Question 10.4: Calculate the threshold voltage of a MOS system using an alu...

From Figure 10.16, we find \phi_{m s} \cong-0.88 \mathrm{~V}. The other parameters are

\phi_{f p}=V_{t} \ln \left(\frac{N_{a}}{n_{i}}\right)=(0.0259) \ln \left(\frac{10^{15}}{1.5 \times 10^{10}}\right)=0.2877 \mathrm{~V}

and

x_{d T}=\left\{\frac{4 \epsilon_{s} \phi_{f p}}{e N_{a}}\right\}^{1 / 2}=\left\{\frac{4(11.7)\left(8.85 \times 10^{-14}\right)(0.2877)}{\left(1.6 \times 10^{-19}\right)\left(10^{15}\right)}\right\}^{1 / 2}=8.63 \times 10^{-5} \mathrm{~cm}

Then

\left|Q_{S D}^{\prime}(\max )\right|=e N_{a} x_{d T}=\left(1.6 \times 10^{-19}\right)\left(10^{15}\right)\left(8.63 \times 10^{-5}\right)=1.381 \times 10^{-8} \mathrm{C} / \mathrm{cm}^{2}

The threshold voltage is now found to be

\begin{aligned}V_{T N}=&\left(\left|Q_{S D}^{\prime}(\max )\right|-Q_{s s}^{\prime}\right)\left(\frac{t_{o x}}{\epsilon_{o x}}\right)+\phi_{m s}+2 \phi_{f p} \\ =&\left[\left(1.381 \times 10^{-8}\right)-\left(10^{10}\right)\left(1.6 \times 10^{-19}\right)\right] \cdot\left[\frac{120 \times 10^{-8}}{(3.9)\left(8.85 \times 10^{-14}\right)}\right] \\ &+(-0.88)+2(0.2877) \end{aligned}

or

V_{T N}=-0.262 \mathrm{~V}

Comment

In this example, the semiconductor is fairly lightly doped, which, in conjunction with the positive charge in the oxide and the work function difference, is sufficient to induce an electron inversion layer charge even with zero applied gate voltage. This condition makes the threshold voltage negative.