Lecture 9

PN junction (I)

1

Introduction

pn junction

– p-region an n-region in intimate contact

Why is the pn junction worth studying?

It is present in virtually every semiconductor device!

2

Example: CMOS cross-section

Gauss’s Law

The relationship of the electric field E and charge density.

where is the electric permittivity (F/cm)

3

dE

dx

- The differential form in one dimension

- The integral form in one

dimension

Charge is the source of electric field.

' '

0

1( ) (0) ( )

x

s

E x E x dx

Poisson’s equation

The electrostatic potential is defined as:

Differentiation of the above

Poisson's equation

4

-potential difference as the integral of electric field.

( )( )

d xE x

dx

2

2

( ) ( )d x dE x

dx dx

' '

0

( ) (0) ( )

x

x E x dx

Semiconductor Electrostatics in

Thermal Equilibrium

Outline

Nonuniformly doped semiconductor in thermal

equilibrium

Relationships between potential, φ(x) and

equilibrium carrier concentrations, po(x), no(x)

–Boltzmann relations & “60 mV Rule”

Quasi-neutral situation

5

6

Nonuniformly doped semiconductor in thermal equilibrium

n-type

⇒ lots of electrons, few holes

⇒ focus on electrons

What is the resulting electron concentration

in thermal equilibrium?

Consider a piece of n-type Si in thermal equilibrium with non-

uniform dopant distribution:

7

OPTION 1: Electron concentration follows doping

concentration EXACTLY ⇒

Gradient of electron concentration

⇒ electron diffusion

⇒ not in thermal equilibrium!

0 ( ) ( )dn x N x

8

OPTION 2: electron concentration uniform in space

Think about space charge density:

0 ( ) ( )aven x n f x

0( ) [ ( ) ( )]dx q N x n x

If Nd(x)≠n0(x)

⇒ρ(x) ≠0

⇒electric field

⇒net electron drift

⇒not in thermal equilibrium

9

OPTION 3: Demand that Jn = 0 in thermal

equilibrium at every x (Jp = 0 too)

What is no(x) that satisfies this condition?

( ) ( ) ( ) 0drift diff

n n nJ x J x J x

Diffusion precisely balances Drift

10

Let us examine the electrostatics implications of

0 ( ) ( )dn x N x

11

Space charge density

0( ) [ ( ) ( )]dx q N x n x

12

Electric Field

Gauss’s law:

Integrate from x = 0:

( )

s

dE x

dx

' '

0

1( ) (0) ( )

x

s

E x E x dx

13

14

Electrostatic Potential

Integrate from x=0:

( )d

E xdx

' '

0

( ) (0) ( )

x

x E x dx

( 0) refSelect x

15

Boltzmann relations

Relationships between potential, φ(x) and

equilibrium carrier concentrations, po(x), no(x)

16

Using Einstein relation:

00

0

0

0

1

n n n

n

n

dnJ qn E qD

dx

dnd

D dx n dx

0lnd nq d

kT dx dx

17

Integrate:

Any reference is good

00 0,

0,

ln ln lnref ref

ref

nqn n

kT n

0 0,

:

expref

ref

Then

qn n

kT

/

0

q kT

in n e 0ref at 0.ref in n

18

We can rewrite as:

0lni

nkT

q n

0lni

pkT

q n

If we do same with holes (starting with Jp=0

in thermal equilibrium, or simply using n0p0=ni2)

/

0

q kT

ip n e

19

“60 mV” Rule

At room temperature for Si:

or 060 logi

nmV

n

0 025 ln 25 ln 10 logi i

n nmV mV

n n

EXAMPLE 1:

18 3

0 10 (60 ) 8 480n cm mV mV

20

With holes:

EXAMPLE 1:

18 3 2 3

0 010 10

(60 ) 8 480

n cm p cm

mV mV

0lni

pkT

q n

21

Relationship between ϕ, n0 and p0 :

22

Quasi-neutral situation

•Small dno/dx implies a small diffusion current. We do not

need a large drift current to balance it.

•Small drift current implies a small electric field and therefore

a small space charge

Then

no(x) tracks Nd(x) well

⇒ minimum space charge

⇒ semiconductor is

quasineutral

0 dn x N x

If Nd(x) changes slowly with x→n0(x) also changes slowly

with x. Why?

Summary of Key Concepts

It is possible to have an electric field inside a

semiconductor in thermal equilibrium

⇒ Nonuniform doping distribution.

In thermal equilibrium, there is a fundamental

relationship between the φ(x) and the equilibrium

carrier concentrations no(x) & po(x)

– Boltzmann relations (or “60 mV Rule”).

In a slowly varying doping profile, majority

carrier concentration tracks well the doping

concentration. 23

Electrostatics of pn junction in equilibrium

24

Doping distribution of an abrupt

pn junction (metallurgical

junction)

What is the carrier concentration distribution in

thermal equilibrium?

First think of the two sides separately:

25Now bring the two sides together. What happens?

Resulting carrier concentration profile in thermal equilibrium:

• Far away from the metallurgical junction:

nothing happens – Two quasi neutral regions

• Around the metallurgical junction:

diffusion of carriers must counterbalance drift

– Space charge region (depletion region)26

27

28

Question？

Compare the number of majority electrons in the

n-type region ( or the number of majority holes in

the p-type region ) before and after the junction

formation.

29

3.The Depletion Approximation

• Assume the QNR’s are perfectly charge neutral

• Assume the SCR is depleted of carriers (complete

ionization)

- depletion region

• Transition between SCR and QNR’s sharp at

- -xp0 and xno (must calculate where to place these)

30

Depletion region: dipole layer

Now, we want to know n0(x), p0(x), ρ(x), E(x) and ϕ(x).

