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ECE 255 Spring 2019

Homework 2 SOLUTIONS

Due 5:00 PM Tuesday, Jan 22 in MSEE 180 Dropbox

1) As shown in the figure below, minority carrier holes are injected into an N-type Si layer of width, W. (We will see later that this can be done with a PN junction.) Assume that

pn x = 0( ) = 1014 cm-3 , pn x =W( ) = 0 , µ p = 300 cm2 V-s , W = 100 nm , and that T = 300

K. Answer the following questions.

1a) Compute the sign and magnitude of the hole current density in A/cm2.

Solution:

Dp =

kBTq

µ p = 0.026× 300 = 7.8 cm2 /s

J p = −qDp

dpdx

= − 1.6×10−19( ) 7.8( ) 0−1014

100×10−7 = +12.5 A/cm2 J p = +12.5 A/cm2

1b) What is the sign and magnitude of the average hole velocity at x = 0? Hint: The current density is proportional to the average velocity.

Solution:

Current is the flow of charge, so it must be proportional to the charge, q, the density of holes, p, and the average velocity at which the holes flow, υx , so we find:

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HW 2 Solutions (continued):

J p = qp x( ) υ x( ) (*) We see from the linear profile of p(x), that the current is independent of position (do you see why?). Let’s check dimensions to be sure that we have the correct expression.

J p = qp x( ) υ x( ) → C× cm-3 × cm/s→ C

s× cm-2 = A/cm2

So we have the right expression for current density in terms of average velocity. Now we also have another expression for the current density

J p = −qDp

dpdx

(**)

By equating (*) and (**), we find: J p = qp x( ) υ x( ) = −qDp

dpdx

, which gives

υ x( ) = −Dp

1p x( )

dpdx

We are asked for the velocity at x = 0:

υ x = 0( ) = −Dp

1p x = 0( )

dpdx x=0

= −Dp

1p x = 0( )

0− p x = 0( )W

= +Dp

W

By checking units, we can see that this is dimensionally correct. Plugging in numbers:

υ x = 0( ) = +

Dp

W= 7.8

100×10−7 = 7.8×105 cm/s

υ x = 0( ) = +7.8×105 cm/s (holes diffuse in the +x direction)

2) Assuming the linear profile of minority carriers as shown in the figure above, derive an expression for the average time it takes for holes to diffuse across a region of thickness, W. This is called the “transit time,” tt . Hint: Remember that current is charge divided by

time, so I = Q tt , where Q is the total minority carrier hole charge in the N-region.

Solution:

I = J p A = −qADp

dpdx

= qADp

p 0( )W

Q = 1

2Wqp 0( )A

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HW 2 solutions (continued):

tt =QI=

12

WAqp 0( )qADp

p 0( )W

= W 2

2Dp

tt =W 2

2Dp

This is a classic expression for the time it takes for particles of any kind to diffuse across a region of width, W.

3) The relationship J =σE is an alternative form of Ohm’s law. Show that this is equivalent to the ECE-201 version (V = IR), by considering a uniformly doped, rectangular bar of semiconductor material with length L (along direction of current flow) and cross-sectional area A (perpendicular to current flow). You may assume that the electric field, E , is uniform along the length of the bar.

Solution:

J =σE → ′I = JA =σ AE (*)

E = − dV

dx= −V

L Here we have assumed that the voltage at x = 0 is 0 V and at x = L, V(L)

= V. Note that the electric field points from the positive voltage to the less positive voltage, i.e. in the –x direction.

Using this expression for the electric field in (*), we find

′I = −σ A V

L or V = − 1

σLA

′I = ρ LA

⎛⎝⎜

⎞⎠⎟

′I = −R ′I

I’ is the current in the x-direction. The negative sign means that the current is flowing in the –x direction, as expected. We usually define the current, I, to be positive when it flows into the most positive terminal, so I = − ′I

V = IR

R = ρ LA

⎛⎝⎜

⎞⎠⎟

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HW 2 solutions (continued):

4) Assume a silicon resistor uniformly doped N-type at N D = 3×1017 cm-3 . The temperature is

300 K and the mobility is µn = 500 cm2 V-s . The resistor is 10 micrometers ( µm ) long. The voltage at x = 0 is 0 V and the voltage at x = 10 µm is 1.5 V. Answer the following questions.

4a) What is the average electron velocity in cm/s in the +x direction?

Solution:

υn = −µnE = −µn −V

L⎛⎝⎜

⎞⎠⎟= 500 1.5

10×10−4 = 7.5×105 V/cm

4b) What is the magnitude of the drift current density in A/cm2?

Solution:

Jn = nqµnE = 3×1017 1.6×10−19( ) 500( ) − 1.5

10−3

⎛⎝⎜

⎞⎠⎟= −3.6×104 A/cm2

Jn = 3.6×104 A/cm2

4c) What is the direction of the drift current (+x or –x)?

