Thermal throttling mechanisms: Exercises — Hinglish deep dive

6.4.6 · D4 · Hardware › Power, Thermal & Reliability › Thermal throttling mechanisms

Ye do engines hain jo tum baar baar use karoge:

Formula Do laws jo tumhe chahiye

Chip jo power banata hai (dynamic): $P_{d y n} = α C V^{2} f$ jahan $α$ = har cycle mein toggling karne wale gates ka fraction (activity factor, unitless), $C$ = switched capacitance (farads), $V$ = supply voltage (volts), $f$ = clock frequency (hertz).

Chip jo temperature reach karta hai (steady state): $T_{j} = T_{a} + P R_{θ}$ jahan $T_{j}$ = junction (chip) temperature, $T_{a}$ = ambient (aas-paas ki hawa ki) temperature, $R_{θ}$ = thermal resistance junction-to-air (°C per watt), $P$ = dissipated power.

Trip point $T_{t r i p}$ : woh safety threshold jo controller dekhta rehta hai. Jab $T_{j}$ $T_{t r i p}$ tak pahunch jaata hai, throttling engage ho jaati hai; yeh destructive limit $T_{j, ma x}$ se thoda neeche set hota hai taaki margin rahe. In poore exercises mein $T_{t r i p}$ woh temperature hai jahan action shuru hota hai — bas itna hi.

Transient (warm/cool hone ke dauran): $T_{j} (t) = T_{\infty} + (T_{0} - T_{\infty}) e^{- t / τ}, τ = R_{θ} C_{θ}$ $T_{0}$ = start temperature, $T_{\infty}$ = woh temperature jahan yeh eventually settle hota, $τ$ = thermal time constant (seconds), $C_{θ}$ = thermal mass (joules per °C).

Level 1 — Recognition

Recall Solution

Yeh dynamic power $P_{d y n} = α C V^{2} f$ reduce karta hai, voltage $V$ aur frequency $f$ knobs ke zariye saath mein (yeh pairing DVFS hai — dekho DVFS Dynamic Voltage and Frequency Scaling). Baaki symbols kyun nahi? $α$ aur $C$ workload aur silicon layout se set hote hain — hardware controller inhe real-time mein change nahi kar sakta. $V$ aur $f$ hi real-time dials hain.

Recall Solution

$T_{j} - T_{a} = P R_{θ}$ , $P$ mein linear hai. $P$ double karo ⇒ rise double hoga. (Absolute $T_{j}$ nahi — sirf ambient se upar wala part. Agar $T_{a} = 30$ aur rise $20$ tha, toh double karne par rise $40$ ho jaata hai, toh $T_{j}$ $50 \to 70$ jaata hai, $50 \to 100$ nahi.)

Level 2 — Application

Recall Solution

Step 1 — junction temp nikalo. $T_{j} = T_{a} + P R_{θ} = 35 + 80 (0.45) = 35 + 36 = 71$ °C. Yeh step kyun: steady-state Ohm-analogue power ko seedha temperature mein convert karta hai. Step 2 — trip se compare karo. $71 < 95$ ⇒ koi throttling nahi, $24$ °C ka margin hai.

Recall Solution

Step 1 — boundary wahan hai jahan $T_{j}$ exactly trip ke barabar hoti hai. Set karo $95 = 35 + P (0.45)$ . Kyun: throttling usi waqt shuru hoti hai jab junction $T_{t r i p}$ ko touch kare; yahi safe region ka edge hai. Step 2 — $P$ ke liye solve karo. Dono sides se ambient offset subtract karo ( $95 - 35 = 60$ ), phir $R_{θ}$ se divide karo us multiplication ko undo karne ke liye jo power ko temperature rise mein badal deta hai: $P = 60/0.45 = 133. \overline{3} \approx 133.3$ W. $R_{θ}$ se kyun divide karo: law kehta hai rise $P R_{θ}$ hai; ek jaani hui rise se $P$ recover karne ke liye hum us scaling ko reverse karte hain, yaani °C/W se divide karte hain. $133.3$ W se upar chip trip cross karti hai aur throttle karti hai.

