Voltage-Controlled Oscillator (VCO)gram.eng.uci.edu/faculty/green/public/courses/270c/materials/... · EECS 270C / Spring 2014 Prof. M. Green / U.C. Irvine 1! Voltage-Controlled Oscillator

EECS 270C / Spring 2014 Prof. M. Green / U.C. Irvine 1

Voltage-Controlled Oscillator (VCO)

VC

fosc

fmin

fmax

slope = Kvco

Desirable characteristics: •  Monotonic fosc vs. VC characteristic

with adequate frequency range •  Well-defined Kvco

€

φout =Kvco

s + KPDKvcoF (s) / N⋅φVC

^ ^

Noise coupling from VC into PLL output is directly proportional to Kvco.

€

φin

€

φout

€

φVC

€

KPD

€

F (s)

€

Kvco

s

^

€

÷N

€

VC

€

VD +


Oscillator Design

€

A(s)

€

f

€

Vin ⇒ 0

€

Vout

€

Vout

Vin

≡HCL (s) =A(s)

1+ f ⋅A(s)

loop gain

Barkhausen’s Criterion:

If a negative-feedback loop satisfies:

€

f ⋅ A jωo( ) ≥1

∠A jωo( ) = −180!

then the circuit will oscillate at frequency ω0.


Inverters with Feedback (1)

V1 V2

V1

V2 1 inverter

feedback

V1

V2

2 inverters

feedback

1 stable equilibrium point

3 equilibrium points: 2 stable, 1 unstable (latch)

1 inverter:

V1 V2

2 inverters:


Inverters with Feedback (2)

3 inverters forming an oscillator:

1 unstable equilibrium point due to phase shift from 3 capacitors

V1 V2

V1

V2

Let each inverter have transfer function

€

Hinv ( jω) =A0

1+ jω p

€

Hloop ( jω) = Hinv ( jω)[ ]3

=A

0

3

1+ jω p( )3

Loop gain:

Applying Barkhausen’s criterion:

€

∠Hloop ( jω) = −3 tan−1 ωp

%

& '

(

) * = −180! ⇒ ωo = 3⋅ p

€

Hloop ( jωo ) =A

0

3

1+ 3[ ]3

2> 1 ⇒ A0 > 2


Ring Oscillator Operation

VA VB VC

tp tp tp

VA

VB

VC

VA

tp

tp

tp

€

12

Tosc

12Tosc = 3tp

⇒Tosc = 6tp


Variable Delay Inverters (1)

VC

Vin Vout

Current-starved inverter: Inverter with variable load capacitance:

Vin Vout

VC


Variable Delay Inverters (2)

R R

Vin+ Vin- Vin+ Vin-

Vout- Vout+

Ifast Islow

RG RG

ISS

VC +

_

Interpolating inverter:

•  tp is varied by selecting weighted sum of fast and slow inverter. •  Differential inverter operation and differential control voltage •  Voltage swing maintained at ISSR independent of VC.

VA

VB

VC

VD

tp

tp

tp

€

12

Tosc

tp

VA


Differential Ring Oscillator

additional inversion (zero-delay)

VA + −

Use of 4 inverters makes quadrature signals available.

VB + −

VC + −

VD + −

VA − +

12Tosc = 4tp

⇒Tosc = 8tp


Resonance in Oscillation Loop

€

Hr (s)

€

Hr (s)

ω ωr

€

Hr ( jω)

€

∠Hr ( jω)

ω ωr

€

+π2

€

−π2

1

At dc: Since Hr(0) < 1, latch-up does not occur.

At resonance:

€

Hr ( jωr ) > 1

∠Hr ( jωr ) = 0

€

⇒ ωo =ω r


LC VCO

Vin Vout

Vin Vout

€

Hr (s)

C L

€

ωr =1LC

€

⇒

realizes negative resistance

2L

C C

€

Hr (s)

€

Hr (s)

€

⇒


A. Reverse-biased p-n junction

+ – VR

VR

Cj

B. MOSFET accumulation capacitance

+

–

VBG

varactor = variable reactance

Variable Capacitance

VBG

Cg

accumulation region

inversion region

p-channel

n diffusion in n-well


LC VCO Variations

2L

C C

2L

C C

2L

C C

ISS

2L

C C

IS IS


1. ideal capacitor load

2. CML buffer load

Effect of CML Loading

1.

3.8 Ω 1 nH

400 fF 400 fF

Cg = 108fF

1 nH 3.8 Ω

400 fF 400 fF 108 fF 108 fF

2.


