Voltage-Controlled Oscillator (VCO)

Voltage-Controlled Oscillator (VCO) Desirable characteristics: Monotonic f osc vs. V C characteristic with adequate frequency range f max f osc Well-defined K vco f min slope = K vco VC V C in V K F(s) ^ D V C K vco PD + s out out = ^ K vco s + K PD K^ vco F(s)/N VC N Noise coupling from V C into PLL output is directly proportional to K vco. 1

Oscillator Design V in 0 A(s) V out V out V in H CL (s) = A(s) 1+ f A(s) loop gain f Barkhausen s Criterion: If a negative-feedback loop satisfies: ( ) 1 f Aj o A( j o )= 180 o then the circuit will oscillate at frequency 0.

Inverters with Feedback (1) 1 inverter: V 1 inverter V 1 V feedback 1 stable equilibrium point V 1 inverters: V V 1 V inverters feedback V 1 3 equilibrium points: stable, 1 unstable (latch) 3

Inverters with Feedback () 3 inverters forming an oscillator: V V 1 V 1 unstable equilibrium point due to phase shift from 3 capacitors V 1 Let each inverter have transfer function H inv ( j) = Loop gain: H loop ( j) = [ H inv ( j) ] 3 = A 0 3 ( 1+ j p) 3 A 0 1+ j p Applying Barkhausen s criterion: H loop ( j) = 3tan 1 = 180 o osc = p H loop ( j osc ) = A 0 3 [ 1+ 3] 3 > 1 A 0 > 3 p 4

Ring Oscillator Operation t p t p t p V A V B V C Total phase shift in loop: Total delay in loop: 3t p V A t p = 3t p T osc T osc = 6 t p V B t p V C t p V A 1 T osc 5

Variable Delay Inverters (1) Inverter with variable load capacitance: Current-starved inverter: V in V out V C V in V out V C 6

Variable Delay Inverters () Interpolating inverter: I SS R R V C + _ V out+ V out- V in+ V in- V in+ V in- R G R G I fast I slow t p is varied by selecting weighted sum of fast and slow inverter. Differential inverter operation and differential control voltage Voltage swing maintained at I SS R independent of V C. 7

Differential Ring Oscillator V A + V B + V C + V D + V A + V A t p additional inversion (zero-delay) V B t p Total phase shift in loop: Total delay in loop: 4t p V C t p = 4t p T osc T osc = 8 t p V D V A t p Use of 4 inverters makes quadrature signals available. 1 T osc 8

Resonance in Oscillation Loop H r ( j) H r (s) 1 H r (s) + H r ( j) r r At dc: Since H r (0) < 1, latch-up does not occur. At resonance: H r ( j r ) > 1 H r ( j r ) = 0 osc = r 9

LC VCO V in H r (s) V out L V in V out C r = 1 LC L C C realizes negative resistance 10

Variable Capacitance varactor = variable reactance A. Reverse-biased p-n junction C j + V R V R B. MOSFET accumulation capacitance C g p-channel V BG + n diffusion in n-well accumulation region inversion region V BG 11

LC VCO Variations L I S L I S C C C C L L C C C C I SS 1

Effect of CML Loading 1. 1. ideal capacitor load 1 nh 3.8 400 ff 400 ff 108 ff 108 ff. C g = 108fF 1 nh 3.8 400 ff 400 ff. CML buffer load 13

CML Buffer Input Admittance (1) Y in = jc gs + jc gd A 0 1+ j / z 1+ j / p A 0 = 1+ g m R where: 1/ p = ( C L +C gd )R 1/ z = C LR A 0 (note p < z) Re( Y in )= A 0 C gd 1 p 1 z 1+ p ( ) Substantial parallel loss at high frequencies weakens VCO s tendency to oscillate 14

CML Buffer Input Admittance () Y in magnitude/phase: Y in real part/imaginary part: magnitude imaginary phase real Contributes k additional parallel resistance 15

CML Buffer Input Admittance (3) 3. CML tuned buffer load 1 nh 3.8 C g = 108 ff imaginary 400 ff 400 ff 3.8 nh real Contributes negative parallel resistance 16

