Calculate the cutoff frequency of an ideal MOSFET with a constant mobility.
Assume that the electron mobility in an n-channel device is \mu_{n}=400 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} and that the channel length is L=4 \mu \mathrm{m}. Also assume that V_{T}=1 \mathrm{~V} and V_{G S}=3 \mathrm{~V}.
From Equation (10.96), the cutoff frequency is
\begin{array}{c}f_{T}=\frac{g_{m}}{2 \pi C_{G}}=\frac{\frac{W \mu_{n} C_{\mathrm{ox}}}{L}\left(V_{G S}-V_{T}\right)}{2 \pi\left(C_{\mathrm{ox}} W L\right)}=\frac{\mu_{n}\left(V_{G S}-V_{T}\right)}{2 \pi L^{2}} \\ \end{array} (10.96)
\begin{array}{c} f_{T}=\frac{\mu_{n}\left(V_{G S}-V_{T}\right)}{2 \pi L^{2}}=\frac{400(3-1)}{2 \pi\left(4 \times 10^{-4}\right)^{2}}=796 \mathrm{MHz}\end{array}
Comment
In an actual MOSFET, the effect of the parasitic capacitance will substantially reduce the cutoff frequency from that calculated in this example.