Page 190 - Phase-Locked Loops Design, Simulation, and Applications
P. 190

DESIGN PROCEDURE FOR MIXED-SIGNAL PLLS   Ronald E. Best                                116
               moment that the series capacitor  C has a very high value; thus, the varactor capacitance
                                                  S
               appears in parallel to the capacitance of the series connection of  C  and  C . Let
                                                                                          1
                                                                                                    2
                              , then the total capacitance of the tank circuit becomes C + C , where C is the
                                                                                         V         V
               capacitance of the varactor. The component values are now chosen such that the oscillator
               generates the output (radian) frequency ω 2min  when the loop filter output voltage is u fmin , and
               the output frequency is ω      when  u  =  u   . For u  =  u   , the varactor capacitance is
                                         2max        f    fmax       f     fmin
               supposed to be C vmax , while for u  = u fmax , the varactor capacitance is supposed to be C vmin .
                                                f
               Afterward, the maximum output frequency is determined by



                                                                                           (5.6)


               and the minimum output frequency is



                                                                                           (5.7)


               Afterward, the ratio of maximum to minimum output frequency is given by



                                                                                           (5.8)



                 Generally, these ratios are close to 1—for example, 1.01. This means the var-actor
               capacitance is small compared with capacitance C. Introducing the variables Δω  = ω       −
                                                                                              2     2max
               ω 2min  and ΔC  = C vmax  − C vmin , we have approximately
                             V


                                                                                           (5.9)




                 Now we introduce the relative quantities                                    , and get the

               expression


                                                                                           (5.10)


                 Because the relative quantities are much smaller than 1, we can apply a Taylor series
               expansion to Eq. (5.10) and finally get



                                                                                           (5.11)
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