Solution manual microelectronics; circuit analysis desing 3rd edition

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Chapter 1
Exercise Problems
EX1.1
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠

GaAs: ni = ( 2.1× 1014 ) ( 300 )
Ge: ni = (1.66 × 1013 ) ( 300 )

3/ 2

3/ 2

⎛
⎞
−1.4
⎟ or ni = 1.8 × 106 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠

⎛
⎞
−0.66
⎟ or ni = 2.40 × 1013 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠

EX1.2
(a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons,
n 2 (1.5 × 10
no = i =
1017
po

)

10 2

(b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes,
n 2 (1.5 × 10 )
= 4.5 × 104 cm −3
po = i =
5 × 1015
no
10 2

EX1.3
For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields

E = 17.9 V / cm

EX1.4
Diffusion current density due to holes:
dp
J p = −eD p
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟
⎜L ⎟
⎜L ⎟
⎝ p⎠
⎝ p⎠
(a) At x = 0

(1.6 ×10 ) (10 ) (10 ) = 16 A / cm
=
−19

Jp

16

2

10−3
−3
(b) At x = 10 cm

⎛ −10−3 ⎞
J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2
⎝ 10 ⎠

EX1.5
⎡N N
Vbi = VT ln ⎢ a 2 d
⎣ ni

⎡ (1016 )(1017 ) ⎤
⎤
⎢
⎥ or Vbi = 1.23 V
=
0.026
ln
(
)
⎥
6 2 ⎥
⎢
⎦
⎣ (1.8 × 10 ) ⎦

EX1.6

⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
and

−1/ 2

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= 2.25 × 103 cm −3


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⎡N N ⎤
Vbi = VT ln ⎢ a 2 d ⎥
⎣ ni ⎦
⎡ (1017 )(1016 ) ⎤
⎥ = 0.757 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
5 ⎞
⎛
Then 0.8 = C jo ⎜ 1 +
⎟
⎝ 0.757 ⎠
or
C jo = 2.21 pF

−1/ 2

= C jo ( 7.61)

−1/ 2

EX1.7

⎡
⎛ v ⎞ ⎤
so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥
⎝ 0.026 ⎠ ⎦
⎣

⎡ 10−3
⎤
Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥
⎣10
⎦
or
vD ≅ ( 0.026 ) ln (1010 )

which yields
vD = 0.599 V
EX1.8
⎛V ⎞
VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
( 4 − VD )
so 4 = I D ( 4 ×103 ) + VD ⇒ I D =
4 ×103
and
⎛ V ⎞

I D = (10 −12 ) exp ⎜ D ⎟
⎝ 0.026 ⎠
By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V

EX1.9

(a)

ID =

(b)

ID =

Then R =
(c)

VPS − Vγ
R
VPS − Vγ
R

5 − 0.7
⇒ I D = 1.08 mA
4
VPS − Vγ
⇒R=
ID
=

8 − 0.7
= 6.79 kΩ
1.075

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⎡
⎛v ⎞ ⎤
iD = I S ⎢exp ⎜ D ⎟ − 1⎥
⎝ VT ⎠ ⎥⎦
⎣⎢


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ID(mA)
Diode curve
1.25
1.08

Load lines
(b)
(a)

0

0.7

2

4
VD(v)

6

8

EX1.10
PSpice analysis

Quiescent diode current I DQ =

VPS − Vγ

=

10 − 0.7
= 0.465 mA
20

R
Time-varying diode current:
V
0.026
We find that rd = T =
= 0.0559 kΩ
I DQ 0.465
Then id =

vI

0.2sin ω t (V )
=
⋅
or id = 9.97sin ω t ( μ A)
rd + R 0.0559 + 20 ( kΩ )

EX1.12
⎛I ⎞
⎛ 1.2 × 10−3 ⎞
or VD = 0.6871 V
For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 4 × 10 ⎠
⎝ IS ⎠
The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V
⎛V ⎞
Now I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
or
1.2 × 10−3
IS =
⇒ I S = 1.07 × 10−10 A
0.4221
⎛
⎞
exp ⎜
⎟
⎝ 0.026 ⎠

EX1.13

P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA

Also I =

10 − 5.6
= 1.79 ⇒ R = 2.46 kΩ
R

Test Your Understanding Exercises
TYU1.1
(a) T = 400K
⎛ − Eg ⎞
Si: ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
ni = ( 5.23 × 1015 ) ( 400 )

or
ni = 4.76 × 1012 cm −3

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

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EX1.11


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Ge: ni = (1.66 × 1015 ) ( 400 )

3/ 2

⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

or
ni = 9.06 × 1014 cm −3
GaAs:
ni = ( 2.1× 1014 ) ( 400 )

3/ 2

⎡
⎤
−1.4
⎥

exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

or
ni = 2.44 × 109 cm −3
(b) T = 250 K
Si: ni = ( 5.23 × 1015 ) ( 250 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦

Ge: ni = (1.66 × 1015 ) ( 250 )

3/ 2

⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

or
ni = 1.42 × 1012 cm −3
GaAs: ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦

or
ni = 6.02 × 103 cm −3
TYU1.2
(a)
n = 5 × 1016 cm −3 , p <<< n, so σ ≅ eμ n n = (1.6 × 10 −19 ) (1350 ) ( 5 × 1016 )

or

σ = 10.8 ( Ω − cm )
(b)

