Week 2 Handout
Remaining Schedule:
6/25 - Week 2:
Thevenin and Norton Equivalents, Superposition Techniques
7/2 Week 3:
Node-Voltage, Mesh-Current,
7/9 Week 4:
Transistors and Op-Amps (Ch 4)
7/16 - Week 5:
Capacitors, Inductors, DC analysis (Ch 6)
7/23 - Week 6:
1st, 2nd order transient analysis (Ch 7)
7/30 - Week 7:
AC analysis (phasors)
8/6 Week 8:
Power Balance
8/13Week 9:
Two-port networks, Transfer functions, s-domain,
8/20Week 10: Filters
9/10 - Final Meeting: Review Practice Test 2 (S08), F08, and any other questions you
have
Today’s topic:
Thevenin and Norton Equivalents, and Superposition Techniques
Superposition Principle
A circuit is linear if it can be adequately modeled using only linear elements and
independent sources (i.e. a Thevenin equivalent exists). Linear circuits have a number of
important properties, which are detailed in section 3-3 (pg 95) in T&R. One of these
properties is Additivity. Additivity means that the output due to two or more inputs can
be found by adding the outputs obtained when each input is applied separately. That is:
f ( x1  x 2 )  f ( x1 )  f ( x 2 )
The superposition principle is a circuit analysis technique that applies additivity to
determine the output of a circuit.
Applying the Superposition Principle
1. Turn off all independent sources (current sources = open circuit, voltage sources =
short circuit) except one, and determine the output from that source acting alone
2. Repeat step 1 for all independent sources
3. The total output with all independent sources turned on will be the sum of the
outputs of each source acting alone
Thevenin and Norton Equivalent Circuits
Any linear circuit can be replaced by a Thevenin or Norton equivalent circuit at a
specified interface. The equivalent circuit produces the same interface signals (v and i) at
the interface as the original circuit
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The parameters of the Thevenin and Norton equivalent circuits at a given interface can be
determined by finding the open-circuit voltage and short-circuit current, as shown below:
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Thevenin and Norton Equivalent Circuits with dependant sources
To find the equivalent of an active circuit, we cannot turn off all of the sources unless we
supply excitation from an external test source. This leaves two solution methods:
1. Determine the open-circuit voltage and short-circuit current with all of the sources
on
2. Turn off all of the independent sources, introduce a test source, and determine the
relationship between the current and voltage to find the Thevenin Resistance:
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Today’s Problems:
1
1.
2.
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3.
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4.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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