Transcript Thevenin & Norton Equivalent

ECE 221
Electric Circuit Analysis I
Chapter 13
Thévenin & Norton Equivalent
Herbert G. Mayer, PSU
Status 11/3/2015
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Syllabus
 Motivation
 Thought Experiment
 Purpose
 Thévenin Problem
 Norton Equivalent
 A Sample
 Thévenin vs. Source Transformations
 Thévenin With Dependent Sources
 Exercise 1
 References
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Motivation
 When working with real electric sources, such as
typical household power supplies, the actual circuit
behind the terminals is unknown
 It should be unknown! Unreasonable to expect users
to know what exactly is behind the electric terminal!
 We only know 1.) that there is a source of constant
voltage at terminals a and b with a complex but finite
resistance internally
 And 2.) that the current supplied depends on external
load, up to some practical limits
 When the limit is exceeded, a fuse flips and we lose
power 
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Motivation
 It is desirable to understand the physical limits of
such a CVS
 Once we know the limits, we can model the whole
complex CVS source with an equivalent, but simpler
alternative
 One such model is the Thévenin equivalent, an
imaginary CVS with identical behavior
 Named after 19th century French telegraph engineer
Léon Charles Thévenin, 1857-1926
 Such a Thévenin equivalent source consists of CVS
with vTh Volt and an internal resistance of RTh Ω in
series; and nothing else!
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Thought Experiment
We only know 1.) that the black (green) box contains
constant voltage- and current sources; 2.) its voltage at
the terminals; 3.) that the current is finite, limited by
some practical maximum
How to model this situation? Answered by the
Thévenin Equivalent
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Thought Experiment
We find the Thévenin parameters, modeling the green
box, by practicing the following experiments:
1. Leaving the circuit open at terminals a and b, allows
measuring the voltage vTh with no load, i.e. load with
resistance RL = ∞
Measuring the voltage yields vTh
2. Short-circuiting the terminals a and b allows us to
measure the maximum load current iSC
Measuring that short-circuit current yields iSC
3. Thus we can compute RTh
RTh = vTh / iSC
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Conclusion: Thévenin Equivalent
A Thévenin Equivalent circuit is a simple, CVS
circuit with a serial resistor, RTH . . .
. . . electrically equivalent to an arbitrary linear
circuit, whose key parameters, idle voltage and
short-circuit current are measurable
Such a circuit’s equivalent resistance is:
RTh = vTh / iSC
A sample Thévenin transformation follows,
taken from [1], p. 113-115
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Sample Thévenin Problem
Looking at the terminals (a) and (b) from the right, we pretend
not to know what’s behind them; only that vab is delivered.
Measure 2 key parameters; in the end replace what’s there with
the Thévenin equivalent:
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Sample Thévenin Problem
 The following circuit has various internal sources
and resistances, 2 external terminals (a) and (b)
 Goal is to model the exact behavior of this circuit via
the RTh and iSC equivalents
 First measure idle voltage VTh at the terminals, here
called vab, then measure the short-circuit current iSC
 Thus we can compute the resulting equivalent
resistance RTh: RTh = vTh / iSC
 With terminals open, no current flows in 4 Ω resistor
 We compute v1, parallel to the CCV and the 20 Ω
resistor
 Note: with open terminals, the 4 Ω resistor in this
circuit might as well be missing! The voltage drop
along the 4 Ω resistor is 0 V
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Thévenin Problem: Open Terminals
First experiment: to measure voltage at
open terminals = vab = v1 identical to vTH
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Thévenin Problem: Open Terminals
(v1 - 25)/5 + v1/20 - 3 = 0
Students compute v1 = vTH!
