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(4) 0.8 \mathrm{dm} process technology \Rightarrow \mathrm{Lmin}=0.8 \mu \mathrm{m}\begin{array}{l}t_{0 x}=15 \mathrm{~nm}, \mu_{n}=55 \mathrm{~km}^{2} \mid V-S, \frac{\omega}{L}=20 . \\i_{D}=0.2 \mathrm{~mA}, \quad V_{\text {ov }}=\text { overdrive vuitage }=V_{n s}-V_{T} \\c_{x}=\frac{f-x}{t 0 x}=\frac{8.85 \times 10^{-12} \mathrm{Flm}}{15 \times 10^{-9} \mathrm{~m}}=0.59 \times 10^{-3} \mathrm{Flm}^{2} \\k_{n}=\mu_{n} \operatorname{lov}\left(\frac{\omega}{L}\right)=550 \times 10^{-4}\left(\frac{\mathrm{m}^{2}}{V-S}\right) \times 0.54 \times 10^{-3} \times 20 . \\\quad L_{n}=0.6490 \times 10^{-3}\end{array}0.59In satulation I_{0}=\frac{1}{2} u_{n}\left(o n,\left(\frac{\omega}{2}\right)\left(v_{n}\right)-v_{T}\right)\begin{array}{c}\left.\Rightarrow \quad 0.2 \times 10^{-3}=\frac{1}{2} \times 06490 \times 10^{-3}\left(V_{u}\right)-V_{T}\right)^{N} \\\Rightarrow \quad\left(V_{u J}-V_{T}\right)=0.78=\text { over drive voltuge. } \\\quad V_{0 V}=0.78 \mathrm{~V}\end{array}minimum value of V_{D S} - is nothing But overdsite\text { voltage }=0.28 \mathrm{~V} ...