For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity...

 

MOSFET Substrate Doping Concentration

Given

Gate oxide thickness: \[ t_{ox} = 10\,\text{nm} = 10^{-8}\,\text{m} \]

Substrate sensitivity: \[ \frac{\partial V_T}{\partial |V_{BS}|} = 50\,\text{mV/V} = 0.05 \]

Substrate voltage: \[ |V_{BS}| = 2\,\text{V} \]

Electron charge: \[ q = 1.6\times10^{-19}\,\text{C} \]

Vacuum permittivity: \[ \epsilon_0 = 8.85\times10^{-12}\,\text{F/m} \]

Relative permittivities: \[ \epsilon_{Si}=12,\qquad \epsilon_{ox}=4 \]

Step 1: Find the body-effect coefficient

For large substrate voltage,

\[ V_T \approx V_{T0}+\gamma\sqrt{|V_{BS}|} \]

Therefore,

\[ \frac{\partial V_T}{\partial |V_{BS}|} = \frac{\gamma}{2\sqrt{|V_{BS}|}} \]

Substituting the given values:

\[ 0.05=\frac{\gamma}{2\sqrt{2}} \]

Hence,

\[ \gamma=0.05(2\sqrt{2}) =0.1414\;\text{V}^{1/2} \]

Step 2: Calculate the oxide capacitance

The oxide permittivity is

\[ \epsilon_{ox} =4\epsilon_0 =4(8.85\times10^{-12}) =3.54\times10^{-11}\,\text{F/m} \]

Therefore,

\[ C_{ox} = \frac{\epsilon_{ox}}{t_{ox}} \]

\[ C_{ox} = \frac{3.54\times10^{-11}} {10^{-8}} = 3.54\times10^{-3}\,\text{F/m}^2 \]

Step 3: Find the substrate doping concentration

For an n-channel MOSFET, the body-effect coefficient is

\[ \gamma = \frac{\sqrt{2q\epsilon_{Si}N_A}} {C_{ox}} \]

Rearranging for \(N_A\):

\[ N_A = \frac{\gamma^2 C_{ox}^2} {2q\epsilon_{Si}} \]

Substituting the values:

\[ N_A = \frac{ (0.1414)^2 (3.54\times10^{-3})^2 } { 2(1.6\times10^{-19}) (12)(8.85\times10^{-12}) } \]

This gives:

\[ N_A \approx 7.37\times10^{21}\,\text{m}^{-3} \]

Step 4: Convert to cm^-3

Since

\[ 1\,\text{m}^{-3}=10^{-6}\,\text{cm}^{-3} \]

Therefore,

\[ N_A = 7.37\times10^{21}\times10^{-6} \]

\[ N_A = 7.37\times10^{15}\,\text{cm}^{-3} \]

Answer: (a) \(7.37\times10^{15}\,\text{cm}^{-3}\)