ECE 606: Spring 2013 Purdue University
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, N S = 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 ) = N C e(
E F ! EC ) k BT
" N D ( x ) " N ( x) %
E F ! EC ( x ) = k BT ln $ D
' $# N C '&
N C = 3.23! 1019 cm-­‐3 " N (0) %
E F ! EC ( 0 ) = k BT ln $ D
' $# N C '&
⎡ 1020 ⎤
E F − EC 0 = k BT ln ⎢
= +1.13k BT
19 ⎥
⎣ 3.23 × 10 ⎦
()
ECE-­‐606 1 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued $ N ( x " #) '
E F ! EC ( x " # ) = k BT ln & D
)
NC
&%
)( ⎡ 1015
⎤
E F − EC x → ∞ = k BT ln ⎢
= −10.4k BT
19 ⎥
⎣ 3.23 × 10 ⎦
(
)
()
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 2 Spring 2013 ECE 606: Spring 2013 Purdue University
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: dn
= 0 dx
k T 1 dn k BT 1 dN D ( x )
E = B
=
q n dx
q N D ( x ) dx
J n = nqµ nE + k BT µ n
! k T $ 1 dN D ( x )
E =# B &
" q % N D ( x ) dx
Another way is to begin with n0 ≈ N D = N C e( 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: ! 1030 $
k T !N N $
Vbi = B ln # A 2 D & = 0.026ln # 20 & = 0.60 V q
" 10 %
" ni %
Vbi = 0.60 2b) Compute xn , x p and W. Solution: 1/2
# 2! "
&
NA
xn = % S 0
Vbi (
%$ q N D ( N A + N D ) ('
= 0.625 µ m xn = x p = 0.625 µ m (because N and P regions are symmetrical) W = xn + x p = 1.25 µ m
ECE-­‐606 3 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 2c) Compute V x = 0 and E x = 0 . (
)
(
)
Solution: By symmetry: V (0) =
Vbi
qN A 2
= 0.30 V or use V ( x = 0 ) =
x 2
2! S " 0 p
E ( x = 0) =
qN A
x p = 9.6 # 103
! S"0
E ( 0 ) = !9.6 " 103 V/cm 2d) Sketch ! x vs. x. ()
Solution: ρ N = +qN D = +1.6 × 10 −4 C/cm 3
! P = "qN A = "1.6 # 10 "4 C/cm 3
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: k T !N N $
Vbi = B ln # D 2 A & no simplification possible q
" ni %
ECE-­‐606 4 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 3b) The total depletion layer depth, W , from Pierret, Eqn. (5.31). Solution: 1/2
1/2
) 2! " # N + N D & ,
W =+ S 0% A
( Vbi . * q $ NDNA ' -
# 2! "
&
N A >> N D  W = % S 0 Vbi (
$ qN D
'
3c) The peak electric field, E 0 , from Pierret, Eqn. (5.19) or (5.21). ()
Solution: E (0) =
2qN DVbi
2Vbi
2qVbi # N D N A &
=
 E ( 0 ) =
%
(
! s" 0
W
! s" 0 $ N A + N D '
3d) The electrostatic potential, V x from Pierret, Eqn. (5.28) ()
Solution: qN D
qN D
2
xn ! x )  V ( x ) = Vbi −
(W − x )2 (
2" S # 0
2κ S ε 0
Now use the expression for W above to find: V ( x ) = Vbi !
2
V ( x ) = Vbi ⎡1 − (1 − x W ) ⎤ ⎣
⎦
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: ! 1025 ' 1019 $
k T !N N $
Vbi = B ln # A 2 D & = 0.026ln #
20
&% = 0.84 V q
" 10
" ni %
Vbi = 0.84 ECE-­‐606 5 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 4b) Compute xn , x p and W. Solution: x p ! 0 1/2
$ 2" #
'
xn ! W = & S 0 Vbi )
% qN D
(
= 1.05 µ m W = 1.05 µ m (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
W = 1.6 # 104 V/cm
! S"0
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 6 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 5) Repeat problem 4) using the “exact” solution to PN junction electrostatics. Solution: ⎛ 1015 ⎞
k BT ⎛ N D ⎞
VN = +
ln ⎜
= 0.026 ln ⎜ 10 ⎟ = 0.30 q
⎝ 10 ⎠
⎝ ni ⎟⎠
VP = !
