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Circuits II
EE221
Unit 2
Instructor: Kevin D. Donohue
Review: Impedance Circuit Analysis with
nodal, mesh, superposition, source
transformation, equivalent circuits and
SPICE analyses.
Equivalent Circuits


Circuits containing different elements are
equivalent with respect to a pair of
terminals, if and only if their voltage and
current draw for any load is identical.
More complex circuits are often reduced
to Thévenin and Norton equivalent circuits.
Equivalent Circuit (Example)
Find and compare the voltages and currents
generated in 3 of the following loads across
terminals AB:



open circuit
resistance RL
short circuit
A
Is
Rth
Norton
B
Vs
Rth
A
Thévenin
B
Results - Equivalent Circuit
Voltage Source
Thévenin Circuit
Current Source
Norton Circuit
VAB
Open
I s Rth
Short
RL
0
Is
IAB
0
Is
RL Rth
Rth
Is
RL  Rth
RL  Rth
What is the Norton equivalent
for the Thévenin circuit?
VAB
Open
Short
RL
Vs
IAB
Vs
0
0
Vs
Rth
RL
RL  Rth
Vs
1
RL  Rth
What is the Thévenin equivalent
for the Norton circuit?
Finding Thévenin and Norton Equivalent
Circuits




Identify terminal pair at which to find the
equivalent circuit.
Find voltage across the terminal pair when no
load is present (open-circuit voltage Voc)
Short the terminal and find the current in the
short (short-circuit current Isc)
Compute equivalent resistance as:
Rth = Voc / Isc
Finding Thévenin and Norton Equivalent
Circuits
The equivalent circuits can then be
expressed in terms of these quantities
V
R th  oc
I sc
A
A
Isc
Rth
B
Rth
Voc
B
Source Transformation
The following circuit pairs are equivalent wrt to terminals AB.
Therefore, these source and resistor combinations can be
swapped in a circuit without affecting the voltages and
currents in other parts of the circuit.
A
Is
Rth
B
Rth
A
Rth
Is Rth
Vs
B
B
A
Is
Vs
B
A
Vs
R th
A
A
Is
B
Rth
A
Rth
B
A
Vs
B
B
Source Transformation



Some equivalent circuits can be determined by
transforming source and resistor combinations and
combining parallel and serial elements around a
terminal of interest.
This method can work well for simple circuits with
source-resistor combinations as shown on the
previous slide.
This method cannot be used if dependent sources
are present.
Source Transformation
Example

Use source transformation to find the phasor
value Vˆc
3 k
50
-j3.5 k
6 k

Vˆ
c


Show Vˆc = 2.8  30 V
Nodal Analysis



Identify and label all nodes in the system.
Select one node as a reference node (V=0).
Perform KCL at each node with an unknown
voltage, expressing each branch current in
terms of node voltages.

(Exception) If branch contains a voltage source


One way: Make reference node the negative end of the
voltage source and set node values on the positive end equal
to the source values (reduces number of equations and
unknowns by one)
Another way: (Super node) Create an equation where the
difference between the node voltages on either end of the
source is equal to the source value, and then use a surface
around both nodes for KCL equation.
Example
Find the steady-state value of vo(t) in
the circuit below, if vs(t) = 20cos(4t):
10 
vs
0.1 F
ix
1H
2 ix
Show: v0(t) = 13.91cos(4t - 161.6º)
+
vo
-
0.5 H
Loop/Mesh Analysis


Create loop current labels that include every
circuit branch where each loop contains a unique
branch (not included by any other loop) and no
loops “crisscross” each other (but they can overlap
in common branches).
Perform KVL around each loop expressing all
voltages in terms of loop currents.

If any branch contains a current source,


One way: Let only one loop current pass through source so loop
current equals the source value (reduces number of equations and
unknowns by one)
Another way: Let more than one loop pass through source and set
combination of loop currents equal to source value (this provides an
extra equation, which was lost because of the unknown voltage drop
on current source)
Analysis Example
Find the steady-state response for vc(t) when vs(t)
= 5cos(800t) V
+ vc(t) 3 k
vs(t)
114.86 nF
6 k
Can be derived with mesh or nodal analysis or source transformation:


Vˆc  2.5000 - j1.4434  2.8868   30   vc (t )  2.8868 cos 800 t   V
6

Linearity and Superposition

If a linear circuit has multiple independent sources, then a
voltage or current anywhere in the circuit is the sum of
the quantities produced by the individual sources (i.e.
activate one source at a time). This property is called
superposition.