CML Buffer Input Admittance (4) ideal capacitor load C g = 108 ff 1 nh 3.8 400 ff 400 ff 3.8 nh CML buffer load Loading VCO with tuned CML buffer allows negative real part at high frequencies more robust oscillation! CML tuned buffer load 17

Differential Control of LC VCO Differential VCO control is preferred to reduce V C noise coupling into PLL output. 18

Oscillator Type Comparison 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) 19

Random Processes (1) Random variable: A quantity X whose value is not exactly known. Probability distribution function P X (x): The probability that a random variable X is less than or equal to a value x. 1 P X (x) Example 1: Random variable X [,+] 0.5 x 0

Random Processes () Probability of X within a range is straightforward: 1 P X (x) 0.5 PX ( [x 1,x ])= P(x ) P(x 1 ) x 1 x x If we let x -x 1 become very small 1

Random Processes (3) Probability density function p X (x): Probability that random variable X lies within the range of x and x+dx. p X (x) dx = P X (x + dx) P X (x) p X (x) = dp X (x) dx 1 P X (x) x ( )= p X (x) dx PX[ x 1,x ] p X (x) x 1 0.5 dx x x

Random Processes (4) Expectation value E[X]: Expected (mean) value of random variable X over a large number of samples. E[X ] X = + x p X (x)dx Mean square value E[X ]: Mean value of the square of a random variable X over a large number of samples. E[X ] = + x p X (x)dx Variance: E (X X ) [ ] = ( x X ) p X (x)dx Standard deviation: = E [(X X ) ] + 3

Gaussian Function 1. Provides a good model for the probability density functions of many random phenomena.. Can be easily characterized mathematically (,X ). 3. Combinations of Gaussian random variables are themselves Gaussian. 1 f (x) f (x) = 1 (x X ) exp 0.607 + f (x)dx = 1 X X + X x 4

Joint Probability (1) Consider random variables: P(x,y) PX ( x and Y y) If X and Y are statistically independent (i.e., uncorrelated): PX ( [ x,x + dx] and Y [ y,y + dy] )= p X (x) p Y (y) dx dy 5

Joint Probability () Consider sum of random variables: y Z = X +Y PZ ( [ z 0, z 0 + dz] )= p X (x)p Y (y) dx dy strip = p X (x)p Y (z 0 x) dx dz x + y = z 0 + dz x + y = z 0 dx dy = dz x p Z (z 0 ) determined by convolution of p X and p Y. 6

Joint Probability (3) Example: Consider the sum of non-gaussian random processes: * 7

3 sources combined: Joint Probability (4) * 8

4 sources combined: Joint Probability (5) * 9

Joint Probability (6) Noise sources Central Limit Theorem: Superposition of random variables tends toward normality. 30

Fourier transform of Gaussians: p X (x) = X 1 (x X ) exp X F P X () = exp 1 X Recall: ( )= p X (x)p Y (z 0 x) dx dz PZ[ z 0,z 0 + dz] p Z (z 0 ) = p Z (z) = 1 p X (x)p Y (z 0 x) dx ( ) exp X + Y (z Z) X + Y ( ) F F -1 P Z () =P X () P Y () = exp 1 X exp 1 Y = exp 1 ( X + X ) Variances of sum of random normal processes add. 31

Autocorrelation function R X (t 1,t ): Expected value of the product of samples of a random variable at times t 1 & t. R X (t 1,t ) = EX(t [ 1 ) X (t )] For a stationary random process, R X depends only on the time difference = t 1 t R X ( ) = EX(t) [ X (t + )] for any t Note R X (0) = Power spectral density S X (): S X () = E + X(t) e jt dt S X () given in units of [dbm/hz] 3

Relationship between spectral density & autocorrelation function: R X ( ) = 1 R X (0) = = 1 S X () e j d S X ()d Example 1: white noise infinite variance (non-physical) S X () R X ( ) S X ( )= K R X ( ) = K t () 33

