Subthreshold leakage current: Worked examples — Hardware deep dive

The two workhorse formulas we lean on all page (both earned in the parent):

Before we use a single symbol, let's re-earn each one on this page so nothing is borrowed on faith:

Definition Every symbol, in plain words

$V_{GS}$ = gate-to-source voltage — the "handle" you turn. Measured from the gate terminal down to the source. Making it more negative (for an n-type device) pushes the switch more OFF.
$V_{t h}$ = threshold voltage — the conventional voltage at which we declare the channel "on." It is a chosen reference line, not a physical cliff (see Threshold voltage Vth).
$V_{D S}$ = drain-to-source voltage — the push from drain to source that drives current along the channel.
$V_{GS} - V_{t h}$ = the overdrive; below threshold it is negative. Only this difference enters the exponent.
$k$ = Boltzmann's constant ( $1.381 \times 1 0^{- 23}$ J/K) — converts temperature into energy.
$T$ = absolute temperature in kelvin (300 K ≈ room temperature).
$q$ = electron charge ( $1.602 \times 1 0^{- 19}$ C).
$V_{T} = k T / q$ = thermal voltage ( $\approx 25.9$ mV at 300 K) — how many volts one "unit" of thermal jiggle is worth. It scales linearly with $T$ . See Thermal voltage kT-q.
$n$ = slope factor, the capacitive-divider ratio $1 + C_{d e p} / C_{o x}$ (typically 1.1–1.5). See Body effect.
$I_{0}$ = technology prefactor: everything that does not depend on the two bias voltages, bundled together. Its known scaling is $I_{0} \propto μ C_{o x} \frac{W}{L} V_{T}^{2}$ — the $V_{T}^{2}$ comes from diffusion transport (carrier density $\propto V_{T}$ times the diffusion coefficient $D_{n} = μ V_{T}$ via the Einstein relation), which is why we track it in the temperature example.

The scenario matrix

Before working anything, let's map out every kind of question this one formula can generate. Think of it as a checklist: if we cover every row, the reader never meets a case we skipped.

#	Case class	The tricky part	Covered by
A	Compute $S$ from capacitances	plug-in, definition of $n$	Ex 1
B	Sign of $Δ V_{GS}$ (deeper OFF vs shallower OFF)	only the difference matters; watch which is more negative	Ex 2
C	Small- $V_{D S}$ regime ( $V_{D S} ≪ V_{T}$ )	the $(1 - e^{- V_{D S} / V_{T}})$ bracket is NOT $\approx 1$	Ex 3
D	Large- $V_{D S}$ / saturation limit ( $V_{D S} ≫ 3 V_{T}$ )	bracket $\to 1$ , current independent of $V_{D S}$	Ex 3
E	Degenerate: $V_{D S} = 0$	current is exactly zero — sanity anchor	Ex 3
F	Ideal limit $n \to 1$ ( $C_{d e p} \to 0$ )	the 60 mV/dec Boltzmann floor	Ex 4
G	Temperature scaling ( $T$ up)	$S \propto T$ and exponent and $I_{0}$ all move	Ex 5
H	Real-world word problem	many transistors, total static power	Ex 6
I	Exam twist: DIBL ( $V_{t h}$ drops with $V_{D S}$ )	$V_{t h}$ is no longer constant	Ex 7
J	Inverse problem ("how much bias to cut leakage 1000×?")	invert the swing definition	Ex 8
K	Body effect (substrate bias raises $V_{t h}$ )	$V_{t h}$ shifts up, cutting leakage	Ex 9

Eleven cells. Nine examples below hit all eleven. Let's go.

Worked examples

Worked example Ex 2 — Cell B: sign of the bias difference

Same device, $n = 1.3333$ , $V_{T} = 25.9$ mV, $V_{D S}$ large (bracket $= 1$ ). Transistor A sits with $V_{GS}$ 80 mV below $V_{t h}$ ; transistor B sits 250 mV below $V_{t h}$ . Which leaks more, and by what ratio?

Forecast: A is less OFF (closer to threshold). Guess: does A leak more or less than B?

Write each overdrive: $V_{GS} - V_{t h} = - 0.080$ V for A, $- 0.250$ V for B. Why this step? The exponent depends on $V_{GS} - V_{t h}$ ; get the signs right — both negative, A less negative.
The ratio kills $I_{0}$ and $V_{t h}$ : $\frac{I _{A}}{I _{B}} = exp (\frac{( - 0.080 ) - ( - 0.250 )}{n V _{T}}) = exp (\frac{0.170}{1.3333 \times 0.0259}) .$ Why this step? Only the difference of exponents survives a ratio — that is why we don't need $I_{0}$ or $V_{t h}$ .
Denominator $= 0.034533$ ; exponent $= 0.170/0.034533 = 4.9228$ ; $e^{4.9228} \approx 137.4$ . Why this step? Numeric evaluation.

