Question

$\text{An ideal MOS capacitor with p-type substrate has substrate doping concentration of}{N}_A=1.2\times {10}^{15}c{m}^2\text{, oxide capcitance}{C}_{ox}=2\times {10}^{-9}F/c{m}^2,{\epsilon }_{si}=1.04\times {10}^{-12}F/cm\text{. If the body terminal is grounded, then the gate voltage required to produce a surface potential of 0.026 V at the semiconductor-oxide layer interface will be\_\_\_\_\_\_V.}$

Answer 1

Substrate doping concentration,
$N_A=1.2\times {10}^{15}c{m}^{-3}$
$\text{Oxide capacitance,}{c}_{ox}=2\times {10}^{-9}F/c{m}^2$
$\text{Surface potential,}{\varphi }_s=0.026V$
$\text{gate voltage,}{V}_G={\varphi }_s+V_{ox}$
$\text{but,}{V}_{ox}=\frac{\sqrt{2q{N}_A{\epsilon }_{si}{\varphi }_s}}{C_{ox}}$

$V_{ox}=\frac{\sqrt{2\times 1.6\times {10}^{-19}\times 1.2\times {10}^{15}\times 1.04\times {10}^{-12}\times 0.026}}{2\times {10}^{-9}}$
$\therefore V_{ox}=1.611V$
$\therefore \text{Gate voltage,}{V}_G={\varphi }_s+V_{ox}=1.611+0.026$
$\therefore V_G=1.637V\approx 1.64V$