solution:# positive - nanohubweek6hwsolutions_s15.pdf · mark%lundstrom% % spring2015% ece8305% %...

Mark Lundstrom Spring 2015

ECE-‐305 Spring 2015 1

ECE 305 Homework SOLUTIONS: Week 6

Mark Lundstrom Purdue University

1) This problem concerns a PN junction doped NA = 2 ×10

18 cm-3 on the P-‐side and ND = 3×1017 cm-3 on the N-‐side. Answer the following questions.

1a) There is a built-‐in potential for this PN junction. Which side is more positive –

the P-‐side or the N-‐side. Explain your answer. Solution:

See the figure below. Holes diffuse from the left to the right. An electric field that points from the right to the left builds up to stop the flow of holes. If the electric field points from the right to the left, the right side, the N-‐side is more positive.

Alternatively, see the figure below, which shows the P and N regions before they are connected. The Fermi level is higher on the N-‐side, so the N-‐side must develop a higher potential to lower the bands on the N-‐side and align the Fermi levels, which must be equal in equilibrium once the two sides are connected.



HW6 Solutions (continued):

1b) Assume that the PN junction is made of silicon. What is the magnitude of the built-‐in potential? You may assume that NC = 3.23×1019 cm-3 ,

NV = 1.83×1019 cm-3 , ni = 1×1010 cm-3 , and EG = 1.125 eV . Solution:

Vbi =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟= 0.026ln 2×1018 × 3×1017

1020

⎛⎝⎜

⎞⎠⎟= 0.95 V

Vbi = 0.95 V

1c) Assume that the PN junction is made of GaAs. What is the magnitude of the

built-‐in potential? You may assume that NC = 4.21×1017 cm-3 ,

NV = 9.52×1018 cm-3 , ni = 2.25×106 cm-3 , and EG = 1.422 eV .

Vbi =kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟= 0.026ln 2×1018 × 3×1017

2.25×106( )2

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 1.38 V

1c) Compare your answers in 1b) and 1c) to the corresponding bandgaps Solution:

Vbi Si( ) = 0.95 V = 0.84 EG eV( ) Vbi GaAs( ) = 1.38 V = 0.97 EG eV( ) The built-‐in potential of a PN junction is typically a little less than the bandgap of the semiconductor (in eV). (Which can be understood from the energy band diagram shown above.)

1d) Derive an expression that relates Vbi to EG . Solution:

Vbi =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟ ni

2 = NCNVe−EG /kBT

Vbi =

kBTq

lnN AN D

NC NV e−EG /kBT

⎛

⎝⎜⎞

⎠⎟=

kBTq

lnN AN D

NC NV

⎛⎝⎜

⎞⎠⎟+

kBTq

ln eEG /kBT( )




Vbi =

kBTq

lnN AN D

NC NV

⎛⎝⎜

⎞⎠⎟+ EG

Vbi =EG

q−

kBTq

lnNC

N D

⎛

⎝⎜⎞

⎠⎟−

kBTq

lnNV

N A

⎛

⎝⎜⎞

⎠⎟

For a non-‐degenerate semiconductor, p0 = N A < NV and n0 = N D < NC , so the built-‐in potential is a little less than the bandgap.

2) For the energy band diagram below, answer the following questions.

2a) Is the potential at x = 0 more or less positive than the potential at x = L ?

Explain your answer.

Solution: A positive potential lowers the bands. At x = 0 , EC is lower than at x = L , so the potential at x = 0 must be more positive than the potential at x = L .

2b) What is the magnitude of the built-‐in potential?

Solution:

Ei x = L( ) q = EF q + 0.025

Ei x = 0( ) q = EF q − 0.030

Vbi = Ei x = L( ) q − Ei x = 0( ) q = 0.025+ 0.030 = 0.055 V

Vbi = 0.055 V



HW6 Solutions (continued): 3) This problem concerns a Si “isotype” junction at room temperature with the doping

profile shown below. Answer the following questions assuming room temperature silicon. Note that the doping is N-‐type on both sides of the junction.

3a) Draw an energy band diagram for this junction.

Solution:

EC − EF

q=

kBTq

lnNC

N D

⎛

⎝⎜⎞

⎠⎟

on the left:

EC − EF

q= 0.026ln 3.23×1019

1015

⎛

⎝⎜⎞

⎠⎟= 0.27 eV

on the right:

EC − EF

q= 0.026ln 3.23×1019

1018

⎛

⎝⎜⎞

⎠⎟= 0.09 eV



HW6 Solutions (continued): 3b) Is the left contact more or less positive in potential than the right contact.

Explain your answer.

Solution: From the energy band diagram, the bands are higher in energy on the left side, so the potential must be negative with respect to the right side. The left contact is less positive in potential than the right contact.

3c) Compute the magnitude of the built-‐in potential for this junction. Compare it to

the bandgap of Si.

