ece606: solid state devices lecture 13 solutions of …ee606/downloads/ece606_f... ·...
Post on 30-Jul-2018
218 Views
Preview:
TRANSCRIPT
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
ECE606: Solid State DevicesLecture 13
Solutions of the Continuity Eqs.Analytical & Numerical
Gerhard Klimeckgekco@purdue.edu
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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)
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Analogously, we solve for our device
6
Solve the equations in different regions independently.
Bring them together by applying boundary conditions.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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…
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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)
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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’
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Outline
15
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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…
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Outline
20
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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 += −
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Outline
24
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
2) Control Volume
25
x
(i)(i -1) (i +1)
“control volume”
3 unknowns at each node:
iii pnV ,,
Need 3 equationsat each node
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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− − − =
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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)
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
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.
top related