The Symmetrical Cascode OTA

Analysis

Christian Enz

Swiss Federal Institute of Technology (EPFL), Lausanne, Switzerland

# Initialization¶

# Introduction¶

Schematic of the symmetrical OTA.

This notebook presents the analysis of the symmetrical cascode OTA which schematic is shown in the above figure. Note that all nMOS transistors have an odd number and all the pMOS transistors an even number which explains the numbering process. We will see below that in differential mode the effects of the source transconductances on the common source node voltage are actually canceled. We will design the circuit with M1a-M1b in a separate well.

# Small-signal analysis¶

Small-signal schematic of the symmetrical cascode OTA.

We start with the small-signal analysis. The small-signal schematic of the symmetrical OTA is shown above. The design of the symmetrical cascode OTA is similar to the design of the Miller OTA, except that there are now more parasitic poles appearing at the various current mirror nodes. The open-loop transfer function is given by \begin{equation*} \Delta V_{out} = \frac{A_{dc}}{1+s\,\tau_0} \cdot \left[\frac{\Delta V_{in+}}{1+s\,\tau_{p4}} - \frac{\Delta V_{in-}}{(1+s\,\tau_{p2})(1+s\,\tau_{p3})} -\frac{\Delta V_{in+}+\Delta V_{in-}}{2(1+s\,\tau_{p1})} \cdot \left(\frac{1}{1+s\,\tau_{p4}} - \frac{1}{(1+s\,\tau_{p2})(1+s\,\tau_{p3}}\right)\right] \end{equation*} where \begin{align*} A_{dc} &= \frac{G_m}{G_o},\\ G_m &= A_3 \cdot G_{m1},\\ G_o &= \frac{G_{ds3b}\,G_{ds7}}{G_{ms7}} + \frac{G_{ds2d}\,G_{ds4}}{G_{ms4}},\\ \tau_0 &= \frac{C_{out}}{G_o},\\ C_{out} &= C_o + C_L,\\ \tau_{p1} &= \frac{C_1}{2 G_{m1}},\\ \tau_{p2} &= \frac{C_2}{G_{m2}},\\ \tau_{p3} &= \frac{C_3}{G_{m3}},\\ \tau_{p4} &= \frac{C_4}{G_{m2}}. \end{align*}

The unity-gain frequency or gain-bandwidth product is defined as $\omega_u = 1/\tau_u \triangleq A_{dc}/\tau_0 = G_m/C_{out}$. In most practical cases, we usually have $\tau_0 \gg \tau_{p2}, \tau_{p3}, \tau_{p4} > \tau_{p1}$.

For a differential input voltage $\Delta V_{id}=\Delta V_{in+}-\Delta V_{in-}$ with a contant common-mode voltage $V_{ic}$ and hence $\Delta V_{ic} = (\Delta V_{in+} + \Delta V_{in-})/2=0$, $\Delta V_{in+} = \Delta V_{in-} = \Delta V_{id}/2$, the differential mode small-signal open-loop gain reduces to \begin{equation*} A_{dm}(s) \triangleq \frac{\Delta V_{out}}{\Delta V_{id}} = \frac{1}{2} \cdot \frac{A_{dc}}{1+s\,\tau_0} \cdot \left[\frac{1}{1+s\,\tau_{p4}} - \frac{1}{(1+s\,\tau_{p2})(1+s\,\tau_{p3})}\right] \end{equation*}

The small-signal output voltage can also be written as \begin{equation*} \Delta V_{out} = \frac{A_{dc}}{1+s\,\tau_0} \cdot \left[F(s) \cdot \Delta V_{in+} - G(s) \cdot \Delta V_{in-}\right], \end{equation*} where \begin{align*} F(s) &= \frac{1}{1+s\,\tau_{p4}} - \frac{1}{2}\,\frac{1}{1+s\,\tau_{p1}} \cdot \left[\frac{1}{1+s\,\tau_{p4}} - \frac{1}{(1+s\,\tau_{p2})(1+s\,\tau_{p3})}\right],\\ G(s) &= \frac{1}{(1+s\,\tau_{p2})(1+s\,\tau_{p3})} + \frac{1}{2}\,\frac{1}{1+s\,\tau_{p1}} \cdot \left[\frac{1}{1+s\,\tau_{p4}} - \frac{1}{(1+s\,\tau_{p2})(1+s\,\tau_{p3})}\right]. \end{align*}

