some%of%the%problems%below%are%taken/adapted%fromchapter%4%in% advanced( semiconductor...

Mark Lundstrom 2/10/2013

ECE-‐606 Spring 2013 1

SOLUTIONS: ECE 606 Homework Week 5 Mark Lundstrom Purdue University (corrected 3/26/13)

Some of the problems below are taken/adapted from Chapter 4 in Advanced Semiconductor Fundamentals, 2nd. Ed. By R.F. Pierret. 1) The conduction band minima in GaP occurs at the boundary of the first Brillouin zone

along [100] directions. The constant energy surfaces are ellipsoidal with m* = 1.12m0

and mt* = 0.22m0 . Determine the density-‐of-‐states effective mass for electrons in GaP.

Solution: The situation here is much like silicon – except that the minima occur at the BZ boundary. The result is that we can only include one-‐half of each ellipsoid inside the BZ. The “6” in enq. (4.25a) of ASF (p. 95) is replaced by 3. We conclude:

mn

* )DOS

= 32/3 m*mt

*2( )1/3

mn

* )DOS

= 0.79m0

2) Assuming that the Si valence bands are spherical with mhh

* = 0.537m0 and mlh* = 0.153m0 ,

determine the fraction of holes that are in the heavy hole band. You should assume non-‐degenerate conditions at T = 300K. Also assume that the effective masses quoted above, which were measured by cyclotron resonance at 4K, can be used at room temperature.

Solution: Begin with:

p = NVF 1/2 ηF( ) m-‐3

NV = 1

42m*kBTπ2

⎛

⎝⎜⎞

⎠⎟

3/2

m-‐3 where: ηF = EV − EF( ) kBT

This expression assumes a single, parabolic energy band (i.e. valley degeneracy is 1). For the heavy and light hole bands:

phh = NVhhF 1/2 ηF( ) plh = NV

lhF 1/2 ηF( )



The fraction of holes in the heavy hole band becomes:

phh

p=

phh

phh + plh

=NV

hh

NVhh + NV

lh =mhh

*( )3/2

mhh*( )3/2

+ mlh*( )3/2

Inserting numbers, we find:

phh

p= 0.5373/2

0.537( )3/2+ 0.153( )3/2

phh

p= 0.868

Most of the holes are in the heavy hole band. This should be expected, because the heavy hole band has a much higher density of states.

NV

hh ∝ mhh*( )3/2

>> NVlh mlh

*( )3/2

Note that we did NOT need to assume MB (non-‐degenerate) statistics.

3) In a Si MOSFET, the source and drain regions are heavily doped. Assuming that the

dopants are fully ionized and that N D = 1020 cm-‐3, compute the location of the Fermi level. Compare the answer you get assuming Maxwell-‐Boltzmann statistics with the answer from Fermi-‐Dirac statistics. You may assume T = 300K and that the effective density of states is

NC = 1

42mD

* kBTπ2

⎛

⎝⎜⎞

⎠⎟

3/2

= 3.23×1019 cm-3

You will need to compute an inverse Fermi-‐Dirac integral. A FD integral tool is available on nanoHUB.org at: nanohub.org/resources/fdical You can also find an iPhone app that does Fermi-‐Dirac integrals.

Solution:

n = NCF 1/2 ηF( ) = 3.23×1019( )F 1/2 ηF( ) = 1020



F 1/2 ηF( ) = 1020

3.23×1019 = 3.10

ηF = F 1/2 3.10( )⎡⎣ ⎤⎦

−1

Using the nanoHUB tool or the iPhone app, we find:

ηF = F 1/2 3.10( )⎡⎣ ⎤⎦

−1= 2.184 =

EF − EC

kBT So the Fermi level is 2.184 kBT above EC for FD statistics. Now repeat for Maxwell Boltzmann statistics:

F 1/2 ηF( ) = 3.10⇒ eηF

ηF = ln 3.10( ) = 1.13=

EF − EC

kBTL

So the Fermi level is 1.13 kBT above EC for MB statistics. In general, we find that the Fermi level is higher when we use FD statistics. This occurs because as states are filled, we need to go higher in the conduction band to find empty state. The implicit assumption for MB statistics is that all states are nearly empty.

