First, we must calculate the various parameters needed. The $1 / f$ noise constants, $B_N$ and $B_P$, are calculated as follows:



$$

B_N=\frac{\mathrm{KF}}{2 C_{\mathrm{ox}} K_N^{\prime}}=\frac{4 \times 10^{-28} \mathrm{~F}-\mathrm{A}}{2 \cdot 24.7 \times 10^{-4} \mathrm{~F} / \mathrm{m}^2 \cdot 110 \times 10^{-6} \mathrm{~A}^2 / \mathrm{V}}=7.36 \times 10^{-22}(V-m)^2

$$



and



$$

B_P=\frac{\mathrm{KF}}{2 C_{\mathrm{ox}} K_P^{\prime}}=\frac{0.5 \times 10^{-28} \mathrm{~F}-\mathrm{A}}{2 \cdot 24.7 \times 10^{-4} \mathrm{~F} / \mathrm{m}^2 \cdot 50 \times 10^{-6} \mathrm{~A}^2 / \mathrm{V}}=2.02 \times 10^{-22}(V-m)^2

$$





Now let us select the geometry of the various transistors that influence the noise performance. To keep $e_{n 1}^2$ small, let $W_1=100 \mu \mathrm{~m}$ and $L_1=1 \mu \mathrm{~m}$. Select $W_3=100 \mu \mathrm{~m}$ and $L_3= 20 \mu \mathrm{~m}$ and let the $W$ and $L$ of M8 be the same as M1 since it has little influence on the noise. Of course, M1 is matched with M2, M3 with M4, and M8 with M9. To check the impact of these choices, let us evaluate Eq. (7.5-8).



First, use Eq. (7.5-6) to calculate $e_{n 1}^2$.



$$

e_{n 1}^2=\frac{B_p}{f W_1 L_1}=\frac{2.02 \times 10^{-22}}{f \cdot 100 \mu \mathrm{~m} \cdot 1 \mu \mathrm{~m}}=\frac{2.02 \times 10^{-12}}{f} \mathrm{~V}^2 / \mathrm{Hz}

$$

Evaluating Eq. (7.5-8) gives



$$

\begin{aligned}

e_{\mathrm{eq}}^2 & =2 \times \frac{2.02 \times 10^{-12}}{f}\left[1+\left(\frac{110 \cdot 7.36}{50 \cdot 2.02}\right)\left(\frac{1}{100}\right)^2\right]=\frac{4.04 \times 10^{-12}}{f} 1.0008 \\

& =\frac{4.043 \times 10^{-12}}{f} \mathrm{~V}^2 / \mathrm{Hz}

\end{aligned}

$$





To help interpret this result, at 100 Hz , the voltage noise in a 1 Hz band is approximately $4 \times 10^{-14}\left(\mathrm{~V}_{\mathrm{rms}}\right)^2$ or $0.202 \mu\left(\mathrm{~V}_{\mathrm{rms}}\right)$.



The thermal noise is calculated from Eqs. (7.5-9) and (7.5-11). From Eq. (7.5-9) let us assume that $I_1=50 \mu \mathrm{~A}$. Therefore, at room temperature, we get



$$

e_{n 1}^2=\frac{8 k T}{3 g_m}=\frac{8 \cdot 1.38 \times 10^{-23} \cdot 300}{3 \cdot 707 \times 10^{-6}}=1.562 \times 10^{-17} \mathrm{~V}^2 / \mathrm{Hz}

$$

Putting this value into Eq. (7.5-11) gives



$$

\begin{aligned}

e_{\mathrm{eq}}^2 & =2 \cdot 1.562 \times 10^{-17}\left[1+\sqrt{\frac{110 \cdot 100 \cdot 1}{50 \cdot 100 \cdot 20}}\right]=3.124 \times 10^{-17} \cdot 1.33 \\

& =4.164 \times 10^{-17} \mathrm{~V}^2 / \mathrm{Hz}

\end{aligned}

$$





The noise corner frequency is found by equating the two expressions for $e_{\mathrm{eq}}^2$ to get



$$

f_c=\frac{4.043 \times 10^{-12}}{4.164 \times 10^{-17}}=97.1 \mathrm{kHz}

$$





This noise corner is indicative of the fact that the thermal noise is much less than the $1 / f$ noise.

To estimate the rms noise in the bandwidth from 1 Hz to 100 kHz , we will ignore the thermal noise and consider only the $1 / f$ noise. Performing the integration gives



$$

\begin{aligned}

V_{\mathrm{eq}}^2(\mathrm{rms}) & =\int_1^{100,000} \frac{4.066 \times 10^{-12}}{f} d f=4.066 \times 10^{-12}[\ln (100,000)-\ln (1)] \\

& =0.468 \times 10^{-10}\left(\mathrm{~V}_{\mathrm{rms}}\right)^2=6.84 \mu \mathrm{~V}_{\mathrm{rms}}

\end{aligned}

$$





The maximum signal in rms is 0.353 V . Dividing this by $6.84 \mu \mathrm{~V}$ gives 51,594 or 94.25 dB , which is equivalent to about 16 bits of resolution.



The design of the remainder of the op amp will have little influence on the noise and is not included in this example.