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Transcript 
Sinusoidal Steady-State
Analysis
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
•
•
•
•
•
•
•
•
Introduction
Nodal Analysis
Mesh Analysis
Superposition Theorem
Source Transformation
Thevenin and Norton Equivalent Circuits
OP-amp AC Circuits
Applications
Introduction
• Steps to Analyze ac Circuits:
– Transform the circuit to the phasor
(frequency) domain.
– Solve the problem using circuit techniques
(nodal/mesh analysis, superposition, etc.).
– Transform the resulting phasor to the time
domain.
Nodal Analysis
• Variables = Node Voltages
• Applying KCL to each node
gives each independent
equation
Supernode
• If supernodes included,
– Applying KCL to each
supernode gives 1 equation.
– Applying KVL at each
supernode gives 1 more
equation.
Example 1
Find ix.
20 cos 4t  200
  4 rad / s
 Z L  j L

Z  1
 C jC
1 H  j4
0.5 H  j 2
0 .1 F   j 2 .5
Example 1 (Cont’d)
Applying KCL at node 1,
20  V1
V1
V  V2

 1
10
 j 2.5
j4
 (1  j1.5)V1  j 2.5V2  20
(1) and (2) can be put in matrix form as
(1)
Applying KCL at node 2,
V1  V2 V2
2I x 

j4
j2
V1
Ix 
 11V1  15V2  0 (2)
 j 2.5
1  j1.5 j 2.5  V1  20
 


 11

15  V2   0 

 V1  18.9718.43
 V2  13.91198.3
V1
Ix 
 7.59108.4
 j 2.5
 ix  7.59 cos( 4t  108.4)
Example 2
• Applying KCL for the
supernode gives 1
equations.
• Applying KVL at the
supernode gives 1
equations.
• 2 variables solved by 2
equations.
Example 2 (Cont’d)
Sol :
Applying KCL at the supernode,
V
V V
3 1  2  2
 j 3 j 6 12
or 36  j 4V1  (1  j 2)V2
But V1  V2  1045
 V2  31.41  87.18
 V1  V2  1045
 25.78  70.48
Mesh Analysis
• Variables = Mesh Currents
• Applying KVL to each
mesh gives each
independent equation
• If supermeshes included,
Excluded
– Applying KVL to each
Supermesh
supermesh gives 1 equation.
– Applying KCL at each
supernode gives 1 more
i2  i1  I S
equation.
Example 1
Sol :
Applying KVL to mesh 1,
(8  j10  j 2)I1
 ( j 2)I 2  j10I 3  0
(1)
For mesh 2,
( 4  j 2  j 2) I 2  (  j 2 ) I 1
 ( j 2)I 3  2090  0 (2)
For mesh 3,  I 3  5
j 2   I1   j 50 
8  8 j



 j2


I
4

j
4

j
30

 2  

 I 2  6.12  35.22
I o  I 2  6.12144.78
Find Io.
Example 2
• Applying KVL for mesh
1 & 2 gives 2 equations.
• Applying KVL for the
supermesh gives 1
equations.
• Applying KCL at node A
gives 1 equations.
• 4 variables solved by 4
equations
Find Vo.
Example 2 (Cont’d)
Sol :
For mesh 1,
 10  (8  j 2)I1  ( j 2)I 2  8I 3  0
 (8  j 2)I1  j 2I 2  8I 3  10
(1)
For mesh 2,  I 2  3
(2)
(8  j 4)I 3  8I1
 (6  j 5)I 4  j 5I 2  0
(3)
Applying KCL at node A gives
I 4  I3  4
(4)
4 variables can be solved by
using 4 equations.
 Vo   j 2(I1  I 2 )
Superposition Theorem
• Since ac circuits are linear, the superposition
theorem applies to ac circuits as it applies to dc
circuits.
• The theorem becomes important if the circuit
has sources operating at different frequencies.
– Different frequency-domain circuit for each
frequency
– Total response = summation of individual responses
in the time domain
– Total response  summation of individual responses
in the phasor domain
Example 1
Find Io.
=
I0  I  I
'
0
"
0
+
Example 1 (Cont’d)
Sol :
Z  ( j 2) || (8  j10)  0.25  j 2.25
j 20
j 20
I '0 

4  j 2  Z 4.25  j 4.25
 2.353  j 2.353
To get I "0 ,
Applying KVL to mesh 1,
(8  j8)I1  j10I 3  j 2I 2  0
(1)
For mesh 2,
(4  j 4)I 2  ( j 2)I1  ( j 2)I 3  0 (2)
For mesh 3, I 3  5
(3)
j 2   I1   j 50 
8  8 j
 j2
 I    j10
4

j
4

 2  

I "0  I 2  2.647  j1.176
I 0  I '0  I "0  (2.353  2.647)  j (2.353  1.176)
 5  j 3.529  6.12144.78
Example 2
v0  v1  v2  v3
v1 is due to the 5 - V dc voltage source

where v2 is due to the 10 cos 2t voltage source
v is due to the 2 sin 5t current source
 3
Example 2 (Cont’d)
Since   0,
jL  0  short - circuit
1
   open  circuit
jC
By voltage division,
1
v1  
(5)  1 V
1 4
Example 2 (Cont’d)
j4 
V2
100 V
 j5 
1
V2 
(100)
10 cos 2t  100,   2 rad/s
1  j 4  ( j 5 || 4)
2 H  jωω  j 4 
10

