MOS transistors with very short channels do not exhibit the square law voltage relation in saturation. The drain current is instead given by

\(I_{D}=W C_{\mathrm{ox}}\left(V_{G S}-V_{T N}\right) v_{\mathrm{sat}}\)

where \(v_{\text {sat }}\) is a saturation velocity. Assuming \(v_{\text {sat }}=2 \times 10^{7} \mathrm{~cm} / \mathrm{s}\) and using the parameters in Problem 3.11, determine the current.

Data From Problem 3.11:-

A particular NMOS device has parameters \(V_{T N}=0.6 \mathrm{~V}, L=0.8 \mu \mathrm{m}\), \(t_{\mathrm{ox}}=200 Å\), and \(\mu_{n}=600 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\). A drain current of \(I_{D}=1.2 \mathrm{~mA}\) is required when the device is biased in the saturation region at \(V_{G S}=3 \mathrm{~V}\). Determine the required channel width of the device.