Unformatted text preview: 1 Lecture 10: Introduction to electronic analog circuits 36113661 7. PositiveFeedback Oscillators
© Eugene Paperno, 2008 7.1. Introduction
Our aim is to develop an electronic circuit that is able to
produce sustained sinusoidal oscillations at a single frequency.
No external ac source will drive the circuit; it should start
from an initial state, and its zeroinput response should
approach a sinewave. We assume that in steady state, the
circuit that we develop can be described by a linear, lumped,
timeinvariant (LLTI) smallsignal equivalent model. We shall
call such a circuit linear sinewave oscillator.
Terminology
Before we start the development of an oscillator, let us pay
some attention to the terminology accepted in basic circuit
theory [1]. We recall that the zeroinput response (ZIR) of an
LLTI circuit is also called the natural response. It is the
response of the circuit to an initial state only, with all the
independent sources suppressed. The response of the circuit to
an external excitation only, with the initial state equals zero, is
called zerostate response (ZSR) or forced response.
The complete response of a circuit is a superposition of its
natural and the forced responses. The complete response of a
stable circuit can also be divided into the transient response
and the steadystate response. The circuit transient response is
contributed by both the initial state and the external excitation.
For a stable circuit, the transient response is decaying and the
circuit steady state depends on the external excitation only.
Note, that the oscillator that we develop should behave in a
different manner compared to a stable circuit: its forced
response is zero, and its steady state should be defined by its
natural response. Hence, the oscillator natural response should
not decay with time. Instead, it should approach a sinewave at
a desired frequency and amplitude.
Applications
In terms of applications, oscillators, either sinusoidal or not,
single or multifrequency, linear or nonlinear, are generally
used to synchronize a system (computers), to carry and detect
information (communication), and to test a system
(instrumentation).
Principles of operation
Depending on the principle of operation, oscillators can be
classified as negative resistance, parametric, relaxation, and
positivefeedback. We shall treat here the latter type only.
Positivefeedback oscillators
Positivefeedback oscillators employ an LLTI passive
frequencyselective electric network (feedback network) β(jω)
and an active, electronic amplifier AOL interconnected in a
positivefeedback loop. We shall assume here that AOL is
frequencyindependent. Passive feedback network
Recall that a passive electric network is that combined of
passive elements: resistors, capacitors, and inductors. A
passive resistor never dissipates and a passive capacitor and
inductor never store negative energy [1]. A circuit or its
elements are active if they are not passive.
It is quite clear that being always stable, a passive network
alone is not able to produce sustained oscillations: it has no
poles in the right half of the Laplace plane, and, therefore, its
natural response approaches zero.
To illustrate this, we recall that the transfer function of a
network in terms of the Laplace transform [1],
∞ L[ f (t )] ≡ ∫ f (t ) e − st dt , (1) 0− is defined as follows [1]:
m H (s) ≡ L[zero − state response]
=K
L[input ] ∏ (s − z j )
j =1
n , (2) ∏ ( s − pi )
i =1 where s is the complex frequency, s=σ+jω, K is a constant, the
zj's are called zeros, since H ( z j ) = 0 , and the pi's are called
poles, since H ( pi ) = ∞ , provided that z j ≠ p i .
The partialfraction extension of (2) can be written as
n H ( s) = ∑ K i
i =1 1
,
s − pi (3) where the Ki's are called the residues of the particular poles
[1]. Applying the inverse Laplace transform L1[H(s)] to (3)—
this is usually done with the help of a look up table—one can
obtain the natural response of the network,
n natural response = ∑K e
i pi t , (4) i =1 where the poles pi's are also called natural frequencies (or
natural modes [2]) of the network. The pi's in (4) depend on 2 Lecture 10: Introduction to electronic analog circuits 36113661 β(jω)
vo(t) iin(t)
R L C
Im[β(s)]=0
(a) jω Q= ∞ jω0
jωd s plane
s =σ+jω [email protected]sLD
β(s) 4 (d) 3
2
1
0
−2 1 0 jω −1 σ −1 0 Q=0.5 1 −α σ 0 −jωd
−jω0 1.5 1 (b)
0.5 Im[β(s)]=0
vo(t) Ke−α t jω 0 2π/ωd −0.5
0 (c) −1 (e)
−1.5 −1 −0.5 0 0.5 1 σ
Fig. 1. (a) A parallel RLC circuit, (b) locus of its natural frequencies, and (c) its natural time response. The characteristic equation: s2+2αs+ω02= s2+(ω0/Q)s+ω02.
The resonant frequency ϖ 0 = 1 / LC , L, and C are kept constant, and R varies. Q=R/(ω0L). Since R is never equal infinity, the time response is always decaying. β(jω0) never equals infinity. (d) Network function  β(s) of the RLC circuit. The red line is the locus of purely real values of β (s). (e) Contour plot of
the network function  β (s). the network topology, and the Ki's (possibly complex) depend,
in addition, on the initial state.