" 1019 %
kBT " N A %
ln $
=
0.026
ln
$# 1010 '& = !0.540 q
# ni '&
Vbi = VN ! VP = 0.84 V V (0) =
Vbi = 0.84 C N ! CP
a N ! aP
aN = N D = 1015 aP = !N A = !1019 C N = aNVN ! 2ni ( kBT q ) cosh ( qVN kBT ) C N = 1015 ! 0.30 " 2 ! 1010 ( 0.026 ) cosh (11.5 ) = 2.74 ! 1014 CP = aPVP ! 2ni ( kBT q ) cosh ( qVP kBT ) CP = ( !1019 ) ( !0.54 ) ! 2 " 1010 ( 0.026 ) cosh ( !20.7 ) = 5.15 " 1018
V (0) =
C N ! CP
= !0.518 a N ! aP
V ( 0 ) ! VP = !0.54 ! 0.512 = 0.028 " k BT q The potential drop across the heavily doped side is about kBT/q. E ( 0 ) = 2q ! S " 0 ( ni ( kBT q ) eqV (0) kBT + ni ( kBT q ) e# qV (0) kBT # aNV (0) + C N ) 1/2
Putting in numbers, we find: E ( 0 ) ! 1.7 " 10 5 V/cm V/cm which is about 10X the electric field we found in prob. 4. ECE-­‐606 7 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued n0 ( 0 ) = ni eqV ( 0 ) kBT = 27 cm-­‐3 p0 ( 0 ) = ni e! qV ( 0 ) kBT = 0.37 " 1017 cm-­‐3 ! 0 " = q #$ p0 ( 0 ) " n0 ( 0 ) + N D %& = q #$ 0.37 ' 1019 + 1015 %& ( q #$ 0.37 ' 1019 %& ( )
! ( 0 ) # q %& 0.37 $ 10
"
19
( )
'( (depletion approximation would give ! 0 " # q $%1015 &' ) ! ( 0 + ) = q #$ p0 ( 0 ) " n0 ( 0 ) " N A %& = 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 8 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 6a) Sketch an energy band diagram for this junction. ( ) 6b) Sketch V x
( ) 6c) Sketch E x
ECE-­‐606 9 Spring 2013 ECE 606: Spring 2013 Purdue University
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
E !E
k T
E !E
k T
E !E
k T
n01 = N C e( F 1 C ) B n02 = N C e( F 2 C ) B = e( F 1 F 2 ) B n02
The built-­‐in potential develops to align these two Fermi levels: "n %
E F1 ! E F 2 = qVbi = k BT ln $ 01 ' # n02 &
(
ECE-­‐606 Vbi =
)
k BT ⎛ N D1 ⎞
ln ⎜
⎟ q
⎝ N D2 ⎠
10 Spring 2013 ECE 606: Spring 2013 Purdue University
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 I D V A 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: sinh #$(WP " x ) / Ln %&
!n ( x ) = !n ( 0 )
sinh (WP / Ln )
Let x = 0 be the edge of the neutral P-­‐region. The electron current is: cosh (WP Ln )
D
d!n
(minus sign means that the electron J n = +qDn
= "q n !n ( 0 )
dx x=0
Ln
sinh (WP Ln )
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. I D = ! AJ n = qA
cosh (WP Ln )
Dn
"n ( 0 )
Ln
sinh (WP Ln )
Finally, use the “Law of the Junction” for the boundary condition: ni2 qVA kBT
!n 0 =
e
" 1 NA
to find: ()
(
)
! Dn ni2 $ cosh (WP Ln ) qVA
I D = # qA
e
&
" Ln N A % sinh (WP Ln )
(
k BT
)
' 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 11 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued cosh ( x ) !
x
e
2
sinh ( x ) !
! D n2 $ qV
I D = # qA n i & e A
" Ln N A %
(
x
e
and we find 2
)
k BT
' 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 ! D n2 $
I D = # qA n i & eqVA
" WP N A %
(
8) )
k BT
' 1 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. (
ni2 qVA
!pn ( xn ) =
e
ND
!pn ( x " # ) = GL$ n
ECE-­‐606 k BT
)
" 1 12 Spring 2013 ECE 606: Spring 2013 Purdue University
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 I D V A 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: D
D
d"p
J p = !qD p
= q p "p ( 0 ) ! q p GL# p dx x=0
Lp
Lp
Use the Law of the Junction: J p = JD = q
D p ni2
Lp N A
(e
qV A k BT
)! q L
Dp
GL" p p
Note that the first term is just the diode current in the dark, J DARK and the second term is the photo-­‐generated current, which is bias-­‐independent and what we measure under short circuit conditions. J D = J DARK V A ! J SC ( )
( )
J DARK V A = q
J SC = q
Dp
Lp
D p ni2
Lp N A
(e
qV A k BT
)
− 1 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 13 Spring 2013 ECE 606: Spring 2013 Purdue University
HW Week 8 Solutions Continued 8d) Sketch I D V A for GL = 0 , GL = G0 and GL = 2G0 . ( )
ECE-­‐606 14 Spring 2013