−1

p = 5 × 1016 cm −3 , n <<< p, so σ ≅ eμ p p = (1.6 × 10−19 ) ( 480 ) ( 5 × 1016 )

or

σ = 3.84 ( Ω − cm )

−1

TYU1.3
J = σ E = (10 )(15 ) or J = 150 A / cm2
TYU1.4

(a)

J n = eDn

⎛ 1015 − 1016 ⎞
dn
Δn
so J n = 1.6 × 10−19 ( 35 ) ⎜
= eDn
−4 ⎟
dx
Δx
⎝ 0 − 2.5 × 10 ⎠

(

)

or
J n = 202 A / cm 2
(b)

J p = −eD p

or
J p = −24.5 A / cm2
TYU1.5

⎛ 1014 − 5 × 1015 ⎞
dp
Δp
so J p = − 1.6 × 10−19 (12.5 ) ⎜
= −eD p
−4 ⎟
dx
Δx
⎝ 0 − 4 × 10 ⎠

(

)

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or
ni = 1.61× 108 cm −3


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no = N d = 8 × 1015 cm −3

(a)

10
n 2 (1.5 × 10 )
po = i =
= 2.81× 10 4 cm −3
no
8 × 1015
2

(b) n = no + δ n = 8 × 1015 + 0.1× 1015
or
n = 8.1×1015 cm−3
p = po + δ p = 2.81 × 10 4 + 1014
or
p ≅ 1014 cm −3

(a)

⎡ (1015 )(1017 ) ⎤
⎡N N ⎤
⎥ = 0.697 V
Vbi = VT ln ⎢ a 2 d ⎥ so Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣ ni ⎦
⎣
⎦

(b)

⎡ (1017 )(1017 ) ⎤
⎥ = 0.817 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦

TYU1.7

⎡
⎛V
I D = I S ⎢exp ⎜ D
⎝ VT
⎣⎢

(a)

⎞ ⎤
⎟ − 1⎥
⎠ ⎦⎥

⎛ 0.5 ⎞
I D ≅ 10−14 exp ⎜
⎟
⎝ 0.026 ⎠
Then, for
VD = 0.5 V, I D = 2.25 μ A
VD = 0.6 V, I D = 0.105 mA
VD = 0.7 V, I D = 4.93 mA
(b)

I D ≅ − I S = −10 −14 A
for both cases.
TYU1.8
ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV
Then VD = 0.650 − 0.20 = 0.450 V
TYU1.9
ID(mA)
Diode
1.0
ഠ0.87

Load line

0

1
ഠ0.54v

2
VD(v)

VD

ID

0.45
0.50
0.55

0.033

0.225
1.54

3

TYU1.10
P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA

4

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TYU1.6


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Now R =

VPS − Vγ
ID

=

10 − 0.7
⇒ R = 6.2 kΩ
1.5

TYU1.11
I

0.8
gd = D =
= 30.8 mS
VT 0.026

TYU1.13
For the pn junction diode,
4 − 0.7
ID =
= 0.825 mA
4

For the Schottky diode, I D =

4 − 0.3
= 0.925 mA
4

TYU1.14
Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10 −3 ) ( 20 ) = 5.18 V

Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V

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TYU1.12
V
0.026
0.026
rd = T ⇒ 50 =

⇒ ID =
ID
ID
50
or
I D = 0.52 mA


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Chapter 1
Problem Solutions
1.1
− E / 2 kT
ni = BT 3 / 2 e g
(a)
Silicon

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
= 2.067 × 1019 exp [ −25.58]
ni = 1.61× 108 cm −3

(i)

ni = ( 5.23 × 1015 ) ( 250 )

(ii)

ni = ( 5.23 × 1015 ) ( 350 )

(b)

GaAs

(i)

ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= 3.425 × 1019 exp [ −18.27 ]
ni = 3.97 ×1011 cm −3

3/ 2

⎡
⎤
−1.4
⎥

exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

= ( 8.301× 1017 ) exp [ −32.56]
ni = 6.02 × 103 cm −3

(ii)

ni = ( 2.10 × 1014 ) ( 350 )

3/ 2

⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦

= (1.375 × 1018 ) exp [ −23.26]
ni = 1.09 × 108 cm −3

1.2

⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
⎛

⎞
−1.1
1012 = 5.23 × 1015 T 3 / 2 exp ⎜
⎟
−6
×
2(86
10
)(
T
)
⎝
⎠

a.

⎛ 6.40 × 103 ⎞
1.91× 10−4 = T 3 / 2 exp ⎜ −
⎟
T
⎝
⎠
By trial and error, T ≈ 368 K
b.
ni = 109 cm −3
⎛
⎞
−1.1
⎟
109 = 5.23 × 1015 T 3 / 2 exp ⎜

⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝
⎠
⎛ 6.40 × 103 ⎞
1.91× 10−7 = T 3 / 2 exp ⎜ −
⎟
T
⎝
⎠
By trial and error, T ≈ 268° K

1.3
Silicon

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3/ 2


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ni = ( 5.23 × 1015 ) (100 )

(a)

3/ 2

⎡
⎤
−1.1

⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦

= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3
ni = ( 5.23 × 1015 ) ( 300 )

(b)

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 300 ) ⎥⎦

= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢

−6
⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦

= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3

Germanium.

ni = (1.66 × 1015 ) (100 )

(a)

3/ 2

⎡
⎤
−0.66
⎥ = (1.66 × 1018 ) exp [ −38.37 ]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦

ni = 35.9 cm −3
ni = (1.66 × 1015 ) ( 300 )

(b)

3/ 2

⎡
⎤

−0.66
⎥ = ( 8.626 × 1018 ) exp [ −12.79]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 300 ) ⎥⎦

ni = 2.40 × 1013 cm −3
ni = (1.66 × 1015 ) ( 500 )

(c)

3/ 2

⎡
⎤
−0.66
⎥ = (1.856 × 1019 ) exp [ −7.674]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦

ni = 8.62 ×1015 cm −3
1.4
a.

