Assuming that the devices in Fig. 7.59(a) operate in saturation and the circuit is symmetric, calculate the input-referred noise voltage.

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Accepted Answer

Since the thermal and 1 / f noise of [latex]M_1 and M_2[/latex] can be modeled as voltage sources in series with the input, we need only refer the noise of [latex]M_3 and M_4[/latex] to the input. Let us calculate the output noise contributed by [latex]M_3[/latex]. The drain noise current of [latex]M_3[/latex] is divided between [latex]r_{O 3}[/latex] and the resistance seen looking into the drain of [latex]M_1[/latex] Fig.7.59( c ). From Chapter 5 , this resistance equals [latex]R_X=r_{O 4}+2 r_{O 1}[/latex]. Denoting the resulting noise currents flowing through [latex]r_{O 3}[/latex] and [latex]R_X by I_{n A} and I_{n B}[/latex], respectively, we have

[latex] 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}} [/latex] (7.133)

and

[latex] I_{n B}=g_{m 3} V_{n 3} \frac{r_{O 3}}{2 r_{O 4}+2 r_{O 1}} [/latex] (7.134)

The former produces a noise voltage of [latex] g_{m 3} V_{n 3} r_{O 3}\left(r_{O 4}+2 r_{O 1} ight) /\left(2 r_{O 4}+2 r_{O 1} ight) [/latex] at node X with respect to ground whereas the latter flows through [latex] M_1, M_2 ext {, and } r_{O 4} ext {, generating } g_{m 3} V_{n 3} r_{O 3} r_{O 4} /\left(2 r_{O 4}+2 r_{O 1} ight) [/latex] at node Y with respect to ground. Thus, the total differential output noise due to [latex]M_3[/latex] is equal to

[latex] \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} [/latex]

(The reader can verify that [latex]V_{nY}[/latex] must be subtracted from [latex]V_{nX}[/latex] .)

Equation (7.136) implies that the noise current of [latex]M_3[/latex] is simply multiplied by the parallel combination of [latex]r_{O 1} and r_{O 3}[/latex] to produce the differential output voltage. This is of course not surprising because, as depicted in Fig. 7.60, the effect of [latex]V_{n 3}[/latex] at the output can also be derived by decomposing [latex]V_{n 3}[/latex] into two differential components applied to the gates of [latex]M_3 and M_4[/latex] 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 [latex]V_{n 3}[/latex] and the circuit as familiar deterministic, linear components.

Applying (7.136) to [latex]M_4[/latex] as well and adding the resulting powers, we have

[latex] \begin{aligned} \left.\overline{V_{n, o u t}^2} ight|_{M 3, M 4} &=g_{m 3}^2\left(r_{O 1} \| r_{O 3} ight)^2 \overline{V_{n 3}^2}+g_{m 4}^2\left(r_{O 2} \| r_{O 4} ight)^2 \overline{V_{n 4}^2} &(7.137)\ &=2 g_{m 3}^2\left(r_{O 1} \| r_{O 3} ight)^2 \overline{V_{n 3}^2} &(7.138)\end{aligned} [/latex]

To refer the noise to the input, we divide (7.138) by [latex] g_{m 1}^2\left(r_{O 1} \| r_{O 3} ight)^2 [/latex], obtaining the total input-referred noise voltage per unit bandwidth as

[latex] \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} [/latex] (7.139)

which, upon substitution for [latex] \overline{V_{n 1}^2} ext { and } \overline{V_{n 3}^2} [/latex], reduces to

[latex] \overline{V_{n, i n}^2}=8 k T \gamma\left(\frac{1}{g_{m 1}}+\frac{g_{m 3}}{g_{m 1}^2} ight)+\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} [/latex] (7.140)

Question 7.22: Assuming that the devices in Fig. 7.59(a) operate in saturat...

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