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EECS 142
Lecture 21: Sinusoidal Oscillators
Prof. Ali M. Niknejad
University of California, Berkeley
Copyright c 2005 by Ali M. Niknejad
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Oscillators
t
T =1
f
V0
An oscillator is an autonomous circuit that converts DCpower into a periodic waveform. We will initially restrictour attention to a class of oscillators that generate asinusoidal waveform.
The period of oscillation is determined by a high-Q LCtank or a resonator (crystal, cavity, T-line, etc.). Anoscillator is characterized by its oscillation amplitude (or
power), frequency, stability, phase noise, and tuningrange.
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Phase Noise
0
Phase Noise
0 +
L() 100 dB
LC Tank Alone
Due to noise, a realoscillator does not have adelta-function power
spectrum, but rather a verysharp peak at theoscillation frequency.
The amplitude drops veryquickly, though, as onemoves away from the cen-ter frequency. E.g. acell phone oscillator has aphase noise that is 100dBdown at an offset of only
0.01% from the carrier!
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An LC Tank Oscillatoret cos0t
Note that an LC tank alone is not a good oscillator. Dueto loss, no matter how small, the amplitude of theoscillator decays.
Even a very high Q oscillator can only sustain
oscillations for about Q cycles. For instance, an LC tankat 1GHz has a Q 20, can only sustain oscillations forabout 20ns.
Even a resonator with high Q 106, will only sustainoscillations for about 1ms.
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Feedback Perspective
vo
von
gmvi n : 1
Many oscillators can be viewed as feedback systems.
The oscillation is sustained by feeding back a fraction ofthe output signal, using an amplifier to gain the signal,and then injecting the energy back into the tank. The
transistor pushes the LC tank with just about enoughenergy to compensate for the loss.
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Negative Resistance Perspective
Active
Circuit
Negative
Resistance
LC Tank
Another perspective is to view the active device as anegative resistance generator. In steady state, the
losses in the tank due to conductance G are balancedby the power drawn from the active device through thenegative conductance G.
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Feedback Approach
si(s) so(s)+
a(s)
f(s)
Consider an ideal feedback system with forward gaina(s) and feedback factor f(s). The closed-loop transferfunction is given by
H(s) =a(s)
1 + a(s)f(s)
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Feedback Example
As an example, consider a forward gain transferfunction with three identical real negative poles withmagnitude
|p
|= 1/ and a frequency independent
feedback factor f
a(s) =a0
(1 + s)3
Deriving the closed-loop gain, we have
H(s) =
a0
(+s)3 + a0f =
K1
(1 s/s1)(1 s/s2)(1 s/s3)
where s1,2,3 are the poles of the feedback amplifier.
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Poles of Closed-Loop Gain
Solving for the poles
(1 + s)3 = a0f
1 + s = ( a0f)1
3 = (a0f)1
3 ( 1)1
3
( 1)
1
3
= 1, ej60
, ej60
The poles are therefore
s1, s2, s3 = 1 (a0f)
1
3
, 1 + (a0f)
1
3ej60
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Root Locus
j
+60
601
a0f= 0
a0f= 8
j
3/
If we plot the poles on thes-plane as a function of theDC loop gain T0 = a0f wegenerate a root locus
For a0f = 8, the poles areon the j-axis with value
s1 = 3/
s2,3 = j
3/
For a0f > 8, the poles moveinto the right-half plane(RHP)
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Natural Response
Given a transfer function
H(s) =K
(s s1)(s s2)(s s3)=
a1
s s1+
a2
s s2+
a3
s s3The total response of the system can be partitioned intothe natural response and the forced response
s0(t) = f1(a1es1t + a2e
s2t + a3es3t) + f2(si(t))
where f2(si(t)) is the forced response whereas the first
term f1() is the natural response of the system, even inthe absence of the input signal. The natural response isdetermined by the initial conditions of the system.
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Real LHP Poles
e
t
Stable systems have all poles in the left-half plane(LHP).
Consider the natural response when the pole is on thenegative real axis, such as s1 for our examples.
The response is a decaying exponential that dies awaywith a time-constant determined by the pole magnitude.
