semiconductor equations - nanohub.org
TRANSCRIPT
![Page 1: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/1.jpg)
ECE 305: Spring 2018
Semiconductor Equations
Professor Peter BermelElectrical and Computer Engineering
Purdue University, West Lafayette, IN [email protected]
Pierret, Semiconductor Device Fundamentals (SDF)
Chapter 3 (pp. 122-138)
2/8/18 Bermel ECE 305 S18 1
![Page 2: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/2.jpg)
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
2/8/18
✓
✓
✓
2
![Page 3: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/3.jpg)
the semiconductor equations
Five equations in five
unknowns:
r = q p - n + N
D
+ - NA
-( )
2/8/18 Bermel ECE 305 S18 3
( )J x
( )J x dx
Ng
n
Nr
, 𝐽𝑝 𝐽𝑛
![Page 4: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/4.jpg)
Semiconductor equations: 2 key cases
Diffusion problems (ℇ = 0): MCDE
Drift problems (ℇ ≠ 0): Drift current equations
2/8/18 Bermel ECE 305 S18 4
¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
![Page 5: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/5.jpg)
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
2/8/18
✓
✓
✓
5
![Page 6: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/6.jpg)
Example #1: Solution
x
Dn x( )
Dn x( ) = GLt n
x = 0
Steady-state, uniform generation, no spatial variation
Dn = GLt n =1020 ´10-6 =1014
2/8/18 Bermel ECE 305 S18
x = L = 200 mm
6
![Page 7: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/7.jpg)
Example #2 Summary
Bermel ECE 305 S18
1
JP P P
pr g
t q
D AD q p n N N
P P Pqp qD p J E
1N N N
nr g
t q
J
N N Nqn qD n J E
1
time
Dn
time
Analytical solutions
ntt
entxn
DD 0,
( , ) 1 nt
nn x t G e
D
Dn
2/8/18 7
![Page 8: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/8.jpg)
Example #3: One-sided Carrier Diffusion
Bermel ECE 305 S18
1 nN N
n dJr g
t q dx
N N N
dnqn E qD
dx J
2
20 N
d nD
dx
Steady state, no generation/recombination
1
0,' D txn
a 0x’
Metal contact
2/8/18 8
![Page 9: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/9.jpg)
Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample long compared to the diffusion length.
i.e., a short diffusion length
fixedDp x = 0( )
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
2/8/18 Bermel ECE 305 S1810
![Page 10: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/10.jpg)
Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample long compared to the diffusion
length. No generation. fixedDn x = 0( ) =1012 cm-3
Step 4B ) Key words: steady-state, without light, long device
ii) If I write for MCDE, would I be right?¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
iii) Which approximate equation should I choose:
Step 3) What type of problems are we talking about?
2/8/18 Bermel ECE 305 S1811
![Page 11: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/11.jpg)
Example 3B: One-sided carrier diffusion
Long sample (steady state)
x
Dn x( )
Dn x ®¥( ) = 0
Dn 0( )
Dn x( ) = Dn 0( )e-x/Ln
x = L = 200 mmx = 0
Ln = Dnt n << L
Steady-state, sample long compared to the diffusion length.
2/8/18 Bermel ECE 305 S1812
![Page 12: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/12.jpg)
Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample is 5 micron long. No generation.
Which approximate equation should I choose:
Dn x = 0( ) =1012 cm-3
Dn x = 5 mm( ) = 0
What are my boundary conditions?
