6.002x CIRCUITS AND ELECTRONICS · 2013. 10. 17. · Method 3: Piecewise Linear Method See Sec 4.4...
Transcript of 6.002x CIRCUITS AND ELECTRONICS · 2013. 10. 17. · Method 3: Piecewise Linear Method See Sec 4.4...
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Nonlinear Circuits
6.002x CIRCUITS AND ELECTRONICS
Reading Chap 4.1 – 4.3
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n Discretize matter à Lumped circuit abstraction
m1 u KVL, KCL, i-v m2 u Composition rules m3 u Node method m4 u Superposition m5 u Thévenin, Norton
Review
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n Discretize value à Digital abstraction u Subcircuits for given “switch” setting are linear! So, all 5 methods (m1 – m5) can be applied
Use SR MOSFET Model
Review
C
Vs
RL
A B
Vs
RL
C
RON RON A=1 B=1
Vs
RL
RON
C
A=1 B=0
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Today n Nonlinear circuits and their analysis
u Analytical method based on m1, m2, m3
u Graphical method
u Piecewise linear method
u Introduction to incremental analysis
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Non-Linear Elements A square law two-terminal element
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Non-Linear Elements Hypothetical nonlinear device (ExpoDweeb J)
(Curiously, this funky device supplies power when vD is negative!)
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How do we analyze nonlinear circuits + -
iD
vD
iD
DbvaeDi =
a
vD 0,0
For example:
D
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Quick Aside: Note this Key Circuits Hack I claim this is a very important circuit pattern! Why?
If you know how to analyze the above circuit pattern, you can analyze any of these more complex circuits below
+ -
+ –
+
-
i
iD
D vD v
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Quick Aside: Note this Key Circuits Hack Thevenin Trick
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Method 1: Analytical Method Using the node method, (remember the node method applies for linear or nonlinear circuits)
+ – D
vD iD
V
R
V
DbvaeDi =
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Solve by trial and error
0=+− DD bvaeRVv
DD
vev411−=
E.g., for V=1V, R=1 Ohm, a=1/4 A, b=1V-1
1V
0.32V 0.65V 0.52V
0.58V
0.56V 0.55V
0.56V
Corresponding Di is 0.44A
0=+−
DD iRVv
1
DbvD aei = 2
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Method 2: Graphical Method Using the node method
0=+−
DD iRVv
1
Can also solve by the graphical method
DbvD aei = 2
+ – D
vD iD
V
R
V
DbvaeDi =
2 unknowns, 2 equations
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Method 2: Graphical Method Notice: the solution satisfies equations and 2 1’
Constraint on on iD and vD imposed by the rest of the circuit (Thevenin equivalent circuit)
Constraint on iD and vD imposed by device
2
0 vD
iD
a
1’
0 vD
iD
DbvD aei = 2
Rv
RVi D
D −= 1’
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Combine the two constraints
0 vD
iD
E.g., for V=1V, R=1 Ohm, a=1/4 A, b=1V-1 Dbv
D aei = 2 Rv
RVi D
D −= 1’
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2
0 vD
iD
a
Combining the two constraints Dbv
D aei =Rv
RVi D
D −=
2 1’
1’
0 vD
iD
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Method 3: Piecewise Linear Method
See Sec 4.4 of the text for details and examples
Insight:
+ -
Determine which of the linear regions applies (e.g., region B) Solve circuit assuming that linear relation (from region B)
iD
vD iD
a
vD 0,0
D
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Piecewise Linear Method
+ -
+ –
iD
vD
For example:
V D Linear Region A
Linear Region B
iD
vD 0,0
Suppose we are given that V is positive
Then D will operate in Linear Region B We can replace D in the circuit with a resistance of value 1/R when D operates in region B
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Method 3: Incremental Analysis (Actually, a disciplined way of using a circuit called small signal method) Motivation: music over a light beam. Can we pull this off?
Remember LED: Light Emitting expoDweep J
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Method 3: Incremental Analysis
+ -
+ – LED AMP vD
iR iD
vI (t)
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Problem: The LED is nonlinear à distortion
vD
t
t
vD
DEMO
t
iD
iD
vD
vD = vI
iD
+
- + – LED AMP vD
iR iD vI
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If only the LED were linear …
it would’ve been ok.
What do we do? Zen is the answer … next sequence!
vD
t
iD iD
vD
+
- + – LED AMP vD
iR iD vI