4) In 3D, the distribution of electrons in the conduction band, n E( ) , is peaked at an energy, E that is very near the bottom of the conduction band, EC . Assume Maxwell-‐

Boltzmann carrier statistics and derive an expression for E for the following cases: 4a) 3D semiconductor with parabolic energy bands 4b) 2D semiconductor with parabolic energy bands

Solution: 4a)

n E( ) = D E( ) f E( ) ≈ D E( )e− E−EF( ) kBT

dn E( )dE

=dD E( )

dEe− E−EF( ) kBT −

D E( )kBT

e− E−EF( ) kBT



dn E( )dE

= 0→ kBTdD E( )

dEE

= D E( )

in 3D, D E( ) = K E − EC( )1/2

(K is a constant), so

kBT2

K E − EC( )−1/2= K E − EC( )1/2

E = EC +

kBT2

4b) Begin with:

kBT

dD E( )dE

E

= D E( ) in 2D, D2 D E( ) = K , so

kBT × 0 = K nonsense. There is no peak to n E( ) .

5) Sometimes we need to know the average kinetic energy for electrons in the

conduction band (or for holes in the valence band). Answer the follow questions:

5a) Derive a general expressions for the average kinetic energy per electron in the conduction band.

Solution:

u =En

=

E − EC( )D E( ) f E( )dEEC

∞

∫

D E( ) f E( )dEEC

∞

∫

5b) Assume parabolic energy bands and evaluate the general expression for a 1D

semiconductor. Simplify your answer for Maxwell-‐Boltzmann carrier statistics.



Solution:

u =EnL

=

E − EC( )D1D E( ) f E( )dEEC

∞

∫

D1D E( ) f E( )dEEC

∞

∫

D1D (E) =

gV

π2mD

*

E − EC You should be able to show:

nL = D1D E( ) f E( )dE

EC

∞

∫ = N1DF −1/2 ηF( ) (*), where: N1D = 2mD

* kBT π . Now, lets work on the numerator:

E = E − EC( ) 1

π2mD

*

E − EC

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

11+ e E−EF( ) kBT dE = 1

π2mD

* E − EC( )1/2dE

1+ e E−EF( ) kBTEC

∞

∫EC

∞

∫

η ≡ E − EC( ) kBT dE = kBTη

ηF ≡ EF − EC( ) kBT

E = 1

π2mD

* kBTη( )1/2kBTdη

1+ eη−ηFEC

∞

∫

E =

2mD* kBT

πkBT η1/2dη

1+ eη−ηF0

∞

∫

η1/2dη1+ eη−ηF

0

∞

∫ = π2

F 1/2 ηF( ) (See “Notes on FD Integrals”)

E =

2mD* kBT π

kBT2

⎛⎝⎜

⎞⎠⎟F 1/2 ηF( ) = N1D

kBT2

⎛⎝⎜

⎞⎠⎟F 1/2 ηF( ) (**)



Combine (*) and (**) to find:

u1D =EnL

=kBT

2⎛⎝⎜

⎞⎠⎟

F 1/2 ηF( )F −1/2 ηF( )

For a nondegenerate semiconductor, all FD integrals reduce to exponentials, so we find:

u1D =

kBT2

⎛⎝⎜

⎞⎠⎟

5c) Assume circular, parabolic energy bands and evaluate the general expression

for a 2D semiconductor. Simplify your answer for Maxwell-‐Boltzmann carrier statistics.

Solution: We begin the same way…

u2 D =EnS

=

E − EC( )D2 D E( ) f E( )dEEC

∞

∫

D2 D E( ) f E( )dEEC

∞

∫

Use the 2D DOS:

D2 D (E) =

mD*

π2 and eventually find:

u2 D =EnS

= kBT( )F 1 ηF( )F 0 ηF( )

For a nondegenerate semiconductor, all FD integrals reduce to exponentials and we find:

u2 D = kBT



5d) Assume spherical, parabolic energy bands and evaluate the general

expression for a 3D semiconductor. Simplify your answer for Maxwell-‐Boltzmann carrier statistics.

Solution: We begin the same way…

u3D =En

=

E − EC( )D3D E( ) f E( )dEEC

∞

∫

D3D E( ) f E( )dEEC

∞

∫

Use the 3D DOS:

D3D E( ) = mD

3/2( ) 2 E − EC( )π 23

and eventually find:

u3D =En

=3kBT

2⎛⎝⎜

⎞⎠⎟F 3/2 ηF( )F 1/2 ηF( )

For a nondegenerate semiconductor, all FD integrals reduce to exponentials and we find:

u3D =

3kBT2

6) Assume Si at T = 300 K, doped with arsenic at N D = 1016 cm-‐3. Make reasonable

assumptions and answer the following questions. 6a) Compute the density of electrons in the conduction band. 6b) Compute the location of the Fermi level. 6c) Compute the fraction of the dopants that are ionized. 6d) Compute the density of holes in the valence band.