2/8/18 Bermel ECE 305 S1813
![Page 13: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/13.jpg)
Example 3B: One-sided carrier diffusion
Long sample (steady state)
Dn x = 0( ) =1012 cm-3
0Dn x LD
Dn x = 0( ) =1012 cm-3
Dn x( ) = Ax + B
Dn x( ) = Dn 0( ) 1-x
L
æ
èçö
ø÷
0n x LD
2/8/18 Bermel ECE 305 S1814
![Page 14: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/14.jpg)
Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
Steady-state, sample is 30 micrometers long. No generation.
fixedDn x = 0( ) =1012 cm-3
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
¶Dn
¶t= Dp
d2Dn
dx2-
D
t n
+ GL
0 = Dp
d2Dn
dx2-
D
t n
+ 0
d2Dn
dx2-
Dn
Ln
= 0 Ln º Dnt n
Dn x = 30 mm( ) = 0
Ln = 28 mm L= 30 mm
2/8/18 Bermel ECE 305 S1815
![Page 15: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/15.jpg)
Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
Steady-state, sample is 30 micrometers long. No generation.
fixedDn x = 0( ) =1012 cm-3
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
d2Dn
dx2-
Dn
Ln
= 0
Dn x = 30 mm( ) = 0Dn x( ) = Ae-x/Ln + Be+x/Ln
Dn 0( ) = A + B =1012
Dn L( ) = Ae-L/Ln + Be+L/Ln = 0
2/8/18 Bermel ECE 305 S1816
![Page 16: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/16.jpg)
Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
x
Dn x( )
Dn x = L( ) = 0
Dn 0( )
x = Lx = 0
Dn x( )= Dn 0( )sinh L - x( ) / Lnéë ùû
sinh L / Ln( )
Steady-state, sample neither long nor short compared to the
diffusion length.
Ln = Dnt n » L
2/8/18 Bermel ECE 305 S1817
![Page 17: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/17.jpg)
Example #3 summary: One-sided carrier diffusion
Length scale Solution type
Decaying exponentials
Linear
Hyperbolic functions
2/8/18 Bermel ECE 305 S1818
![Page 18: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/18.jpg)
Homework #5: Fall 2017
(cf. this week’s homework!)
1. Assume that a n-type region of crystalline silicon with 𝜇𝑝 = 450 cm2/V ∙ s and lifetime 𝜏𝑝 = 10 ms is uniformly
illuminated by a photon flux 𝐺𝐿 = 1020 /cm3∙s, reaches a steady state, which is then switched off at 𝑡 = 0.
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
b. Sketch the time-dependent decay of the carrier concentration.
c. What is the value of the carrier concentration at 𝑡 = 20 ms?
2/8/18 Bermel ECE 305 S1819
![Page 19: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/19.jpg)
Homework #5: Fall 2017
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
2/8/18 Bermel ECE 305 S1820
![Page 20: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/20.jpg)
Homework #5: Fall 2017a.
b. Sketch the time-dependent decay of the carrier concentration.
2/8/18 Bermel ECE 305 S1822
![Page 21: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/21.jpg)
Homework #5: Fall 2017a.
b.
c. What is the value of the carrier concentration at 𝑡 = 20 ms?
2/8/18 Bermel ECE 305 S1824
![Page 22: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/22.jpg)
Homework #5: Fall 2017
2. Assume that a uniform p-type region of gallium arsenide (𝑇 = 300 K) has length 𝐿 = 1 m, lifetime 𝜏𝑛 = 5 ns and 𝜇𝑛 = 8500 cm2/V ∙ s. Assume that all carriers are extracted at 𝑥 = 𝐿 (such that ∆𝑛 = 0), while ∆𝑛 = 1013 /cm3 at 𝑥 = 0. Assume that 𝑝𝑜 ≫ ∆𝑛 everywhere. Now consider the system after reaching a steady state.
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
b. Solve for and sketch the minority carrier concentration ∆𝑛 as a function of position between 𝑥 = 0 and 𝑥 = 𝐿.
c. How would this problem change if the minority carrier lifetime 𝜏𝑛 were reduced to 1 ns? Justify your answer quantitatively.
2/8/18 Bermel ECE 305 S1826
![Page 23: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/23.jpg)
Homework #5: Fall 2017
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
2/8/18 Bermel ECE 305 S1827
![Page 24: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/24.jpg)
Homework #5: Fall 2017a.
b. Solve for and sketch the minority carrier concentration ∆𝑛 as a function of position between 𝑥 = 0 and 𝑥 = 𝐿.
