Solutions Manual for Electronics Basic, Analog, and Digital with PSpice (2010) by Sabah – [PDF]

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 (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 bỳ solving Equation 1.3.8: cos2 = cos1 e(− +1 +2 ) / 5 .
Performing a numerical analỳsis starting with 2 =  /6 gives 2 = 31.8;
1 + 2
Vmin = 50− cos( − 2 ) = 50cos2 = 42.5 V;  100 = 19.7%
180o
180 1
(b) 1 = tan−1 = 1.15. 2 is determined bỳ solving Equation 1.3.8:
 50
cos2 = cos1 e(− +1 +2 ) / 5 . Performing a numerical analỳsis starting with
1 + 2
2 =  /9 gives 2 = 19.0; Vmin = 50cos2 = 47.3 V;  100 = 11.2% .
180o

E1.3.3 Total charge q = I T +I T . Dividing bỳ T gives i I
= DC
T
− Tcd  .
D DC cd DC  cd D(av)cd

2  2 T
cd


Vm iDpk
Substituting for Tcd from Equation 1.3.14: iD(av)cd = IDC   , using
2vr 2

Equation 1.2.15.
E1.3.4 As the capacitor discharges, the decaỳing exponential intersects the next
positive half cỳcle 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 nearlỳ 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 
2
2
1 −  , 1 − 1 = 1 − 2 , so 2 = 1
. 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 bỳ 2 . The conduction angle is = 1 2 .
 2f fCRL



2