Exercise 6 (30.10.2024)
Christian Enz
Swiss Federal Institute of Technology (EPFL), Lausanne, Switzerland

Schematic of an elementary gain stage with diode connected load.

Small-signal schematic of an elementary gain stage with diode connected load including all the noise sources.
The equivalent small-signal schematic is shown in the above figure. We have neglected the output conductances of M1 and M2 because it comes in parallel with $G_{m2}$ and usually we can consider that $G_{ds1},G_{gds2} \ll G_{m2}$.
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{G_{m1}}{G_{m2}},\\ \omega_c &= \frac{G_{m2}}{C_L}. \end{align*}
The DC voltage gain is maximized when M1 is biased in weak inversion and M2 in strong inversion. In this case we have \begin{align*} G_{m1} &= \frac{I_b}{n_1\,U_T},\\ G_{m2} &= \frac{2 I_b}{V_{P2}} = \frac{2 n_2 \, I_b}{V_{G2}-V_{T0p}}, \end{align*} which results in \begin{equation*} A_{dc} = -\frac{V_{BG2}-V_{T0p}}{2 n_1 \, n_2 \, U_T}. \end{equation*} Since M1 is biased in weak inversion its saturation voltage is about $V_{Dsat1}=4\,U_T$. The maximum voltage gain is then given by \begin{equation*} A_{dc,max} \cong -\frac{V_{DD}-4\,U_T-V_{T0p}}{2 n_1 \, n_2 \, U_T}. \end{equation*}
The voltage gain is ultimately limited by the supply voltage (and the pMOS threshold voltage).
The input-referred noise resistance is given by \begin{equation*} R_{nin} = \frac{G_{n1} + G_{n2}}{G_{m1}^2}, \end{equation*} where \begin{equation*} G_{ni} = \gamma_{ni} \cdot G_{mi} + G_{mi}^2 \cdot \frac{\rho_i}{W\,L\,f} \quad \textsf{for $i=1,2$} \end{equation*}
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{\gamma_{n2}}{\gamma_{n1}} \cdot \frac{G_{m2}}{G_{m1}} = \frac{\gamma_{n2}}{\gamma_{n1}} \cdot \frac{1}{|A_v|}. \end{equation*}
Since the gain is set by setting the quiescent output voltage, we can estimate the contribution of M2 to the input thermal noise relative to that of M1
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} = \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*}
The maximum DC gain is given by
We choose a DC voltage $A_{dc}$
The transconductances $G_{m1}$, $G_{m2}$ and the minimum bias current $I_{b,min}$ for achieving the required cut-off frequency are 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
Having $G_{m2}$ and $I_b$, we can deduce
We can now size M2
Choosing $W_2=W_{min}$, we get
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 are given by
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
The DC gain is achieved and the cut-off frequency is slightly higher than the specs.
The simulated input-referred noise perfectly match the theoretical prediction.