Solution: At x = 0 , we have

n0 x = 0( ) = nieEF −Ei 0( )( ) kBT

Ei 0( ) = EF − kBT ln

n0 0( )ni

⎛

⎝⎜

⎞

⎠⎟

Ei 0( ) eV( ) = EF eV( )− 0.026ln 1015

1010

⎛⎝⎜

⎞⎠⎟

eV( ) = EF eV( )− 0.299

At x = L , we have

n0 x = L( ) = nieEF −Ei L( )( ) kBT

Ei L( ) = EF − kBT ln

n0 L( )ni

⎛

⎝⎜

⎞

⎠⎟

Ei L( ) eV( ) = EF eV( )− 0.026ln 1018

1010

⎛⎝⎜

⎞⎠⎟

eV( ) = EF eV( )− 0.479

Vbi = Ei x = 0( ) q − Ei x = L( ) q = −0.299+ 0.479 = 0.18 V

Vbi = 0.18 V

Note that Vbi for an isotype junction is much less than the bandgap of the semiconductor in eV.

3d) Derive a general formula for the built-‐in potential in terms of the doping density

on the left, N DL and on the right, N DR .

Solution: We can follow the approach of prob. 3c), but without putting in specific numbers.




qVbi = Ei x = 0( )− Ei x = L( ) = EF − kBT ln

n0 x = 0( )ni

⎛

⎝⎜

⎞

⎠⎟ − EF + kBT ln

n0 x = L( )ni

⎛

⎝⎜

⎞

⎠⎟

qVbi = kBT ln

n0 x = L( )n0 x = 0( )

⎛

⎝⎜

⎞

⎠⎟

N DL = n0 x = 0( ) N DR = n0 x = L( )

Vbi =kBTq

lnN DR

N DL

⎛

⎝⎜⎞

⎠⎟

Note that we might have guessed this from the formula for the PN junction:

Vbi =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟=

kBTq

lnN D

ni2 N A

⎛

⎝⎜⎞

⎠⎟=

kBTq

lnn0 N

n0 P

⎛⎝⎜

⎞⎠⎟

The last result says that Vbi is related to the log of the ratio of the electron concentration on the N-‐side, n0 N , to the electron concentration on the P-‐side, n0 P . So the PN junction expression can be given in terms of the electron concentration on the two sides, just as we got for the nN junction.

4) The electric field vs. position for a Si NP junction is sketched below. (The depletion

approximation is assumed). Given the information provided, answer the following questions.

4a) What is the doping density on the P-‐side?



HW6 Solutions (continued): Solution:

According to the Poisson equation:

d εSiE( )dx

= ρ x( ) = −qN A

dEdx

=−qN A

εSi

N A = −

εSi

qdEdx

From the figure:

dEdx

= −1.12×104 − 0( )V/cm

0.73×10−4 = −1.53×108 V/cm2

N A = −11.7 ×8.854×10−14 F/cm

1.6×10−19 C× −1.53×108 V/cm2( ) = 9.91×1014 cm−3

N A = 9.91×1014 cm−3

4b) What is the doping density on the N-‐side?

Solution: We could do this just as in 4a), but we can also recognize that charge must balance.

N Dxn = N Axp

N D = N A

xp

xn

= 9.91×1014 × 0.730.37

= 1.95×1015

N D = 1.95×1015 cm−3

4c) If the voltage is zero at

x = xp , then what is the voltage at x = 0 ?

Solution:

Remember the definition of the electric field: E = − dV

dx, so

V xp( )−V 0( ) = − E dx

0

xp

∫ =

Since V xp( ) = 0 , we find

V 0( ) = E dx

0

xp

∫



HW6 (continued): The voltage at x = 0 is just the area under the E x( ) vs. x curve between x = 0 and x = xP .

V 0( ) = 1

2E 0( )xp = 0.5×1.12×104 × 0.73×10−4 = 0.41

V 0( ) = 0.41 V

5) Consider the problem sketched below. A sheet of negative charge is imbedded in an

N-‐type semiconductor. The number of negative charges per cm2 is NS , so the charge in Coulombs per cm2 is QS = −q × NS . Answer the following questions.

5a) Sketch the space charge density vs. position in the semiconductor assuming the

depletion approximation in the semiconductor.

Solution: The negative charge pushes electrons away in both directions. The semiconductor is depleted for −xn < x < 0 and for 0 < x < +xn .



HW6 (continued):

5b) Sketch the electric field vs. position in the semiconductor assuming the depletion approximation in the semiconductor.

Solution: The electric field points from the positive depletion charge, qN D

+ , to the negative charge on the sheet, so E > 0 for −xn < x < 0 and E < 0 and for

0 < x < xn . The electric field is zero outside of the depletion regions, so E =0 for x ≤ −xn and E =0 for for

x ≥ xp .

dE dx is given from the Poisson equation as

dEdx

=+qN D

εSi

. The slope is

positive and the same value for −xn < x < 0 and for 0 < x < xn .

We conclude that E x( ) looks as in the figure below. Note the discontinuity at x − 0 .

Note that there is a discontinuity in the electric field, which is caused by the delta function of charge. This can be seen from the Poisson equation:

d εSiE( )dx

= ρ x( )

dEdx

=ρ x( )εSi

dEdx

=QSδ x( )εSi

integrate across x = 0 :

dEE 0−( )

E 0+( )∫ =

QS

εSi

δ x( )dx0−

0+

∫ =QS

εSi

E 0+( )−E 0−( ) = QS

εSi

E 0+( ) =E 0−( ) + QS

εSi

E 0+( ) =E 0−( )− q ×1013

εSi

So there is an abrupt change in the electric field across the sheet of charge.

solution:# positive - nanohubweek6hwsolutions_s15.pdf · mark%lundstrom% % spring2015% ece8305% %...

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