Solution:



For problems 6) -‐ 10), let’s assume at T = 300 K (from Pierret, ASF, 2nd Ed., p. 113)

NC = 1

42mn

*kBTπ2

⎛

⎝⎜⎞

⎠⎟

3/2

= 3.23×1019 cm-3

NV = 1

42mp

* kBTπ2

⎛

⎝⎜

⎞

⎠⎟

3/2

= 1.83×1019 cm-3

It should be understood that the effective masses in these expression are the density-‐of-‐states effective masses. For problems 6) – 10), the starting point is space charge neutrality: p − n + ND

+ − NA− = 0

NV F 1/2

EV − EF

kBT⎛⎝⎜

⎞⎠⎟− NC F 1/2

EF − EC

kBT⎛⎝⎜

⎞⎠⎟+ ND

1+ gDeEF−ED( ) kBT − NA

1+ gAeEA−EF( ) kBT = 0

For this problem, we can assume that N A = 0 . Since the doping is moderate, we assume non-‐degenerate carrier statistics. Because Arsenic is a shallow dopant and we expect the Fermi level to be well below ED, we can assume that the dopants are fully ionized. Because the semiconductor is extrinsic, n >> p . We conclude that

n = N D to a very good approximation. All these assumptions can be checked as we proceed. 6a)

n = N D = 1016 cm-3

6b) Determine EF from the electron density:

n = NCe EF −EC( ) kBT → EF − EC = kBT ln n NC( )

EF = EC + kBT ln 1016 3.23×1019( ) = EC −8.1kBT

EF = EC −8.1kBT

So the assumptions that the semiconductor is non-‐degenerate is good. 6c) Begin with:



ND+

ND

= 11+ gDe

EF−ED( ) kBT = 11+ gDe

EF−EC+EC−ED( ) kBT

According to Fig. 4.14, p. 111 of Pierret, ASF: EC − ED = 0.054 eV = 2.08kBT and we also know that for donors, gD = 2 . ND

+

ND

= 11+ 2e −8.1kBT +2.08kBT( ) kBT = 1

1+ 2e−6.02= 0.995

ND+

ND

= 0.995

So our assumptions that the dopants are fully ionized is good. 6d) Use the Law of Mass action: np = ni

2

p =

ni2

n=

1010( )2

1016 = 104

p = 104 cm-3

So our assumption that n >> p is good.

7) Assume Si at T = 640 K, doped with arsenic at N D = 1016 cm-‐3. Make reasonable

assumptions and answer the following questions. 7a) Compute the density of electrons in the conduction band. 7b) Compute the location of the Fermi level. 7c) Compute the fraction of the dopants that are ionized. 7d) Compute the density of holes in the valence band. Solution: Dopants should be fully ionized and semiconductor should be nondegenerate, but

ni may be large, which means that p may not be negligible. From Fig. 4.17, p. 117 of Pierret, ASF we find:

ni 640 K( ) ≈1016cm-3 The space charge neutrality condition becomes:


ECE-‐606 Spring 2013 10

p − n + ND

+ − NA− = 0 = p − n + ND = ni

2 n − n + ND = 0 which can be solved for

n = ND

2+ ND

2⎛⎝⎜

⎞⎠⎟2

+ ni2⎡

⎣⎢

⎤

⎦⎥

1/2

7a) Solve for n:

n = ND

2+ ND

2⎛⎝⎜

⎞⎠⎟2

+ ni2⎡

⎣⎢

⎤

⎦⎥

1/2

= 5 ×1015 + 5 ×1015( )2 + 1016( )2⎡⎣

⎤⎦1/2

n = 5 ×1015 +1.12 ×1016 n = 1.62 ×1016 cm-3


n = NCe EF −EC( ) kBT → EF − EC = kBT ln n NC( ) but NC is temperature dependent.

