1. Chapter 2
Problem Solutions
2.1
␯I ␯0
Rϭ1K
␯I
10
0
Ϫ10
Vγ = 0.6 V, rf = 20 Ω
⎛ R ⎞
⎜ R + r ⎟(
For vI = 10 V, v0 = ⎜ 10 − 0.6 )
⎟
⎝ f ⎠
⎛ 1 ⎞
=⎜ ⎟ ( 9.4 )
⎝ 1 + 0.02 ⎠
v0 = 9.22
10
0.6
␯0
9.22
2.2
ϩ ␯D Ϫ
␯I ␯0
D
R
v0 = vI − vD
⎛i ⎞ v
vD = VT ln ⎜ D ⎟ and iD = 0
⎝ IS ⎠ R
⎛ v ⎞
v0 = vI − VT ln ⎜ 0 ⎟
⎝ IS R ⎠
2.3
80sin ω t
(a) vs = = 13.33sin ω t
6
13.33
Peak diode current id (max) =
R

2. (b) PIV = vs (max) = 13.3 V
(c)
To π
1 1
vo ( avg ) = ∫ v (t )dt = 2π ∫ 13.33sin × dt
o
To o o
13.33 13.33 13.33
[ − cos x ]o =
π
= ⎡ − ( −1 − 1) ⎤ =
⎣ ⎦
2π 2π π
vo ( avg ) = 4.24 V
(d) 50%
2.4
v0 = vS − 2Vγ ⇒ vS ( max ) = v0 ( max ) + 2Vγ
a. For v0 ( max ) = 25 V ⇒ vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V
N1 160 N
= ⇒ 1 = 6.06
N 2 26.4 N2
b. For v0 ( max ) = 100 V ⇒ vS ( max ) = 101.4 V
N1 160 N
= ⇒ 1 = 1.58
N 2 101.4 N2
From part (a) PIV = 2vS ( max ) − Vγ = 2 ( 26.4 ) − 0.7
or PIV = 52.1 V or, from part (b) PIV = 2 (101.4 ) − 0.7 or PIV = 202.1 V
2.5
(a)
vs (max) = 12 + 2(0.7) = 13.4 V
13.4
vs ( rms ) = ⇒ vs (rms) = 9.48 V
2
(b)
VM VM
Vr = ⇒C =
2 f RC 2 f Vr R
12
C= ⇒ C = 2222 μ F
2 ( 60 )( 0.3)(150 )
(c)
VM ⎡ 2VM ⎤
id , peak = ⎢1 + π ⎥
R ⎢⎣ Vr ⎥⎦
12 ⎡ 2 (12 ) ⎤
= ⎢1 + π ⎥
150 ⎢ 0.3 ⎥
⎣ ⎦
id , peak = 2.33 A
2.6
(a)
vS ( max ) = 12 + 0.7 = 12.7 V
vS ( max )
vS ( rms ) = ⇒ vS ( rms ) = 8.98 V
2
VM V 12
(b) Vr = ⇒C = M = or C = 4444 μ F
fRC fRVr ( 60 )(150 )( 0.3)

3. VM ⎛ VM ⎞ 12 ⎛ 12 ⎞
(c) For the half-wave rectifier iD , max = ⎜ 1 + 4π
⎜ ⎟= ⎜ 1 + 4π
⎟ 150 ⎜ ⎟ or
R ⎝ 2Vr ⎠ ⎝ 2 ( 0.3) ⎟
⎠
iD , max = 4.58 A
2.8
(a)
vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V
16.4
vs ( rms ) = = 11.6 V
2
VM 15
(b) C= = = 2857 μ F
2 f RVr 2 ( 60 )(125 )( 0.35 )
2.9
␯S
26
0.6
Ϫ0.6
Ϫ26
25.4
ϩ
␯O
0
0
Ϫ
␯O
Ϫ25.4
2.10
R
iD
ϩ
␯S ϭ ϩ VB ϭ 12 V
Ϫ
Ϫ
␯S
iD
␻t
x1 x2
vS ( t ) = 24sin ω t
T
1
Now iD ( avg ) = iD ( t ) dt
T∫0
24sin x − 12.7
We have for x1 ≤ ω t ≤ x2 iD =
R
To find x1 and x2, 24sin x1 = 12.7
x1 = 0.558 rad
x2 = π − 0.558 = 2.584 rad
Then