Recall Solution

Step 1 — scaling likho. $f \propto V$ ke saath, $P_{d y n} = α C V^{2} f \propto V^{2} \cdot V = V^{3}$ . $V^{3}$ aur $V^{2}$ nahi kyun: voltage do baar enter hoti hai — ek baar explicitly (squared) aur ek baar $f$ ke andar chupi hui (kyunki is region mein $f$ $V$ ke saath chalti hai). Yahi double entry DVFS ke powerful hone ki poori wajah hai. Step 2 — plug in karo. Factor $= (0.85)^{3} = 0.614$ . Power lagbhag $61.4%$ tak gir jaati hai — $15%$ voltage trim ke liye $\sim 38.6%$ ki cut.

Level 3 — Analysis

Recall Solution

Step 1 — fixed $V$ par, power $f$ mein linear hai. Toh $\frac{P _{n e w}}{P _{n o w}} = \frac{f _{n e w}}{f _{n o w}}$ . Kyun: $P = α C V^{2} f$ mein, $α, C, V$ fixed rakhne par sirf $f$ variable bachta hai — seedha proportion. Step 2 — solve karo. $f_{n e w}$ isolate karne ke liye hum current frequency ko us power ratio se multiply karte hain jo hamein chahiye (kyunki $f$ aur $P$ fixed $V$ par saath chalte hain, power ek fraction tak cut karna $f$ usi fraction tak cut karna hai): $f_{n e w} = 4.2 \times \frac{133.3}{150} = 4.2 \times 0.8889 = 3.733$ GHz. Ratio se multiply kyun: $f_{n e w} = f_{n o w} \cdot (P_{n e w} / P_{n o w})$ bas Step 1 ka proportion hai jo unknown ke liye rearrange kiya gaya hai. Safety ke liye neeche round karo: $\approx 3.73$ GHz par chalao.

Recall Solution

Step 1 — voltage pehle cut ka kuch hissa handle karta hai. $V$ kam karne se power $(1.10/1.20)^{2} = (0.9167)^{2} = 0.8403$ factor se scale hoti hai $f$ touch kiye bina. Yahan $V^{2}$ kyun ( $V^{3}$ nahi): hum $f$ independently choose kar rahe hain, toh voltage sirf apne explicit square ke zariye enter hoti hai. Step 2 — V-drop ke baad purani $f$ par power. $150$ W se yeh $150 \times 0.8403 = 126.05$ W ho jaati hai — pehle se hi $133.3$ W target se neeche hai! Step 3 — interpret karo. $1.10$ V par hum actually poore $4.2$ GHz rakh sakte hain (isse sirf $126$ W chahiye < $133.3$ W). Compare karo: sirf clock throttling ne humein $3.73$ GHz tak le jaaya tha. Wohi heat budget, zyada performance — yahi wajah hai DVFS jeetata hai. (Dekho Dynamic vs Static Power: yeh sirf dynamic term ko shrink karta hai.)

Recall Solution

(a) Disengage temperature $= 95 - 8 = 87$ °C. $87$ °C se neeche chip un-throttle hoti hai; $87$ aur $95$ ke beech woh jis state mein thi ussi mein rehti hai. (b) $Δ = 0$ ke saath engage aur disengage points exactly $95$ °C par coincide karte hain. Junction seedhi us line par baith jaati hai; kyunki $T_{j} (t) = T_{\infty} + (T_{0} - T_{\infty}) e^{- t / τ}$ continuously move karta hai, chhoti si fluctuations isse $95$ °C se thoda upar dhakkelti hain (throttle ON hoti hai, heat girta hai, yeh $95$ °C se neeche cool hoti hai) phir thoda neeche (throttle OFF hoti hai, heat badhta hai, waapas upar). Period quantify karna. $t = 0$ ke paas exponential nearly linear hai: $T_{j} (t) \approx T_{0} + (T_{\infty} - T_{0}) \frac{t}{τ}$ (Taylor: $e^{- t / τ} \approx 1 - t / τ$ ). Trip ke across ek fixed chhoti band $δ$ move karne mein time lagta hai $Δ t \approx \frac{δ}{∣ T _{\infty} - T _{0} ∣} τ$ — yaani chatter period $τ$ ke proportional aur is baat ke inversely proportional hai ki chip ko kitna zyada ya kam drive kiya ja raha hai. Zyada thermal mass ( $τ = R_{θ} C_{θ}$ bada) flip slow karta hai; ek real gap $Δ$ chhoti $δ$ ki jagah $8$ °C le leta hai, har half-cycle ko enormously stretch karta hai aur clean, slow switching deta hai.

