Correct Answer - Option 1 : 7.37 × 10
15 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