Lundstrom ECE 305 S15

ECE-305: Spring 2015

Metal-Semiconductor Junctions: II

Professor Mark Lundstrom

Electrical and Computer Engineering Purdue University, West Lafayette, IN USA

lundstro@purdue.edu

3/11/15

Pierret, Semiconductor Device Fundamentals (SDF) pp. 477-487

Question 1)

2

EC

EV

EFN

metal

EFM

E0

ΦM = 4.5 eV χ = 4.0 eV

EG = 1.0 eV

0.2 eV

V = 0 V

V = ? V

What is the voltage on the semiconductor?

Question 1)

3

What is the voltage on the semiconductor? a)  +0.6 V b)  -0.6 V c)  +0.3 V d)  -0.3 V e)  +0.2 V

Question 1)

4

EC

EV

EFN

metal

EFM

E0

ΦM = 4.5 eV χ = 4.0 eV

V =Vbi =ΦM −ΦS

q

EG = 1.0 eV

0.2 eV

V = 0 V

V = +0.3V

= 4.5 − 4 + 0.2( ) = +0.3V

Question 1)

5

EC

EV

EFN

metal

EFM

E0

ΦM = 4.5 eV χ = 4.0 eV

EG = 1.0 eV

0.2 eV

V = 0 V

V = ? V

Draw the energy band diagram

Question 2)

6

EC

EVEFP

E0

χ = 4.0 eV

EG = 1.0 eV0.2 eV

metal

EFM

ΦM = 4.5 eV

V = 0 V

V = ? VWhat is the voltage on the metal?

Question 2)

7

What is the voltage on the metal? a)  +0.6 V b)  -0.6 V c)  +0.3 V d)  -0.3 V e)  +0.2 V

Question 2)

8

EC

EVEFP

E0

χ = 4.0 eV

EG = 1.0 eV0.2 eV

metal

EFM

ΦM = 4.5 eV

V = 0 V

V = 0.3VV =Vbi =ΦS −ΦM

q= 4.0 + 1− 0.2( )− 4.5 = +0.3V

Question 2)

9

EC

EVEFP

E0

χ = 4.0 eV

EG = 1.0 eV0.2 eV

metal

EFM

ΦM = 4.5 eV

V = 0 V

V = 0.3VDraw the energy band diagram

Question 3)

10

EC

EV

EFP

E0

χ = 4.0 eV

EG = 1.0 eV0.2 eV

V = 0 V

EC

EV

EFN

χ = 4.0 eV

EG = 1.0 eV

0.2 eV

V = ? V

What is the voltage on the P-type semiconductor?

Question 2)

11

What is the voltage on the P-type semiconductorl? a)  +0.6 V b)  -0.6 V c)  +0.3 V d)  -0.3 V e)  +0.2 V

Question 3)

12

EC

EV

EFP

E0

χ = 4.0 eV

EG = 1.0 eV0.2 eV

V = 0 V

EC

EV

EFN

χ = 4.0 eV

EG = 1.0 eV

0.2 eV

V = −0.6 V

Vbi =ΦP −ΦS

q= 4.0 + 1− 0.2( )− 4.0 + 0.2( ) = 0.6 V

V = −0.6 V

What is the voltage on the right contact?

13

N PV = 0 V = ?

V = +0.3 V = −0.3

V = 0

We can’t measure the built-in voltage of the NP junction.

14

E-band diagram

EF

EC

EV

outline

15

1)  Electrostatics

2)  (IV next lecture)

Lundstrom ECE 305 S15

A metal – P-type semiconductor junction

16

EC

EVEFS

Ei

metal

EFM

E0

ΦM = 4.5 ΦS

χ = 4.05

ΦBP

ΦM <ΦS

energy band diagram

17

EC

EVEF

Ei

metal

EF

ΦBP

qVbi

W

p x( ) << NA

ρ x( ) ≈ −qNA

electrostatics

18

ρ

x

metal P

W

ρ = −qNA

dEdx

=ρ x( )KSε0

dEdx

< 0

electric field

19

E x( )

x

metal P

W

ρ = −qNA

dEdx

=ρ x( )KSε0

E x( ) > 0

dEdx

< 0

electrostatic potential

20

E x( )

x

metal P

Vbi = E x( )∫ dx

W

final answers

21

E x( )

x

metal P

dEdx

= − qNA

KSε0→E 0( ) = qNA

KSε0W

12E 0( )W = Vbi

E 0( )

W

W = 2KSε0qNA

Vbi⎡

⎣⎢

⎤

⎦⎥

1/2

E 0( ) = 2qNAVbi

Ksε0

one-sided NP junction

22

W = 2KSε0q

NA + ND

NDNA

⎛⎝⎜

⎞⎠⎟Vbi

⎡

⎣⎢

⎤

⎦⎥

1/2

→ 2KSε0qNA

Vbi⎡

⎣⎢

⎤

⎦⎥

1/2

E 0( ) = 2Vbi

W

xn =NA

NA + ND

W ≈ 0

xp =ND

NA + ND

W ≈W

N P

ND = 1018

depletion region xp−xn0

E x( )

Vbi =kBTqln NDNA

ni2

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

W NA = 1015

Vbi =ΦM −ΦS( )

q

summary

23

1)  Built-in potential depends on workfunction differences.

2)  Electrostatics is just like a one-sided NP junction. 3)  Energy barrier in the e-band diagram we have

considered implies a rectifying characteristic.

equilibrium energy band diagram

24

EC

EVEF

Ei

metal

EF

ΦBPq Vbi −VA( )

W

P = e−ΔE kBT = e−q Vbi−VA( ) kBT ∝ eqVA kBT

Only majority carriers are involved!