Recall that a network function (a particular transfer function
from an input variable to an output variable) and the natural
frequencies of a network can be determined by applying an
excitation that does not change the natural structure of the
circuit [1], [2]. This can be done either with an independent
current source connected in parallel to a network branch, or
with an independent voltage source connected in series with a
network branch. As far as the natural response is concerned,
the above independent sources turn into an open or closed
circuit, correspondingly; thus, not altering the natural structure
of the network.
Consider for example a passive RLC network of Fig. 1(a).
Its network function, β ( s) = L[vo (t )]
1
s
,
≡ Z ( s) =
C ( s − p )( s − p*)
L[iin (t )] p = −α + jϖ d , p* = −α − jϖ d α= 1
, ϖ0 =
2 RC , (5) 1
2
, ϖd ≡ ϖ0 −α2
LC can be found by connecting an independent current source
iin(t), as shown in Fig. 1(a).
Locus of the natural frequencies for the RLC network is
shown in Fig. 1(b). This figure and (5) clearly demonstrate
that the poles of a practical network (with a finite and positive
R and, hence, α) are always located to the left of the jω axis.
Or in other words, the denominator of (5) is never zero for any
s with a nonnegative σ. As a result, a practical RLC network
cannot generate a sustained sinewave. Its output is always
decaying, as shown in Fig. 1(c), and its natural response
approaches zero. 3 Lecture 10: Introduction to electronic analog circuits 36113661 One way to reach nondecaying oscillations is to
compensate for the network positive (passive) resistance R
(read energy losses) with an element having negative (active)
resistance –R (read energy pumping into the circuit), as shown
in Fig. 2(a). Negativeresistance oscillators are based just on
this idea. They employ nonlinear electronic elements, such as
the tunnel diode, having negative slope in a part of their
voltagecurrent characteristic, to obtain—for relatively small
signals—negative resistance.
Negativeresistance approach to oscillators is generally
used at radio (RF) and especially at microwave frequencies,
where it becomes difficult to construct a feedback circuit
without introducing excess phase shift. At relatively low
frequencies, from kilohertz up to hundreds of megahertz,
positivefeedback approach is generally used. β (jω)
iin(t) vo(t)
−R C jω
jω0 0 s plane
s =σ+jω σ −jω0 Another way to make a passive network to oscillate
continuously is to integrate it in a feedback loop, as shown in
Fig. 3(a), such that the total network function becomes
So
AOL
P( s)
.
=
=K
Q( s )
S in 1 − AOL β ( s ) L (a) Feedback loop H ( s) = R vo(t) 0 Ke−0 t (b) 2π/ω0 t (6) Note that contrary to (5), the denominator of (6) can reach
zero for an s1 with a nonnegative σ, namely, it happens when
the loop gain AOLβ(s1)=1.
Barkhausen criterion
To reach sustained oscillations in steady state, the loop gain
AOLβ(s1) should be equal to unity at a physical, ω —not
complex, s —frequency, namely,
⎧
⎪ AOL 1 β ( jω1 ) = 1 , amplitude criterion
. (7)
AOL 1 β ( jω1 ) = 1 ⇒ ⎨
⎪∠AOL 1 β ( jω1 ) = 0, phase criterion
⎩ where AOL 1 β ( jϖ 1 ) and ω1 are the oscillator's steadystate
loop gain and frequency; index 1 emphasizes that in steady
state the loop gain AOL 1β ( jω1 ) is unity.
Since for a physical frequency, σ =0, the oscillator's steadystate poles (natural frequencies) are located directly on the jω
axis of the s plane, as shown in Fig. 3(b), where according to
(6) and (7), H ( jω1 ) = ∞ . Equations (7) determine the steadystate oscillation conditions and are called the Barkhausen
criterion.
Initial pole location
We now have to decide on the pole location in the initial
state. Let us first suppose that there should be no difference
between the initial and final pole locations; in both the cases
the poles can be located directly on the jω axis. (c)
Fig. 2. A parallel RLC network employing negativeresistance (a), its natural
frequencies (b), and its natural time response (c). Since RII(−R)=∞, H(jω0)
does equal infinity. Magnitude of the natural response (4) in this case [see Fig.
3(b)] is a function of the initial state. It is an undesirable
situation for the three following reasons. First, we do not
intend to supply the circuit with additional means to control
the initial state. Second, any circuit is subject to disturbances
and noise. In the case of an oscillator, the uncontrolled initial
state, disturbances, and noise will make its steady state
unpredictable. Third, it is impossible to provide the initial pole
locations exactly on the jω axis because of the nonzero
tolerances of the circuit components. Since it is undesirable to
locate the oscillator's initialstate poles directly on the jω axis,
they should be located in the right half of the s plane at a
distance from the jω axis. This distance [see Fig. 3(b)] should
be large enough to ensure the desirable pole location for any
deviations of the circuit element values within their tolerances
and temperature ranges.