From \eqref{eq:DVout_F_G} we see that the time constant $\tau_0$ introduces a $-90^\circ$ phase shift. Hence if the desired phase margin is at least $60^\circ$, the the additional phase shift introduced by the smaller time constants $\tau_{p1}$, $\tau_{p2}$, $\tau_{p3}$ and $\tau_{p4}$ should not exceed $30^\circ$. In this range, the product $(1+s\,\tau_{p1})(1+s\,\tau_{p2})(1+s\,\tau_{p3})(1+s\,\tau_{p4})$ can be expanded and approximated by neglecting all the order terms higher than $s$ \begin{equation*} \prod_{k=1}^{n} (1+s\,\tau_{pk}) \cong 1 + s \, \sum_{k=1}^{n} \tau_{pk}. \end{equation*} The above approximation gives a good estimation of the phase shift $\phi$ for $\phi < 30^\circ$, which correspond to the following frequency range \begin{equation*} \omega \leq \left(2\sum_{k=1}^{n} \tau_{pk}\right)^{-1}. \end{equation*} This leads to \begin{equation*} F(s) \cong G(s) \cong \frac{1+s\,\tau_n}{1 + s\,\tau_d}, \end{equation*} where \begin{align*} \tau_n &= \tau_{p1} + \frac{\tau_{p2}+\tau_{p3}+\tau_{p4}}{2},\\ \tau_d &= \tau_{p1}+\tau_{p2}+\tau_{p3}+\tau_{p4}. \end{align*}

For $\omega \leq 1/(2 \tau_d)$, the above expression can be further simplified considering that \begin{equation*} \frac{1+s\,\tau_n}{1 + s\,\tau_d} \cong \frac{1}{1 + s\,(\tau_d-\tau_n)} \end{equation*} where \begin{equation*} \tau_d-\tau_n = \frac{\tau_{p2}+\tau_{p3}+\tau_{p4}}{2} = \frac{\tau_p}{2}. \end{equation*}

$F(s)$ and $G(s)$ become \begin{equation*} F(s) \cong G(s) \cong \frac{1}{1+s\,\frac{\tau_p}{2}}, \end{equation*} which is valid for $\omega \leq 1/(2 \tau_d) =1/(2(\tau_{p2}+\tau_p))$.

The simplified differential transfer function is then given by \begin{equation*} A_{dm}(s) \cong \frac{A_{dc}}{\left(1+s\,\tau_0\right)\left(1+s\,\frac{\tau_p}{2}\right)} = \frac{A_{dc}}{\left(1+\frac{s}{\omega_0}\right)\left(1+\frac{s}{\omega_p}\right)} \end{equation*} with $\omega_0 \triangleq 1/\tau_0$ and $\omega_p \triangleq 2/\tau_p$ where it is assumed that $\tau_p = \tau_{p2}+\tau_{p3}+\tau_{p4} \ll \tau_0$. $\omega_0 \triangleq 1/\tau_0 \ll \omega_p$ is the dominant pole which is set by the load capacitance at the OTA output, whereas $\omega_p \triangleq 2/\tau_p$ represents the non-dominant pole due to all the time constants due to the parasitic capacitances at the current mirror nodes.

