振荡

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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.

电路一

Figure 15.4

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.

电路二

Figure 15.5 Two-pole feedback system

  • 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 $V_{DD}$)

电路三

Figure 15.6 Two-pole feedback system with additional signal inversion

  • 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.

电路四

Figure 15.8 Three-stage ring oscillator

  • 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.

从上面几个案例,我们可以总结如下分析步骤:

  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
Figure 15.10 Linear model of three-stage ring oscillator

下面我们继续分析最后一个 3 级振荡器,我们还需要考虑增益。设每一级的传输函数为 $\dfrac{-A_0}{1+s/\omega_0}$,则 loop gain 为:

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

要使相移达到 $180^\circ$,则每一级贡献 $60^\circ$,即:

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

代入 loop gain,得到:

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

总结, a three-stage ring oscillator requires a low-frequency gain of 2 per stage, and it oscillates at a frequency of $\sqrt{3} \omega_0$, where $ω_0$ is the 3-dB bandwidth of each stage

LC Oscillators

Basic

我们先来回顾一下电路与高频的知识。

对于 LC 并联电路,我们有:

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

当 $j\omega C = \dfrac{1}{j\omega L}$,即 $\omega = 1/\sqrt{LC}$ 时,$Z=\infty$,此时为谐振状态(电路的知识)


考虑上电感的电阻

Figure 15.21 Conversion of a tank to three parallel components

对于 Fig. 15.21a,

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

当 $\omega = 1/\sqrt{LC}$ 时,$Z$ 达到最大值 $\dfrac{L}{CR}$。完整的幅频特性如下图所示:

Figure 15.22 (a) Magnitude and (b) phase ofthe impedance ofan LC tank as a function of frequency

有时候我们希望电阻也是并联的,这样好算一点。转换的方法如下(具体推导过程略)

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

转换成并联后,我们可以写出其 $Q$ 值为 $Q_p = \dfrac{R_p}{\omega_p L}=R_p \omega_p C= R_p \sqrt{\dfrac{C}{L}}$

Cross-Coupled Oscillator

我们将 LCR 振荡网络作为负载。当只有一级时,CS 贡献 $180^\circ$,RLC 贡献 $-90^\circ\sim 90^\circ$,加起来不够 $360^\circ$

Figure 15.23 (a) Tuned gain stage; (b) stage of (a) in feedback

如果再增加一级,CS 贡献 $360^\circ$,RLC 贡献 $-180^\circ\sim 180^\circ$,那么在谐振频率处刚好是 $360^\circ$,可以起振。注意对比之前 2 级反相的情况,由于之前的 0 相移在 DC,所以被锁定,无法振荡。

Figure 15.25 Two tuned stages in a feedback loop

我们从另一个角度来理解为什么上面这种解法可以振荡。我们将电路整理一下就会发现,下面两个MOS组成了负阻(前面某道练习题有说),负阻抵消了振荡的电阻损耗,使得振荡可以维持下去。

Figure 15.27 (a) Redrawing of the oscillator shown in Fig. 15.25; (b) another redrawing of the circuit; (c) addition of tail current source to lower supply sensitivity

下面我们来详细研究一下这个负阻结构。

Figure 15.33 (a) Source follower with positive feedback to create negative input impedance; (b) equivalent circuit of (a) to calculate the input impedance

考虑 Fig. 15.33b 的小信号模型,有:

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

若 $g_{m1}=g_{m2}=g_m$,则这个电路具有一个负阻 $-\dfrac{2}{g_m}$。

要让 $-\dfrac{2}{g_m}$ 抵消两个 $R_P$,则 $g_m = \frac{1}{R_P}$,此时的差分增益为 $g_m R_P = 1$,这也是差分增益的最小要求。


另一种负阻结构如下图所示:

Figure 15.37 (a) Circuit topology providing negative resistance; (b) equivalent circuit of (a); (c) oscillator using (a)

根据小信号模型容易写出:

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

注意到第一项是负的($s^2=-\omega^2$),所以 Fig. 15.17c 的结构是可以振荡的。

Colpitts Oscillator

Figure 15.31 (a) Colpitts oscillator; (b) equivalent circuit of (a) with input stimulus

详细分析看书,我这里只是简单说一下。首先我们需要把这个看作一个 CG Stage,输出的电压经过电容分压后再放大 $g_m R_D$ 倍(忽略 $g_{mb}$),以弥补 $R_P$ 的损耗。其振荡频率为:

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

而放大倍数需要满足:

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

关于这个式子的来源嘛,书上列举了一大堆公式,你可以去看一下。我们一般把这个表示为 $\frac{C_1}{C_2}$ 的函数,即:

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

最小值可以在 $\frac{C_1}{C_2} = 1$ 时取得.

相关论文:Output voltage analysis for the MOS Colpitts oscillator

Voltage-Controlled Oscillators

现在我们想要让电压能控制频率,而要改变谐振频率,则必须改变 L 或 C,由于电感很难改变,所以我们必须用0可变电容。在半导体器件中,我们知道二极管两边的反偏电容会随着电压改变,即:

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

电路图见 Fig. 15.53

Figure 15.53 LC oscillator using varactor diodes


对于 Ring Oscillators,对于 N 个反相器组成的延时电路,其振荡频率是 $(2NT_D)^{-1}$。我们可以通过改变反相器的延时时间来改变周期,具体电路如下:

Figure 15.42 Differential pair with variable output time constant

其中,M3、M4 处于深度线性区,对应的 RC 延时为:

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

Mathematical Model of VCOs

我们知道,相位是频率的积分,频率是相位的微分(就像速度和路程一样),那么,当频率变快时,相位的积累速度就会变快。可以参考下面这幅图:

Figure 15.62

用公式表述如下:

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

从公式中可以看出,VCO 实际上是一个积分器!为什么要说这个呢?因为后续我们可能需要对系统进行建模,需要求出系统函数(也就是信号与系统中学的拉普拉斯变换啦!忘了的同学可以去看本站的相关笔记)。