Download Problems Solutions for Microelectronics: Circuit Analysis and Design | ECE 2204 and more Study notes Basic Electronics in PDF only on Docsity!
Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
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
Silicon
(a) ( )( )
( ) [^ ]
15 3 / 2 6
18
10 3
5.23 10 100 exp 2 86 10 100
5.23 10 exp 63.
8.79 10 cm
i
i
n
n
−
− −
= × ⎢^ ⎥
⎢ × ⎥
= × −
= ×
(b) ( )( )
( )(^ )
( ) [ ]
15 3 / 2 6
19
10 3
5.23 10 300 exp 2 86 10 300
2.718 10 exp 21.
1.5 10 cm
i
i
n
n
−
−
= × ⎢ ⎥
⎢ × ⎥
= × −
= ×
(c) ( )( )
( ) [^ ]
15 3 / 2 6
19
14 3
5.23 10 500 exp 2 86 10 500
5.847 10 exp 12.
1.63 10 cm
i
i
n
n
−
−
= × ⎢ ⎥
⎢ × ⎥
= × −
= ×
Germanium.
(a) ( )( )
( )(^ )^
( ) [^ ]
15 3 / 2 18 6
3
1.66 10 100 exp 1.66 10 exp 38. 2 86 10 100
35.9 cm
i
i
n
n
−
−
= × ⎢^ ⎥= × −
⎢ × ⎥
(b) ( )( )
( )(^ )^
( ) [ ]
15 3 / 2 18 6
13 3
1.66 10 300 exp 8.626 10 exp 12. 2 86 10 300
2.40 10 cm
i
i
n
n
−
−
= × ⎢^ ⎥= × −
⎢ × ⎥
= ×
(c) ( )( )
( ) [ ]
15 3 / 2 19 6
15 3
1.66 10 500 exp 1.856 10 exp 7. 2 86 10 500
8.62 10 cm
i
i
n
n
−
−
= × ⎢ ⎥= × −
⎢ × ⎥
= ×
______________________________________________________________________________________
(a) n-type; cm ;
15 no = 10
− 3 (^ )^11
15
2 132
- 76 10 10
= ×
×
o
i o n
n p cm
− 3
(b) n-type; cm ;
15 no = 10
− 3 (^ )^5
15
2 102
- 25 10 10
= ×
×
o
i o n
n p cm
− 3
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
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
- 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
cm
2
=( 0. 026 )( 1250 ) = 32. 5 /s;
n
D =( 0. 026 )( 450 ) = 11. 7
p D cm /s
2
16 12 19 =− ⎟
= = ×
−
dx
dn J eD n n A/cm
2
12 16 19 =− ⎟
= − =− ×
−
dx
dp J eD p p A/cm
2
Total diffusion current density
A/cm
2 J =− 52 − 18. 72 =− 70. 7
______________________________________________________________________________________
15
19 15
4
/
10 exp
exp 10 10
p
p p
p p p
p p
x L p
dp J eD dx
x eD L L
x J L
J e
−
−
−
× ⎛ − ⎞
×
(a) x = 0
2 J (^) p =2.4 A/cm
(b) x = 10 μm
1 2 J (^) p 2.4 e 0.883 A/cm
− = =
(c) x = 30 μm
3 2 2.4 0.119 A/cm p J e
− = =
______________________________________________________________________________________
a.
17 3 17 3 N (^) a 10 cm po 10 cm
− − = ⇒ =
2 2 6 5 3 17
3.24 10 cm 10
i o o o
n n n p
− −
×
= = ⇒ = ×
b.
5 15 15 3
17 15 17 3
3.24 10 10 10 cm
10 10 1.01 10 cm
o
o
n n n n
p p p p
− −
−
= + = × + ⇒ =
= + = + ⇒ = ×
______________________________________________________________________________________
2 ln
i
a d bi T n
NN
V V
(a) (i) ( )
- 026 ln 102
15 15
= ⎥
×
× ×
Vbi = V
(ii) ( )
- 026 ln 102
17 15
=
⎥
×
×
bi
V V
(iii) ( )
- 026 ln 102
18 18
=
⎥
×
bi
V V
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
j
C avg. pF
t / C C C C v t v final v initial v final e
− τ = + −
where 3 1
τ RC RC j ( avg ) (47 10 )(0 00962. 10 )
− = = = × ×
2
or 10
τ 4 52. 10 s
− = ×
Then ( ) 1 5 0 ( 5 0 )
t /i vC t. e
− τ = = + −
1 / 1
ln 1.5 1.
r e t
τ τ
10 1 t 5.44 10 s
− = ×
(b) For VR = 0 V, Cj = Cjo = 0.02 pF
For VR = 3.5 V, ( )
1/ 2
0.02 1 0.
j C p
− ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠
F
j C avg pF
10 6.72 10 j
τ RC avg s
− = = ×
t / C C C C v t v final v initial v final e
− τ = + −
2 /^2 / 3.5 5 (0 5) 5 1
t t e e
− τ − τ = + − = −
so that
10 2 t 8.09 10 s
− = ×
______________________________________________________________________________________
1 / 2
−
bi
R j jo V
V
C C ; ( )
- 026 ln 102
15 17
=
⎥
×
×
Vbi = V
For VR = 1 V,
C (^) j = pF
For VR = 3 V,
j C pF
For = 5 V, R
V
j C pF
(a)
3 12
× ×
− − o o f LC
f
MHz
(b)
f f MHz o o
3 12
× ×
− −
Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
(iv) (^ )^
13 14 1 5. 37 10
- 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
- 026 ln 11
6
= ⎟
⎛ ×
−
−
V (^) D V
- 026 ln 11
6
= ⎟
⎛ ×
−
−
V (^) D V
- 026 ln 11
3
= ⎟
−
−
V (^) D V
(ii) 1 0. 018
- 026
5 10 10 exp
12 11 ⎥⇒ =− ⎦
− × =
− − D
D V
V
V
(b) (i) ( ) 0. 479
- 026 ln 13
6
= ⎟
⎛ ×
−
−
V (^) D V
- 026 ln 13
6
= ⎟
⎛ ×
−
−
V (^) D V
- 026 ln 13
3
= ⎟
−
−
V (^) D V
(ii) 1 0. 00274
- 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
(a) 12 10 S
I A
−
VD (v) ID (A) log 10 ID
11 4 68_._ 10
− × −10 3_._
0.20 (^) 2 19_._ × 10 −^9 −8 66_._
0.30 1 03. × 10 −^7 −6 99.