N d = 5 × 1015 cm −3 ⇒ n − type

n0 = N d = 5 × 1015 cm −3
n 2 (1.5 × 10 )
⇒ p0 = 4.5 × 10 4 cm −3
p0 = i =
5 × 1015

n0
10 2

N d = 5 × 1015 cm −3 ⇒ n − type

b.

no = N d = 5 × 1015 cm −3
ni = ( 2.10 × 1014 ) ( 300 )
= ( 2.10 × 1014 ) ( 300 )

3/ 2

3/ 2

⎛
⎞
−1.4
exp ⎜
⎟
−6
⎝ 2(86 × 10 )(300) ⎠

(1.65 ×10 )
−12

= 1.80 × 106 cm −3
6
ni2 (1.8 × 10 )
=

⇒ p0 = 6.48 × 10 −4 cm −3
p0 =
15
5 × 10
n0
2

1.5

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ni = ( 5.23 × 1015 ) ( 500 )

(c)


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(a)
(b)

n-type
no = N d = 5 × 1016 cm −3
10
n 2 (1.5 × 10 )
po = i =
= 4.5 × 103 cm −3
no
5 × 1016
2

no = N d = 5 × 1016 cm −3

(c)

From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3

( 3.97 × 10 )
=

11 2

po

= 3.15 × 106 cm −3

5 × 1016

1.6
a.

N a = 1016 cm −3 ⇒ p − type

p0 = N a = 1016 cm −3
n 2 (1.5 × 10 )
⇒ n0 = 2.25 × 10 4 cm −3
n0 = i =
1016
p0
b.

Germanium
N a = 1016 cm −3 ⇒ p − type
p0 = N a = 1016 cm −3
ni = (1.66 × 1015 ) ( 300 )
= (1.66 × 1015 ) ( 300 )

3/ 2

3/ 2

⎛
⎞
−0.66
⎟
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠

( 2.79 × 10 )
−6

= 2.4 × 1013 cm −3
13
n 2 ( 2.4 × 10 )
n0 = i =
⇒ n0 = 5.76 × 1010 cm −3
p0
1016
2

1.7
(a)
(b)

p-type
po = N a = 2 × 1017 cm −3
10
ni2 (1.5 × 10 )
no =
=
= 1.125 × 103 cm −3
po
2 × 1017
2

(c)

po = 2 × 1017 cm −3

From Problem 1.1(a)(i) ni = 1.61 × 108 cm −3

(1.61×10 )
=

8 2

no

1.8

(a)

2 × 10

17

= 0.130 cm −3

no = 5 × 1015 cm −3
10
ni2 (1.5 × 10 )
po =
=
⇒ po = 4.5 × 104 cm −3
no
5 × 1015
2

(b)

no  po ⇒ n-type

(c)

no ≅ N d = 5 × 1015 cm −3

1.9
Add Donors
a.
N d = 7 × 1015 cm −3

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10 2


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Want po = 106 cm −3 = ni2 / N d

b.

So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21
⎛ − Eg ⎞
= B 2T 3 exp ⎜
⎟
⎝ kT ⎠

⎛
⎞
2
−1.1
⎟
7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜
6
−
⎜ ( 86 × 10 ) (T ) ⎟
⎝
⎠
By trial and error, T ≈ 324° K

1.10
I = J ⋅ A = σ EA

I = ( 2.2 )(15 ) (10−4 ) ⇒ I = 3.3 mA
1.11
J 85
=
E 12
σ = 7.08 (ohm − cm) −1

1.12
g≈

1
1
1
⇒ Na =
=
eμ p N a
eμ p g (1.6 × 10 −19 ) ( 480 )( 0.80 )

N a = 1.63 × 10 16 cm −3

1.13
σ = eμ n N d
Nd =

σ
eμ n

=

( 0.5)

(1.6 ×10 ) (1350 )
−19

N d = 2.31× 1015 cm −3

1.14
(a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d
For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm )
(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 × 103 A / cm2
1.15
J n = eDn

dn
Δn
= eDn
dx
Δx

⎡10 15 −10 2 ⎤
= (1.6 × 10 −19 ) (180 ) ⎢
−4 ⎥
⎣ 0.5 × 10 ⎦
J n = 576 A/cm 2
1.16

−1

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J =σE ⇒σ =


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J p = −eD p

dp
dx

⎛ −1 ⎞
⎛ −x ⎞
= −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟
⎜ Lp ⎟
⎜ Lp ⎟
⎝ ⎠
⎝ ⎠

(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞
−19

Jp =

15

⎜⎜ ⎟⎟

⎝ Lp ⎠

10 × 10 −4

J p = 2.4 e

− x / Lp

J p = 2.4 A/cm2

(a)

x=0

(b)

x = 10 μ m

J p = 2.4 e−1 = 0.883 A/cm 2

(c)

x = 30 μ m

J p = 2.4 e−3 = 0.119 A/cm 2

1.17
a.

