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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



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.



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




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




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




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.




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.




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.




6a) Sketch an energy band diagram for this junction.

6b) Sketch V x( )

6c) Sketch E x( )


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 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 12


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 13


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 14

HW Week 8 Solutions Continued 8d) Sketch ID VA( ) for GL = 0 , GL = G0 and GL = 2G0 .

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