 currents jp(x) and jn(x) are in this case: jp(0)=0 and jn(tox)=0 (we thus assume that the electrodes do not inject any carrier). The traps are supposed to be uniformly distributed in the oxide. This limitation is not compulsory, but it greatly simplifies the calculations. With these boundary conditions, we obtain the following solutions to Eqs. (24) and (27): (34) i = The above expressions have been simplified by introducing the concepts of mean free path before capture for electrons ( n), and holes ( p), which can respectively be written: The currents are represented schematically in fig. 17 above. Because of the uniform distribution of traps, the continuity equations for trapped holes and trapped electrons can be written in this case: III-27 (38) (x,t) [NTp-pt(x,t)] - nre (39) = - pr(E)_jp(x,t) n,(x,t) These expressions contain cross-coupled terms jp(x,t) and jn(x,t), which makes it very complex to find an analytical solution. An additional simplifying assumption is to neglect the free carrier-trapped charge recombination. In this case, the expressions of the mean free paths before capture (i.e. (27) and (28)) become: The continuity equations for trapped carriers then reduce to: In the next paragraphs, the calculation of the voltage shifts associated with charge trapping will be detailed, first by taking into account only one type of trapped carrier, then by taking into account both types, using the superposition theorem. III-28 CASE OF A SINGLE TYPE OF TRAP 3.9.1. SOLUTION FOR POSITIVE BIAS Let us consider for the moment that the material only contains hole traps, uniformly distributed in the oxide, and a positive gate bias. Replacing jp(x,t) by expression (35) into Eqn. (43), we get: (44) Pp where D is the dose rate. This equation can be rewritten: In the low dose regime, the trapped charge density remains small, and we can develop the exponential term to the first order. It yields: (46) PtD- By integrating over time, we find the following expression of the positive charge trapped in the oxide as a function 