Intuition The ONE core idea
A MOSFET turned "OFF" is a leaky tap , not a closed one: a tiny current keeps dripping and shrinks exponentially — by a fixed factor of ten for every fixed step you push the gate voltage down. Everything on the parent page is just naming the pieces of that exponential and the physics that fixes its steepness at 60 millivolts per ten-fold drop .
This page assumes you know nothing . We build every letter, ratio, and picture the parent note leans on, in an order where each one only uses things already defined. When you finish, re-read the parent — every symbol will feel like an old friend.
Before any symbol, picture the object. A MOSFET is a tiny sandwich sitting on a slab of silicon.
The source (S) and drain (D) are two terminals we want current to flow between .
The gate (G) is a metal plate hovering just above the silicon, separated by a paper-thin insulator called the oxide .
Between source and drain lies the channel — the strip of silicon the current must cross.
Intuition The gate is a valve you never touch the water with
The gate does not carry the current. It sits behind glass (the oxide) and influences the silicon below by its voltage — like a magnet moving iron filings through a table. Raise the gate voltage → you coax electrons up to the surface → the channel becomes conductive.
We link the geography of these regions to the parent's MOSFET operating regions .
Definition A "voltage difference" symbol
V X Y
V X Y means "the voltage at terminal X minus the voltage at terminal Y ." It is a push — how hard charge is urged to move from X toward Y .
V GS = gate voltage minus source voltage = how hard we press the valve .
V D S = drain voltage minus source voltage = how hard we pull current across the channel .
I D
I D is the current that flows into the drain terminal — the actual charge-per-second crossing the channel from drain to source. It is the output we care about. When the transistor is OFF and this drain current is just the tiny leakage, we give it the special name I s u b (subthreshold current). So I s u b is I D in the OFF region.
Why the topic needs these. The whole subthreshold story is "how does the drain current I D (the drip) depend on the two knobs?" The parent's headline formula has exactly these two knobs: V GS inside the exponential (the strong control) and V D S inside a separate bracket (the weak control). You cannot read that formula without knowing which letter is which knob.
Definition Threshold voltage
V t h
The gate push V GS at which we declare the channel "on enough." Below it we say the transistor is OFF; above it, ON.
The key mind-shift: V t h is a convention , a line we drew , not a physical cliff. Look at the figure — the current curve is smooth and continuous through V t h . Nothing snaps at that line. Below it there is still a real, measurable trickle. That trickle is the subthreshold current.
V t h is where current becomes zero."
Why it feels right: we're taught "below V t h the transistor is off."
The fix: OFF means small , not zero . The current below V t h is tiny but genuinely there — that's the entire subject of this topic.
We build on Threshold voltage Vth . The quantity that matters most turns out to be the distance below threshold , V GS − V t h (a negative number in the OFF state) — that difference is what sits in the exponential.
Current is just charge in motion. There are exactly two reasons charge moves, and the topic swaps from one to the other at threshold.
Charges pushed by an electric field , like marbles rolling downhill. Above threshold (strong inversion), the channel is a good conductor and the drain current I D is carried by drift — this gives the familiar square-law expression for I D .
Charges spreading from crowded places to empty places, purely by random jostling — like a drop of ink spreading in still water, no push needed. Below threshold the channel is nearly empty and there's no field along it, so the only way carriers cross is diffusion.
Intuition Why this matters for the whole page
Because subthreshold transport is diffusion , a MOSFET-below-threshold behaves like the base of a bipolar transistor — see BJT diffusion current . That is why the parent formula looks BJT-ish (exponential in voltage) instead of MOSFET-ish (square-law).
Intuition Ink in a corridor
Picture a narrow corridor of length L . At the left end (source) there is a dense crowd of electrons, n ( 0 ) ; at the right end (drain) a sparser crowd, n ( L ) . Everyone wanders randomly. In any second, more people happen to wander rightward out of the dense end than wander leftward back into it — simply because there are more of them on the left. The net rightward flow is proportional to how big the crowd difference is and how short the corridor is (a steeper thin-out drives a stronger net drift).
That "flow proportional to the concentration slope" is Fick's law . In symbols the slope is d x d n ≈ L n ( 0 ) − n ( L ) , and multiplying by the electron charge q , the flow area A , and a proportionality constant D n (how mobile the carriers are) turns a flow of particles into a current :
I s u b ≈ q A D n d x d n ≈ q A D n L n ( 0 ) − n ( L ) .
We now unpack every letter in that expression.
q , A , L , D n , n ( x )
q = charge of one electron ≈ 1.6 × 1 0 − 19 coulombs — the "size of one carrier."
A = cross-sectional area the current flows through.
L = channel length, source-to-drain distance the carriers cross.
