1. Chapter 2
Exercise Problems
EX2.1
30 − 12 − 0.6
iD ( peak ) = = 87 mA
0.2
vR ( max ) = 30 + 12 = 42 V
⎛ 12.6 ⎞
ω t1 = sin −1 ⎜ ⎟ ⇒ 24.8°
⎝ 30 ⎠
By symmetry
ω t2 = 180 − 24.8 = 155.2°
155.2 − 24.8
% time = × 100% = 36.2%
360
EX2.2
(a) vO = 12sin θ1 − 1.4 = 0
1.4
or sin θ1 = = 0.1166
12
which yields
θ1 = 6.7°
By symmetry, θ 2 = 180 − 6.7 = 173.3°
173.3 − 6.7
Then % time = × 100% = 46.3%
360
1.4
(b) sinθ1 = = 0.35
4
which yields
θ1 = 20.5°
By symmetry, θ 2 = 180 − 20.5 = 159.5°
159.5 − 20.5
Then % time = × 100% = 38.6%
360
EX2.3
VM 24
C= = ⇒ or C = 500 μ F
2 fRVr 2 ( 60 ) (103 ) ( 0.4 )
EX2.4
VM VM 75
Vr = ⇒R= or R =
f RC f CVr ( 60 ) ( 50 × 10−6 ) ( 4 )
Then R = 6.25 k Ω
EX2.5
10 ≤ VPS ≤ 14 V , VZ = 5.6 V , 20 ≤ RL ≤ 100 Ω
5.6
I L ( max ) = = 0.28 A,
20
5.6
I L ( min ) = = 0.056 A
100
⎡VPS ( max ) − VZ ⎤ ⋅ I L ( max ) ⎡VPS ( min ) − VZ ⎤ ⋅ I L ( min )
I Z (max) = ⎣ ⎦ − ⎣ ⎦
VPS ( min ) − 0.9VZ − 0.1VPS ( max ) VPS ( min ) − 0.9VZ − 0.1VPS ( max )
(14 − 5.6 )( 280 ) − (10 − 5.6 )( 56 )
or I L ( max ) =
10 − ( 0.9 )( 5.6 ) − ( 0.1)(14 )

2. or I L ( max ) = 591.5 mA
Power(min) = I Z ( max ) ⋅ VZ = ( 0.5915)( 5.6 )
So Power(min) = 3.31 W
VPS ( max ) − VZ 14 − 5.6
Now Ri = = or Ri ≅ 13 Ω
I Z ( max ) + I L ( min ) 0.5915 + 0.056
EX2.6
13.6 − 9
For vPS = 13.6 V , I Z = = 0.2383 A
15.3 + 4
vL ,max = 9 + ( 4 )( 0.2383) = 9.9532 V
11 − 9
For vPS = 11 V , I Z = = 0.1036 A
15.3 + 4
vL ,min = 9 + ( 4 )( 0.1036 ) = 9.4144 V
ΔvL 9.9532 − 9.4144
Source Reg = × 100% = × 100%
ΔvPS 13.6 − 11
or Source Reg = 20.7%
13.6 − 9
For I L = 0, I Z = = 0.2383 A
15.3 + 4
vL , noload = 9 + ( 4 )( 0.2383) = 9.9532 V
13.6 − ⎡9 + I Z ( 4 ) ⎤
⎣ ⎦
For I L = 100 mA, I Z = − 0.10
15.3
which yields
I Z = 0.1591 A
vL , full load = 9 + ( 4 )( 0.1591A ) = 9.6363 V
vL , noload − vL , full load
Load Reg = × 100%
vL , full load
9.9532 − 9.6363
= × 100%
9.6363
or Load Reg = 3.29%
EX2.7
R1
VO
␯I ϩ
Ϫ D1 D2
ϩ Ϫ
V1 V2
Ϫ ϩ
For vI < 5 V , D2 on ⇒ VO = −5 V
Then, V2 = 4.3 V.
D1 turns on when v1 = 2.5 V,
Then, V1 = 1.8 V.

3. ΔvO 1 R2 1
For vI > 2.5 V , = ⇒ =
ΔvI 3 R1 + R2 3
So that R1 = 2R2
EX2.8
For Vγ = 0, vO ( max ) = −2 V
Now, ΔvO = 8 V , so that vO ( min ) = −10 V
EX2.9
10 − 4.4
vO = 4.4 V , I = = 0.5895 mA
9.5
Set I = ID1, then vI = 4.4 − 0.6 − ( 0.5895 )( 0.5 ) = 3.505 V
Summary: For 0 ≤ vI ≤ 3.5 V , vO = 4.4 V
For vI > 3.5 V , D2 turns on and when vI ≥ 9.4 V , vO = 10 V
␯O(V)
10
4.4
0
3.505 9.4 ␯I (V)
EX2.10
VO = −0.6 V , I D1 = 0
−0.6 − ( −10 )
ID2 = I = ⇒ I D 2 = I = 4.27 mA
2.2
EX2.11
At the VA node:
15 − VA V
= ID2 + A (1)
5 15
At the VB node:
15 − (VB + 0.7 ) V
+ I D 2 = B (2)
5 10
We see that
VB = VA − 0.7, so Equation (2) becomes
15 − VA V − 0.7
+ ID2 = A (2’)
5 10
Solving for ID2 from Equation (1) and substituting into Equation (2’), we find
⎛ 15 − VA ⎞ VA VA − 0.7
2⎜ ⎟− =
⎝ 5 ⎠ 15 10
Then VA = 10.71 V and VB = 10.01 V
15 − 10.71
Solving for the diode currents, we obtain I D1 = ⇒ I D1 = 0.858 mA
5

