Electronics: Basic, Analog & Digital with PSpice (2010) – Solutions Manual – Sabah

ALL 13 CHAPTERS COVERED




SOLUTIONS MANUAL

, Contents


Page

Chapter 1 Basic Diode Circuits 1

Chapter 2 Basic Principles of Semiconductors 39

Chapter 3 pn Junction and Semiconductor Diodes 51

Chapter 4 Semiconductor Fabrication 65

Chapter 5 Field Effect Transistors 67

Chapter 6 Bipolar Junction Transistor 93

Chapter 7 Two-Port Circuits, Amplifiers, and Feedback 109

Chapter 8 Single-Stage Transistor Amplifiers 129

Chapter 9 Multistage and Feedback Amplifiers 185

Chapter 10 Differential and Operational Amplifiers 215

Chapter 11 Power Amplifiers and Switches 245

Chapter 12 Basic Elements of Digital Circuits 273

Chapter 13 Digital Logic Circuit Families 293

Companion CD: Classroom Presentations

Figures

, Chapter 1 Basic Diode Circuits


Solutions to Exercises
E1.1.1 ( )
(a) -0.99IS = IS eVD / 2VT − 1 ; vD = 0.052ln0.01 = -0.24 V.

( )
(b) 100IS = IS eVD / 2VT − 1 ; vD = 0.052ln101 = 0.24 V.

0.052
E1.1.2 (a) rd = = 5.2 Ω.
10 − 10 − 8
−2


iD 10 −2
(b) vD = 0.052 ln = 0.052 ln − 8 = 0.718 V.
IS 10

(c) vD = 0.052ln(2×106) = 0.754 V.
E1.2.1 (a) Vγ ≅ 0.18 V, 10

VDO ≅ 0.42 V.
8
(b) Vγ ≅ 0.5 V,
IS = 1 μA IS = 1 nA
VDO ≅ 0.78 V.
6
For both diodes: iD
mA
rD ≅ 0.052 V/10 mA 4

= 5.2 Ω.
2




0
0 0.2 0.4 0.6 0.8


( )
vD V
E1.2.3 i D = IS e v D /ηVT − 1 .
−6
/(10 −6 ×0.026 )
The exponential becomes e10 = e1/ 0.026 and

10 −12 e1/ 0.026 = 5.1× 10 4 A.
dv L dv L
E1.3.1 From Equation 1.3.4, = -ωVmsinθ1 . From Equation 1.3.7,
dt ωt =θ1 dt ωt =θ1


1 1
= -ωVmcosθ1 × . Equating these slopes gives tanθ1 = . Note that
ωCRL ωCRL
vL is continuous because there are no current impulses to change the
capacitor voltage at ωt = θ1. Moreover, because vL appears across a resistor,
the capacitor current must also be continuous at this instant.




1

, 180 1
E1.3.2 (a) ωCRL = 100π × 5 × 10 −6 × 10 4 = 5π = 15.71, θ1 = tan −1 = 3.64°. θ2
π 5π

is determined by solving Equation 1.3.8: cosθ2 = cosθ1 e ( −π +θ1 +θ2 ) / 5π .

Performing a numerical analysis starting with θ2 = π /6 gives θ2 = 31.8°;
θ1 + θ 2
Vmin = 50[− cos(π − θ 2 )] = 50cosθ2 = 42.5 V; × 100 = 19.7%
180 o
180 1
(b) θ1 = tan −1 = 1.15°. θ2 is determined by solving Equation 1.3.8:
π 50

cosθ2 = cosθ1 e ( −π +θ1 +θ2 ) / 5π . Performing a numerical analysis starting with
θ1 + θ 2
θ2 = π /9 gives θ2 = 19.0°; Vmin = 50cosθ2 = 47.3 V; × 100 = 11.2% .
180 o

⎛T ⎞ I T
E1.3.3 Total charge qD = IDCTcd + I DC ⎜ − Tcd ⎟ . Dividing by Tcd gives iD(av)cd = DC .
⎝2 ⎠ 2 Tcd

Vm′ i Dpk
Substituting for Tcd from Equation 1.3.14: iD(av)cd = I DC π ≅ , using
2v r 2

Equation 1.2.15.
E1.3.4 As the capacitor discharges, the decaying exponential intersects the next
positive half cycle at 2π – θ2 instead of π – θ2. Hence, Equation 1.3.8 is

modified to cosθ2 = cosθ1 e −[2π − (θ1 +θ2 )] / ωCRL ≅ e − 2π / ωCRL = 1 −
1
.
fCRL

Vm′
Assuming that during discharge, IDC is constant and equals , the
vr

discharge period is nearly T or 1/f. The charge in capacitor voltage is
I DC I
therefore = v r . Equation 1.3.11 becomes VDC = Vm′ – D . In other
fC 2fC
words, the peak-to-peak ripple is doubled. Using the approximation, cosα ≈

α2 1 θ2 1
1− ,1− = 1 − 2 , so θ2 = . Because the ripple is doubled,
2 fCR L 2 fCR L

θ 22
Vm′ (1 – cosθ2) is doubled. Using the approximation 1 – cosθ2 = , this
2
θ2 1 2
means that θ2 is multiplied by 2 . The conduction angle is = .
ω 2πf fCR L




2