MOSFET Degradation Time In a short-channel MOSFET with xox = 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

Question

MOSFET Degradation Time

In a short-channel MOSFET with [latex] x_{ox} [/latex] = 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[latex]^{-6} [/latex] of the injected electrons becomes trapped at an average distance of 0.1 [latex] x_{ox} [/latex] from the [latex] Si-SiO_2 [/latex] 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.

Accepted Answer

The gate current of 1 nA is carried by a current density across the injection area of

[latex] 10^{-9}/2 imes10^{-8}=5 imes10^{-2} \ A \ cm^{-2} [/latex]

Because 1 in 10[latex]^6 [/latex] electrons become trapped, the rate of charge trapping in the [latex] SiO_2 [/latex] is [latex] 5 imes 10^{-2} imes 10^{-6} = 5 imes 10^{-8} ext{ Coulombs } S^{-1} cm^{-2} [/latex]. 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 [latex] \Delta V_{FB} [/latex] from Equation 8.5.3 to be

[latex] \Delta V_{FB}=-\frac{1}{C_{ox}} \int_{0}^{x_{ox}}{\frac{x}{x_{ox}} ho (x) } dx [/latex] (8.5.4)

[latex] V_T=V_{FB}+V_C+2\left|\phi _p ight| +\frac{1}{C_{ox}} \sqrt{2\epsilon _sqN_a(2\left|\phi _p ight|+V_C-V_B )} [/latex] (8.3.18)

[latex] \Delta V_{FB}=-\frac{Q_{ox}x_1}{C_{ox}x_{ox}} [/latex] (8.5.3)

[latex] \Delta V_{FB}=\Delta V_T=\left(\frac{1}{C_{ox}} ight) \left(\frac{0.9x_{ox}}{x_{ox}} ight) \Delta Q_{ot} [/latex]

or

[latex] \Delta Q_{ot}=\frac{C_{ox}\Delta V_T}{0.9} =\frac{1.15 imes10^{-7} imes0.1}{0.9} =1.28 imes 10^{-8} \ C \ cm^{-2} [/latex]

The time required to trap this quantity of charge is

[latex] t=\frac{\Delta Q_{ot}}{J_{ot}} =\frac{1.28 imes 10^{-8}}{5 imes 10^{-8}} =0.256 \ s [/latex]

Question 10.2: MOSFET Degradation Time In a short-channel MOSFET with xox ...

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