4. 15 − 10.71 10.71
Also I D 2 = − ⇒ I D 2 = 0.144 mA
5 15
EX2.12
Assuming all diodes are conducting, we have VB = −0.7 V and VA = 0
Summing currents, we have
5 − VA
= I D1 + I D 2 (1)
5
V − ( −5 )
I D 2 + I D3 = B (2)
5
V − ( −10 ) 10 − 0.7
I D1 = B = (3)
10 10
From ( 3) , we find
I D1 = 0.93 mA
From (1) , we obtain
I D 2 = 0.07 mA
From ( 2 ) , we have
I D 3 = 0.79 mA
EX2.13
(a) I ph = η eΦ
⎡ 6.4 × 10 −2 ⎤
so I = ( 0.8 ) (1.6 × 10−19 ) ⎢ ⎥ ( 0.5 )
⎢ ( 2 ) (1.6 × 10−19 ) ⎥
⎣ ⎦
or I ph = 12.8 mA
(b) We have vO = (12.8)(1) = 12.8 V .
The diode must be reverse biased so that VPS > 12.8 V
EX2.14
The equivalent circuit is
ϩ5 V
ϩ
Vr ϭ 1.7 V
Ϫ
rf ϭ 15 ⍀
I
R
0.2 V
5 − 1.7 − 0.2
So I = = 15 mA
rf + R
15 − 1.7 − 0.2 3.1
Or rf + R = = = 0.207 kΩ
15 15
Then R = 207 − 15 ⇒ R = 192 Ω
2.15
40 − 12
(a) IZ = = 0.233 A
120
P = ( 0.233)(12 ) = 2.8 W

5. (b) I R = 0.233 A, I L = ( 0.9 )( 0.233) = 0.21 A
12
So 0.21 = ⇒ RL = 57.1 Ω
RL
(c) P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W
Test Your Understanding Exercises
TYU2.1
vI = 120sin ( 2π 60t ) , Vγ = 0.7 V , and R = 2.5 kΩ
Full-wave rectifier: Turns ratio 1:2 so that
vS = v I
VM = 120 − 0.7 = 119.3 V
Vr = 119.3 − 100 = 19.3 V
VM 119.3
So C = = or C = 20.6 μ F
2 f RVr 2 ( 60 ) ( 2.5 x103 ) (19.3)
TYU2.2
vI = 50sin ( 2π 60t ) , Vγ = 0.7 V , and R = 10 kΩ. Full-wave rectifier
VM ( 50 − 1.4 )
C= =
2 f RVr 2 ( 60 ) (10 × 103 ) ( 2 )
or C = 20.3 μ F
TYU2.3
Using Equation (2.10)
2Vr 2 ( 4)
(a) ω Δt = = = 0.327
VM 75
⎛ 0.327 ⎞
Percent time = ⎜ ⎟ × 100% = 5.2%
⎝ 2π ⎠
2Vr 2 (19.3)
(b) ω Δt = = = 0.569
VM 119.3
⎛ 0.569 ⎞
Percent time = ⎜ ⎟ × 100% = 18.1%
⎝ π ⎠
2Vr 2 ( 2)
(c) ω Δt = = = 0.287
VM 48.6
⎛ 0.287 ⎞
Percent time = ⎜ ⎟ × 100% = 9.14%
⎝ π ⎠
TYU2.4
V − VZ
I Z = PS − IL
Ri
11 − 9
For VPS (min) and IL (max), then I Z ( min ) = − 0.1 = 0 (Minimum Zener current is zero.)
20
13.6 − 9
For VPS (max) and IL (min), then I Z ( max ) = − 0 ⇒ I Z ( max ) = 230 mA
20
The characteristic of the minimum Zener current being one-tenth of the maximum value is violated.
The proper circuit operation is questionable.
TYU2.5

6. VPS ( min ) − VZ
I Z ( min ) = − I L ( max )
Ri
10 − 9
so 30 = − I L ( max ) which yields I L ( max ) = 35.4 mA
0.0153
TYU2.6
␯O(V)
4.35
2.7
Ϫ4.7
2.7 6 ␯I (V)
Ϫ4.7
TYU2.7
As vS goes negative, D turns on and vO = +5 V . As vS goes positive, D turns off. Output is a square
wave oscillating between +5 and +35 volts.
TYU2.8
␯O(V)
4.4
0 0.6 5 ␯I (V)
(a)
␯O(V)
4.4
2.4
0 3 5 ␯I (V)
(b)

7. TYU2.9
(a) VO = 0.6 V for all V1.
VO (V)
3.6
0.6
0 3 VI (V)
(b)
PSpice Exercises