Abstract
Low residual voltages (70–95 V) were observed in our experiments with Si n +–p–n 0–n + avalanche transistors at current pulses of a few nanoseconds with an amplitude of ∼100 A. The voltages are much lower than that predicted by a simple theory of avalanche transistor switching. A physical explanation is suggested and a numerical model is produced which explains the low residual voltages by a strong rebuilding of the electric field domain in the n 0 collector. This reconstruction takes place when the current density significantly exceeds a critical value, which is associated with a drift of equilibrium carriers in the collector at a saturated velocity. The final electric field distribution across the collector region was shown to be greatly dependent on both total current density and the ratio of the injection current component to the total current. A voltage drop of less than 50 V was calculated at high total currents (∼10 5 A/cm 2) provided that the ratio of the electron injection current to the total current exceeded 0.7. The maximum possible value of this ratio is determined by the fundamental properties of the semiconductor material and plays an essential role in the phenomenon. By contrast, we did not succeed in obtaining any appreciable reduction in the residual voltage for p +–n–p 0–p + transistors either experimentally or numerically. The physical reasons for this behaviour were found to be mainly determined by the difference in the electron and hole mobilities.
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