CIRCUITS AND
6.002 ELECTRONICS
Basic Circuit Analysis Method
(KVL and KCL method)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Review
Lumped Matter Discipline LMD:
Constraints we impose on ourselves to simplify
our analysis
∂φ B Outside elements
=0
∂t
∂q Inside elements
=0
∂t
wires resistors sources
Allows us to create the lumped circuit
abstraction
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Review
LMD allows us to create the
lumped circuit abstraction
i
+
v Lumped circuit element
-
power consumed by element = vi
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Review
Review
Maxwell’s equations simplify to
algebraic KVL and KCL under LMD!
KVL:
∑ jν j = 0
loop
KCL:
∑jij = 0
node
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Review
a
R1 R4
R3
+ b d
–
R2 R5
c
vca + vab + vbc = 0 KVL
DEMO
ica + ida + iba = 0 KCL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
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
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Method 1: Basic KVL, KCL method of
Circuit analysis
Element Relationships
R
For R, V = IR
For voltage source, V = V0 +–
V0
For current source, I = I 0
Io
3 lumped circuit elements
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
KVL, KCL Example
a
+ +
ν1 R1 ν4 R4
– R3 –
+ + b
ν 0 = V0 – d
– +ν 3 –
+ +
ν2 R2 ν5 R5
– –
c
The Demo Circuit
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Associated variables discipline
i
+
ν Element e
-
Current is taken to be positive going
into the positive voltage terminal
Then power consumed = νi is positive
by element e
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
KVL, KCL Example
a
i1 L 2 i4
+ +
ν1 R1 ν 4 R4
i0 – R3 –
+ + L1 b i3
ν 0 = V0 – d
– i2 +ν 3 – i5
+ +
ν2 R2 ν 5 R5
– L3 –
c
L4
The Demo Circuit
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Analyze
ν 0 …ν 5 ,ι0 …ι5 12 unknowns
1. Element relationships (v, i )
v0 = V0 given v3 = i3 R3 6 equations
v1 = i1 R1 v4 = i4 R4
v2 = i2 R2 v5 = i5 R5
2. KCL at the nodes
a: i0 + i1 + i4 = 0 3 independent
b: i2 + i3 − i1 = 0 equations
d: i5 − i3 − i4 = 0
e: − i0 − i2 − i5 = 0 redundant
3. KVL for loops
L1: − v0 + v1 + v2 = 0 3 independent
L2: v1 + v3 − v4 = 0 equations
L3: v3 + v5 − v2 = 0 s
on
L4: − v0 + v4 + v5 = 0 redundant ati
u owns
eq u nkn
1 2 12
ugh @#!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
R1 R2 R3 RN R1 + R2 + + RN
A … ⇔
B G1 G2 GN ⇔ G1 + G2 + GN
1
Gi =
Ri
V1 V2 V1 + V2
C +– +– ⇔ +–
D
I1 I2 ⇔ I1 + I 2
Surprisingly, these rules (along with superposition, which
you will learn about later) can solve the circuit on page 8
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
Example I =?
V + R1
–
R2 R3
I I
R1
V +
– V +
– R
R2 R3
R2 + R3
R2 R3
R = R1 +
R2 + R3
V
I=
R
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Method 3—Node analysis
Particular application of KVL, KCL method
1. Select reference node ( ground)
from which voltages are measured.
2. Label voltages of remaining nodes
with respect to ground.
These are the primary unknowns.
3. Write KCL for all but the ground
node, substituting device laws and
KVL.
4. Solve for node voltages.
5. Back solve for branch voltages and
currents (i.e., the secondary unknowns)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Example: Old Faithful
plus current source
V0
R1 R R4
3 e2
+ V e1
– 0
R2 R5 I1
Step 1
Step 2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Example: Old Faithful
plus current source
V0
R1 R R4
3 e2
+ V e1
– 0
for
R2 R5 I1 convenience,
write
1
Gi =
Ri
KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0
KCL at e2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
Step 3
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Example: Old Faithful
plus current source
V0
R1 R R4
3 e2
+ V e1
– 0
R2 R5 I1
1
Gi =
Ri
KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0
KCL at l2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
move constant terms to RHS & collect unknowns
e1 (G1 + G2 + G3 ) + e2 (−G3 ) = V0 (G1 )
e1 (−G3 ) + e2 (G3 + G4 + G5 ) = V0 (G4 ) + I1
2 equations, 2 unknowns Solve for e’s
(compare units) Step 4
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
In matrix form:
⎡G1 + G2 + G3 − G3 ⎤ ⎡ e1 ⎤ ⎡ G1V0 ⎤
⎢ = ⎢
⎣ − G3 G3 + G4 + G5 ⎥ ⎢e2 ⎥
⎦ ⎣ ⎦
⎥
⎣G4V0 + I1 ⎦
conductivity unknown sources
matrix node
voltages
Solve
⎡G3 + G4 + G5 G3 ⎤ ⎡ G1V0 ⎤
⎡ e1 ⎤ ⎢
⎣ G3 G1 + G2 + G3 ⎥ ⎢G4V0 + I1 ⎥
⎦ ⎣ ⎦
⎢e ⎥ = (G1 + G2 + G3 )(G3 + G4 + G5 ) − G3 2
⎣ 2⎦
e =
G +G +G G V + G G V + I
3 4
(
5 1 0 3 4 0 1
)( ) ( )( )
1 G G +G G +G G +G G +G G +G G +G 2 +G G +G G
1 3 1 4 1 5 2 3 2 4 2 5 3 3 4 3 5
(G3 )(G1V0 ) + (G1 + G2 + G3 )(G4V0 + I 1 )
e2 =
G1G3 + G1G4 + G1G5 + G2G3 + G2G4 + G2 G5 + G3 + G3G4 + G3G5
2
(same denominator)
Notice: linear in V0 , I1 , no negatives
in denominator
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000 Lecture 2
Solve, given
G1 ⎫ 1 G2 ⎫ 1 1
⎬= ⎬= G3 =
G5 ⎭ 8.2 K G4 ⎭ 3.9 K 1.5 K
I1 = 0
G G V + G +G +G G V + I
e = 3 10 1 2 3 40 1
( )( )
1 2 3
(
2 G + G + G + G + G + G −G 2
3 4 5 3
)( )
1 1 1
G +G +G = + + =1
1 2 3 8.2 3.9 1.5
1 1 1
G3 + G4 + G5 = + + =1
1.5 3.9 8.2
1 1 1
× + 1×
e2 = 8.2 1.5 3.9 V
0
1
1− 2
1.5
Check out the
e2 = 0.6V0 DEMO
If V0 = 3V , then e2 = 1.8V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000 Lecture 2