Question 1. Consider the circuit in below. The transistor parameters are \( V_{T N}=0.4 \mathrm{~V} \) and \( k_{n}^{\prime}=100 \mu \mathrm{A} / \mathrm{V}^{2} \). Determine the transistor width-to-length ratio such that \( V D S=1.6 \mathrm{~V} \). (40 pts) 1. Consider the circuit in below. The transistor parameters are VTN​=0.4 V and kn′​=100μA/V2. Determine the transistor width-to-length ratio such that VDS=1.6 V. (40 pts)

So1n Given V_{T_{N}}=0.4 \quad K_{n}^{\prime}=100 \mathrm{~mA} / \mathrm{V}^{2}V_{D S}=1.6 \mathrm{~V}, \quad \mathrm{Km}^{\prime}=\mu_{m} C_{0 x}Problem 1Here V_{G}=V_{D}\because For saturation region V_{D s}>V_{G s}-V_{t}Aecording to given ckt ArrougementV_{D}-V_{S}>V_{D}-V_{S}-V_{t} \quad\left(\because V_{G}=V_{D}\right)\therefore OUR MOSFET is in sat { }^{4}I_{D} \text { (sat) }=\frac{1}{2} \min \operatorname{Cox} \frac{\omega}{L}\left(V_{\text {Gs }}-v_{t}\right)^{2}here I_{D}=\frac{V_{S}}{R_{S}}=\frac{V_{S}}{10} \Rightarrow V_{S}=10 I_{D}Also \quad V_{D}=3.3 \mathrm{~V}\begin{array}{c}S V_{D}-V_{S}=1.6 \Rightarrow 3.3-10 I_{D}=1.6 \\3.3-1.6=10 I_{D} \\I_{D}=0.17 \mathrm{~mA}\end{array}ทow\begin{array}{l}0.17 \times 10^{-3}=\frac{1}{2} \times 100 \times 10^{-6} \times \frac{\omega}{L}\left(V_{\text {Gs }}-V_{t}\right)^{2} \\0.34 \times 10^{-3}=10^{-4} \times \frac{\omega}{L}(1.6-0.4)^{2} \quad\left(\because V_{U S}=V_{D S}\right) \\\frac{0.34 \times 10^{-3}}{10^{-4} \times 1.44}=\frac{w}{L} \Rightarrow 0.236 \times 10 \\\frac{\omega}{L}=2.36 \quad \text { Aws } \\\end{array}Problam 2.- converting the ckt into thererin equivehele V_{\text {th }}=\frac{U_{D D} R_{2}}{R_{1}+R_{2}} \quad is R_{+4}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}Let the trautistor is in saturation region\therefore \quad V_{G}=V_{t h}, \quad V_{S}=I_{D} R_{S} \quad V_{D}=3.3-I_{D} R_{D}Now,I_{D(\text { sat })}=\frac{1}{2} \times 100 \times 10^{-6} \times \frac{\omega}{L}\left(V_{4 S}-V_{t}\right)^{2}Aiso. VDS =1.6 (given)\begin{array}{l}\Rightarrow \quad 3.3-I_{D R_{D}}-I_{D R_{S}}=1.6 \\3.3-I_{D}\left(R_{D}+R_{S}\right)=1.6 \\\frac{3.3-1.6}{R_{D}+R_{S}}=I_{D} \\\frac{1.7}{R_{D}+R s}=I_{D} \\\end{array}Yow\begin{array}{l}I_{D(\text { sat })}=\frac{1}{2} \times 10^{-4} \times \frac{\omega}{L}\left(V_{\text {th }}-I_{D} R_{S}\right)^{2} \\\frac{1.7}{R_{D}+R_{S}}=\frac{1}{2} \times 10^{-4}\left(V \text { th }-I_{D} R_{S}\right)^{2} \times \frac{\omega}{L}\end{array}\begin{array}{l}\frac{\omega}{L}=\frac{1.7 \times 2}{10^{-4}\left(V_{t h}-I_{\left.D R_{S}\right)^{2}\left(R_{D}+R_{S}\right)}\right.} \\ \frac{\omega}{L}=\frac{3.4}{10^{-4}\left(R_{D}+R_{S}\right)\left(V_{+U}-I_{D} R_{S}\right)^{2}} \text { Aus } \\\end{array} ...