Example : band-limited white noise S X () K R X ( ) = 1 K p p S X p K ( )= 1+ p R X ( ) = e p p X (x) For parallel RC circuit capacitor voltage noise: K = i n f R = k B TR p = 1 RC VC = k B T C + x 34

Random Jitter (Time Domain) Experiment: CLK data source DATA CDR (DUT) RCK analyzer 35

Noise Spectral Density (Frequency Domain) Power spectral density of oscillation waveform: S v () [ ] dbm Hz Single-sideband spectral density: L total ( ) [ ] dbc Hz 1/ 3 region (-30dBc/Hz/decade) 1/ region (-0dBc/Hz/decade) osc ( ) osc + L total () = 10 log P 1Hz osc + P total (log scale) L total () given in units of [dbc/hz] L total includes both amplitude and phase noise 36

Jitter Accumulation (1) Experiment: Observe N cycles of a free-running VCO on an oscilloscope over a long measurement interval using infinite persistence. NT Free-running oscillator output trigger T osc = 1 f osc 1 3 4 1 3 4 Histogram plots 37

Jitter Accumulation () proportional to Observation: As increases, rms jitter increases. proportional to 38

Noise Analysis of LC VCO (1) + noise from resistor r = C L R -R 1 LC Q = R r L active circuitry v c _ C L i nr jl Z( j) = 1 r Consider frequencies near resonance: Zj [ ( r + ) ]= j( r + )L 1 r + r + r ( ) jl r r L = R Q Zj [ ( + r )] j R Q r r r + 39

Noise Analysis of LC VCO () + v c _ C L i nr Noise current from resistor: i nr = 4kT R f v c = i nr Z( j) = 4kT R f R r Q = 4kTR r Q f Leeson s formula (taken from measurements): L{ }= 10 log F kt 1+ r P sig Q 1+ 3 1/f dbc/hz spot noise relative to carrier power Where F and 1/f 3 are empirical parameters. 40

Oscillator Phase Disturbance Current impulse q/t i p (t) i p (t) i p (t) _ t t V 1 osc + V osc (t) V osc (t) = 0 < 0 V osc jumps by q/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 41

Impulse Sensitivity Function (1) The phase response for a particular noise source can be determined at each point over the oscillation waveform. Impulse sensitivity function (ISF): ( ) ( ) q q max change in phase charge in impulse = C V max (normalized to signal amplitude) Example 1: sine wave V osc (t) Example : square wave V osc (t) V max t t ( ) ( ) Note has same period as V osc. 4

Impulse Sensitivity Function () Recall from network theory: H(s) h(t) i in out LaPlace transform: Impulse response: out (s) I in (s) out (t) = = H(s) t 0 h(t, ) i in ( ) d Recall: ( ) ( ) q q max ( ) = ( ) q max q time-variant impulse response ISF convolution integral: t ( ) (t) = u(t ) [ i( ) d ]= 0 q max from q t 0 ( ) q max i( ) d can be expressed in terms of Fourier coefficients: = 1for (0,t) ( ) = c k cos k osc + k k=0 ( ) 43

Impulse Sensitivity Function (3) Case 1: Disturbance is sinusoidal: i(t) = I 0 cos[ ( m osc + )t], m = 0, 1,, (Any frequency can be expressed in terms of m and.) (t) = I t 0 c k cos( k osc + k )cos ( m osc + )t q max { [ ]}d 0 = {[ ]t + k } + sin [ (k m) + osc ]t + k (k + m) osc + (k m) osc + I 0 sin (k + m) c + osc k q max k=0 I 0 k=0 ( ) ( ) c m sin t + k q max negligible { } significant only for m = k 44

Impulse Sensitivity Function (4) (t) I 0 [ ] For i(t) = I 0 cos ( m osc + )t ( ) c m sin t + k q max = I 0 8q max c m ( ) Current-to-phase frequency response: I osc osc 1 osc 1 osc + 1 osc 1 osc + 1 I 0 q max c 0 1 I 0 q max c 1 1 I 0 q max c 1 1 45