4. x2
1 ⎡ 24sin x − 12.7 ⎤
iD ( avg ) = 2 = ∫⎢ ⎥dx
2π ⎣
x1 R ⎦
1 ⎛ 24 ⎞ 1 ⎛ 12.7 ⎞ x2 6.482 4.095
⎜ ⎟ ( − cos x ) x1 − ⇒ R = 1.19Ω
x2
= ⎜ ⎟ ⋅ xx or 2 = −
2π ⎝ R ⎠ 2π ⎝ R ⎠ 1 R R
x −x 2.584 − 0.558
Fraction of time diode is conducting = 2 1 × 100% = × 100% or Fraction = 32.2%
2π 2π
Power rating
x 2
R 2 ⎡ 24sin x − 12.7 ⎤
T
R 2
∫ iD dt = 2π x∫ ⎢
2
Pavg = R ⋅ irms = ⎥ dx
T 0 1
⎣ R ⎦
x2
1 ⎡( 24 ) sin 2 x − 2 (12.7 )( 24 ) sin x + (12.7 ) ⎤dx
2π R ∫ ⎣
2 2
=
⎦
x1
x
( 24 ) ⎡ −
1 ⎡ 2 x sin 2 x ⎤ 2 x2
⎥ − 2 (12.7 )( 24 )( − cos x ) x1 + (12.7 ) xx1 ⎦
2 x2
= ⎢2 ⎤
2π R ⎣ ⎣ 4 ⎦x 1
For R = 1.19 Ω, then Pavg = 17.9 W
2.11
(a)
15
R= = 150Ω
0.1
vS ( max ) = vo ( max ) + Vγ = 15 + 0.7 or vS ( max ) = 15.7 V
15.7
Then vS ( rms ) = = 11.1 V
2
N1 120 N
Now = ⇒ 1 = 10.8
N 2 11.1 N2
VM VM 15
(b) Vr = ⇒C = = or C = 2083 μ F
2 fRC 2 fRVr 2 ( 60 )(150 )( 0.4 )
(c) PIV = 2vS ( max ) − Vγ = 2 (15.7 ) − 0.7 or PIV = 30.7 V
2.12
␯0
D2
␯i ϩ
Ϫ R2 RL
R1
␯0
␯i ϩ
Ϫ RL
R2
R1
For vi > 0
Vγ = 0
Voltage across RL + R1 = vi
⎛ RL ⎞ 1
Voltage Divider ⇒ v0 = ⎜ ⎟ vi = vi
⎝ RL + R1 ⎠ 2

5. ␯0
20
2.13
For vi > 0, (Vγ = 0 )
R1
␯i ϩ
Ϫ ␯0
R2 RL
a.
⎛ R2 || RL ⎞
v0 = ⎜ ⎟ vi
⎝ R2 || RL + R1 ⎠
R2 || RL = 2.2 || 6.8 = 1.66 kΩ
⎛ 1.66 ⎞
v0 = ⎜ ⎟ vi = 0.43 vi
⎝ 1.66 + 2.2 ⎠
␯0
4.3
v0 ( max )
b. v0 ( rms ) = ⇒ v0 ( rms ) = 3.04 V
2
2.14
3.9
IL = ⇒ I 2 = 0.975 mA
4
20 − 3.9
IR = = 1.3417 mA
12
I Z = 1.3417 − 0.975 ⇒ I Z = 0.367 mA
PT = I Z ⋅ VZ = ( 0.367 )( 3.9 ) ⇒ PT = 1.43 mW
2.15
(a)
40 − 12
IZ = = 0.233 A
120
P = ( 0.233)(12 ) = 2.8 W
(b) IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A
12
So 0.21 = ⇒ RL = 57.1Ω
RL
(c) P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W
2.16
Ri
VI V0
II ϩ
VZ RL
Ϫ
IZ IL
VI = 6.3 V, Ri = 12Ω, VZ = 4.8

