Subthreshold leakage current: Visual walkthrough — Hardware deep dive

We are chasing this final destination (don't panic, we build every piece): $I_{s u b} = I_{0} exp (\frac{V _{GS} - V _{t h}}{n V _{T}}) (1 - e^{- V_{D S} / V_{T}})$

Step 1 — What a MOSFET actually is (the tap)

WHAT. A MOSFET has four terminals we care about. Two are the source and drain — think of them as the two ends of a pipe where current can enter and leave. Between them lies the channel, the strip of silicon just under the surface. Sitting on top, separated by a thin insulating layer of oxide, is the gate — a metal plate that controls the pipe without touching it.

WHY start here. Before we can talk about "leakage below threshold" we must agree on the geometry and name every voltage. $V_{GS}$ = gate voltage measured relative to the source. $V_{D S}$ = drain voltage relative to the source. Nothing exotic yet — just labels on a picture.

PICTURE. The gate is a capacitor plate hovering over the channel. When it pushes hard enough it pulls electrons up to the surface, forming a conducting layer. When it pushes weakly, only a whisper of electrons gathers — that whisper is our whole story.

Step 2 — Below threshold there is still a whisper of electrons

WHAT. Set $V_{GS}$ below $V_{t h}$ . The textbook square-law says the current is exactly zero here. That is a lie of convenience. In reality, electrons at room temperature are jiggling with thermal energy, and a tiny Boltzmann fraction of them always has enough energy to sit at the surface.

WHY this matters. The number of these thermally-lucky electrons is set by how far the gate has bent the energy bands at the surface. We call that bending the surface potential $ψ_{s}$ (Greek "psi", just a name for "how much the surface energy has dropped"). The count of electrons rises as: $n (0) \propto e^{q ψ_{s} / k T}$

$n (0)$ = electron concentration at the source end of the surface (the "0" means position $x = 0$ ).
$q$ = charge on one electron.
$ψ_{s}$ = the surface band-bending. Bigger bend → deeper energy well → more electrons.
$k T$ = the thermal energy scale ( $k$ = Boltzmann's constant, $T$ = temperature). This is how hard the electrons jiggle.
$e^{(\dots)}$ = the exponential. This is where the entire exponential character of leakage is born.

WHY an exponential and not a straight line? Because populating a higher energy state costs energy $q ψ_{s}$ , and statistical physics says the odds of paying an energy cost fall off exponentially with (cost / thermal energy). That single fact — Boltzmann statistics — is the seed of everything below.

PICTURE. An energy hill. Only electrons on the high tail of the jiggle-distribution clear it. Raise $ψ_{s}$ a little and the hill lowers — exponentially more electrons pour over.

Step 3 — Those electrons diffuse (this is a BJT in disguise)

WHAT. We now have a few electrons at the source end and even fewer at the drain end. A concentration difference across a strip means particles drift from crowded to empty — that is diffusion. There is no strong electric field pushing them along (weak inversion has no proper channel), so diffusion is the only game in town, exactly like the base of a BJT.

WHY diffusion, not drift? Drift needs a field pulling carriers along the channel. In strong inversion that field dominates. But below threshold the along-channel field is negligible; the imbalance of how many carriers sit at each end is what drives current. Diffusion answers the question "what current flows when carriers spread from many to few?"

$I_{s u b} = q A D_{n} \frac{d n}{d x} \approx q A D_{n} \frac{n ( 0 ) - n ( L )}{L}$

$q$ = electron charge (each carrier moves this much charge).
$A$ = cross-sectional area of the flow.
$D_{n}$ = diffusion constant — how fast electrons spread.
$\frac{d n}{d x}$ = the slope of concentration along the channel; steeper slope → more current.
$n (0) - n (L)$ = crowd at source minus crowd at drain.
$L$ = channel length. Longer pipe → gentler slope → less current.

PICTURE. A ramp of electron density falling from the source end (many, from Step 2) to the drain end (few). The steepness of that ramp is the current.

Step 4 — Only a fraction of the gate voltage reaches the surface

WHAT. We need $ψ_{s}$ in terms of the handle $V_{GS}$ . Here's the catch: not all of $V_{GS}$ shows up as band-bending. The gate voltage splits between two capacitors in series — the oxide capacitance $C_{o x}$ (gate-to-surface) and the depletion capacitance $C_{d e p}$ (surface-to-body). It is a capacitive voltage divider.

