ECEN 301 Lecture #14

Report
Obedience = Happiness
Mosiah 2:41
41 And moreover, I would desire that ye should consider on the
blessed and happy state of those that keep the commandments
of God. For behold, they are blessed in all things, both temporal
and spiritual; and if they hold out faithful to the end they are
received into heaven, that thereby they may dwell with God in a
state of never-ending happiness. O remember, remember that
these things are true; for the Lord God hath spoken it.
ECEN 301
Discussion #14 – AC Circuit Analysis
1
Lecture 14 – AC Circuit Analysis
Phasors allow the use of familiar
network analysis
ECEN 301
Discussion #14 – AC Circuit Analysis
2
RLC Circuits
Linear passive circuit elements: resistors (R), inductors
(L), and capacitors (C) (a.k.a. RLC circuits)
• Assume RLC circuit sources are sinusoidal
R1
vs(t)
+
~
–
ia(t)
ix
C
Time domain
ECEN 301
ZL
ZR1
L
ib(t)
R2
Ix
+
–
Vs(jω)
Ia(jω)
ZC
Ib(jω)
ZR2
Frequency (phasor) domain
Discussion #14 – AC Circuit Analysis
3
RLC Circuits - Series Impedances
• Series Rule: two or more circuit elements are said to be in
series if the current from one element exclusively flows into
the next element.
• Impedances in series add the same way resistors in series add
N
Z EQ 
Z
n
n 1
Z1
Z2
Zn
Z3
∙∙∙
ZN
∙∙∙
ZEQ
ECEN 301
Discussion #14 – AC Circuit Analysis
4
RLC Circuits - Parallel Impedances
• Parallel Rule: two or more circuit elements are said to be in
parallel if the elements share the same terminals
• Impedances in parallel add the same way resistors in parallel add
1

Z EQ
1
Z1

1
Z2

1

Z3
1
Z EQ 
ZN
1
1
Z1
Z1
ECEN 301
Z2
Z3
Zn

1

Z2
ZN
Discussion #14 – AC Circuit Analysis
1
Z3
 
1
ZN
ZEQ
5
RLC Circuits
AC Circuit Analysis
1. Identify the AC sources and note the excitation
frequency (ω)
2. Convert all sources to the phasor domain
3. Represent each circuit element by its impedance
4. Solve the resulting phasor circuit using network
analysis methods
5. Convert from the phasor domain back to the time
domain
ECEN 301
Discussion #14 – AC Circuit Analysis
6
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
R1
is(t)
vs(t)
+
~
–
ECEN 301
R2
C
Discussion #14 – AC Circuit Analysis
7
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
1.
R1
is(t)
vs(t)
+
~
–
ECEN 301
Note frequencies of AC sources
Only one AC source - ω = 377 rad/s
R2
C
Discussion #14 – AC Circuit Analysis
8
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
1.
2.
Note frequencies of AC sources
Convert to phasor domain
Z1 = R1
R1
is(t)
vs(t)
+
~
–
ECEN 301
Vs=10ej0
R2
C
Is(jω)
+
~
–
Discussion #14 – AC Circuit Analysis
Z2=R2
Z3=1/jωC
9
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
1.
2.
3.
Z1 = R1
Vs=10ej0
+
~
–
Note frequencies of AC sources
Convert to phasor domain
Represent each element by its impedance
Is(jω)
v s ( t )  10 cos(  t )
Z2=R2
Z3=1/jωC
V s ( j  )  10  0  10 e
Z 1  R1
ECEN 301
Discussion #14 – AC Circuit Analysis
Z 2  R2
Z3 
j0
1
j C
10
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
Z1 = R1
Vs=10ej0
+
~
–
Node a
Is(jω)
Z2=R2
1.
2.
3.
4.
Note frequencies of AC sources
Convert to phasor domain
Represent each element by its impedance
Solve using network analysis
•
Use node voltage and Ohm’s law
Z3=1/jωC
Ohm' s Law :
KCL at Node a :
I s ( j ) 
V a ( j )
I s ( j ) 
Z 2 || Z 3

V1 ( j  )
Z1
V s ( j )  V a ( j )
Z1
ECEN 301
Discussion #14 – AC Circuit Analysis
11
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
4.
Z1 = R1
Node a
Solve using network analysis
•
Use node voltage and Ohm’s law
Vs  V a
Vs=10ej0
+
~
–
Z1
Is(jω)
Z3=1/jωC
Va
Z 2 || Z 3
Vs
Z2=R2

