F2011-3EJ4 Set 06 Oscillators Students

8/3/2019 F2011-3EJ4 Set 06 Oscillators Students

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6-15-1

ECE 3EJ4Electronic Devices & Circuits II

Lecture Set 6Lecture Set 6 OscillatorsOscillators

Prof. M. Jamal DeenProf. M. Jamal Deen

Professor and Senior Canada Research Chair

Dept. of Electrical and Computer EngineeringMcMaster University Hamilton, ON L8S 4K1, Canada

6-25-2

IntroductionIntroduction OscillatorsOscillators


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6-3

Oscillators

Text - Sections 17.1, 17.2, 17.3, 17.4

Lecture notes

Review frequency analysis and op-amps

Review small-signal models of BJTs and FETs

Practice problems related to class lectures and materialin textSolve individually, worked examples and compare your answers with

those in text

Solve as many exercises as possible check with answers in text

Suggestion - 17.1, 17.2, 17.3-17.4, 17.5-17.6, 17.7, 17.8-17.9, 17.10, 17-13-17.15

Solve as many problems at end of chapter 17

Suggestion 17.1, 17.4, 17.8, 17.10, 17.15, 17.16, 17.21, 17.22, 17.23, 17.27

Some specific suggestions on exercises and problems to attemptwill be provided in class during lectures

6-4

Early Circuit with Oscillator

Oscillators - technique of combining gain circuit with feedback circuit

Together have phase/time delay required for system to oscillate atspecific frequency

http://pdfserv.maxim-ic.com/en/an/AN1768.pdf

Vintage 1929 Hartley-style transmitter


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6-5

Simple Oscillator1

0 i dtC

iL Ri

d

dt

= + +

( )expi t=

2 4

2

R R L C

L

=

24

0

Take complex root

R L C

1 oscillations grow

Need mechanism to force A = 1 at desired value ofoutput amplitude

Use non-linear circuit for gain control

Start-up A > 1 poles - RH of s-plane Stable oscillations A = 1 poles - LH

of s-plane

If A < 1, amplitude - detected by non-linear circuit &then loop gain until it becomes 1

Limiter circuit oscillation grows until amp. reacheslevel to which limiter is set

Use variable R

Element whose R is controlled by output

Put in feedback circuit so R determines

loop gain Diodes or JFETs

( )( )

( ) ( )

( )

( )( )

1

11

v

A sFrom A s

A s s

A swant L s

L s

=

=

j (rad/s)

(Np/s)

-


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6-9

Limiter CircuitLimiter Circuit -- Amplitude ControlAmplitude Control Consider small vi & small vo- vA +ve, vB -ve

D1 and D2 are off

Input I=v1/R1 thru Rf and

Voltages at nodes A and B by superposition

As vi +ve, vo -ve, vB more ve, D2 off vi +ve, vo -ve until vA = -0.7V & D1 on For VD=D1 voltage drop, using vA

expression, we get

If vi further, vA ~ VD & more I flows throughD1 and R3, but I(R2) constant

Thus R3 appears // to Rf

Similarly, we can derive

Removing Rf results in comparator-like I-V

6-10

WienWien--bridge Oscillatorbridge Oscillator

2 2 1

1

( ) 1( ) 1

( ) ( ) 1 ( ) ( )

P

P S P S

R Z s R RL s

R Z s Z s Z s Y s

+= + = + +

( ) 21

11

3 1

RL s

R sCR sCR

= + + +

( )( )

1

3 1 L j Gain

j CR CR

=+

Condition for oscillation ( )( )

( ) ( )( )

( )( ) 1

1 1v

A s A sFrom A s want L s

A s s L s= =

+

-

Vs

Va

ZP

ZS

V0

Put Vs=0


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6-11

Amplitude StabilizationD1, D2, R1-R4 form

amplitude controlnetwork

When vo +ve, D1 on

for v(R3) > VD(on)Now R4//R3 , effective

loop gain is reduced

6-12

Amplitude Stabilization

1 1

3 4

o o Dv v v v V iR R

= +


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6-13

Wien Bridge OscillatorWien Bridge OscillatorWien Bridge Oscillator with limiters for amplitude control

( )21

2 0 -1-L s Roots of in RH s plaR

nR

e= + =

6-14

Tuned Oscillator

High-Q bandpass filter f0

Positive feedback loop withhard limiter

Assume oscillation has started

Output of filter sinewave @ f0

Sine wave into limiter square

wave of frequency f0

Square wave fed to bandpassfilter which filters outharmonics and producessinewave