Case : Disturbance is stochastic: Impulse Sensitivity Function (5) MOSFET current noise: f i n f 8q max c m ( ) i n (f ) f = 4kTg m + g m K f C g f thermal noise 1/f noise A /Hz i n i n f 1/f noise K g f m C g c 0 4kTg m i n f thermal noise c 0 c 1 c osc osc osc osc S ( ) 46

Impulse Sensitivity Function (6) Total phase noise: i n f S () = 1 8q max c k 0 4kTg m due to thermal noise ( ) + g m K f c 0 C g ( ) 3 due to 1/f noise c 0 c 1 c n osc osc S ( ) c 0 = ( ) c k k=0 = rms ( ) 47

Impulse Sensitivity Function (7) S () = 1 8q max 4kTg m ( rms ) ( ) + g m K f C g ( ) ( ) 3 4kTg m ( rms ) ( ) = g m K f C g ( ) ( ) 3 n,phase = S ( ) (dbc/hz) C g kt g m noise corner frequency n rms 1/() 3 region: 30 dbc/hz/decade 1/() region: 0 dbc/hz/decade n,phase (log scale) 48

Impulse Sensitivity Function (8) Example 1: sine wave V osc (t) Example : square wave V osc (t) t t ( ) ( ) Example 3: asymmetric square wave V osc (t) rms is higher will generate more 1/() phase noise t ( ) > 0 will generate more 1/() 3 phase noise 49

Impulse Sensitivity Function (9) Effect of current source in LC VCO: Due to symmetry, ISF of this noise source contains only even-order coefficients c 0 and c are dominant. + V osc _ Noise from current source will contribute to phase noise of differential waveform. 50

Impulse Sensitivity Function (10) I D varies over oscillation waveform Same period as oscillation Let i n f = 4kTg (t) m W = (4kT) μc ox L V GS (t) V t i n0 f = (4kT) μc ox ( ) W L V GS(DC ) V t ( ) Then i n f = i n0 f (t) where (t) = V GS(t) V t V GS(DC ) V t We can use eff ( ) = ( ) ( ) 51

ISF Example: 3-Stage Ring Oscillator R 1A R 1B R A R B R 3A R 3B + V out M 1A M 1B M A M B M 3A M 3B M S1 M S M S3 f osc = 1.08 GHz PD = 11 mw 5

ISF of Diff. Pairs M1A ISF by tx1 for 3stage differential ring osc MA ISF by tx3 for differential ring osc M3A ISF by tx5 for differential ring osc 3 1 3 1 3 1 ISF by tx1 0 0-1 1 3 4 5 6 7 - ISF by tx3 0 0-1 1 3 4 5 6 7 - ISF by tx5 0 0-1 1 3 4 5 6 7 - -3-3 -3-4 -4-4 -5 M1B 3 Radian ISF by tx for differential ring osc -5 MB 3 Radian ISF by tx4 for differential ring osc -5 M3B 3 Radian ISF by tx6 for differential ring osc 1 1 1 ISF by tx 0 0-1 1 3 4 5 6 7 - ISF by tx4 0 0 1 3 4 5 6 7-1 - ISF by tx6 0 0-1 1 3 4 5 6 7 - -3-3 -3-4 -4-4 -5 Radian -5 Radian -5 Radian rms = 1.86 = 0.6 for each diff. pair transistor 53

ISF of Resistors R1A RA R3A rms = 1.7 = 0.16 for each resistor 54

ISF of Current Sources ISF by tail tx1 for differential ring osc ISF by tail tx for differential ring osc MS1 MS MS3 ISF by tail tx3 for differential ring osc 1.5 1 1.5 1 1.5 1 ISF by tail tx1 0.5 0 0-0.5 1 3 4 5 6 7 ISF by tail tx 0.5 0 0-0.5 1 3 4 5 6 7 ISF by tail tx3 0.5 0 0-0.5 1 3 4 5 6 7-1 -1-1 -1.5-1.5-1.5 - Radian - Radian - Radian rms = 1.00 = 0.1 for each current source transistor ISF shows double frequency due to source-coupled node connection. 55