Substantial parallel loss at high frequencies ⇒ weakens VCO’s tendency to oscillate

(note p < z) where:

€

Yin = jωCgs + jωCgd A0 ⋅1+ jω / z1+ jω / p

€

A0 = 1+ gmR

€

1/ p = CL +Cgd( )R

€

1/ z =CLRA0

€

Re Yin( ) = A0Cgdω2 ⋅

1 p −1 z

1+ ω p( )2

CML Buffer Input Admittance (1)


Yin magnitude/phase: Yin real part/imaginary part:

magnitude

phase

imaginary

real

Contributes 2kΩ additional parallel resistance



imaginary

real

Contributes negative parallel resistance

Cg = 108 fF

3.8 nH

3.8 Ω 1 nH

400 fF 400 fF


3. CML tuned buffer load


Loading VCO with tuned CML buffer allows negative real part at high frequencies ⇒ more robust oscillation!

ideal capacitor load

CML buffer load

CML tuned buffer load


Cg = 108 fF

3.8 nH

3.8 Ω 1 nH

400 fF 400 fF


Differential Control of LC VCO

Differential VCO control is preferred to reduce VC noise coupling into PLL output.


Ring Oscillator LC Oscillator

– slower – low Q ⇒ more jitter generation + Control voltage can be applied differentially + Easier to design; behavior more predictable + Less chip area

+ faster + high Q ⇒ less jitter generation – Control voltage applied single-ended – Inductors & varactors make design more difficult and behavior less predictable – More chip area (inductor)

Oscillator Type Comparison


Random Processes (1)

Random variable: A quantity X whose value is not exactly known. Probability distribution function PX(x): The probability that a random variable X is less than or equal to a value x.

0.5

1

x

PX(x)

Example 1:

€

X ∈ [−∞,+∞]Random variable


0.5

1

x

PX(x)

x1 x2

€

P X ∈ [x1,x2]( ) = P(x2) −P(x1)

Probability of X within a range is straightforward:

If we let x2-x1 become very small …



Probability density function pX(x): Probability that random variable X lies within the range of x and x+dx.

€

pX (x) ⋅dx = PX (x + dx) −PX (x)

€

⇒ pX (x) =dPX (x)

dx

0.5

1

x

PX(x)

x

pX(x)

dx

€

P X ∈ x1,x2[ ]( ) = pX (x) dxx1

x2∫



Expectation value E[X]: Expected (mean) value of random variable X over a large number of samples.

€

E[X ] ≡ X = x ⋅pX (x)dx−∞

+∞

∫

Mean square value E[X2]: Mean value of the square of a random variable X2 over a large number of samples.

€

E[X 2] = x2 ⋅pX (x)dx−∞

+∞

∫

Variance:

€

E (X − X )2[ ] ≡σ 2 = x − X( )2pX (x)dx

−∞

+∞

∫

Standard deviation:

€

σ = E (X − X )2[ ]



Gaussian Function

€

f (x) =1

σ 2πexp −(x − X )2

2σ 2

%

& ' '

(

) * *

x

€

f (x)

€

X

€

1σ 2π

€

0.607σ 2π

€

X −σ

€

X +σ

2σ

€

f (x)dx = 1−∞

+∞

∫

1.  Provides a good model for the probability density functions of many random phenomena.

2.  Can be easily characterized mathematically . 3.  Combinations of Gaussian random variables are themselves

Gaussian.

€

σ,X ( )


Joint Probability (1)

€

P X ∈ x,x + dx[ ] and Y ∈ y,y + dy[ ]( ) = pX (x) ⋅pY (y) ⋅dx dy

€

P(x,y) ≡ P X ≤ x and Y ≤ y( )

If X and Y are statistically independent (i.e., uncorrelated):

Consider 2 random variables:


Consider sum of 2 random variables:

€

Z = X +Y

x

y

€

x + y = z0

€

x + y = z0 + dz

dx

dy = dz

€

P Z ∈ z0, z0 + dz[ ]( ) = pX (x)pY (y) dx dystrip∫∫

€

= pX (x)pY (z0 − x) dx−∞

∞

∫% & ' (

) * dz

€

pZ (z0)

determined by convolution of pX and pY.



*

Example: Consider the sum of 2 non-Gaussian random processes:



3 sources combined:

*



4 sources combined:

*



Central Limit Theorem: Superposition of random variables tends toward normality.

