ECE606: Solid State Devices Lecture 13 Solutions of …ee606/downloads/ECE606_f... · 2012-10-02 · ECE606: Solid State Devices Lecture 13 Solutions of the Continuity Eqs. Analytical

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ECE606: Solid State DevicesLecture 13

Solutions of the Continuity Eqs.Analytical & Numerical

Gerhard [email protected]


Outline

2

Analytical Solutions to the Continuity Equations

1) Example problems

2) Summary

Numerical Solutions to the Continuity Equations

1) Basic Transport Equations

2) Gridding and finite differences

3) Discretizing equations and boundary conditions

4) Conclusion


Consider a complicated real device example

3

x

1 2 3

UnpassivatedsurfaceMetal contact

� Acceptor doped

� Light turned on in the middle section.

� The right region is full of mid-gap traps because of dangling bonds due

to un-passivated surface.

� Interface traps at the end of the right region

(That’s where the dangling bonds are…)

� The left region is trap free.

� The left/right regions contacted by metal electrode.


Recall: Analytical Solution of Schrodinger Equation

4

( ) 0

( ) 0

x

x

ψψ

= −∞ == +∞ =

d 2ψdx2

+ k2ψ = 0

B B

B B

x x x x

x x x x

d d

dx dx

ψ ψ

ψ ψ

− +

− +

= =

= =

=

=

1) 2N unknowns for N regions

2) Reduces 2 unknowns

3)Set 2N-2 equations for 2N-2 unknowns (for continuous U)

Det(coefficient matix)=0And find E by graphical or numerical solution

4) 2( , ) 1x E dxψ

∞

−∞=∫5)

for wave function


Recall: Bound-levels in Finite well

5

( ) 0

( ) 0

x

x

ψψ

= −∞ == +∞ =

U(x)

E

2) Boundary Conditions …

0 a

sin cosA kx B kxψ = +

x xMe Ceα αψ − ++=x xDe Neα αψ − ++=

1)


Analogously, we solve for our device

6

Solve the equations in different regions independently.

Bring them together by applying boundary conditions.


Region 2: Transient, Uniform Illumination, Uniform doping

7

1N N N

nr g

t q

∂ = ∇ • − +∂

J

2

J µ= + ∇N N Nqn E qD n

(uniform)

0( )

n

n n nG

t τ∂ + ∆ ∆= − +

∂ Acceptor doped

1∂ −= ∇ • − +∂

J p P p

pr g

t qµ= − ∇J p p Pqp E qD p

(uniform)

0( )

τ∂ + ∆ ∆= − +

∂ p

p p pG

tMajority carrier

( ) ( )0 0 0+ − + −∇ • = − + − = + ∆− − + − =∆D A D AD q p n N N q p n N Npn

Recall Shockley-Read-Hall

Electric field still zero because new carriers balance


Example: Transient, Uniform Illumination, Uniform doping, No applied electric field

8

2

( )

n

n nG

t

∂ ∆ ∆= − +∂ τ

( , ) ntn x t A Be−∆ = + τ

( )( , ) 1 ntnn x t G e ττ −∆ = −

Acceptor doped

0, ( ,0) 0

, ( , ) n

t n x A B

t n x G A

= ∆ = ⇒ = −→ ∞ ∆ ∞ = =τ

time

No carriers yet generated…

Steady state, no change in carriers with time…


Region 1: One sided Minority Diffusion at steady state

9

∂n∂t

= 0(steady-state)

rN = 0(trap free)

gN

= 0(nogeneration)1

E = 0

DN

dndx

≠ 0 (due to insertion of electrons from central region)

2

20 = N

d nD

dx

Steady stateAcceptor doped

Trap-free

1 nN N

n dJr g

t q dx

∂ = − +∂

N N N

dnqn E qD

dxµ= +J


Example: One sided Minority Diffusion

10

, ( ' ) 0

(Metal has high electron density

as compared to semiconductor)

x a n x a C Da= ∆ = = ⇒ = −

'( , ) ( 0 ') 1

∆ = ∆ = −

xn x t n x

a

2

20 N

d nD

dx=

( , ) 'n x t C Dx∆ = +

x’

a

Metal contact

x = 0', ∆n (x ' = 0')=C

Just substitute x=0 in above eqn.