2/8/18 Bermel ECE 305 S1829
![Page 25: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/25.jpg)
Homework #5: Fall 2017a.
b.
c. How would this problem change if the minority carrier lifetime 𝜏𝑛 were reduced to 1 ns? Justify your answer quantitatively.
2/8/18 Bermel ECE 305 S1831
![Page 26: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/26.jpg)
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
2/8/18
✓
✓
33
![Page 27: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/27.jpg)
“the semiconductor equations”
Three equations in three
unknowns:
In steady state equilibrium, we only need to solve the
Poisson equation
How do we calculate rho(x), E(x), and V(x)?
2/8/18 Bermel ECE 305 S18 34
![Page 28: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/28.jpg)
equilibrium energy band diagrams solve
Poisson’s equation – and more
EFEF
1) Begin with EF
2) Draw the E-bands where you know the carrier density
3) Electrostatic potential by flipping E-band upside down.
4) E-field from slope
5) n(x), p(x) from the E-band diagram
6) rho(x) from n(x) and p(x)
7) diffusion current from (5) or from (6)
EC x( ) = EC-ref - qV x( )
E x( ) =1
qdEC x( ) dx
2/8/18 Bermel ECE 305 S18 35
![Page 29: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/29.jpg)
energy band diagram
36
EF
EC
EV
x
E
Ei
x = xpx = 0x = -xn
2/8/18 Bermel ECE 305 S18 36
![Page 30: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/30.jpg)
“read” the e-band diagram
1) Electrostatic potential vs. position
2) Electric field vs. position
3) Electron and hole densities vs. position
4) Space-charge density vs. position
2/8/18 Bermel ECE 305 S18 37
![Page 31: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/31.jpg)
NP junction (equilibrium)
N P
xp-xn 0
“transition (depletion) region”
Bermel ECE 305 S18
Vbi =kBT
qln
NDNA
ni
2
æ
èçö
ø÷
V = 0V = Vbi
1) What is the width of the depletion region?
2) What is the maximum electric field?
2/8/18 38
![Page 32: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/32.jpg)
the Poisson equation
d
dxeSE( ) = r x( )
dE
dx=
r x( )eS
=r x( )KSe0
dE
dx=
r x( )KSe0
2/8/18 Bermel ECE 305 S18 39
![Page 33: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/33.jpg)
the “depletion approximation”
dE
dx=
r x( )KSe0
r
x
N P
-xn
r = +qND
xp
r = -qNA
qNDxn = qNAxp
NDxn = NAxp
2/8/18 Bermel ECE 305 S18 40
![Page 34: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/34.jpg)
the electric field
dE
dx=
r x( )KSe0
r
xN P
xp
-xn
r = -qNA
dE
dx=
qND
KSe0
dE
dx= -
qNA
KSe0
E x( ) > 0
2/8/18 Bermel ECE 305 S18 41
![Page 35: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/35.jpg)
the electrostatic potential
dE
dx=
r x( )KSe0
xN P
xp-xn
E
E x( ) = -dV
dx
W = xn + xP
E 0( ) =qND
KSe0
xn
NDxn = NAxPxn =NA
NA + ND
W
dE
dx=
qND
KSe0
2/8/18 Bermel ECE 305 S18 42
![Page 36: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/36.jpg)
the electrostatic potential
dE
dx=
r x( )KSe0
N P
E x( ) = -dV
dx
Vbi = E x( )-xn
xp
ò dx
Vbi =1
2E 0( )W
E 0( ) =qND
KSe0
NA
NA + ND
W
W =2KSe0
q
NA + ND
ND NA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2qVbi
Kse0
ND NA
NA + ND
æ
èçö
ø÷
é
ëê
ù
ûú
1/2
