振荡


General Considerations

In summary, if a negative-feedback circuit has a loop gain that satisfies two conditions:

\begin{aligned} \vert H(j\omega_0) \vert &\geq 1\\ \angle H(j\omega_0) &= 180^\circ \end{aligned}

For the oscillation to begin, a loop gain of unity or greater is necessary.

Ring Oscilators

A ring oscillator consists of a number of gain stages in a loop. We start from single-stage feedback to multi-stage feedback.

From Fig. 15.4, the open-loop circuit contains only one pole, thereby providing a maximum frequency dependent phase shift of $90^\circ$ (at a frequency of infinity). Added by a dc phase shift of $180^\circ$, the max total phase shift is $270^\circ< 360^\circ$. The loop therefore fails to sustain oscillation growth.

• two pole → $180^\circ$
• dc phase shift → $360^\circ$
the circuit achieve a total phase shift of $360^\circ$ at DC. As a result, it simply “latches up” rather than oscillates.（即反馈导致 $V_E$ 电压不断上升，直到固定到 $V_{DD}$）

• dc phase shift → $540^\circ$
• two pole → $180^\circ$
The frequency-dependent phase shift can reach $180\circ$ at a frequency of infinity. Since the loop gain vanishes at very high frequencies, we observe that the circuit does not satisfy both of Barkhausen’s criteria at the same frequency.

• dc phase shift → $540^\circ$
• two pole → $270^\circ$
The total phase shift around the loop $\phi$ equals $-180^\circ$ at $\omega<\infty$. This circuit indeed oscillates if the loop gain is sufficient.

1. 判断 dc phase shift
2. 判断 pole 带来的 phase shift
3. 把 1 和 2 相加，看在 $(0,\infty)$ 内 total phase shift 能否够 360 （并且 360 时不能在 0 或 $\infty$）
4. 判断在 360 时 gain 是否大于等于 1

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

$\arctan \frac{\omega_\tx{osc}}{\omega_0} = 60^\circ\\ \Rightarrow \omega_\tx{osc} = \sqrt{3} \omega_0$

$\frac{A_0^3}{\left[\sqrt{1+(\dfrac{\omega_\tx{osc}}{\omega_0})}\right]^3} = 1\\ \Rightarrow A_0 = 2$

LC Oscillators

Basic

$Z = \frac{1}{j\omega C+\dfrac{1}{j\omega L}}$

$Z = \frac{(R + j \omega L) \frac{1}{j\omega C} }{ R + j(\omega L - \frac{1}{\omega C})}$

$C_p=C_1\\ L_p=L_1\\ R_p \approx \frac{L_1^2 \omega^2}{R_S} =Q^2 R_S\\$

Cross-Coupled Oscillator

$\begin{cases} V_X - V_1 + V_2 = 0\\ I_X = g_{m2}V_2 = - g_{m1}V_1 \end{cases}\\ \Rightarrow V_X + \frac{I_X}{g_{m1}} + \frac{I_X}{g_{m2}} = 0\\ \Rightarrow \frac{V_X}{I_X} = -\frac{1}{1/g_{m1}+1/g_{m2}}$

$V_X = (I_X - \frac{-I_X}{C_1 s}g_m) \frac{1}{C_2 s}+\frac{I_X}{C_1 s}\\ \Rightarrow \frac{V_X}{I_X}=\frac{g_m}{C_1 C_2 s^2}+\frac{1}{C_2 s}+\frac{1}{C_1 s}$

Colpitts Oscillator

$\omega_R^2 = \frac{1}{L_P C}\\ C=\frac{1}{1/C_1+1/C_2}$

$g_m R_P = \frac{(C_1+C_2)^2}{C_1C_2}$

$g_m R_P = \frac{C_1}{C_2}(1+\frac{C_2}{C_1})^2 \geq 4\\$

Voltage-Controlled Oscillators

$C_\tx{var} = \frac{C_0}{(1+\dfrac{V_R}{\phi_B})^m}$

\begin{aligned} \tau &= R_{3,4} C_L\\ &=\frac{C_L}{\mu_p C_\tx{ox} (\frac{W}{L})_{3,4}(V_{DD}-V_\tx{cont}-\vert V_\tx{THP}\vert)} \end{aligned}

Mathematical Model of VCOs

$\begin{cases} \phi = \int \omega \dif t + \phi_0\\ \omega_\tx{out}=\omega_0 + K_{VCO}V_\tx{cont} \end{cases}$ \begin{aligned} V_\tx{out}(t) &= V_m \cos\left( \int \omega_\tx{out} + \phi_0 \right)\\ &=V_m \cos (\omega_0 t+K_{VCO}\int V_\tx{cout} \dif t+\phi_0) \end{aligned}\\ \Rightarrow\phi_{ex} = K_{VCO} \int V_\tx{cont} \dif t\\ \Rightarrow\frac{\phi_{ex}}{V_\tx{cont}}(s) = \frac{K_{VCO}}{s}