Level 4 — Synthesis

Recall Solution

Step 1 — starting temperature (sustained 65 W). $T_{0} = 30 + 65 (0.5) = 62.5$ °C. Step 2 — 95 W run jis target ki taraf ja raha hai. $T_{\infty} = 30 + 95 (0.5) = 77.5$ °C. Step 3 — rigorous rehne ke liye transient set up karo. $τ = R_{θ} C_{θ} = 0.5 \times 20 = 10$ s, aur $T_{j} (t) = 77.5 + (62.5 - 77.5) e^{- t /10} = 77.5 - 15 e^{- t /10}$ . Step 4 — trip tak pahunchne ki koshish karo. Jab $t \to \infty$ , $e^{- t /10} \to 0$ , toh $T_{j} \to 77.5$ °C — curve ki ceiling. $95 = 77.5 - 15 e^{- t /10}$ solve karne ke liye $e^{- t /10} = (77.5 - 95) /15 = - 1.17$ chahiye, jo impossible hai (ek exponential kabhi negative nahi hota). Impossible kyun: junction kabhi apne asymptote $T_{\infty} = 77.5$ °C se upar nahi ja sakta, aur $77.5 < 95$ . Toh is cooler par chip $95$ W forever hold kar sakti hai trip hit kiye bina: boost time unlimited hai (thermally). Real limit power delivery hogi, temperature nahi. Lesson: hamesha pehle $T_{\infty}$ compute karo — agar $T_{\infty} < T_{t r i p}$ , toh koi finite crossing exist nahi karti aur koi algebra ki zaroorat nahi.

Recall Solution

Step 1 — start temp. $T_{0} = 30 + 65 (0.8) = 82$ °C. Step 2 — 95 W ke liye asymptote. $T_{\infty} = 30 + 95 (0.8) = 106$ °C. Kyunki $106 > 95$ , chip trip hit karegi — ek finite boost window exist karti hai. Step 3 — time constant. $τ = R_{θ} C_{θ} = 0.8 \times 20 = 16$ s. Step 4 — $T_{j} = 95$ ke liye transient invert karo. $95 = 106 + (82 - 106) e^{- t /16} \Rightarrow - 11 = - 24 e^{- t /16} \Rightarrow e^{- t /16} = 0.4583.$ Aage logarithm kyun lein: unknown $t$ ek exponent ke andar trapped hai; natural log $e^{(\cdot)}$ ka exact inverse hai, toh dono sides par $ln$ apply karne se $t$ free ho jaata hai. $t = - 16 ln (0.4583) = 16 \times 0.7802 = 12.48 s .$ Toh yeh throttling engage hone se pehle lagbhag $12.5$ s tak boost kar sakta hai. (Yahi wajah hai boost ek kuch-seconds wala feature hai — dekho TDP note TDP Thermal Design Power.) Yahi woh run hai jo page ke upar figure mein drawn hai.

