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Section 06
Impedance
 treat all passive components as resistors
 but with complex resistances
Electrical Engineering
Umm Al-Qura University
Slide 2
Impedance
 What is the impedance of a 10mF capacitor when
operated at 60Hz?
j
j
ZC  

  j 265.25
6
wC
2   60  10  10
 What is the impedance of a 2mH inductor when
operated at 60Hz?
ZL  jwL  j 2   60  2  10 3   j 0.754
Electrical Engineering
Umm Al-Qura University
Slide 3
Laplace Domain
 When
 Mixing AC and DC sources
 Multiple different frequencies
 use Laplace instead of Fourier
 jw  s
 Initial conditions
Electrical Engineering
Umm Al-Qura University
Slide 4
Complex AC Source
 AC Volt or Current has:
 Amplitude
 Frequency
 Phase
 Phase can be expressed
in Complex Number
A cos 2  f  t   
Electrical Engineering
Umm Al-Qura University
 A
Slide 5
Complex AC Source
A cos 2 f  t   
A
 A  cos   j sin  
11030  110  cos 30  j sin 30 
 95.26  j55
Electrical Engineering
Umm Al-Qura University
Slide 6
Example
Electrical Engineering
Umm Al-Qura University
Slide 7
Solution
KCL : I1  I 2  I 3
KVL :
 110  20,000  I1  j531  I 3
0
 j531  I 3  j11.3  I 2  1000  I 2
0
I1
Electrical Engineering
 5.44  1.2
mA
 I2
 2.56  61.4 mA
I3
 4.83  29.3 mA
Umm Al-Qura University
Slide 8
Simulation Solution
I1
Electrical Engineering
Umm Al-Qura University
 5.44  1.2
mA
 I2
 2.56  61.4 mA
I3
 4.83  29.3 mA
Slide 9
Section 07
Contents
 Generators
 Square, Sine, Triangle, Pulse Generators
 Converters
 AC/DC
 Analog/Digital
 Protection Circuits
 Voltage/Current Limiter
 Reverse Polarity
 ESD Protection
Electrical Engineering
Umm Al-Qura University
Slide 11
Contents
 Math Circuits
 General adders (mixers)
 Integrators, Differentiator
 Transfer Functions
 Filters
 Low/High Pass Filters
 Band Pass/Stop Filters
Electrical Engineering
Umm Al-Qura University
Slide 12
Note!!
 Some are Conceptual Designs
 Consider Ready-Made IC’s available
 Or even PCBs
Electrical Engineering
Umm Al-Qura University
Slide 13
AC/DC Converter
VD
n2
vdc  vac
 2  VD
n1
v dc
r
2  f  C  Rload
Electrical Engineering
Umm Al-Qura University
Slide 14
Example
 design an AC/DC converter to:
 Produce 8.4 VDC
 Ripples 1%
 Diodes available VD=0.6V
 Maximum load expected 2k
n2
v dc  v ac
 2  VD
n1
v dc
r
2  f  C  Rload
Electrical Engineering
Umm Al-Qura University
Slide 15
Clipping Protection
Electrical Engineering
Umm Al-Qura University
Slide 16
ESD Protection
 Simple Parallel Discharging Zener Diode
 Special Diodes for High speed Lines
Electrical Engineering
Umm Al-Qura University
Slide 17
General Adder (Mixer)
n
m
i 1
i 1
Vo   Ai v pi   Bi v ni
Ai 
Let
R pi
, Bi 
Rf
R ni
A   Ai , B   Bi , C  A  B  1

C  0

C  0

Electrical Engineering
Rf
Umm Al-Qura University
Rx  
Rx  
Ry 
Rf
C
Rf
C
Ry  
Slide 18
Integrator & Differentiator
the integrator produces a voltage output
proportional to the product (multiplication)
of the input voltage and time
the differentiator produces a voltage
output proportional to the input voltage's
rate of change.
Slide 19
Integrator


vo  


R
RS
1
vs (t ) dt

R SC
f  f0
f  f0
1
f0 
2 RC
Electrical Engineering
Umm Al-Qura University
Slide 20
Differentiator
d

