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Exam (elaborations)

Solutions Manual for Microelectronic Circuit Design (6th Edition, 2023) by Jaeger and Blalock

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This comprehensive solutions manual provides detailed, step-by-step answers to end-of-chapter exercises from Microelectronic Circuit Design, 6th Edition (2023) by Richard C. Jaeger and Travis N. Blalock. It covers key topics such as semiconductor physics, diodes, bipolar junction transistors (BJTs), MOSFETs, amplifier design, frequency response, digital logic, analog and mixed-signal circuits, and integrated circuit layout principles. Ideal for students in electrical and computer engineering, this manual helps solidify theoretical understanding while preparing learners for lab work, exams, and real-world circuit design challenges. microelectronic circuit design solutions, jaeger blalock 6th edition answers, analog circuit problems solved, mosfet and bjt exercises, diode circuit analysis, amplifier design solutions, frequency response problems, digital logic circuit answers, mixed signal design manual, semiconductor device solutions, electronics textbook problems, electrical engineering homework help, ic design principles solved, microelectronics practice problems, jaeger microelectronic circuit solutions

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Computer Engineering
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Institution
Computer engineering
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Computer engineering

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Uploaded on
September 17, 2025
Number of pages
11
Written in
2025/2026
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Exam (elaborations)
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SOLUTIONS MANUAL

,1.1
Answering machine Model airplanes
Alarm clock MP3 player
Automatic door Musical greeting cards
Automatic lights Musical tuner
ATM Pagers
Automobile: Personal computer
Engine controller Personal planner/organizer (PDA)
Temperature control Radar detector
ABS Broadcast Radio (AM/FM/Shortwave)
Electronic dash Razor
Navigation system Satellite radio
Automotive tune-up receiver Security
equipment Baggage scanner systems Sewing
Bar code scanner machine Smoke
Battery charger detector Sprinkler
Cable/DSL Modems and system Stereo
routers Calculator system
Camcorder Amplifier
Carbon monoxide detector CD/DVD player
Cash register Receiver
CD and DVD players Tape player
Ceiling fan (remote) Stud sensor
Cellular phones Talking toys
Coffee maker Telephone
Compass Telescope controller
Copy machine Thermostats
Cordless phone Toy robots
Depth finder Traffic light controller
Digital Camera TV receiver & remote control
Digital watch Variable speed appliances
Digital voice recorder Blender
Digital scale Drill
Digital thermometer Mixer
Electronic dart board Food processor
Electric guitar Fan
Electronic door bell Vending machines
Electronic gas pump Video game controllers
Elevator Wireless headphones & speakers
Exercise Wireless thermometer
machine Fax Workstations
machine Fish
finder Electromechanical Appliances*
Garage door opener Air conditioning and heating systems
GPS Clothes washer and dryer
Hearing aid Dish washer Electrical
Invisible dog timer Iron, vacuum
fences Laser cleaner, toaster
pointer Oven, refrigerator, stove, etc.
LCD projector
Light dimmer *These appliances are historically based only
Keyboard synthesizer upon on-off (bang-bang) control. However,
Keyless entry system many of the high end versions of these
Laboratory appliances have now added sophisticated
instruments Metal electronic control.
detector Microwave
oven

1-1 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.2
B 19.97 x 100.1997 2020 1960
14.5 x 1012 14.5 Tb/chip

1.3
(a)
0.1977 Y2 1960

B2 19.97 x10 0.1977 Y2 Y1 0.1977 Y2 Y1
10
0.1977 Y1 1960 
so 2 10
B1 19.97x10
log 2
Y2 Y1 1.52 years
0.1977 log10
(b) Y2 Y1 5.06 years
0.1977

1.4
0.1548 2020 1970
N 1610x10 8.85 x 1010 transistors/ P

1.5
0.1548 Y 1970
2
N2 1610 x10 0.1548 Y2 Y1
N1
0.1548 Y1 1970 10
1610x10
log2
(a) Y2 Y1 1.95 years
0.1548
log10
(b) Y Y 6.46 years
2 1
0.1548

1.6 F 8.00x10
0.05806 2020 1970
m 10 nm .

No, this distance corresponds to the diameter of only a few atoms. Also, the wavelength of the
radiation needed to expose such patterns during fabrication is represents a serious problem.

1.7
From Fig. 1.4, there are approximately 600 million transistors on a complex Pentium
IV microprocessor in 2004. From Prob. 1.4, the number of transistors/ P will be 8.85 x
1010. in 2020. Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV
processors.




