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EE415/515 Fundamentals of Semiconductor Devices

Fall 2012

Lecture 9:MS & Hetero-Junctions

(Chapter 9)

Schottky Barrier

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Reference level↓

MetalN-type semiconductor

Metal work function→

SC work ←function

In contact: Fermi levels align

Electron affinity→

Schottky barrier : φB0 = φm – χ

Built-in potential barrier: Vbi = φB0 - φn

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Schottky barrier bias

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Reverse Forward

Ideal Schottky junction

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metalin charge surface interface negative Law) (Gauss' so metal,in zero is )E(0

0Max

)()( and

C so xat x 0

doping uniformfor

)(region charge spaceIn

-

1n

1

ns

d

ns

dn

s

d

s

nd

s

d

s

d

xeN

)E(

xxeN

xxeN

E

xeN

CxeN

dxeN

E

x

dx

dE

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Ideal Schottky junction

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FCnnbi

ds

Rbi

Rbi

ds

R

nd

d

Rbisn

EEV

Ne

VV

C

VV

Ne

dV

dxeNC

eN

VVxW

as calculated whereThen

(slope) N and )(intercept V gives V vs21

Plot

2

2

junction n p sided-onefor As

B0

dbiR

2

/

/

Ex 9.1 Ideal W to n-type GaAs M-S junction; Nd=5x1015/cm3. Find theoretical barrier height, built-in potential barrier, and maximum electric field for zero applied bias.

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48.007.455.40 mB V

15

17

105

107.4ln0259.0ln

d

ctn

N

NV

1177.0 V 3623.01177.048.00 nBbiV V

2/12

d

bisn eN

Vx

2/1

1519

14

105106.1

3623.01085.81.132

51024.3 cm

s

nd xeN

max

14

51519

1085.81.13

1024.3105106.1

41024.2 V/cm

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Ex 9.2 Find the GaAs doping concentration and Schottky barrier height for the W-GaAs diode in Fig 9.3 (slide 5).

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From Figure 9.3, 64.0biV V

1212

2

10335.264.03

105.8

1

RV

C

Then

R

sd

V

C

eN

21

12

121419 10335.21085.81.13106.1

2

or 181062.4 dN cm 3

Schottky Effect (barrier lowering)

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ss

s

s

xs

s

eE

E

e

dx

x

Exx

ex

x

e

dxx

ex

x

eeEF

4 where

16at x

0))(d(e

en barrier wh Maximum

16)(

:E field appliedWith

0)(for 16

)(16)( :Potential

)2(4

force restoring exerts charge image

surface,metalleaveselectron As

m

/2/

2

2

8

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Ex 9.3 Calculate the Schottky barrier lowering for a GaAs M-S contact for which the electric field in the GaAs is E=6.8x104V/cm, for reverse biases of (a) VR=1V & (b) VR=5V.

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2/1

2

d

Rbisn eN

VVx

2/1

1619

14

10106.1

1085.87.112

Rbi VV

2/1910294.1 Rbi VV

(a) 1RV V, 334.0biV V

510155.4 nx cm

s

nd xeN

max

14

51619

1085.87.11

10155.410106.1

41042.6 V/cm

Then s

e

4

2/1

14

419

1085.87.114

1042.6106.1

0281.0 V

(a) 5RV V,

510309.8 nx cm

14

51619

max1085.87.11

10309.810106.1

510284.1 V/cm

2/1

14

519

1085.87.114

10284.1106.1

0397.0 V

Interface States

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φB0 is not exactly φm – χ

More complete model:Thin oxide (electron transparent)Dit/cm2.eV surface states in semiconductor

Donor states below eφAcceptor states above eφ

mBnitgBnit

Bnmit

inBnds

itBng

DeED

eDNe

eDeeE

0 If )/e( If

)()(21

)(

0

0

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I-V characteristics

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kT

eexpexp*J loweringbarrier with and

4* and exp*J where

1exp

1exp.exp*

exp24

before As

2sT

3

2*2

sT

2

3

2

3*

kT

eTA

h

kemA

kT

eTA

kT

eVJ

kT

eV

kT

eTAJJJ

dEkT

EEEE

h

mdn

Bn

nBn

asT

aBnsmms

FC

n

Device characteristics

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Reverse current increases with VR due to barrier lowering

Breakdown→

Plot gives JsT and hence (Si) A*=114A/K2.cm2 for φBn=0.67V

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Ex 9.4 Calculate the ideal Richardson constant for a free electron.

