Question

A voltage $V_G$ is applied across a MOS capacitor with metal gate and p-type silicon substrate at $T=300K$. The inversion carrier density (in number of carriers per unit area) for $V_G=0.8V$ is $2\times {10}^{11}c{m}^{-2}$. For $V_G=1.3V$, the inversion carrier density is $4\times {10}^{11}c{m}^{-2}$. What is the value of the inversion carrier density for $V_G=1.8V$?

• A. $4.5\times {10}^{11}c{m}^{-2}$
• B. $7.2\times {10}^{11}c{m}^{-2}$
• C. $6.0\times {10}^{11}c{m}^{-2}$
• D. $8.4\times {10}^{11}c{m}^{-2}$

Answer 1

The correct option is C $6.0\times {10}^{11}c{m}^{-2}$

Ans : (b)

In a MOS- Capacitor $(V_{G1}-V_T)\propto Q$
$\frac{(V_{G1}-V_T)}{(V_{G_2}-V_T)}=\frac{Q_1}{Q_2}$
$\frac{0.8-V_T}{1.3-V_T}=\frac{2\times {10}^{11}}{4\times {10}^{11}}$
On solving, $V_T=0.3V$
Now Consider, $\frac{V_{G_2}-V_T}{V_{G_3}-V_T}=\frac{Q_1}{Q_2}$
$\frac{1.3-0.3}{1.8-0.3}=\frac{4\times {10}^{11}}{Q_3}$ Inverse charge density with $V_G=1.8V$
$Q_3=6\times {10}^{11}/c{m}^2$