3.439  j 2.049
1
0.1 F 
  j5 
 2.498  30.79
jC
 v2  2.498 cos( 2t  30.79) V
Example 2 (Cont’d)
j10 
V3
2  90 A
2 sin 5t  2  90,   5 rad/s
2 H  jL  j10 
1
0.1 F 
  j2 
jC
 j2 
By current division,
j10
I1 
(2  90)
j10  1  ( j 2 || 4)
j10
V3  I1 1 
( j 2)
1.8  j8.4
 2.33  80
 v3  2.33 cos(5t  80) V
Source Transformation
Example 1
Find Vx.
Is
20  90
Is 
 4  90   j 4
5
5(3  j 4)
Vs  I s  5 || (3  j 4)   j 4
8  j4
  j 4(2.5  j1.25)  5  j10
By voltage division,
10
Vx 
(5  j10)
2.5  j1.25  4  j13  10
 5.519  28 V
Vs
Thevenin & Norton Equivalent Circuits
Z Th
VTh
 ZN 
IN
Example 1
Z Th  (8 ||  j 6)  (4 || j12)
 6.48  j 2.64
 8
j12 
VTh  

(12075)

 8  j 6 4  j12 
 37.95220.31 V
Example 2
Example 2 (Cont’d)
Applying KVL at node 1 gives
15  I 0  0.5I 0  I 0  10
Set I s  3 for simplicity ,
Applying KVL to the loop gives
 I 0 (2  j 4)  0.5I 0 (4  j 3)  VTh  0
Applying KVL gives
Vs  I 0 (4  j 3  2  j 4)  2(6  j )
 VTh  10(2  j 4)  5(4  j 3)
  j 55
 55  90 V
I s  3  I 0  0.5I 0  I 0  2
Z Th
Vs 2(6  j )


Is
3
Example 3
By current division,
ZN
I0 
IN
Z N  (20  j15)
Example 3 (Cont’d)
(1) Z N can be found easily , Z N  5
(2) Apply mesh analysis to get I N .
Applying KVL for mesh 1 gives
 j 40  (18  j 2)I1  (8  j 2)I 2  (10  j 4)I 3  0 (1)
Applying KVL for the supermesh gives
(13  j 2)I 2  (10  j 4)I 3  (18  j 2)I1  0 (2)
Applying KCL at node a gives
I 3  I 2  3 (3)
(1), (2), and (3) give
I N  I 3  3  j8
5
 I0 
I N  1.46538.48 A
5  20  j15
OP AMP AC Circuits: Example 1
• Ideal op amps assumed
– Zero input current & zero differential input voltage
Applying KCL at node 1,
3  V1
V
V  0 V1  Vo
 1  1

10
 j5
10
20
 6  (5  j 4)V1 - Vo (1)
Applying KCL at node 2,
V1  0 0  Vo

10
 j10
 V1   jVo
(2)
vs  3 cos1000t V
(1) and (2) give
6
Vo 
 1.02959.04
3  j5
 vo (t )  1.029 cos(1000t  59.04) V
Example 2
Sol :
1
Zf
Vo
jC 2
G


Vs
Zi R  1
1
jC1
R2 ||

 jC1 R2
(1  jR1C1 )(1  jR2C2 )

 j4
 0.434130.6
(1  j 4)(1  j 2)
Open  loop gain : 0.434

130.6
Phase shift :
Find the close - loop
gain and phase shift .
R1  R2  10 k
C1  2 F
C2  1 F
  200 rad/s
Applications: Capacitance Multiplier
Vi  Vo
Ii 
 jC (Vi  Vo )
1 j C
V
Vi
1
Zi  i 

Ii
jC (Vi  Vo )
 V 
jC 1  o 
 Vi 
Vo
R2
But

Vi
R1
1
 Zi 
jCeq
 R2 
where Ceq  1  C
 R1 
Applications: Oscillators
• Barkhausen criteria must be meet for oscillators
– (1) Overall gain  1
– (2) Overall phase shift = 0
V2
R2 || 1 jC2

Vo R1  1 jC1  R2 || 1 jC2 
If R1  R2  R and C1  C2  C ,
V2
RC

Vo 3RC  j ( 2 R 2C 2  1)
(2) : 02 R 2C 2  1  0  0 

1
RC
V2 1
 for   0
Vo 3
1  R f
(1) : 1 
3  Rg

  1  R f  2 Rg


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