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Overview
• This chapter applies the circuit analysis
introduced in the DC circuit analysis for AC
circuit analysis.
• Nodal and mesh analysis are discussed.
• Superposition and source transformation for
AC circuits are also covered.
• Applications in op-amps and oscillators are
reviewed.
1
Steps to Analyze an AC
Circuit
• There are three steps to analyzing an AC
circuit.
• They make use of the fact that frequency
domain analysis is simpler because it can
make use of the nodal and mesh techniques
developed for DC
1. Transform the circuit to the phasor or
frequency domain
2. Solve the problem using circuit techniques
3. Transform back to time domain.
2
Nodal Analysis
• It is possible to use KCL to analyze a circuit
in frequency domain.
• The first step is to convert a time domain
circuit to frequency domain by calculating
the impedances of the circuit elements at the
operating frequency.
• Note that AC sources appear as DC sources
with their values expressed as their
amplitude.
3
Nodal Analysis II
• Impedances will be expressed as complex
numbers.
• Sources will have amplitude and phase
noted.
• At this point, KCL analysis can proceed as
normal.
• It is important to bear in mind that complex
values will be calculated, but all other
treatments are the same.
4
Nodal Analysis III
• The final voltages and current calculated are
the real component of the derived values.
• The equivalency of the frequency domain
treatment compared to the DC circuit
analysis includes the use of supernodes.
5
Mesh Analysis
• Just as in KCL, the KVL analysis also applies
to phasor and frequency domain circuits.
• The same rules apply: Convert to frequency
domain first, then apply KVL as usual.
• In KVL, supermesh analysis is also valid.
6
Superposition
• Since AC circuits are linear, it is also
possible to apply the principle of
superposition.
• This becomes particularly important is the
circuit has sources operating at different
frequencies.
• The complication is that each source must
have its own frequency domain equivalent
circuit.
7
Superposition II
• The reason for this is that each element has
a different impedance at different
frequencies.
• Also, the resulting voltages and current must
be converted back to time domain before
being added.
• This is because there is an exponential factor
ejωt implicit in sinusoidal analysis.
8
Source Transformation
• Source transformation in frequency domain
involves transforming a voltage source in
series with an impedance to a current
source in parallel with an impedance.
• Or vice versa:
Vs  Z s I s

Is 
Vs
Zs
9
Thevenin and Norton
Equivalency
• Both Thevenin and Norton’s theorems are
applied to AC circuits the same way as DC.
• The only difference is the fact that the
calculated values will be complex.
VTh  Z N I N
ZTh  Z N
10
Op Amp AC Circuits
• As long as the op amp is working in the
linear range, frequency domain analysis can
proceed just as it does for other circuits.
• It is important to keep in mind the two
qualities of an ideal op amp:
– No current enters ether input terminals.
– The voltage across its input terminals is zero with
negative feedback.
11
Application
• The op amp circuit shown here is known as
a capacitance multiplier.
• It is used in integrated circuit technology to
create large capacitances out of smaller
ones.
12
Capacitor Multiplier
• The first op amp is acting as a voltage
follower, while the second one is an inverting
amplifier.
• At node 1:
Ii 
Vi  Vo
 jC Vi  Vo 
1/ jC
• Applying KCL at node 2 gives
Vo  
R2
Vi
R1
13
Capacitor Multiplier II
• Combining these two gives:
 R2 
Ii
 j 1   C
Vi
 R1 
• Which can be expressed as an input
impedance.
Vi
1
Zi  
Ii
jCeq
14
Capacitor Multiplier III
• Where:
 R2 
Ceq  1   C
 R1 
• By selecting the resistors, the circuit
can produce an effective capacitance
between its terminal and ground.
15
Capacitor Multiplier IV
• It is important to understand a limitation of
this circuit.
• The larger the multiplication factor, the
easier it is for the inverting stage to go out of
the linear range.
• Thus the larger the multiplier, the smaller the
allowable input voltage.
• A similar op amp circuit can simulate
inductance, eliminating the need to have
physical inductors in an IC.
16
Oscillators
• It is straightforward to imagine a DC voltage
source.
• One typically thinks of a battery.
• But how to make an AC source?
• Mains power comes to mind, but that is a
single frequency.
• This is where an oscillator comes into play.
• They are designed to generate an oscillating
voltage at a frequency that is often easily
changed.
17
Barkhausen Criteria
• In order for a sine wave oscillator to sustain
oscillations, it must meet the Barkhausen
criteria:
1. The overall gain of the oscillator must be
unity or greater. Thus losses must be
compensated for by an amplifying device.
2. The overall phase shift (from the output and
back to the input) must be zero
• Three common types of sine wave oscillators
are phase-shift, twin T, and Wein-bridge
oscillators.
18
Wein-bridge
• Here we will only consider
the Wein-bridge oscillator
• It is widely used for
generating sinusoids in the
frequency range below 1
MHz.
• It consists of a RC op amp
with a few easily tunable
components.
19
Wein-bridge II
• The circuit is an amplifier in a non-inverting
configuration.
• There are two feedback paths:
– The positive feedback path to the non-inverting input
creates oscillations
– The negative feedback path to the inverting input controls
the gain.
• We can define the RC series and parallel
combinations as Zs and Zp.
j
Z s  R1 
C1
R2
Zp 
1  j R2C2
20
Wein-bridge III
• The resulting feedback ratio is:
Zp
V2

Vo Z s  Z p
• Expanded out:
V2
 R2C1

Vo   R2C1  R1C1  R2C2   j  2 R1C1R2C2  1


• To satisfy the second Barkhausen criterion,
V2 must be in phase with Vo.
• This means the ratio of V2/Vo must be real
21
Wein-bridge IV
• This requires that the imaginary part be set
to zero:
o2 R1C1 R2C2  1  0
• Or
o 
1
R1 R2C1C2
• In most practical cases, the resistors and
capacitors are set to the same values.
22
Wein-bridge V
• In most cases the capacitors and resistors
are made equal.
• The resulting frequency is thus:
1
RC
1
fo 
2 RC
o 
• Under this condition, the ratio of V2/Vo is:
V2 1

Vo 3
23
Wein-bridge VI
• Thus in order to satisfy the first Barkhausen
criteria, the amplifier must provide a gain of 3
or greater.
• Thus the feedback resistors must be:
R f  2 Rg
24