N a = 1017 cm −3 ⇒ po = 1017 cm −3

6
ni2 (1.8 × 10 )
no =
=
⇒ no = 3.24 × 10−5 cm −3
po
1017

b.

n = no + δ n = 3.24 × 10 −5 + 1015 ⇒ n = 1015 cm −3
p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3

1.18
(a)

⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
⎡ (10 16 )(10 16 ) ⎤
⎥ = 0.697 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦

(b)

⎡ (10 18 )(10 16 ) ⎤

⎥ = 0.817 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦

(c)

⎡ (10 18 )(10 18 ) ⎤
⎥ = 0.937 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦

1.19
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠

a.

⎡ (1016 )(1016 ) ⎤
⎥ ⇒ Vbi = 1.17 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦

b.

⎡ (1018 )(1016 ) ⎤
⎥ ⇒ Vbi = 1.29 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦

c.

⎡ (1018 )(1018 ) ⎤
⎥ ⇒ Vbi = 1.41 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦

1.20

⎡ N a (1016 ) ⎤
⎛ Na Nd ⎞
⎥
Vbi = VT ln ⎜ 2 ⎟ = ( 0.026 ) ln ⎢
10 2
⎢⎣ (1.5 × 10 ) ⎥⎦
⎝ ni ⎠

www.elsolucionario.net

2


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For N a = 1015 cm −3 , Vbi = 0.637 V
For N a = 1018 cm −3 , Vbi = 0.817 V
Vbi (V)
0.817

0.637

1015

1016

1018 Na(cmϪ3)

1017

⎛ T ⎞
kT = (0.026) ⎜
⎟
⎝ 300 ⎠
T
kT
(T)3/2
200
0.01733
2828.4
250
0.02167
3952.8
300

0.026
5196.2
350
0.03033
6547.9
400
0.03467
8000.0
450
0.0390
9545.9
500
0.04333
11,180.3
⎛
⎞
−1.4
⎟
ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜
−
6
⎜ 2 ( 86 × 10 ) (T ) ⎟
⎝
⎠
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠

T
200

250
300
350
400
450
500

ni
1.256
6.02 × 103
1.80 × 106
1.09 × 108
2.44 × 109
2.80 × 1010
2.00 × 1011

Vbi
1.405
1.389
1.370
1.349
1.327
1.302
1.277

Vbi (V)
1.45
1.35
1.25

200

250

300

1.22
⎛ V ⎞
C j = C jo ⎜ 1 + R ⎟
⎝ Vbi ⎠

−1/ 2

350

400

450

500 T(ЊC)

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1.21


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⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤
⎥ = 0.684 V

Vbi = ( 0.026 ) ln ⎢
⎢⎣
(1.5 ×10 10 ) 2 ⎥⎦
1 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠

−1/ 2

(a)

3 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠

−1/ 2

(b)

5 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
0.684
⎝
⎠

−1/ 2

(c)

= 0.255 pF
= 0.172 pF
= 0.139 pF

1.23

⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠

−1 / 2

5 ⎞
⎛
For VR = 5 V, C j = (0.02) ⎜ 1 +
⎟
⎝ 0. 8 ⎠

−1 / 2

= 0.00743 pF
−1 / 2

⎛ 1. 5 ⎞
For VR = 1.5 V, C j = (0.02) ⎜1 +

= 0.0118 pF
⎟
⎝ 0. 8 ⎠
0.00743 + 0.0118
C j (avg ) =
= 0.00962 pF
2
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

where
τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 )
or

τ = 4.52 ×10−10 s
Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e − ti / τ

5
+ r /τ
⎛ 5 ⎞
= e 1 ⇒ t1 = τ ln ⎜ ⎟
1.5
⎝ 1.5 ⎠
−10
t1 = 5.44 × 10 s
For VR = 0 V, Cj = Cjo = 0.02 pF
−1/ 2
⎛ 3.5 ⎞
For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 +
= 0.00863 pF
⎟

⎝ 0.8 ⎠
0.02 + 0.00863
C j (avg ) =
= 0.0143 pF
2
τ = RC j ( avg ) = 6.72 ×10−10 s

(b)

vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

(

3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ

so that t2 = 8.09 × 10

−10

)

s

1.24
⎡ (1018 )(1015 ) ⎤
⎥ = 0.757 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦

a.
VR = 1 V
1 ⎞
⎛
C j = (0.25) ⎜ 1 +
⎟
⎝ 0.757 ⎠

−1/ 2

= 0.164 pF

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(a)


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f0 =

1
2π LC

=

1

( 2.2 ×10 )( 0.164 ×10 )

2π

−3

−12

f 0 = 8.38 MHz
b.

