+ v i F02E2P2 I. Solution (a.) The small-signal transfer function of the stages can be written as, V out (s) V in (s) = g m1 /g m3.

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1 ECE 6440 Summer 003 Page 1 Homework Assgnment No. 7 s Problem 1 (10 ponts) A fourstage rng oscllator used as the VCO n a PLL s shown. Assume that M1 and M are matched and M3 and M4 are matched. Also assume that g m = K' W L I D where K N = 100µA/V and K P = 50µA/V and that r ds =. The parastc capactors to ground at the outputs are 0.1pF each. (a.) If I =ma, fnd the frequency of oscllaton n Hertz. (b.) Fnd the W/L rato of M1 (M) necessary for oscllaton when I =ma. (c.) If the current I s used to vary the frequency, express the relatonshp between ω osc and I. In otherwords, fnd ω osc = f(i). V DD v + v o + v + v o + v + v o + v + v o + v v o v v o v v o v v o 10µm 1µm v o M3 M4 10µm 1µm v o + v M1 M + v F0EP I (a.) The smallsgnal transfer functon of the stages can be wrtten as, V out (s) V n (s) = g m1 /g m3 C Arg V out (jω) s g +1 = tan V n (jω) 1 ωc g m3 m3 From the above, we see that each stage must contrbute 45 of phase to oscllate. Therefore, ω osc = g m3 C = K' I 50x C = = rads/s f osc = 1.59GHz (b.) The gan of the 4stage rng oscllator at ω osc should be equal to 1 so we can wrte, 1 = g m1 /g m (g m1 /g m3 ) 4 = 4 g m1 = g m3 = g m3 = ms ms = K' N (W/L) 1mA = 100x10 6 (W/L) 1mA (W/L) 1 = ms 0.mS = 10 (W/L) 1 = 10 (c.) From part (a.) we get, ω osc = g m3 C = K' I C = 50x I =.36x10 11 I ω osc =.36x10 11 I

2 ECE 6440 Summer 003 Page Problem (10 ponts) How does the oscllaton frequency depend on I SS for a rng oscllator usng the stage shown? Express your answer n terms of V DD, V REF, I SS, the smple large sgnal model parameters of the MOSFETs (K, V T, λ) and the W/L values of the MOSFETs. Ths topology uses a replca basng crcut to defne the onresstance of M3 and M4 based on the onresstance of M5. The onresstance of M5 s V REF I SS W 8 L 8 + V DD M8 M5 0.5I SS M7 M3 v o1 v 1 M1 W 7 L 7 V DD M4 W 3 L 3 W 1 L 1 W L M6 W 4 L 4 v o M v ISS W 5 L 5 SU03H07P3 R on5 = V DD V REF 0.5I SS We can ether assume that the W/Ls of M3, M4 and M5 are equal or snce we know that R on s nversely proportonal to the W/L rato, we can wrte that, R on3 = R on4 = W 5 /L 5 W 3 /L R 3 on5 where W 3 /L 3 = W 4 /L 4. Assumng a capactance at each output of C L, allows us to wrte the transfer functon of the rng oscllator stage as, V o V o1 g m1 R on3 V 1 V = sr on3 C L + 1 The phase shft due to a stage can be wrtten as, θ(jω) = tan 1 (ωr on3 C L ) To oscllate, ths phase shft needs to be equal to some value, say k (n degrees). Therefore we can wrte that, ω osc = k 0.5I SS k R on3 C = L W 5 /L 5 W 3 /L (V 3 DD V REF )C L Therefore, the oscllaton frequency vares lnearly wth I SS.

