If we position two electrodes in free air, 1Centimeter apart, and if we then apply 10000Volts across them, the air’s ability to resist electric current will break down, and an electric arc will appear across it.
Because of this simple observation, the question could (and probably, should) be asked, ‘Can a MOSFET transistor the size of a micron, on an Integrated Circuit, withstand 15Volts of Source-Drain voltage, at all?’
A suggestion to the contrary would be, that 10000Volts /Centimeter, is equal to 1 Volt /Micron. Thus, if the two electrodes were 1Micron apart, and standing free in air, it would take only 1 Volt to cause the air to break down, and for a microscopic arc to appear. Yet, Integrated Circuits are known to exist, which operate at 2 Volts, and which use ‘nanometer technologies’. And so in an effort to answer my own question, I would take two further observations into consideration:
- I already recall reading elsewhere, that the breakdown voltage of high-quality, semiconductor silicon, is considerably greater than that of air !
- I possess a suite of programs named “SPICE”, which, when performing a Level-8 simulation of MOSFET transistors, only needs to be given the width and the length of a transistor-instance, and which will, on that basis, compute all the other properties of the resulting transistor, making certain assumptions about its design.
This use of SPICE has been commented on, on the following Bulletin-Board:
The part of the thread, which I’ve linked to before, and which I want to call the reader’s attention to, is the part where Holger Vogt writes:
“A 0.18µm process however should run at lower supply voltage, e.g. 1.8-2 V.”