From Paper
VCO Phase Noise and Benchmarking
\[\begin{align} FOM_{VCO} &= 10 \log \bigg(L_{VCO}(f_m) \cdot \frac{f_m^2}{f_{VCO}^2} \cdot \frac{P_{VCO}}{1mW}\bigg)\\ L_{VCO}(f_m) &= \frac{10^{FOM_{VCO}/10}}{P_{VCO}/1mW} \cdot \frac{f_{VCO}^2}{f_m^2} \end{align}\]where \(L\) is single sideband noise power to carrier power ratio.
Loop Phase Noise and Benchmarking
To calculate the output phase noise due to the reference, frequency divider, phase detector and the charge pump. Since they are low-pass filtered, at low frequency the phase noise level is
\[\begin{align} L_{loop} = \frac{S_{\phi,loop}}{2} = \frac{1}{2} \cdot N^2 \cdot \bigg( S_{\phi,ref} + S_{\phi,div} + S_{\phi, PD} + \frac{S_{i,CP}}{K_d^2} \bigg) \end{align}\]Then at high frequency
\[L_{loop}(f) = L_{loop} \cdot \big|\frac{G(s)}{1+G(s)}\big|^2\]Phase Noise Due to the Reference Path, Divider, and PD
According to this paper, the time jitter can be written in terms of the integral of the single-sided power spectral density (PSD) of the phase \(S_{\phi}\) within the Nyquist band (why?)
\[\begin{align} \sigma_{t_0}^2 = \frac{1}{4 \pi^2 f_{out}^2} \int_{0}^{f_{out}/2} S_{\phi}(f) df \end{align}\]So
\[\begin{align} S_{\phi} = 8 \pi^2 \cdot f_{ref} \cdot \sigma_{t}^2 \end{align}\]time jitter depending on the output voltage slope \(\alpha_{out}\)
\[\begin{align} \sigma_{t}^2 = \frac{\bar{v_{n}^2}}{\alpha_{out}^2} = \frac{F_{n}\cdot kT/C_{out}}{\alpha_{out}^2} \end{align}\]the total power in for the reference path, divider and PD can be expressed as
\[P = f_{ref} \cdot C_{tot} \cdot V_{dd}^2\]then time jitter can be expressed as
\[\sigma_{t}^2 = \frac{f_{ref}}{P} \cdot \bigg( \frac{F_{n} \cdot kT \cdot V_{dd}^2 \cdot C_{tot}/C_{out}}{\alpha_{out}^2} \bigg)\]To minimize the output jitter, designer can optimize the circuit by choosing the relative sizes of components, e.g., to maximize \(\alpha_{out}\). Once this optimization has been done, the jitter can always be reduced on a system level via admittance level scaling. Then the voltage slope at every node does not change. Thus, \(C_{tot}/C_{out}\) as well as \(F_{n}\) remains the same as all the node admittances scale toether. Therefore, on the system level, we can treat the bracketed part as a design-dependent constant and get
\[\sigma_{t}^2 \propto f_{ref}/P\]So
\[S_{\phi} \propto f_{ref}^2/P\]Phase Noise Due to the CP
\[\begin{align} S_{i} &= 8 kT \gamma \cdot (2 I_{CP}/ V_{ov})\\ S_{i,CP} &= S_{i} \cdot (\tau_{PD} / T_{ref}) \end{align}\]Power
\[P_{CP} = I_{CP} V_{dd} \tau_{PD} \cdot f_{ref}\]and using \(K_{d} = I_{CP}/2\pi\)
\[\frac{S_{i,CP}}{K_{d}^2} = \frac{f_{ref}^2}{P_{CP}} \cdot \bigg( \tau_{PD}^2 \cdot \frac{64 \pi^2 \gamma \cdot kT \cdot V_{dd}}{V_{ov}} \bigg) \propto \frac{f_{ref}^2}{P_{CP}}\]Loop Phase Noise Benchmarking
Thus
\[L_{loop} \propto N^2 \cdot \frac{f_{ref}^2}{P_{loop}} = \frac{f_{out}^2}{P_{loop}}\]define
\[\begin{align} FOM_{loop} &= 10 \log \bigg( L_{loop} \cdot \big( \frac{1 Hz}{f_{out}} \big)^2 \cdot \frac{P_{loop}}{1mW} \bigg)\\ L_{loop} &= 10^{FOM_{loop}/10} \cdot (\frac{f_{out}}{1 Hz})^2 \cdot \frac{1mW}{P_{loop}} \end{align}\]PLL Jitter and Benchmarking
To calculate the long-term PLL absolute jitter due to VCO
\[\begin{align} \sigma_{t,VCO}^2 = \frac{1}{2\pi^2 f_{out}^2} \cdot \int_{0}^{\infty} L_{VCO}(f) \cdot |H_{VCO}(s)|^2 d f \end{align}\] \[\sigma_{t,VCO}^2 = \frac{2 L_{VCO} (f_r) \cdot f_r^2}{f_{out}^2} \cdot \frac{f_{c,0}}{f_c} \cdot \int_{0}^{\infty} \bigg|\frac{1}{s \cdot [1 + G_0(s)]}\bigg|^2 df\] \[\sigma_{t,loop}^2 = \frac{L_{loop}}{2\pi^2 f_{out}^2} \cdot \frac{f_{c}}{f_{c,0}} \cdot \int_{0}^{\infty} \bigg| \frac{G_{0}(s)}{1+G_0(s)} \bigg|^2 df\] \[\sigma_{t,PLL} = \sigma_{t,VCO} + \sigma_{t,loop}\] \[\sigma_{t,PLL}^2 = \frac{1}{P_{PLL}} \cdot \bigg( 10^{\frac{FOM_{loop} + FOM_{VCO}}{20}} \cdot \frac{4}{\pi} \cdot \frac{1 mW}{1 Hz} \bigg) \cdot \square\] \[\mathrm{FOM}_{\mathrm{PLL}} = 10 \log \Big( \big(\frac{\sigma_{\mathrm{t,PLL}}}{1 \mathrm{ s}}\big)^2 \cdot \frac{P_{\mathrm{PLL}}}{1 \mathrm{ mW}} \Big)\] \[\mathrm{FOM}_{\mathrm{PLL}} = 10 \log \Big( \big(\frac{\sigma_{\mathrm{t,PLL}}}{1 \mathrm{ s}}\big)^2 \cdot \frac{\color{red}P_{\mathrm{PLL}}}{1 \mathrm{ mW}} \Big)\] \[\mathrm{FOM} = 10 \log \Big( \big(\frac{\sigma_{\mathrm{t,out}}}{1 \mathrm{ s}}\big)^2 \cdot \frac{P_{\mathrm{total}}}{1 \mathrm{ mW}} \Big)\]