Solution:

The current is negative, which means that it is flowing in the -x direction (electrons are moving in the +x direction).

5) Consider the equilibrium energy band diagram shown below. The semiconductor is silicon at 300 K with ni = 1010 cm-3 . The wafer is doped P-type at N A = 1018 cm-3 , so that for x >> 0

pB = 1018 cm-3 . Answer the following questions assuming that the dopants are fully ionized.

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HW 2 solutions (continued):

5a) What is the equilibrium electron density at x = 0?

Solution:

The electron concentration in the bulk is nB = ni2 pB = 1020 1018 = 102

The electron concentration at x = 0 is exponentially larger because the bands are bent down.

n 0( ) = nBeqV kBT = 102 e0.4 0.026 = 4.8×108

n 0( ) = 4.8×108 cm-3

5b) What is the equilibrium hole density at x = 0? Solution:

We could use: p 0( ) = ni2 n 0( )

or

p 0( ) = pBe−qV kBT = 1018 e−0.4 0.026 = 2.08×1011

p 0( ) = 2.1×1011 cm-3

Note that p 0( ) << N A . The surface layer is depleted of majority carrier holes.

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HW 2 solutions (continued): 5c) What is the space charge density in C/cm3 at x = 0? Be sure to include the sign.

Solution:

ρ 0( ) = q p 0( )− N A − n 0( )⎡⎣ ⎤⎦ = q 2.1×1011 −1018 − 4.8×108⎡⎣ ⎤⎦ = −qN A

ρ 0( ) = −qN A = −1.6×10−19 1018( ) = −0.16 C

cm3

ρ 0( ) = −0.16 C

cm3

6) Consider a PN junction in equilibrium with N A = 1017 cm-3 and N D = 1018 cm-3 at T = 300

K. Answer the following questions.

5a) If the material is Si with ni = 1×1010 cm-3 , what is the built-in potential, Vbi , of the PN junction?

Solution:

Vbi =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟= 0.026ln10171018

1020 = 0.90 V Vbi = 0.90 V

6b) If the material is Si with ni = 1×1010 cm-3 , but the doping on both sides of the

junction is increased by 10X, what is the built-in potential, Vbi , of the PN junction?

Solution:

Vbi =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟= 0.026ln10181019

1020 = 1.02 V Vbi = 1.02 V

6c) If the material is GaAs at T = 300 K with ni = 2×106 cm-3 , with N A = 1017 cm-3 and

N D = 1018 cm-3 , what is the built-in potential, Vbi , of the PN junction?

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HW 2 solutions (continued):

Solution:

Vbi =kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟= 0.026ln 10171018

2×106( )2 = 1.35 V Vbi = 1.35 V

6d) Compare your answers to the above questions to the band gap of the semiconductor.

Solution:

The built-in potential is comparable to, but typically a little less than, the bandgap of the semiconductor in eV.

7) Consider a silicon PN junction with N A = 1016 cm-3 and N D = 1019 cm-3 at T = 300 K, so

that ni = 1×1010 cm-3 . The thickness of the undepleted N-region is 100 nm and of the undepleted P-region is 200 nm. The electron mobility is 1200 cm2/V-s, and the hole mobility is 70 cm2/V-s. The area of the PN junction is 500 nm X 500 nm. Answer the following questions.

7a) What is the saturation current density, IS , in A/cm2?

Solution:

IS = qA

ni2

N A

Dn

Wp

+ qAni

2

N D

Dp

Wn

≈ qAni

2

N A

Dn

Wp

Because N D >> N A , we can ignore the second term. This is called a “one-sided” PN junction.

IS ≈ qA

ni2

N A

Dn

Wp

= 1.6×10−19 500×10−7( )2 1020

1016

0.026 1200( )200×10−7 = 6.24×10−18 A

IS = 6.24×10−18 A

7b) What voltage must be applied so that the diode current is 10-12 A?

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HW 2 solutions (continued):

Solution:

ID = IS eqVA kBT −1( ) ≈ ISeqVA kBT

VA =

kBTq

lnID

IS

⎛

⎝⎜⎞

⎠⎟= 0.026ln 10−12

6.24×10−18

⎛⎝⎜

⎞⎠⎟= 0.31 V

VA = 0.31 V

7c) What voltage must be applied so that the diode current is 10-5 A?

Solution:

VA =

kBTq

lnID

IS

⎛

⎝⎜⎞

⎠⎟= 0.026ln 10−5

6.24×10−18

⎛⎝⎜

⎞⎠⎟= 0.73 V

VA = 0.73 V

Note that the “turn on” voltage of a diode depends on what we define the “on-current” to be. Many books define the diode voltage to be 0.6-0.7 V when the diode is on, but that is only a rule of thumb for typical silicon diodes at typical currents.