Noise sources



Fourier transform of Gaussians:

€

pX (x) =1

σ X 2πexp −(x − X )2

2σ X2

%

&

' '

(

)

* *

€

PX (ω) = exp −12σ X

2ω2%

& '

(

) * F

€

P Z ∈ z0,z0 + dz[ ]( ) = pX (x)pY (z0 − x) dx−∞

∞

∫& ' ( )

* + dz

Recall:

€

pZ (z0) = pX (x)pY (z0 − x) dx−∞

∞

∫ F

€

PZ (ω) =PX (ω) ⋅PY (ω)

€

= exp −12σ X

2ω2%

& '

(

) * ⋅exp −

12σY

2ω2%

& '

(

) *

€

= exp −12

(σ X2 +σ X

2 )ω2%

& '

(

) *

F -1

€

pZ (z) =1

2π σ X2 +σY

2( )exp −(z − Z)2

2 σ X2 +σY

2( )

%

&

' '

(

)

* *

Variances of sum of random normal processes add.


Autocorrelation function RX(t1,t2): Expected value of the product of 2 samples of a random variable at times t1 & t2.

€

RX (t1,t2 ) = E X (t1) ⋅X (t2 )[ ]

For a stationary random process, RX depends only on the time difference

€

RX (τ ) = E X (t) ⋅X (t +τ )[ ] for any t

€

RX (0) =σ 2Note €

τ = t1 − t2

Power spectral density SX(ω):

€

SX (ω) = E X (t) ⋅e− jωtdt−∞

+∞

∫2'

(

) ) )

*

+

, , ,

SX(ω) given in units of [dBm/Hz]


€

RX (τ ) =1

2πSX (ω) ⋅e jωτdω

−∞

∞

∫

Relationship between spectral density & autocorrelation function:

€

⇒ RX (0) =σ 2 =1

2πSX (ω)dω

−∞

∞

∫

Example 1: white noise

ω

€

SX (ω)

€

RX (τ )

τ

infinite variance (non-physical)

€

SX ω( ) = K

€

RX (τ ) =K2π

⋅δ t( )


Example 2: band-limited white noise

€

SX (ω)

ω

€

ωp

€

−ωp

€

RX (τ )

€

σ 2 =12

Kωp

€

RX (τ ) =σ 2e−ω p τ τ

€

K

€

SX ω( ) =K

1+ω2

ωp2

x

€

pX (x)

€

−σ

€

+σ

For parallel RC circuit capacitor voltage noise:

€

K =in

2

Δf⋅R2 = 2kBTR

€

ωp =1

RC

€

σVC2 =

kBTC


Random Jitter (Time Domain)

Experiment:

data source

CDR (DUT) analyzer

CLK

DATA RCK


Jitter Accumulation (1)

€

Tosc =1

fosc

Free-running oscillator output

Histogram plots

Experiment: Observe N cycles of a free-running VCO on an oscilloscope over a long measurement interval using infinite persistence.

NT

τ1 τ2 τ3 τ4

trigger

€

σ1

€

σ 2

€

σ 3

€

σ 4


Observation: As τ increases, rms jitter increases.

€

τ

€

στ2

proportional to τ2

proportional to τ

€



Noise Spectral Density (Frequency Domain)

fosc fosc+Δf

Sv(f)

Ltotal (Δf ) =10 logP1Hz fosc +Δf( )

Ptotal

"

#

$$

%

&

''

€

dBcHz[ ]

Δf (log scale)

Ltotal Δf( )

1/Δf2 region (-20dBc/Hz/decade)

Power spectral density of oscillation waveform:

€

dBmHz[ ]

Single-sideband spectral density:

Ltotal includes both amplitude and phase noise

Ltotal(Δf) given in units of [dBc/Hz]

1/Δf3 region (-30dBc/Hz/decade)


€

ωr + Δω

€

ωr

Noise Analysis of LC VCO (1)

active circuitry

C L R -R C L

+

_ vc

inR

€

Z( jω) =jωL

1− ωωr

$

% &

'

( )

2

€

ωr =1LC

€

Z j ωr + Δω( )[ ] =j ωr + Δω( )L

1−ωr

2 + 2ωrΔω + Δω( )2

ωr2

≈ jL ⋅ ωr2

2Δω

Consider frequencies near resonance:

€

Q =RωrL

noise from resistor

€

ωrL =RQ⇒ Z j ωr + Δω( )[ ] ≈ j R

2Q⋅ωr

Δω


Spot noise current from resistor:

