of X: Here, D being the (integrated) absorbed dose (D dt where tirr is the irradiation duration). The spatial distribution of this positive trapped charge after a given irradiation time t is represented schematically in fig. 18 below: Figure 18. Distribution of the positive charge trapped in the bulk of the oxide (at a given dose), under positive gate bias (see text for assumptions made). It is now necessary to calculate the voltage shift associated with this trapped charge. This voltage shift corresponds in fact to the voltage one must apply to the gate, to compensate the image III-29 charge induced in the silicon substrate by the charge trapped in the oxide. This image charge is given by the first moment of the space charge distribution. The voltage shift associated with this charge can then be written: ( Voth is the voltage shift due to the oxide-trapped hole charge, under positive gate voltage.) Replacing Pt(x,tirr) by expression (47) yields : (49) dx After integration, it becomes: Voth+= If we note that the mean free path before capture, in the low dose regime (neglecting the electron trapped hole recombination), is: then the voltage shift can be written: Expression (51) gives the voltage shift associated with the positive charge trapped on hole traps uniformly distributed in the oxide, in the case of a positive applied electric field. The ratio =tox! p evidently plays a critical role as representing the capture efficiency. The equation can be rewritten: III-30 NTPS representing the total trap density per unit area (uniform distribution). This equation can be simplified in two limit cases: when the mean free path is much greater than the oxide thickness (case of a low trap density), or much smaller than it. It yields: When the mean free path is larger than the oxide thickness, i.e. when = tpNTPS <<1, then developing the exponential term to the third order we find that +() tends towards When the mean free path is smaller than the oxide thickness, i.e. when = tpNTps>>1, then +() tends towards 1/2. In summary V