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3

24

h

kemA n

Assume on mm , then

334

2233119

10625.6

1038.11011.9106.14

A

61020.1 A/K 2 -m 2

120 A A/K 2 -cm 2

Ex 9.5 Use the results of Example 9.5 to determine the forward bias voltages required to produce a current of 10μA in a W-Si diode of measured barrier height φBn=0.67eV and a P-N junction of Js=3.66x10-11A/cm2. Assume junction areas of 10-4cm2.

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t

as

V

VAJI exp

so that

sta

AJ

IVV ln

For the pn junction:

114

6

1066.310

1010ln0259.0aV

5628.0 V For the Schottky junction:

54

6

1098.510

1010ln0259.0aV

1922.0 V

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Ex 9.6 A PN diode and a Schottky diode have equal c/s areas and forward bias currents of 0.5mA. The Schottky reverse saturation current is 5x10-7A. The difference between the forward bias voltages is 0.30V. Find the reverse saturation current of the PN diode.

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tS

ST

t

aS

t

aST

VI

I

V

VI

V

VI

3.0exp1

exp

3.0exp

Then

0259.0

3.0exp105 7

SI

121066.4 SI A

M-S Ohmic Contacts:Ideal non-rectifying barrier (N-type)

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Previous case: φm>φs, (Schottky diode.) Consider now φm<φs (below).No barrier to electron flow semiconductor → metalEffective barrier height for electron flow metal →semiconductor is φBn=φn (small)Hence ohmic contact.

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Biased N-type Ohmic contacts

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M-S Ohmic Contacts:Ideal non-rectifying barrier (P-type with φm>φs)

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Electrons flow metal → sc valence band, (or no barrier to hole flow sc valence band→ metal)

Barrier small for electrons → metal

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Tunneling barrier

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2

expJ

through.can tunnel electrons and

ry thin,barrier ve so charge Space

heavily.tor very semiconduc thedope

) with type-por with type-(n

result, otherwise uldbarrier woSchottky aWhen

*s

t

2

1

m

d

nBn

d

ssm

N

m

e

e

N

Ex 9.7 Calculate the space charge width of a rectifying metal-GaAs junction. Assume Nd=7x1018/cm3 in the GaAs and that the built-in potential barrier is Vbi=0.80V.

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We have

2/12

d

bisn eN

Vx

2/1

1819

14

107106.1

80.01085.81.132

610287.1 cm

or o

n Ax 7.128

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Contact resistance

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d

Bnms

Bnc

Bnn

V

c

N

m

e

e

kT

TAR

kT

eV

kT

eTAJ

V

JR

.2

expR

)N (largecontact ingFor tunnelkT

exp..*

1 so

1expexp*

:dominatescurrent c thermionicontact, recifyingFor

cm-in

*

c

d

2

2

2

0

1

Heterojunctions

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Homojunctions → same semiconductor (e.g. GaAs) on both sidesHeterojunctions → different materials

e.g. GaAlAs system …. Vary composition to vary band-gap

“Straddling” most common, and considered below

“Anisotype” → different doping → Np or Pn (where “N” or “P” indicates the larger bandgap material

“Isotype” → Nn or Pp

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e.g. nP (Ge/GaAs)

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EEEEE

eE

gngPvc

Pnc

Np heterojunction

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pP heterojunction

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nN GaAs-AlGaAs heterojunction

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Electrons flow to align EF’sn-type GaAs ← N-type AlGaAs

Electrons in interface potential well

“2-D electron gas”Quantized energy levels _|_ j’nFree to move in 2D ║ junction

GaAs AlGaAs

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Approximate by triangular potential well

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V(z)=eEz x>0

V(z)=∞ x<0

E0, E1

quantized energy levels

2D electron gas

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2D currents ║junction high mobility since doping light

See also 9.3.4 junction electrostatics

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Ex 9.8 Find ΔEc, ΔEv, Vbi for n-Ge P-GaAs heterojunction at 300K using the electron affinity rule. (ni=2.4x1013/cm3 for Ge.)N-Ge: Nd=1015/cm3 P-GaAs: Na=1015/cm3.

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From Example 9.8, 70.0 E eV.

We find

11

15

2132

1076.510

104.2

d

ino N

np cm 3

Now

P

n

no

Pobi N

N

p

pkTEeV

ln

1811

1815

1071076.5

10610ln0259.070.0

or 889.0biV V