Z1

1
1 

 V a 

Z 1 
 Z 2 || Z 3
 R  (1 / j  C )
1 

 V a  2


R

(
1
/
j

C
)
R
1 
 2
 j  CR 2  1
1 

 V a 

R2
R1 

 ( j  CR 1 R 2  R1 )  R 2 

 V a 

R
R
1
2


ECEN 301
Discussion #14 – AC Circuit Analysis
12
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
4.
Z1 = R1
Solve using network analysis
•
Use node voltage and Ohm’s law
Node a
 ( j  CR 1 R 2  R1 )  R 2 

 V a 

Z1
R1 R 2


Vs
Vs=10ej0
+
~
–
Is(jω)
Z2=R2
Z3=1/jωC
Va 


Vs 
R1 R 2


R1  ( j  CR 1 R 2  R1 )  R 2 

10  0 
( 50 )( 200 )


4
50  j ( 377 )( 10 )( 50 )( 200 )  ( 50 )  ( 200 ) 
 4 . 421  (  0 . 9852 )
ECEN 301
Discussion #14 – AC Circuit Analysis
13
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
4.
Z1 = R1
Vs=10ej0
+
~
–
Node a
Is(jω)
Solve using network analysis
•
Use node voltage and Ohm’s law
I s ( j ) 
Z2=R2
Z3=1/jωC

V s ( j )  V a ( j )
Z1
10  0  4 . 421  (  . 9852 )
50
 0 . 1681  0 . 4537
ECEN 301
Discussion #14 – AC Circuit Analysis
14
RLC Circuits
• Example 1: find is(t)
• vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF
4.
Z1 = R1
Vs=10ej0
+
~
–
Node a
5.
Solve using network analysis
•
Use node voltage and Ohm’s law
Convert to time domain
Is(jω)
Z2=R2
Z3=1/jωC
I s ( j  )  0 . 1681  0 . 4537
i s ( t )  0 . 1681 cos( 377 t  0 . 4537 )
ECEN 301
Discussion #14 – AC Circuit Analysis
15
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω,
L1 = 0.1H, L2 = 20mH
Rs
is(t)
vs(t)
+
~
–
ECEN 301
R1
i1(t)
R2
i2(t)
L1
L2
Discussion #14 – AC Circuit Analysis
16
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω,
L1 = 0.1H, L2 = 20mH
Rs
1.
is(t)
vs(t)
+
~
–
ECEN 301
R1
i1(t)
R2
Note frequencies of AC sources
Only one AC source - ω = 377 rad/s
i2(t)
L1
L2
Discussion #14 – AC Circuit Analysis
17
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω,
L1 = 0.1H, L2 = 20mH
1.
2.
Rs
is(t)
vs(t)
+
~
–
R1
i1(t)
Note frequencies of AC sources
Convert to phasor domain
R2
Zs = Rs
L2 Vs=155ej0
+
~
–
Is(jω)
i2(t)
L1
I1(jω)
ZR1=R1
I2(jω)
ZL1=jωL1
ECEN 301
Discussion #14 – AC Circuit Analysis
ZR2=R2
ZL2=jωL2
18
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω,
L1 = 0.1H, L2 = 20mH
1.
2.
3.
Note frequencies of AC sources
Convert to phasor domain
Represent each element by its impedance
ZR2=R2
v s ( t )  155 cos(  t )
Zs = Rs
Vs=155ej0
+
~
–
Is(jω)
I1(jω)
ZR1=R1
I2(jω)
ZL1=jωL1
Z R 1  R1
 2
ECEN 301
V s ( j  )  155  0  155 e
j0
ZL2=jωL2
Z L 1  j  L1
 j ( 377 )( 0 . 1)
 j 37 . 7 
Z R 2  R2
 0 .2 
Discussion #14 – AC Circuit Analysis
Z L 2  j L 2
 j ( 377 )( 0 . 02 )
 j 7 . 54 
19
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1
= 0.1H, L2 = 20mH
4.
Zs = Rs
Vs=155ej0
+
~
–
Solve using network analysis
•
Ohm’s law
Is(jω)
I1(jω)
I2(jω)
Z2=ZR2+ZL2
Z1=ZR1+ZL1
Z 1  Z R1  Z L1
ECEN 301
Z 2  Z R2  Z L2
 2  j 37 . 7
 0 . 2  j 7 . 54
 37 . 75  1 . 52 
 7 . 54  1 . 54 
Discussion #14 – AC Circuit Analysis
20
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1
= 0.1H, L2 = 20mH
4.
Zs
Vs=155ej0
+
~
–
V(jω)
Is(jω)
I1(jω)
I s ( j )  I1 ( j )  I 2 ( j )  0
Z2
Z1
I2(jω)
Z 1  37 . 75  1 . 52 
Z 2  7 . 54  1 . 54 
Solve using network analysis
•
KCL
V ( j )

