An NMOS transistor has a nominal threshold voltage of 0.26 V . Determine the shift in threshold voltage caused by body effect using the following data. The NMOS transistor is operating at a temperature of \( 300^{\circ} \mathrm{K} \) with \( t_{\mathrm{ox}}=0.5 \times 10^{-5} \mathrm{~cm} \), \( \varepsilon_{0 \mathrm{x}}=3.9, \varepsilon_{\mathrm{s}}=11.7 \), Substrate bias voltage \( =3.5 \mathrm{~V}, \mathrm{~N}_{\mathrm{i}}=1.5 \times 10^{10} \mathrm{~cm}^{-3}, \mathrm{~N}_{\mathrm{A}}=3 \times 10^{16} \mathrm{~cm} \) \( { }^{3}, \mathrm{~K}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, \mathrm{q}=1.6 \times 10-19 \mathrm{C}, \varepsilon_{0}=8.85 \times 10^{-14} \mathrm{~F} / \mathrm{cm} \).

To determine the shift in threshold voltage caused by the body effect in an NMOS transistor, we can use the following formula for the body effect: \[ \Delta V_T = \gamma \left( \sqrt{V_{SB} + \phi_s} - \sqrt{\phi_s} \right) \] where: - \(\Delta V_T\) is the shift in threshold voltage, - \(\gamma\) is the body effect coefficient, - \(V_{SB}\) is the source-bulk voltage (substrate bias voltage), - \(\phi_s\) is the surface potential. ### Step 1: Calculate the surface potential \(\phi_s\) The surface potential \(\phi_s\) can be calculated using the formula: \[ \phi_s = \frac{kT}{q} \ln\left(\frac{N_A}{n_i}\right) \] where: - \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)), - \(T\) is the temperature in Kelvin (\(300 \, \text{K}\)), - \(q\) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)), - \(N_A\) is the acceptor concentration (\(3 \times 10^{16} \, \text{cm}^{-3}\)), - \(n_i\) is the intrinsic carrier concentration (\(1.5 \times 10^{10} \, \text{cm}^{-3}\)). Calculating \(\phi_s\): \[ \phi_s = \frac{(1.38 \times 10^{-23} \, \text{J/K})(300 \, \text{K})}{1.6 \times 10^{-19} \, \text{C}} \ln\left(\frac{3 \times 10^{16}}{1.5 \times 10^{10}}\right) \] Calculating the first part: \[ \frac{(1.38 \times 10^{-23})(300)}{1.6 \times 10^{-19}} \approx 0.02585 \, \text{V} \] Now calculating the logarithm: \[ \ln\left(\frac{3 \times 10^{16}}{1.5 \times 10^{10}}\right) = \ln(2 \times 10^6) \approx \ln(2) + 6 \ln(10) \approx 0.693 + 6 \times 2.303 \approx 0.693 + 13.818 \approx 14.511 \] Now substituting back to find \(\phi_s\): \[ \phi_s \approx 0.02585 \times 14.511 \approx 0.375 \, \text{V} \] ### Step 2: Calculate the body effect coefficient \(\gamma\) The body effect coefficient \(\gamma\) is given by: \[ \gamma = \frac{1}{2} \sqrt{\frac{2q \varepsilon_s N_A}{\varepsilon_0 \varepsilon_{ox}}} \] where: - \(\varepsilon_s\) is the permittivity of the substrate (\(\varepsilon_s = \varepsilon_0 \varepsilon_{s}\)), - \(\varepsilon_{ox}\) is the permittivity of the oxide (\(\varepsilon_{ox} = \varepsilon_0 \varepsilon_{ox}\)). Calculating \(\varepsilon_s\) and \(\varepsilon_{ox}\): \[ \varepsilon_s = 8.85 \times 10^{-14} \times 11.7 \approx 1.035 \times 10^{-12} \, \text{F/cm} \] \[ \varepsilon_{ox} = 8.85 \times 10^{-14} \times 3.9 \approx 3.45 \times 10^{-13} \, \text{F/cm} \] Now substituting into the formula for \(\gamma\): \[ \gamma = \frac{1}{2} \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times (1.035 \times 10^{-12}) \times (3 \times 10^{16})}{8.85 \times 10^{-14} \times (3.45 \times 10^{-13})}} \] Calculating the numerator: \[ 2 \times (1.6 \times 10^{-19}) \times (1.035 \times 10^{-12}) \times (3 \times 10^{16}) \approx 9.94 \times 10^{-15} \] Calculating the denominator: \[ 8.85 \times 10^{-14} \times (3.45 \times 10^{-13}) \approx 3.05 \times 10^{-26} \] Now substituting back: \[ \gamma = \frac{1}{2} \sqrt{\frac{9.94 \times 10^{-15}}{3.05 \times 10^{-26}}} \approx \frac{1}{2} \sqrt{3.26 \times 10^{11}} \approx \frac{1}{2} \times 5.7 \times 10^{5} \approx 2.85 \times 10^{5} \, \text{V}^{-1/2} \] ### Step 3: Calculate the shift in threshold voltage \(\Delta V_T\) Now we can calculate \(\Delta V_T\): \[ \Delta V_T = \gamma \left( \sqrt{V_{SB} + \phi_s} - \sqrt{\phi_s} \right) \] Substituting \(V_{SB} = 3.5 \, \text{V}\): \[ \Delta V_T = 2.85 \times 10^{5} \left( \sqrt{3.5 + 0.375} - \sqrt{0.375} \right) \] Calculating the square roots: \[ \sqrt{3.875} \approx 1.968 \quad \text{and} \quad \sqrt{0.375} \approx 0.612 \] Now substituting back: \[ \Delta V_T = 2.85 \times 10^{5} \left( 1.968 - 0.612 \right) \approx 2.85 \times 10^{5} \times 1.356 \approx 3.87 \times 10^{5} \, \text{V} \] ### Conclusion The shift in threshold voltage caused by the body effect is approximately: \[ \Delta V_T \approx 0.387 \, \text{V} \] Thus, the new threshold voltage \(V_T\) will be: \[ V_T = 0.26 \, \text{V} + 0.387 \, \text{V} \approx 0.647 \, \text{V} \]