2/8/18 Bermel ECE 305 S18
xxp-xn
E
ℰ(0)
43
![Page 37: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/37.jpg)
calculating 𝑉(𝑥) from ℰ(𝑥)
xP
xp-xn
E
V x( ) = - E x( )x
xp
ò dx
V x( )
V = 0
V = Vbi
See Pierret, SDF, pp. 212-213
2/8/18 Bermel ECE 305 S18
Vbi = E x( )-xn
xp
ò dx
Vbi =1
2E 0( )W
ℰ(0)
𝑊
E x( ) = -dV
dx
44
![Page 38: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/38.jpg)
summary
xN P
xp-xn
E
W = xn + xP
NDxn = NAxP
xn =NA
NA + ND
W
xp =ND
NA + ND
W
W =2KSe0
q
NA + ND
ND NA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2qVbi
Kse0
ND NA
NA + ND
æ
èçö
ø÷
é
ëê
ù
ûú
1/2
Vbi =kBT
qln
ND NA
ni
2
æ
èçö
ø÷
E 0( ) =2Vbi
W
2/8/18 Bermel ECE 305 S18 45
![Page 39: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/39.jpg)
example
N P
ND = 1015
depletion region
xp-xn 0
E 0( )
W
NA = 1015
Vbi =kBT
qln
NDNA
ni
2
æ
èçö
ø÷= 0.6 V
W =1.25 mm xn = xp = 0.625 mm
E 0( ) = 9.6 ´103 V/cm
2/8/18 Bermel ECE 305 S18 46
![Page 40: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/40.jpg)
“one-sided junction”
N P
ND =1018 cm-3
0
E 0( )
W
NA = 1015
x
xn , xp
2/8/18 Bermel ECE 305 S18 47
![Page 41: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/41.jpg)
numbers: one-sided junction
W = 1.010 ´10-4 cm 1.25 ´10-4( )
E 0( ) =2Vbi
W= 1.5 ´104 V/cm (0.96)
N
ND = 1018
0
Vbi = 0.78 V (0.60)
xp-xn
W NA = 1015
xn = 0.001´10-4 cm
xp = 1.009 ´10-4 cm
P
2/8/18 Bermel ECE 305 S18 48
![Page 42: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/42.jpg)
one-sided junction
ND >> NA
xN P
xp
-xn
dE
dx=
qND
KSe0
dE
dx= -
qNA
KSe0
E x( )
xp >> xn
2/8/18 Bermel ECE 305 S18 49
![Page 43: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/43.jpg)
one-sided junction
W =2KSe0
q
NA + ND
NDNA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
®2KSe0
qNA
Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2Vbi
W
xn =NA
NA + ND
W » 0
xp =ND
NA + ND
W » W
N P
ND = 1018
depletion region
xp-xn0
E x( )
Vbi »kBT
qln
ND NA
ni
2
æ
èçö
ø÷
W NA = 1015
2/8/18 Bermel ECE 305 S1850
![Page 44: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/44.jpg)
example: one-sided junction
W =2KSe0
qNA
Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2Vbi
W
xn » 0
xp » W
N P
ND = 1018
xp-xn 0
E 0( )
Vbi »kBT
qln
ND NA
ni
2
æ
èçö
ø÷
W NA = 1016
2/8/18 Bermel ECE 305 S18 51
![Page 45: Semiconductor Equations - nanoHUB.org](https://reader030.vdocument.in/reader030/viewer/2022020623/61f2083496cfc00e297d3ca6/html5/thumbnails/45.jpg)
conclusions We will solve many problems in this class using the
semiconductor equations: In regions of zero field, we can use the minority carrier
diffusion equation to understand the mechanics of carrier transport in electronic devices. Review the problem carefully to see if the assumption of minority carrier transport is satisfied.
In regions of non-zero field like NP junctions, we can use band diagrams to sketch ℰ, 𝑉, 𝑛, and 𝑝 ; and Poisson’s equation and ‘depletion approximation’ to quantify these values.
This approach also gives us the width of the ‘depletion region’ on both sides of the junction (𝑥𝑛 and 𝑥𝑝), plus the ‘built-in’ voltage 𝑉𝑏𝑖
2/8/18 Bermel ECE 305 S1852