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Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

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Chapter 1

3 / 2 Eg^ / 2 kT ni BT e

−

(a) Silicon

(i) (^) ( )( ) ( )( )

[ ]

15 3 / 2 6

19

8 3

5.23 10 250 exp 2 86 10 250

2.067 10 exp 25.

1.61 10 cm

i

i

n

n

−

−

= × ⎢ ⎥

⎢ × ⎥

= × −

= ×

(ii) (^) ( )( )

( )(^ )

[ ]

15 3 / 2 6

19

11 3

5.23 10 350 exp 2 86 10 350

3.425 10 exp 18.

3.97 10 cm

i

i

n

n

−

−

= × ⎢^ ⎥

⎢ × ⎥

= × −

= ×

(b) GaAs

(i) (^) ( )( )

( )(^ )

( ) [ ]

14 3 / 2 6

17

3 3

2.10 10 250 exp 2 86 10 250

8.301 10 exp 32.

6.02 10 cm

i

i

n

n

−

−

= × ⎢^ ⎥

⎢ × ⎥

= × −

= ×

(ii) (^) ( )( )

( )(^ )

( ) [ ]

14 3 / 2 6

18

8 3

2.10 10 350 exp 2 86 10 350

1.375 10 exp 23.

1.09 10 cm

i

i

n

n

−

−

= × ⎢^ ⎥

⎢ × ⎥

= × −

= ×

______________________________________________________________________________________

a.

3 / 2 exp 2

i

Eg n BT kT

12 15 3 / 2 6

10 5.23 10 exp 2(86 10 )( )

T

T

−

= ×

⎝ × ⎠

3 4 3 / 2 6.40^10 1.91 10 T exp T

− ⎛^ × ⎞

× = −

By trial and error, T ≈368 K

b.

9 3 ni 10 cm

−

( )( )

9 15 3 / 2 6

10 5.23 10 exp 2 86 10

T

T

−

= × ⎜^ ⎟

⎜ × ⎟

3 7 3 / 2 6.40^10 1.91 10 T exp T

− ⎛^ × ⎞

× = ⎜ − ⎟

By trial and error, T ≈ 268 °K

---

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(a) p-type; cm ;

16 po = 10

− 3 (^ )^4

16

2 62

3. 24 10 10

= ×

×

o

i o p

n n cm

− 3

(b) p-type; cm ;

16 = 10 o p

− 3 (^ )^10

16

2 132

5. 76 10 10

= ×

×

o

i o p

n n cm

− 3

______________________________________________________________________________________

(a) n -type

(b)

16 3

2 2 10 3 3 16

5 10 cm

4.5 10 cm 5 10

o d

i o o

n N

n p n

−

−

= = ×

×

= = = ×

×

(c)

16 3 no Nd 5 10 cm

− = = ×

From Problem 1.1(a)(ii)

11 3 ni 3.97 10 cm

− = ×

2 11 6 3 16

3.15 10 cm 5 10

p o

−

×

= = ×

×

______________________________________________________________________________________

(a) p-type; cm ;

16 po = 5 × 10

− 3 (^ )^3

16

2 102

4. 5 10 5 10

= ×

×

×

o

i o p

n n cm

− 3

(b) p-type; cm ;

16 po = 5 × 10

− 3 (^ )^5

16

2 62

6. 48 10 5 10

= ×

×

×

o

i o p

n n cm

− 3

______________________________________________________________________________________

(a) Add boron atoms

(b) cm

17 = = 2 × 10 a o N p

− 3

(c)

17

2 102

1. 125 10 2 10

= ×

×

×

o

i o p

n n cm

− 3

______________________________________________________________________________________

(a)

15 3

2 2 10 4 3 15

o

i o o o

n cm

n p p n

−

cm

−

= ×

×

= = ⇒ = ×

×

(b) > ⇒n-type o o n p

(c)

15 3 5 10 o d n N cm

− ≅ = ×

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

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a. Add Donors 15 3 7 10 cm d

N

− = ×

6 3 2 10 cm / o i d n N

− b. Want p = =

So ( )( )

