2
\$\begingroup\$

I was recently reading up on vacuum tube like devices that could serve as a possible replacement for traditional CMOS technology. How could vacuum tube like technology replace CMOS if vacuum tubes don't have two different types of switching (In CMOS this would be N-channel and P-channel transistors). Vacuum tubes only stop current if there is a negative charge on the gate, in contrast MOSFETs have two different kinds of transistors (N and P channel) that switch with different voltages. So my question is: Is it possible to implement two different kinds of switching in vacuum tubes? And if not, why do people think this could serve as a possible replacement for traditional MOSFETs?

Here is the article I was reading: http://www.nytimes.com/2016/06/06/technology/smaller-chips-may-depend-on-technology-from-grandmas-radio.html?_r=0

\$\endgroup\$
1
  • \$\begingroup\$ Not an answer to your question, but the idea was presented a couple of years ago in IEEE spectrum amongst other places which actually gives somewhat more information about the approach. \$\endgroup\$
    – Tom Carpenter
    Commented Jun 15, 2016 at 1:25

3 Answers 3

Reset to default
3
\$\begingroup\$

Much of the history of semiconductor IC logic was done with only N-channel or P-channel transistors. It is only in more recent generations that designers had the luxury of having access to both types. Certainly logic can be designed with only one type of switch. As it was back in the original days of real "firebottle" vacuum tubes (BrEnglish: "valves") and the germanium and then silicon transistors that replaced them.

\$\endgroup\$
3
  • \$\begingroup\$ But without two kinds of transistors don't you end up with a lot of useless current flowing through your logic circuits? It seems like an enormous waste of energy. \$\endgroup\$
    – Derek Farkas
    Commented Jun 15, 2016 at 1:16
  • 1
    \$\begingroup\$ @DerekFarkas: Sure, which is why CMOS IC fabrication was such a huge leap over previous NMOS IC fabrication. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jun 15, 2016 at 1:31
  • \$\begingroup\$ "more recent" meaning "since the 1980s" ... but even before N-MOS logic, bipolar RTL and later TTL was almost exclusively NPN transistors which were faster than PNP in the technology of the time. Designers simply got used to asymmetric logic levels (strong pulldowns, weak pullups) to minimise that wasted power and that's why Reset and Enable signals are still active low today - old habits die hard. \$\endgroup\$
    – user16324
    Commented Jun 15, 2016 at 11:40
3
\$\begingroup\$

So my question is: Is it possible to implement two different kinds of switching in vacuum tubes?

Yes, it is possible.

For example, you could build a tube consisting of one cathode, two anodes, one acceleration grid (like in a triode) and one deflection grid (like in a old display/CRT). The acceleration grid is used to establish electron flow between the cathode and the anodes. The deflection grid is used to select which anode gets hit by the electron beam.

This gives you the two polarities you're asking for.

\$\endgroup\$
2
  • \$\begingroup\$ Never heard of a beam deflection device used for binary logic! References? They make great RF mixers though, and I suppose you could call the Williams Tube (2048 bit DRAM) a beam deflection device of sorts... \$\endgroup\$
    – user16324
    Commented Jun 15, 2016 at 11:43
  • \$\begingroup\$ @BrianDrummond neither have I, but you can certainly build one. That aside there exist dual control pentodes like the 6888 which are directly usable as NAND gates. \$\endgroup\$
    – Nils Pipenbrinck
    Commented Jun 15, 2016 at 19:25
1
\$\begingroup\$

It is similar to RTL or PMOS, which uses a single type of BJT or MOSFET except that vacuum tubes are more like depletion mode n-channel MOSFETs. You can find many schematics online if you look.

As far as wasted power in the load goes, the tube filaments use a lot of power too (several watts for a dual in many cases) the load dissipation was not that huge in comparison, maybe 10-20% more.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Actually more like JFETs in that if you pull the grid positive, you see grid current... \$\endgroup\$
    – user16324
    Commented Jun 15, 2016 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.