6.002x CIRCUITS AND ELECTRONICS - courses.edx.orgMITx+6.002.1x_1+2... · 1 Basic Circuit Analysis...
Transcript of 6.002x CIRCUITS AND ELECTRONICS - courses.edx.orgMITx+6.002.1x_1+2... · 1 Basic Circuit Analysis...
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Basic Circuit Analysis Methods (KVL and KCL method, Node method)
6.002x CIRCUITS AND ELECTRONICS
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Remember, our EECS playground
Review
Observe the lumped matter discipline LMD
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Lumped circuit element +
-‐
power consumed by element =
Review
i
v
vi
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LMD allows us to create the lumped circuit abstraction
Review
+!–! R1
R3 R2
V
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KVL: For all loops
KCL: For all nodes
0=∑ j jν
0=∑ j ji
Review Review
Maxwell’s equations simplify to algebraic KVL and KCL under LMD!
tdlE B
∂∂
−=⋅∫φ
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DEMO
+!–!
Review
R1 R4
R3
R2 R5
d
c
b
a
V0
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Let’s Begin by Building aToolchest of Analysis Techniques
+!–!
R1 R4
R3
R2 R5
Analyzing a circuit means:
Find all the element v’s and i’s
V0
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Method 1: Basic KVL, KCL method of Circuit analysis
Goal: Find all element v’s and i’s
1. write element v-i relationships (from lumped circuit abstraction) 2. write KCL for all nodes 3. write KVL for all loops
lots of unknowns lots of equations lots of fun solve
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i +
-
Element e
Current is taken to be positive going into the positive voltage terminal
Method 1: Basic KVL, KCL method of Circuit analysis
Then power consumed by element e
is positive = vi
This convention is called: Associated variables discipline
Goal: Find all element v’s and i’s Labeling element v’s and i’s
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Method 1: Basic KVL, KCL method of Circuit analysis
For R For voltage source For current source
You will need this for step 1: Element Relationships
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Let’s Apply KVL, KCL Method to this Example
The Demo Circuit
+!–!
R1 R4
R3
R2 R5
V0
Goal: Find all element v’s and i’s
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KVL, KCL Example
a
d
c
+!–!
Note the use of associated variables…
R4
R5 R2
R1
R3 b V0
Label all v’s and i’s Goal: Find all element v’s and i’s
5050 , ii ……νν12 unknowns
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Step 1 of KVL, KCL Method 12 unknowns
1. Element relationships ( )iv, 5id
c
1ν+ –
5ν+ –
+ – 2ν+ –
4ν+ – 0i
4i
+!–!
R4
R5 R2
R1
R3 b
v3 V0
i1
i3
i2 0ν+ –
a
5050 , ii ……ννL2
L3
L1
L4
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Step 2 of KVL, KCL Method 12 unknowns
2. KCL at the nodes
(use convention, e.g., sum currents leaving the node)
5050 , ii ……νν5id
c
1ν+ –
5ν+ –
+ – 2ν+ –
4ν+ – 0i
4i
+!–!
R4
R5 R2
R1
R3 b
v3 V0
i1
i3
i2 0ν+ –
a
L2
L3
L1
L4
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Step 3 of KVL, KCL Method 12 unknowns
3. KVL for loops
(use convention, e.g., as you go around loop, assign first encountered sign to each voltage)
5050 , ii ……νν5id
c
1ν+ –
5ν+ –
+ – 2ν+ –
4ν+ – 0i
4i
+!–!
R4
R5 R2
R1
R3 b
v3 V0
i1
i3
i2 0ν+ –
a
L2
L3
L1
L4
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KVL, KCL Method
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ugh @#!
1. Element v, i relationships v0 = V0
v1 = i1R1
v2 = i2R2
v3 = i3R3
v4 = i4R4
v5 = i5R5
2. KCL at the nodes
redundant
0410 =++ iii0132 =−+ iii0435 =−− iii0520 =−−− iii
a: b: d:
c:
3. KVL for loops
0431 =−+ vvv0210 =++− vvv
0253 =−+ vvv0540 =++− vvv redundant
L1:
L2:
L3: L4:
Method 3 – the node method will be much better!
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Other Analysis Methods Method 2— Apply element combination rules
A …R1 R2 R3 RN
B G2 G1 GN
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Method 2 — Apply element combination rules
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Method 2— Apply element combination rules
+!–!
Example
R1
R3 R2
V
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1. 2. 3. 4. 5.
Select reference node ( ground) from which voltages are measured. Label voltages of remaining nodes with respect to ground. These are the primary unknowns. Write KCL for all but the ground node, substituting device laws and KVL. Solve for node voltages. Back solve for branch voltages and currents (i.e., the secondary unknowns).
Particular application of KVL, KCL method
Method 3 — Node analysis
6.002x workhorse!
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Example: Old Faithful, plus current source
+!–!
R1 R4 R3
R2 R5
V0
Method 3 — Node analysis
1. Select reference ground node
2. Label node voltages with respect to ground.
g
I1
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Step 3 of Node Method
For convenience, write i
i RG 1
=
To avoid mistakes, use convention – E.g., always sum the currents leaving a node
g
+!–!
I1
e1
R4 R3 R1
R2 R5
e2 V0
V0
3. Write KCL for nodes, substituting device laws and KVL.
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Step 4 of Node Method
Move constant terms to RHS & collect unknowns
2 equations, 2 unknowns Solve for e’s (compare units)
0)()()( 21321101 =+−+− GeGeeGVe
0)()()( 152402312 =−+−+− IGeGVeGee
KCL at e1
KCL at e2
4. Solve for node voltages
g
+!–!
I1
e1
R4 R3 R1
R2 R5
e2 V0
V0
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Step 5 of Node Method
5. Back solve for branch voltages and currents
g
+!–!
I1
e1
R4 R3 R1
R2 R5
e2 V0
V0
e1 e2 Once you have solved for and , easy to find branch v’s and i’s
For example:
i1 + – v1
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In matrix form:
⎥⎦
⎤⎢⎣
⎡+
=⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡++−
−++
104
01
2
1
5433
3321
IVGVG
ee
GGGGGGGG
conductivity matrix
unknown node
voltages sources
Revisit Step 4 of Node Method for Cultural Interest )()()( 10323211 GVGeGGGe =−+++
140543231 )()()( IGVGGGeGe +=+++−
4. Solve for node voltages
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⎥⎦
⎤⎢⎣
⎡+
=⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡++−
−++
104
01
2
1
5433
3321
IVGVG
ee
GGGGGGGG
( )( ) 23543321
104
01
3213
3543
2
1
GGGGGGGIVG
VGGGGG
GGGG
ee
−++++
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡++
++
=⎥⎦
⎤⎢⎣
⎡
Solve
( )( ) ( )( )5G3G4G3G
23G5G2G4G2G3G2G5G1G4G1G3G1G1I0V4G3G0V1G5G4G3G
1e ++++++++
++++=
( )( ) ( )( )5343
23524232514131
1043210132 GGGGGGGGGGGGGGGGG
IVGGGGVGGe++++++++
++++=
(same denominator)
Notice: linear in , , no negatives in denominator – we will use this later
V0 I1
Step 4 of Node Method 4. Solve for node voltages
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E.g., solve for , given
K2.81
GG
5
1 =⎭⎬⎫
K9.31
GG
4
2 =⎭⎬⎫
K5.11G3 = 01 =I
02 6.0 Ve =
If , then VV 30 = 02 8.1 Ve =
Step 4 of Node Method
g
+!–!
I1
e1
R4 R3 R1
R2 R5
e2 V0
V0
e2