2 6 15

2 3

exp

i n

Eg B T kT

= × = ×

21

( )(^ )

2 21 15 3 6

7 10 5.23 10 exp 86 10

T

T

−

× = × ⎜^ ⎟

⎜ × ⎟

By trial and error, T ≈ 324 °K

______________________________________________________________________________________

(a) ( 10 )( 1. 5 )( 10 ) 0. 15 mA

5 = Ε= ⇒ =

−

I A σ I

(b)

4

3

= ×

×

−

−

A

A I

I

ρ

ρ

V/cm

______________________________________________________________________________________

1

6. 67 18

= Ε⇒ = cm

J

J σ σ

16 19

= ×

×

− n

n d d e

e N N μ

σ σ μ cm

− 3

______________________________________________________________________________________

(a)

15 19

= ×

×

− μ μρ

ρ

n

d n d e

N

e N

cm

− 3

(b) ⇒Ε= =( 0. 65 )( 160 ) = 104

J = ρ J ρ

V/cm

______________________________________________________________________________________

(a)

15 19

= ×

×

− n

n d d e

e N N μ

σ σ μ cm

− 3

(b)

16 19

= ×

×

− p

a e

N

cm

− 3

______________________________________________________________________________________

(a) For n-type, ( )( )

19 1.6 10 8500 n d d σ e μ N N

− ≅ = ×

For ( )

15 19 3 4 1 10 10 1.36 1.36 10 d N cm σ cm

− − ≤ ≤ ⇒ ≤ ≤ × Ω −

(b) ( )

3 2

J = σ E = σ 0.1 ⇒ 0.136 ≤ J ≤ 1.36 × 10 A cm /

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(b) (i) ( )

1. 026 ln 62

15 15

=

⎥

×

× ×

bi

V V

(ii) ( )

1. 026 ln 62

17 15

=

⎥

×

×

bi

V V

(iii) ( )

1. 026 ln 62

18 18

=

⎥

×

Vbi = V

______________________________________________________________________________________

2

ln

i

a d bi T n

NN

V V

or

16

2 2

1. 76 10
2. 026

exp 10

exp ⎟= × ⎠

× ⎛

T

bi

d

i a V

V

N

n N cm

− 3

______________________________________________________________________________________

16

2 1

ln 0.026 ln (1.5 10 )

a d^ a bi T i

N N N

V V

n

⎢ × ⎥

0 2

V

V

For

15 3 Na 10 cm , Vbi 0.

− = =

For

18 3 Na 10 cm , Vbi 0.

− = =

______________________________________________________________________________________

T

kT

kT ( T )

3/

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

( )( ) ( )( )

14 3 / 2 6

2.1 10 exp 2 86 10

n i T T

−

= × ⎜^ ⎟

⎜ × ⎟

2 ln

a d bi T i

N N

V V

n

T ni Vbi

200 1.256 1.

250 6.02 × 10

3

(^300) 1.80 × 10

6

(^350) 1.09 × 10

8

(^400) 2.44 × 10 9

450 2.80 × 10

10

(^500) 2.00 × 10

11

______________________________________________________________________________________

1/ 2

R j jo bi

V

C C

V

− ⎛ ⎞ = (^) ⎜ + ⎟

⎝ ⎠

( )

( )( )

( )

16 15

10 2

0.026 ln 0.684 V 1.5 10

bi

V

⎡ × × ⎤

⎢ × ⎥

(a) (^) ( )

1/ 2 1 0.4 1 0.255 pF

j

C

− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠

(b) (^) ( )

1/ 2 3 0.4 1 0.172 pF

j

C

− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠

(c) (^) ( )

1/ 2 5 0.4 1 0.139 pF

C j

− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠

---

(a)

1 2

/

R j jo bi

V

C C

V

− ⎛ ⎞ = (^) ⎜ + ⎟

⎝ ⎠

For VR = 5 V ,

1 2 5 (0 02) 1 0 00743 0 8

/

j C.. p .

− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠

F

For VR = 1.5 V ,

1 2 1 5 (0 02) 1 0 0118 0 8

/

j

C.. p .

− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠

F

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(c)

3 12

× ×

− −

fo fo

MHz

______________________________________________________________________________________

a. exp 1 0.90 exp 1

D D S T T

V V

I I

V V

exp 1 0.90 0.

D

T

V

V

⎜ ⎟=^ −^ =

VD = VT ln 0.10 ( )⇒ VD = −0.0599 V

b.

exp 1 exp 1

exp 1 exp^1

F

F S T

R S (^) R

T

F

R

V

I I V

I I V

V

I

I

⎣ ⎝^ ⎠ ⎦

______________________________________________________________________________________

= exp 1

T

D D S V

V

I I

(a) (i) ( ) 1. 03 μ

10 exp

11 ⎟⇒ ⎠

− I (^) D A

(ii) ( ) 2. 25

10 exp

11 ⎟⇒ ⎠

− D I mA

(iii) (^ )^4.^93

10 exp

11 ⎟⇒ ⎠

− D

I A

(iv) ( )

11 12 1 5. 37 10

1. 026

10 exp

− − =− × ⎥ ⎦

D

I A

(v) ( )

11 11 1 10

1. 026

10 exp

− − ≅− ⎥ ⎦

D

I A

(vi) ( )A

11 10

− =− D

I

(b) (i) ( ) 0. 0103 μ

10 exp

13 ⎟⇒ ⎠

− I (^) D A

(ii) ( ) 22. 5 μ

10 exp

13 ⎟⇒ ⎠

− D

I A

(iii) (^ )^49.^3

10 exp

13 ⎟⇒ ⎠

− D I (^) mA

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(iv) (^ )^

13 14 1 5. 37 10

1. 026

10 exp

− − ⎥=− × ⎦

I D = A

(v) A

13 10

− ID ≅−

(vi) A

13 10

− ID ≅−

______________________________________________________________________________________

S

D D T I

I

V V ln

(a) (i) ( ) 0. 359

1. 026 ln 11

6

= ⎟

⎛ ×

−

−

V (^) D V

1. 026 ln 11

6

= ⎟

⎛ ×

−

−

V (^) D V

1. 026 ln 11

3

= ⎟

−

−

V (^) D V

(ii) 1 0. 018

1. 026

5 10 10 exp

12 11 ⎥⇒ =− ⎦

− × =

− − D

D V

V

V

(b) (i) ( ) 0. 479

1. 026 ln 13

6

= ⎟

⎛ ×

−

−

V (^) D V

1. 026 ln 13

6

= ⎟

⎛ ×

−

−

V (^) D V

1. 026 ln 13

3

= ⎟

−

−

V (^) D V

(ii) 1 0. 00274

1. 026

10 10 exp

14 13 ⎥⇒ =− ⎦

− − D

D V

V

V

______________________________________________________________________________________

(a)

10 exp

I S

− ⎛^ ⎞

15 I (^) S 2.03 10 A

− = ×

(b)

V D I D ( A ) ( n = 1) I D ( A ) ( n = 2 )

0.1 9.50 × 10 −^14 1.39 × 10 −^14

12 4.45 10

− ×

14 9.50 10

− ×

10 2.08 10

− ×

13 6.50 10

− ×

0.4 (^) 9.75 × 10 −^9 4.45 × 10 −^12

0.5 4.56 × 10 −^7 3.04 × 10 −^11

5 2.14 10

− ×

10 2.08 10

− ×

3 10

− 9 1.42 10

− ×

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

3

14

3

12

ln (0 026) ln 0 6347 V 5 10

(0 026) ln 0 5150 V 5 10

0 5150 0 6347 V

D D t S

D

D

I

V V..

I

V..

. V.

