Upload
others
View
4
Download
0
Embed Size (px)
Citation preview
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 1
SOLUTIONS: ECE 606 Homework Week 8 Mark Lundstrom Purdue University March 7, 2013
1) The doping profile for an n-‐type silicon wafer ( N D = 1015 cm-‐3) with a heavily doped
thin layer at the surface (surface concentration, NS = 1020 cm-‐3) is sketched below. Answer the following questions.
1a) Assume approximate space charge neutrality ( n x( ) ! N D x( ) ) and equilibrium
conditions and compute the position of the Fermi level with respect to the bottom of the conduction band at x = 0 and as x !" .
Solution:
n0 x( ) = NCe EF !EC( ) kBT " N D x( )
EF ! EC x( ) = kBT ln
N D x( )NC
"
#$$
%
&'' NC = 3.23!1019 cm-‐3
EF ! EC 0( ) = kBT ln
N D 0( )NC
"
#$$
%
&''
EF − EC 0( ) = kBT ln 1020
3.23×1019
⎡
⎣⎢
⎤
⎦⎥ = +1.13kBT
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 2
HW Week 8 Solutions Continued
EF ! EC x "#( ) = kBT ln
N D x "#( )NC
$
%&&
'
())
EF − EC x →∞( ) = kBT ln 1015
3.23×1019
⎡
⎣⎢
⎤
⎦⎥ = −10.4kBT
1b) Using the above information, sketch EC x( ) vs. x. Be sure to include the Fermi level.
1c) Sketch the electrostatic potential vs. position.
1d) Sketch the electric field vs. portion.
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 3
HW Week 8 Solutions Continued 1e) Derive an expression for the position dependent electric field, E x( ) , in terms of
the position-‐dependent doping density, N D x( ) . HINT: Use the electron current equation and assume equilibrium conditions.
Solution:
Jn = nqµnE + kBTµn
dndx
= 0
E =
kBTq
1n
dndx
=kBTq
1N D x( )
dN D x( )dx
E =kBTq
!"#
$%&
1N D x( )
dN D x( )dx
Another way is to begin with n0 ≈ ND = NCe
EF −EC( ) kBT and differentiate.
2) A silicon diode is symmetrically doped at N D = N A = 1015 cm-‐3. Answer the following questions assuming room temperature, equilibrium conditions, and the depletion approximation. 2a) Compute Vbi .
Solution:
Vbi =
kBTq
lnN AN D
ni2
!
"#$
%&= 0.026ln 1030
1020
!"#
$%&= 0.60 V
Vbi = 0.60
2b) Compute
xn ,xp and W. Solution:
xn =
2! S"0
qN A
N D N A + N D( )Vbi
#
$%%
&
'((
1/2
= 0.625 µm
xn = xp = 0.625 µm (because N and P regions are symmetrical)
W = xn + xp = 1.25 µm
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 4
HW Week 8 Solutions Continued
2c) Compute V x = 0( ) and E x = 0( ) .
Solution: By symmetry:
V 0( ) = Vbi
2= 0.30 V or use
V x = 0( ) = qN A
2! S"0
xp2
E x = 0( ) = qN A
! S"0
xp = 9.6#103
E 0( ) = !9.6"103 V/cm
2d) Sketch ! x( ) vs. x.
Solution: ρN = +qND = +1.6 ×10−4 C/cm3
!P = "qNA = "1.6 #10"4 C/cm3
3) Your textbook (Pierret, SDF) presents the “classic” expressions for PN junction
electrostatics. Simplify these expressions for a “one-‐sided” P+N junction for which
N A >> N D . Present simplified expressions (when possible) for: 3a) The built-‐in potential, Vbi , from Pierret, Eqn. (5.10). Solution:
Vbi =kBTqln NDNA
ni2
!"#
$%& no simplification possible
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 5
HW Week 8 Solutions Continued
3b) The total depletion layer depth, W , from Pierret, Eqn. (5.31).
Solution:
W = 2! S"0q
NA + ND
NDNA
#$%
&'(Vbi
)
*+
,
-.
1/2
NA >> N D W = 2! S"0qND
Vbi#
$%
&
'(
1/2
3c) The peak electric field, E 0( ) , from Pierret, Eqn. (5.19) or (5.21).
Solution:
E 0( ) = 2Vbi
W= 2qVbi
! s"0NDNA
NA + ND
#$%
&'(
E 0( ) = 2qNDVbi
! s"0
3d) The electrostatic potential, V x( ) from Pierret, Eqn. (5.28)
Solution:
V x( ) =Vbi !qND
2" S#0xn ! x( )2 V x( ) =Vbi −
qND
2κ Sε0W − x( )2
Now use the expression for W above to find:
V x( ) =Vbi 1− 1− x W( )2⎡⎣
⎤⎦
4) A silicon diode is asymmetrically doped at N A = 1019 cm-‐3 and N D = 1015 cm-‐3 Answer
the following questions assuming room temperature, equilibrium conditions, and the depletion approximation.