D n = diffusion constant for electrons — how fast crowding spreads out (bigger = faster spreading).
n ( x ) = electron concentration (carriers per volume) at position x along the channel. n ( 0 ) = at the source end, n ( L ) = at the drain end.
The slope L n ( 0 ) − n ( L ) is just "how steeply the crowd thins out from source to drain" — the steeper the thinning, the stronger the diffusion push. Hold on to the two ends n ( 0 ) and n ( L ) : the source end will give the V GS exponential, and the drain end will give the V D S term.
Here is the single most important idea on the whole page: where does the exponential come from?
Definition Boltzmann factor
In a warm system, the number of particles that have enough energy to sit at a raised "shelf" of height E is proportional to
e − E / k T .
Small energy raise → many make it; large raise → exponentially fewer. Here k is Boltzmann's constant (energy per degree of temperature) and T is absolute temperature.
Intuition The staircase picture
Imagine electrons as a jiggling crowd at the bottom of a staircase. Heat gives each a random kick. The higher the step, the exponentially fewer electrons can jump it. The surface potential ψ s (next section) lowers the step the electrons must climb to reach the channel surface — so the population there grows exponentially with ψ s :
n ( 0 ) ∝ e q ψ s / k T .
That single line is the seed of every exponential on the parent page.
The combination k T / q shows up so often we give it a name and a unit of volts .
Definition Thermal voltage
V T
V T = q k T .
It converts "thermal energy k T " into "volts" by dividing out one electron's charge. It is the natural voltage scale of thermal jiggling : a voltage of V T changes electron populations by a factor of e .
Using V T , the Boltzmann factor e q ψ s / k T becomes the tidier e ψ s / V T . This is the parent's V T ; see Thermal voltage kT-q . Notice V T ∝ T — hotter chip, bigger V T — which will explain why leakage worsens with heat.
Definition Surface potential
ψ s
The voltage the gate manages to create at the silicon surface just below the oxide. It "bends the energy bands," lowering the step electrons must climb (Section 4). More ψ s → exponentially more surface electrons → more current.
The catch: not all of V GS reaches the surface. Some of your gate push is "wasted" pushing back the depletion layer. That leads directly to the next symbols.
C
A capacitor stores charge; C measures "how much charge per volt." Two things in the MOSFET act like capacitors in a stack:
C o x = oxide capacitance : the gate plate + oxide insulator.
C d e p = depletion capacitance : the depleted silicon layer below the surface.
They form a capacitive voltage divider — the gate voltage splits between them. Here is why series capacitors split a small voltage change in the ratio the parent uses.
Intuition Why series capacitors share a voltage bump this way
When you nudge the gate voltage up by a tiny d V GS , a tiny extra charge d Q flows onto the stack. Because the two capacitors are in series , the same d Q lands on each (charge has nowhere else to go). But each capacitor turns that charge into a voltage change by its own rule d V = d Q / C : the oxide takes d Q / C o x , the depletion layer takes d Q / C d e p . The surface voltage change d ψ s is the bit across the depletion capacitor — but it is easier to write the fraction of the total. The total bump splits so that:
d V GS d ψ s = C o x + C d e p C o x = n 1 .
A big C o x (thin oxide) grabs little voltage for itself, leaving most for the surface → n → 1 . A big C d e p steals more → n grows.
Why the topic needs n . It is the "efficiency loss" that makes the exponential lazier : instead of e V GS / V T you get e V GS / ( n V T ) . That single n is exactly why the subthreshold swing (below) is ≥ 60 mV/decade, never less. The same capacitor stack drives the Body effect .
We built n ( 0 ) (source end) from the gate. What about n ( L ) , the drain end? This is where V D S enters — and where the parent's bracket ( 1 − e − V D S / V T ) comes from.
Intuition The drain sucks the crowd down
The drain sits at a higher voltage V D S than the source. By the same Boltzmann rule (Section 4), a higher voltage at the drain end lowers the electron population there by a Boltzmann factor:
n ( L ) ∝ e − V D S / V T .
Pulling V D S up drains that end emptier, so the crowd difference n ( 0 ) − n ( L ) grows.
Now feed both ends into the diffusion current from Section 3. The current is proportional to n ( 0 ) − n ( L ) , i.e. to
n ( 0 ) − n ( L ) ∝ e ( V GS − V t h ) / ( n V T ) ( 1 − e − V D S / V T ) .
The source term e ( V GS − V t h ) / ( n V T ) factors out front; what's left in the bracket is the drain's contribution. That reproduces the parent's boxed formula:
I s u b = I 0 e ( V GS − V t h ) / ( n V T ) ( 1 − e − V D S / V T ) .