−

−

−

−

⎛ ⎞ ⎛ × ⎞

⎝ ⎠ ⎝ × ⎠

⎛ × ⎞

⎝ × ⎠

______________________________________________________________________________________

(a) I^ S IS A

3 8

1. 46 10
2. 026

1. 5 10 exp

− − ⎟⇒ = × ⎠

× =

(b) (i) ( ) 10. 3

1. 462 10 exp

8 ⎟⇒ = ⎠

= ×

− I (^) D I D mA

(ii) ( ) 0. 219

1. 462 10 exp

8 ⎟⇒ = ⎠

= ×

− I (^) D I D mA

______________________________________________________________________________________

(a) ( ) 2. 31

10 exp

22 ⎟⇒ ⎠

− I (^) D nA

10 exp

22 ⎟⇒ ⎠

− I (^) D A

10 exp

22 ⎟⇒ ⎠

− D I mA

22 23 1 5. 37 10

1. 026

10 exp

− − =− × ⎥ ⎦

D

I A

For V (^) D =− 0. 20 V, A

22 10

− ID =−

For =− 2 V, A D

V

22 10

− =− D

I

(b)

5 10 exp

24 ⎟⇒ ⎠

= ×

− I (^) D pA

5 10 exp

24 ⎟⇒ ⎠

= ×

− I (^) D A

5 10 exp

24 ⎟⇒ ⎠

= ×

− D I mA

24 24 1 2. 68 10

1. 026

5 10 exp

− − =− × ⎥ ⎦

= ×

D

I A

For =− 0. 20 V, A D

V

24 5 10

− =− × D

I

For VD =− 2 V, A

24 5 10

− ID =− ×

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

IS doubles for every 5C increase in temperature. 12 I (^) S 10 A

− = at T = 300K

For

12 I (^) S 0.5 10 A T 295 K

− = × ⇒ =

For

12 50 10 , (2) 50 5.

n S I A n

− = × = ⇒ =

Where n equals number of 5C increases.

Then Δ T = ( 5.64)( 5 ) =28.2 K

So 295 ≤ T ≤328.2 K

______________________________________________________________________________________

2 , 155 C

S T

S

I T

T

I

Δ = Δ = −

S

S

I

I

= = ×

@100 C 373 K 0.

T T

V ° ⇒ ° ⇒ V =

@ 55 C 216 K 0.

T T

V − ° ⇒ ° ⇒ V =

9

9 8

13

3

exp (100) 0. (2.147 10 ) ( 55) 0. exp

D

D

D

D

I

I

I

I

= × ×

× ×

×

= ×

______________________________________________________________________________________

(a) PS D D

V = I R + V

= I D ( ) + VD ;

6

= ×

−

1. 026

5 10 exp

11 D D

V

I

By trial and error,

VD = 0. 282 V, I D = 2. 52 μA

(b)

A, V

11 5 10

− ≅−× D

I =− 2. 8

D

V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(a) ID 1 = ID 2 = 1 mA

(i) ( ) 0. 599

1. 026 ln 13

3

1 2

−

−

D D

V V V

(ii) ( ) 0. 617

1. 026 ln 14

3

1

×

−

−

D

V V

1. 026 ln 13

3

2

×

−

−

D

V V

(b) VD (^) 1 = VD 2

(i) 0. 5 2

1 2

i D D

I

I I mA

1. 026 ln 13

3

1 2

⎛ ×

−

−

D D

V V V

(ii) 0. 10 5 10

13

14

2

1

2

1

×

×

−

−

S

S

D

D

I

I

I

I

So I (^) D 1 = 0. 10 ID 2

ID 1 + ID 2 = 1. 1 ID 2 = 1 mA

So I (^) D 2 = 0. 909 mA, ID 1 = 0. 0909 mA

Now

1. 026 ln 14

3

1

×

×

−

−

D

V V

1. 026 ln 13

3

2

×

×

−

−

D

V V

______________________________________________________________________________________

(a) ( ) 2. 426

6 10 exp

14 3 ⎟⇒ ⎠

= ×

− I (^) D mA

R I mA

2. 426 0. 635 3. 061 mA 1 2

D D

I I

1. 026 ln 14

3

1 2

×

×

−

−

D D

V V V

= 2 ( 0. 641 ) + 0. 635 = 1. 917 V

I

V

(b) 2. 426 mA 3

D

I

R I (^) mA

ID 1 = ID 2 = 2. 426 + 1. 27 = 3. 696 mA

1. 026 ln 14

3

×

×

−

−

VD V D V

VI = 2 ( 0. 6459 )^ + 0. 635 = 1. 927 V

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

______________________________________________________________________________________

(a) Assume diode is conducting.

