震荡的条件 (Barkhausen’s criteria)

一个系统在$\omega_0$ 附近Oscillate的充分条件是

$$ \begin{align} \angle H(s) &= 180 ^\circ \\ |H(s)| &\ge 0 \end{align} $$

Ring Oscillators

One-Pole & Two-Pole Feedback System

One-Pole

当$\omega_0 \longrightarrow \infin$，Pole会产生$90 ^\circ$的Phase Shift
CS Stage 产生$180 ^\circ$的Phase Shift
故$\angle H(s) = 270 ^\circ$，无法Oscillate

Two-Pole

$$ \begin{equation} \begin{aligned} V_E &= \frac{R_{on}}{R_{on} + R_D} V_{DD} \\ R_{on} &= \frac{V_E}{I_D} \\ I_D &= \frac{1}{2} k_n (V_E - V_t)^2 \\ \therefore V_E &= \frac{V_{DD}}{1 + k_n R_D (V_{in} - V_t)} \end{aligned} \end{equation} $$

分析电路

2 CS Stage $\omega_0 \longrightarrow 0 \Longrightarrow \angle H(s) = 360 ^\circ$
根据Equation (3), $V_E \uparrow \Longrightarrow V_F \downarrow \Longrightarrow V_E \uparrow \Longrightarrow M_1 \ Closed$

电路增益曲线

Three-Stage Ring Oscillator

Oscillating Frequency and Transfer Function

$$ \begin{aligned} \angle H(s) &= 180 ^\circ \\ \therefore \angle H(s){each \ stage} &= 60 ^\circ \\ \therefore tan^{-1}\frac{\omega{osc}}{\omega_0} &= 60 ^\circ \\ \omega_{osc} &= \sqrt{3} \omega_0 \end{aligned} $$

$$ \begin{equation} H(s) = -\frac{A_0^3}{(1 + \dfrac{s}{\omega_0})^3} \end{equation} $$

Loop Gain = 1

$$ \begin{aligned} |H(s)| &= 0 \\ A_0 &= 2 \end{aligned} $$

Gain

$$ \begin{align} A_v &= \frac{H(s)}{1 - H(s)} \\ &= - \frac{A_0^3}{A_0^3 + (1+s/\omega_0)^3} \end{align} $$

Time Delay

$$ \begin{align}

\end{align} $$