V D S small (near zero): the "linear-like" edge
For tiny V D S , use e − x ≈ 1 − x , so 1 − e − V D S / V T ≈ V D S / V T . The current is then proportional to V D S — it rises linearly from zero as you first raise the drain. At exactly V D S = 0 there is no crowd difference, so I s u b = 0 : with no pull, diffusion has no preferred direction.
V D S large (≳ 3 V T ≈ 78 mV): saturation
Once V D S exceeds a few V T , e − V D S / V T → 0 and the bracket → 1 . The drain end is already essentially empty; pulling harder can't remove electrons that aren't there. So I s u b stops depending on V D S — it saturates . This is why the parent says the dominant knob is V GS (exponentially), not V D S .
Common mistake "Subthreshold current keeps growing with
V D S ."
Why it feels right: in strong-inversion saturation we tie I D to V D S via channel-length modulation.
The fix: in weak inversion the bracket ( 1 − e − V D S / V T ) flattens out past ≈ 78 mV. Beyond that, V GS is the only real control.
The current spans many factors of ten , so we plot log 10 I , not I .
ln 10
Natural log and base-10 log differ by a constant conversion factor:
ln 10 ≈ 2.303.
It appears whenever we translate the physics (which lives in base e , from e − E / k T ) into engineering "per-decade" language.
Because I s u b ∝ e V GS / ( n V T ) , taking log 10 gives a straight line in V GS — its inverse slope is the swing:
Definition Subthreshold swing
S
The millivolts of V GS you must remove to divide the current by 10:
S = n V T ln 10.
With n = 1 , V T = 25.9 mV, ln 10 = 2.303 : S m i n ≈ 60 mV/decade — the Boltzmann floor.
MOSFET structure S G D channel
Voltage knobs Vgs and Vds and Id
Diffusion current Fick law
Boltzmann factor e to minus E over kT
Drain end nL shrinks exp with Vds
Thermal voltage Vt equals kT over q
Slope factor n equals 1 plus Cdep over Cox
Subthreshold current exponential
Subthreshold swing S equals n Vt ln10
Every arrow is a "you need this before that." Follow any path top-to-bottom and you re-derive a piece of the parent page.
Worked example The 60 mV floor, checked
Ideal case n = 1 , T = 300 K:
S = 1 × 25.9 mV × 2.303 ≈ 59.6 mV/decade ≈ 60.
This is the number you can never beat with an ordinary MOSFET.
Worked example A realistic swing
With n = 1.5 : S = 1.5 × 25.9 × 2.303 ≈ 89.5 mV/decade — matches the parent's worked Example 1. The extra 30 mV over ideal is entirely the price of the C d e p / C o x voltage loss.
Test yourself — cover the right side. If any fails, re-read that section before the parent page.
What does V GS physically mean? The voltage push gate-minus-source; the main knob that controls the channel.
What is I D , and what is it called in the OFF region? The drain current (charge/second across the channel); in the OFF region it is the leakage I s u b .
Is V t h a physical cliff where current becomes zero? No — it's a chosen convention; current is smooth and nonzero below it.
Name the two transport mechanisms and which one rules below threshold. Drift (field-pushed) above; diffusion (crowd-spreading) below.
State Fick's law in words. Diffusion current is proportional to the concentration slope ( n ( 0 ) − n ( L )) / L times q A D n .
Where does the exponential in I s u b come from? The Boltzmann factor e − E / k T : surface electron population grows exponentially with surface potential.
What is V T = k T / q and its 300 K value? Thermal voltage, the volt-scale of thermal jiggling; ≈ 25.9 mV.
Why do series capacitors split the gate bump as C o x / ( C o x + C d e p ) ? Same charge d Q lands on both; each turns it to voltage by d V = d Q / C , so the surface gets the fraction C o x / ( C o x + C d e p ) .
What does the slope factor n = 1 + C d e p / C o x represent? The fraction of gate voltage lost to the depletion capacitor; the divider inefficiency, typically 1.1–1.5.
Where does the ( 1 − e − V D S / V T ) term come from? The drain end population n ( L ) ∝ e − V D S / V T ; the bracket is the drain's share of the crowd difference.
What happens to I s u b for small vs large V D S ? Small V D S : linear in V D S (bracket ≈ V D S / V T ); large V D S ≳ 78 mV: saturates (bracket → 1 ).
Why do we plot log 10 I and what is a "decade"? Because current spans many factors of 10; one decade = one factor of 10.
State the subthreshold swing and its ideal floor. S = n V T ln 10 ; ideal S m i n ≈ 60 mV/decade at room temperature.
Why is a MOSFET-below-threshold like a BJT base? Both move minority carriers by diffusion, giving an exponential voltage dependence.