GATE 2021 EC – Question 40
For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity $(\partial V_T/\partial|V_{BS}|)$ is found to be 50 mV/V at a substrate voltage $|V_{BS}|=2$ V, where $V_T$ is the threshold voltage of the MOSFET. Assume that, $|V_{BS}|\gg2\Phi_B$, where $q\Phi_B$ is the separation between the Fermi energy level $E_F$ and the intrinsic level $E_i$ in the bulk. Parameters given are
Electron charge $(q)=1.6\times10^{-19}$ C
Vacuum permittivity $(\varepsilon_0)=8.85\times10^{-12}$ F/m
Relative permittivity of silicon $(\varepsilon_{si})=12$
Relative permittivity of oxide $(\varepsilon_{ox})=4$
The doping concentration of the substrate is
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Correct answer: (A) $7.37\times10^{15}\ \text{cm}^{-3}$
Explanation
The body factor is $\gamma=\frac{\sqrt{2q\varepsilon_{si}\varepsilon_0N}}{C_{ox}}$ and $\frac{\partial V_T}{\partial|V_{BS}|}=\frac{\gamma}{2\sqrt{2\Phi_B+|V_{BS}|}}\approx\frac{\gamma}{2\sqrt{|V_{BS}|}}$. So $\gamma=2\times0.05\times\sqrt2=0.1414\ \text{V}^{1/2}$. The oxide capacitance is $C_{ox}=\frac{4\times8.85\times10^{-12}}{10\times10^{-9}}=3.54\times10^{-3}$ F/m$^2$. Then $N=\frac{\gamma^2C_{ox}^2}{2q\varepsilon_{si}\varepsilon_0}=\frac{0.02\times(3.54\times10^{-3})^2}{2(1.6\times10^{-19})(12)(8.85\times10^{-12})}\approx7.37\times10^{21}\ \text{m}^{-3}=7.37\times10^{15}\ \text{cm}^{-3}$.