WHY a divider? Two capacitors in series share an applied voltage inversely to their sizes. The surface only "feels" the fraction that lands across $C_{d e p}$ . So: $\frac{d ψ _{s}}{d V _{GS}} = \frac{C _{o x}}{C _{o x} + C _{d e p}} = \frac{1}{n}, n \equiv 1 + \frac{C _{d e p}}{C _{o x}}$

$C_{o x}$ = oxide capacitance. A thinner, better oxide → bigger $C_{o x}$ .
$C_{d e p}$ = depletion-layer capacitance under the surface.
$\frac{1}{n}$ = the fraction of gate-wiggle that reaches the surface.
$n$ = the slope factor (also the body-effect factor). If $C_{d e p} = 0$ , all voltage reaches the surface and $n = 1$ (perfect). Real devices: $n \approx 1.1$ – $1.5$ .

PICTURE. Two capacitors stacked; the input $V_{GS}$ arrives at the top, and the surface sits at the midpoint tap. The plum-coloured slice of the voltage is what the surface actually sees.

Step 5 — Assemble the exponential in $V_{GS}$

WHAT. Chain the last two facts. Step 2 said electrons grow as $e^{q ψ_{s} / k T}$ . Step 4 said $ψ_{s}$ grows as $V_{GS} / n$ . Substitute: $n (0) \propto e^{q ψ_{s} / k T} and ψ_{s} \approx \frac{V _{GS}}{n} \Rightarrow n (0) \propto e^{q V_{GS} / (nk T)}$

Define the thermal voltage $V_{T} = k T / q$ — this bundles the two physical constants into one voltage (the natural voltage scale of jiggling electrons). Then $q / (nk T) = 1/ (n V_{T})$ , and the source crowd becomes $\propto e^{V_{GS} / (n V_{T})}$ . Referencing $V_{GS}$ to the convention $V_{t h}$ folds the prefactors into $I_{0}$ :

$I_{s u b}_{gate part} = I_{0} exp (\frac{V _{GS} - V _{t h}}{n V _{T}})$

$V_{T} = k T / q \approx 25.9$ mV at 300 K — see Thermal voltage kT-q.
$\frac{V _{GS} - V _{t h}}{n V _{T}}$ = "how many thermal-voltage steps below threshold, softened by the divider $n$ ."
$I_{0} \propto μ C_{o x} (W / L) V_{T}^{2}$ = a technology prefactor lumping mobility $μ$ , geometry $W / L$ , and constants.

WHY it's a straight line on a log plot. Take $lo g_{10}$ : the exponent becomes linear in $V_{GS}$ . So plotting $lo g_{10} I_{s u b}$ against $V_{GS}$ gives a straight tail — that is the signature you measure in the lab.

Step 6 — The drain factor and its degenerate cases

WHAT. So far only the source end. The drain end also holds a Boltzmann crowd, but pulled down by $V_{D S}$ : $n (L) \propto e^{- q V_{D S} / k T}$ . Subtracting drain from source (Step 3's $n (0) - n (L)$ ) gives the bracket: $(1 - e^{- V_{D S} / V_{T}})$

Now walk every case of this bracket so you never meet a surprise:

$V_{D S} = 0$ : bracket $= 1 - 1 = 0$ . No current. Correct — with equal ends there is no concentration ramp, nothing to diffuse. (Degenerate case.)
Small $V_{D S}$ (a few mV): bracket $\approx V_{D S} / V_{T}$ , linear — the device looks like a tiny resistor.
$V_{D S} ≳ 3 V_{T} \approx 78$ mV: $e^{- 3} \approx 0.05$ , bracket $\to 1$ . Current saturates in $V_{D S}$ — pushing the drain harder does nothing. The drain end is already emptied.
$V_{D S} < 0$ (drain below source): current reverses sign — the ramp tilts the other way. Symmetric device.

WHY this shape. The drain can only remove carriers down to (nearly) zero; once it's empty, extra drain voltage has nothing left to take. Hence saturation. This is why $V_{GS}$ , not $V_{D S}$ , is the exponential control knob.