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Complex Conjugate LHP Poles
Since s2,3 are a complexconjugate pair
s2, s3 = j0We can group theseresponses since a3 = a2
into a single term
a2es2t+a3e
s3t = Kaet cos 0t
et cos0t
When the real part of the complex conjugate pair isnegative, the response also decays exponentially.
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C l C j P l (RHP)
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Complex Conjugate Poles (RHP)
When is positive (RHP),the response is anexponential growing
oscillation at a frequencydetermined by theimaginary part 0
Thus we see for any am-plifier with three identicalpoles, if feedback is appliedwith loop gain T0 = a0f > 8,the amplifier will oscillate.
te cos0t
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F D i P i
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Frequency Domain Perspective
1
a0f= 8
Closed Loop Transfer Function
In the frequencydomain perspective,we see that a
feedback amplifierhas a transferfunction
H(j) = a(j)1 + a(j)f
If the loop gain a0f = 8, then we have with purely
imaginary poles at a frequency x = 3/ where thetransfer function a(jx)f = 1 blows up. Apparently, thefeedback amplifier has infinite gain at this frequency.
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O ill ti B ild U
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Oscillation Build Upstart-up region
steady-state region
In a real oscillator, the amplitude of oscillation initiallygrows exponentially as our linear system theorypredicts. This is expected since the oscillator amplitude
is initially very small and such theory is applicable. Butas the oscillations become more vigorous, thenon-linearity of the system comes into play.
We will analyze the steady-state behavior, where thesystem is non-linear but periodically time-varying.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 21 p. 17/2
E l LC O ill t
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Example LC Oscillator
vovi
n : 1
The emitter resistor isbypassed by a large
capacitor at ACfrequencies.
The base of thetransistor is convenientlybiased through thetransformer windings.
The LC oscillator uses a transformer for feedback.Since the amplifier has a phase shift of 180, thefeedback transformer needs to provide an additional
phase shift of 180
to provide positive feedback.
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AC E i l t Ci it
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AC Equivalent Circuit
vovi
vo
von
At resonance, the AC equivalent circuit can besimplified. The transformer winding inductance Lresonates with the total capacitance in the circuit. R
Tis
the equivalent tank impedance at resonance.
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S ll Si l E i l t Ci it
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Small Signal Equivalent Circuit
roCin gmvin+vin
Rin Co
CLRL
L
n : 1
The forward gain is given by a(s) = gmZT(s), wherethe tank impedance ZT includes the loading effectsfrom the input of the transistor
R = R0||RL||n2Ri
C = CL + Cin2
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Open Loop Transfer F nction
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Open-Loop Transfer Function
The tank impedance is therefore
ZT(s) =1
sC+1
R +1
Ls
=Ls
1 + s2
LC+ sL/R
The loop gain is given by
af(s) = gmRn
LRs1 + LRs + s
2LC
The loop gain at resonance is the same as the DC loopgain
A =gmR
n
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Closed Loop Transfer Function
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Closed-Loop Transfer Function
The closed-loop transfer function is given by
H(s) =gmRLRs
1 + s2LC+ sLR(1 gmRn )Where the denominator can be written as a function ofA
H(s) =gmRLRs
1 + s2LC+ sLR
(1 A)
Note that as n , the feedback loop is broken andwe have a tuned amplifier. The pole locations aredetermined by the tank Q.
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Oscillator Closed Loop Gain vs A
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Oscillator Closed-Loop Gain vs A
A < 1
A = 1
0 =
1
LC
Closed Loop Transfer Function
If A = 1, then the denominator loss term cancels out
and we have two complex conjugate imaginary axispoles
1 + s2LC = (1 + sj
LC)(1
sj
LC)
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Root Locus for LC Oscillator
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Root Locus for LC Oscillator
For a second order transfer function, notice that themagnitude of the poles is constant, so they lie on acircle in the s-plane
s1, s2 =a2b a
2b
1 4b
a2=a2bj a
2b
4b
a2 1
|s1,2| =
a2
4b2+
a2
4b2(
4b
a2+ 1) =
1
b= 0
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Root Locus (cont)
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Root Locus (cont)
0
A < 1 A > 1
A = 1
We see that for A = 0, the poles are determined by the
tank Q and lie in the LHP. As A is increased, the actionof the positive feedback is to boost the gain of theamplifier and to decrease the bandwidth. Eventually, as
A = 1, the loop gain becomes infinite in magnitude.
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