Question 19.5: A PLL is to operate with a reference clock frequency of 10 M......
A PLL is to operate with a reference clock frequency of 10 MHz and an output frequency of 200 MHz. Design a loop filter for the charge-pump phase comparator so that the loop bandwidth is approximately 1 / 20th of the reference clock frequency and Q = 0.1. Assume the VCO has K_{osc} =2\pi \times 10^{7} rad/Vs and the charge-pump current is I_{ch} = 20\mu A .
From (19.32)
\omega _{pll} = \sqrt{\frac{K_{pd}K_{lp}K_{osc}}{N} } (19.32)
and (19.40),
\omega _{3dB} = \omega ^{2}_{pll}/\omega _{z} = \omega _{pll}/Q (19.40)
it is known that with Q « 0.5 the loop bandwidth is approximately
\omega _{3dB} \cong \frac{\omega _{pll} }{Q} = \frac{1}{Q} \sqrt{\frac{K_{pd} K_{lp} K_{osc} }{N} }
\Rightarrow K_{lp} = \frac{Q^{2} \omega ^{2}_{3db} N }{K_{pd} K_{osc} } (19.45)
Substituting (19.23)
K_{pd} = \frac{I_{ch}}{2\pi } (19.23)
and (19.27)
K_{lp} = 1/C_{1} (19.27)
for a charge-pump PLL into (19.45) and rearranging,
C_{1} = \frac{ I_{ch} K_{osc} }{ 2\pi Q^{2} \omega ^{2}_{3dB} N } (19.46)
In this example with \omega _{3dB} = 2\pi \cdot 10 MHz / 20 , substitution of the known values into (19.46) yields a value very close to C_{1} = 100 pF . Also from (19.40), with Q « 0.5 it is known that w_{z} \cong \omega _{3dB} Q^{2} . Hence, using (19.28)
\omega _{z} = 1/(RC_{1}) (19.28)
for a charge-pump loop filter reveals,
R = \frac{1}{C_{1} \omega _{3dB} Q^{2} } (19.47)
which in this example yields in a resistor value of R = 314 KΩ.