Phase Noise Calculation Using: C out = 1.13 pf V out = 601 mv p-p q max = 679 fc L{ f }= 6 rms(dp) 8 f 4kT g m(dp) + 6 rms(res) 4kT R + 3 rms(cs) 8 f 8 f 4kT g m(cs) q max q max 3 1 70 f f f q max L{ f }= 514 f = 11 dbc/hz @ f = 10 MHz 56

Phase Noise vs. Amplitude Noise (1) How are the single-sideband noise spectrum L total () and phase spectral density S () related? [ ( )] V osc (t) = [ V c +v (t)]exp j osc t +(t) osc t v v Spectrum of V osc would include effects of both amplitude noise v(t) and phase noise (t). 57

Phase Noise vs. Amplitude Noise () Recall that an input current impulse causes an enduring phase perturbation and a momentary change in amplitude: i(t) i(t) t t V c (t) V c (t) t = 0 q t = osc Amplitude impulse response exhibits an exponential decay due to the natural amplitude limiting of an oscillator... 58

Phase Noise vs. Amplitude Noise (3) L amp ( ) L ( ) + c Q L total ( ) Phase noise dominates at low offset frequencies. 59

Phase Noise vs. Amplitude Noise (4) V osc (t) = ( V c +v (t))cos( osc t +(t) ) ( V c +v (t)) cos( osc t) (t) sin( osc t) [ ] = V c cos( osc t) (t) V c sin( osc t) +v (t) cos( osc t) S v () phase noise amplitude noise noiseless oscillation waveform phase noise component amplitude noise component Amplitude limiting will decrease amplitude noise but will not affect phase noise. osc Phase & amplitude noise can t be distinguished in a signal. 60

Sideband Noise/Phase Spectral Density ( ) [ ] V osc (t) = V c cos osc t +(t) V c cos( osc t) (t) sin( osc t) V c cos( osc t) V c (t) sin( osc t) noiseless oscillation waveform phase noise component P phase noise P signal = 1 V c 1 V c = L phase ( )= 1 S ( ) 61

Jitter/Phase Noise Relationship (1) NT T osc = 1 f osc 1 1 3 3 4 4 autocorrelation functions 1 osc = 1 osc E (t + ) (t) [ ] { [ ]} E[ (t + )]+ E[ (t)] E (t) (t + ) R (0) R (0) R ( ) = osc [ R (0) R ( )] Recall R and S () are a Fourier transform pair: R ( ) = 1 S () e j ( ) d() 6

Jitter/Phase Noise Relationship () R (0) = 1 R ( ) = 1 S ()d() S () e j ( ) d() = = = 1 osc 1 osc 4 osc S () 1 e j ( ) ( ) d() S ()[ 1 cos() j sin() ] S () sin d() 0 d() 63

Jitter/Phase Noise Relationship (3) Jitter from 1/() noise: Let S () = = = 4 osc 4 osc 0 ^ a () a 4 ^ a^ () () sin d() 3 Jitter from 1/() noise: Let S () = = 4 osc = b () 3 b () () 3 sin d() ^ = a osc = a f osc ^ where a ( ) a Consistent with jitter accumulation measurements! 64

Jitter/Phase Noise Relationship (4) S ( f ) (dbc/hz) -100 MHz -0dBc/Hz per decade f Let f osc = 10 GHz Assume phase noise dominated by 1/() S ( f )= S ( 10 6 )= a (f ) Setting f = X 10 6 and S =10-10 : a ( 10 6 ) = 1010 a = 400 Accumulated jitter: = a f = 400 c 10 10 9 ( ) [ ] = 4 1018 [ ] = 10 9 Let = 100 ps (cycle-to-cycle jitter): = 0.0ps rms (0. mui rms) 65

Jitter/Phase Noise Relationship (5) More generally: S ( f ) (dbc/hz) N m f m -0 dbc/hz per decade f S ( f )= = a f osc = a (f ) = (f m ) 10 Nm 10 (f ) f m f osc = f m f osc 10 N m 0 T osc = f m 10 N m 0 10 N m 10 [ UI] [ ps] Let phase noise increase by 10 dbc/hz: ( f m 10 Nm+10 ) 0 T osc = f m 10 N m 0 10 0.5 rms jitter increases by a factor of 3. 66