V ( j )
Z1
Z2

V s ( j )  V ( j )
Zs
 1
1
1  V s ( j )


V ( j )



Zs
 Z1 Z 2 Z s 
V ( j  )  154 . 1 0 . 079 V
Z s  0 .5 
ECEN 301
Discussion #14 – AC Circuit Analysis
21
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 =
0.1H, L2 = 20mH
4.
Zs
Vs=155ej0
+
~
–
V(jω)
I 1 ( j ) 
Is(jω)
I1(jω)
Z2
Z1
I2(jω)
Z 1  37 . 75  1 . 52 
Z 2  7 . 54  1 . 54 
Solve using network analysis
•
Ohm’s Law
V ( j )
Z1
 4 . 083   1 . 439
I 2 ( j ) 
V ( j )
Z2
 20 . 44   1 . 465
Z s  0 .5 
ECEN 301
Discussion #14 – AC Circuit Analysis
22
RLC Circuits
• Example 2: find i1 and i2
• vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 =
0.1H, L2 = 20mH
5.
Zs
Vs=155ej0
+
~
–
V(jω)
I 1 ( j  )  4 . 083   1 . 439
Is(jω)
I1(jω)
Convert to Time domain
Z2
Z1
i1 ( t )  4 . 083 cos( 377 t  1 . 439 ) A
I2(jω)
I 2 ( j  )  20 . 44   1 . 465
i 2 ( t )  20 . 44 cos( 377 t  1 . 465 ) A
ECEN 301
Discussion #14 – AC Circuit Analysis
23
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)
+
~
–
ECEN 301
L
R2
C
ia(t)
ib(t)
Discussion #14 – AC Circuit Analysis
24
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)
+
~
–
ECEN 301
L
1.
R2
C
ia(t)
Note frequencies of AC sources
Only one AC source - ω = 1500 rad/s
ib(t)
Discussion #14 – AC Circuit Analysis
25
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)
+
~
–
1.
2.
L
ZR1
R2
C
ia(t)
Note frequencies of AC sources
Convert to phasor domain
ib(t)
Vs(jω)
ECEN 301
ZL
ZC
+
~
–
Ia(jω)
Discussion #14 – AC Circuit Analysis
ZR2
Ib(jω)
26
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
ZR1
Vs(jω)
ZL
ZC
+
~
–
Ia(jω)
Z R 1  R1
 100 
ECEN 301
1.
2.
3.
ZR2
Ib(jω)
Z L  j L
 j (1500 )( 0 . 5 )
 j 750 
Note frequencies of AC sources
Convert to phasor domain
Represent each element by its impedance
v s ( t )  15 cos(  t )
V s ( j  )  15  0  15 e
Z R 2  R2
 75 
Discussion #14 – AC Circuit Analysis
j0
Z C  1 / j C
 1 / j (1500 )( 10
6
)
  j 667 
27
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
+ZR1–
Vs(jω)
+
~
–
+
ZC
Ia(jω) –
+ZL–
4.
Solve using network analysis
•
Mesh current
KVL at a :
Ib(jω)
+
ZR2
–
 V s ( j )  V R1 ( j )  VC ( j )  0
I a Z R1  ( I a  I b ) Z C  V s
I a ( Z R1  Z C )  I b Z C  V s
Z R 1  100 
Z R 2  75 
Z L  j 750 
Z C   j 667 
ECEN 301
KVL at b :
 V C ( j )  V L ( j )  V R 2 ( j )  0
 (I a  Ib )Z C  IbZ L  IbZ R2  0
 I a Z C  Ib (Z C  Z L  Z R2 )  0
Discussion #14 – AC Circuit Analysis
28
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
+ZR1–
Vs(jω)
+
~
–
+
ZC
Ia(jω) –
+ZL–
Ib(jω)
4.
+
ZR2
–
Solve using network analysis
•
Mesh current
I a (100  j 667 )  I b ( j 667 )  15
I a ( j 667 )  I b ( 75  j 83 )  0
I 1  0 . 0032  0 . 917 A
I 2  0 . 019   1 . 49 A
ECEN 301
Discussion #14 – AC Circuit Analysis
29
RLC Circuits
• Example 3: find ia(t) and ib(t)
• vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
+ZR1–
Vs(jω)
+
~
–
+
ZC
Ia(jω) –
+ZL–
Ib(jω)
5.
+
ZR2
–
Convert to Time domain
I 1  0 . 0032  0 . 917 A
i1 ( t )  3 . 2 cos( 1500 t  0 . 917 ) mA
I 2  0 . 019   1 . 49 A
i 2 ( t )  19 cos( 1500 t  1 . 49 ) mA
ECEN 301
Discussion #14 – AC Circuit Analysis
30
AC Equivalent Circuits
Thévenin and Norton equivalent circuits apply in AC analysis