PICTURE. The bracket rising from 0 at $V_{D S} = 0$ , bending over, and flattening at 1 past $\sim 78$ mV, with the three regimes shaded.

Step 7 — The subthreshold swing $S$ and the 60 mV/dec wall

WHAT. How many millivolts of $V_{GS}$ must you remove to cut $I_{s u b}$ by a factor of ten? Because the log-current is a straight line (Step 5), its inverse slope is a constant we name $S$ :

\;\Rightarrow\; S \equiv \frac{dV_{GS}}{d(\log_{10}I_{sub})} = n\,V_T\ln 10$$ - $\ln 10 \approx 2.303$ = the price of converting "$e$-fold" into "10-fold." - $n V_T$ = the softened thermal step. - $S$ = millivolts of gate per **decade** (factor of 10) of current. **WHY there is a floor.** The best possible device has $n=1$ (perfect divider, Step 4). Even then: $$S_{min} = V_T\ln 10 = \frac{kT}{q}\ln 10 \approx 60\ \text{mV/decade at }300\text{ K}$$ Both $kT/q$ and $\ln 10$ are fixed by physics. No fabrication trick beats it at room temperature — the "Boltzmann tyranny." (See [[Short-channel effects / DIBL]] for how bad devices go the *wrong* way.) **Case — temperature.** $S\propto T$: heat the chip 300 K → 360 K and $S$ grows by $1.2\times$. Leakage worsens super-linearly. Hot chips leak → heat more → risk thermal runaway. **PICTURE.** The log-current line with a shaded triangle: horizontal leg = $S$, vertical leg = one decade. Steeper line = smaller $S$ = better switch. The dashed 60 mV wall you cannot cross. ![[deepdives/dd-hardware-2.4.17-d2-s07.png]] --- ## The one-picture summary Every arrow of this walkthrough compressed: gate handle → divider → surface Boltzmann crowd → diffusion ramp → drain saturation → the straight log tail with its 60 mV/dec floor. ![[deepdives/dd-hardware-2.4.17-d2-s08.png]] > [!formula] The full result we built > $$\boxed{\,I_{sub} = I_0\,\exp\!\left(\frac{V_{GS}-V_{th}}{n\,V_T}\right)\left(1 - e^{-V_{DS}/V_T}\right)\,},\qquad S = nV_T\ln10 \ge 60\ \tfrac{\text{mV}}{\text{dec}}$$ > [!recall]- Feynman: the whole walkthrough in plain words > You have a tap that's supposed to be off. But the water molecules are always jiggling, and a tiny few always have enough kick to sneak past — that's Step 2, the Boltzmann whisper. Once a few sneak in at the top of the pipe and almost none at the bottom, water spreads from the crowded end toward the empty end just by bumping around: that's diffusion, Step 3. When you turn the handle (the gate), only *part* of your turn reaches the water because there's slack in the mechanism — that slack is the capacitor divider $n$, Step 4. Chain those together and the drip changes *ten-fold* for every little turn of the handle — a perfectly straight line if you plot the drip on a "×10, ×100, ×1000" ruler, Step 5. The drain at the far end can only suck out so much; once it's empty, cranking it does nothing — that's the saturation bracket, Step 6. And the cruel rule: because the electrons' jiggle sets the scale, each ten-fold-smaller drip always costs you at least 60 millivolts of handle. You can approach 60, never beat it, at room temperature. That's Step 7, the wall. > [!mnemonic] The chain > **Gate → Divide → Boltzmann → Diffuse → Drain-saturate → 60-floor.** > Read it forwards and you've re-derived $I_{sub}$. --- ## Connections - [[2.4.17 Subthreshold leakage current (Hinglish)|Parent · Subthreshold leakage current]] — the boxed result this page derives. - [[MOSFET operating regions]] — where this weak-inversion tail sits relative to the square law. - [[Threshold voltage Vth]] — the convention $V_{th}$ our exponent is measured against. - [[BJT diffusion current]] — the identical diffusion physics of Step 3. - [[Thermal voltage kT-q]] — the $V_T=kT/q$ that fixes the 60 mV/dec floor. - [[Body effect]] — same capacitive divider that produces $n$. - [[Static vs dynamic power dissipation]] — why this leakage is the static-power villain. - [[Short-channel effects / DIBL]] — how real devices degrade $S$ and $V_{th}$.