Question 10.6: Calculate Cox, C min′ , and C FB′ for a MOS capacitor. Consi...

The oxide capacitance is

C_{o x}=\frac{\epsilon_{o x}}{t_{o x}}=\frac{(3.9)\left(8.85 \times 10^{-14}\right)}{180 \times 10^{-8}}=1.9175 \times 10^{-7} \mathrm{~F} / \mathrm{cm}^{2}

To find the minimum capacitance, we need to calculate

\phi_{f p}=V_{t} \ln \left(\frac{N_{a}}{n_{i}}\right)=(0.0259) \ln \left(\frac{10^{16}}{1.5 \times 10^{10}}\right)=0.3473 \mathrm{~V}

and

\begin{aligned}x_{d T} &=\left\{\frac{4 \epsilon_{s} \phi_{f f}}{e N_{a}}\right\}^{1 / 2}=\left\{\frac{4(11.7)\left(8.85 \times 10^{-14}\right)(0.3473)}{\left(1.6 \times 10^{-19}\right)\left(10^{16}\right)}\right\}^{1 / 2} \\ & \cong 0.30 \times 10^{-4} \mathrm{~cm}\end{aligned}

Then

\begin{aligned} C_{\text {min }}^{\prime} &=\frac{\epsilon_{o x}}{t_{o x}+\left(\frac{\epsilon_{o x}}{\epsilon_{s}}\right) x_{d T}}=\frac{(3.9)\left(8.85 \times 10^{-14}\right)}{180 \times 10^{-8}+\left(\frac{3.9}{11.7}\right)\left(0.30 \times 10^{-4}\right)} \\ &=2.925 \times 10^{-8} \mathrm{~F} / \mathrm{cm}^{2} \end{aligned}

We may note that

\frac{C_{\min }^{\prime}}{C_{o x}}=\frac{2.925 \times 10^{-8}}{1.9175 \times 10^{-7}}=0.1525

The flat-band capacitance is

\begin{aligned}C_{F B}^{\prime} &=\frac{\epsilon_{o x}}{t_{o x}+\left(\frac{\epsilon_{o x}}{\epsilon_{s}}\right) \sqrt{\frac{V_{t} \epsilon_{s}}{e N_{a}}}} \\ &=\frac{(3.9)\left(8.85 \times 10^{-14}\right)}{180 \times 10^{-8}+\left(\frac{3.9}{11.7}\right) \sqrt{\frac{(0.0259)(11.7)\left(8.85 \times 10^{-14}\right)}{\left(1.6 \times 10^{-19}\right)\left(10^{16}\right)}}} \\ &=1.091 \times 10^{-7} \mathrm{~F} / \mathrm{cm}^{2} \end{aligned}

We also note that

\frac{C_{F B}^{\prime}}{C_{o x}}=\frac{1.091 \times 10^{-7}}{1.9175 \times 10^{-7}}=0.569

Comment

The ratios of C_{\min }^{\prime} / C_{o x} and C_{F B}^{\prime} / C_{o x} are typical values obtained in C-V plots.