The transfer function can be further simplified as \begin{equation*} A_{dm}(s) \cong \frac{1}{s\,\tau_u\left(1+s\,\frac{\tau_p}{2}\right)} = \frac{1}{\frac{s}{\omega_u}\left(1+\frac{s}{\omega_p}\right)} \end{equation*}

where $\omega_u = 1/\tau_u = A_{dc} \cdot \omega_0 = G_m/C_o$ is the unity gain frequency or the gain-bandwidth product $GBW$ which is set by the OTA transconductance $G_m$ and the load capacitance $C_o$.

# Noise analysis¶

Neglecting the contribution of the cascode transistors M9 and M10 and remembering that $A_3=A_1 \cdot A_2$, the PSD of the output noise current is given by \begin{equation*} S_{I_{nout}} = A_3^2\,\left(S_{I_{n1a}}+S_{I_{n1b}}+S_{I_{n2a}}+S_{I_{n2b}}\right) + A_2^2\,\left(S_{I_{n2c}}+S_{I_{n3a}}\right)+S_{I_{n2d}}+S_{I_{n3b}} \end{equation*} or if we express the output PSD in terms of the output noise conductance \begin{equation*} S_{I_{nout}} =4 kT \cdot G_{nout} \end{equation*} where \begin{equation*} G_{nout} = 2 A_3^2 \cdot (G_{n1}+G_{n2}) + A_2^2 \cdot (G_{n2c}+G_{n3a}) + G_{n2d} + G_{n3b} \end{equation*} where we have assumed that M1a and M1b and M2a and M2b are identical. The $G_{ni}$ are given by \begin{equation*} G_{ni} = \gamma_{ni} \cdot G_{mi} G_{mi}^2 \cdot \frac{\rho_i}{W_i\,L_i\,f} \qquad \text{for all transistors}. \end{equation*}

The input-referred noise is then given by \begin{equation*} R_{nin} \triangleq \frac{G_{nout}}{A_3^2 \cdot G_{m1}^2} = \frac{2(G_{n1}+G_{n2})}{G_{m1}^2} + \frac{G_{n2c}+G_{n3a}}{A_1^2 \cdot G_{m1}^2} + \frac{G_{n2d}+G_{n3b}}{A_3^2 \cdot G_{m1}^2} \end{equation*} where we have used $A_3=A_1 \cdot A_2$. As expected, we see that the noise contribution to the input-referred noise of M2d and M3b is divided by $A_3^2$ and that from M2c and M3a is divided by $A_1^2$.

## Input-referred thermal noise¶

The input-referred thermal noise resistance is given by \begin{equation*} R_{nth} = \frac{2 \gamma_{n1}}{G_{m1}} \cdot (1 + \eta_{th}) \end{equation*} where \begin{equation*} \eta_{th} = \frac{\gamma_{n2}}{\gamma_{n1}}\,\frac{G_{m2}}{G_{m1}} + \frac{1}{2 A_1^2}\,\left(\frac{\gamma_{n2c}}{\gamma_{n1}}\,\frac{G_{m2c}}{G_{m1}} +\frac{\gamma_{n3a}}{\gamma_{n1}}\,\frac{G_{m3a}}{G_{m1}}\right) +\frac{1}{2 A_3^2}\,\left(\frac{\gamma_{n2d}}{\gamma_{n1}}\,\frac{G_{m2d}}{G_{m1}} +\frac{\gamma_{n3b}}{\gamma_{n1}}\,\frac{G_{m3b}}{G_{m1}}\right) \end{equation*} represents the contributions to the input-referred thermal noise of the current mirrors relative to that of the differential pair. Now $G_{m2c}=A_1 \cdot G_{m2a}=A_1 \cdot G_{m2}$, $G_{m3b}=A_2 \cdot G_{m3a}=A_2 \cdot G_{m3}$ and $G_{m2d}=A_3 \cdot G_{m2b} = A_3 \cdot G_{m2}$, and assuming that $\gamma_{n2a}=\gamma_{n2b}\cong\gamma_{n2c}\cong\gamma_{n2d}=\gamma_{n2}$ and $\gamma_{n3a}\cong\gamma_{n3b}=\gamma_{n3}$, then $\eta_{th}$ reduces to \begin{equation*} \eta_{th} = \frac{\gamma_{n2}}{2\gamma_{n1}}\,\frac{G_{m2}}{G_{m1}} \left(2+\frac{1}{A_1}+\frac{1}{A_3}\right) +\frac{1}{2 A_1^2}\,\frac{\gamma_{n3}}{\gamma_{n1}}\,\frac{G_{m3}}{G_{m1}} \left(1+\frac{1}{A_2}\right) \end{equation*}

