Calculate the body-effect coefficient and the change in the threshold voltage due to an applied source-to-body voltage.
Consider an n-channel silicon MOSFET at T=300 \mathrm{~K}. Assume the substrate is doped to N_{a}=3 \times 10^{16} \mathrm{~cm}^{-3} and assume the oxide is silicon dioxide with a thickness of t_{o x}=20 \mathrm{~nm}=200 \mathring{A} . Let V_{S B}=1 \mathrm{~V}.
We can calculate that
\phi_{f p}=V_{t} \ln \left(\frac{N_{a}}{n_{i}}\right)=(0.0259) \ln \left(\frac{3 \times 10^{16}}{1.5 \times 10^{10}}\right)=0.3758 \mathrm{~V}
and
C_{o x}=\frac{\boldsymbol{\epsilon}_{o x}}{t_{o x}}=\frac{(3.9)\left(8.85 \times 10^{-14}\right)}{200 \times 10^{-8}}=1.726 \times 10^{-7} \mathrm{~F} / \mathrm{cm}^{2}
From Equation (10.82), we find the body-effect coefficient to be
\begin{aligned}\gamma=& \frac{\sqrt{2 e \epsilon_{s} N_{a}}}{C_{o x}} \\ \end{aligned} (10.82)
\begin{aligned}\gamma &=\frac{\sqrt{2 e \epsilon_{s} N_{a}}}{C_{o x}}=\frac{\left[2\left(1.6 \times 10^{-19}\right)(11.7)\left(8.85 \times 10^{-14}\right)\left(3 \times 10^{16}\right)\right]^{1 / 2}}{1.726 \times 10^{-7}}\end{aligned}
or
\gamma=0.5776 \mathrm{~V}^{1 / 2}
The change in threshold voltage for V_{S B}=1 \mathrm{~V} is found to be
\begin{aligned}\Delta V_{T} &=\gamma\left[\sqrt{2 \phi_{f p}+V_{S B}}-\sqrt{2 \phi_{f p}}\right] \\&=(0.5776)[\sqrt{2(0.3758)+1}-\sqrt{2(0.3758)}] \\&=(0.5776)[1.3235-0.8669]=0.264 \mathrm{~V}\end{aligned}
Comment
Figure 10.51 shows plots of \sqrt{I_{D}(\text { sat })} versus V_{G S} for various applied values of V_{S B}. The original threshold voltage is assumed to be V_{T O}=0.64 \mathrm{~V}.