Peak-peak amplitude of sinewave VPP

From Fourier analysis, sinewave at f0 will have amplitude

4VPP/Purity of sine wave depends on

Q of filter

Band pass filter - 2nd order active op-amp + RC

Wave shaping circuit diodes as hard limiters

V1V2

tt

f0

Bandpass filter

V2

V1

Wave shaping circuit


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6-15

Active-filter Tuned Oscillator Break the circuit at the blue line

Circuit in red rectangle is Leq Equiv. circuit is shown - Leq = R

2C

Derive the expression for AV(s)

v2

v3 v1

v4

Show v2 = 2*v1 = 2*AV(s)*v3 v2 is fed to diode pair to create square wave v3

with p-p amplitude 1.4V (for VD,on = 0.7V)

v3 = 2*4VPP/= 3.57V

QR

C Leq

3 1

4 4

6-16

LCLC--tuned Oscillatorstuned Oscillators

Colpitts Hartley

( )1 21 2

1 21 ; 1o, ,HC oC CL

C C L L C = = ++


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6-17

Complete CircuitComplete Circuit ColpittsColpitts OscillatorOscillator

6-18

Colpitts Oscillator( ) 3 21 ;S o GSCG C CR r = +=

[ ]

( ) ( )1

3

3 3

3 ( )0

0

1

( )

GD g

sm ms

s C v ssC

v ss C C g

CL

g

s

C G

+ +

= + + + +VDD

RS

L

vs

VDD

RS

Nodal eqns

for Vg & Vs

L

No excitation = 0


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6-19

Colpitts Oscillator

Put real & imag parts = 0

( )

( )

21 3 1

3

1 33

0

GD

mm GD

C C C C C C

j g G C gC G

C

G

L

L

=

+ + +

+

+

+ +

=

i

Oscillation

condition set byL //CTotal

Feedback set byC ratio must be

large enough tomeet gainrequirement

L

6-20

VCO

VCO voltage controlled oscillator

Want to vary fVCO with vinput, fo = Function(vinput) ideally linearfunction

Typically fo = 1/RC

Let C vary with bias varactor or reverse biased pn diode

Let R change with bias voltage varistor use FET or BJT to makeequivalent R that depends on biasing

We will look at case of BJT (try FET as exercise)

CmT

Ig

V r

= =

( )bi

T

C as

V constant at fixe

I function V

d T

=

( )T Cr V I =i


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6-21

VCO

Let VCC or VEE both change with vi ICchanges, r changes

However, mode of BJT may change from active to cut-off or saturation

Better way use vi to control a currentsource keep BJT is same mode

Then use current source to bias BJTand r will follow IBias

6-22

VCO Circuit

Requiv

( )ir Function v =

( )( )

1o i

i

f G vF v C

= =

( )1

2of

r C=

Requiv


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6-23

LC OscillatorLC Oscillator Negative gNegative gmm

Simple topology for both circuits, determine the resistance Rxy

Differential implementation - two outputs are 180 degrees out of phase very usefulfor many applications driving a Gilbert cell mixer

Good phase noise performance can be achieved

Ibias oscillation amplitude control but it adds noise

Variable capacitor (varactor) controls foscillation by adjusting Vcont Much fixed capacitance cannot be removed - lowers frequency tuning range

LC Oscillator

x yx y

Voltage-controlled Oscillator

VDD

VDD VDD VDD VDD

6-24

PLL

Basically is a feedback control system

Is an electronic circuit used for frequency control or is a frequencyselective circuit

Synchronize with incoming signal

Maintain synchronization in presence of noise or frequency variations

Configured as frequency multipliers. Demodulators, tracking generators,clock recovery circuits

Has three basic componentsPhase-frequency detector differences in phase/frequency in two signals

Loop filter removes hf components form VCO

VCO

PDLoop

Filter

InputVCO

LOvd vC

vd = Kde


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6-25

PLL

6-26

PLL


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6-27

PLL

6-28

PLL


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6-29

Use of PLL Sample Problem

Design an 8-note keyboard following the pure tone pitch:

Pitch C D E F G A B C*

Fraction 1 9/8 5/4 4/3 3/2 5/3 15/8 2

Freq. (Hz) 520 585 650 693 780 867 975 1040

Available components are: ONE ideal op-amp, TWO diodes, 1/2, 1/3 and1/5 frequency dividers, PLLs, 8 touch switchers (represent the 8 keys),

any resistors and capacitors with 10% accuracy, 15V DC voltagesources.

The required output voltage amplitude for every frequency (Vpp/2) is3.5V (assuming frequency dividers and PLLs are scalable components

without any insertion loss).

End of Lectures

on Oscillators

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F2011-3EJ4 Set 06 Oscillators Students

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