To deactivate a voltage source, set the voltage equal to
zero (equivalent to replacing it with a short circuit).

To deactivate a current source, set the current equal to
zero (equivalent to replacing it with an open circuit).
Analysis Example
Find the steady-state response for vc(t) when
vs(t) = 4cos(200t) V and is(t) = 8cos(500t) A.
10 mH
4
vs(t)
is(t)
6
5 mF
+
vc(t)
-
Can be derived with superposition:
vc (t )  2.1cos500t  135  .92cos(200t - 94.4) V
SPICE Solution
Steady-State Analysis in SPICE is performed using the
.AC (frequency sweep) option in the simulation set up.
It will perform the analysis for a range of frequencies.
You must indicate the:
1. Starting frequency
2. Ending frequency
3. Number of stepping increments and scale (log or linear)
4. Scale for the results (linear or Decibel, Phase or radians)
Sources in the AC analysis must be set up in “edit
simulation model” menu to:
1. Identify source as sinusoidal through the small signal AC and
distortion tab.
2. Provide a magnitude and phase and check the USE box
SPICE Example
Find the phasor for vc(t) for vs(t)= 5cos(2ft) V in the circuit below
for f = 100, 200, 300, 400, 500, …..1000 Hz. Note that 400 Hz was
the frequency of the original example problem.
IVm
C
114.86n
R1
3k
V
R2
6k
ex16-Small Signal AC-2-Table
FREQ
MAG(V(IVM))
(Hz)
(V)
+100.000 +3.299
+200.000 +3.203
+300.000 +3.059
+400.000 +2.887
+500.000 +2.703
+600.000 +2.520
+700.000 +2.345
+800.000 +2.182
+900.000 +2.033
+1.000k
+1.898
PH_DEG(V(IVM))
(deg)
-8.213
-16.102
-23.413
-30.000
-35.817
-40.893
-45.295
-49.107
-52.411
-55.285
Plotting Frequency Sweep Results
Choices for AC (frequency sweep simulation)
 For frequency ranges that include several orders
of magnitude, a logarithmic or Decade (DEC) scale
is more practical than a linear scale

The magnitude results can also be computed on a
logarithmic scale referred to a decibels or dB
defined as:
M dB  20 log 10 ( M )
Plot of Magnitude
Linear Magnitude, Log Frequency
Circuit1-Small Signal AC-5
+1.000
+10.000
+100.000
dB Magnitude, Log Frequency
Frequency (Hz)
+1.000k
Circuit1-Small Signal AC-6
+10.000k
+1.000
+10.000
+100.000
Frequency (Hz)
+1.000k
+10.000k
+3.000
+0.000e+000
+2.000
+1.000
-20.000
+0.000e+000
MAG(V(IVM))
DB (V(IVM))
Linear Magnitude, Linear Frequency
u14ex1.ckt-Small Signal AC-7
+10.000k
+20.000k
+30.000k
dB Magnitude, Linear Frequency
u14ex1.ckt-S mall S ignal A C-8
Frequency (Hz)
+40.000k
+10.000k
+50.000k
+3.000
+0.000e+000
+2.000
+1.000
-20.000
+0.000e+000
MAG(V(IVM))
DB (V(IVM))
+20.000k
+30.000k
Frequency (Hz)
+40.000k
+50.000k
Plot of Phase
Linear Frequency, in Degrees
u14ex1.ckt-S mall S ignal A C-8
+10.000k
+20.000k
+30.000k
Frequency (Hz)
+40.000k
+50.000k
+0.000e+000
-50.000
PH_DEG(V(IVM))
Log Frequency, in Degrees
u14ex1.ckt-Small Signal AC-9
+1.000
+0.000e+000
-50.000
PH_DEG(V(IVM))
+10.000
+100.000
Frequency (Hz)
+1.000k
+10.000k
SPICE Example
Find the phasor for vc(t) when vs(t)= 20cos(4t) V in
the circuit below (note f = 2/ =0.6366)
R
10
L
1
vAm
V
0
FCCCS
L0
.5
Voltage of interest
IVm
C
.1
Extract Line from table of interest
FREQ
MAG(I(VAM))
PH_DEG(I(VAM))
MAG(V(IVM))
PH_DEG(V(IVM))
+108.440
+13.912
-161.560
(Hz)
+636.600m
+7.589