VR = 10 V
−1/ 2
10 ⎞
⎛
= 0.0663 pF
C j = (0.25) ⎜ 1 +
⎟
⎝ 0.757 ⎠
1
f0 =
−3
2π ( 2.2 × 10 )( 0.0663 × 10−12 )

f 0 = 13.2 MHz
1.25
⎞ ⎤
⎛ VD
⎟ − 1⎥ − 0.90 = exp ⎜
⎠ ⎦
⎝ VT

⎛V ⎞
exp ⎜ D ⎟ = 1 − 0.90 = 0.10
⎝ VT ⎠
VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V

b.
IF
IR

⎡
⎛ VF
⎢ exp ⎜
I
⎝ VT
= S ⋅⎣
IS ⎡
⎛ VR
⎢exp ⎜
⎝ VT
⎣
=

⎞ ⎤
⎛ 0.2 ⎞
⎟ − 1⎥ exp ⎜
⎟ −1
⎠ ⎦
⎝ 0.026 ⎠
=
⎛ −0.2 ⎞

⎞ ⎤
⎟ − 1⎥ exp ⎜⎝ 0.026 ⎟⎠ − 1
⎠ ⎦

2190
−1

IF
= 2190
IR

1.26
a.

⎛ 0.5 ⎞
I ≅ (10−11 ) exp ⎜
⎟ ⇒ I = 2.25 mA
⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−11 ) exp ⎜
⎟ ⇒ I = 0.105 A
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−11 ) exp ⎜
⎟ ⇒ I = 4.93 A
⎝ 0.026 ⎠
b.
⎛ 0.5 ⎞
I ≅ (10−13 ) exp ⎜
⎟ ⇒ I = 22.5 μ A

⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−13 ) exp ⎜
⎟ ⇒ I = 1.05 mA
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−13 ) exp ⎜
⎟ ⇒ I = 49.3 mA
⎝ 0.026 ⎠
1.27
(a)
150 × 10

(

)

I = I S eVD / VT − 1
−6

= 10

−11

(e

VD / VT

)

− 1 ≅ 10−11 eVD / VT

⎞
⎟ −1
⎠

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⎡
⎛V
I = I S ⎢ exp ⎜ D
⎝ VT
⎣

a.


www.elsolucionario.net

⎛ 150 × 10−6
Then VD = VT ln ⎜
−11
⎝ 10
Or VD = 0.430 V

⎞
⎛ 150 × 10−6 ⎞
⎟ = (0.026) ln ⎜
⎟
−11

⎠
⎝ 10
⎠

(b)
⎛ 150 × 10−6 ⎞
VD = VT ln ⎜
⎟
−13
⎝ 10
⎠
Or VD = 0.549 V

1.28

⎛ 0.7 ⎞
10−3 = I S exp ⎜
⎟
⎝ 0.026 ⎠
I S = 2.03 × 10 −15 A
(b)
VD
I D ( A ) ( n = 1)

I D ( A )( n = 2 )

0.1
0.2
0.3
0.4

0.5
0.6
0.7

9.50 ×10 −14
4.45 ×10 −12
2.08 ×10 −10
9.75 ×10 −9
4.56 ×10 −7
2.14 ×10 −5
10 −3

1.39 ×10 −14
9.50 ×10 −14
6.50 ×10 −13
4.45 ×10 −12
3.04 ×10 −11
2.08 ×10 −10
1.42 ×10 −9

ID(A)
4.68 ×10−11
2.19 ×10−9
1.03 ×10−7
4.80 ×10−6
2.25 ×10−4
1.05 ×10−2
4.93 ×10−1

log10ID

−10.3
−8.66
−6.99
−5.32
−3.65
−1.98
−0.307

ID(A)
4.68 ×10−13
2.19 ×10−11
1.03 ×10−9
4.80 ×10−8
2.25 ×10−6
1.05 ×10−4
4.93 ×10−3

log10ID
−12.3
−10.66
−8.99
−7.32
−5.65
−3.98
−2.31

1.29
(a)
I S = 10 −12 A
VD(v)

0.10
0.20
0.30
0.40
0.50
0.60
0.70
(b)
I S = 10 −14 A
VD(v)
0.10
0.20
0.30
0.40
0.50
0.60
0.70

1.30
a.
⎛ V − VD1 ⎞
ID2
= 10 = exp ⎜ D 2
⎟
I D1
⎝ VT
⎠
ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV

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(a)


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b.

ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV

1.31

(ii)

⎛I
VD = Vt ln ⎜ D
⎝ IS
VD = 0.669 V

⎞
⎛ 150 × 10−6 ⎞
⎟ = ( 0.026 ) ln ⎜
⎟
−15
⎝ 10
⎠
⎠

⎛ 25 × 10−6 ⎞
VD = ( 0.026)ln ⎜

⎟
−15
⎝ 10
⎠
VD = 0.622 V

(ii)

⎛ 0.2 ⎞
−12
I D = (10−15 )exp ⎜
⎟ = 2.19 × 10 A
⎝ 0.026 ⎠
ID = 0

(iii)

I D = −10 −15 A

(iv)

I D = −10 −15 A

(b) (i)

1.32
⎛I
VD = Vt ln ⎜ D
⎝ IS

⎞
⎛ 2 × 10−3 ⎞
=
= 0.6347 V
.
(0
026)
ln
⎟
⎜
−14 ⎟
⎝ 5 × 10 ⎠
⎠

⎛ 2 × 10−3 ⎞
= 0.5150 V
VD = (0.026) ln ⎜
−12 ⎟
⎝ 5 × 10 ⎠
0.5150 ≤ VD ≤ 0.6347 V

1.33
⎛V ⎞
I D = I S exp ⎜ D ⎟
⎝ Vt ⎠
⎛ 1.10 ⎞
−21
12 ×10−3 = I S exp ⎜
⎟ ⇒ I S = 5.07 × 10 A
⎝ 0.026 ⎠

(a)

(b)

