The Simple Differential OTA
Design
Christian Enz
Swiss Federal Institute of Technology (EPFL), Lausanne, Switzerland

Schematic of the simple differential OTA.
This notebook presents the design of the simple differential OTA shown in the above figure. The OTA is analyzed in details in the Analysis Notebook, leading to all the design equations that are then used in this Notebook to size all transistors and set the bias currents in order to achiev a give set of specifications
We will design the circuit with M1a-M1b in a separate well for the specifications given below. In this example, we don't give any specification on the slew-rate. However the sew-rate specification can determine a bias current that is way above the one derived in this example ignoring the SR. If the SR is too high, we can move to adaptive biasing OTAs.
All the parameters required for the design are given below. They correspond to a generic $0.18\,\mu m$ process.
M1a-M1b are biased in weak inversion in order to minimize the input-referred offset. They are sized according to the specification on the GBW and the load capacitance and the required slew-rate.
Recalling that \begin{equation*} GBW = \frac{G_{m1}}{2 \pi C_{out}}, \end{equation*} where $C_{out}$ is the total output capacitance which includes the parasitic capacitance and the load capacitance $C_L$. Since we do not yet know the sizes of M1a-M2b, we can not estimate the total output capacitance. We will start assuming $C_{out}=C_L$.
$G_{m1}$ is the gate transconductance of M1a-M1b which in weak inversion is given by \begin{equation*}\ G_{m1} = \frac{I_b}{n U_T}. \end{equation*} The bias current $I_b$ is the current flowing in each transistor M1a-M1b when the input differential voltage is zero. The bias current provided by M3b is therefore $2 I_b$. The bias current must satisfy the following inequality: \begin{equation*} I_b \geq 2\pi n_{0n} U_T C_L GBW_{min}. \end{equation*} which for the given specifications gives
The corresponding slew-rate is given by
which we will consider as sufficient. If this is not the case, the bias current $I_b$ should be increased or a dynamic or adaptive biasing OTA should be used. To have some margin to account for the additional parasitic capacitance at the output due to the junction capacitances and that add to the load capacitance $C_L$, we set $I_b$ to
and the inversion coefficient $IC$ to
The transconductance can be calculated from the $G_m/I_D$ function as:
which leads to the following $GBW$
which is slightly higher than the target specification offering some margin.
Knowing the drain current $I_{D1}$ and the inversion coefficient, we can calculate the $W/L$ aspect ratio for M1a-M1b as
The degree of freedom left ($W_1$ or $L_1$) can be determined by constraints either on the DC gain, the offset voltage or the flicker noise. In this example we will set a minimum DC gain. The latter is given by \begin{equation*} A_{dc} =\frac{G_{m1}}{G_o} \end{equation*} where $G_o = G_{ds1b} + G_{ds2b}$ is the conductance to the AC ground at the output node. The output conductances are estimated with the following simple model \begin{equation*} G_{dsi} = \frac{I_{Di}}{V_{Mi}} \end{equation*} with $V_{Mi} = \lambda_i \cdot L_i$. Note that this model of the output conductance is only a very rough approximation and that the gain should be checked by simulation. Some margin can be taken to ensure a sufficient DC gain.
A good optimum for the output conductance is to set $G_{ds1b}=G_{ds2b}$ or since M1a-M2b share the same bias current $V_{M1b} = V_{M2b}$. The minimum length of M1a-M2b is then given by
which we round to
which gives the width of M1a-M1b
which we round to
We can recompute the transconductance and gain-bandwidth product for the chosen dimensions of M1a-M1b
The gate voltage of M2a should be set as low as possible for a given maximum common mode input voltage still keeping M1a in saturation. For a maximum input common-mode voltage $V_{ic,max}$ given by
The minimum source-to-gate voltage of M2a $V_{SG2a}$ for this maximum input common mode voltage $V_{ic,max}$ is given by \begin{equation*} V_{SG2a} = V_{DD} - V_{icmax} + V_{GS1a} - V_{DSsat1a}. \end{equation*} The gate-to-source voltage $V_{GS1a}$ and the saturation voltage $V_{DSsat1a}$ of M1a-M1b are given by
We set $V_{SG2}$ to
This corresponds to an inversion coefficient given by
We choose
The saturation voltage of M2a-M2b is then given by
The specific current $I_{spec}$, transconductance $G_m$ and $W/L$ are then given by