Answer: the shallower-OFF transistor A leaks about 137× more. Sign confirmed: less OFF ⇒ more leakage.

Verify: $0.170$ V $/ S = 0.170/0.0795 = 2.138$ decades; $1 0^{2.138} = 137$ . ✓ Same number two ways — internally consistent. ✓

Worked example Ex 3 — Cells C, D, E: the

$V_{D S}$ bracket in three regimes Fix $V_{GS}, V_{t h}, n$ so the exponential prefactor $I_{0} e^{(V_{GS} - V_{t h}) / n V_{T}} = 10$ nA. Compute $I_{s u b}$ at three drain voltages at 300 K: (E) $V_{D S} = 0$ , (C) $V_{D S} = 10$ mV, (D) $V_{D S} = 200$ mV. $V_{T} = 25.9$ mV.

Forecast: guess the shape — does current rise linearly, then flatten? Where does it flatten?

The figure below plots $I_{s u b}$ vs $V_{D S}$ : a curve that starts at the origin (0,0), climbs steeply for small $V_{D S}$ , then bends over and flattens onto a horizontal ceiling at 10 nA. Three coloured dots mark our three cases — lavender at the origin (E, 0 nA), butter at 10 mV (C, ≈3.2 nA on the steep part), mint at 200 mV (D, ≈10 nA on the flat ceiling). A dashed lavender vertical line at $3 V_{T} \approx 78$ mV shows where the curve is already 95% up to the ceiling.

(E) $V_{D S} = 0$ : bracket $= 1 - e^{0} = 0$ . So $I_{s u b} = 0$ exactly. Why this step? Degenerate anchor: no drain–source voltage means no net diffusion gradient, so no current — a physical sanity check the formula must pass.
(C) $V_{D S} = 10$ mV: bracket $= 1 - e^{- 10/25.9} = 1 - e^{- 0.3861} = 1 - 0.6797 = 0.3203$ . $I_{s u b} = 10 nA \times 0.3203 = 3.20$ nA. Why this step? Small $V_{D S}$ (below $\sim 3 V_{T}$ ): the bracket is genuinely fractional — you may NOT approximate it as 1. This is the most-missed cell.
(D) $V_{D S} = 200$ mV: bracket $= 1 - e^{- 200/25.9} = 1 - e^{- 7.722} = 1 - 0.000442 = 0.99956$ . $I_{s u b} \approx 10.0$ nA. Why this step? $V_{D S} ≫ 3 V_{T} \approx 78$ mV ⇒ bracket $\to 1$ ⇒ current saturates in $V_{D S}$ . This is why $V_{GS}$ , not $V_{D S}$ , is the real control knob.

Verify: the three values $0, 3.20, 10.0$ nA rise then flatten exactly as the curve shows. At $V_{D S} = 3 V_{T} = 77.7$ mV the bracket is $1 - e^{- 3} = 0.950$ — already 95% saturated, matching the " $3 V_{T}$ rule." ✓

Worked example Ex 4 — Cell F: the ideal

$n \to 1$ Boltzmann floor A magically perfect gate has $C_{d e p} \to 0$ (all of $V_{GS}$ reaches the surface). Find $S$ at 300 K and at 350 K.

Forecast: guess the 300 K number — this is the famous "wall."

$C_{d e p} \to 0 \Rightarrow n = 1 + 0 = 1$ . Why this step? With no depletion capacitance the divider is trivial — the gate fully controls the surface. This is the best a MOSFET can do.
At 300 K: $S = 1 \times 25.9 \times ln 10 = 25.9 \times 2.3026 = 59.6 \approx 60$ mV/dec. Why this step? This is $S_{min}$ , set purely by $k T / q$ and $ln 10$ — no fabrication trick lowers it.
At 350 K: recompute $V_{T} = k T / q$ . Since $V_{T} \propto T$ , $V_{T} (350) = 25.9 \times (350/300) = 30.22$ mV. $S = 30.22 \times 2.3026 = 69.6$ mV/dec. Why this step? The "60 mV" floor is really $60 \times (T /300)$ — it is only 60 at room temperature. Higher $T$ raises even the ideal floor.

Verify: $S (350) / S (300) = 69.6/59.6 = 1.168 = 350/300$ . ✓ The floor scales linearly with $T$ , exactly as $S \propto T$ predicts. ✓ Thermal voltage kT-q sets this wall.

Worked example Ex 5 — Cell G: temperature blows leakage up (all three effects)

A device with $n = 1.3333$ is heated from 300 K to 360 K. It stays OFF with $V_{GS} - V_{t h} = - 0.150$ V and $V_{D S}$ large (bracket $= 1$ ). By what factor does the exponential term change, and then the total including $I_{0}$ ?