4a) Compute Vbi .
Solution:
Vbi =
kBTq
lnN AN D
ni2
!
"#$
%&= 0.026ln 1025 '1019
1020
!"#
$%&= 0.84 V
Vbi = 0.84
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 6
HW Week 8 Solutions Continued
4b) Compute
xn ,xp and W. Solution:
xp ! 0
xn !W =
2" S#0
qN D
Vbi
$
%&
'
()
1/2
= 1.05 µm
W = 1.05 µm (depletion region mostly on the N-‐side, the lightly doped side)
4c) Compute V x = 0( ) and E x = 0( ) .
Solution:
V 0( ) ! 0 V
E 0( ) = qN D
! S"0
W = 1.6#104 V/cm
E 0( ) = 1.6!104 V/cm (plus sign assumes N region is on the left)
4d) Sketch ! x( ) vs. x.
Solution:
The charge on the P-‐side is essentially a delta function with the total charge in C/cm2
equal in magnitude and opposite in sign to the charge on the N-‐side.
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 7
HW Week 8 Solutions Continued
5) Repeat problem 4) using the “exact” solution to PN junction electrostatics.
Solution:
VN = + kBTqln ND
ni
⎛⎝⎜
⎞⎠⎟= 0.026 ln 1015
1010⎛⎝⎜
⎞⎠⎟= 0.30
VP = ! kBTqln NA
ni
"#$
%&'= 0.026 ln 1019
1010"#$
%&'= !0.540
Vbi =VN !VP = 0.84 V Vbi = 0.84
V 0( ) = CN !CP
aN ! aP
aN = ND = 1015 aP = !NA = !1019 CN = aNVN ! 2ni kBT q( )cosh qVN kBT( ) CN = 1015 ! 0.30 " 2 !1010 0.026( )cosh 11.5( ) = 2.74 !1014 CP = aPVP ! 2ni kBT q( )cosh qVP kBT( ) CP = !1019( ) !0.54( )! 2 "1010 0.026( )cosh !20.7( ) = 5.15 "1018
V 0( ) = CN !CP
aN ! aP= !0.518
V 0( )!VP = !0.54 ! 0.512 = 0.028 " kBT q
The potential drop across the heavily doped side is about kBT/q.
E 0( ) = 2q ! S"0 ni kBT q( )eqV (0) kBT + ni kBT q( )e#qV (0) kBT # aNV (0)+CN( )1/2 Putting in numbers, we find:
E 0( ) !1.7 "105 V/cm V/cm
which is about 10X the electric field we found in prob. 4.
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 8
HW Week 8 Solutions Continued
n0 0( ) = nieqV 0( ) kBT = 27 cm-‐3 p0 0( ) = nie!qV 0( ) kBT = 0.37 "1017 cm-‐3 ! 0"( ) = q p0 0( )" n0 0( ) + ND#$ %& = q 0.37 '1019 +1015#$ %& ( q 0.37 '1019#$ %&
! 0"( ) # q 0.37 $1019%& '( (depletion approximation would give ! 0"( ) # q 1015$% &' )
! 0+( ) = q p0 0( )" n0 0( )" NA#$ %& = q 0.37 '1019 "1019#$ %&
! 0+( ) = "q 0.63#1019$% &' (depletion approximation would give ! 0+( ) " #q 1019$% &' )
6) Semiconductor devices often contain “high-‐low” junctions for which the doping density
changes magnitude, but not sign. The example below shows a high-‐low step junction. Answer the questions below.
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 9
HW Week 8 Solutions Continued
6a) Sketch an energy band diagram for this junction.
6b) Sketch V x( )
6c) Sketch E x( )
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 10
HW Week 8 Solutions Continued 6d) Sketch ! x( ) vs. x.
6e) Name the charged entities responsible for ! x( ) in 6d).
Solution: For x < 0, the charge is a depletion charge. Mobile electrons leave the heavily doped side of the junction leaving behind a concentration, ND1, of ionized donors. For x > 0, the charge is due to the additional mobile electrons that have spilled over from the heavily doped side. This is NOT a depletion region.
6f) Explain why the depletion approximation cannot be used for this problem.
Solution: Because, as explained above, there is a depletion region on only ONE side of the junction. We could use the depletion approximation there, but not on the lightly doped side.
6g) Calculate Vbi for this high-‐low junction assuming silicon at room temperature.
