6.002
CIRCUITS AND
ELECTRONICS
Incremental Analysis
6.002 Fall 2000
Lecture 7
1
Review
Nonlinear Analysis
X Analytical method
X Graphical method
Today
X Incremental analysis
Reading: Section 4.5
6.002 Fall 2000
Lecture 7
2
Method 3: Incremental Analysis
Motivation: music over a light beam
Can we pull this off?
iD
+
vD LED
light
intensity
I D ∝ iD
vI music signal
vI (t ) +
–
iR
t
vI (t )
iD (t )
light
AMP
iR ∝ I R
light intensity IR
in photoreceiver
LED: Light
Emitting
expoDweep ☺
iR (t )
sound
nonlinear
linear
problem! will result in distortion
6.002 Fall 2000
Lecture 7
3
Problem:
The LED is nonlinear
distortion
iD
iD
vD
vD = vI
t
vD
t
iD
6.002 Fall 2000
vD
Lecture 7
t
4
Insight:
iD
small region
looks linear
(about VD , ID)
ID
VD
vD
DC offset
or DC bias
Trick:
vI
vi (t ) +
–
VI
+
–
iD = I D + id
+
vD LED
vD = VD + vd
VI
6.002 Fall 2000
Lecture 7
vi
5
Result
iD
id
ID
vD
VD
6.002 Fall 2000
Lecture 7
vd
very small
6
Result
vD = vI
vd
vD
VD
t
iD
id
iD
~linear!
ID
t
Demo
6.002 Fall 2000
Lecture 7
7
The incremental method:
(or small signal method)
1. Operate at some DC offset
or bias point VD, ID .
2. Superimpose small signal vd
(music) on top of VD .
3. Response id to small signal vd
is approximately linear.
Notation:
iD = I D + id
total
DC
small
variable offset superimposed
signal
6.002 Fall 2000
Lecture 7
8
What does this mean
mathematically?
Or, why is the small signal response
linear?
nonlinear
iD = f (vD )
We replaced
vD = VD + ∆vD
large DC
vd
increment
about VD
using Taylor’s Expansion to expand
f(vD) near vD=VD :
iD = f (VD ) +
+
df (vD )
⋅ ∆vD
dvD vD =VD
1 d 2 f (v D )
2! dvD 2 v
2
⋅ ∆vD + "
D =VD
neglect higher order terms
because ∆vD is small
6.002 Fall 2000
Lecture 7
9
iD ≈ f (VD ) +
constant
w.r.t. ∆vD
d f (v D )
⋅ ∆vD
d vD vD =VD
constant w.r.t. ∆vD
slope at VD, ID
We can write
X : I D + ∆iD ≈ f (VD ) +
d f (v D )
⋅ ∆ vD
d vD vD =VD
equating DC and time-varying parts,
I D = f (VD )
operating point
d f (v D )
∆iD =
⋅ ∆vD
d vD vD =VD
constant w.r.t. ∆vD
so, ∆ iD ∝ ∆vD
6.002 Fall 2000
Lecture 7
By notation,
∆ iD = id
∆ v D = vd
10
In our example,
iD = a e
bv D
From X : I D + id ≈ a e bVD + a e bVD ⋅ b ⋅ vd
Equate DC and incremental terms,
I D = a ebVD
operating point
aka bias pt.
aka DC offset
id = a ebVD ⋅ b ⋅ vd
id = I D ⋅ b ⋅ vd
constant
6.002 Fall 2000
Lecture 7
small signal
behavior
linear!
11
Graphical interpretation
operating point
I D = a ebVD
id = I D ⋅ b ⋅ vd
A
slope at
VD, ID
iD
ID
id
B
VD
operating
point
vd
vD
we are
approximating
A with B
6.002 Fall 2000
Lecture 7
12
graphically
mathematically
now, circuit
We saw the small signal
Large signal circuit:
VI
ID
+
LED VD
-
+
–
I D = a ebVD
Small signal response: id = I D b vd
+ vd -
behaves like:
id
R=
small signal circuit:
vi
+
–
+
vd
-
1
ID b
id
1
I Db
Linear!
6.002 Fall 2000
Lecture 7
13