Determine the inversion carrier mobility from experimental results.
Consider an n-channel MOSFET with W=15 \mu \mathrm{m}, L=2 \mu \mathrm{m}, and C_{\mathrm{ox}}=6.9 \times 10^{-8} \mathrm{~F} / \mathrm{cm}^{2}.
Assume that the drain current in the nonsaturation region for V_{D S}=0.10 \mathrm{~V} is I_{D}=35 \mu \mathrm{A} at V_{G S}=1.5 \mathrm{~V} and I_{D}=75 \mu \mathrm{A} at V_{G S}=2.5 \mathrm{~V}
From Equation (10.68), we can write
\begin{aligned}I_{D}=\frac{W \mu_{n} C_{\mathrm{ox}}}{L}\left(V_{G S}-V_{T}\right) V_{D S} \\ \end{aligned} (10.68)
\begin{aligned}I_{D 2}-I_{D 1}=& \frac{W \mu_{n} C_{\mathrm{ox}}}{L}\left(V_{G S 2}-V_{G S 1}\right) V_{D S}\end{aligned}
so that
75 \times 10^{-6}-35 \times 10^{-6}=\left(\frac{15}{2}\right) \mu_{n}\left(6.9 \times 10^{-8}\right)(2.5-1.5)(0.10)
which yields
\mu_{n}=773 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}
We can then determine
V_{T}=0.625 \mathrm{~V}
Comment
The mobility of carriers in the inversion layer is less than that in the bulk semiconductor due to the surface scattering effect. We will discuss this effect in the next chapter.