• Equivalent voltage/current is complex and frequency dependent
I
Thévenin Equivalent
Source
+
V
–
Load
Norton Equivalent
I
I
VT(jω)
+
–
ZT
IN(jω)
+
V
Load
–
ECEN 301
+
ZN
V
Load
–
Discussion #14 – AC Circuit Analysis
31
AC Equivalent Circuits
Computation of Thévenin and Norton Impedances:
1.
2.
Remove the load (open circuit at load terminal)
Zero all independent sources
•
•
3.
Voltage sources
Current sources
Compute equivalent impedance across load terminals (with load removed)
Z3
Z1
+
–
Vs(jω)
short circuit (v = 0)
open circuit (i = 0)
ZL
Z2
Z3
Z1
a
Z4
a
ZT
Z2
Z4
b
b
NB: same procedure as equivalent resistance
ECEN 301
Discussion #14 – AC Circuit Analysis
32
AC Equivalent Circuits
Computing Thévenin voltage:
1.
2.
3.
•
4.
Remove the load (open circuit at load terminals)
Define the open-circuit voltage (Voc) across the load terminals
Chose a network analysis method to find Voc
node, mesh, superposition, etc.
Thévenin voltage VT = Voc
Z3
Z1
+
–
Vs(jω)
+
–
Vs(jω)
Z2
Z4
Z3
Z1
a
Z2
b
Z4
a
+
VT
–
b
NB: same procedure as equivalent resistance
ECEN 301
Discussion #14 – AC Circuit Analysis
33
AC Equivalent Circuits
Computing Norton current:
1.
2.
3.
•
4.
Replace the load with a short circuit
Define the short-circuit current (Isc) across the load terminals
Chose a network analysis method to find Isc
node, mesh, superposition, etc.
Norton current IN = Isc
Z3
Z1
+
–
Vs(jω)
+
–
Vs(jω)
Z2
Z4
Z3
Z1
a
a
IN
Z2
b
Z4
b
NB: same procedure as equivalent resistance
ECEN 301
Discussion #14 – AC Circuit Analysis
34
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
L
Rs
vs(t)
+
~
–
ECEN 301
C
RL
+
vL
–
Discussion #14 – AC Circuit Analysis
35
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
1.
Note frequencies of AC sources
L
Rs
vs(t)
+
~
–
ECEN 301
Only one AC source - ω = 103 rad/s
C
RL
+
vL
–
Discussion #14 – AC Circuit Analysis
36
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
L
1.
2.
Rs
Note frequencies of AC sources
Convert to phasor domain
ZL
vs(t)
+
~
–
ECEN 301
C
RL
+
vL
–
Zs
Vs(jω)
+
~
–
Discussion #14 – AC Circuit Analysis
ZC
ZLD
37
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
ZL
Zs
ZC
1.
2.
3.
Note frequencies of AC sources
Convert to phasor domain
Find ZT
•
Remove load & zero sources
Z T  Z S  Z C || Z L
 RS 
( j  L )( 1 / j  C )
( j  L )  (1 / j  C )
 RS  j
L
1   LC
2
 50  j 65 . 414
 82 . 33  0 . 9182
ECEN 301
Discussion #14 – AC Circuit Analysis
38
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
ZL
1.
2.
3.
Zs
Vs(jω)
+
~
–
ZC
+
VT(jω)
–
4.
Note frequencies of AC sources
Convert to phasor domain
Find ZT
•
Remove load & zero sources
Find VT(jω)
•
Remove load
NB: Since no current flows in the
circuit once the load is removed:
Z T  82 . 33  0 . 9182
ECEN 301
VT  V S
Discussion #14 – AC Circuit Analysis
39
AC Equivalent Circuits
• Example 4: find the Thévenin equivalent
• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
ZL
Zs
Vs(jω)
+
~
–
ZT
ZC
ZLD
VT  V S
ECEN 301
VT(jω)
+
~
–
ZLD
Z T  82 . 33  0 . 9182
Discussion #14 – AC Circuit Analysis
40

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