1-2 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.8
8
P 75x106 tubes 1.5W tube 113 MW! I 511 kA!
  1.13 x 10 W
220V

1.9 D, D, A, A, D, A, A, D, A, D, A

1.10 10.24V
VLSB 10.24V 4096bit 10.24V
2.500 mV VMSB = 5.120V
212 bits s 2
1001001001102 2 11 2 2 2 2 234210 VO 2342 2.500mV 5.855 V
8 5 2



1.11
VLSB 5V 5V mV an 2.77V
19.53 mV 142 LSB
28
256bit bit d 19.53
bits s
bit
14210 128 8 4
10
2 100011102

1.12 2.5V
VLSB 2.5V 1024 bits 2.4 m
10
2 bits 4 V
bit
8 6 5 3 2 0 ⎛ 2.5V ⎞
01011011012 2 2 2 2 36510 VO 365 ⎟ 0.891 V
2 2 ⎜
10 ⎝ 1024⎠

1.13 10V 21 bits 11191 bits
VLSB 0.610 mV an 6.83V 4
214 bits
4 bit d 10V
1119110 8192 2048 512 256 128 32 16 4 2 1 10
1119110 101011101101112

1.14
A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000. The
number of bits must satisfy 2B ≥ 10,000 where B is the number of bits. Here B = 14 bits.

1.15 5.12V bit
VLSB 212 bits 5.12V 1.25 m
and VO 1011101110112 VLSB
4096 bits V

, S
V B 2
L

VO 2 9 2 8 2 7 2 5 2 4 2 3 2
11

2 1 1.25mV
10
0.0625V
1.8
VO 3.754 0.000625 or 3.753V VO 3.755V




1-3
6/9/06

, 1.16
IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A

1.17
VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000 t Volts

1.18

vCE = [5 + 2 cos (5000t)] V

1.19

vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V

1.20

V = 10 V, R1 = 22 k , R2= 47 k and R3 = 180 k .
V
+ 1 -
I I
R1 2 + 3
R R
V 2 V
2 3

-


V1 10V 22k 10V 22k 3.71 V
22k 47k 180k 22k 37.3k
37.3k
V2 10V 6.29 V Checking : 6.29 + 3.71 = 10.0 V
I2 I1 22k180k37.3k ⎛ 10V ⎞ 180k 134 A
47k 180k ⎝ 22k 37.3k ⎠ 47k 180k
⎜ ⎟
47k ⎛ ⎞ 47k
⎜ 10V ⎟
I3 I1 34.9 A
47k 180k ⎝ 22k 37.3k ⎠ 47k 180k
10V
Checking : I1 169 A and I1 = I2 I3
22k 37.3k




1-4 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.21
V = 18 V, R1 = 56 k , R2= 33 k and R3 = 11 k .
V
+ 1 -
I I
R1 2 + 3
R R
V 2 V
2 3

-


V1 18V 56k 15.7 V V2 18V 33k 11k
2.31 V
56k 33k 11k 56k 33k 11k
Checking : V1 V2 15.7 2.31 18.0 V which is correct.
I1 18V 280 A I2 I1 11k 280 11k 70.0 A
A
56k 33k 11k 33k 11k 33k 11k
33k
33k
I3 I1 280 A 210 A Checking : I2 I3 280 A
33k 11k 33k 11k


1.22 5.6k 3.6k
I1 5mA 3.97 mA I2 2.4k 1.03 mA
5.6k 3.6k 2.4k
5mA 9.2k 2.4k

3.6k
V
3 
5mA 2.4k 9.2k
5.6k 3.6k
3.72V

Checking : I1 I2 5.00 mA and I2 R2 1.03mA 3.6k 3.71 V

1.23
I2 250 A 150k 150k
125 A I3 250 A 125 A
150k 150k 82k 150k 150k
V
3 
250 A 150k 150k
68k 82k
10.3V

Checking : I1 I2 250 A and I2 R2 125 A 82k 10.3 V




1-5
6/9/06

,1.24
+
R
v 1


vs
-
+
gm v v
th
-




Summing currents at the output node yields :
v + .002v = 0 so v = 0 and vth v  v vs
s
5x104

+
v R
1
i
- x

g v
m vx




Summing currents at the output node :
v
ix 5x104 0.002v = 0 but v = vx
ix vx vx 495
0.002v = 0 R 1
5x104 x th
ix 1
gm
R1

Thévenin equivalent circuit:



495 
vs




1-6 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.25 The Thévenin equivalent resistance is found using the same approach as Problem
1.24, and

1
⎛R 1 ⎞ 39.6
th
.02
5
⎝ 4k
⎜ ⎟

+

R
v 1

vs -

gmv in




The short circuit current is :
v
in  0.025v and v = vs
4k
vs
in 0.025vs 0.0253vs
4k

Norton equivalent circuit:




0.0253v s 39.6




1-7
6/9/06

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