Jitter Accumulation (1) in fb phase detector K pd loop filter F(s) VCO K vco + vco out N Open-loop characteristic: out = G(s) = K pd F(s) K vco s 1 N Closed-loop characteristic: out = NG(s) 1+G(s) in + 1 1+G(s) vco 67

Jitter Accumulation () Recall from Type- PLL: -40 db/decade G(s) = I chk vco N 1 s (C +C p ) 1+ scr 1+ sc eq R S ( ) (dbc/hz) 1 + G 1/() 3 region: 30 dbc/hz/decade z 0 p G 1/() region: 0 dbc/hz/decade 1 out vco ( j) n,phase 80 db/decade As a result, the phase noise at low offset frequencies is determined by input noise... 0 68

Jitter Accumulation (3) S ( f ) (dbc/hz) -100-0dBc/Hz per decade f osc = 10 GHz Assume 1-pole closed-loop PLL characteristic S ( f )= a ( f 0 ) 1+ f f 0 a ( f 0 ), f << f 0 a ( f ), f >> f 0 f 0 = MHz f = R ( ) = S (f ) e j (f ) d(f ) = f osc [ R (0) R ( )] a ( f 0 ) ef0 = a f osc 1 ef 0 f 0 69

Jitter Accumulation (4) = a f osc 1 ef 0 f 0 a f osc a f osc 1 (f 0 ) 1, << (f 0 ) 1, >> (f 0 ) a = 4 10 f 0 = MHz f osc = 10 GHz (log scale) For small : = 0.0 ps rms cycle-to-cycle jitter For large : = 1.4 ps rms Total accumulated jitter slope = a f osc 1 ( ) ( MHz) 70

Jitter Accumulation (5) (log scale) proportional to (due to 1/f noise) proportional to (due to thermal noise) The primary function of a PLL is to place a bound on cumulative jitter: (log scale) 71

Closed-Loop PLL Phase Noise Measurement L() for OC-19 SONET transmitter 7

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. 73

Jitter/Bit Error Rate (1) Eye diagram from sampling oscilloscope Histogram showing Gaussian distribution near sampling point L R L R 1UI Bit error rate (BER) determined by and UI 74

Jitter/Bit Error Rate () p L (t) = 1 exp t p R (t) = 1 exp T t ( ) t T t 0 0 0 T T R Probability of sample at t > t 0 from lefthand transition: Probability of sample at t < t 0 from righthand transition: P L = P R = 1 exp x dx t 0 1 ( exp T x ) dx t 0 75

Jitter/Bit Error Rate (3) P L = 1 exp x dx t 0 P R = 1 ( exp T x ) dx = t 0 1 exp x Tt 0 Total Bit Error Rate (BER) given by: 1 BER = P L + P U = exp x dx + t 0 = 1 erfc t 0 + erfc T t 0 1 exp x dx Tt 0 where erfc(t) exp( x )dx t 76

Jitter/Bit Error Rate (4) Example: T = 100ps log BER =.5 ps log(0.5) = 5ps =.5 ps : BER 10 1 for t 0 18ps, 8ps [ ] = 5ps: BER 10 1 for t 0 36ps, 74ps [ ] (64 ps eye opening) (38 ps eye opening) t 0 (ps) 77

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. RJ ( J rms ) measured ( ) actual RJ = J rms ( ) trigger RJ + J rms 78

Bathtub Curves () Bathtub curve can easily be numerically extrapolated to very low BERs (corresponding to random jitter), allowing much lower measurement times. Example: 10-1 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! t 0 (ps) 79

Equivalent Peak-to-Peak Total Jitter p(t) BER 10-10 RJ J PP 1.7 Areas sum to BER 10-11 13.4 10-1 14.1 10-13 14.7 10-14 15.3, T determine BER BER determines effective Total jitter given by: RJ J PP 1 n 1 n J TJ = ( DJ n )+ J PP 80