6. a.
6.3 − 4.8
II = ⇒ 125 mA
12
I L = I I − I Z = 125 − I Z
25 ≤ I L ≤ 120 mA ⇒ 40 ≤ RL ≤ 192Ω
b.
PZ = I Z VZ = (100 )( 4.8 ) ⇒ PZ = 480 mW
PL = I LV0 = (120 )( 4.8 ) ⇒ PL = 576 mW
2.17
a.
20 − 10
II = ⇒ I I = 45.0 mA
222
10
IL = ⇒ I L = 26.3 mA
380
I Z = I I − I L ⇒ I Z = 18.7 mA
b.
400
PZ ( max ) = 400 mW ⇒ I Z ( max ) = = 40 mA
10
⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40
10
⇒ I L ( min ) = 5 mA =
RL
⇒ RL = 2 kΩ
(c) For Ri = 175Ω I I = 57.1 mA I L = 26.3 mA I Z = 30.8 mA
I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA
10
RL = ⇒ RL = 585Ω
17.1
2.18
a. From Eq. (2-31)
500 [ 20 − 10] − 50 [15 − 10]
I Z ( max ) =
15 − ( 0.9 )(10 ) − ( 0.1)( 20 )
5000 − 250
=
4
I Z ( max ) = 1.1875 A
I Z ( min ) = 0.11875 A
20 − 10
From Eq. (2-29(b)) Ri = ⇒ Ri = 8.08Ω
1187.5 + 50
b.
PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W
PL = I L ( max ) V0 = ( 0.5 )(10 ) ⇒ PL = 5 W
2.19
(a) As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18.
V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ
= 10 + (1.1875)(2) = 12.375 V
V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ
= 10 + (0.11875)(2) = 10.2375 V

7. 12.375 − 10.2375
b. % Reg = × 100% ⇒ % Reg = 21.4%
10
2.20
VL ( max ) − VL ( min )
% Reg = × 100%
VL ( nom )
VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz )
=
VL ( nom )
⎡ I Z ( max ) − I Z ( min ) ⎤ ( 3)
=⎣ ⎦ = 0.05
6
So I Z ( max ) − I Z ( min ) = 0.1 A
6 6
Now I L ( max ) = = 0.012 A, I L ( min ) = = 0.006 A
500 1000
VPS ( min ) − VZ
Now Ri =
I Z ( min ) + I L ( max )
15 − 6
or 280 = ⇒ I Z ( min ) = 0.020 A
I Z ( min ) + 0.012
VPS ( max ) − VZ
Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri =
I Z ( max ) + I L ( min )
VPS ( max ) − 6
or 280 = ⇒ VPS ( max ) = 41.3 V
0.12 + 0.006
2.21
Using Figure 2.21
a. VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V
For VPS ( min ) :
I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA
VPS ( min ) − VZ 15 − 10
Ri = = ⇒ Ri = 200Ω
II 25
25 − 10
b. For VPS ( max ) ⇒ I I ( max ) = ⇒ I I ( max ) = 75 mA
Ri
For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA
VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V
V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25
V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90
ΔV0 = 0.35 V
ΔV0
c. % Reg = × 100% ⇒ % Reg = 3.5%
V0 ( nom )
2.22
From Equation (2.29(a))
VPS ( min ) − VZ 24 − 16
Ri = = or Ri = 18.2Ω
I Z ( min ) + I L ( max ) 40 + 400
VM VM
Also Vr = ⇒C =
2 fRC 2 fRVr
R ≅ Ri + rz = 18.2 + 2 = 20.2Ω
Then

8. 24
C= ⇒ C = 9901 μ F
2 ( 60 )(1)( 20.2 )
2.23
VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V
8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V
VS ( max ) − VZ ( nom ) 12 − 8
Ii = = = 1.333 A
Ri 3
For I L = 0.2 A ⇒ I Z = 1.133 A
For I L = 1 A ⇒ I Z = 0.333 A
VL ( max ) = VZ 0 + I Z ( max ) rZ
= 7.95 + (1.133)( 0.5 ) = 8.5165
VL ( min ) = VZ 0 + I Z ( min ) rZ
= 7.95 + ( 0.333)( 0.5 ) = 8.1165
ΔVL = 0.4 V
ΔVL 0.4
% Reg = = ⇒ % Reg = 5.0%
V0 ( nom ) 8
VM VM
Vr = ⇒C =
2 fRC 2 fRVr
R = Ri + rz = 3 + 0.5 = 3.5Ω
12
Then C = ⇒ C = 0.0357 F
2 ( 60 )( 3.5 )( 0.8 )
2.24
(a) For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0
For 0 ≤ vI ≤ 3, Zener not in breakdown, so i1 = 0, vO = 0
vI − 3
For vI > 3 i1 = mA
20
⎛ v −3⎞ 1
vo = ⎜ I ⎟ (10 ) = vI − 1.5
⎝ 20 ⎠ 2
At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA
␯O(V)
4
3.5
Ϫ10 3.0 10 ␯I(V)
(b) For vI < 0, both diodes forward biased
0 − vI
−i1 = . At vI = −10 V , i1 = −1 mA
10
v −3
For vI > 3, i1 = I . At vI = 10 V , i1 = 0.35 mA
20