The oscillator in this case will generate an increasing
sinewave. When the sinewave magnitude will become
sufficiently large, a negative feedback should stabilize it by
shifting the poles towards the jω axis, as shown in Figs. 3(b)
and (c). It is undesirable to locate the poles in the left half of
the s plane because, in this case, the oscillator steady state will
be zero.
To summarize the above: we want the oscillator to be
unstable in the initial state and conditionally stable in the 4 Lecture 10: Introduction to electronic analog circuits 36113661 Natural response at ω1 vo
Sε Σ So AOL t 0 Ss (a)
Sf β(jω) ϕ=0 Forced response at ω vo ω1 H(jω) t 0 Initial pole location
Undesirable
Desirable
jω
jω1 Forced response at ω vo s plane ω1
ϕ=π AOLβ(s) =1
AOLβ(jω) =1 t 0 AOLβ(s) =1 ϕ=0 σ (b) 0 Natural and forced responses at ω vo ω1 σ >0 −jω1
vo t 0 ϕ=0
0 ϕ=0 (c) t Natural and forced responses at ω vo ϕ=π +VCC
Sε 0 So AOL ϕ=0 Fig. 4. The effect of disturbances and noise on the oscillator output.
AOL max
So=Sε AOL max
+VCC
So< Sε AOL max
So ~ Sε AOL ~ (1/3)VCC
0 t (d) −VCC
So(ω1) ω1 max Sε (ω1) −VCC Fig. 3. Positivefeedback oscillator (note that suppressing Ss returns the
network to its original topology; would a "" used at the summing point, we
would add an inverter to the original topology) (a), locus of its natural
frequencies as a function of the lop gain AOLβ(jω) (b), its natural time response
(c), and nonlinear properties of the amplifier AOL (d). steady state. We never want it to be stable.
The simplest—but not always the best (see the concluding
part of this Section)—way to determine the initial poles
location in the right half of the s plane and to provide their
movement to the jω axis, when the oscillations approach their
steadystate magnitude, is to use the nonlinear properties of
the amplifier AOL, as shown in Fig. 3(d).
In the initial state, the smallsignal value of AOL is simply
set to be greater than it is needed to satisfy the Barkhausen
criterion (7), AOL > AOL 1 . As a result, the initial pole location
will be either to the right or to the left of the jω axis [see Fig.
3(b)], depending on the specific behavior of the feedback
transmission β(s).
For the reasons discussed above, β(s) for which the initial
poles are in the left half of the s plane should not be employed
in our design. Below (see Section III), we will show how one
can easily decide whether AOL > AOL 1 shifts poles to the right
or to the left of the jω axis. 5 Lecture 10: Introduction to electronic analog circuits 36113661 Negative feedback
Meanwhile, let us suppose that AOL > AOL 1 shifts the poles as
desired, to the right of the jω axis. This forces unstable
operation of the oscillator. It starts from an initial state and
builds up its oscillations, as shown in Fig. 3(c). When the
oscillations became large enough and reach nonlinear part of
the amplifier transfer characteristic AOL=So/Sε in Fig. 3(d), the
amplifier gain AOL decreases (negative feedback) and
approaches AOL 1 . Accordingly, the poles move towards their
steadystate location on the jω axis, as shown in Fig. 3(c). The
steady state is reached when the amplifier gain at frequency
ω1 equals AOL 1 , and the Barkhausen criterion (7) is satisfied.
The effect of disturbances and noise
A disturbance and noise affecting the circuit can be
translated to the input of the oscillator and represented by an
equivalent source Sin, as shown in Fig. 3(a). The source Sin
will contribute to the oscillator output: it will either decrease
or increase the oscillation caused by the initial state (the
natural response).
Recall that a circuit having a single couple of purely
imaginary poles is conditionally stable. (A circuit having more
than a couple of equal imaginary poles has linearly increasing
natural response and, hence, is unstable.) A conditionally
stable circuit has a bounded steady state only if it is not
excited at the frequency of the poles (see Fig. 4). If the circuit
is excited at the frequency of the poles then its forced
response linearly increases with time.
Let us suppose now that the spectrum of the disturbance
and/or noise is continuous and always includes a sinusoidal
excitation at the frequency of the oscillator poles. This
excitation will contribute a linearly increasing part (forced
response) to the oscillator natural response as shown in Fig. 4.
Depending on the phase difference between the natural and
forced responses, the oscillator output will either increase
immediately after adding the external excitation or first
decrease and after that increase (see Fig. 4).
