Osc Lect21

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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

A. M. Niknejad University of California, Berkeley EECS 142 Lecture 21 p. 1/25

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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.


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


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.


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.


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.


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


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

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.


Oscillator Closed Loop Gain vs A

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Oscillator Closed-Loop Gain vs A

A < 1

A = 1

0 =

1

LC


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)


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


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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