⎛ 1.0 ⎞
I D = ( 5.07 × 10−21 ) exp ⎜
⎟
⎝ 0.026 ⎠
I D = 2.56 × 10−4 A = 0.256 mA

1.34
(a)
(b)
(c)

⎛ 1.0 ⎞
−7
I D = 10−23 exp ⎜
⎟ = 5.05 × 10 A
⎝ 0.026 ⎠
⎛ 1.1 ⎞
−5
I D = 10−23 exp ⎜
⎟ = 2.37 × 10 A
0
.
026
⎝
⎠

1
.
2
⎛
⎞
−3
I D = 10−23 exp ⎜
⎟ = 1.11× 10 A
⎝ 0.026 ⎠

1.35
IS doubles for every 5C increase in temperature.
I S = 10 −12 A at T = 300K
For I S = 0.5 × 10 −12 A ⇒ T = 295 K
For I S = 50 × 10 −12 A, (2) n = 50 ⇒ n = 5.64
Where n equals number of 5C increases.
Then ΔT = ( 5.64 )( 5 ) = 28.2 K
So 295 ≤ T ≤ 328.2 K

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(a) (i)


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1.36
I S (T )
= 2ΔT / 5 , ΔT = 155° C
I S (−55)

I S (100)
= 2155 / 5 = 2.147 × 109
I S (−55)
VT @100°C ⇒ 373°K ⇒ VT = 0.03220
VT @− 55°C ⇒ 216°K ⇒ VT = 0.01865

I D (100)
= (2.147 × 109 ) ×
I D (−55)

⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.0322 ⎠
⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.01865 ⎠

( 2.147 ×10 )(1.237 ×10 )
( 9.374 ×10 )
9

8

13

I D (100)
= 2.83 × 103
I D (−55)

1.37
3.5 = ID (105) + VD

⎛ V ⎞
⎛ ID ⎞
I D = 5 ×10−9 exp ⎜ D ⎟ ⇒ VD = 0.026 ln ⎜
−9 ⎟
⎝ 0.026 ⎠
⎝ 5 × 10 ⎠
Trial and error.
VD
ID
VD
−5
0.50
0.226
3 ×10
0.40
0.227
3.1×10−5
−
5
0.250
0.228
3.25 ×10
−5
0.229
0.2284
3.271×10
−

5
0.2285
0.2284
3.2715 ×10
(a)

So VD ≅ 0.2285 V
I D ≅ 3.272 × 10−5 A

(b)

I D = I S = 5 × 10−9 A
VR = ( 5 × 10−9 )(105 ) = 5 × 10−4 V
VD = 3.4995 V

1.38
⎛ I ⎞
10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D−12 ⎟
⎝ 10 ⎠
Trial and error.
VD(v)
ID(A)
VD(v)
0.50
0.5194
4.75 ×10−4
−4
0.517
0.5194
4.7415 ×10

0.5194
0.5194
4.740 ×10−4

VD = 0.5194 V
I D = 0.4740 mA
1.39

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=


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I s = 5 × 10 −13 A
R1 ϭ 50 K

ϩ
1.2 V
Ϫ

ϩ
VD
Ϫ

R2 ϭ 30 K
ID

RTH ϭ R1 ͉͉ R2 ϭ 18.75 K

ϩ
VD
Ϫ

ID

⎛ R2
VTH = ⎜
⎝ R1 + R2

⎞
⎛ 30 ⎞
⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V
⎝ 80 ⎠
⎠

⎛I ⎞
0.45 = I D RTH + VD , VD = VT ln ⎜ D ⎟
⎝ IS ⎠
By trial and error:
I D = 2.56 μ A, VD = 0.402 V

1.40
ϩ VDϪ ϩ VDϪ

ϩ
VI
Ϫ

I1

1K
IR

I2

ϩ
V0
Ϫ

I S = 2 × 10 −13 A

V0 = 0.60 V

⎛V ⎞
⎛ 0.60 ⎞
I 2 = I S exp ⎜ 0 ⎟ = ( 2 × 10−13 ) exp ⎜
⎟
⎝ 0.026 ⎠
⎝ VT ⎠
= 2.105 mA
0.6
= 0.60 mA
1K
I1 = I 2 + I R = 2.705 mA

IR =

⎛I

VD = VT ln ⎜ 1
⎝ IS
= 0.6065

⎞
⎛ 2.705 × 10−3 ⎞
(0.026)
ln
=
⎟
⎜
⎟
−13
⎝ 2 × 10
⎠
⎠

VI = 2VD + V0 ⇒ VI = 1.81 V
1.41
(a) Assume diode is conducting.
Then, VD = Vγ = 0.7 V
So that I R 2 =

0. 7
⇒ 23.3 μ A
30

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ϩ

VTH
Ϫ


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1.2 − 0.7
⇒ 50 μ A
10
Then I D = I R1 − I R 2 = 50 − 23.3
I R1 =

Or I D = 26.7 μ A
(b) Let R1 = 50 k Ω Diode is cutoff.
30
⋅ (1.2) = 0.45 V
30 + 50
Since VD < Vγ , I D = 0
VD =

1.42
ϩ5 V

2 k⍀
ID

VB ϭVAϪVr

VA

2 k⍀

2 k⍀

A&VA:

5 − VA
V
= ID + A
2
2
A& VA − Vr

(1)

(2)

5 − (VA − Vr )

