Question 6.16: Consider the cascode stage shown in Fig. 6.35(a), where the ...
Consider the cascode stage shown in Fig. 6.35(a), where the load resistor is replaced by an ideal current source. Neglecting the capacitances associated with M_1, representing V_{in} and M_1 by a Norton equivalent as in Fig. 6.35(b), and assuming γ = 0, compute the transfer function.
Since the current through C_X is equal to −V_{out}C_Y s − I_{in}, we have V_X = −(V_{out}C_Y s + I_{in})/(C_X s), and the smallsignal drain current of M_2 is −g_{m2}(−V_{out}C_Y s − I_{in})/(C_X s). The current through r_{O2} is then equal to −V_{out}C_Y s − g_{m2}(V_{out}C_Y s + I_{in})/(C_X s). Noting that V_X plus the voltage drop across r_{O2} is equal to V_{out}, we write
−r_{O2} \left[(V_{out}C_Y s + I_{in})\frac{g_{m2}}{C_X s} + V_{out}C_Y s\right] − (V_{out}C_Y s + I_{in}) \frac{1}{C_X s} = V_{out} (6.82)
That is
\frac{V_{out}}{I_{in}}= \frac{−g_{m2}r_{O2} + 1}{C_X s} · \frac{1}{1 + (1 + g_{m2}r_{O2})\frac{C_Y}{C_X}+ C_Y r_{O2}s} (6.83)
which, for g_{m2}r_{O2} \gg 1 and g_{m2}r_{O2}C_Y /C_X \gg 1 (i.e., C_Y > C_X ), reduces to
\frac{V_{out}}{I_{in}}≈ \frac{− g_{m2}}{C_X s} \frac{1}{\frac{C_Y}{C_X}g_{m2} + C_Y s} (6.84)
and hence
\frac{V_{out}}{V_{in}}=\frac{−g_{m1}g_{m2}}{C_YC_X s}\frac{1}{g_{m2}/C_X + s} (6.85)
The magnitude of the pole at node X is still given by g_{m2}/C_X . This is because at high frequencies (as we approach this pole), C_Y shunts the output node, dropping the gain and suppressing the Miller effect of r_{O2}.