Thevenin Ex

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Chapter 10Sinusoidal Steady State Analysis

Huseyin Bilgekul

Eeng224 Circuit Theory IIDepartment of Electrical and Electronic Engineering

Eastern Mediterranean University

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, Thevenins and Nortons 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.


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Transform a voltage source in series with an impedance to a current source in

parallel with an impedance for simplification or vice versa.

Source Transformation


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Source Transformation

If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a).

Practice Problem 10.4: Calculate the currentIo


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Source TransformationPractice Problem 10.4: Calculate the currentIo


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Thevenin Equivalent CircuitThvenins theorem, as stated for sinusoidal ACcircuits, is changed only to

include the term impedanceinstead 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 CircuitThe 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:

Th N N Th N V Z I Z Z


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Thevenin Equivalent CircuitP.P.10.8 Thevenin Equivalent At terminals a-b


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Thevenin Equivalent CircuitP.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 CircuitP.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources


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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 CircuitsPractice Problem 10.11: Calculate voand current io

The frequency domain equivalent circuit.


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P l l


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Capacitance multiplier: The circuit acts as an equivalent capacitance Ceq

2

1

11ii eq

i eq

V RZ C C

I j C R

OP Amp Capacitance Multiplier Circuit

( )1

i oi i o

V VI j C V V

j C

0 2

0

1 2 1

0 0ii

V V R

V VR R R

2 2

1 1

Substituting, (1 ) or (1 )ii ii

IR RI j C V j C

R V R

O ll


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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.

O ill


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OscillatorsAn 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).

OUTPUT

+ INPUT

- INPUT

Phase shift circuit to

produce 180 degree

shift

Produce overall gain

greater than 1

A i b S b i d


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Assignment to be Submitted

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 V2and Vo to show the phase

relationship.

Submit the printout of your circuit schemmatic and the oscilloscope waveforms

of V2and Voas shown in the next page for a similar circuit.

Do you obtain the required phase shift and the oscillation frequency? If not it willnot 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.

VoV2

A i t (A l ti S l ti )


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Assignment (Analytic Solution)

Chapter 10, Solution 91.

voltage at the noninverting terminal of the op amp2V

oV output voltage of the op amp

110p o sk R R j L

j C

Z Z

2 2

2

) )( ( 1

p o o

o s p o oo

R CR

j C R RR R j LC

j LC

ZV V

V Z Z V

For this to be purely real,

2

-3 -9

1 1 11 0

2 2 (0.4 10 )(2 10180kHz Osc. Fr

)eq.o o oLC f

LC LC

o

o

oo

oo

o

2

RR

R

)RR(C

CR

V

VAt oscillation,

This must be compensated for by

2

80

40

1

1 5 420 k5

o o

v oo

R

R RR R

VA

V

Si il O ill t th A i t


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Similar Oscillator as the Assignment

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