Exercise 6 (30.10.2024)
Christian Enz
Swiss Federal Institute of Technology (EPFL), Lausanne, Switzerland
Initialization¶
Problem 2¶
Schematic of an elementary gain stage with resistive feedback.
- Draw the small-signal schematic including all the noise sources.
- Derive the small-signal transfer function $A_v(s) \triangleq \Delta V_{out}/\Delta V_{in}$. Give the DC gain $A_{dc}$ and cut-off frequency $f_c$ assuming that $G_{m1} \cdot R_f \gg 1$.
- Calculate the input-referred noise resistance $R_{nin}$ and split it in terms of the input-referred thermal noise resistance $R_{nt}$ and flicker noise resistance $R_{nf}$.
- Calculate the input-referred thermal noise excess factor $\gamma_{neq} \triangleq G_{m1} \cdot R_{nt}$.
- How should M1, M2a and M2b be biased to minimize the input-referred thermal noise excess factor?
- Design the amplifier for a DC gain $A_{dc}=10$ a cut-off frequency $f_c=1\,MHz$. Use the following parameters \begin{align*} &V_{T0n}=0.455\,V \quad n_{0n}=1.27 \quad I_{spec\Box n}=715\,nA\\ &V_{T0p}=0.445\,V \quad n_{0p}=1.31 \quad I_{spec\Box p}=173\,nA \end{align*} with $U_T=25.875\,mV$.
Solution to Problem 2¶
Small-signal equivalent circuit¶
Small-signal schematic of the elementary gain stage with resistive feedback including all the noise sources.
The equivalent small-signal schematic is shown in the above figure. Conductance $G_{out}$ is the sum of the output conductances of M1 and M2b $G_{out} = G_{ds1} + G_{ds2}$. Note that the noise current source $I_{n1}$ includes the noise of M1, M2a and M2b.
Small-signal voltage gain¶
The small-signal voltage gain assuming there is no load is given by \begin{equation*} A_v(s) \triangleq \frac{\Delta V_{out}}{\Delta V_{in}} = \frac{A_{dc}}{1+s/\omega_c}. \end{equation*} with \begin{align*} A_{dc} &= \frac{1-G_{m1}\,R_f}{1+G_{out}\,R_f},\\ \omega_c &= \frac{1+G_{out}\,R_f}{R_f\,C_L}. \end{align*} The DC gain is maximized by setting $G_{m1}\,R_f \gg 1$ and $G_{out}\,R_f \ll 1$, resulting in \begin{align*} A_{dc} &\cong G_{m1}\,R_f,\\ \omega_c &\cong \frac{1}{R_f\,C_L}. \end{align*}
Input-referred noise resistance¶
The input-referred noise resistance is given by \begin{equation*} R_{nin} = \left(\frac{R_f}{G_{m1}\,R_f-1}\right)^2 \cdot (G_{n1} + G_{n2}), \end{equation*} where $G_{n1}$ includes the thermal and flicker noise coming from M1, M2a and M2b and $G_{n2}=1/R_f$.
Thermal noise¶
Assuming that $G_{m1}\,R_f \gg 1$, the input-referred thermal noise resistance is given by \begin{equation*} R_{nt} = \frac{\gamma_{n1}}{G_{m1}} \cdot \left(1+\eta_{th}\right) = \frac{\gamma_{neq}}{G_{m1}} \end{equation*} and the amplifier thermal excess noise factor by \begin{equation*} \gamma_{neq} = \gamma_{n1} \cdot \left(1+\eta_{th}\right), \end{equation*} where \begin{equation*} \eta_{th} = \frac{1}{\gamma_{n1}\,G_{m1}\,R_f} + \frac{2 \gamma_{n2}}{\gamma_{n1}} \cdot \frac{G_{m2}}{G_{m1}}. \end{equation*} The first term corresponds to the contribution of the feedback resistance relative to that of M1, while the second term corresponds to the contribution of M2a and M2b relative to that of M1. We see that the contribution of $R_f$ can be made negligible by making $G_{m1}\,R_f$ (which is actually what was assuming above). The contribution of M2a and M2b can be minimized by making $G_{m1} \gg G_{m2}$. This can be realized by biasing M1 in weak inversion and M2a-M2b in strong inversion.
Flicker noise¶
The input-referred flicker noise resistance is given by \begin{equation*} R_{nf} = \frac{\rho_n}{W_1\,L_1\,f} \cdot \left(1 + \eta_{fl}\right), \end{equation*} with \begin{equation*} \eta_{fl} = 2\,\frac{\rho_p}{\rho_n} \cdot \left(\frac{G_{m2}}{G_{m1}}\right)^2 \cdot \frac{W_1\,L_1}{W_2\,L_2} \end{equation*} Finally the corner frequency is given by \begin{equation*} f_k = \frac{\rho_n}{W_1\,L_1} \cdot \frac{G_{m1}}{\gamma_{n1}} \cdot \frac{1+\eta_{fl}}{1+\eta_{th}} \end{equation*}
Design¶
Process parameters¶
DC gain¶
The DC gain is set to
Assuming that $G_{out}\,R_f \ll 1$, the DC gain depends on both $G_{m1}$ and $R_f$. However if the load capacitance is set, then $R_f$ is given by
Choosing the inversion coefficient of M1 as $IC_1=0.1$, we can deduce the bias current $I_b$ as
We can deduce $I_{spec}$ and $W/L$ for M1
Choosing $L_1=L_{min}$, we get
In order to minimize the contribution of M2a-M2b to the input-referred noise we choose $G_{m2}/G_{m1}=1/10$
We can now size M2
Choosing $W_2=W_{min}$, we get
Noise¶
The thermal noise excess factors are given by
The input-referred thermal noise resistance and PSD are given by
The flicker noise PSd and corner frequency become
To reduce the corner frequency we need to increase both $W_1\,L_1$. In order to keep $\eta_{fl}$ constant, we hence need to increase $W_2\,L_2$ at the same time. In order to bring the corner frequency down to 1 MHz we need to change $W_1$, $L_1$, $W_2$ and L_2$ according to
Simulations¶
Transfer function¶
The DC gain is achieved and the cut-off frequency is slightly large than the spcecs.
Noise¶
The noise simulation perfectly match the theoretical results.