Exercises — Thermal throttling mechanisms

It reduces dynamic power $P_{dyn}=\alpha C{V}^{2}f$, through the voltage $V$ and frequency $f$ knobs together (this pairing is DVFS — see DVFS Dynamic Voltage and Frequency Scaling). Why not the other symbols? $\alpha$ and $C$ are set by the workload and the silicon layout — the hardware controller cannot change them on the fly. $V$ and $f$ are the only real-time dials.

$T_j-T_a=P{R}_{\theta }$ is linear in $P$. Double $P$ ⇒ double the rise. (Not the absolute $T_j$ — only the part above ambient. If $T_a=30$ and rise was $20$, doubling makes rise $40$, so $T_j$ goes $50\to 70$, not $50\to 100$.)

Step 1 — get the junction temp. $T_j=T_a+P{R}_{\theta }=35+80(0.45)=35+36=71$ °C. Why this step: steady-state Ohm-analogue converts power straight into temperature. Step 2 — compare to trip. $71<95$ ⇒ no throttling, with $24$ °C of margin.

Step 1 — the boundary is where $T_j$ exactly equals the trip. Set $95=35+P(0.45)$. Why: throttling begins the instant the junction touches $T_{trip}$; that is the edge of the safe region. Step 2 — solve for $P$. Subtract the ambient offset from both sides ($95-35=60$), then divide by $R_{\theta }$ to undo the multiplication that turned power into a temperature rise: $P=60/0.45=133.\overline{3}\approx 133.3$ W. Why divide by $R_{\theta }$: the law says a rise of $P{R}_{\theta }$; to recover $P$ from a known rise we reverse that scaling, i.e. divide out the °C/W. Above $133.3$ W the chip crosses the trip and throttles.

Step 1 — write the scaling. With $f\propto V$, $P_{dyn}=\alpha C{V}^{2}f\propto V^2\cdot V=V^3$. Why $V^3$ and not $V^2$: voltage enters twice — once explicitly (squared) and once hidden inside $f$ (because in this region $f$ rides along with $V$). That double entry is the entire reason DVFS is powerful. Step 2 — plug in. Factor $=(0.85{)}^3=0.614$. Power drops to about $61.4\%$ — a $\sim 38.6\%$ cut for a $15\%$ voltage trim.

Step 1 — at fixed $V$, power is linear in $f$. So $\frac{P_{new}}{P_{now}}=\frac{f_{new}}{f_{now}}$. Why: in $P=\alpha C{V}^{2}f$, holding $\alpha ,C,V$ fixed leaves $f$ as the only variable — straight proportion. Step 2 — solve. To isolate $f_{new}$ we multiply the current frequency by the power ratio we need (because $f$ and $P$ move together at fixed $V$, cutting power to a fraction cuts $f$ to that same fraction): $f_{new}=4.2\times \frac{133.3}{150}=4.2\times 0.8889=3.733$ GHz. Why multiply by the ratio: $f_{new}=f_{now}\cdot (P_{new}/P_{now})$ is just the proportion of Step 1 rearranged for the unknown. Round down for safety: run at $\approx 3.73$ GHz.

Step 1 — voltage handles part of the cut first. Lowering $V$ scales power by $(1.10/1.20{)}^2=(0.9167{)}^2=0.8403$ before touching $f$. Why $V^2$ here (not $V^3$): we are choosing $f$ independently, so voltage only enters through its explicit square. Step 2 — power still at old $f$ after the V-drop. From $150$ W it becomes $150\times 0.8403=126.05$ W — already below the $133.3$ W target! Step 3 — interpret. At $1.10$ V we may actually keep the full $4.2$ GHz (it needs only $126$ W < $133.3$ W). Compare: pure clock throttling forced us down to $3.73$ GHz. Same heat budget, more performance — that is why DVFS wins. (See Dynamic vs Static Power: this only shrinks the dynamic term.)

(a) Disengage temperature $=95-8=87$ °C. Below $87$ °C the chip un-throttles; between $87$ and $95$ it stays in whatever state it was. (b) With $\Delta =0$ the engage and disengage points coincide at exactly $95$ °C. The junction sits right on the line; since $T_j(t)=T_{\infty }+(T_0-T_{\infty })e^{-t/\tau }$ moves continuously, tiny fluctuations push it a hair above (throttle ON, heat drops, it cools below) then a hair below (throttle OFF, heat rises, back above). Quantifying the period. Near $t=0$ the exponential is nearly linear: $T_j(t)\approx T_0+(T_{\infty }-T_0)\frac{t}{\tau }$ (Taylor: $e^{-t/\tau }\approx 1-t/\tau$). To move a fixed small band $\delta$ across the trip takes a time $\Delta t\approx \frac{\delta }{|T_{\infty }-T_0|}\tau$ — i.e. the chatter period is proportional to $\tau$ and inversely proportional to how hard the chip is being over/under-driven. A larger thermal mass ($\tau =R_{\theta }{C}_{\theta }$ bigger) slows the flip; a real gap $\Delta$ replaces the tiny $\delta$ with $8$ °C, stretching each half-cycle enormously and giving clean, slow switching.

