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LMX2301TM bảng dữ liệu(PDF) 11 Page - National Semiconductor (TI) |
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LMX2301TM bảng dữ liệu(HTML) 11 Page - National Semiconductor (TI) |
11 / 14 page Application Information LOOP FILTER DESIGN A block diagram of the basic phase locked loop is shown TLW12458 – 24 FIGURE 1 Basic Charge Pump Phase Locked Loop An example of a passive loop filter configuration including the transfer function of the loop filter is shown in Figure 2 TLW12458 – 25 Z(s) e s (C2 # R2) a 1 s2 (C1 # C2 # R2) a sC1 a sC2 FIGURE 2 2nd Order Passive Filter Define the time constants which determine the pole and zero frequencies of the filter transfer function by letting T2 e R2 # C2 (1a) and T1 e R2 # C1 # C2 C1 a C2 (1b) The PLL linear model control circuit is shown along with the open loop transfer function in Figure 3 Using the phase detector and VCO gain constants Kw and KVCO and the loop filter transfer function Z(s) the open loop Bode plot can be calculated The loop bandwidth is shown on the Bode plot (0p) as the point of unity gain The phase margin is shown to be the difference between the phase at the unity gain point and b180 TLW12458 – 27 Open Loop Gain e ii ie e H(s) G(s) e Kw Z(s) KVCO Ns Closed Loop Gain e io ii e G(s) 1 a H(s) G(s) TLW12458 – 26 FIGURE 3 Open Loop Transfer Function Thus we can calculate the 3rd order PLL Open Loop Gain in terms of frequency G(s) # H(s) ls e j # 0 e b Kw # KVCO (1 a j0 # T2) 02C1 # N(1 a j0 # T1) # T1 T2 (2) From equation 2 we can see that the phase term will be dependent on the single pole and zero such that w (0) e tanb1 (0 # T2) b tanb1 (0 # T1) a 180 (3) By setting dw d0 e T2 1 a (0 # T2)2 b T1 1 a (0 # T1)2 e 0 (4) we find the frequency point corresponding to the phase in- flection point in terms of the filter time constants T1 and T2 This relationship is given in equation 5 0p e 1 0T2 # T1 (5) For the loop to be stable the unity gain point must occur before the phase reaches b180 degrees We therefore want the phase margin to be at a maximum when the magni- tude of the open loop gain equals 1 Equation 2 then gives C1 e Kw # KVCO # T1 0p2 # N # T2 (1 a j0p # T2) (1 a j0p # T1) (6) http www nationalcom 11 |
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