€

inR2 =

4kTR

⋅ ΔfC L

+

_ vc

inR

€

vc2 = inR

2 ⋅ | Z( jω) |2

=4kTR

Δf ⋅ R ωr Δω2Q

%

& '

(

) *

2

= 4kTR ⋅ωr Δω

2Q%

& '

(

) *

2

⋅ Δf

Noise Analysis of LC VCO (2)

€

L Δω{ } = 10 ⋅ log F ⋅kTPsig

1+ωr

2Q ⋅ Δω

%

& '

(

) *

2+ , -

. -

/ 0 -

1 - 1+

ω1/ f 3

Δω

%

&

' '

(

)

* *

2

3

4 4

5

6

7 7

Leeson’s formula (taken from measurements):

Where F and ω1/f3 are empirical parameters.

dBc/Hz

spot noise relative to carrier power


Oscillator Phase Disturbance Current impulse Δq/Δt

_ + Vosc

t t

ip(t)

Vosc(t) Vosc(t)

Vosc jumps by Δq/C

•  Effect of electrical noise on oscillator phase noise is time-variant. •  Current impulse results in step phase change (i.e., an integration). ⇒ current-to-phase transfer function is proportional to 1/s

ip(t)

€

τ 1

€

τ 2

€

Δφ = 0

€

Δφ < 0

ip(t)


Impulse Sensitivity Function (1) The phase response for a particular noise source can be determined at each point τ over the oscillation waveform.

€

Γ(τ ) ≡ Δφ(τ )Δq

⋅qmaxImpulse sensitivity function (ISF):

€

= C ⋅Vmax(normalized to signal amplitude)

change in phase charge in impulse

t

τ

€

Vosc (t)

€

Γ(τ )

€

Vmax

Example 1: sine wave

t

τ

€

Vosc (t)

€

Γ(τ )

Example 2: square wave

Note Γ has same period as Vosc.


Impulse Sensitivity Function (2)

€

H(s)h(t)

€

iin

€

φout

Recall from network theory:

€

Φout (s)Iin (s)

= H(s)LaPlace transform:

€

φout (t) = h(t,τ ) ⋅ iin (τ ) dτ0

t

∫Impulse response:

time-variant impulse response

€

Γ(τ ) ≡ Δφ(τ )Δq

⋅qmax ⇒ Δφ(τ ) =Γ(τ )qmax

⋅ ΔqRecall:

ISF convolution integral:

€

φ(t) =Γ(τ )qmax0

t

∫ ⋅u(t −τ ) ⋅ i (τ ) ⋅dτ[ ] =Γ(τ )qmax0

t

∫ ⋅ i (τ ) ⋅dτ

from Δq

€

= 1 for τ ∈ (0,t)

€

Γ(τ ) = ck cos kωoscτ +θk( )k=0

∞

∑

Γ can be expressed in terms of Fourier coefficients:


Case 1: Disturbance is sinusoidal:

€

i (t) = I0 cos mωosc + Δω( )t[ ] , m = 0, 1, 2, …

€

=I0

2qmax

ck

sin (k + m)ωosc + Δω[ ]t +θk{ }(k + m)ωosc + Δω

+sin (k −m)ωosc + Δω[ ]t +θk{ }

(k −m)ωosc + Δω

& ' (

) (

* + (

, ( k=0

∞

∑

negligible significant only for m = k

(Any frequency can be expressed in terms of m and Δω.)

€

φ(t) =I0

qmax

ck cos kωoscτ +θk( ) ⋅cos mωosc + Δω( )t[ ]{ }dτ0

t

∫k=0

∞

∑

€

Γ(τ )

€

≈I0

2qmax

⋅cm ⋅sin Δω t +θk( )

Δω



ω

I

2ωosc

Δω

φ

€

×I0

2qmax

c0

ω1

€

×I0

2qmax

c1

ω1

€

×I0

2qmax

c2

ω1


€

φ(t) ≈ I02qmax

⋅cm ⋅sin Δω t +θk( )

Δω⇒φ2 =

I02

8qmax2

⋅cm

2

Δω( )2

Current-to-phase frequency response:

ωosc ωosc-ω1

ω1

ω1 ωosc+ω1 2ωosc-ω1 2ωosc+ω1

For

€

i (t) = I0 cos mωosc + Δω( )t[ ]


€

×c0

€

×c1

€

in2

Δf

€

4kTγgm

ω ωosc 2ωosc

€

×c2

Case 2: Disturbance is stochastic:


