Lundstrom ECE 305 S16

ECE-305: Spring 2016

Intro to PN Junctions: II

Professor Mark Lundstrom Electrical and Computer Engineering

Purdue University, West Lafayette, IN USA lundstro@purdue.edu

2/18/15

Pierret, Semiconductor Device Fundamentals (SDF) pp. 195-209

NP junction (equilibrium)

2

N P

p0 ! NA

ρ ! 0 n0 ! ND

ρ ! 0

xp−xn 0

“transition region”

Lundstrom ECE 305 S16

p0 < NAn0 < ND

energy band diagram

3

EF

EC

EV

x

E

Ei

x = xpx = 0x = −xn

qVbi

p0 < NAn0 < ND

Lundstrom ECE 305 S16

electrostatics: V(x)

4

V

x

N P

xp−xn

qVbi = kBT lnNAND

ni2

Lundstrom ECE 305 S16

electrostatics: E (x)

5

E

xN P xp−xn

Lundstrom ECE 305 S16

carrier densities vs. x

6

log10 n x( ), log10 p x( )

xN P xp−xn

p0P = NA

p0N = ni2 ND

n0N = ND

n0 p = ni2 NA

n0N << ND p0P << NA

Lundstrom ECE 305 S16

electrostatics: rho(x)

7

ρ

x

N P

ρ = q p0 x( ) − n0 x( ) + ND+ x( ) − NA

− x( )⎡⎣ ⎤⎦

xp−xn

qND

−qNA

Lundstrom ECE 305 S16

n0N << ND

p0P << NA

NP junction electrostatics

8

How do we calculate rho(x), E(x), and V(x)?

Lundstrom ECE 305 S16

Gauss’s Law

9

+Q n̂

“Gaussian surface”

!D = ε0

!E

!D = KSε0

!E

!D i d

!S"∫ = Q

Lundstrom ECE 305 S16

Gauss’s Law in 1D

10

!D i d

!S"∫ = Q

xx x + dx

ρ x( )C/cm3

D x + dx( )D x( )

n̂n̂

Area = A

−D x( )A+ D x + dx( )A = Q

Q = ρ x( )Adx

D x + dx( )− D x( )dx

= ρ x( )

dDdx

= ρ x( )Lundstrom ECE 305 S16

the Poisson equation

11

dEdx

=ρ x( )KSε0

dDdx

= ρ x( )

∇ i!D = ρ x( )

!D i d

!S"∫ = Q

D = KSε0E

Lundstrom ECE 305 S16

electrostatics: rho(x)

12

ρ

x

N P

ρ = q p0 x( ) − n0 x( ) + ND+ x( ) − NA

− x( )⎡⎣ ⎤⎦

xp−xn

qND

−qNA

“depletion approximation”

Lundstrom ECE 305 S16

n < ND

p < NA

the “depletion approximation”

13

dEdx

=ρ x( )KSε0

ρ

x

N P

−xn

ρ = +qND

xp

ρ = −qNA

qNDxn = qNAxp

NDxn = NAxpLundstrom ECE 305 S16

but first

14

E

x

V = 0V > 0

d

E = - dV

dx

V = − E

x1

x2

∫ dx

E = V

d

dEdx

=ρ x( )KSε0

= 0 →E is constant

Lundstrom ECE 305 S16

electric field between the plates

15

E

x

+ d2

− d2

E = V

d V = − E

−d /2

+d /2

∫ dx

Lundstrom ECE 305 S16

the NP junction

16

dEdx

=ρ x( )KSε0

is not constant!

ρ

x

+qND

ρ = −qNA

xn + xp

P

V = 0

N

V =Vbi > 0

1)  Make depletion approximation

2)  Solve

3)  Find

NP junction electrostatics

17

dEdx

=ρ x( )KSε0

ρ

x

N P

−xn

ρ = +qND

xp

ρ = −qNA

E x( ),V x( ), xn , xp