In the case of unity current gains $A_1=A_2=A_3=1$, $\eta_{th}$ simplifies to \begin{equation*} \eta_{th} = 2 \frac{\gamma_{n2}}{\gamma_{n1}}\,\frac{G_{m2}}{G_{m1}} + \frac{\gamma_{n3}}{\gamma_{n1}}\,\frac{G_{m3}}{G_{m1}} \end{equation*}

We can also introduce the OTA thermal noise excess factor as \begin{equation*} \gamma_{ota} \triangleq G_m \cdot R_{nth}, \end{equation*} where $G_m = A_3 \cdot G_{m1}$ is the OTA transconductance. This results in \begin{equation*} \gamma_{ota} = 2 A_3 \cdot \gamma_{n1} \cdot (1 + \eta_{th}). \end{equation*} We see that introducing some current gains larger than one in the current mirrors allows to reduce the contributions of transistors M2c-M2d and M3a-M3b, but on the other hand increases the contribution of the differential pair.

For $G_{m1} \gg G_{m2}$ and $G_{m1} \gg G_{m3}$, the noise is dominated by the differential pair and \begin{equation*} \gamma_{ota} \cong 2 A_3 \cdot \gamma_{n1}. \end{equation*}

For unity current gains $A_1=A_2=A_3=1$, the OTA thermal noise excess factor $\gamma_{ota}$ reduces to \begin{equation*} \gamma_{ota} \cong 2 \gamma_{n1}. \end{equation*}

## Input-referred flicker noise¶

The input-referred flicker noise resistance is given by \begin{equation*} f \cdot R_{nfl} = \frac{2 \rho_n}{W_1 L_1} + \rho_p\,\left(\frac{G_{m2}}{G_{m1}}\right)^2 \left(\frac{2}{W_2 L_2}+\frac{1}{W_{2c} L_{2c}}+\frac{1}{W_{2d} L_{2d}}\right) + \frac{\rho_n}{A_1^2}\left(\frac{G_{m3}}{G_{m1}}\right)^2\left(\frac{1}{W_{3a} L_{3a}}+\frac{1}{W_{3b} L_{3b}}\right) \end{equation*} which can be written as \begin{equation*} R_{nfl} = \frac{2 \rho_n}{W_1 L_1 f} \cdot (1 + \eta_{fl}) \end{equation*} where $\eta_{fl}$ represents the contributions to the input-referred flicker noise resistance of the current mirrors relative to that of the differential pair \begin{equation*} \eta_{fl} = \frac{\rho_p}{2\rho_n}\,\left(\frac{G_{m2}}{G_{m1}}\right)^2 \left(\frac{2 W_1 L_1}{W_2 L_2}+\frac{W_1 L_1}{W_{2c} L_{2c}}+\frac{W_1 L_1}{W_{2d} L_{2d}}\right) + \frac{1}{2 A_1^2}\left(\frac{G_{m3}}{G_{m1}}\right)^2\left(\frac{W_1 L_1}{W_{3a} L_{3a}}+\frac{W_1 L_1}{W_{3b} L_{3b}}\right). \end{equation*}