MOSFET current noise:

thermal noise 1/f

noise

€

in2(f )Δf

= 4kTγgm + gm2 Kf

CgfA2/Hz

€

in

€

φ2 Δf ≈ in2 Δf

8qmax2

⋅cm

2

Δω( )2

€

Sφ Δω( )

Δω

€

in2

Δf

ω ωosc 2ωosc

€

×c0

€

gm2 2π ⋅Kf

Cgωthermal noise

1/f noise



€

Sφ Δω( )

Δω

€

×c0

€

×c1

€

in2

Δf

ω ωosc 2ωosc

€

×c2

€

Sφ (Δω) =1

8qmax2

4kTγgm ⋅

ck2

0

∞

∑

Δω( )2

+ 2π gm2 Kf

Cg

⋅c0

2

Δω( )3

*

+

, , , , ,

-

.

/ / / / /

due to 1/f noise

due to thermal noise

Total phase noise:

€

c02 = Γ ( )

2

ck2

k=0

∞

∑ = Γrms( )2

ωn



€

Sφ (Δω) =1

8qmax2

4kTγgm ⋅Γrms( )

2

Δω( )2

+ 2π gm2 Kf

Cg

⋅Γ ( )

2

Δω( )3

)

*

+ + +

,

-

.

.

.

€

4kTγgm ⋅Γrms( )

2

Δω( )2

= 2π gm2 Kf

Cg

⋅Γ ( )

2

Δω( )3

€

⇒

€

Δωn ,phase =π

2kT⋅

gm

γCg

⋅ΓΓrms

(

) * *

+

, - -

2

noise corner frequency ωn

Δω (log scale)

€

Sφ Δω( ) (dBc/Hz)

€

Δωn ,phase

1/(Δω)3 region: −30 dBc/Hz/decade



t

τ

€

Vosc (t)

€

Γ(τ )

t

τ

€

Vosc (t)

€

Γ(τ )

Example 1: sine wave Example 2: square wave


Example 3: asymmetric square wave

t

τ

€

Vosc (t)

€

Γ(τ )

€

Γ > 0 ⇒ will generate more 1/(Δω)3 phase noise

€

Γrms is higher ⇒ will generate more 1/(Δω)2 phase noise



Effect of current source in LC VCO:

Vosc + _

Due to symmetry, ISF of this noise source contains only even-order coefficients − c0 and c2 are dominant.

⇒ Noise from current source will contribute to phase noise of differential waveform.



ID varies over oscillation waveform

€

in2

Δf= 4kTγgm (t)

= (4kTγ ) ⋅ µCoxWL⋅ VGS (t) −Vt( )

&

' (

)

* +

Same period as oscillation

€

in02

Δf= (4kTγ ) ⋅ µCox

WL⋅ VGS(DC ) −Vt( )

&

' (

)

* + Let

Then

€

in2

Δf=

in02

Δf⋅α(t)

€

α(t) =VGS (t) −Vt

VGS(DC ) −Vtwhere

€

Γeff (τ ) = Γ(τ ) ⋅α(τ )We can use


ISF Example: 3-Stage Ring Oscillator

M1A M1B M2A M2B M3A M3B

MS1 MS2 MS3

R1A R1B R2A R2B R3A R3B + Vout −

fosc = 1.08 GHz PD = 11 mW


ISF of Diff. Pairs

ISF by tx6 for differential ring osc

-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x6


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x5


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x4


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x3


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x2

ISF by tx1 for 3stage differential ring osc

-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF

by t

x1

€

ΓM1A

€

ΓM1B

€

ΓM2A

€

ΓM2B

€

ΓM3A

€

ΓM3B

€

Γrms = 1.86Γ = −0.26

for each diff. pair transistor


ISF of Resistors

€

ΓR1A

€

ΓR2A

€

ΓR3A

€

Γrms = 1.72Γ = −0.16

for each resistor


ISF of Current Sources

ISF by tail tx3 for differential ring osc

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF

by t

ail t

x3


-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF

by t

ail t

x2


-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF

by t

ail t

x1

€

ΓMS1

€

ΓMS2

€

ΓMS3

ISF shows double frequency due to source-coupled node connection.

