Question 9.4: Short-Channel Model Use the short-channel model to find valu...
Short-Channel Model
Use the short-channel model to find values for V_{Dsat} and I_{Dsat} for a MOSFET with the following parameters: x_{ox} = 20 \ nm, W = 50 \ μm, L = 0.5 \ μm, V_T = 0.7 \ V when its source is at zero potential, and with bias voltages V_G = 3 \ V and V_D = 1.5 \ V . Compare these results to the predictions of the long-channel model.
From Equation 9.2.3 with V_Z = 0.5 V ,
\xi _{eff}=\frac{(V_G-V_T)}{6x_{ox}} +\frac{(V_T+V_Z)}{3x_{ox}} =3.92\times10^5 \ Vcm^{-1}From Equation 9.2.4 and Table 9.3
\mu _{eff}=\frac{\mu _0}{1+(\xi _{eff}/\xi _0)^\nu } (9.2.4)
TABLE 9.3 Parameters for effective mobility © 1986 IEEE [13]
| Unit | Electron (surface) | Hole (surface) | Hole (subsurface) | |
| \mu _0 | cm²/V-s | 670 | 160 | 290 |
| \xi _0 | MV/cm | 0.67 | 0.7 | 0.35 |
| \nu | 1.6 | 1.0 | 1.0 |
Taking v_{sat} = 8 \times 10^6 cm s^{-1} , we calculate \xi _{sat} = 2\frac{v_{sat}}{\mu _{eff}} = 3.4 \times 10^4 \ Vcm^{-1}
From Equation 9.2.11, V_{Dsat} = \frac{\xi _{sat}L(V_G-V_T)}{\xi _{sat}L+(V_G-V_T)} =0.98 \ V
Because the applied drain voltage is 1.5 V (> V_{Dsat} ), the MOSFET is in the saturated-drain-current region; hence, using Equation 9.2.10,
I_{Dsat}=WC_{ox}(V_G-V_T-V_{Dsat})v_{sat}=9 \ mATo compare these results to those from the long-channel theory, we first calculate V_{Dsat} = V_G – V_T = 2.3 \ V so that the MOSFET is operating below the saturation region. From Equation 9.1.5 we find
I_D=\mu _nC_{ox}\frac{W}{L} \left[\left(V_G-V_T-\frac{1}{2}V_D \right)V_D \right] \ \ \ \ \ \text{ or } \ \ \ \ \ I_D=24 \ mAusing a typical low-field mobility value of 600 cm^2 V^{-1} s^{-1} . The very different results obtained using these two models show the degree to which the MOSFET behavior is influenced by short-channel effects.