4
\$\begingroup\$

Insulated gate field effect transistors are actually 4-terminal devices, because the body voltage can act as a second gate or cause unwanted changes to the threshold voltage.

So now I'm taking a look at the GAAFET, also known as the SGT or TUBEFET. As far as I can tell the body is totally surrounded by the gate, and then the two terminals of source and drain are fully connected to the sides of the gate/body tube.

So this leaves an interesting question. How can you connect the body to ground or source? The body is totally surrounded by the gate and two other terminals. There appears to be no location to add a 4th connection into the body.

Is that somehow not necessary in GAAFET, and if so, why? I don't want my GAAFET acting as a floating gate like a flash cell with memory.

My interest is digital circuits for CPUs, but it would be interesting to know if the answer is different for other stuff like power transistors.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Not all transistors are 4 terminal, (for example BJTs and JFETs) so I modified your question. \$\endgroup\$ Commented May 9, 2023 at 22:25
  • \$\begingroup\$ The body of a GAAFET is below the gate(s) like on a FINFET. Similar to a FINFET where body where the body voltage doesn't do anything since it's so far from the channel, I think on GAAFET it probably doesn't matter. Maybe someone with more experience can comment on that. \$\endgroup\$ Commented May 10, 2023 at 12:58

1 Answer 1

1
+50
\$\begingroup\$

The channel still has an "anchored" potential in a GAAFet when off (assume N type):

  • It can't float negative because the gate would turn it on, shorting it to the source.
  • It can't float positive because it would get clamped to the source voltage + few 100 mV via their pn-junction.

So while off, the channel potential will naturally reside near the source voltage. As you begin turning it on by raising the gate voltsge, the channel will start to get inverted near the source, making that source-body-clamp stronger (lower impedance) and providing a stable threshold voltage for when the inversion reaches the drain end of the channel.

\$\endgroup\$

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.