€

Γrms = 1.00Γ = −0.12

for each current source transistor


Phase Noise Calculation

Using: Cout = 1.13 pF Vout = 601 mV p-p qmax = 679 fC

€

L Δf{ } = 6 ⋅Γrms(dp)

2

8π 2Δf 2⋅4kTγ gm(dp)

qmax2

+ 6 ⋅Γrms(res)

2

8π 2Δf 2⋅4kT R

qmax2

+ 3 ⋅Γrms(cs)

2

8π 2Δf 2⋅4kTγ gm(cs)

qmax2

€

322Δf 2

€

122Δf 2

€

70Δf 2

€

⇒ L Δf{ } =514Δf 2 = −112 dBc/Hz @ Δf = 10 MHz


Phase Noise vs. Amplitude Noise (1)

€

Vosc (t) = Vc +v (t)[ ] ⋅exp j ωosct +φ(t)( )[ ]

ωosct

φ v vφ Spectrum of Vosc would

include effects of both amplitude noise v(t) and phase noise φ(t).

How are the single-sideband noise spectrum Ltotal(Δω) and phase spectral density Sφ(ω) related?



t t

i(t) i(t)

Vc(t) Vc(t)

0=Δtosc

qt

ωΔ

=Δ

Recall that an input current impulse causes an enduring phase perturbation and a momentary change in amplitude:

Amplitude impulse response exhibits an exponential decay due to the natural amplitude limiting of an oscillator ...


Δω

€

Lamp Δω( )

€

ωcQ

Δω

+

Δω

€

Ltotal Δω( )

Phase noise dominates at low offset frequencies.


€

Lφ Δω( )

Δω


€

Vosc (t) = Vc +v (t)( ) ⋅cos ωosct +φ(t)( )≈ Vc +v (t)( ) ⋅ cos(ωosct) −φ(t) ⋅sin(ωosct)[ ]= Vc cos(ωosct) −φ(t) ⋅Vc sin(ωosct) +v (t) ⋅cos(ωosct)

ωosc Phase & amplitude noise can’t be distinguished in a signal.

Sv(ω)

Amplitude limiting will decrease amplitude noise but will not affect phase noise.


noiseless oscillation waveform

phase noise

component

amplitude noise

component

phase noise

amplitude noise

ω


Sideband Noise/Phase Spectral Density

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Vosc (t) = Vc ⋅cos ωosct +φ(t)( )≈Vc ⋅ cos(ωosct) −φ(t) ⋅sin(ωosct)[ ]

€

Vc ⋅cos(ωosct) −Vc ⋅φ(t) ⋅sin(ωosct)

€

Pphase noise

Psignal

=

12

Vc2 ⋅φ2

12

Vc2

=φ2

€

Lphase Δω( ) =12⋅Sφ Δω( )

noiseless oscillation waveform

phase noise

component


Jitter/Phase Noise Relationship (1)

€

στ2 ≡

1ωosc

2⋅E φ(t +τ ) −φ(t)[ ]

2) * +

, - .

€

=1

ωosc2

⋅ E φ2(t +τ )[ ] + E φ2 (t)[ ] −2E φ(t) ⋅φ(t +τ )[ ]{ }

€

Rφ (0)

€

Rφ (0)

€

2Rφ (τ )autocorrelation functions

€

Rφ (τ ) =1

2πSϕ (Δω) ⋅e j (Δω )τd(Δω)

−∞

∞

∫Recall Rφ and Sφ(Δω) are a Fourier transform pair:

€

⇒στ2 =

2ωosc

2⋅ Rφ (0) −Rφ (τ )[ ]

NT


€

Rφ (0) =1

2πSφ (Δω)d(Δω)

−∞

∞

∫

Rφ (τ ) =1

2πSφ (Δω) ⋅e j (Δω )τd(Δω)

−∞

∞

∫

€

στ2 =

1πωosc

2⋅ Sφ (Δω) 1−e j (Δω )τ( )−∞

∞

∫ d(Δω)

=1

πωosc2

⋅ Sφ (Δω) 1−cos(Δωτ ) − j sin(Δωτ )[ ]−∞

∞

∫ d(Δω)

=4

πωosc2

⋅ Sφ (Δω) ⋅sin2 Δωτ2

,

- .

/

0 1

0

∞

∫ d(Δω)



Let

€

Sφ (Δω) =a

(Δω)2

€

στ2 =

4πωosc

2⋅

a(Δω)2

⋅sin2 (Δω)τ2

(

) *

+

, -

0

∞

∫ d(Δω)

€

=4

πωosc2

⋅aπτ

4

€

=a

ωosc2

⋅τ

Let

€

Sφ (Δω) =b

(Δω)3

€

στ2 =

4πωosc

2⋅

b(Δω)3

⋅sin2 (Δω)τ2

(

) *

+

, -

ε

∞

∫ d(Δω)

€

= ζ ⋅ τ 2

Consistent with jitter accumulation measurements!