0x’

υ= m mJ q n


Region 3: Steady state Minority Diffusion with recombination

11

20

2

( )0

τ+ ∆ − ∆=

nND

n

dx

nd n

3

Steady stateAcceptor doped

Flux2

20

τ∆= ∆−N

n

d n

d

n

xD

Trap-filled

∂n∂t

= 0(steady-state)

rN

≠ 0(not trap free)

g N = 0(nogeneration)

E = 0

DN

dndx

≠ 0 (due to insertion of electrons from central region)


Diffusion with Recombination …

12

( , ) n nx L x Ln x t Ee Fe−∆ = +

2

20

τ∆ ∆− =N

n

d n nD

dx

3

2⇒ = − nb LF Ee

x

b

22

(0)( , ) ( )

(1 )−∆∆ = −

−n n n

n

x L b L x Lb L

nn x t e e e

e

∆n, ( ) 0= ∆ = =x b n x b

0, ( 0) ( 0)= ∆ = = + = ∆ =x n x E F n x

0 X

Metal contactFunctionally similar to Schrodinger eqn.


Combining them all ….

13

1 ( 0)'

( ') 1∆ =∆ = −

x

anx xn

1 2 322

2

(0

( )

0 ')( )

τ∆ = =∆ = ∆

n

n n

n x G

22

(0 ')( ) ( )

(1 )−∆ = −

−∆

n n n

n

x L b L x Lb L

n x e e en

e

'1n

xG

a = −

τ

2

2

( )

(1 )

n n n

n

x L b L x Ln

b L

G e e e

e

−−=−

τ

Match boundary condition

Calculating current N N N

dnqn E qD

dxµ= +J

∆n

b0x’ 0’


Analytical Solutions Summary

14

1) Continuity Equations form the basis of analysis

of all the devices we will study in this course.

2) Full numerical solution of the equations are

possible and many commercial software are

available to do so.

3) Analytical solutions however provide a great deal

of insight into the key physical mechanism

involved in the operation of a device.


Outline

15


1) Example problems

2) Summary





4) Conclusion


Preface

16

• The 5 equations we derived the past few lectures have been used for the longest time in the industry and in academia to understand carrier transport in devices.

• It is useful to know the essentials of how these equations are implemented on a modern computer so that one understands some of the finer details involved in creating tools that simulate thee phenomena.

• Understanding some of these details helps one become a ‘power user’ of the simulation tools that implement the physics. One also understands the limitations re. numerical issues and applicability ranges of results.


Equations to be solved – derived last time…

17

( )+ −∇ • = − + −D AD q p n N N

1∂ −= ∇ • − +∂

JP P P

pr g

t q

P P Pqp E qD pµ= − ∇J

1N N N

nr g

t q

∂ = ∇ • − +∂

J

J µ= + ∇N N Nqn E qD n

Band-diagram

Diffusion approximation,Minority carrier transport,

Ambipolar transport


1) The Semiconductor Equations

18

∇•�

D = ρ∇•

�J

n−q( ) = g

N− r

N( )∇•

�J

pq( ) = g

P− r

P( )

VED ∇−==��

00 κεκε

(steady-state)

( )−+ −+−= AD NNnpqρ�

Jn

= nqµn

�E +qD

n

�∇n

�J

p= pqµ

p

�E − qD

p

�∇p

, ( , ) etc.=N Pg f n p

Conservation Laws: not specific to a particular problem - Universal

Constitutive relations: specific to problem at hand – reflect physics of the problem


1) The Mathematical Problem

19

∇•�

D = ρ∇•

�J n −q( ) = g N − rN( )

∇•�

J p q( ) = g N − rN( )

The “Semiconductor Equations”

3 coupled, nonlinear,second order PDE’sfor the 3 unknowns:

V (�r ) n (

�r ) p (

�r )

Conservations laws: exactTransport eqs. (drift-diffusion): approximate

Why are these equations coupled?Potential�Field�current�changespotential�changes field and so on…

Why are these equations coupled?Potential�Field�current�changespotential�changes field and so on…


Outline

20


1) Example problems

2) Summary





4) Conclusion


2) The Grid

21

(ii) “exact” numerical solutions

i

i

i

p

n

V

N nodes3N unknowns

a

‘Gridding‘– total length divided into ‘N’ parts- equal (uniform gridding) , or - unequal (adaptive and non-uniform gridding)

Variables described at each point ‘i’.

Vo and Vn+1 is known because these are voltages at source and drain.


