Exercises — Subthreshold leakage current

WHAT we do: plug into $S=n{V}_T\ln 10$. WHY: $S$ is defined as that product — nothing to derive. $S=1.2\times 25.9\text{ mV}\times 2.3026=71.6\text{ mV/decade}.$ Meaning: removing ~72 mV of gate voltage cuts leakage by a factor of 10.

Since $V_{GS}=0.15<V_{th}=0.40$, the gate is below threshold → the channel is only weakly inverted → this is the weak-inversion (subthreshold) region. Carriers cross by diffusion (a Boltzmann-small population sliding from source to drain), exactly like minority carriers in a BJT base — not by drift. See also MOSFET operating regions.

The minimum is at $n=1$ (when $C_{dep}/C_{ox}\to 0$): $S_{min}=1\times 25.9\text{ mV}\times 2.3026=59.6\approx 60\text{ mV/decade}.$ It is pinned by $kT/q$ — the Boltzmann factor governing how thermal energy spreads electrons over energy. This is the "60 mV/dec wall."

Step 1 (invert the swing formula): we know $S=n{V}_T\ln 10$; solving for $n$ just divides both sides by $V_T\ln 10$. WHY divide: $V_T\ln 10$ is a fixed number ($59.64$ mV) at 300 K, so $n$ is simply "how many of those units the measured swing contains." $n=\frac{S}{V_T\ln 10}=\frac{85}{25.9\times 2.3026}=\frac{85}{59.64}=\mathrm{1.425.}$ Step 2 (recover the capacitance ratio): since $n=1+C_{dep}/C_{ox}$ by definition, subtract 1. WHY subtract 1: the "1" is the ideal gate-only part; whatever exceeds it is exactly the depletion-layer share. $\frac{C_{dep}}{C_{ox}}=n-1=\mathrm{0.425.}$ Meaning: the depletion capacitance is about $43\%$ of the oxide capacitance — a fair chunk of $V_{GS}$ is being "stolen" by the depletion layer instead of reaching the surface.

WHAT: take ${\log }_{10}$ of $I_{sub}/I_0=e^{(V_{GS}-V_{th})/n{V}_T}$. WHY divide the exponent by $\ln 10$: a "decade" means a factor of 10, i.e. we measure size in powers of ten. But the formula is a power of $e$. To convert "how many $e$'s" into "how many 10's," use $e^x=10^{x/\ln 10}$ — so ${\log }_{10}(e^x)=x/\ln 10$. Dividing the exponent by $\ln 10$ is exactly that base change. ${\log }_{10}\frac{I_{sub}}{I_0}=\frac{V_{GS}-V_{th}}{n{V}_T\ln 10}=\frac{-0.300}{1.3\times 0.0259\times 2.3026}.$ Denominator $=1.3\times 0.0259\times 2.3026=0.07753$. ${\log }_{10}\frac{I_{sub}}{I_0}=\frac{-0.300}{0.07753}=-\mathrm{3.87.}$ So $I_{sub}$ is about 3.87 decades (≈ 7400×) below $I_0$.

WHY only the bracket: the $V_{GS}$-dependent exponential and $I_0$ are the same at both drain voltages (only $V_{DS}$ changes), so they cancel — the current ratio equals the ratio of the two brackets. $V_T=25.9$ mV.

• At 30 mV: $1-e^{-30/25.9}=1-e^{-1.158}=1-0.314=\mathrm{0.686.}$
• At 80 mV: $1-e^{-80/25.9}=1-e^{-3.089}=1-0.0456=\mathrm{0.954.}$ Ratio $=0.954/0.686=\mathrm{1.39.}$ Meaning: climbing from 30 mV to 80 mV raises current only ~$39\%$; beyond $\sim 3V_T=78$ mV the bracket is basically 1 — the current has saturated in $V_{DS}$. The real knob is $V_{GS}$.

(a) WHY it linearises: for a small exponent $u=V_{DS}/V_T\ll 1$, the Taylor expansion $e^{-u}=1-u+\frac{u^2}{2}-\cdots$ gives $1-e^{-u}\approx u$ (dropping $u^2$ and higher, which are negligible when $u$ is small). So the bracket $\approx V_{DS}/V_T$. WHY expand at all: near $V_{DS}=0$ the bracket must vanish (no drain-source difference → no diffusion imbalance → no current), and the first surviving term of the expansion tells us how it climbs out of zero — linearly. (b) With $u=5/25.9=0.1931$:

• Exact: $1-e^{-0.1931}=1-0.8244=\mathrm{0.1756.}$
• Linear approx: $u=\mathrm{0.1931.}$ The linear estimate overshoots by only ~$10\%$ — good for a first pass at such small $V_{DS}$. (c) Meaning: for $V_{DS}\lesssim V_T$, $I_{sub}\propto V_{DS}$ — the device behaves like a small resistor (Ohmic/linear subthreshold regime), current rising in proportion to $V_{DS}$. Only once $V_{DS}\gtrsim 3V_T$ does the bracket flatten to 1 and the current saturate (as in L2.3). So the bracket smoothly interpolates: linear near 0, constant far out.