For $A_1=A_2=A_3=1$, $W_{2a}=W_{2b}=W_{2c}=W_{2d}=W_2$, $L_{2a}=L_{2b}=L_{2c}=L_{2d}=L_2$, $W_{3a}=W_{3b}=W_3$ and $L_{3a}=L_{3b}=L_3$ reduces to \begin{equation*} \eta_{fl} = 2\frac{\rho_p}{\rho_n}\,\left(\frac{G_{m2}}{G_{m1}}\right)^2\,\frac{W_1 L_1}{W_2 L_2} + \left(\frac{G_{m3}}{G_{m1}}\right)^2\,\frac{W_1 L_1}{W_3 L_3}. \end{equation*}

# Input-referred offset voltage¶

The input-referred voltage can be obtained in a similar way than noise. The variance of the input-referred offset voltage is given by \begin{equation*} \sigma_{V_{os}}^2 = \sigma_{V_{T1}}^2 \cdot (1 + \xi_{V_T}) + \left(\frac{I_b}{G_{m1}}\right)^2 \cdot \sigma_{\beta_1}^2 \cdot (1 + \xi_{\beta}) \end{equation*} where $\xi_{V_T}$ represents the $V_T$ mismatch contributions to the input-referred offset of the current mirror relative to that of the differential pair \begin{equation*} \xi_{V_T} = 2 \left(\frac{G_{m2}}{G_{m1}}\right)^2 \cdot \frac{\sigma_{V_{T2}}^2}{\sigma_{V_{T1}}^2} + \left(\frac{G_{m3}}{A_1\,G_{m1}}\right)^2 \cdot \frac{\sigma_{V_{T3}}^2}{\sigma_{V_{T1}}^2} \end{equation*} and $\xi_{\beta}$ represents the $\beta$ mismatch contributions to the input-referred offset of the current mirror relative to that of the differential pair \begin{equation*} \xi_{\beta} = 2 \frac{\sigma_{\beta_2}^2}{\sigma_{\beta_1}^2} + \frac{\sigma_{\beta_3}^2}{A_1^2\,\sigma_{\beta_1}^2} \end{equation*} with \begin{align*} \sigma_{V_{T1}}^2 &= \frac{A_{VTn}^2}{W_1 L_1},\\ \sigma_{V_{T2}}^2 &= \frac{A_{VTp}^2}{W_2 L_2},\\ \sigma_{V_{T3}}^2 &= \frac{A_{VTn}^2}{W_3 L_3}, \end{align*} and \begin{align*} \sigma_{\beta_1}^2 &= \frac{A_{\beta n}^2}{W_1 L_1},\\ \sigma_{\beta_2}^2 &= \frac{A_{\beta p}^2}{W_2 L_2},\\ \sigma_{\beta_3}^2 &= \frac{A_{\beta n}^2}{W_3 L_3}. \end{align*}

Replacing \begin{equation*} \xi_{V_T} = 2 \left(\frac{G_{m2}}{G_{m1}}\right)^2 \cdot \left(\frac{A_{VTp}}{A_{VTn}}\right)^2 \cdot \frac{W_1 L_1}{W_2 L_2} + \left(\frac{G_{m3}}{A_1\,G_{m1}}\right)^2 \cdot \frac{W_1 L_1}{W_3 L_3} \end{equation*} and \begin{equation*} \xi_{\beta} = 2 \left(\frac{A_{\beta p}}{A_{\beta n}}\right)^2 \cdot \frac{W_1 L_1}{W_2 L_2} + \frac{1}{A_1^2}\,\frac{W_1 L_1}{W_3 L_3} \end{equation*}

Similarly to the flicker noise, the input-referred offset (variance or standard deviation) can be reduced by increasing the M1a-M1b area $W_1 L_1$ but at the same time also increasing the area of the current mirrors $W_2 L_2$ (and hence $W_{2c} L_{2c}$ and $W_{2d} L_{2d}$) and depending on $A_1$ also the area of M3a-M3b $W_{3a} L_{3a}$ and $W_{3b} L_{3b}$.

# Conclusion¶

This notebook presented the detailed analysis of the symmetrical OTA. The analysis allowed to derive the design equations that will be used in the Desin Notebook to achieve given specifications.