The increased oscillations will decrease the amplifier gain
AOL below its steady state value AOL 1 , and the decreased gain
will shift the oscillator poles in the left half of the s plane. The
oscillator natural response will decay, and the oscillator steady
state will be defined by its steady state response to the input
signal Sin in accordance with the network function (6).
Since in the new steady state, caused by a disturbance or
noise, the loop gain AOL β ( jω1 ) will be smaller than unity,
the network function (6) will be finite, and so will be the
oscillator output. The specific value of the amplifier gain AOL
in the new steady state and thus the specific value of the
network function and the oscillator output depend on the
magnitude of the disturbance or noise and on the nonlinear
properties of the amplifier. Note that under influence of a
disturbances or noise, the oscillator operates as a frequencyselective amplifier with the total gain described by the
network function (6). At the frequency of oscillation, the total
gain is at maximum and, at the other frequencies, it decreases in accordance with the frequency behavior of the feedback
transmission β ( jω ) . The disadvantage of the use of the
nonlinear properties of the amplifier is that its output signal
becomes distorted.
To avoid distortions, it is better to control with the help of a
negative feedback the value of β(jω1) in such a way that the
amplifier output is always within one third of its full scale [see
Fig. 3(d)]. This can be done, for example, by using in the
feedback network photo or thermoresistors, which values
depend on average, not instant, magnitude of oscillations.
The Nyquist stability criterion
We shall finally illustrate how one can easily decide
whether AOL > AOL 1 shifts the oscillator poles from their
steadystate position on the jω axis to the right or to the left of
this axis.
Let us first assume that the denominator in (6) has only two
zeros:
Q ( s ) = 1 − AOL β ( s ) = ( s − z )( s − z*),
Q ( s ) = ( s − z )( s − z*) . (8) ∠Q ( s ) = ∠( s − z ) + ∠( s − z*) Let us also assume that the two zeros of Q(s) lie in the right
half of the s plane, as shown in Fig. 5. We consider as well
that for a passive β(jω), Q(s) has no poles in the right half of
the s plane.
If we now encircle the entire right half of the s plane with a
closed contour (so called Nyquist Dcontour), as shown in
Fig. 5(a), another closed contour, Q(jω), is created in the
complex Q(s) plane around its origin, Q(s)=(0, j0), as shown
in Fig. 5(b).
The function Q(s) is said to map the right half of the s plane
inside the contour Q(jω) in the Q(s) plane. [Note also that
contour Q(jω) in Fig. 5(b) encircles the origin of the Q(s)
plane in the same direction as the right half of the s plane is
encircled in Fig. 5(a).]
The above result is known as the Principle of the Argument.
This principle can be easily understood if one considers an
opposite case. Consider, for example, that Q(jω) does not
encircle the origin of the Q(s) plane, as sown in Fig. 5(c). In
this case, the vector Q(jω) will not rotate by 720o for
−∞ < jϖ < ∞ as Fig. 5(a) and (8) state.
We are now ready to formulate the Nyquist stability
criterion for a positivefeedback oscillator combined of a
stable feedback network β(jω) and a stable amplifier AOL:
such an oscillator is unstable if its loop gain AOLβ(jω) does
encircle the point (1, j0), as shown in Fig. 6. 6 Lecture 10: Introduction to electronic analog circuits 36113661 jω s Im s plane s 360 R→∞ σ 0 z* = Q(s) = (s−z) (s−z*) Q(s)=(s−z)(s−z*) s−z
z Q(s) plane
Q(s)= (s−z) + (s−z*) Q( jω) σ >0 σ >0 1−AOLβ(s) plane Im s plane
Q(s)=1−AOLβ(s) Q(jω)
0 720 jω H(s)= ∞
1−AOLβ(s)=
0
AOLβ(s)=1 R→∞ Re σ 0 −1 0 Re Nyquist Dcontour (b) (a)
H(s) = AOL
P(s)
=K
1−AOLβ (s)
Q(s) Im Im Q(s) plane −AOLβ(s) plane AOLβ(s) plane Im AOLβ(jω) Q ( s)
< 720
0 σ >0 −1 0 Re 0 Re 1 Re Q(jω)
(c) Fig. 5. Mapping of the right half of the s plane: the Principle of the Argument. Fig. 6 illustrates that AOLβ(jω1)>1 can cause both unstable
and stable operations of a circuit. Or in other words,
AOLβ(jω1)>1 can shift the circuit's poles either to the right or
to the left of the jω axis, depending on the specific behavior of
the feedback transmission β(jω). Fig. 6. The Nyquist stability criterion. H(s) = AOL
1−AOLβ(s) A β(jω )>1
OL
1 AOLβ (jω)
H(s) is unstable 7.2. Oscillators for low frequencies: RC oscillators 0 Our current aim is to develop oscillators for low
frequencies. Since the value L=1/(ω02C) and, therefore, the
size and cost of inductances should be high at a low frequency
of oscillation ω0, we will not use them. Capacitors are much
smaller and less expensive; therefore, our feedback networks
and oscillators at low frequencies will be of the RC type.