WHY only the difference matters: $I_0$, $V_{th}$ and the $V_{DS}$ bracket are identical at A and B, so they cancel in the ratio. Only $\Delta V_{GS}=(-0.100)-(-0.250)=+0.150$ V survives. $\frac{I_A}{I_B}=\exp \left(\frac{0.150}{1.4\times 0.0259}\right)=\exp \left(\frac{0.150}{0.03626}\right)=e^{4.137}=\mathrm{62.7.}$ Meaning: biasing 150 mV deeper "off" (from A to B) cuts leakage ~63×. On the log-current plot (figure) this is exactly the vertical red gap between the two dots — a drop of ${\log }_{10}62.7\approx 1.8$ decades on a line of slope $1/S$.

Step 1 (evaluate the exponent): $\frac{-0.200}{1.4\times 0.0259}=\frac{-0.200}{0.03626}=-\mathrm{5.516.}$ Step 2 (invert to isolate $I_0$): from $I_{sub}=I_0{e}^{-5.516}$, divide both sides by $e^{-5.516}$. WHY multiply by $e^{+5.516}$: dividing by $e^{-5.516}$ is the same as multiplying by its reciprocal $e^{+5.516}$; this "undoes" the exponential shrink and pushes the measured 5 nA back up to what it would be at threshold, which is by definition $I_0$. $I_0=\frac{5\text{ nA}}{e^{-5.516}}=5\text{ nA}\times e^{5.516}=5\times 248.7\text{ nA}=1243\text{ nA}\approx 1.24\mu \text{A}.$ Meaning: extrapolating the exponential straight line back to $V_{GS}=V_{th}$ gives $I_0\approx 1.2\mu$A — the "notional on-current" of the exponential model.

Step 1 (total current): the transistors are identical and in parallel (all draining the same supply), so their leakage currents add. WHY add: currents in parallel branches sum by Kirchhoff's current law. $I_{tot}=2\times 10^9\times 5\text{ nA}=10\text{ A}.$ Step 2 (power): static power is voltage times the current it drives. WHY $P=VI$: each coulomb of leakage falls through $V_{DD}$ volts, delivering $V_{DD}$ joules; multiply by coulombs-per-second (current) to get joules-per-second (watts). $P=V_{DD}{I}_{tot}=0.8\times 10=8\text{ W}.$ Meaning: billions of "off" transistors, each dripping 5 nA, add up to 8 W of pure static waste — this is why subthreshold leakage is a first-order chip-design problem, not a footnote.

(a) Swing scales linearly with $T$ because $S=n{V}_T\ln 10$ and $V_T=kT/q\propto T$. WHY multiply by $375/300$: with $n$ and $\ln 10$ fixed, $S$ is directly proportional to $V_T$, which is directly proportional to $T$; so scaling $T$ up by $375/300$ scales $S$ by the same factor. $S(375)=75\times \frac{375}{300}=93.75\text{ mV/dec}.$ (b) Two effects multiply. WHY multiply the two factors: $I_{sub}=I_0\cdot e^{(\dots )}$ is a product, so the ratio of currents is the product of (ratio of prefactors)×(ratio of exponentials).

• Prefactor: $V_T(300)=25.9$ mV, $V_T(375)=25.9\times \frac{375}{300}=32.375$ mV. Ratio of $I_0$: $(32.375/25.9{)}^2=1.25^2=\mathrm{1.5625.}$
• Exponential: exponent is $\frac{-0.200}{n{V}_T}$. WHY the exponent shrinks in magnitude: a bigger $V_T$ in the denominator makes the (negative) exponent less negative, so $e^{(\dots )}$ is larger — the hotter device leaks more for the same off-bias.
  • At 300 K: $\frac{-0.200}{1.15\times 0.0259}=\frac{-0.200}{0.029785}=-\mathrm{6.715.}$
  • At 375 K: $\frac{-0.200}{1.15\times 0.032375}=\frac{-0.200}{0.037231}=-\mathrm{5.372.}$
  • Ratio of exponentials: $e^{-5.372}/e^{-6.715}=e^{(6.715-5.372)}=e^{1.343}=\mathrm{3.831.}$ Total factor: $1.5625\times 3.831=5.99\approx \mathrm{6.0.}$ Meaning: a 75 K rise makes this leakage ~6× worse — the exponential (colder exponent shrinks) dominates, the prefactor helps. This positive feedback (hotter → leakier → hotter) is the seed of thermal runaway.

