Care to explain why the gate wrap around the active channel allows for a thicker Tox ??
I'm gonna stumble around a bit here with my words as I'm trying to distill down into a few sentences the basics of something that encompasses thousands of pages of dissertations.
(ignoring the 3rd gate for the moment) By having the channel impacted by a gate on both the top and bottom (think 2D) you have dueling e-fields propagating from the gate into the channel.
In effect what you have is a channel that is half the thickness (depth) as a planar channel, requires half the time to "turn on", requires half the time to reach any given point on the Ion curve (basically switching speed is faster).
Lemme see if this half-assed diagram schematic communicates anything:

You see the red arrows in Step 1? The time it takes for those to "fully develop" is dependent on the gate oxide thickness (and the voltage).
Now look at the following:

Here I have intentionally made the fin thicker for illustration purposes, but you see how the channel is "red arrowed" all the quicker from all three sides because the gate wraps around the channel?
Because of this, if the gate-oxide were the same thickness for both the planar transistor and the finfet then the finfet xtor would reach fully "turn on" stage much much quicker (it would switch faster).
If you instead made the gate oxide thicker for the finfet, just enough such that the red arrows (from all sides) came to be in the same amount of time that the red arrows require for the planar xtor then you'd have a xtor that was the same speed but has less leakage because the gate-oxide was thicker.
(I'm intentionally cutting a lot of device physics corners here to simplify the discussion to get this point across)
edit: a couple good Intel papers to dive into for more -
Advanced Depleted-Substrate Transistors: Single-gate, Double-gate and Tri-gate
Tri-Gate Fully-Depleted CMOS Transistors: Fabrication, Design and Layout
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