This page assumes you know nothing. Before you can read the parent note, you need a small toolbox of ideas. We build each one from a picture, define its symbol, and say exactly why the topic needs it. Each idea depends only on the ones before it.
Forget electronics jargon. A transistor here is just a controllable gate for electric current: a tiny bridge between two wires that is either open (current flows) or shut (no current). A third terminal — the gate — decides which.
Why the topic needs it: a flash cell reads its stored bit by applying a fixed read push VG and watching whether current flows. No notion of "push," no notion of "read."
Every such switch has a tipping point — the smallest push that flips it from shut to open. Below it: nothing. Above it: current flows.
Why the topic needs it: the entire bit-storage trick is "shift Vth up or down." If you understand this one variable, you understand flash. See MOSFET and threshold voltage for the device physics.
Now add the flash-only part. Between the pushing gate and the current bridge, we bury an extra piece of metal completely wrapped in glass-like insulator (oxide). Nothing electrically connects to it — hence floating.
Why the topic needs it: this is the storage element. "Charge present" and "charge absent" are the two physical states that become the two logical bit values.
Now we glue physics to logic. Pick one fixed read push. Two cases:
Bucket
Vth
At read push, current...
Reads as
Full (charged)
high
blocked
0
Empty (erased)
low
flows
1
Why the topic needs it: every "program = write 0," "erase = back to 1" statement in the parent note only makes sense once this inverted mapping is fixed.
If the bucket is sealed in insulator, how do marbles ever get in or out? In everyday physics they can't — glass blocks them. But electrons are tiny enough to obey quantum tunnelling: with a strong enough electric field, an electron can appear on the other side of a thin insulating wall it should not be able to cross.
Why the topic needs it: this is the mechanism for both write and erase, and its wall-damage is the root cause of limited lifetime.
So far the bucket has two settings: full or empty = 1 bit. But what if we can fill it to quarter, half, three-quarter, full — four distinct weights? Then one cell stores more than one bit.
First, what is a bit? A bit is one yes/no answer — enough to pick between 2 options. Two bits pick between 2×2=4. Three bits pick between 2×2×2=8. In general, n bits distinguish 2n options.
We want the reverse question: given L distinguishable levels, how many bits? That inverse of "raise 2 to a power" is the logarithm base 2, written log2.
Why log2 and not division or L itself? Because information is multiplicative: each extra bit doubles the choices. Counting choices needs 2n; counting bits needs its inverse, log2. Nothing else answers "how many yes/no answers hide in L options."
Why the topic needs it: density claims (MLC/TLC/QLC) and the capacity-vs-endurance tradeoff are all stated in these terms.
To compare NOR and NAND cost, we compare how small a cell is. We measure length in units of the smallest line the factory can print, called the feature sizeF.
Why the topic needs it: the parent's ANAND≈4F2 vs ANOR≈10F2 comparison is meaningless without knowing F2 is an area in process units, and that a smaller number means a denser, cheaper chip.
From §5, every write forces electrons through the oxide wall, and each crossing chips at it. Eventually the wall leaks and the cell becomes unreliable.
Why the topic needs it: endurance, wear leveling, and the whole "why can't I edit one byte" story rest on this.