+ ID =

(VA − Vr )

2
2
5 − (VA − Vr ) ⎡ 5 − VA VA ⎤ VA − Vr
+⎢
− ⎥=
So
3

2⎦
2
⎣ 2
Multiply by 6:
10 − 2 (VA − Vr ) + 15 − 6VA = 3 (VA − Vr )

25 + 2Vr + 3Vr = 11VA
Vr = 0.6 V
(a)

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

5 − VA VA
−
= 2.5 − VA ⇒ I D Neg. ⇒ I D = 0
2
2
Both (a), (b) I D = 0

From (1) I D =

VA = 2.5, VB =

2
⋅ 5 = 2 V ⇒ VD = 0.50 V
5

1.43
Minimum diode current for VPS (min)
I D (min) = 2 mA, VD = 0.7 V

I2 =

0.7
5 − 0.7 4.3
, I1 =
=
R2
R1
R1

We have I1 = I 2 + I D

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3 k⍀


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4.3 0.7
=
+2
R1
R2
Maximum diode current for VPS (max)
P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA

so (1)

I1 = I 2 + I D

or
9.3 0.7
=
+ 14.3
(2)
R1
R2

Using Eq. (1),

9.3 4.3
=
− 2 + 14.3 ⇒
R1
R1

R1 = 0.41 kΩ

Then R2 = 82.5Ω 82.5Ω

Vo = 0.7 V
5 − 0.7
I=
⇒ I = 0.215 mA
20
10 − 0.7
(b)
I=
⇒ I = 0.2325 mA
20 + 20

Vo = I (20 K) − 5 ⇒ Vo = −0.35 V
(c)
(d)

10 − 0.7
⇒ I = 0.372 mA
5 + 20
Vo = 0.7 + I (20) − 8 ⇒ Vo = +0.14 V
I =0
Vo = I (20) − 5 ⇒ Vo = −5 V
I=

1.45
(a)
VD
0.6
0.482
(b)
Vo
0.5
0.484
(c)

5 = I ( 2 × 109 ) + VD

→

ID
→
2.2 ×10−4

2.259 ×10−4

10 = I ( 4 × 10 4 ) + VD

→

I

→

I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 ×10 ⎠
VD
Vo = VD = 0.482 V
0.481
0.482
I = 0.226 mA
I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 ×10 ⎠
VD
VD = 0.483 V

2.375 ×10−4
2.379 ×10−4

I = 0.238 mA
Vo = −0.24 V

I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
×
2
10
⎝
⎠
VD
VD = 0.496 V
I
→
0.496
I = 0.380 mA
3.808 ×10−4
−4
0.496
Vo = −0.10 V
3.802 ×10

10 = I ( 2.5 × 10 4 ) + VD

Vo
0.480
0.496

→

(d)

I = − I S ⇒ I = 2 × 10−12 A
Vo ≅ −5 V

1.46
(a)

0.4834
0.4834

Diode forward biased VD = 0.7 V

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1.44
(a)


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5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V
P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω

(b)
1.47

0.65
= 0.65 mA = I D1
1
= 2(0.65) = 1.30 mA

I R 2 = I D1 =

(a)

ID2

ID2 =

VI − 2Vr − V0 5 − 3(0.65)
=
= 1.30 ⇒ R1 = 2.35 K
R1
R1

0.65
= 0.65 mA
1
8 − 3(0.65)
ID2 =
⇒ I D 2 = 3.025 mA
2
I D1 = I D 2 − I R 2 = 3.025 − 0.65

I D1 = 2.375 mA

1.48

τd =

a.

VT
(0.026)
=
= 0.026 kΩ = 26Ω
1
I DQ

id = 0.05 I DQ = 50 μ A peak-to-peak
vd = idτ d = (26)(50) μ A ⇒ vd = 1.30 mV peak-to-peak
For I DQ = 0.1 mA ⇒ τ d =

b.

(0.026)
= 260Ω
0. 1

id = 0.05 I DQ = 5 μ A peak-to-peak

vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak
1.49
RS

␯S

a.

ϩ
␯d
Ϫ

ϩ
Ϫ

diode resistance rd = VT /I

⎛
⎜ VT /I
⎛ rd ⎞
vd = ⎜
⎟ vS = ⎜ V
⎝ rd + RS ⎠
⎜⎜ T + RS
⎝ I
⎛ VT
⎞
vd = ⎜
⎟ vs = vo
⎝ VT + IRS ⎠

b.

RS = 260Ω

⎞
⎟
⎟ vS
⎟⎟
⎠

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IR2 =

(b)


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⎞
v0 ⎛ VT
v
0.026
=⎜
⇒ 0 = 0.0909
⎟=
vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26)
vS
v
v
0.026
I = 0.1 mA, 0 =

⇒ 0 = 0.50
vs 0.026 + ( 0.1)( 0.26 )
vS
I = 1 mA,

I = 0.01 mA.

v0
v
0.026
=
⇒ 0 = 0.909
vS 0.026 + (0.01)(0.26)
vS

1.50
⎛V
I ≅ I S exp ⎜ a
⎝ VT

⎛ I ⎞
⎞
⎟ , Va = VT ln ⎜ ⎟
⎠
⎝ IS ⎠

⎛ 100 × 10−6 ⎞
Schottky diode, Va = (0.026) ln ⎜
⎟
−9

⎝ 10
⎠
Va = 0.299 V

1.51
Schottky
pn junction
I

⎛V ⎞
Schottky: I ≅ I S exp ⎜ a ⎟
⎝ VT ⎠

⎛ I ⎞
⎛ 0.5 × 10−3 ⎞
Va = VT ln ⎜ ⎟ = (0.026) ln ⎜
−7 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
= 0.1796 V
Then
Va of pn junction = 0.1796 + 0.30