Jitter from 1/(Δω) noise: 2 Jitter from 1/(Δω) noise: 3

^

^

^

^

€

=a

fosc2⋅τ where a ≡ (2π )2 ⋅a^



Δf

€

Sφ Δf( ) (dBc/Hz)

-100

-20dBc/Hz per decade

•  Let fosc = 10 GHz •  Assume phase noise dominated by 1/(Δω)2

€

Sφ Δf( ) =a

(Δf )2

€

Sφ 2 ⋅106( ) =a

2 ⋅106( )2

= 10−10 ⇒ a = 400

Setting Δf = 2 X 106 and Sφ =10-10:

€

στ2 =

afc

2⋅τ =

400

10 ⋅109( )2⋅τ = 4 ⋅10−18[ ] ⋅τ

€

στ = 2 ⋅10−9[ ] ⋅ τ

Let τ = 100 ps (cycle-to-cycle jitter): ⇒ στ = 0.02ps rms (0.2 mUI rms)

Accumulated jitter:

2 MHz


More generally:

Δf

€

Sφ Δf( ) (dBc/Hz)

Δfm

Nm

-20 dBc/Hz per decade

€

στ2 =

afosc

2⋅τ =

fmfosc

%

& '

(

) *

2

⋅10Nm 10 ⋅τ

€

στ =fmfosc

$

% &

'

( ) ⋅10Nm 20 ⋅ τ ps[ ]

€

στ

Tosc

= fm ⋅10Nm 20 ⋅ τ UI[ ]

€

στ

Tosc

→ fm ⋅10 Nm+10( ) 20⋅ τ = fm ⋅10Nm 20 ⋅ τ&

' ( )

* + ⋅100.5

⇒ rms jitter increases by a factor of 3.2

€

Sφ Δf( ) =a

(Δf )2=

(Δfm )2 ⋅10Nm 10

(Δf )2


Let phase noise increase by 10 dBc/Hz:



Kpd

phase detector

loop filter

€

F (s) Kvco

VCO

€

÷N€

+φin

φout φvco

φfb

€

φout

φε= G(s) = Kpd ⋅F (s) ⋅Kvco

2πs⋅

1N

Open-loop characteristic:

€

φout =NG(s)1+G(s)

⋅φin +1

1+G(s)⋅φvcoClosed-loop characteristic:



€

G(s) =IchKvco

N⋅

1s2 (C +Cp )

⋅1+ sCR

1+ sCeq RRecall from Type-2 PLL:

Δω

|G| z p ω0

|1 + G|

-40 dB/decade

Δω

€

Sφ Δω( ) (dBc/Hz)

€

Δωn ,phase



Δω

€

φout

φvco

jΔω( )2

1

80 dB/decade

ω0

As a result, the phase noise at low offset frequencies is determined by input noise...


•  fosc = 10 GHz •  Assume 1-pole closed-loop PLL characteristic


Δf

€

Sφ Δf( ) (dBc/Hz)

Δf0 = 2 MHz

-100 -20dBc/Hz per decade

€

Sφ Δf( ) =

a

Δf0( )2

1+ΔfΔf0

$

% &

'

( )

2≈

a

Δf0( )2

, Δf << Δf0

a

Δf( )2

, Δf >> Δf0

+

,

- -

.

- -

€

⇒στ2 =

22π ⋅ fosc

2⋅ Rφ (0) −Rφ (τ )[ ]

=a

fosc2⋅1−e−2π⋅f0τ

2π ⋅Δf0

€

Rφ (τ ) = Sφ (Δf ) ⋅e j (2πΔf )τ ⋅d(Δf ) =a

2π ⋅Δf0( )⋅e−2π⋅f0τ

−∞

∞

∫


€

a = 4×102

For large τ:

στ = 0.02 ps rms cycle-to-cycle jitter


Δf0 = 2 MHz

fosc = 10 GHz

€

στ2 =

afosc

2⋅1−e−2π⋅f0τ

2π ⋅Δf0≈

afosc

2⋅τ

afosc

2⋅

12π (Δf0)

)

* + +

, + +

€

, τ <<1

2π (Δf0)

, τ >>1

2π (Δf0)

For small τ:

€

στ2 (log scale)

τ

€

1(2π ) ⋅ (2 MHz)

€

slope =a

fosc2

στ = 1.4 ps rms Total accumulated jitter


The primary function of a PLL is to place a bound on cumulative jitter:

τ

€

στ2 (log scale)

€

στ2 (log scale)

proportional to τ (due to thermal noise)

proportional to τ2 (due to 1/f noise)

τ



L(Δω) for OC-192 SONET transmitter

Closed-Loop PLL Phase Noise Measurement


Other Sources of Jitter in PLL

•  Clock divider

•  Phase detector Ripple on phase detector output can cause high-frequency jitter. This affects primarily the jitter tolerance of CDR.


