Question

For an n-channel silicon MOSFET with $10nm$ gate oxide thickness. the substrate sensitivity $\left(\frac{\delta V_T}{\delta |V_{BS}|}\right)$ is found to be $50mV/V$ at a substrate voltage $V_{BS}=2V$, where $V_T$ is the threshold voltage of the MOSFET. Assume that, $|V_{BS}|>>2{\Phi }_B$, 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) = $1.6\times {10}^{-19}C$
Vaccum permittivity $\left({ϵ}_0\right)=8.85\times {10}^{12}F/m$
Relative permittivity of silicon ${ϵ}_{Si})=12$
Relative permittivity of oxide (${ϵ}_{ox})=4$
The doping concentration of the substrate is

Answer 1

Given, N-channel MOSFET
$t_{ox}=10nm=10\times {10}^{-7}cm$
$\frac{\delta V_T}{\delta |V_{BS}|}=50mV/V,|V_{BS}|=2V$
$q=1.6\times {10}^{-19}C,|V_{BS}|>>2{\varphi }_B$

${ϵ}_o=8.85\times {10}^{-14}F/cm$
Ans : (d)

${ϵ}_{r_{si}}=12$
${ϵ}_{r_{ox}}=4$
Threshold voltage, including body effect,
$V_T={\varphi }_{ms}+\frac{\sqrt{2{ϵ}_{si}q{N}_A(2{\varphi }_B-V_{BS})}}{C_{ox}}+2{\varphi }_B$
In question, we need, $|V_{BS}|=|V_{SB}|$ $\Rightarrow V_T={\varphi }_{ms}+\frac{\sqrt{2{ϵ}_{si}q{N}_A(2{\varphi }_B+|V_{SB}|)}}{C_{ox}}+2{\varphi }_B$ $\therefore \frac{\delta V_T}{\delta |V_{BS}|}=0+\frac{\sqrt{2{ϵ}_{si}q{N}_A}}{C_{ox}}.\frac{1}{2\sqrt{2{\varphi }_B+|V_{SB}|}}+0$ $50\times {10}^{-3}=\frac{\sqrt{2\times 8.85\times {10}^{-14}\times 12\times 1.6\times {10}^{-19}{N}_A}}{{ϵ}_{ox}/t_{ox}}.\frac{1}{2\sqrt{|V_{SB}|}}$ $\Rightarrow {\left(\frac{50\times {10}^{-3\times 4\times 8.85\times {10}^{-14}}}{10\times {10}^{-7}}\right)}^2$ $\frac{2\times 8.85\times {10}^{-14}\times 12\times 1.6\times {10}^{-19}}{4\times 2}\times N_A$ $[\because |V_{SB}|>>2{\varphi }_B]$ $\Rightarrow N_A=7.375\times {10}^{15}c{m}^{-3}$