Question 10.2: MOSFET Degradation Time In a short-channel MOSFET with xox ...
MOSFET Degradation Time
In a short-channel MOSFET with x_{ox} = 30 nm, a gate current of 1 nA is momentarily caused by hot-electron injection. This gate current is estimated to flow through a section of oxide measuring 200 nm × 10 μm near the drain end of the channel. Assume that a fraction equal to 10^{-6} of the injected electrons becomes trapped at an average distance of 0.1 x_{ox} from the Si-SiO_2 interface. Calculate the time needed for gate-current flow (under the conditions stated) to change the threshold voltage by 100 mV in the region where the injection is taking place.
The gate current of 1 nA is carried by a current density across the injection area of
10^{-9}/2\times10^{-8}=5\times10^{-2} \ A \ cm^{-2}Because 1 in 10^6 electrons become trapped, the rate of charge trapping in the SiO_2 is 5 \times 10^{-2} \times 10^{-6} = 5 \times 10^{-8} \text{ Coulombs } S^{-1} cm^{-2} . From Equation 8.5.4, trapped charge in the oxide shifts the flat-band voltage and thereby the threshold voltage (Equation 8.3.18). Assuming that the charge is concentrated in a sheet, we calculate \Delta V_{FB} from Equation 8.5.3 to be
\Delta V_{FB}=-\frac{1}{C_{ox}} \int_{0}^{x_{ox}}{\frac{x}{x_{ox}}\rho (x) } dx (8.5.4)
V_T=V_{FB}+V_C+2\left|\phi _p\right| +\frac{1}{C_{ox}} \sqrt{2\epsilon _sqN_a(2\left|\phi _p\right|+V_C-V_B )} (8.3.18)
\Delta V_{FB}=-\frac{Q_{ox}x_1}{C_{ox}x_{ox}} (8.5.3)
\Delta V_{FB}=\Delta V_T=\left(\frac{1}{C_{ox}} \right) \left(\frac{0.9x_{ox}}{x_{ox}} \right) \Delta Q_{ot}or
\Delta Q_{ot}=\frac{C_{ox}\Delta V_T}{0.9} =\frac{1.15\times10^{-7}\times0.1}{0.9} =1.28 \times 10^{-8} \ C \ cm^{-2}The time required to trap this quantity of charge is
t=\frac{\Delta Q_{ot}}{J_{ot}} =\frac{1.28\times 10^{-8}}{5\times 10^{-8}} =0.256 \ s