7.2.1. Phaseshift oscillator
According to the above, our aim is to design a circuit that is
able to produce sustained oscillations as a response to the
initial conditions only. We would also like this circuit to
oscillate at a single frequency. As a result we would like the
Barkhausen criterion to hold at this frequency only.
We start the design from choosing a CS configuration for
the amplifier (see. Fig. 7). We suppose that the CS amplifier
has a real negative gain in the entire frequency band of
interest. As a result, the feedback network has to have a real
negative transmission at a single finite frequency to satisfy the
phase conditions of the Barkhausen criterion. One of the
simplest networks satisfying this request is a threestage RC
highpass filter shown in Figs. 7 and 8. We know that an
elementary RC stage is able to rotate the phase by an angle AOLβ (s) plane Im 1 Re (a) AOLβ(jω1) =1
AOLβ(jω1) >1 Im AOLβ (jω)
H(s) is stable 0 1 Re (b)
Fig. 6. Examples for an unstable (a) and stable (b) networks. Note that in the
both cases the loop gain AOLβ(jω1) can be greater than unity. less than 90○. Therefore, to obtain a real negative
transmission, reed a 180○ phase shift, we need at least three
such stages. 7 Lecture 10: Introduction to electronic analog circuits 36113661 VDD β(jω) RD 0.1C C vo 0.01C
10R R
CC RS 10R R vo
AOL 0.01C 0.1C C 100R 100R R>>RD Im[β(s)]=0
100 β(jω) AOL
G D 0.1C C 0.01C
vo gmvgs 10R R RDro [email protected]sLD
β(s) 1 0
−2 0 jω 100R σ 0
−1 (a)
iD AOL=5 iD VDS
V'DS<VDS AOL=8 1
3 −1/RD 0.2
0.125
0.1
0.08 0.5 Q Im[β(s)]=0 0.06 0.25
Vt vGS VGS V'DS VDS VDD vDS 0.04
0.02 jω 0
− 0.25 Linear gain − 0.5 Nonlinear gain t AOL=25 0.3 1 0.75 gm
Q 2 t Fig. 7. Phaseshift oscillator. Note that the transistor transfer characteristic,
iD−vGS, drops off with decreasing VDS. To find the solution for AOLβ(jω1)=1 in a simple way, we
decouple the stages of the feedback network from each other
and also from the amplifier. To do this, we increase the input
impedance of each following stage compared to the output
impedance of the amplifier and the output impedance of the
preceding stages (see Fig. 7). This helps us to find the
oscillator return ratio as: − 0.75 −2 − 1.5 −1 − 0.5 0 0.5 σ
(b) Fig. 8. (a) Network function β(s) of the phaseshift oscillator feedback
network. The red line is the locus of purely real values of β(s). (b) Contour
plot of the network function. The pole of β(s) is shown in blue, and the poles
of the network function H(s) of the oscillator are shown in red for AOL>8,
green for AOL=8 [see also Fig. 8(c)], and brawn for AOL<8. 8 Lecture 10: Introduction to electronic analog circuits 36113661 AOL β ( jϖ ) ≡ + v ′′
Sε′′
= + gs
Sε
v gs ′′
= v gs
vgs =1 ⎛
⎞
1
≈ − g m ( RD  ro ) ⎜
1 + 1/ jϖ RC ⎟
⎝
⎠
≈ − g m RD 3 , (9)
Abs@ AfHs(s)
H LD 1
3
3
1
−
−
1+
jϖ RC (ϖ RC )2 j (ϖ RC )3 Note the "+" in the definition of the return ratio AOLβ(jω) in
(9): we write "+" and not "−" because the feedback in the
present case is positive and not negative.
Solving (9) for AOLβ(jω) zerophase condition, we find the
frequency of oscillations, ω1:
∠AOL β ( jϖ 1 ) = 0
⇒ 3
1
−
=0
jϖ RC j (ϖ RC )3
, ⇒ 1 (ϖ 1RC ) 2 (10) =3 1
3 RC ⇒ ϖ1 = Solving (9) for the AOLβ(jω1)=1 condition, we find the
transconductance gain of the transistor, gm:  AOL β ( jϖ 1 ) = 1
⇒ − g m RD 1
1− = − g m RD 3 (ϖ 1RC ) 1
1 − 3⋅ 3 2 ,
= − g m RD ⇒ gm = (11) 1
1
= g m RD = 1
1 − 3⋅ 3
8 8
RD Fig. 8 shows the locus of real values of β(s) in the Laplace
domain. One can see that for some s=jω1, β(s)=0.125. This
means that for AOLβ(s)=1 or AOL=1/β(s)=8 the poles of the
oscillator transfer function Η(s) [see also Fig. 8(c) on this 50
1 0
−2 0 σ jω 0
−1 Fig. 8. (c) Network function of the phaseshift oscillator for AOL=8. page] will be located at the jω axis. One can also see that
β(s)<0.125 on the right of the jω axis and β(s)>0.125 on the
left of the jω axis. This means that for AOL>8, the oscillator
poles will be in the right half of the Laplace plane, and for
AOL<8, the oscillator poles will be in the left half of the
Laplace plane.
