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### Transcript of Intro to PN Junctions: II - nanoHUB PNJunctionsII... · PDF file Lundstrom ECE 305 S16...

• 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 [email protected]

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 ln NAND ni 2

Lundstrom ECE 305 S16

• electrostatics: E (x)

5

E

x N P xp−xn

Lundstrom ECE 305 S16

• carrier densities vs. x

6

log10 n x( ), log10 p x( )

x N P xp−xn

p0P = NA

p0N = ni 2 ND

n0N = ND

n0 p = ni 2 NA

n0N

• 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

• 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( )

dD dx

= ρ x( ) Lundstrom ECE 305 S16

• the Poisson equation

11

dE dx

= ρ x( ) KSε0

dD dx

= 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

dE dx

= ρ x( ) KSε0

ρ

x

N P

−xn

ρ = +qND

xp

ρ = −qNA

qNDxn = qNAxp

NDxn = NAxp Lundstrom ECE 305 S16

• but first

14

E

x

V = 0V > 0

d

E = - dV

dx

V = − E

x1

x2

∫ dx

E = V

d

dE dx

= ρ x( ) KSε0

= 0 →E is constant

Lundstrom ECE 305 S16

• electric field between the plates

15

E

x

+ d 2

− d 2

E = V

d V = − E

−d /2

+d /2

∫ dx

Lundstrom ECE 305 S16

• the NP junction

16

dE dx

= 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

dE dx

= ρ x( ) KSε0

ρ

x

N P

−xn

ρ = +qND

xp

ρ = −qNA

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