9. i1(mA)
0.35
Ϫ10 3 10 ␯I(V)
Ϫ1
2.25
(a)
1K
␯I ␯0
1K 2K
ϩ15
V1
1
V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI
3
For vI > 5.7 V
vI − (V1 + 0.7 ) 15 − V1 V1
+ = , v0 = V1 + 0.7
1 2 1
vI − v0 15 − ( v0 − 0.7 ) v0 − 0.7
+ =
1 2 1
vI 15.7 0.7 ⎛ 1 1 1⎞
+ + = v0 ⎜ + + ⎟ = v0 ( 2.5 )
1 2 1 ⎝ 1 2 1⎠
1
vI + 8.55 = v0 ( 2.5 ) ⇒ v0 = vI + 3.42
2.5
vI = 5.7 ⇒ v0 = 5.7
vI = 15 ⇒ v0 = 9.42
␯0(V)
9.42
5.7
0 5.7 15 ␯I (V)
(b) iD = 0 for 0 ≤ vI ≤ 5.7
Then for vI > 5.7 V
⎛ v ⎞
vI − ⎜ I + 3.42 ⎟
vI − vO ⎝ 2.5 ⎠ or i = 0.6vI − 3.42 For v = 15, i = 5.58 mA
iD = = D I D
1 1 1

10. iD(mA)
5.58
5.7 15 ␯I (V)
2.26
⎛ 20 ⎞
(a) For D off, vo = ⎜ ⎟ (20) − 10 = 3.33 V
⎝ 30 ⎠
Then for vI ≤ 3.33 + 0.7 = 4.03 V ⇒ vo = 3.33 V
For vI > 4.03, vo = vl − 0.7;
For vI = 10, vo = 9.3
␯O(V)
9.3
3.33
0 4.03 10 ␯I (V)
(b) For vI ≤ 4.03 V , iD = 0
10 − vo vo − ( −10 )
For vI > 4.03, iD + =
10 20
3
Which yields iD = vI − 0.605
20
For vI = 10, iD = 0.895 mA
iD(mA)
0.895
0 4.03 10 ␯I(V)
2.27
␯O
12.5
10.7
Ϫ30
10.7 30 ␯I
Ϫ30
30 − 10.7
For vI = 30 V, i = = 0.175 A
100 + 10
v0 = i(10) + 10.7 = 12.5 V

11. b.
␯O
12.5
10.7
0
Ϫ30
2.28
ϩ 5Ϫ
␯I ␯O
R ϭ 6.8 K
Vγ = 0.6 V
vI = 15sin ω t
␯O
␯O
Ϫ4.4
Ϫ19.4
2.29
a.
Vγ = 0
0
Ϫ3 V
Vγ = 0.6
0
Ϫ2.4
b.
Vγ = 0
20
5
Vγ = 0.6
19.4
5
2.30

12. 10
6.7
0
Ϫ4.7
Ϫ10
2.31
One possible example is shown.
Ri D L
Ii
ϩ ϩ ϩ
Ϫ VZ ϭ 14 V Ϫ RADIO
DZ Ϫ
Vign VRADIO
L will tend to block the transient signals
Dz will limit the voltage to +14 V and −0.7 V.
Power ratings depends on number of pulses per second and duration of pulse.
2.32
␯O(V)
40
(a) 0
␯O(V)
35
0
(b) Ϫ5
2.33
C
␯I ␯O
ϩ
Vx
Ϫ
a. For Vγ = 0 ⇒ Vx = 2.7 V
b. For Vγ = 0.7 V ⇒ Vx = 2.0 V
2.34
C
ϩ ϩ
␯I Ϫ ␯O
10 V
ϩ
Ϫ Ϫ
2.35

13. 20
␯O
10
0 VB ϭ 0
␯I
Ϫ10
20 ␯O
13
10
3
0 VB ϭ ϩ3 V
Ϫ7 ␯I ϩ VB
20
␯O
10
7
0 VB ϭ Ϫ3 V
Ϫ3
␯ I ϩ VB
Ϫ13
2.36
For Figure P2.32(a)
10
␯I
0
Ϫ10 ␯O
Ϫ20
2.37
a.
10 − 0.6
I D1 = ⇒ I D1 = 0.94 mA ID2 = 0
9.5 + 0.5
V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V
b.
5 − 0.6
I D1 = ⇒ I D1 = 0.44 mA ID2 = 0
9.5 + 0.5
V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V
c. Same as (a)
d.
(I )
10 = ( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA
2
V0 = I ( 9.5 ) ⇒ V0 = 9.16 V
I
I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.482 mA
2
2.38
a. I = I D1 = I D 2 = 0 V0 = 10
b.

14. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) ⇒ I = I D 2 = 0.94 mA I D1 = 0
V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V
c.
10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 ⇒ I = I D 2 = 0.44 mA I D1 = 0
V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V
d.
I
10 = I ( 9.5 ) + 0.6 + ( 0.5) ⇒ I = 0.964 mA
2
I
I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.482 mA
2
V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V
2.39
a.
V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V
10 − 4.4
I= ⇒ I = 0.589 mA
9.5
4.4 − 0.6
I D1 = ID2 = ⇒ I D1 = I D 2 = 7.6 mA
0.5
I D3 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA
b.
V1 = V2 = 5 V D1 and D2 on, D3 off
I
10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA
2
I
I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA
2
I D3 = 0
V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V
c. V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V
10 − 4.4
I= ⇒ I = 0.589 mA
9.5
4.4 − 0.6
I D2 = ⇒ I D 2 = 7.6 mA
0.5
I D1 = 0
I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA
d. V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V
10 − 4.4
I= ⇒ I = 0.589 mA
9.5
4.4 − 0.6 − 2
I D2 = ⇒ I D 2 = 3.6 mA
0.5
I D1 = 0
I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA
2.40
(a) D1 on, D2 off, D3 on
So I D 2 = 0
10 − 0.6 − ( −0.6 ) 10
Now V2 = −0.6V , I D1 = = ⇒ I D1 = 1.25 mA
R1 + R2 2+6

15. V1 = 10 − 0.6 − (1.25 )( 2 ) ⇒ V1 = 6.9 V
−0.6 − ( −5 )
I R3 = = 2.2 mA
2
I D3 = I R 3 − I D1 = 2.2 − 1.25 ⇒ I D 3 = 0.95 mA
(b) D1 on, D2 on, D3 off
So I D 3 = 0
10 − 0.6 − 4.4 5
V1 = 4.4 V , I D1 = =
R1 6
or I D1 = 0.833 mA
4.4 − ( −5 ) 9.4
I R2 = = = 0.94 mA
R2 + R3 10
I D 2 = I R 2 − I D1 = 0.94 − 0.833 ⇒ I D 2 = 0.107 mA
V2 = I R 2 R3 − 5 = ( 0.94 )( 5 ) − 5 ⇒ V2 = −0.3 V
(c) All diodes are on V1 = 4.4V , V2 = −0.6 V
10 − 0.6 − 4.4
I D1 = 0.5 mA = ⇒ R1 = 10 k Ω
R1
4.4 − ( −0.6 )
I R 2 = 0.5 + 0.5 = 1 mA = ⇒ R2 = 5 k Ω
R2
−0.6 − ( −5 )
I R 3 = 1.5 mA = ⇒ R3 = 2.93 k Ω
R3
2.41
⎛ 0.5 ⎞
For vI small, both diodes off vO = ⎜ ⎟ vI = 0.0909vI
⎝ 0.5 + 5 ⎠
When vI − vO = 0.6, D1 turns on. So we have vI − 0.0909vI = 0.6 ⇒ vI = 0.66, vO = 0.06
vI − 0.6 − vO vI − vO vO 2v − 0.6
For D1 on + = which yields vO = I
5 5 0.5 12
2vI − 0.6
When vO = 0.6, D2 turns on. Then 0.6 = ⇒ vI = 3.9 V
12
v − 0.6 − vO vI − vO vO vO − 0.6
Now for vI > 3.9 I + = +
5 5 0.5 0.5
2vI + 5.4
Which yields vO = ; For vI = 10 ⇒ vO = 1.15 V
22
2.42
ϩ10 V
10 K
D1 D2
␯I ␯0
D3 D4 10 K
10 K
Ϫ10 V