WHAT: solve $I_0{e}^{(V_{GS}-V_{th})/n{V}_T}\le 1$ nA for $(V_{th}-V_{GS})$. WHY take $\ln$: the unknown is trapped inside an exponential; the natural logarithm is the exact inverse of $e^{(\dots )}$, so applying $\ln$ to both sides frees the exponent for algebra. $\frac{I_{sub}}{I_0}=\frac{1\text{ nA}}{1000\text{ nA}}=10^{-3}.$ $\frac{V_{GS}-V_{th}}{n{V}_T}=\ln (10^{-3})=-\mathrm{6.9078.}$ WHY multiply by $n{V}_T$ and flip sign: multiplying both sides by $n{V}_T$ isolates $V_{GS}-V_{th}$; negating gives the magnitude of off-bias $V_{th}-V_{GS}$. $V_{th}-V_{GS}=6.9078\times n{V}_T=6.9078\times 1.3\times 0.0259=0.2326\text{ V}.$ Meaning: you need at least 233 mV of off-margin. Equivalently, in "decades" language: $3$ decades $\times S=3\times (1.3\times 25.9\times 2.3026\text{ mV})=3\times 77.5=232.6$ mV — same answer, sanity checked.

Step 1 (how much $V_{th}$ drops): going from $V_{DS}=0.1$ to $1.0$ V, $\Delta V_{DS}=0.9$ V. WHY $\Delta V_{th}=-\eta \Delta V_{DS}$: the model says $V_{th}$ falls by $\eta$ volts for each extra volt of $V_{DS}$; multiply the coefficient by the drain-voltage change. $\Delta V_{th}=-\eta \Delta V_{DS}=-0.1\times 0.9=-0.09\text{ V}.$ Step 2 (effect on exponent): $I_{sub}\propto e^{(V_{GS}-V_{th})/n{V}_T}$. WHY the numerator rises: the exponent contains $-V_{th}$; lowering $V_{th}$ by 0.09 V raises $V_{GS}-V_{th}$ by exactly $+0.09$ V (a fixed $V_{GS}$ minus a smaller $V_{th}$ is bigger). The current ratio is then $e^{(+0.09)/n{V}_T}$ since all else cancels. $\frac{I(1.0\text{V})}{I(0.1\text{V})}=\exp \left(\frac{+0.090}{1.3\times 0.0259}\right)=\exp \left(\frac{0.090}{0.03367}\right)=e^{2.673}=\mathrm{14.5.}$ Meaning: DIBL alone makes an "off" short-channel device leak ~14.5× more at high $V_{DS}$ — even though the $V_{DS}$ bracket has already saturated. This is why $V_{DS}$ appears to control leakage in nanoscale nodes: it does, but through $V_{th}$, not through the clean bracket.

Step 1 (what $n$ is implied): invert the swing formula. WHY divide by $V_T\ln 10$: same base change as before — it converts the claimed swing into the number of $n$-units it represents. $n=\frac{S}{V_T\ln 10}=\frac{45}{25.9\times 2.3026}=\frac{45}{59.64}=\mathrm{0.755.}$ Step 2 (why impossible): $n=1+C_{dep}/C_{ox}$. WHY $n\ge 1$: physical capacitances are non-negative, so $C_{dep}/C_{ox}\ge 0$, forcing $n\ge 1$ always. An $n=0.755<1$ would need a negative capacitance ratio — physically forbidden for a conventional MOSFET. The floor: the best "perfect oxide" achieves $n\to 1$, giving $S\to V_T\ln 10=59.6$ mV/dec. You approach 60 from above, never cross it. Breaking it needs a different transport mechanism (band-to-band tunnelling in a TFET), or lowering $T$.

Step 1: for $n=1$, $S=V_T\ln 10$, so solve for the required $V_T$. WHY divide by $\ln 10$: $\ln 10$ is a fixed constant; dividing the target swing by it gives the thermal voltage that would produce that swing. $V_T=\frac{S}{\ln 10}=\frac{40}{2.3026}=17.37\text{ mV}.$ Step 2 (scale $T$): $V_T=kT/q\propto T$, so temperature scales in proportion to the required $V_T$. WHY multiply by $17.37/25.9$: since $V_T$ and $T$ are directly proportional, the temperature ratio equals the thermal-voltage ratio. $T=300\text{ K}\times \frac{17.37}{25.9}=201.2\text{ K}\approx 201\text{ K }(\approx -72^{\circ }\text{C}).$ Meaning: $S$ is linear in $T$, so cooling to ~201 K buys 40 mV/dec. Cryogenic operation genuinely sharpens the switch — the "60 mV wall" is a room-temperature wall, set by $kT/q$, not a universal one. Cool the electrons and they crowd more tightly in energy, letting the gate shut them off with fewer millivolts per decade.