Then, VD = V (^) γ=0.7 V

So that 2

R

I = ⇒. μ A

1

R

I μ A

Then 1 2

D R R

I = I − I = −.

Or 26 7 D

I =. μ A

(b) Let R 1 (^) = 50 k ΩDiode is cutoff.

D

V = ⋅. =

. V

Since , 0 D D

V V I

γ

______________________________________________________________________________________

At node VA :

A A D

V V

I

At node V (^) B = VA − V γ

(^5) ( ) ( )

A r A r D

V V V V

I

So

(^5) ( ) 5

A r (^) A A A

− V − V ⎡ − V V ⎤ V − V

r

)

A

Multiply by 6:

10 − (^2) ( VA − Vr (^) ) + 15 − 6 V (^) A = (^3) ( VA − Vr

r r

+ V + V = V

(a) 0.6 V r

V =

11 V (^) A = 25 + 5 0.6 ( ) = 28 ⇒ VA =2.545 V

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

(c) (i)

I = mA, = 2 −( 0. 372 )( 5 ) = 0. 14

O

V V

(ii)

I = mA, = 2 −( 0. 376 )( 5 ) = 0. 12

VO V

(d) (i) I = 0 , VO =− 5 V

(ii) I = 0 , VO =− 5 V

______________________________________________________________________________________

(a) 20

O

V

I

= ×

−

5 10 exp

14 VD

I

By trial and error, V (^) D = VO = 0. 5775 V, I = 0. 221 mA

(b) 40

D

V

I

= , VO = 5 − I ( 20 ) − VD

I = 0. 2355 mA, V (^) D = 0. 579 V, VO =− 0. 289

(c) 25

D

V

I

= , VO = 2 − I ( ) 5

I = 0. 3763 mA, VD = 0. 5913 V, VO = 0. 1185

(d) A, V

14 5 10

− I =−× ≅− 5 O

V

______________________________________________________________________________________

(a) Diode forward biased VD = 0.7 V

5 = (0.4)(4.7) + 0.7 + V ⇒ V =2.42 V

(b) P = I V ⋅ (^) D = (0.4)(0.7) ⇒ P =0.28 mω

______________________________________________________________________________________

(a) 2 1 1

2

0 2 1 1 1

0.65 mA 1

2(0.65) 1.30 mA

1.30 2.35 K

R D D

D

I r D

I I I

I

V V V

I R

R R

(b) (^2)

2 2

1 2 2

1

0.65 mA 1

3.025 mA 2

2.375 mA

R

D D

D D R

D

I

I I

I I I

I

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1

By D. A. Neamen Problem Solutions

---

a.

0 026 k 26 1

0 05 50 A peak-to-peak

(26)(50) A 1 30 mV peak-to-peak

T d DQ

d DQ

d d d d

V.

I

i. I

v i v.

τ

μ

τ μ

b. For

0 1 mA 260 0 1

DQ d

I.

= ⇒ τ = = Ω

0 05 5 A peak-to-peak d DQ i =. I = μ

(260)(5) V 1 30 mV peak-to-peak d d d d

v = i τ = μ ⇒ v =.

______________________________________________________________________________________

(a) 1

1. 026

DQ

T d I

V

r k Ω

(b) = ⇒ 100 Ω

1. 26

r d

(c) = ⇒ 10 Ω

2. 6

d r

______________________________________________________________________________________

a. diode resistance d T r = V / I

d T d S d S T S

T d s o T S

r V I v v r R V R I

V

v v v V IR

⎛ ⎞ ⎜^ / ⎟

v S

b. RS = 260 Ω

( )( )

0 0

0 0

0 0

1 mA, 0 0909 0 026 (1)(0 26)

0 1 mA, 0 50 0 026 0 1 0 26

0 01 mA. 0 909 0 026 (0 01)(0 26)

T

S T S S

s S

S S

v V. v I. v V IR.. v

v. v I.. v... v

v. v I.. v... v

⎝ +^ ⎠ +

______________________________________________________________________________________