NC = 1

42mn

*kBTπ2

⎛

⎝⎜⎞

⎠⎟

3/2

= NC 300K( ) T300

⎛⎝⎜

⎞⎠⎟

3/2

= 3.23×1019 640300

⎛⎝⎜

⎞⎠⎟

3/2

= 1.01×1020 cm-3

EF − EC = kBT ln 1.62×1016 1.01×1020( ) = −8.74kBT

EF = EC −8.74kBT (but realize that kBT is bigger now). 7c) Proceeding as in 6c), we find:

EC − ED = 0.054 eV = 0.98kBT and ND

+

ND

= 11+ 2e −8.74 kBT +0.98kBT( ) kBT = 1

1+ 2e−6.02= 0.999

ND+

ND

= 0.999

Dopants are fully ionized as assumed. 7d) Use the Law of Mass Action: np = ni

2


ECE-‐606 Spring 2013 11

p =

ni2

n=

1016( )2

1.62×1016 = 6.2×1015

p = 6.2×1015 cm-3

So in this case, the hole concentration is not negligible. The semiconductor is nearly intrinsic.

8) Assume Si at T = 77 K, doped with arsenic at N D = 1016 cm-‐3. Make reasonable assumptions and answer the following questions. 8a) Compute the density of electrons in the conduction band. 8b) Compute the location of the Fermi level. 8c) Compute the fraction of the dopants that are ionized. 8d) Compute the density of holes in the valence band. Solution: In this case, we worry that there may not be enough thermal energy to ionize the donors. The intrinsic concentration should be very small, so holes can be ignored. The space charge neutrality condition becomes:

p − n + ND+ − NA

− = 0 = −n + ND+

−NC F 1/2

EF − EC

kBT⎛⎝⎜

⎞⎠⎟+ ND

1+ gDeEF−ED( ) kBT = 0

Assume non-‐degenerate carrier statistics (check later):

−NC expEF − EC

kBT⎛⎝⎜

⎞⎠⎟+ ND

1+ gDeEF−ED( ) kBT = 0

This equation can be solved. The answer is eqn. (4.88) in Pierret, ASF

EF = EC + kBT ln Nζ 2NC( ) 1+ 4ND Nζ( )1/2 −1⎡⎣⎢

⎤⎦⎥{ }

where

Nζ = NC gD( )e− EC−ED( ) kBT

Now lets put numbers in:


ECE-‐606 Spring 2013 12

NC = NC 300K( ) T

300⎛⎝⎜

⎞⎠⎟

3/2

= 3.23×1019 77300

⎛⎝⎜

⎞⎠⎟

3/2

= 4.20×1018 cm-3

kBT = 6.64×10−3 eV

Nζ = NC gD( )e− EC−ED( ) kBT = 4.20×1018 2( )e− 54( ) 6.64 = 6.17 ×1014

EF = EC + kBT ln 6.17 ×1014 2 × 4.20 ×1018( ) 1+ 4 ×1016 6.17 ×1014( )1/2 −1⎡⎣

⎤⎦{ }

EF = EC + kBT ln 7.35 ×10−5( ) 65.8( )1/2 −1⎡⎣ ⎤⎦{ }

EF = EC − 7.65kBT

Note that ED = EC −8.13kBT So the Fermi level is above the donor level, but far enough below EC that we can use non-‐degenerate carrier statistics. 8a) Since we know the Fermi level, we can get the electron density:

n = NC expEF − EC

kBT⎛⎝⎜

⎞⎠⎟= 4.20 ×1018 exp −7.65( )

n = 2.00 ×1015 cm-3

8b) We already solved this problem.

EF = EC − 7.65kBT

8c) Since n = N D

+ = 2.00×1015 , we deduce

N D+

N D

= 0.2

We could also get this directly from ND

+

ND

= 11+ gDe

EF−ED( ) kBT

8d) The number of holes is essentially zero. We could get this from the law of mass

action, but then we need to figure out ni at 77 K. Another way is:


ECE-‐606 Spring 2013 13

p = NV expEV − EF

kBT⎛⎝⎜

⎞⎠⎟= NV exp

EV − EC + EC − EF

kBT⎛⎝⎜

⎞⎠⎟= NV exp − EG

kBT⎛⎝⎜

⎞⎠⎟exp EC − EF

kBT⎛⎝⎜

⎞⎠⎟

NV = 1

42mp

* kBTπ2

⎛

⎝⎜

⎞

⎠⎟

3/2

= NV 300K( ) T300

⎛⎝⎜

⎞⎠⎟

3/2

= 1.83×1019 77300

⎛⎝⎜

⎞⎠⎟

3/2

= 2.38×1018 cm-3

Also need the bandgap of Si at 77 K. From pp. 82-‐83 of Pierret, ASF, we have

EG T( ) = EG 0( )− αT 2

T + β

For Si:

