Electronics: Basic, Analog, and Digital with PSpice 1st Edition (2010) - Nassir H. Sabah - Solutions Manual (PDF)

ALL 13 CHAPTERS COVERED

, 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 e
VD / 2VT

 1 ; vD  0.052ln101  0.24 V.

0.052
E1.1.2 (a) rd   5.2 .
10  108
2

iD 102
(b) vD  0.052 ln  0.052 ln  0.718 V.
IS 108

(c) vD  0.052ln(2106)  0.754 V.
E1.2.1 (a) V  0.18 V, 10

VDO  0.42 V.
(b) V  0.5 V, 8
IS = 1 A IS = 1 nA
VDO  0.78 V.
6
For both diodes: iD
mA

rD  0.052 V/10 mA 4

= 5.2 .
2




0
0 0.2 0.4 0.6 0.8


 
vD V
E1.2.3 iD  IS evD /VT  1 .
6
/(106 0.026 )
The exponential becomes e10  e1/ 0.026 and

1012 e1/0.026  5.1 104 A.

E1.3.1 From Equation 1.3.4, dvL  -Vmsin1 . From Equation 1.3.7, dv L
dt t 1 dt t 1


1 1
 -Vmcos1  . Equating these slopes gives tan1  . Note that
CRL CRL
vL is continuous because there are no current impulses to change the
capacitor voltage at t = 1. Moreover, because vL appears across a resistor,
the capacitor current must also be continuous at this instant.




1

, 180 1
E1.3.2 (a) CR  100  5  106  104  5  15.71, 1  tan1  3.64. 2
L
 5
is determined by solving Equation 1.3.8: cos2  cos1 e( 1 2 ) / 5 .
Performing a numerical analysis starting with 2   /6 gives 2  31.8;
1  2
Vmin  50 cos(  2 )  50cos2  42.5 V;  100  19.7%
180∘
180 1
(b) 1  tan1  1.15. 2 is determined by solving Equation 1.3.8:
 50
cos2  cos1 e( 1 2 ) / 5 . Performing a numerical analysis starting with
1  2
2   /9 gives 2  19.0; Vmin  50cos2  47.3 V;  100  11.2% .
180∘

E1.3.3 Total charge q  I T +I T . Dividing by T gives i I
 DC
T
 Tcd  .
D DC cd DC  cd D(av)cd
 2  2 Tcd

Vm iDpk
Substituting for Tcd from Equation 1.3.14: i D(av)cd  IDC   , using
2vr 2

Equation 1.2.15.
E1.3.4 As the capacitor discharges, the decaying exponential intersects the next
positive half cycle at 2 – 2 instead of  – 2. Hence, Equation 1.3.8 is
2 (1 2 )/ CRL 1
modified to cos2  cos1 e  e 2 / CRL  1 .
fCRL
Vm
Assuming that during discharge, IDC is constant and equals v , the
r

discharge period is nearly T or 1/f. The charge in capacitor voltage is
I ID
therefore DC  v r . Equation 1.3.11 becomes VDC  Vm – . In other
fC 2fC
words, the peak-to-peak ripple is doubled. Using the approximation, cos 

1
2 , 1  1  1
22 , so   1
2 . Because the ripple is doubled,
2 fCRL 2 fCRL

22
Vm (1 – cos2) is doubled. Using the approximation 1 – cos2  , this
2
2
means that 2 is multiplied by . The conduction angle is  1 2 .
 2f fCRL




2