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For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity (∂VT/∂|VBS|) is found to be 50 mV/V at a substance voltage |VBS| = 2 V, where VT is the threshold voltage of the MOSFET. Assume that, |VBS| >> 2ϕB, where qϕB is the separation between the Fermi energy level EF and the intrinsic level Ei in the bulk. Parameters given are

Electron charge (q) = 1.6 × 10-19 C

Vacuum permittivity (ε0) = 8.85 × 10-12 F/m

Relative permittivity of silicon (εSi) = 12

Relative permittivity of oxide (εox) = 4

The doping concentration of the substrate is      


1. 7.37 × 1015 cm-3
2. 9.37 × 1015 cm-3
3. 2.37 × 1015 cm-3
4. 4.37 × 1015 cm-3

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Correct Answer - Option 1 : 7.37 × 1015 cm-3

Concept:

The threshold value for the transistor is given by:

\({V_t} = {V_{t0}} + γ \left[ {√ {{V_{SB}} + \left| {2{\phi _F}} \right|} - √ {\left| {2{\phi _F}} \right|} } \right]\)

VBS for the n channel transistor

Calculation:

Given that substrate, sensitivity is 50 mV/V

\(\frac{{\partial {V_t}}}{{\partial \left| {{V_{BS}}} \right|}} = \frac{γ }{{2√ {{V_{BS}} + 2\left| {{\phi _F}} \right|} }}\)

\(\frac{{50mV}}{V} = \frac{γ }{{2√ 2 }}\left( {\because\left| {{V_{BS}}} \right| \gg 2{\phi _B}} \right)\)

Consider ϕB = ϕF 

γ = 0.1 √2

γ is defined as:

\(γ = \frac{{√ {2q{N_A}{ \in _{Si}}} }}{{{C_{ox}}}}\)

\({C_{ox}} = \frac{{{ \in _{ox}}}}{{{t_{ox}}}} = \frac{{4{ \in _0}}}{{10nm}}\)

\(0.1√ 2 = \frac{{√ {2q{N_A}{ \in _{Si}}} }}{{{C_{ox}}}}\)

Squaring on both sides, we get:

\(0.01 × \frac{{16 \in _0^2}}{{100 × {{10}^{ - 18}}}} = 1.6 × {10^{ - 19}}{N_A} × 12{ \in _0}\)

\({N_A} = \frac{{16 × 8.85 × {{10}^2}}}{{1.6 × {{10}^{ - 19}} × 12}}\)

\({N_A} = \frac{{8.85 × {{10}^{22}}}}{{12\;{m^3}}}\)

NA = 0.7375 × 1022 m-3

NA = 7.375 × 1015 cm-3

Hence option (1) is correct

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