3 ECE 6440 Summer 003 Page 3 Problem 3 (10 ponts) In every practcal oscllator, the LC tank s not the only source of phase shft. Hence, the actual oscllaton frequency may dffer somewhat from the resonant frequency of the tank. Usng the tmevaryng model, explan why the oscllators s phase nose can degrade f such offfrequency oscllatons occur. If there s any offfrequency oscllatons that are close to the actual oscllaton frequency or harmoncs of t, we know from the LTV theory that these frequences and ther assocated nose wll fold nto the nose spectrum around the actual frequency and degrade the oscllator s phase nose. The followng dagram llustrates the process. n f (ω) 1/f nose ω o ω S o φ (ω) c 0 c 1 c c 3 3ω o ω S v (ω) ω Phase Modulaton ω o ω oωo + ω o 3ω o ω Fg. 3.43

4 ECE 6440 Summer 003 Page 4 Problem 4 (10 ponts) Assume that the steadystate output ampltude of the followng oscllator s 1V. Calculate the phase nose n dbc/hz at an offset of 100kHz from the carrer from the sgnal comng out of the deal comparator. Assume that L 1 = 5nH, L = 100nH, M = 10nH, and C = 100pF. Further assume that the nose current s n1 k = M L 1 L L 1 L C Comparator v out + SU03H07P4 n1 = 4kTG eff f where 1/G eff = 50Ω. The temperature of the crcut s 300 K. Frst of all, several assumptons must be made to work ths problem. They are: 1.) The load on the secondary of the transformer approxmates a short..) The output of the comparator s a square wave of ampltude 0.5V. Our objectve s to fnd the value of L{f m } = 10log 10 n / f Γrms q max (f m ) Frst, the nfluence of the transformer. The equatons of a general transformer are, V 1 = sl 1 I 1 + smi and V = smi 1 + sl I If we assume that V 0, then I M L I 1 = 0.1I 1. Snce we are lookng at the square of the current, we can wrte that the nose njected nto the tank s n f n = 0.01 f = 0.04kTG eff = 0.04(1.381x103 ) = 3.314x10 4 A /Hz Next, we wll evaluate Γ rms. From the notes (page 1601), we see that Γ rms = 1 c n where c n are the coeffcents of the ISF represented by a Fourer seres. n=0 What are the c n? We shall assume that the ISF of the LC tank s a snusod of the same perod. Therefore, only the c 1 coeffcent s mportant. If the peak value of the ISF s 1V (a questonable assumpton) then the rms value s Thus Γ rms 0.5. q max = Cv max = 100pF(1V) = coulombs. L{f m } = 10log x104 (0.5) 10 0 (10 5 ) = 10log 10 (4.14x10 13 ) = 13.8 dbc/hz L{f m } = 13.8 dbc/hz

5 ECE 6440 Summer 003 Page 5 Problem 5 (10 ponts) A crystal reference oscllator and ts assocated transstor have the followng specfcatons at 90 K. Output frequency: 6.4MHz Power output: +10 dbm Nose fgure:.0 db Flcker corner: 15 khz Loaded Q: 1x10 3 (a.) Determne and plot the SSB phase nose n dbc as a functon of the frequency offset from the carrer. Include the frequency range from 10Hz to 10MHz. (b.) Suppose that ths reference oscllator s used wth a frequency syntheszer whose transfer functon from the reference to the output s θ n,o (s) θ n,ref (s) = N Nref ζω n s + ω n s + ζω n s + ω n where N = 19,000, N ref = 56, ζ = 0.7, and ω n = 908 sec. 1. Make a plot of the SSB reference nose n the output of the syntheszer. (a.) NF =.0dB, F = 10.0/10 = 1.585, and P o = 10 dbm = 0.01W L{f m } = 10 log FkT P s Q f o f m 1 + f c f m = 10 log x (1x10 3 ) 6.4x10 6 f kHz m f m L{f m } = 10 log 6.348x x103 f m x104 f m dbc Offset from carrer, f m (Hz) SU03H07S5A

6 ECE 6440 Summer 003 Page 6 Problem 5 Contnued (b.) The VCO phase nose transfer functon s θ n,o (s) θ n,ref (s) = N ζω n s + ω n Nref s + ζω n s + ω = s x10 5 n s s x10 5 θ n,ref (dbc) = 10 log θ n,o (jω) L{f θ n,ref (jω) m } Below s a plot of the above equaton as well as the transfer functon, θ n\,o (s)/θ n\,ref (s), and the nput reference nose dbc Nose Floor Input Reference Nose Output Nose Transfer Functon Ths regon s not possble f m SU03H07S5B