6 4 80_._ 10
− × −5 32_._
4 2 25_._ 10
− × −3 65_._
0.60 (^) 1 05_._ × 10 −^2 −1 98_._
0.70 4 93. × 10 −^1 −0 307.
(b) 14 10 S
I A
−
VD (v) ID (A) log 10 ID
13 4 68_._ 10
− × −12 3_._
0.20 (^) 2 19_._ × 10 −^11 −10 66_._
− ×
8 4 80_._ 10
− × −7 32_._
6 2 25_._ 10
− × −5 65_._
0.60 (^) 1 05_._ × 10 −^4 −3 98_._
− ×
______________________________________________________________________________________
a.
2 2 1
1
10 exp
ln (10) 59.9 mV 60 mV
D D D
D T
D T D
I V V
I V
V V V
b. Δ V D = VT ln 100( ) ⇒ Δ V D = 119.7 mV ≈120 mV
______________________________________________________________________________________
(a) (i) (^ )^0.^539
- 026 ln 9
×
D −
V V
(ii) ( ) 0. 599
- 026 ln 9
×
D −
V V
(b) (i) ( ) 9. 60
2 10 exp
9 ⎟⇒ ⎠
= ×
− I (^) D mA
(ii) ( ) 144
2 10 exp
9 ⎟⇒ ⎠
= ×
− D
I A
______________________________________________________________________________________
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
= ×
−
- 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
( )
4 10 2 10 D D = I × + V and (^) ( ) 12
0.026 ln 10
D D
I
V
−
Trial and error.
VD (v) ID (A) VD (v)
0.50 (^) 4.75 × 10 −^4 0.
4 4.7415 10
− × 0.
4 4.740 10
− × 0.
0.5194 V
0.4740 mA
D
D
V
I
______________________________________________________________________________________
13 I (^) s 5 10 A
− = ×
2
1 2
(1.2) (1.2) 0.45 V
TH
R
V
R R
⎝ +^ ⎠ ⎝^ ⎠
0.45 , ln
D D TH D D T S
I
I R V V V
I
By trial and error:
2.56 A, 0.402 V D D
I = μ 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
From (1)
A A D A
V V
D
I V I
= − = − ⇒ Neg. ⇒ ID = 0
Both (a), (b) 0 D
I =
VA = 2.5,
5 2 V 0.50 V
B D
V = ⋅ = ⇒ V =
______________________________________________________________________________________
(a) ( ) 1 ; ; for mA
O i
V = I = 0
D
I 0 ≤ ≤ 0. 7
i
I
VO = 0. 7 V; I D = ( Ii − 0. 7 ) mA; for Ii ≥ 0. 7 mA
(b) VO = Ii ( ) 1 ; I D = 0 ; for 0 ≤ I i ≤ 1. 7 mA
VO = 1. 7 V; I D = ( Ii − 1. 7 ) mA; for Ii ≥ 1. 7 mA
(c) VO = 0. 7 V; I (^) D 1 = Ii ; I (^) D 2 = 0 ; for 0 ≤ I (^) i ≤ 2 mA
______________________________________________________________________________________
Minimum diode current for VPS (min)
(min) 2 , 0. D D I = mA V = V
2 1 2 1
I I
R R
1
R
We have I 1^ =^ I^ 2 + ID
so (1)
1 2
R R
Maximum diode current for VPS (max)
P = I VD D 10 = ID (^) ( 0 7_._ (^) )⇒ I (^) D =14 3_. mA_
1 2 D
I = I + I
or
1 2
R R
Using Eq. (1), (^1)
1 1
. R. k R R
Then R 2 (^) = 82 5_._ Ω 82 5_._ Ω
______________________________________________________________________________________
(a) (i) 0. 215 20
I = mA, VO = 0. 7 V
(ii) 0. 220 20
I = mA, VO = 0. 6 V
(b) (i)
I = mA, =( 0. 2325 )( 20 ) − 5 =− 0. 35
O
V V
(iii)
I = mA, =( 0. 235 )( 20 ) − 5 =− 0. 30
O
V V
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
- 026
DQ
T d I
V
r k Ω
(b) = ⇒ 100 Ω
- 26
r d
(c) = ⇒ 10 Ω
- 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
⎝ +^ ⎠ +
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
pn junction diode
- 026 ln 13
3
= ⎟
×
×
−
−
V D V
Schottky diode
- 026 ln 8
3
= ⎟
×
×
−
−
D
V V
______________________________________________________________________________________
Schottky: exp
a S T
V
I I
V
3
7
ln (0.026) ln 5 10
a T S
I
V V
I
V
−
−
⎛ ⎞ ⎛ × ⎞
×
Then
of pn junction 0.1796 0.
a
V = +
3 0 5 10
exp exp 0 026
S a
T
I.
I
V.
V.
− × = = ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜^ ⎟ ⎝ ⎠ ⎝ ⎠
12 4 87 10 A S
I.
− = ×
______________________________________________________________________________________
(a)