EG T( ) = 1.170−

4.730×10−4 77( )2

77 + 636= 1.166 eV

so

p = 2.38 ×1018 exp −11666.64

⎛⎝⎜

⎞⎠⎟ exp 7.65( ) = 2.38 ×1018 5.46 ×1077( ) 2.10 ×103( )

p = 2.73×10−55 cm-3

This is essentially zero… hardly worth the effort of computing it. Since we have n and p, we can deduce ni = 2.34 ×10

−20 cm-3

9) Assume Si at T = 300 K, doped with arsenic at ND =1018 cm-‐3. Make reasonable assumptions and answer the following questions.

9a) Compute the density of electrons in the conduction band. 9b) Compute the location of the Fermi level. 9c) Compute the fraction of the dopants that are ionized. 9d) Compute the density of holes in the valence band. Solution:


ECE-‐606 Spring 2013 14

This requires a little thought… the temperature is high enough so the donors may be fully ionized, but the number of dopants is high, so the Fermi level might be too close to the donor level to assume complete ionization. It might also be too close to the conduction band to assume non-‐degenerate carrier statistics. Let assume full ionization and nondegenerate carrier statistics and see where it leads. 9a) Assume complete ionization:

n = N D = 1018 cm-3


n = NCe EF −EC( ) kBT → EF − EC = kBT ln n NC( )

EF = EC + kBT ln 1018 3.23×1019( ) = EC − 3.48kBT

EF = EC − 3.48kBT

So the assumptions that the semiconductor is non-‐degenerate is OK. 9c) As in problem 6c):

ND

+

ND

= 11+ 2e −3.48kBT +2.08kBT( ) kBT = 1

1+ 2e−1.40= 0.67

ND+

ND

= 0.67

So our assumptions that the dopants are fully ionized is not justified. We need to start over and solve the problem using the procedure of problem 8). The rest of this problem is left as an exercise for the interested student!

10) Assume Si at T = 300 K, doped with arsenic at N D = 1020 cm-‐3. Make reasonable assumptions and answer the following questions. (In this case, you should assume that “heavy doping effects” cause the dopants to be fully ionized and the bandgap to shrink by 0.1 eV.)

10a) Compute the density of electrons in the conduction band. 10b) Compute the location of the Fermi level. 10c) Compute the fraction of the dopants that are ionized. 10d) Compute the density of holes in the valence band.

Solution:


ECE-‐606 Spring 2013 15

We expect the Fermi level to be inside the conduction band. We can assume full ionization and ignore holes, but we cannot use nondegenerate statistics.

n = N D to a very good approximation. 10a)

n = N D = 1020 cm-3


n = NC F 1/2

EF − EC

kBT⎛⎝⎜

⎞⎠⎟ NC = 3.23×1019 cm-3

EF − EC

kBT= F 1/2

nNC

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−1

= F 1/21020

3.23×1019⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−1

= F 1/2 3.10( )⎡⎣ ⎤⎦−1= 2.184

(from Fermi Dirac Integral Calculator on nanoHUB.org https://nanohub.org/tools/fdical)

EF = EC + 2.184kBT

So the semiconductor is, indeed degenerate with EF inside the conduction band. 10c) By assumption, we are above the metal-‐insulator transition and ND

+

ND

= 1

10d) Use the Law of Mass Action: np = ni

2 But bandgap narrowing increases ni.

ni

2 = 1020 exp ΔEG / kBT( ) = 1020 exp 100 / 26( ) = 4.68×1021

ni = 6.84×1010

p =

ni2

n=

4.68×1021( )1018 = 4.68×103

p = 4.68×103 cm-3

So our assumption that n >> p is good.


ECE-‐606 Spring 2013 16

11) For the energy band sketched below, provide sketches of the following:

11a) the carrier densities, n(x), and p(x) vs. position. Solution:

11b) the electrostatic potential, ψ x( ) , vs. position. Solution:


ECE-‐606 Spring 2013 17

11c) the electric field E vs. position.

11d) the space charge density, ρ x( ) vs. position Solution:

some%of%the%problems%below%are%taken/adapted%fromchapter%4%in% advanced( semiconductor...

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