Step 1 — starting temperature (sustained 65 W). $T_0=30+65(0.5)=62.5$ °C. Step 2 — the target the 95 W run is heading toward. $T_{\infty }=30+95(0.5)=77.5$ °C. Step 3 — set up the transient anyway, to be rigorous. $\tau =R_{\theta }{C}_{\theta }=0.5\times 20=10$ s, and $T_j(t)=77.5+(62.5-77.5)e^{-t/10}=77.5-15e^{-t/10}$. Step 4 — try to reach the trip. As $t\to \infty$, $e^{-t/10}\to 0$, so $T_j\to 77.5$ °C — the curve's ceiling. Solving $95=77.5-15e^{-t/10}$ needs $e^{-t/10}=(77.5-95)/15=-1.17$, which is impossible (an exponential is never negative). Why impossible: the junction can never rise above its own asymptote $T_{\infty }=77.5$ °C, and $77.5<95$. So at this cooler the chip could hold $95$ W forever without hitting the trip: the boost time is unlimited (thermally). The real limit would be power delivery, not temperature. Lesson: always compute $T_{\infty }$ first — if $T_{\infty }<T_{trip}$, no finite crossing exists and no algebra is needed.

Step 1 — start temp. $T_0=30+65(0.8)=82$ °C. Step 2 — asymptote for 95 W. $T_{\infty }=30+95(0.8)=106$ °C. Since $106>95$, the chip will hit the trip — a finite boost window exists. Step 3 — time constant. $\tau =R_{\theta }{C}_{\theta }=0.8\times 20=16$ s. Step 4 — invert the transient for $T_j=95$. $95=106+(82-106)e^{-t/16}\Rightarrow -11=-24e^{-t/16}\Rightarrow e^{-t/16}=\mathrm{0.4583.}$ Why take a logarithm next: the unknown $t$ is trapped inside an exponent; the natural log is the exact inverse of $e^{(\cdot )}$, so applying $\ln$ to both sides frees $t$. $t=-16\ln (0.4583)=16\times 0.7802=12.48\text{s}.$ So it can boost for about $\boxed{12.5}$ s before throttling engages. (This is exactly why boost is a few-second feature — see the TDP note TDP Thermal Design Power.) This is the run drawn in the figure at the top of the page.

(a) Power. $P=\alpha C{V}^{2}f=0.7\times 8\times 10^{-8}\times (1.0{)}^2\times 3.0\times 10^9$. Group: $0.7\times 8\times 10^{-8}=5.6\times 10^{-8}$; times $3.0\times 10^9=168$ W. So $P=168$ W. (b) Required cooler. At the boundary $T_j=T_{trip}$: $90=40+168R_{\theta }$. Why solve here: the worst allowed cooler is the one that lands $T_j$ exactly on the trip; any weaker (higher $R_{\theta }$) overshoots. Subtract the ambient ($90-40=50$) then divide by the power to free $R_{\theta }$: $R_{\theta }=50/168=0.2976$ °C/W. The cooler must have $R_{\theta }\le 0.298$ °C/W — a beefy heatsink. A worse (higher $R_{\theta }$) cooler forces throttling; the fix is a better heatsink, not blaming the chip.

(a) Sustainable power. Set $T_j=T_{trip}$ and reverse the temperature law (subtract ambient, divide by $R_{\theta }$): $P_{max}=(90-40)/0.45=50/0.45=111.1$ W. (b) DVFS scaling. We need $P_{new}=P_{max}$. With $P\propto V^3$ under $f\propto V$: $k^3=\frac{P_{max}}{P_{now}}=\frac{111.1}{168}=0.6614\Rightarrow k=0.6614^{1/3}=\mathrm{0.8712.}$ Why the cube root: power scales as the cube of the voltage factor, so to invert for $k$ we take the cube root of the power ratio — the exact undo of "raise to the third power." New voltage: $V_{new}=0.8712\times 1.0=0.871$ V. New frequency: since $f\propto V$, $f_{new}=0.8712\times 3.0=2.614$ GHz. Performance loss: frequency fell by $1-0.8712=12.9\%$, yet we shed $(168-111.1)/168=33.9\%$ of the power. A $\sim 13\%$ speed cut for a $\sim 34\%$ heat cut — the $V^3$ leverage in action.

Boundary of throttling occurs when the junction temp equals what? ::: The trip point $T_{trip}$; at that boundary the sustainable power is $P_{max}=(T_{trip}-T_a)/R_{\theta }$. Under DVFS with $f\propto V$, to change power by a factor $r$ you scale voltage by what? ::: $r^{1/3}$ (because $P\propto V^3$). When is a turbo-boost window infinite in time? ::: When the boost asymptote $T_{\infty }=T_a+P{R}_{\theta }$ is already below the trip. What determines how fast the chip heats toward that asymptote? ::: The time constant $\tau =R_{\theta }{C}_{\theta }$. Which two knobs and which quantity does throttling act on? ::: Voltage and frequency, cutting dynamic power $\alpha C{V}^{2}f$.