16. For vI > 0. when D1 and D4 turn off
10 − 0.7
I= = 0.465 mA
20
v0 = I (10 kΩ ) = 4.65 V
␯0
4.65
Ϫ10 Ϫ4.65
4.65 10 ␯I
Ϫ4.65
v0 = vI for − 4.65 ≤ vI ≤ 4.65
2.43
a.
R1 D2
ϩ10 V V0
D1
R2
ID1
Ϫ10 V
R1 = 5 kΩ, R2 = 10 kΩ
D1 and D2 on ⇒ V0 = 0
10 − 0.7 0 − ( −10 )
I D1 = − = 1.86 − 1.0
5 10
I D1 = 0.86 mA
b.
R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0
10 − 0.7 − ( −10 )
I= = 1.287
15
V0 = IR2 − 10 ⇒ V0 = −3.57 V
2.44
If both diodes on
(a)
VA = −0.7 V, VO = −1.4 V
10 − ( −0.7 )
I R1 = = 1.07 mA
10
−1.4 − ( −15 )
IR2 = = 2.72 mA
5
I R1 + I D1 = I R 2 ⇒ I D1 = 2.72 − 1.07
I D1 = 1.65 mA
(b) D1 off, D2 on
10 − 0.7 − ( −15 )
I R1 = I R 2 = = 1.62 mA
5 + 10
VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V
VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off ,
I D1 = 0
2.45

17. (a) D1 on, D2 off
10 − 0.7
I D1 = = 0.93 mA
10
VO = −15 V
(b) D1 on, D2 off
10 − 0.7
I D1 = = 1.86 mA
5
VO = −15 V
2.46
15 − (V0 + 0.7 )
V0 + 0.7 V0
= +
10 20 20
15 0.7 0.7 ⎛ 1 1 1 ⎞ ⎛ 4.0 ⎞
− − = V0 ⎜ + + ⎟ = V0 ⎜ ⎟
10 10 20 ⎝ 10 20 20 ⎠ ⎝ 20 ⎠
V0 = 6.975 V
V0
ID = ⇒ I D = 0.349 mA
20
2.47
10 K Ϫ VD ϩ 10 K
Va Vb
V1 V2
ID
10 K 10 K
a. V1 = 15 V, V2 = 10 V Diode off
Va = 7.5 V, Vb = 5 V ⇒ VD = −2.5 V
ID = 0
b. V1 = 10 V, V2 = 15 V Diode on
V2 − Vb Vb Va Va − V1
= + + ⇒ Va = Vb − 0.6
10 10 10 10
15 10 ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞
+ = Vb ⎜ + ⎟ + Vb ⎜ + ⎟ − 0.6 ⎜ + ⎟
10 10 ⎝ 10 10 ⎠ ⎝ 10 10 ⎠ ⎝ 10 10 ⎠
⎛ 4⎞
2.62 = Vb ⎜ ⎟ ⇒ Vb = 6.55 V
⎝ 10 ⎠
15 − 6.55 6.55
ID = − ⇒ I D = 0.19 mA
10 10
VD = 0.6 V
2.48 vI = 0, D1 off, D2 on
10 − 2.5
I= = 0.5 mA
15
vo = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V
For vI > 7.5 V , Both D1 and D2 on
vI − vo vo − 2.5 vo − 10
= + or vI = vo ( 5.5 ) − 33.75
15 10 5
When vo = 10 V, D2 turns off
vI = (10 )( 5.5 ) − 33.75 = 21.25 V
For vI > 21.25 V, vo = 10 V

18. 2.49
a. V01 = V02 = 0
b. V01 = 4.4 V, V02 = 3.8 V
c. V01 = 4.4 V, V02 = 3.8 V
Logic “1” level degrades as it goes through additional logic gates.
2.50
a. V01 = V02 = 5 V
b. V01 = 0.6 V, V02 = 1.2 V
c. V01 = 0.6 V, V02 = 1.2 V
Logic “0” signal degrades as it goes through additional logic gates.
2.51
(V1 AND V2 ) OR (V3 AND V4 )
2.52
10 − 1.5 − 0.2
I= = 12 mA = 0.012
R + 10
8.3
R + 10 = = 691.7Ω
0.012
R = 681.7Ω
2.53
10 − 1.7 − VI
I= =8
0.75
VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V
2.54
ϩ VR Ϫ
ϩ
VPS Rϭ 2 K
Ϫ
VR = 1 V, I = 0.8 mA
VPS = 1 + ( 0.8 )( 2 )
VPS = 2.6 V
2.55
I Ph = η eΦA
0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A
A = 3.75 × 10−2 cm 2