We choose therefore the static state of the CD amplifier in
such a way that it provides a AOL=gmRD>8. As a result, the
oscillator will generate an increasing sinewave at a real
frequency approaching ω1. Due to the nonlinearity of the
transistor transfer characteristic (see Fig. 7) and its
dependence on VDS, the amplifier gain AOL=gMRD will
decrease with increasing the amplitude of the oscillations, and
at an AOL=gMRD>8, the steady state will be reached. (Note
that gM is the largesignal transconductance gain.) 9 Lecture 10: Introduction to electronic analog circuits 36113661 ϕ Frequency stability ϕβ 270 Let us now use the phase response (see Fig. 9) of the
feedback network to analyze the frequency stability of the
phaseshift oscillator. According to the Barkhausen criterion,
at the oscillation frequency the phase of the return ratio
∠ AOLβ(jω1)=ϕA1+ϕβ1 =0 or 360○, were ϕA1 is the phase of the
smallsignal openloop gain, and ϕβ1 is the phase of the
feedback transmission. In the above example, we assumed that
both ϕA=180○ and ϕβ=180○.
Let us assume now that ϕA1 in Fig. 9 increases, for example,
due to a temperature change, by ΔϕA1, ϕ'A1=ϕA1+ΔϕA1. ϕβ1
ϕA1 180 such that ϕA Δϕβ1 (a) Δω1
0 ω ω1 ω'1 HRLC(jω)
vo (12) (13) ϕ'β1 dϕβ /dω ΔϕA1 90 iin
R The phase condition of the Barkhausen criterion will now
hold
true
for
a
different
frequency
of
oscillations, ω'1=ω'1+Δω1, at which ϕ'β1=ϕβ1−ΔϕA1, ϕ'A1 L C
(b) HRLC(jω) normalized
1 0.707 ϕ'A1+ϕ'β1=ϕA1+ΔϕA1+ϕβ1−ΔϕA1=360○. (14) Note in Fig. 8, that the translation of ΔϕA1 into Δω1 depends
on the slope of the feedbacknetwork phase response at ω1:
the steeper the slope, the less sensitive the oscillation
frequency is to the amplifier phase fluctuations.
To describe the slope dϕβ/dω at ω1 quantitatively it is a
common practice to substitute the phase response of an RLC
resonant circuit with the same dϕRLC/dω at ω0 for the
feedbacknetwork phase response and to use the quality factor
of the resonant circuit QRLC, which is proportional to dϕ/dω
(see Appendix)
QRLC = − ϖ 0 dϕ RLC
2 dϖ , ϖ 0 = ϖ1 , (15) R1
(c)
R2>R1
0 ω ω0 ϕRLC Δω ω0 90 Q= Δω 45 Q= ω0L 0 R ω0 (d) ω −45 ϖ =ϖ 0 as a measure for dϕβ/dω at ω1 (see Fig. 9). For example, an
RLC resonant circuit with QRLC=0.65 has dϕ/dω at ω0, which
equals dϕβ/dω at ω1 of the phase response of the phaseshift
oscillator feedback network. We say, therefore, that the phaseshift oscillator feedback network has the equivalent quality
factor Qequiv=0.65. For such a small Qequiv the frequency
stability is very low a 1% relative fluctuation of the amplifier
phase corresponds to a 1.3% fluctuation of the frequency of
oscillations [see (A2) in Appendix]. For example, the
frequency fluctuations are as large as 13 Hz for a frequency of
oscillations of 1 kHz. −90 dϕ RLC/dω ∝ QRLC ϕβ
270 dϕ RLC /dω = dϕβ /dω 180 ϕRLC
90 0 −90 90 Qequiv= QRLC
0 ω0= ω1 Fig. 9. Equivalent quality factor as a measure for the slope of the phase
response of the oscillator feedbacknetwork at ω1. 10 Lecture 10: Introduction to electronic analog circuits 36113661 va Im[H(jω)] AOL vb vo
2R Zs βb HHPF (jω) ω vR HPF βa(jω) Zp
R* vin=1
vR R
C 0 vin=1 C R vC ϕHPF vC vC
LPF 1/3 R*=R ω βb R*<R ω1 0 HPF ϕβ (b) 0 βa(jω) ω1 ϕβ RC Re[β(jω)] ∞ ∞ 1/3 ω ω βa(jω) β (jω) ω R*=R ω β (jω) 0 −1/3 1/3−β R vR HLPF (jω) βb βb Re[H(jω)]
0 Im[β(jω)] Wien bridge β (jω) ∞ vin ϕLPF ∞ (a) ω
Im[β (jω)] LPF
R*=R ω ω=ω1
R ϕβ 0 ∞ ZC Zp  β RC (jϖ(jω) =
βa 1 ) Zp
Z p + Zs = Zp
Z p + 2Z p = 1
3 −1/3 (c) R*<R β (jω) ω1 Re[β (jω)] ω1
ω β ( jϖ 1 ) = 1
− βb
R
3 = 1
R*
−
3 2 R + R* Zs Fig. 11. Finding quantitatively the Wienbridge phase response. (a) Graphical
solution (hodograph of Nyquist) for the phase response of elementary RC