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Thévenin Problem: Open Terminals
(v1 - 25)/5 + v1/20 - 3 = 0
4*v1 -100 -60 + v1
= 0
5*v1
= 160
v1
= vab
= vTh
= 32 V
vTh = 32 V
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Thévenin Problem: Short-Circuit
 In the next experiment terminals (a) and (b)
are short-circuited
 A real current flows through the 4 Ω R, now
parallel to the 3 A CCS
 We compute v2, the voltage along the 3 A
CCS, using the NoVoMo
 Once v2 is known, then all other units,
specifically iSC can be computed
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Sample Thévenin Problem
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Thévenin Problem: Short-Circuit
v2/20 + (v2-25)/5 - 3 + v2/4
Students Compute v2
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= 0
Thévenin Problem: Short-Circuit
v2/20 + (v2-25)/5 - 3 + v2/4
= 0
v2 + 4*v2 + 5*v2
= 160
10*v2
= 160
v2
= 16 V
vTH = 32 V
iSC = v2/4 = 16/4 = 4 A
RTh = vTh/iSC = 32/4 = 8 Ω
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Final Thévenin Equivalent Circuit
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Short-Cut For Thévenin Resistance
A simpler way to compute the Thévenin Equivalence
resistance for a linear circuit with only CCS, CVS,
and resistances --does not work with dependent
sources!:
Open all CCS, and short-circuit all CVS, compute the
resulting resistance REQ which is the final RTH
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Short-Cut For Thévenin Resistance
All CCS are left open, all CVS are short-circuited,
resulting equivalent R = 8 Ω
20 Ω and 5 Ω in parallel = 4 Ω, plus 4 Ω in series = 8 Ω
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Norton Equivalent
 Method 1: Earlier we covered transforming some
CVS into an equivalent CCS
 Method 2: The Thévenin transformation, introduced
here, yields a CVS with R in series
 We can combine the 2 methods and generate a CCS
equivalent to any CVS-circuit by just measuring idle
voltage and short-circuit current
 The equivalent Norton current iN is 4 A, with the 8 Ω
resistor in parallel:
iN = iTh = 32/8 = 4 A
iN = 4 A
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Norton Equivalent Sample for RTH = RL
VLCVS
= 32 * 8 / ( 8 + 8 ) = 16 V
iLCVS
= 32 / ( 8 + 8 )
=
2 A
iLCCS
= 4 * 8 / ( 8 + 8 )
=
2 A
VLCCS
= 4 * 8 / 2
= 16 V
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Thévenin- vs. Source Transformation
Instead of using Thévenin equivalent, use successive source-tosource transformations to reduce a complex circuit into one as
simple as Thévenin equivalent. What will it look like?
Transformation 1: starting on the left side, substitute 25 V CVS
with R in series into equivalent CCS with R in parallel:
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Thévenin- vs. Source Transformation
Students, transform 25 V CVS with 5 Ohm in series into
equivalent CCS with 5 Ohm in parallel
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Source Transformation 1
Transformation 1: as a result, we have 2 resistors or 5 Ohm and 20
Ohm and 2 CCS in parallel; can be combined trivially
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Source Transformation 2
Combined 20 Ω and 5 Ω into an equivalent 4 Ω. Combined 5 A and
parallel 3 A CCS into an equivalent 8 A CCS.
Now convert back into CVS:
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Source Transformation 3
As a result, we have two 4 Ω resistors in series. Trivial
transformation. See next simple replacement!
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Source Transformation 4
Simple transformation, two 4 Ω become one 8 Ω
Now convert CVS back into equivalent CCS
Happens  to be Thévenin Equivalent
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Source Transformation 4
Students, transform 32 V CVS and 8 Ohm in series into
equivalent CCS with 8 Ohm in parallel
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Source Transformation 5
Final transformation shows that Thévenin equivalent at terminals
(a) and (b) is the same as source-to-source transformation down
to minimal number of components!
Thévenin equivalent is: vTH = 32 V, iSC = 4 A, and RTH = 8 Ω
With CCS, happens  to be the Norton Equivalent
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Thévenin Equivalent
With
Dependent Sources
(Taken from [1])
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Find Thévenin: Dependent Sources
Consider circuit C1 below: what is the Thévenin Equivalent at
terminals (a) and (b)?
Can it be simplified by totally omitting the left part, since ix = 0 ?
To answer that we exercise the Thévenin transformation: open
terminals and short-circuited terminals (a) and (b)
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Find Thévenin: Dependent Sources
Can circuit C1 be simplified by omitting left part, since ix = 0 A?