Solution: First, consider the two sides of the junction separately:
n01 = NCe EF 1!EC( ) kBT n02 = NCe EF 2!EC( ) kBT
n01
n02
= e EF 1!EF 2( ) kBT
The built-‐in potential develops to align these two Fermi levels:
EF1 ! EF 2( ) = qVbi = kBT lnn01
n02
"
#$%
&'
Vbi =kBTq
lnN D1
N D2
⎛
⎝⎜⎞
⎠⎟
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 11
HW Week 8 Solutions Continued 7) Consider an N+P diode with the length of the quasi-‐neutral P-‐region being, WP. Answer the
following questions assuming that recombination in the space-‐charge region can be neglected. 7a) Derive a general expression for ID VA( ) valid for a P region of any length, WP.
Solution: In HW7, problem 12c, we solved the minority carrier diffusion equation for a region of any length and found:
!n x( )= !n 0( ) sinh WP " x( ) / Ln#$ %&sinh WP / Ln( )
Let x = 0 be the edge of the neutral P-‐region. The electron current is:
Jn = +qDn
d!ndx x=0
= "qDn
Ln
!n 0( )cosh WP Ln( )sinh WP Ln( ) (minus sign means that the electron
current is flowing in the minus x direction. Since this is a one-‐sided junction, and we are ignoring recombination in the space-‐charge region, this is the total diode current, ID. Let’s define the forward biased current to be positive.
ID = !AJn = qA
Dn
Ln
"n 0( )cosh WP Ln( )sinh WP Ln( )
Finally, use the “Law of the Junction” for the boundary condition:
!n 0( ) = ni
2
N A
eqVA kBT "1( ) to find:
ID = qADn
Ln
ni2
N A
!
"#
$
%&
cosh WP Ln( )sinh WP Ln( ) eqVA kBT '1( )
7b) Simplify the expression derived in 7a) for a “long diode”. Explain what “long” means
(i.e. WP is long compared to what?)
Solution: A “long diode” is one with the quasi-‐neutral region is much longer than the diffusion length, WP >> Ln .
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 12
HW Week 8 Solutions Continued
cosh x( )! ex
2
sinh x( )! ex
2 and we find
ID = qADn
Ln
ni2
N A
!
"#
$
%& eqVA kBT '1( )
7c) Simplify the expression derived in 7a) for a “short diode”. Explain what “short”
means.
Solution: A “short diode” is one with the quasi-‐neutral region is much shorter than the diffusion length, WP << Ln .
cosh x( )!1 sinh x( )! x and we find
ID = qADn
WP
ni2
N A
!
"#
$
%& eqVA kBT '1( )
8) Consider a P+N diode that is illuminated with light, which produces a uniform generation, GL,
of electron-‐holes pairs per cm3 per second. The N-‐region is long compared to a diffusion length. 8a) Consider first a uniform, infinitely long N-‐type semiconductor with a uniform
generation rate and solve for the steady-‐state excess minority carrier density, !p .
Solution: We have solved this problem before, in HW7. The answer is:
!p = GL" p
8b) Now consider the illuminated P+N diode. What are the boundary conditions at
!pn xn( ) and !pn x "#( ) ? Solution: Assume that the Law of the Junction still applies.
!pn xn( ) = ni
2
N D
eqVA kBT "1( )
!pn x "#( ) = GL$ n
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 13
HW Week 8 Solutions Continued
8c) Use the boundary conditions developed in 8b), neglect recombination-‐generation in the SCR and in the P+ layer, and solve for ID VA( ) for this illuminated diode.
Solution: Having solved the MDE so many times, we can see that the solution is:
!p x( ) = Ae" x/ Lp +GL# p This satisfies the b.c. for x !"
!p 0( ) = A+GL" p
A = GL! p " #p 0( ) so the solution is:
!p x( ) = !p 0( )e" x/ Lp +GL# p 1" e" x/ Lp( ) The current is:
J p = !qDp
d"pdx x=0
= qDp
Lp
"p 0( )! qDp
Lp
GL# p
Use the Law of the Junction:
J p = JD = qDpni
2
Lp N A
eqVA kBT( )! qDp
Lp
GL" p
Note that the first term is just the diode current in the dark, JDARK and the second term is the photo-‐generated current, which is bias-‐independent and what we measure under short circuit conditions.
JD = JDARK VA( )! JSC
JDARK VA( ) = q
Dpni2
Lp N A
eqVA kBT −1( )
JSC = q
Dp
Lp
GL! p
This result is the “classical” way of describing a solar cell – the approach is called “superposition” – we add the dark current and the current due to collection of photo-‐generated carriers. Note that superposition assumes that the collected photocurrent is independent of bias and that the Law of the Junction is valid under illumination.
ECE 606: Spring 2013 Purdue University
ECE-‐606 Spring 2013 14
HW Week 8 Solutions Continued 8d) Sketch ID VA( ) for GL = 0 , GL = G0 and GL = 2G0 .