Forecast: guess — does leakage go up or down when hot, and by roughly how much?

$V_{T} (300) = 25.9$ mV; $V_{T} (360) = 25.9 \times (360/300) = 31.08$ mV. Why this step? $V_{T} \propto T$ — the thermal voltage widens the exponential's "denominator."
Exponent at 300 K: $\frac{- 0.150}{1.3333 \times 0.0259} = \frac{- 0.150}{0.034533} = - 4.3437$ . Exponent at 360 K: $\frac{- 0.150}{1.3333 \times 0.03108} = \frac{- 0.150}{0.041440} = - 3.6197$ . Why this step? Same overdrive, but a bigger $V_{T}$ makes the (negative) exponent less negative — a smaller penalty for being OFF.
Exponential-term ratio $\frac{I ( 360 )}{I ( 300 )}_{e x p} = e^{- 3.6197 - (- 4.3437)} = e^{+ 0.7240} = 2.063$ . Why this step? The exponential alone more than doubles the leakage for a 60 K rise.
Now bring in $I_{0} \propto V_{T}^{2}$ (derived in the definitions box via the Einstein relation $D_{n} = μ V_{T}$ ): $I_{0}$ -ratio $= (360/300)^{2} = 1.44$ . Total $= 2.063 \times 1.44 = 2.97$ . Why this step? Both the exponent and the prefactor push the same way, so leakage rises super-linearly with $T$ .

Verify: total leakage factor $2.97$ — nearly 3× leakage for 60 K. This super-linear jump is the thermal-runaway seed the parent warned about. ✓ Sign: hotter ⇒ more leakage. ✓

Worked example Ex 6 — Cell H: real-world static power on a chip

A chip has $1.2 \times 1 0^{9}$ transistors. On average $40%$ are OFF at any instant, each leaking $I_{s u b} = 8$ nA at supply $V_{D D} = 0.9$ V. Estimate the static leakage power. (See Static vs dynamic power dissipation.)

Forecast: guess — milliwatts, or full watts?

Number leaking: $N = 0.40 \times 1.2 \times 1 0^{9} = 4.8 \times 1 0^{8}$ . Why this step? Only OFF transistors leak subthreshold current; scale the count.
Total leakage current: $I_{t o t} = N \times I_{s u b} = 4.8 \times 1 0^{8} \times 8 \times 1 0^{- 9} = 3.84$ A. Why this step? Currents in parallel branches add — a billion tiny drips make a real river.
Static power: $P = V_{D D} \times I_{t o t} = 0.9 \times 3.84 = 3.456$ W. Why this step? Static power is just $V \times I$ with the leakage current; this is wasted even when the chip computes nothing.

Verify: units: A × V = W. ✓ $3.46$ W of pure leakage explains why datacenter chips fight so hard over $V_{t h}$ and $S$ . Sanity: per-transistor $0.9 \times 8 nA = 7.2$ nW; × $4.8 \times 1 0^{8} = 3.456$ W. ✓

Worked example Ex 7 — Cell I (exam twist): DIBL makes

$V_{t h}$ move A short-channel device shows DIBL: its threshold drops with drain voltage as $V_{t h} = 0.40 - 0.10 V_{D S}$ (volts). At $V_{GS} = 0.20$ V, $n = 1.3333$ , $V_{T} = 25.9$ mV, compare leakage at $V_{D S} = 0.5$ V vs $V_{D S} = 1.0$ V. (Bracket $\approx 1$ in both.) See Short-channel effects / DIBL.

Forecast: naively $V_{D S}$ shouldn't matter (bracket saturates). Guess — does DIBL break that?

At $V_{D S} = 0.5$ : $V_{t h} = 0.40 - 0.05 = 0.35$ V; overdrive $= 0.20 - 0.35 = - 0.15$ V. At $V_{D S} = 1.0$ : $V_{t h} = 0.40 - 0.10 = 0.30$ V; overdrive $= 0.20 - 0.30 = - 0.10$ V. Why this step? DIBL lowers $V_{t h}$ , so the transistor is less OFF at higher $V_{D S}$ — $V_{t h}$ is no longer a constant we can cancel.
Ratio $\frac{I ( V _{D S} = 1.0 )}{I ( V _{D S} = 0.5 )} = exp (\frac{( - 0.10 ) - ( - 0.15 )}{n V _{T}}) = exp (\frac{0.05}{0.034533})$ . Why this step? Same trick as Ex 2, but now the difference comes from $V_{t h}$ shifting, not from $V_{GS}$ .
$= e^{1.4479} = 4.254$ . Why this step? Numeric evaluation.

Answer: despite the $V_{D S}$ -bracket being saturated, leakage grows 4.25× from 0.5 V to 1.0 V — purely because DIBL lowered $V_{t h}$ .

