Chapter 10
Sinusoidal Steady State Analysis
Chapter Objectives:
 Apply previously learn circuit techniques to sinusoidal steady-state
analysis.
 Learn how to apply nodal and mesh analysis in the frequency domain.
 Learn how to apply superposition, Thevenin’s and Norton’s theorems
in the frequency domain.
 Learn how to analyze AC Op Amp circuits.
 Be able to use PSpice to analyze AC circuits.
 Apply what is learnt to capacitance multiplier and oscillators.
Huseyin Bilgekul
Eeng224 Circuit Theory II
Department of Electrical and Electronic Engineering
Eastern Mediterranean University
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Source Transformation
 Transform a voltage source in series with an impedance to a current source in
parallel with an impedance for simplification or vice versa.
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Source Transformation
 Practice Problem 10.4: Calculate the current Io
If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a).
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Source Transformation
 Practice Problem 10.4: Calculate the current Io
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Thevenin Equivalent Circuit
 Thévenin’s theorem, as stated for sinusoidal AC circuits, is changed only to
include the term impedance instead of resistance.
 Any two-terminal linear ac network can be replaced with an equivalent
circuit consisting of a voltage source and an impedance in series.
 VTh is the Open circuit voltage between the terminals a-b.
 ZTh is the impedance seen from the terminals when the independent sources are
set to zero.
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Norton Equivalent Circuit
 The linear circuit is replaced by a current source in parallel with an impedance.
IN is the Short circuit current flowing between the terminals a-b when the
terminals are short circuited.
 Thevenin and Norton equivalents are related by:
VTh  Z N I N
ZTh  Z N
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Thevenin Equivalent Circuit
P.P.10.8 Thevenin Equivalent At terminals a-b
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Thevenin Equivalent Circuit
P.P.10.9 Thevenin and Norton Equivalent
for Circuits with Dependent Sources
To find Vth , consider the circuit in Fig. (a).
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Thevenin Equivalent Circuit
P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
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Thevenin Equivalent Circuit
P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
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Thevenin Equivalent Circuit
P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
Since there is a dependent source, we can find the impedance by inserting a voltage source
and calculating the current supplied by the source from the terminals a-b.
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OP Amp AC Circuits
 Practice Problem 10.11: Calculate vo and current io
The frequency domain equivalent circuit.
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OP Amp AC Circuits
 Practice Problem 10.11: Calculate vo and current io
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OP Amp AC Circuits
 Practice Problem 10.11: Calculate vo and current io
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OP Amp Capacitance Multiplier Circuit
 Capacitance multiplier: The circuit acts as an equivalent capacitance Ceq
Ii 
Vi  Vo
 jC (Vi  Vo )
1
j C
Substituting, I i  j C (1 
Vi
1
Zi  
Ii
jCeq
Vi  0 0  V0
R

 V0   2 Vi
R1
R2
R1
R2
)Vi
R1
or
Ii
R
 j (1  2 )C
Vi
R1
 R2 
Ceq  1   C
 R1 
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Oscillators
 An oscillator is a circuit that produces an AC waveform as output when
powered by a DC input (The OP AMP circuit needs DC to operate).
 A circuit will oscillate if the following criteria (BARKHAUSEN) is satisfied.
 The overall gain of the oscillator must be unity or greater.
 The overall phase shift from the input to ouput and back to input must be
zero.
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Oscillators
 An oscillator is a circuit that produces an AC waveform as output when powered by a
DC input (The OP AMP circuit needs DC to operate).
Produce overall gain
greater than 1
- INPUT
OUTPUT
+ INPUT
Phase shift circuit to
produce 180 degree
shift
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Assignment to be Submitted
V2
Vo
Construct the PSpice schemmatic of the oscillator shown Prob. 10.91 from the
textbook which is also shown above.
 Display the oscilloscope AC waveforms of V2 and Vo to show the phase
relationship.
 Submit the printout of your circuit schemmatic and the oscilloscope waveforms
of V2 and Vo as shown in the next page for a similar circuit.
 Do you obtain the required phase shift and the oscillation frequency? If not it will
not oscillate to produce a pure sine wave.
 Submission date 21 March 2007.
 The analytic solution is given in the next page to help your simulation.
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Assignment (Analytic Solution)
Chapter 10, Solution 91.
V2 
voltage at the noninverting terminal of the op amp
Vo 
output voltage of the op amp
Z p  10 k   Ro Z s  R  j L 
1
jC
CRo
V2

Vo C ( R  Ro )  j ( 2 LC  1)
Zp
Ro
V2


Vo Z s  Z p R  R  j L  j
o
C
For this to be purely real,
o2 LC  1  0 
 o 
At oscillation,
1
LC
fo 
1
2 LC

1
2 (0.4 10-3 )(2 10-9 )
 180kHz Osc. Freq.
o CR o
Ro
V2


Vo o C (R  R o ) R  R o
This must be compensated for by
Av 
Vo
80
 1
5
V2
20
Ro
1

R  Ro 5

 R  4 Ro  40 k
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Similar Oscillator as the Assignment
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