Clearly not! See C2, which is NOT an equivalent option, since
both parts, left and right, depend on electric units of their
mutual other half
The dependent current source is a function of i namely 20 * i,
flowing through the 2 kΩ resistor on the left part
And the dependent voltage source is a function of voltage vab
along the 25 Ω resistor, namely 3*v = 3*vab
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Find Thévenin: Dependent Sources
Is C3 equivalent to C1? Yes, since ix = 0 A, but circuit analysis
needs both parts due to mutual dependencies
Goal to find Thévenin Equivalent of circuit C1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b)
open
Step 2: Compute current iSC with terminals (a) and (b) shortcircuited
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Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b)
open:
i = ( 5 – 3 * v ) / 2000
=
( 5 – 3 * vTH ) / 2000
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Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b)
open:
i
= ( 5 – 3 * v ) / 2000
AKA
i
v
= -20 * 25 * i
AKA
vTH = -500 * i
=
( 5 – 3 * vTH ) / 2000
Students, compute v = vTH
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Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b)
open:
i
= ( 5 – 3 * v ) / 2000
AKA
i
v
= -20 * 25 * i
AKA
vTH = -500 * i
=
( 5 – 3 * vTH ) / 2000
vTH = -500 * ( 5 – 3 * vTH ) / 2000= -5 * ( 5 – 3 * vTH ) / 20
vTH = -25 / 20 + vTH * 15 / 20
-- multiply by * 20
20 * vTH = -25 + vTH * 15
5 * vTH = -25
vTH
= -5 V
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Thévenin Equivalent Step 2
Step 2: Compute i = iSC with terminals (a) and (b) short-circuited
Short-circuit bypasses 25 Ω resistor; that means the voltage
controlling the dependent voltage source is also shortcircuited, i.e. the voltage drop is 0 V. That dependent source
can be removed.
Note that the 20 * i and iSC run in opposite directions! Sign!!
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Thévenin Equivalent Step 2
Step 2: Compute i = iSC with terminals (a) and (b) short-circuited
Note that the 20 * i and iSC run opposite directions!
i
=
iSC =
5 / 2000
=
2.5 mA
- 20 * i
iSC = -50 mA
With iSC
RTH
= -50 mA,
vTH = -5 V we get
= - 5 / -50 * 1000 [ V / A ]
RTH = 1/10 k Ω = 100 Ω
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Thévenin Equivalent With Dependent
Below we have C5, the Thévenin Equivalent circuit of the
original C1 circuit
With dependent sources the convenient short-cut (p. 19 above)
for computing RTH does not work!
Note reversed polarity of 5 V CVS in C5!
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Norton Equivalent of C1
The Norton Equivalent has the 100 Ω resistor parallel to the
constant current source of 5 V / 100 Ω = 50 mA
Note the direction of the CCS, the tip pointing to the + sign of
the equivalent CVS!
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Exercise 1
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Exercise 1: Thévenin Equivalent
 Given the circuit below, generate the
Thevénin equivalent
 First leave terminals (a) and (b) open and
compute vTH
 Then short-circuit terminals (a) and (b), and
compute iSC
 Yielding the Thevénin equivalent vTH, iSC,
including resistance RTH
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Exercise 1 Open: Thévenin Equivalent
With open plugs, the voltage drop vTH at terminals (a) and (b) is
the same as along the CCS with 4 A
CCS pushes 4 A through 6 Ω resistor, creating 4 * 6 = 24 V
voltage drop
Ohms Law: yields 6 * 4 = 24 V:
vTH = 24 + 12 = 36 V
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Exercise 1 SC: Thévenin Equivalent
With terminals (a) and (b) short-circuited, iSC runs through the
14 Ω resistor, and i6 through the 6 Ω resistor
KCL:
iSC – 4 + i6
=
0
KVL:
14 * iSC – 12 – 6 * i6 =
0
iSC
1.8 A
RTH
=
= vTH / iSC
=
44
36 / 1.8 = 20 Ω
Exercise 1: Thévenin Equivalent VTH
Easier to use the Node Voltage Methodology to compute iSC:
KCL:
v / 14 – 4 + ( v – 12 ) / 6
=
0
v
=
14*36 / 20
iSC = v / 14
=
14 * 36 / ( 20 * 14 )
iSC
=
36 / 20 = 1.8 A
Yields iSC:
RTH
= vTH / iSC
=
45
36 / 1.8 = 20 Ω
Exercise 1: A Simpler Way
To identify RTH, all we need to do is short-circuit all CVSs,
eliminate all CCSs, and compute the equivalent REQ, which is RTH
This just leaves the 2 resistors in series, resulting RTH = 20 Ω
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Finally: Thévenin Equivalent
The circuit below shows the equivalent Thévenin arrangement
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References
1. Electric Circuits, James W. Nielsson and
Susan A. Riedel, Pearson Education Inc.,
publishing as as Prentice Hall, © 2015, ISBN13: 978-0-13-376003-3
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