Jitter/Bit Error Rate (1)

Histogram showing Gaussian distribution

near sampling point

1UI

Bit error rate (BER) determined by σ and UI …

L R €

2σR

€

2σ L

Eye diagram from sampling oscilloscope


R

€

2σ

0 T

€

T2

€

t0

€

T − t0

€

PL =1

σ 2π⋅ exp −

x2

2σ 2

&

' (

)

* +

t0

∞

∫ dx

€

PR =1

σ 2π⋅ exp −

T − x( )2

2σ 2

&

'

( ( (

)

*

+ + +

t0

∞

∫ dx

€

pL (t) =1

σ 2π⋅exp −

t2

2σ 2

&

' (

)

* +

€

pR (t) =1

σ 2π⋅exp −

T − t( )2

2σ 2

&

'

( ( (

)

*

+ + +

Probability of sample at t > t0 from left-hand transition:

Probability of sample at t < t0 from right-hand transition:

€

2σ



Total Bit Error Rate (BER) given by:

€

BER = PL + PU =1

σ 2π⋅ exp −

x2

2σ 2

&

' (

)

* +

t0

∞

∫ dx +1

σ 2π⋅ exp −

x2

2σ 2

&

' (

)

* +

T−t0

∞

∫ dx

€

=12

erfc t02σ

#

$ % %

&

' ( ( + erfc T − t0

2σ

#

$ % %

&

' ( (

*

+ , ,

-

. / /

€

where erfc(t) ≡2

π⋅ exp

t

∞

∫ −x2( )dx

€

PL =1

σ 2π⋅ exp −

x2

2σ 2

&

' (

)

* +

t0

∞

∫ dx

€

PR =1

σ 2π⋅ exp −

T − x( )2

2σ 2

&

'

( ( (

)

*

+ + +

t0

∞

∫ dx =1

σ 2π⋅ exp −

x2

2σ 2

&

' (

)

* +

T−t0

∞

∫



€

•

€

•

€

•

€

•

t0 (ps)

log BER

€

σ = 5 ps

€

σ = 2.5 ps

€

σ = 2.5 ps :

€

BER ≤10−12 for t0 ∈ 18ps, 82ps[ ]

€

σ = 5 ps :

€

BER ≤10−12 for t0 ∈ 36ps, 74ps[ ]

Example: T = 100ps

(64 ps eye opening)

(38 ps eye opening)

log(0.5)



Bathtub Curves (1)

The bit error-rate vs. sampling time can be measured directly using a bit error-rate tester (BERT) at various sampling points.

Note: The inherent jitter of the analyzer trigger should be considered.

€

JrmsRJ( )

measured

2= Jrms

RJ( )actual

2+ Jrms

RJ( )trigger

2


Bathtub Curves (2)

Bathtub curve can easily be numerically extrapolated to very low BERs (corresponding to random jitter), allowing much lower measurement times.

Example: 10-12 BER with T = 100ps is equivalent to an average of 1 error per 100s. To verify this over a sample of 100 errors would require almost 3 hours!

€

•

€

•

€

•

€

•

t0 (ps)


Equivalent Peak-to-Peak Total Jitter

BER

10-10

10-11

10-12

10-13

10-14

€

JPPRJ

σ, T determine BER BER determines effective Total jitter given by:

€

JPPRJ

€

JTJ = n ⋅σ( ) + JPPDJ

€

12

nσ

€

p(t)

€

12

nσ

Areas sum to BER

€

12.7 ⋅σ

€

13.4 ⋅σ

€

14.1⋅σ

€

14.7 ⋅σ

€

15.3 ⋅σ

Documents

Voltage-Controlled Oscillator (VCO)gram.eng.uci.edu/faculty/green/public/courses/270c/materials/... · EECS 270C / Spring 2014 Prof. M. Green / U.C. Irvine 1! Voltage-Controlled Oscillator