The length of M2a-M2b is set by the dc gain similarly to the length of M1a-M1b
We keep
We can now derive M2a-M2b transistor width
which is smaller than the minimum width $W_{min}$. We then set $W_2$ to the minimum width and deduce the length
which we round to
Since $L_2$ is longer than the initial value it will not affect the DC gain which should actually be slightly larger. We can evaluate the transconductance $G_{m2}$
Since $G_{m2}$ may become small, we need to check whether the non-dominant pole $f_{p2}$ lies sufficiently highabove the GBW to insure the desired phase margin. The non-dominant pole is given \begin{equation*} \omega_{p2} = \frac{G_{m2}}{C_2}, \end{equation*} where $C_2$ is given by \begin{equation*} C_2 = 2(C_{GS2} + C_{GB2}) \end{equation*} Assuming M2 is in strong inversion and saturation, we have \begin{equation*} C_{GS2} \cong \frac{2}{3}\,W_2\,L_2\,C_{ox} + W_2 \cdot C_{GSOp} \end{equation*} and \begin{equation*} C_{GB2} \cong \frac{n_{0p}-1}{3\,n_{0p}}\,W_2\,L_2\,C_{ox} + W_2 \cdot C_{GBOp}. \end{equation*}
The capacitance at node 2 scales with $W_2$ and $L_2$ according to \begin{equation*} C_2 = W_2\,L_2 \cdot C_{WL} + W_2 \cdot C_W, \end{equation*} with \begin{align*} C_{WL} &= 2\,C_{ox} \left(\frac{2}{3} +\frac{n_{0p}-1}{3\,n_{0p}}\right),\\ C_W &= 2(C_{GSop} + G_{GBop}). \end{align*}
which is sufficient.
In case it would not be, we then need to reduce $IC_2$ which will increase the $G_m/I_D$ and since the current is set by the bias current $I_b$, it will increase $G_{m2}$. If $W_2$ is set to its minimum value $W_{min}$, decreasing $IC_2$ keeping the same current, will increase the $W_2/L_2$. With $W_2$ set, this leads to a decrease of $L_2$ and hence a decrease of $C_2$. The increase of $G_{m2}$ an decrease of $C_2$ leads to an increase of $f_{p2}$, as required.
Therefore, if $f_{p2}$ is not suffciently higher than $GBW$, we can use the following script to find the required $IC$ for having the non-dominant pole at 10 times the $GBW$.
We see that $IC_2$ has been lowered, increasing $G_{m2}$ and $W_2/L_2$ for the given current $I_b$. Since $W_2$ hits $W_{min}$, the length $L_2$ has been decreased, reducing $C_2$ at the same time to keep $f_{p2}$ at 10 times the GBW.
We finally choose
We now need to size M3a-M3b according to the minimum common mode input voltage $V_{ic,min}$ to be handled
which leads to the saturation voltage of M3b
We can derive the corresponding inversion coefficient $IC_3$ directly from the normalized saturation voltage
from which we get the $W/L$ ratio
We need to set $W_3$ to $W_{min}$ and recalculate the length $L_3$
which we round to
We will summarize the result of the sizing procedure and save all the information in a dataframe and write it to an excel sheet.
The specifications are recalled below.
The bias information are summarized below.
The transistor sizes are summarized below.
In this section, we read the results of the sizing procedure from the Excel file without having to go through it again and check whether the specs are achieved. We need to run the first intialization before starting the notebook at this point.
The $GBW$ given above is only an estimation. We can find the actual $GBW$ accounting for the non-dominant pole using the following script.
In this case the actual $GBW$ is very close to the estimated one.
We can plot the magnitude and phase of the open-loop gain.
We can now calculate the noise excess factor of the OTA and the input-referred thermal noise resistance.
We see that the OTA thermal noise excess factor is only slightly larger than that of the differential pair. This is due to the rather large $G_{m1}/G_{m2}$ ratio.
We can now compute the input-referred flicker noise and the corner frequency.
We can plot the input-reffered noise
The variance of the input-referred offset is given by \begin{equation*} \sigma_{V_{os}}^2 = \left(\frac{I_b}{G_{m1}}\right)^2 \left(\sigma_{\beta_1}^2 + \sigma_{\beta_2}^2\right) + \left(\frac{G_{m2}}{G_{m1}}\right)^2 \sigma_{V_{T02}}^2 + \sigma_{V_{T01}}^2, \end{equation*} where \begin{align} \sigma_{\beta_i}^2 &= \frac{A_{\beta}^2}{W_i L_i} \qquad i=1,2,\\ \sigma_{V_{T0i}}^2 &= \frac{A_{VT}^2}{W_i L_i} \qquad i=1,2. \end{align} From the values calculated above we get
This notebook showed how to design the simple OTA from given specifications using the design equations derived in the Analysis Notebook. The result of the design has been checked. The next step is to validate the design by simulations. This is done in the Verification Notebook.