= 0.4796
I
0.5 × 10−3
=
⎛V ⎞
⎛ 0.4796 ⎞
exp ⎜ a ⎟ exp ⎜

⎟
⎝ 0.026 ⎠
⎝ VT ⎠
I S = 4.87 × 10 −12 A

IS =

1.52
(a)

ϩ VD Ϫ
I1

0.5 mA

I2

I1 + I 2 = 0.5 × 10 −3

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⎛ 100 × 10−6 ⎞
pn junction, Va = (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
Va = 0.599 V

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⎛V
5 × 10−8 exp ⎜ D
⎝ VT

⎞
⎛ VD ⎞
−12
−3
⎟ + 10 exp ⎜ ⎟ = 0.5 × 10
V
⎠
⎝ T ⎠

⎛V ⎞
5.0001× 10 −8 exp ⎜ D ⎟ = 0.5 × 10 −3
⎝ VT ⎠
⎛ 0.5 × 10−3 ⎞
⇒ VD = 0.2395
VD = (0.026) ln ⎜
−8 ⎟
⎝ 5.0001× 10 ⎠
Schottky diode, I 2 = 0.49999 mA

pn junction, I1 = 0.00001 mA
(b)
ϩ VD1 Ϫ
I

ϩ

ϩ VD2 Ϫ

Ϫ

0.90 V

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⎛V ⎞
⎛V ⎞
I = 10 −12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ D 2 ⎟
⎝ VT ⎠
⎝ VT ⎠
VD1 + VD 2 = 0.9

⎛V ⎞
⎛ 0.9 − VD1 ⎞
10−12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜
⎟
⎝ VT ⎠
⎝ VT
⎠
⎛ 0.9 ⎞
⎛ −VD1 ⎞
= 5 ×10−8 exp ⎜
⎟
⎟ exp ⎜
⎝ VT ⎠

⎝ VT ⎠
⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞
⎛ 0.9 ⎞
exp ⎜ D1 ⎟ = ⎜
exp ⎜
⎟
−12 ⎟
⎝ 0.026 ⎠
⎠
⎝ VT ⎠ ⎝ 10
⎛ 5 × 10−8 ⎞
+ 0.9 = 1.1813
2VD1 = VT ln ⎜
−12 ⎟
⎝ 10
⎠
VD1 = 0.5907 pn junction

VD 2 = 0.3093 Schottky diode
⎛ 0.5907 ⎞
I = 10−12 exp ⎜
⎟ ⇒ I = 7.35 mA
⎝ 0.026 ⎠
1.53
R ϭ 0.5 K
V0
ϩ
VPS ϭ 10 V
Ϫ

I

ϩ
VZ
Ϫ
IZ

VZ = VZ 0 = 5.6 V at I Z = 0.1 mA
rZ = 10Ω

I Z rZ = ( 0.1)(10 ) = 1 mV

VZ0 = 5.599
a.
RL → ∞ ⇒
IZ =

10 − 5.599
4.401
=
= 8.63 mA
R + rZ
0.50 + 0.01

VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 )
VZ = V0 = 5.685 V

RL
IL

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11 − 5.599
= 10.59 mA
0.51
VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V

b.

VPS = 11 V ⇒ I Z =

9 − 5.599
= 6.669 mA
0.51
VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V

VPS = 9 V ⇒ I Z =

ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V
I = IZ + IL
V0
V − V0
V − VZ 0
IL =
, I = PS
, IZ = 0
RL
R
rZ

10 − V0 V0 − 5.599 V0
=
+
0.50
0.010
2
10 5.599
1
1⎤
⎡ 1
+
= V0 ⎢
+
+ ⎥
0.50 0.010
⎣ 0.50 0.010 2 ⎦
20.0 + 559.9 = V0 (102.5)
V0 = 5.658 V

1.54
a.

9 − 6.8
⇒ I Z = 11 mA
0.2
PZ = (11)( 6.8 ) ⇒ PZ = 74.8 mW
IZ =

12 − 6.8

⇒ I Z = 26 mA
0.2
b.
26 − 11
%=
× 100 ⇒ 136%
11
PZ = ( 26 )( 6.8 ) = 176.8 mW
IZ =

%=

176.8 − 74.8
× 100 ⇒ 136%
74.8

1.55
I Z rZ = ( 0.1)( 20 ) = 2 mV
VZ 0 = 6.8 − 0.002 = 6.798 V
a.
RL = ∞
10 − 6.798
⇒ I Z = 6.158 mA
0.5 + 0.02
V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 )
IZ =

V0 = 6.921 V
b.

I = IZ + IL

10 − V0 V0 − 6.798 V0
=
+
0.50
0.020
1
10 6.798
1
1⎤
⎡ 1
+
= V0 ⎢
+
+ ⎥
0.30 0.020
0
.
50
0
.
020
1⎦
⎣
359.9 = V0 (53)
V0 = 6.791 V
ΔV0 = 6.791 − 6.921

ΔV0 = −0.13 V

1.56

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c.