 RC dt vs (t )
vo  
R

 r
f  f0
f  f0
1
f0 
2 rC
Electrical Engineering
Umm Al-Qura University
Slide 21
Filters
Electrical Engineering
Umm Al-Qura University
Slide 22
A differentiator circuit produces a
constant output voltage for a
steadily changing input voltage.
An integrator circuit produces a
steadily changing output voltage for
a constant input voltage.
Slide 23
Progress Check
 What circuits did you face in your field?
 Do you need more details?
 Where to find more Designs?
Electrical Engineering
Umm Al-Qura University
Slide 24
Section 08
Power Transmission
Electrical Engineering
Umm Al-Qura University
Slide 26
Slide 27
Transformer
 A transformer is an electrical device that transfers power from one
electrical circuit to another by magnetic coupling
 It is most often used to convert between high and low voltages
 Must be AC current; a transformer does not convert DC
 Uses Faraday’s law and ferromagnetic properties
Slide 28
Some Transformers
Slide 29
Ideal Transformer
n1
n2
n1
n2
v1

v2
i2

i1
Electrical Engineering
n1
n
n2
p1  p 2
Umm Al-Qura University
R  n  RL
'
L
2
Slide 30
Example
2
R eq
i1
Electrical Engineering
 10 
    10  1000 
 1 
115

 0.1147 A
3  1000
Umm Al-Qura University
Slide 31
Example

v1

v1
v2
i2
Electrical Engineering
 115  i1  3  114.7V
1

 v 1  11.47
10
11.47

 1.147 A
10
Umm Al-Qura University
Slide 32
Activity
n1  10
n2  5
v1  220v
RL  2
v2  ?
i1  ?
i2  ?
p?
Slide 33
Section 09
DC Machines
 Can we use the magnetic force to rotate something?
Slide 35
Basic Concept
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.html
Slide 36
DC Machines Examples










microwave fan
hi-fi tape deck
fridge
mixer
washing machine
tumble dryer
vacuum
computers
electric saw
drill










screwdriver
leaf blower
toothbrush
hair dryer
razor
CD player
video player
clocks
pond pumps
toys
Slide 37
Electric Machine Parts
Slide 38
Electric Machine Parts
Slide 39
Two-Pole DC Machine
Slide 40
Four-Pole DC Machine
Slide 41
Commutator
Slide 42
Generated Voltage
Slide 43
DC Motor Variables
EA
VT
Tdev
Tload
wm
Ra
IA
(volt)
(volt)
(N.m)
(N.m)
(rad/s)
()
(A)
is the back EMF
is the applied voltage
is the torque developed by DC Motor
is the opposing load torque
is the armature shaft speed = 2 rpm/60
is the motor internal resistance
is the motor current
Slide 44
DC Generator Variables
EA
VT
Tpm
Tdev
wm
Ra
IA
(volt)
(volt)
(N.m)
(N.m)
(rad/s)
()
(A)
is the generated voltage
is the load voltage
is the prime-mover generated torque
is the opposing motor torque
is the armature shaft speed = 2 rpm/60
is the motor internal resistance
is the motor current
Slide 45
DC Machine Equations
EA
Tdev
wm
IA
K

EA
 K    wm
Tdev
 K   IA
(volt)
(N.m)
(rad/s)
(A)
is the generated voltage
is the motor torque
is the armature shaft speed = 2 rpm/60
is the motor current
(Wb)
is the machine constant
is the magnetic flux per pole
Slide 46
Power
 Electric Power:
P  I V
 Mechanical Power:
P  T w
Slide 47
Ideal DC Machine
 Motor
 IN:
Electric Power
 OUT: Mechanical Power
Pelec
 EA  I A
 K    wm  I A
 Tdev  wm
 Pmech
Slide 48
Ideal DC Machine
 Generator
 IN:
Mechanical Power
 OUT: Electric Power
Pmech
 Tdev  wm
 K    I A  wm
 EA  I A
 Pelec
Slide 49
Motor Example
 A DC motor having
 Ra = 0.2, IA=5A, VT=220V, wm = 1200 rpm
 What is:
 back EMF voltage?
VT  EA  I A  R a
E A  220  5  0.2  219
 developed torque?
Tdev  wm  EA  I A
 developed power?
P  Tdev  wm  E A  I A
219  5
 8.7 N .m
2
1200 
60
2
 Tdev  w m  8.7  1200 
60
 EA  I A
 219  5
Tdev 
P
 1095W
 1095W
Slide 50
Generator Example
 A DC generator having
 Pmech=1kW, Ra = 0.3 , RL = 10
 What is:
 electric current drawn?
P
EA
 EA  I A

2
  P  I A  R a  R L 
 I A  R a  RL 
 terminal voltage?
VT  I A  RL
IA 
1000
 9.85 A
10  0.3
VT  9.85  10  98.5V
Slide 51
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