Question bank — Subthreshold leakage current

False. The channel is only weakly inverted, so a Boltzmann-small population of carriers still diffuses across — current is exponentially small but never zero.

False. That square law is a strong-inversion model only; below $V_{th}$ transport is diffusion-dominated and the current is exponential in $V_{GS}$, not quadratic.

False. Weak-inversion current is diffusion-dominated because the carrier gradient is huge while the lateral field is weak; there is still a small lateral field so $J_{drift}\ne 0$, but it is negligible next to $J_{diff}$ — the same balance seen in a BJT base.

False. Thinner oxide raises $C_{ox}$ so $n\to 1$, which only pushes $S$ toward $60$ mV/dec. The floor $S_{min}=(kT/q)\ln 10$ is set by physics, not fabrication.

True. $I_0$ is a multiplicative prefactor; $S=n{V}_T\ln 10$ depends only on $n$ and temperature, so scaling $I_0$ shifts the log-current curve vertically without changing its slope.

False. The factor $(1-e^{-V_{DS}/V_T})\to 1$, so $I_{sub}$ saturates in $V_{DS}$; the exponential control knob is $V_{GS}$, not $V_{DS}$ (ignoring DIBL).

True. $S=n{V}_T\ln 10\propto T$, $V_T$ sits in the exponent, and the prefactor $I_0\propto \mu C_{ox}(W/L)V_T^2$ carries both $V_T^2$ and mobility $\mu \propto T^{-m}$, so the net effect is a strong super-linear drop on cooling.

False. $n=1+C_{dep}/C_{ox}$, and $C_{dep}>0$ always, so $n>1$ (typically 1.1–1.5); $n=1$ is an unreachable ideal.

False. $V_{th}$ is a convention marking strong inversion; the surface potential and current vary smoothly through it — the exponential tail extends well below $V_{th}$.

The exponent is 0 so $I_{sub}=I_0$, but $I_0$ is a small subthreshold prefactor ($\propto \mu C_{ox}\frac{W}{L}{V}_T^2$), not the full strong-inversion ON current. The formula only describes the weak-inversion tail up to threshold.

60 mV per decade means a 10× cut, not 2×. To halve, you need $n{V}_T\ln 2$ of extra bias — roughly 27 mV at $n=1.5$ ($1.5\times 25.9\times 0.693$), well under one decade.

The factor is $\ln 10\approx 2.303$, not $10$. Using $10$ inflates $S$ by ~4.3×; the correct value is $\approx 89.5$ mV/dec, not $\approx 389$.

Wrong scaling. Both $S\propto T$ and the $V_T$ inside the exponent plus $I_0\propto V_T^2$ change (mobility $\mu \propto T^{-m}$ partly offsets $I_0$), so leakage rises super-linearly and by far more than 20%.

Diffusion current is set by a concentration gradient, not a linear $V$–$I$ relation. The dependence on $V_{GS}$ is exponential and on $V_{DS}$ it saturates — nothing Ohmic about it.

They matter greatly. DIBL lowers the effective $V_{th}$ as $V_{DS}$ rises, shifting the entire exponential curve up and boosting leakage — the OFF state leaks more.

No — $V_T=kT/q\approx 25.9$ mV is the thermal voltage Thermal voltage kT-q, an energy-scale, entirely different from the threshold voltage $V_{th}$ (often hundreds of mV). Confusing the two is a classic slip.

Because $V_{GS}$ moves the surface potential ${\psi }_s$, and the carrier population depends on ${\psi }_s$ through the Boltzmann factor $e^{q{\psi }_s/kT}$ — the exponential band-bending physics of Picture 2.

The gate voltage splits across a capacitive divider between $C_{ox}$ and $C_{dep}$ (Picture 3); the surface sees $d{\psi }_s/d{V}_{GS}=C_{ox}/(C_{ox}+C_{dep})=1/n$, the rest dropping across the depletion layer.

Because $S=n{V}_T\ln 10$ with $n\ge 1$, and $V_T\ln 10=(kT/q)\ln 10\approx 60$ mV is fixed by thermal statistics — the "Boltzmann tyranny." Beating it needs a non-thermal transport like band-to-band tunneling.

Both move minority carriers by diffusion across a region where the concentration gradient dominates over drift, giving an exponential dependence on the controlling voltage — the weak-inversion MOSFET is essentially a lateral BJT.

The drain end depletes carriers as $V_{DS}$ pulls them away; when $V_{DS}$ is a few $V_T$ the drain concentration is negligible and the gradient — hence the current — stops depending on $V_{DS}$.

$S=n{V}_T\ln 10$ is directly proportional to $V_T=kT/q$, and $I_0\propto V_T^2$ (partly countered by $\mu \propto T^{-m}$); heat literally jiggles more electrons over the barrier, so the net leakage climbs.

Both arise from the same $C_{dep}/C_{ox}$ ratio — the depletion capacitance steals gate control, which simultaneously sets $n$ and shifts $V_{th}$ under body bias.

It flows continuously while the device is merely powered and "off," with no switching required — so it dissipates energy even in idle, unlike the charge/discharge losses of dynamic power.

The bracket $(1-e^{-V_{DS}/V_T})\to 0$, so $I_{sub}\to 0$: with no drain–source voltage there is no net carrier gradient, hence no diffusion current.

The exponent is zero, so $I_{sub}=I_0$ — the prefactor value marking the boundary between the weak-inversion tail and strong inversion.

$n\to 1$ so $S\to V_T\ln 10\approx 60$ mV/dec at 300 K — the ideal floor, approached but never crossed.

Since $S\propto T$, the limit shrinks toward $0$; the Boltzmann floor is a room-temperature statement, and cryogenic operation gives steeper switching.

Identically — $S$ depends only on $n$ and $T$, not on geometry. A wider device leaks more (bigger $I_0$) but with the same slope.

It approaches zero exponentially but never exactly reaches it; the tail is asymptotic, so a real switch always has a nonzero, if vanishingly small, floor.

Expanding $1-e^{-V_{DS}/V_T}\approx V_{DS}/V_T$, so leakage becomes linear in $V_{DS}$ — a near-ohmic regime only in this tiny-$V_{DS}$ corner, before saturating.