Level 5 — Mastery

Recall Solution

(a) Power. $P = α C V^{2} f = 0.7 \times 8 \times 1 0^{- 8} \times (1.0)^{2} \times 3.0 \times 1 0^{9}$ . Group karo: $0.7 \times 8 \times 1 0^{- 8} = 5.6 \times 1 0^{- 8}$ ; times $3.0 \times 1 0^{9} = 168$ W. Toh $P = 168$ W. (b) Required cooler. Boundary par $T_{j} = T_{t r i p}$ : $90 = 40 + 168 R_{θ}$ . Yahan kyun solve karo: sabse bura allowed cooler woh hai jo $T_{j}$ ko exactly trip par land karta hai; koi bhi kamzor (zyada $R_{θ}$ ) overshoot karega. Ambient subtract karo ( $90 - 40 = 50$ ) phir $R_{θ}$ free karne ke liye power se divide karo: $R_{θ} = 50/168 = 0.2976$ °C/W. Cooler ka $R_{θ} \leq 0.298$ °C/W hona chahiye — ek beefy heatsink. Ek bura (zyada $R_{θ}$ ) cooler throttling force karta hai; fix ek better heatsink hai, chip ko blame karna nahi.

Recall Solution

(a) Sustainable power. $T_{j} = T_{t r i p}$ set karo aur temperature law reverse karo (ambient subtract karo, $R_{θ}$ se divide karo): $P_{ma x} = (90 - 40) /0.45 = 50/0.45 = 111.1$ W. (b) DVFS scaling. Hamein $P_{n e w} = P_{ma x}$ chahiye. $f \propto V$ ke saath $P \propto V^{3}$ se: $k^{3} = \frac{P _{ma x}}{P _{n o w}} = \frac{111.1}{168} = 0.6614 \Rightarrow k = 0.661 4^{1/3} = 0.8712.$ Cube root kyun: power voltage factor ke cube se scale hoti hai, toh $k$ ke liye invert karne ke liye hum power ratio ka cube root lete hain — "teen ke power tak uthao" ka exact undo. Naya voltage: $V_{n e w} = 0.8712 \times 1.0 = 0.871$ V. Nayi frequency: kyunki $f \propto V$ , $f_{n e w} = 0.8712 \times 3.0 = 2.614$ GHz. Performance loss: frequency $1 - 0.8712 = 12.9%$ giri, lekin humne $(168 - 111.1) /168 = 33.9%$ power shed ki. $\sim 13%$ speed cut ke liye $\sim 34%$ heat cut — $V^{3}$ leverage in action.

Active recall

Recall Pehle answers cover karo

Throttling ki boundary tab hoti hai jab junction temp kiske barabar ho? ::: Trip point $T_{t r i p}$ ; us boundary par sustainable power $P_{ma x} = (T_{t r i p} - T_{a}) / R_{θ}$ hoti hai. $f \propto V$ ke saath DVFS mein, power ko factor $r$ se change karne ke liye voltage ko kisse scale karte hain? ::: $r^{1/3}$ se (kyunki $P \propto V^{3}$ ). Turbo-boost window time mein kabhi infinite hoti hai? ::: Jab boost asymptote $T_{\infty} = T_{a} + P R_{θ}$ already trip se neeche ho. Chip us asymptote ki taraf kitni tez heat hoti hai, yeh kya determine karta hai? ::: Time constant $τ = R_{θ} C_{θ}$ . Throttling kaun se do knobs par aur kaun si quantity par act karta hai? ::: Voltage aur frequency, dynamic power $α C V^{2} f$ cut karte hue.

Connections

Thermal throttling mechanisms — woh parent jise yeh exercise set drill karta hai.
DVFS Dynamic Voltage and Frequency Scaling — $V^{3}$ lever jo L2.3, L3.2, L5.2 mein use hua.
TDP Thermal Design Power — sustained budget vs boost, L4.1–L4.2.
Thermal Resistance and Heatsinks — L5.1 mein design kiya gaya $R_{θ}$ .
Dynamic vs Static Power — throttling sirf dynamic term ko touch karta hai.
Clock Gating and Power Gating — coarser sibling mechanisms.
Reliability and Electromigration — kyun $T_{j, ma x}$ exist karta hai.
6.4.06 Thermal throttling mechanisms (Hinglish) — wohi content Hinglish mein.