high and lowfilters. (b) Graphical solution for the phase response of βRC(jω)
and (b) its translation to the phase response of β(jω)=βRC(jω)−βR. ϕβ
180 R*=R
90 0 R*<R ω1 ω −90 −180 Fig. 10. Wienbridge oscillator. For R*=0.913, β(jω1) =1/3−R*/(2R*+
R).=0.02. 7.2.2. Wienbridge oscillator
Our current aim is to develop an oscillator with much better
frequency stability than that of the treated above phase shift
oscillator. To achieve this aim, we will utilize a very
interesting property of the Wien bridge (developed by Max
Wien in 1891), to provide the approaching infinity slope of its
phase response (see Fig. 10). Note that this is obtained due to
the differential output of the bridge.
The Wien bridge comprises two voltage dividers, one of them is purely resistive, with the transmission βb, and the
other is reactive, with the transmission βa(jω). The total
transmission of the bridge equals the difference
β(jω)=βa(jω)−βb. The bridge is said to be in equilibrium when
R*= R (we will see below why).
When the bridge is in equilibrium, we can easily find its
amplitude response for ω→0, ω1=1/RC (ωC=R at this
frequency), and ω→∞ (see Fig. 10). To find quantitatively the
amplitude response for 0<ω<ω1 and ω1<ω<∞, we note that in
the first frequency range, where ωC>R, the reactive voltage
divider behaves like the elementary RC highpass filter, and
in the other frequency range, where ωC<R, it behaves like the
elementary RC lowpass filter.
To find quantitatively the bridge phase response, we first
refer to Fig. 11. One can see in Fig. 11(a) that in the complex
plane the loci of the tips of the output phasors of the
elementary RC high and lowfilters rotates making a circle.
It is reasonable to expect a similar behavior from βa(jω) [see
Fig. 11(b)]. To translate βa(jω) to β(jω), we simply subtract 11 Lecture 10: Introduction to electronic analog circuits 36113661 from the phasor βa(jω) the phasor βb. The result is shown in
Fig. 11(c) for R*= R and for R*< R. According to Fig. 11(c),
we plot in Fig. 10 the phase response ϕβ(jω) of the Wien
bridge.
One can see from Fig. 10 that the slope of ϕβ(jω)
approaches infinity at ω1 when R* approaches R. Note,
however, that simultaneously the bridge amplitude response
β(jω), measured at ω1, approaches zero. It becomes clear,
therefore, that the frequency stability of an oscillator based on
the Wien bridge depends on the available voltage gain of the
amplifier: the higher the amplifier gain, the lower β(jω1),
which provides AOLβ(jω1)=1, the higher slope of ϕ(jω) at ω1
is, and the higher frequency stability is of the oscillator. In a
home exercise you will see that having an amplifier with a
voltage gain of 2000 provides a Qequiv=223, 300 times as much
compared to Qequiv=0.65 of the phaseshift oscillator. Now a
1% relative fluctuation of the amplifier phase corresponds to a
0.004% fluctuation of the frequency of oscillations, or to 0.04
Hz, instead of 13 Hz in the above case, at a f1=1 kHz.
The first Wienbridge oscillator was developed by William
Hewlett (the cofounder of the HewlettPackard company) in
1939. To stabilize the amplitude of the oscillation before the
amplifier approaches nonlinear region, Hewlett replaced the
resistor R* in Fig. 10 with a small lamp having at room
temperature a resistance Rinitial* that is small enough to
provide the oscillator return ratio AOLβ(jω1)>1. This causes the
oscillator instability (see Appendix) and its natural response is
a growing sinewave. Due to heating, the lamp resistance
increases with increasing the amplitude of oscillation. When
the lamp resistance reaches the value R*>Rinitial*, for which
AOLβ(jω1)=1, the oscillation amplitude stops to grow.
