorial 2EE143 S06

-Electron Energy Band

- Fermi Level

-Electrostatics of device charges





The Simplified Electron Energy Band Diagram





x

ElectronEnergy

E-field +-

EC

EV

21

x

ElectronEnergy

E-field + -

EC

EV

21

Electric potentialφ(2) < φ(1)

Electric potentialφ(2) > φ(1)

Energy Band Diagram with E -field

Electron concentration n

kT/)]1()2([q

kT/)1(q

kT/)2(q

ee

e

)1(n

)2(n φ

φ

φ

=





Probability of available states at energy E being occupied

f(E) = 1/ [ 1+ exp (E- Ef ) / kT]where Ef is the Fermi energy and k = Boltzmann constant=8.617 ∗ 10-5 eV/K

The Fermi-Dirac Distribution (Fermi Function)

T=0K

0.5

E -Ef

f ( E )





(2) Probability of available states at energy E NOT being occupied

1- f(E) = 1/ [ 1+ exp (Ef -E) / kT]

Properties of the Fermi-Dirac Distribution

(1) f(E) ≅ exp [- (E- Ef ) / kT] for (E- Ef ) > 3kT Note:

At 300K,

kT= 0.026eV•This approximation is called Boltzmann approximation

Probability

of electron state

at energy Ewill be occupied


http://slidepdf.com/reader/full/semiconductor-tutorial-2 6/206Professor N Cheung, U.C. Berkeley


Ec

Ev

Ei

Ef

(n-type)

Ef

(p-type)

q|ΦF|

Let qΦF ≡ Ef - Ei

n = ni exp [(Ef - Ei)/kT]

∴n = ni exp [qΦF /kT]

How to find Ef when n(or p) is known




Dependence of Fermi Level with Doping Concentration

Ei ≡ (EC+EV)/2 Middle of energy gap

When Si is undoped, Ef = Ei ; also n =p = ni




At thermal equilibrium ( i.e., no external perturbation),

The Fermi Energy must be constant for all positions

The Fermi Energy at thermal equilibrium

Material A Material B Material C Material D

EF

Position x

Electron energy




After contact

formation

Before

contactformation

Electron Transfer during contact formation

E

System 1 System 2

EF1

EF2e

System 1 System 2

EF

-

--

+

++

E

System 1 System 2

EF1

EF2

e

System 1 System 2

EF

-

-

-

+

+

+

Net

negativecharge

Net

positive

charge




Fermi level of the side which has a relatively higher electric potential will have arelatively lower electron energy ( Potential Energy = -q • electric potential.)

Only difference of the E 's at both sides are important, not the absolute position

of the Fermi levels.

Side 2Side 1

Ef1

f 2E

Va > 0

qVaVa

Side 2Side 1

Ef1

f 2E

Va < 0

q| |

+ - -+

Potential difference across depletion region

= Vbi - Va

Applied Bias and Fermi Level




PN junctions

Complete Depletion Approximation used for charges inside depletion region

r(x) ≈ ND+(x) – NA

-(x)

n-Sip-Si

Depletion region

ND+ onlyNA

- only

ρ(x) is +ρ(x) is -

ND+ and n

ρ(x) is 0

NA- and p

ρ(x) is 0

E -field

++

++

- -

- -Quasi-neutral

region

Quasi-neutral

region

n-Sip-Si

Depletion region

ND+ onlyNA

- only

ρ(x) is +ρ(x) is -

ND+ and n

ρ(x) is 0

NA- and p

ρ(x) is 0

E -field

++

++

- -

- -Quasi-neutral

region

Quasi-neutral

region

http://jas.eng.buffalo.edu/education/pn/pnformation2/pnformation2.html

Thermal Equilibrium

S i d T i l 2EE143 S06




1) Summation of all charges = 0

Electrostatics of Device Charges

ρ1

Semiconducto(x)

ρ2

xd1xd2

Semiconductor

-

-

x=0

p-typen-type

2)E

-field =0 outside depletion regions

ρ2 • xd2 = ρ1 • xd1

E = 0E = 0E ≠0

S i d t T t i l 2EE143 S06




3) Relationship between E -field and charge density ρ(x)

d [ε E (x)] /dx = ρ(x) “Gauss Law”

4) Relationship between E -field and potential φ

E (x) = - dφ(x)/dx

S i d t T t i l 2EE143 S06




Example Analysis : n+/ p-Si junction

x

-qNa

xd

x=0

+Q'

n+ Sip-Si

ρ (x)

E (x)

xd

1) ⇒ Q’ = qNaxd

3) ⇒

Slope = qNa/εs

4) ⇒ Area under E-field curve

= voltage across depletion region

= qNaxd2/2εs

Emax =qNaxd/εs

Depletion

region

2) ⇒ E = 0

Depletion regionis very thin and is

approximated as

a thin sheet charge





x

-

x=0

-qNA

+qND

ρ(x)

-xp

+xn

x=0

x

ρ1(x)

-xp

-qNA

Q=+qNAxp

x

x=0

ρ2(x)

+qND

+xn

Q=-qNAxp

Superposition Principle

If ρ1(x) ⇒ E1(x) and V1(x)ρ2(x) ⇒ E2(x) and V2(x)

then

ρ1(x) + ρ2(x) ⇒ E1(x) + E2(x) andV1(x) + V2(x)

+





x

-

x=0

Slope =

- qNA /εs

E1(x)

-xp

x=0

x

ρ1(x)

-xp

-qNA

Q=+qNAxp

x

x=0

ρ2(x)

+qND

+xn

Q=-qNAxp

+ x

-

x=0

+xn

E2(x)

Slope =

+ qND /εs





x

-

x=0

Slope = - qNA /εs

E(x) = E1(x)+ E2(x)

-xp +xn

Slope = + qND /εs

Emax = - qNA xp /εs

= - qND xn /εs

Sketch of E(x)





x

x=0

Q'

Metal

ρ(x)

Oxide

x=xo

x

−ρ

xd

x=0

Q'

Metal Semiconductor

ρ(x)

Oxide

x=xo

x=xo+

Q'-

x

−ρ

xd

x=0

Q'

Metal Semiconducto

ρ(x)

Oxide

x=xo

x=xo+

Semiconductor

Depletion Mode :Charge and Electric Field Distributions

by Superposition Principle of Electrostatics

x

xd

x=0

Metal Semiconductor

E(x)Oxide

x=xo

x=xo+

x

xd

x=0

Metal Semiconductor

E(x)

Oxide

x=xo

x=xo+ x

xd

x=0

Metal Semiconductor

E(x)

Oxide

x=xo

x=xo+

=

+





Approximation assumes

VSi does not change much

OX

OX

C

nV

∆∝∆

SiOX FBG V V V V ++=

Picks up all

the changesin VG

Justification:

If surface electron density

changes by ∆nEi

Ef

( )

kTen

nln

δ

∝

∆δ but the change of VSi changes

only by kT/q [ ln (∆n)] – small!

Why xdmax ~ constant beyond onset of strong inversion ?

Higher than VT




20


VSi n-surface

0 2.10E+040.1 9.84E+05

0.2 4.61E+07

0.3 2.16E+09

0.4 1.01E+110.5 4.73E+12

0.6 2.21E+14

0.7 1.04E+160.8 4.85E+17

0.9 2.27E+19

n-surface

p-Si

Na=1016/cm3

Ef

Ei

0.35eV

n-bulk = 2.1 • 104/cm3

qVSi

Onset of strong inversion

(at VT)

N (surface) = n (bulk) • exp [ qVSi/kT]

Documents

Semiconductor Tutorial 2