Verify: this is the exam trap — "Subthreshold current is $V_{D S}$ -independent" is only true without DIBL. Here the $V_{D S}$ -dependence sneaks in through $V_{t h}$ . ✓ $0.05$ V $/ S = 0.05/0.0795 = 0.629$ decades; $1 0^{0.629} = 4.25$ . ✓

Worked example Ex 8 — Cell J (inverse problem): how much bias to cut leakage 1000×?

A device has swing $S = 79.5$ mV/dec (from Ex 1). By how many millivolts must you push $V_{GS}$ deeper OFF to reduce $I_{s u b}$ by a factor of 1000? And confirm via the natural-log form.

Forecast: 1000× is three decades. Guess the millivolts before computing.

Decades to cut: $lo g_{10} (1000) = 3$ . Why this step? $S$ is defined per decade, so first count decades.
$Δ V_{GS} = 3 \times S = 3 \times 79.5 = 238.5$ mV. Why this step? Inverting the swing definition: each decade costs $S$ millivolts.
Cross-check with the natural-log form: cutting by $1000$ means exponent change of $- ln 1000 = - 6.9078$ ; $Δ V_{GS} = 6.9078 \times n V_{T} = 6.9078 \times 34.533 mV = 238.5$ mV. Why this step? The exponent uses $ln$ , the swing uses $lo g_{10}$ ; they must agree, since $ln 10 \times 3 = 6.9078$ .

Answer: about 238 mV deeper OFF-bias.

Verify: both methods give $238.5$ mV. ✓ Consistency: $Δ V_{GS} / S = 238.5/79.5 = 3.0$ decades $=$ factor $1 0^{3} = 1000$ . ✓

Worked example Ex 9 — Cell K (body effect): substrate bias raises

$V_{t h}$ Reverse-biasing the source–body junction (applying $V_{S B} > 0$ ) raises $V_{t h}$ — the body effect. Suppose $V_{S B}$ raises $V_{t h}$ by $Δ V_{t h} = + 0.070$ V. With $n = 1.3333$ , $V_{T} = 25.9$ mV, $V_{D S}$ large, by what factor does leakage change? See Body effect.

Forecast: raising $V_{t h}$ pushes the device more OFF. Guess: does leakage go up or down, and by how much?

Overdrive shift: the exponent is $(V_{GS} - V_{t h}) / n V_{T}$ . Raising $V_{t h}$ by $+ 0.070$ V lowers the overdrive by $0.070$ V. Why this step? This is the opposite sign to DIBL (Ex 7) — body effect raises $V_{t h}$ , DIBL lowers it.
Ratio $\frac{I _{after}}{I _{before}} = exp (\frac{- 0.070}{n V _{T}}) = exp (\frac{- 0.070}{0.034533}) = e^{- 2.0271}$ . Why this step? Only the $V_{t h}$ change enters; $V_{GS}$ , $I_{0}$ cancel exactly as in Ex 2.
$= 0.1317$ . Why this step? Numeric evaluation — leakage drops to about 13% of its original value.

Answer: body effect cuts leakage by a factor of ~7.6 here. This is a real design lever ("reverse body bias" for standby power).

Verify: decades cut $= 0.070/0.0795 = 0.880$ ; $1 0^{- 0.880} = 0.132$ . ✓ Same as $e^{- 2.0271}$ . ✓ Sign: more OFF ⇒ less leakage — opposite of the DIBL cell. ✓

Recall Which cell is this? (self-test)

" $V_{D S}$ increases but leakage is flat" ::: Cell D — saturation, bracket $\to 1$ (no DIBL). " $V_{D S}$ increases and leakage rises even in saturation" ::: Cell I — DIBL shifting $V_{t h}$ down. " $V_{S B}$ applied and leakage drops" ::: Cell K — body effect shifting $V_{t h}$ up. " $V_{D S} = 15$ mV, must I approximate the bracket as 1?" ::: No — Cell C, $15 < 3 V_{T}$ , bracket is fractional. "The 60 mV/dec number" ::: Cell F — ideal $n = 1$ floor at 300 K. "Chip-level watts of leakage" ::: Cell H — multiply per-device $I_{s u b}$ by count, then $\times V_{D D}$ .

Connections

Parent: Subthreshold leakage current — the formulas drilled here.
MOSFET operating regions — every example lives below the square-law regime.
Threshold voltage Vth — the anchor of the exponent; moved by DIBL (Ex 7) and body effect (Ex 9).
Thermal voltage kT-q — the 60 mV/dec floor in Ex 4.
Static vs dynamic power dissipation — the real-world Ex 6.
Short-channel effects / DIBL — the exam twist Ex 7.
Body effect — the substrate-bias cell Ex 9.
BJT diffusion current — the diffusion transport behind the $V_{T}^{2}$ prefactor.