We need to solve Poisson’s equation using a simple

but powerful approximation

31

0 0 d aq p n N N

2

2

: ( ) , ( )

0 : ( ), ( )

0 : ( ), ( )

: ( ) , ( )

ipo o a o

a

po o o a

no o o d

ino o d o

d

nx x p x N n x

N

x x p x n x N

x x n x p x N

nx x n x N p x

N

Space Charge Density

32

Uncovering of

impurity ions near

the junction on

both sides

( ) 0; ;

; 0;

; 0 ;

0; ;

po

a po

d no

no

x x x

qN x x

qN x x

x x

Electric Field

Integrate Poisson’s equation

33

2

E

2

1

2 1

1x

s x

E x E x x dx

34

Build-in field

'

; ( ) 0

10; ( ) ( )

=

0 ; ( ) ( )

; ( ) 0

po

po

po

x

po po a

s x

x

a apo

s sx

dno no

s

no

x x E x

x x E x E x qN dx

qN qNx x x

qNx x E x x x

x x E x

Electrostatic Potential

35

－intrinsic semiconductor

From Boltzman relationship0lni

nkT

q n 0ln

i

pkT

q n

(with ϕ=0 @ n0=p0=ni)

In QNR’s, n0 and p0 are known ⇒ can determine ϕ

0

0

: ln

: ln

aa p

i

dd n

i

NkTin p QNR p N

q n

NkTin n QNR n N

q n

36

The built-in potential can be looked as the

potential hill or barrier that keeps electrons on the

n-type side and keeps holes on the p-type sides.

Built-in potential（内建电势差）:

2ln d a

B n p

i

N NkT

q n

37

From energy band diagram

Energy-band diagram of a pn junction in

thermal equilibrium

•In thermal equilibrium, the

Fermi energy level is constant

throughout the entire system.

0lnFi F

i

pE E kT

n

• EFi, Ec, Ev bend with the

distribution of carrier

concentration.

Built-in potential : a potential barrier

for electrons in the n region trying to

move into the conduction band of the p

region.

0

ln CC F

NE E kT

n

0

ln VF v

NE E kT

p

3/2

2

22 n

C

m kTN

h

3/2

2

22

p

v

m kTN

h

38

0lnFi F

i

pE E kT

n

bi Fn FpV

Fn Fi Fe E E

0

ln iFi F

nE E kT

n

Fp Fi Fe E E

ln dFn

i

NkT

e n

ln aFp

i

NkT

e n

2 2ln lna d a d

bi T

i i

N N N NkTV V

e n n

Built-in potential

39

To obtain ϕ(x) in between, integrate E(x)

2

1

' '

2 1( ) ( ) ( )

x

x

x x E x dx

40

'

2

2

2

; ( )

0; ( ) ( ) ( )

=2

( )2

0 ; ( )2

;

po

po p

x

apo po po

sx

apo

s

ap po

s

ano n no

s

no

x x x

qNx x x x x x dx

qNx x

qNx x x

qNx x x x x

x x

( )

Almost done...

nx

1. Require overall charge neutrality:

0 0p A n Dqx AN qx AN

The depletion region exists in both the p and n materials and

that equal amounts of charge exist on both side. In order to

uncover the same amount of charge, the depletion layer will

extend deeper into the more lightly doped material.

where xp0 and xn0 are the width of depletion region in the p

side and in the n side, respectively, and A is the cross-

sectional area of the junction.

0

0

p D

n A

x N

x N

41

Still do not know xn0 and xp0 ⇒ need two more equations

2. Require φ(x) to be continuous at x=0;

Two equations with two unknowns — obtain solution:

42

Now problem is completely solved!

2 200

2 2

a dnp p n

s s

qN qNx x

0

2 s B dp

a d a

Nx

q N N N

0

2 s B an

a d d

Nx

q N N N

The total width of the depletion region W is

n pW x x

43

Solution Summary

Width of the space charge region:

Field at the metallurgical junction is maximum:

44

0 0 0

2 s B a d

d p n

a d

N Nx x x

qN N

0

2 B a d

s a d

q N NE

N N

Three Special Cases

Symmetric junction: Na = Nd

45

Asymmetric junction: Na > Nd

Strongly asymmetric junction

0 0p nx x

0 0p nx x

0 0

2 s Bp n

d

x xqN

0

2 B d

s

q NE

p+n junction: Na >>Nd

The lightly-doped side controls the

electrostatics of the pn junction

Contact Potential

Potential distribution in thermal equilibrium so far:

46

Question 1: If I apply a

voltmeter across the pn

junction diode, do I measure

φB?

yes ;no it depends

Question 2: If I short terminals

of pn junction diode, does

current flow on the outside

circuit?

yes ; no ; sometimes

47

We are missing contact

potential at the metal

semiconductor contacts:

Metal-semiconductor contacts: junction of dissimilar materials

⇒ built-in potentials at contacts φ mn and φ mp .

Potential difference across structure must be zero

At equilibrium, the diffuse current equals to the drift current, so there is

no net current.

⇒ Cannot measure φB.

Net current: macroscopical

view

Diffusion current and drift

current: microcosmic view.

Summary of Key Concepts

•Electrostatics of pn junction in equilibrium

– A space charge region surrounded by two quasi-neutral

regions formed.

•To first order, carrier concentrations in space charge region

are much smaller than the doping level

⇒ can use depletion approximation

•From contact to contact, there is no potential buildup

across the pn junction diode

– Contact potential(s).

48

Homework8

A silicon pn junction in thermal equilibrium at T =300 K

is doped such that EF -EFi =0.365 eV in the n region

and EFi - EF =0.330 eV in the p region.

(a) Sketch the energy-band diagram for the pn junction.

(b) Find the impurity doping concentration in each

region.

(c) Determine Vbi

Homework9