Question 7.22: Assuming that the devices in Fig. 7.59(a) operate in saturat...
Assuming that the devices in Fig. 7.59(a) operate in saturation and the circuit is symmetric, calculate the input-referred noise voltage.
Since the thermal and 1 / f noise of M_1 and M_2 can be modeled as voltage sources in series with the input, we need only refer the noise of M_3 and M_4 to the input. Let us calculate the output noise contributed by M_3. The drain noise current of M_3 is divided between r_{O 3} and the resistance seen looking into the drain of M_1 Fig.7.59( c ). From Chapter 5 , this resistance equals R_X=r_{O 4}+2 r_{O 1}. Denoting the resulting noise currents flowing through r_{O 3} and R_X by I_{n A} and I_{n B}, respectively, we have
I_{n A}=g_{m 3} V_{n 3} \frac{r_{O 4}+2 r_{O 1}}{2 r_{O 4}+2 r_{O 1}} (7.133)
and
I_{n B}=g_{m 3} V_{n 3} \frac{r_{O 3}}{2 r_{O 4}+2 r_{O 1}} (7.134)
The former produces a noise voltage of g_{m 3} V_{n 3} r_{O 3}\left(r_{O 4}+2 r_{O 1}\right) /\left(2 r_{O 4}+2 r_{O 1}\right) at node X with respect to ground whereas the latter flows through M_1, M_2 \text {, and } r_{O 4} \text {, generating } g_{m 3} V_{n 3} r_{O 3} r_{O 4} /\left(2 r_{O 4}+2 r_{O 1}\right) at node Y with respect to ground. Thus, the total differential output noise due to M_3 is equal to
\begin{aligned} V_{n X Y} &=V_{n X}-V_{n Y} & (7.135)\\ &=g_{m 3} V_{n 3} \frac{r_{O 3} r_{O 1}}{r_{O 3}+r_{O 1}} & (7.136) \end{aligned}(The reader can verify that V_{nY} must be subtracted from V_{nX} .)
Equation (7.136) implies that the noise current of M_3 is simply multiplied by the parallel combination of r_{O 1} and r_{O 3} to produce the differential output voltage. This is of course not surprising because, as depicted in Fig. 7.60, the effect of V_{n 3} at the output can also be derived by decomposing V_{n 3} into two differential components applied to the gates of M_3 and M_4 and subsequently using the half-circuit concept. Since this calculation relates to a single noise source, we can temporarily ignore the random nature of noise and treat V_{n 3} and the circuit as familiar deterministic, linear components.
Applying (7.136) to M_4 as well and adding the resulting powers, we have
\begin{aligned} \left.\overline{V_{n, o u t}^2}\right|_{M 3, M 4} &=g_{m 3}^2\left(r_{O 1} \| r_{O 3}\right)^2 \overline{V_{n 3}^2}+g_{m 4}^2\left(r_{O 2} \| r_{O 4}\right)^2 \overline{V_{n 4}^2} &(7.137)\\ &=2 g_{m 3}^2\left(r_{O 1} \| r_{O 3}\right)^2 \overline{V_{n 3}^2} &(7.138)\end{aligned}To refer the noise to the input, we divide (7.138) by g_{m 1}^2\left(r_{O 1} \| r_{O 3}\right)^2 , obtaining the total input-referred noise voltage per unit bandwidth as
\overline{V_{n, i n}^2}=2 \overline{V_{n 1}^2}+2 \frac{g_{m 3}^2}{g_{m 1}^2} \overline{V_{n 3}^2} (7.139)
which, upon substitution for \overline{V_{n 1}^2} \text { and } \overline{V_{n 3}^2} , reduces to
\overline{V_{n, i n}^2}=8 k T \gamma\left(\frac{1}{g_{m 1}}+\frac{g_{m 3}}{g_{m 1}^2}\right)+\frac{2 K_N}{C_{o x}(W L)_1 f}+\frac{2 K_P}{C_{o x}(W L)_3 f} \frac{g_{m 3}^2}{g_{m 1}^2} (7.140)