Finite Difference Expression for Derivative

22

f(x)

xxi xi+1xi-1

( )1/ 2

1

i

i i

x

fdf

dx

f

a+

+ −=

a

“centered difference”

0( )2 2( )o

af xfd

da

a f

xx += ++((

2)) 2oo

a df

dxaff x x += −


The Second Derivative …

23

( ) ( )0 0

0

2 2

20 2= =

= ++ + +x a x a

df a d ff a ...

dx dxx a xf

( ) ( )0 0

0

2 2

20 2= =

= −− + −x a x a

df a d ff a ...

dx dxx a xf

( ) ( ) ( )0

00 0

22

22

=

−+ − =+x a

xd f

f f adx

x af x a

21 1

2 2

2i i i

i

f f fd f

dx a

− +− +=3 point formula, could be extended to N points depending on the number of derivatives we carry in our expansion


Outline

24


1) Example problems

2) Summary





4) Conclusion


2) Control Volume

25

x

(i)(i -1) (i +1)

“control volume”

3 unknowns at each node:

iii pnV ,,

Need 3 equationsat each node


Discretizing Poisson’s Equation

26

2 / os

V Kρ ε∇ = −

( ) 0,,,, 11 =+− iiiiii

V pnVVVF

(i)(i -1) (i +1)

DRDL

V ( i − 1) − 2V ( i ) +V ( i + 1)

a2

= − q

Ksε

0

(pi −ni +ND ,i+ −N

A,i− )

0 0s sKD KD Vρ ε ε∇ • = = ∇E = -

Since Vo and V-1 are known, as are carrier concentration on doping (or lack thereof) in contacts, we find V1 and iterate from this point to solve for potential. Once this potential is

found, solve continuity equation to obtain new carrier concentrations


Discretizing Continuity Equations

27

nL n n

dV

dJ q k

dn

x

n

xT

dµ µ= − +

∇•�

Jn

= −q gN

− rN( )

The simplest approach…..

( )1 11

/2i iin iiL

n

iJ

kT kT q

n n V

a

n nV

aµ−−−

= − +

−

−+

(i)(i -1) (i +1)JnL

( )1 1 1, , , , , 0i

n i i i i i iF V V n n p p− − − =


Three Discretized Equations

28

0

0

0

=

=

=

ip

in

iV

F

F

F

3 unknowns at each nodeN nodes3N unknowns and 3N equations (coupled to each other)

x

(i)(i -1) (i +1, j)


Numerical Solution – Poisson Equation Only

29

• Have a system of 3N nonlinear equations to solve

• Recall Poisson’s equation at node (i):

( ) 0,,,, 11 =+− iiiiii

V pnVVVF

linear if ni and pi are known A[ ]�

V =�b

[A]:�

V =

V1

V2

⋮

VN


Boundary conditions

30

20 0 in p n= 2

1 1N N in p n+ + =

a

AV V= 0V =

Dopant density

Contacts are assumed large and in equilibrium � detailed balance and law of mass-action apply!!

One could have unequal materials on the two contact sides, one must be careful to use the right intrinsic concentration <-> material.


Numerical Solution…

31

11 11 11 1 0

2

1

1

11 11 11 0

1

11 1

2

1

1

2

( )

( )

( )

( )

( )

( )

( )

....

( )

..

...

( )

m m

mm

mm

m m

m mm

V x

V x

V x

N R V n x n

n x

n x n

R P V p

P p

V Q

p x

p x

V Q

p x

+

+

=

p=>Qn=>Q

Poissonp-continuityn-continuity

Off-diagonal terms are Poisson-Continuity equations talking to each other. Recombination-generation terms also feed into continuity equations.

V=>n

V=>p


3) Uncoupled Numerical Solution

32

The semiconductor equations are nonlinear!(but they are linear individually)

Uncoupled solution procedure

Guess V,n,p

Solve Poissonfor new V

Solve electroncont for new n

Solve holecont for new p

repeatuntilsatisfied


Summary

33

1) Two methods to solve drift-diffusion equation consistently – analytical and numerical.

2) Analytical solution provides great insight and the solution methodology is similar to that of Schrodinger equations.

3) Numerical solution is more versatile. One begins with a set of equations and boundary conditions, discretize the equations on a grid with N nodes to obtain 3N nonlinear equations in 3N unknowns, and solve the system of nonlinear equations by iteration.

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ECE606: Solid State Devices Lecture 13 Solutions of …ee606/downloads/ECE606_f... · 2012-10-02 · ECE606: Solid State Devices Lecture 13 Solutions of the Continuity Eqs. Analytical