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
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当$\omega_0 \longrightarrow \infin$,Pole会产生$90 ^\circ$的Phase Shift
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CS Stage 产生$180 ^\circ$的Phase Shift
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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
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2 CS Stage $\omega_0 \longrightarrow 0 \Longrightarrow \angle H(s) = 360 ^\circ$
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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
$$ \begin{align}
\end{align} $$