Question $9$: MOS, solve

Assume our usual nMOS model where:

Ohmic region

$I_D=k(V_{GS}-V_{TH}{)}^2$ when

$V_{DS}{V}_{GS}-V_{TH}{V}_{TH}=1V,V_{DD}=3V{R}_D=25I_D-V_{DS}k{V}_{GS}{V}_{DS}=0\mathrm{.}92Vk=\{\begin{array}{ccc}\text{:}[& \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}(2\text{s}\text{i}\text{g}\text{n}\text{i}\text{f}\text{i}\text{c}\text{a}\text{n}\text{t}\text{f}\text{i}\text{g}\text{u}\text{r}\text{e}\text{s})m\frac{A}{V^2},& ?]\end{array}$

$V_{GS}=\{\begin{array}{ccc}\text{:}[& \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}(2\text{s}\text{i}\text{g}\text{n}\text{i}\text{f}\text{i}\text{c}\text{a}\text{n}\text{t}\text{f}\text{i}\text{g}\text{u}\text{r}\text{e}\text{s})V& ]\end{array}{I}_D=k{V}_{DS}(V_{GS}-V_{TH})$ when

$V_{DS}$

Active region

$I_D=k(V_{GS}-V_{TH}{)}^2$ when

$V_{DS}{V}_{GS}-V_{TH}$

Consider the following circuit and the family $ofIV$ characteristic curves.

You are given that $V_{TH}=1V,V_{DD}=3V,$ and $R_D=25.$

Use the $I_D-V_{DS}$ characteristics $to$ find the transitor parameter $k$ and the value $of{V}_{GS}$ that produces $V_{DS}=0\mathrm{.}92V.$

$k=\{\begin{array}{ccc}\text{:}[& \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}(2\text{s}\text{i}\text{g}\text{n}\text{i}\text{f}\text{i}\text{c}\text{a}\text{n}\text{t}\text{f}\text{i}\text{g}\text{u}\text{r}\text{e}\text{s})m\frac{A}{V^2},& ?]\end{array}$

$V_{GS}=\{\begin{array}{ccc}\text{:}[& \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}(2\text{s}\text{i}\text{g}\text{n}\text{i}\text{f}\text{i}\text{c}\text{a}\text{n}\text{t}\text{f}\text{i}\text{g}\text{u}\text{r}\text{e}\text{s})V& ]\end{array}$