Hacker News Reader

Comment by jakobegger | Hacker News Reader

> Does this mean that chips would have less resistive heating?

No. The parts of the chip that conduct electricity to the active areas in current chips are already made from metal. Basic chips are made from layers of metal / oxide / semiconductor (MOS).

Current flows through silicon only in small areas, and the way chips work is that you can "turn off" the high resistivity in those areas and make them conducting (by doping the material + applying electric currents & fields)

In the presented work, an air gap is used for these "active areas" instead of a semiconductor. An air gap does have resistance, so there will be resistive heating just like in normal chips.

View on HN

The current does not flow through a CMOS device in a conventional sense. Single nanoamperes, at most, but when you have billions of transistors that stacks up quickly.

In CMOS, only one device of a PMOS/NMOS pair is open at the time. In simplest terms: electric field gets through a CMOS device, but not so much of current.

And what happens during the switching transition then? Answer: both conduct for a very short period, hence the dependency of power consumption on frequency.

Yep, just enough to flip the device state.

There are two factors to that. Cross conduction is one, the other is the current pulse needed to charge the capacitance of the switched signal (the trace capacitance but also the gate capacitance of the next stage(s); of course this simple model is only an idealized concept with current process nodes).

> An air gap does have resistance, so there will be resistive heating just like in normal chips.

They also mentioned putting these in a vacuum. Presumably that won't have as much resistance as an air gap.

A perfect vacuum is a perfect insulator. There are capacitors that use vacuum.

Electrons can still pass through perfect vacuums.

I think that’s implied by his pointing out that capacitors use vacuum. A cap that doesn’t allow electron flow wouldn’t be a cap.

A capacitor shouldn't allow electron flow between its (dielectric separated) plates. The charges should build up on the plates, then exit through the terminals when discharging. If charges cross the dielectric, then you have leakage.

Hm. Apparently my Phys101-level understanding of electronics has caught up to me.

Caps don’t allow electrons through, do they? They resist DC, they only conduct AC?

That is correct. The dielectric ‘conducts’ an electric field, which attracts and repels electrons on the opposing plates.

Vacuum is a special case of an insulator. Others will conduct beyond some threshold voltage; vacuum will not. On the other hand, vacuum will conduct at any voltage as soon as free charges are injected into it.

But electrons can still pass through perfect vacuums. Whether or not a vacuum conducts (obviously it doesn’t) is irrelevant.

> A perfect vacuum is a perfect insulator.

If that were true, vacuum tubes (used for radio and early computers) wouldn't work.

It is better to call vacuum "the perfect semiconductor," then - it only conducts when free charges are injected into it.

An insulator by definition is matter that doesn't have freely flowing electrons so vacuums aren't an insulator. However, all insulators have a "breakdown voltage" at which point the electrons start "ripping" through the material so a vacuum has all of the properties of an insulator except the mass: the breakdown voltage is the the amount of energy required to eject electrons from the materials surrounding the vacuum and until you reach that point, no current will flow through the vacuum.

Vacuum tubes (thermionic valves) work by heating a tungsten filament-- the cathode-- until the filament is hot enough to emit electrons, which are then attracted to the positively-charged plate (anode), traveling through the vacuum (and thus flowing current). Current only flows in one direction in this setup, which has two electrodes, leading to the moniker "diode." Adding a third electrode in between the cathode and anode-- called the "grid" based on its physical shape-- allowed a vacuum tube (now a 'triode') to use a small signal to control a large one. An amplifier!

The key to the function is the heating of the cathode filament. Vacuum tube designs that use a separate heating current (which is, today, most of them) do not conduct if the heating current is not applied. "Cold cathode" type tubes are not vacuum tubes-- they are typically filled with a low-pressure gas.

a vacum diode have characteristics more similar to a ideal diode that a semiconductor diode.

Valves work because the heating filament fills the vacuum with free electrons.

Vacuum: isolator.

Vacuum filled with electrons: not so good isolator.

Vacuum dielectric strength is 20 megavolts per metre.

Diamond is almost perfect though, at 20 gigavolts per metre, and that only given if the diamond has at least some microscopic imperfections.

I think those might be reversed...

IIRC diamonds are good isolators because their crystal structure is both very close to perfect (few defects) and because their crystal lattice has all electrons tied up.

You are correct. Except, this quality of the vacuum is very fragile. "What's perfect is imperfect, it is the imperfect that is perfect." (Example: a perfectly rigid system will collapse under stress while an imperfect one might withstand it.)

The vacuum has a finite impedance of ~375ohms so it is not a perfect insulator. This comes from the fundamental magnetic permeability constant mu naught divided by the speed of light. Also interesting is the fact that the vacuum impedance is an exact number because it’s solely based on fundamental constants.

Given a small enough distance, arcing is possible between 2 surfaces in vacuum as charges are ejected from the surface (that’s how vacuum tubes and these transistors work).

It is essential to distinguish between characteristic impedances (of free space or other) and resistances. Although both are real values measured in ohms, characteristic impedances are non-dissipative.

Just because EM waves can propagate in vacuum does not mean you can induce a current density.

First part is correct. Second part: ...that's the basis of all RF systems, and exactly how antennas work, isn't it?

Obviously you can't induce a current in vacuum, but you can definitely convert EM waves to current flux in a receiving structure.

Right. I see now that I was unclear in that second part - I should have written something like "just because EM waves can propagate in vacuum does not mean you can induce a current density in vacuum." Of course EM waves can induce current density in antennas and other structures.

> Obviously you can't induce a current in vacuum ...

Yes. I wrote that second part because 'mmmBacon seemed to be arguing that one could induce a current in vacuum; that because vacuum has a characteristic impedance, it isn't an insulator (i.e. can support current because it has characteristic impedance).

Ah, gotcha, appreciate the clarification.

@twtw was not arguing that one can induce a current in the vacuum. However, I did say that a surface can arc in vacuum. This is because a surface can emit charges either thermally or in presence of a large electric field. This is correct and didstinct from current induction in media. Please note this phenomenon should not be confused with arc discharge in gas.

See: https://en.m.wikipedia.org/wiki/Vacuum_arc

Right, but that effect is pretty much unrelated to the characteristic impedance of free space and doesn't mean that vacuum is not an insulator.

I’m sorry but I fundamentally disagree. GP said vacuum is a perfect insulator. Comment was in response to that demonstrating that it is not.

Reply on news.ycombinator.com