Correct Answer - Option 3 : Q and R
Concept:
For VDS < VGS – VT, n-channel enhancement type MOSFET will operate in triode/linear region:
\({I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]\)
For VDS ≥ VGS – VT, the MOSFET will be in saturation with the current given by:
\({I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {{{\left( {{V_{GS}} - {V_T}} \right)}^2}} \right]\)
L = Channel length
Conclusion: Since the Current is inversely proportional to Channel length, the current will increase as the length will decrease. ∴ Statement P and S are correct.
Resistance:
The current in an NMOS in the linear region is:
\({I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]\)
The transistor operates in the linear region for small values of VDS, i.e. neglecting higher powers of VDS, the above expression can be approximated by:
\({I_D} \approx \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{L}}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]\)
\(R_d=\frac{V_{DS}}{I_{DS}}\)
\(R_d=\frac{1}{\frac{{{\rm{W\mu_n }}{C_{ox}}}}{{L}}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]}\)
\(R_d=\frac{L}{W\mu_nC_{ox}{{{}}}{}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]}\)
Conclusion: As the output resistance is directly proportional to the channel length, it reduces with a decrease in channel length. ∴ Statement Q is incorrect.
Threshold Voltage:
The threshold voltage of a MOSFET is given as:
\({V_T} = 2{\phi _F} - \frac{{{Q_d}}}{{{C_i}}} + {\phi _{ms}} - \frac{{{Q_i}}}{{{C_i}}}\)
From the above expression, we can conclude that the threshold voltage is also dependent on the channel length. ∴ Statement R is incorrect.