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Q. 6.16

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.

Verified Solution

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 $r_{O2}$.