For a given AOL, the resistance R* can be found as follows Rjϖ L
1
R + jϖ L jϖ C
β RLC ( jϖ ) = R Z L Z C =
Rjϖ L
1
+
R + jϖ L jϖ C
= * ϕ RLC ( jϖ ) = arctan R⎞
⎛
1 + j ⎜ Rϖ C −
ϖL⎟
⎝
⎠ − Rϖ C + R
ϖL 1
ϕ RLC ( jϖ ) → 0 = R ϖL − Rϖ C and then we obtain dϕRLC/dω as a function of QRLC ⎛R
⎞
d⎜
− Rϖ C ⎟
dϕ RLC ( jϖ )
⎝ϖ L
⎠ = − R − RC
=
2
dϖ
dϖ
ϖ 2L
ϖ 2 =ϖ 0
=− =− R ϖ 02 L R − RC LC
LC ϖ 2 =
0 R − ϖ 02 L ϖ 02 L =− . 1
LC 2R ϖ 02 L Q =−
R
RLC ≡
ϖ 0L (A2) 2QRLC ϖ0 Finally we get ϖ0
. (16) * ⇒ AOL R − AOL R* = 1.5R * +3R
AOL − 3
1.5 + AOL =− dϕ RLC
dϕ o RLC π
dϕ o RLC
. (A2)
100% ≈ −0.87
=−
2QRLC
2QRLC 180
2QRLC c= dϖ R * +2 R − 3 R *
2R − 2R *
= AOL
=1
3R * + 6 R
3R * +6 R ⇒ R* = R R (A1) R* ⎞
⎛1
AOL β ( jϖ 1 ) = AOL ⎜ −
⎟ =1
⎝ 3 R * +2 R ⎠
⇒ AOL Rjϖ L
=
− Rϖ LC + R + jϖ L
2 1
1 a − jb
a
b
;
=
=
−j
2 + b2
2 + b2
a + jb a + jb a − jb a
a c= 1
a + b2
2 , APPENDIX ∠c = arctan Relationship between dϕRLC/dω and QRLC
To find the dependence of the slope of the phase response
dϕRLC/dω of the parallel resonant circuit at ω0 (see Fig. 9) on
its quality factor QRLC, we first find the phase response −b
a
REFERENCES [1]
[2] C. A. Desoer, E. S. Kuh, Basic circuit theory, McGrawHill, 1969.
S. Sedra, K. C. Smith, Microelectronic Circuits, 4th ed. New York:
Oxford University Press, 1998. 12 Lecture 10: Introduction to electronic analog circuits 36113661 Wienbridge oscillator in the Laplace domain 1
0.8
[email protected] sLD
0.6
0.4
0.2
0
−2 1
0 jω 100
80
[email protected] sLD 60
40
20
0
−2 −1 1
0 σ −1 0 σ −1 0
1 1 Fig. A1. Abs[β(s)]. Fig. A2. Abs[H(s)] for AOLβ( jω1) =1. 1.5 1.5 1 1 0.5 jω 0.5 jω 0 0 − 0.5 − 0.5 −1 −1 − 1.5 −1 − 0.5 0 0.5 − 1.5 1 −1 σ − 0.5 0 0.5 1 σ Fig. A3. Abs[β(s)]. Fig. A4. Abs[H(s)] for AOLβ( jω1) =1. 1.5 1.5 1 1 0.5 jω jω −1 0.5 jω 0 0 − 0.5 − 0.5 −1 −1 − 1.5 −1 − 0.5 0 0.5 σ
Fig. A5. Abs[H(s)] for AOLβ( jω1) =0.5. 1 − 1.5 −1 − 0.5 0 0.5 σ
Fig. A6. Abs[H(s)] for AOLβ( jω1) =2. 1 13 Lecture 10: Introduction to electronic analog circuits 36113661 Illustration
Note that the instant shift of the oscillator poles to the imaginary axis of the Laplace plane by reducing the gain from 8.1 to 8.0 results in the immediate stabilization of the amplitude of the oscillations. There is no transient. The signals at the poles
shifting time, 1 ms, are the initial conditions for the new time response. From 1 ms on, the poles are located on the imaginary
axis, and the envelope of the oscillations becomes horizontal. 1.0V 0.5V 0V 0.5V 1.0V
0s 0.5ms 1.0ms 1.5ms 2.0ms V(GAIN12:OUT)
Time 1.0V 0.5V 0V 0.5V 1.0V
0.990ms
0.995ms
V(GAIN12:OUT) 1.000ms
Time 1.005ms 1.010ms ...
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This note was uploaded on 01/14/2012 for the course EE 36113711 taught by Professor Prof.eugenepaperno during the Fall '11 term at BenGurion University.
 Fall '11
 Prof.EugenePaperno

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