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EECS 40 Fall 2002
Copyright Regents of University of California
S. Ross and W. G. Oldham
LECTURE 14
Today:
• Summary of Linear and Nonlinear Rules
• Load-line graphical method for circuits with
nonlinear elements
• Three terminal devices, solution methods and power
• Unsolicited Advice: Make learning your priority
1
EECS 40 Fall 2002
Copyright Regents of University of California
S. Ross and W. G. Oldham
Art, Michelle and Rob
(Be Like Michelle)
Reduce or eliminate:
• Over-focusing on policy and grades
– Better off devoting energy towards understanding
• Perfectionism
– Don’t expect perfection of yourself or your instructors
• Competition
– Cooperation, helping others benefits you too
Big don’ts for success at school and at work
2
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Circuit Laws and Tricks
Linear Circuits:
Nonlinear Circuits:
• KCL, KVL
• KCL, KVL
• Nodal Analysis
• Nodal Analysis
• Thévenin and Norton
equivalents
Special Graphical Technique for Circuits containing
linear and nonlinear elements (Load-line Method):
• Combine all linear parts into a simple Thévenin
equivalent Circuit attached to the nonlinear element.
• Plot the I-V Characteristics of NLE and Thévenin
circuit on same axes, recognizing the intersection as
the solution.
3
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Example of Load-Line Method
We have a circuit containing a two-terminal non-linear element
“NLE”, and some linear components.
First replace the entire linear part of the circuit by its Thevenin
equivalent.
Then define I and V at the NLE terminals (typically associated signs)
D
Nonlinear
9mA
element
S
D
250K 1M
+
-
1V
N
L
E
ID
+
VDS + 200K
2V
S
-
4
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Example of Load-Line method (con’t)
Given the graphical properties of two terminal non-linear circuit
(i.e. the graph of a two terminal device)
And have this connected to a linear
(Thévenin) circuit
D
N
L
E
Whose I-V can also be graphed
on the same axes (“load line”)
+
VDS + 200K
2V
S
-
Application of KCL, KVL gives circuit solution
ID (mA)
D ID
N
L
E
10
The solution !
200K
S
ID
+
-
2V
1
2
VDS (V)
5
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Load-Line method
The method is graphical, and therefore approximate
But if we use equations instead of graphs, it could be accurate
It can also be use to find solutions to circuits with three terminal
nonlinear devices (like transistors)
Application of KCL, KVL gives circuit solution
ID (mA)
D ID
10
The solution !
200K
S
+
-
2V
1
2
VDS (V)
6
EECS 40 Fall 2002
Copyright Regents of University of California
S. Ross and W. G. Oldham
Three-Terminal Devices
ID
G
3-Terminal
Device
D
S
With three terminals we have four independent variables (two
voltages and two currents)
We can set one variable at each terminal and the I-V behavior of the
device will determine the value of the two unknowns.
For example apply a VGS and VDS, then measure IG and ID.
Similarly we could apply a fixed VGS and IG and plot the
two-terminal characteristic ID versus VDS.
7
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Three-Terminal Parametric Graphs
ID
3-Terminal
Device
G
VGS
ID (mA)
D
10
+
-
VGS = 3
VGS = 2
VGS = 1
S
Concept of 3-Terminal Parametric Graphs:
1
2
VDS (V)
We set a voltage (or current) at one set of
terminals (here we will apply a fixed VGS, IG=0)
and conceptually draw a box around the device
with only two terminals emerging so we can
again plot the two-terminal characteristic
(here ID versus VDS).
But we can do this for a variety of values of VGS
with the result that we get a family of curves.
8
EECS 40 Fall 2002
S. Ross and W. G. Oldham
Copyright Regents of University of California
Graphical Solutions for 3-Terminal Devices
ID
G
V
S
+
-
+
-
ID (mA)
D
200K
10
VGS = 3
2V
VGS = 2
VGS = 1
We can only find a solution for
one input (VGS) at at time:
First select VGS (e.g. 2V) and draw
ID vs VDS for the 3-Terminal device.
Now draw ID vs VDS for the 2V 200KW Thevenin source.
ID (mA)
1
10
VDS (V)
2
The solution !
The only point on the I vs V
plane which obeys KCL and
KVL is ID = 5mA at VDS = 1V.
1
2
VDS (V)
9
EECS 40 Fall 2002
Copyright Regents of University of California
S. Ross and W. G. Oldham
Three-Terminal Devices – Power Flow
IG
G
ID
3-Terminal
Device
D
S
With three terminals we have two independent measures of
power (e.g. one at GS and one at DS).
We can set any two variables and the I-V behavior of the device will
determine the value of the other two.
Defining all currents as inward, the power dissipated is simply the
algebraic sum of VGS IG and VDS ID, i. e.,
P (in) = VGS IG + VDS ID .
See the textbook, p. 117 for more generality.
10
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