Exercises — Thermal design power (TDP)

Iska matlab: cooling solution (heatsink + fan) ko at least 65 W heat dissipate karne ki capability honi chahiye ek sustained realistic heavy workload ke dauran, bina chip ka maximum junction temperature cross karaye.

Yeh kya promise NAHI karta: yeh average power draw nahi hai, idle power nahi hai, aur absolute peak bhi nahi hai — chip Turbo Boost ke dauran briefly 65 W se zyada pull kar sakti hai. TDP ek thermal design target hai, power-consumption guarantee nahi.

• $T_{\text{j,max}}$ ::: maximum safe junction (die) temperature, jiske upar chip throttle karta hai ya fail ho jaata hai (typically 90–105 °C).
• $T_{\text{ambient}}$ ::: surrounding air temperature jisme heat finally dump hoti hai.
• $R_{\theta ,\text{SA}}$ ::: sink-to-ambient thermal resistance — chain ka heatsink + airflow stage.
• $\Delta T$ ::: temperature difference jitna heat ko climb karna hota hai, yahan $T_{\text{j,max}}-T_{\text{ambient}}$.

Jhooth. TDP heat measure karta hai, speed nahi. Ek newer 65 W chip jo chhote process node par hai, ek older 125 W chip ko better IPC (instructions per cycle) aur architecture ke through beat kar sakti hai. Performance $\approx$ IPC $\times$ frequency $\times$ cores — benchmarks compare karo, watts nahi.

Step 1 — temperature rise (WHAT/WHY): heat jo resistance ke through push hoti hai woh ek rise create karti hai, $\Delta T=P\times R_{\theta }$. $\Delta T=80\times 0.5=40\text{°C}$ Step 2 — starting point add karo: die jo bhi air mein dump karti hai usse $40\text{°C}$ upar baith hai. $T_{\text{j}}=T_{\text{ambient}}+\Delta T=30+40=70\text{°C}$ Answer: $T_{\text{j}}=70\text{°C}$. Typical 100 °C limit se comfortably neeche.

Step 1 — degrees ka budget jo hum spend kar sakte hain: $\Delta T_{\text{max}}=T_{\text{j,max}}-T_{\text{ambient}}=95-25=70\text{°C}$ Step 2 — Ohm's-law analogy ko invert karo (hume $\Delta T$ aur $P$ pata hai, $R_{\theta }$ chahiye): $R_{\theta ,\text{total}}\le \frac{\Delta T_{\text{max}}}{\text{TDP}}=\frac{70}{95}\approx 0.737\text{°C/W}$ Answer: $R_{\theta ,\text{total}}\le 0.737\text{°C/W}$. Isse zyada kuch bhi aur die 95 °C cross kar lega.

Step 1 — shrunken budget: $\Delta T_{\text{max}}=95-40=55\text{°C}$ Step 2 — new limit: $R_{\theta ,\text{total}}\le \frac{55}{95}\approx 0.579\text{°C/W}$ Interpretation: zyada garm air tumhara temperature budget chura leti hai, isliye same chip ke liye tumhe better (lower-resistance) cooler chahiye. Yahi reason hai ki warm rooms mein servers pehle throttle karte hain.

Step 1 — temperature budget: $\Delta T_{\text{max}}=100-25=75\text{°C}$. Step 2 — total allowed resistance: $R_{\theta ,\text{total}}=\frac{75}{125}=0.6\text{°C/W}$ Step 3 — fixed stages subtract karo (WHY subtract): series mein resistances add hote hain, isliye heatsink ko sirf wahi milta hai jo chip package aur paste apna share lene ke baad bachta hai. $R_{\theta ,\text{SA}}=0.6-0.2-0.05=0.35\text{°C/W}$ Answer: $R_{\theta ,\text{SA}}\le 0.35\text{°C/W}$ — ek solid tower cooler ya AIO. Dekho Heatsink design and thermal resistance.

Har stage ka rise hai $\Delta T_{\text{stage}}=P\times R_{\theta ,\text{stage}}$ jahan $P=125\text{W}$:

• Junction→Case: $125\times 0.2=25\text{°C}$
• Case→Sink (paste): $125\times 0.05=6.25\text{°C}$
• Sink→Air: $125\times 0.35=43.75\text{°C}$

Sabse bada jump: sink-to-air stage, $43.75\text{°C}$ — kyunki iska resistance sabse bada hai aur yeh poore 125 W carry karta hai. Teeno $75\text{°C}$ mein add hote hain, exactly hamaara budget, jo confirm karta hai ki die 100 °C par land karti hai.

Step 1 — new total: $R_{\theta ,\text{total}}=0.2+0.30+0.35=0.85\text{°C/W}$ Step 2 — actual rise: $\Delta T=125\times 0.85=106.25\text{°C}$. Step 3 — junction temp: $T_{\text{j}}=25+106.25=131.25\text{°C}$. Answer: die 131.25 °C tak pahunch jaati — 100 °C limit se 31.25 °C upar, isliye Thermal throttling hard kick karta hai. Ek 0.25 °C/W paste error $125\times 0.25=31.25\text{°C}$ ka cost hai.

Step 1 — scaling law: $P_{\text{dyn}}\propto V^{2}f$; $\alpha$ aur $C$ ratio mein cancel ho jaate hain. $\frac{P_{\text{new}}}{P_{\text{old}}}=\frac{V_{\text{new}}^2{f}_{\text{new}}}{V_{\text{old}}^2{f}_{\text{old}}}=\frac{(1.35{)}^2\times 4.2}{(1.20{)}^2\times 3.5}$ Step 2 — calculate: $=\frac{1.8225\times 4.2}{1.44\times 3.5}=\frac{7.6545}{5.04}\approx 1.519$ Step 3 — baseline par apply karo: $P_{\text{new}}\approx 1.519\times 125\approx 190\text{W}$. Answer: $\approx 190\text{W}$. Notice karo voltage sabse zyada hurt karta hai: sirf 12.5 % voltage bump power ko $1.35^2/1.20^2\approx 1.27$ se multiply karta hai.

Step 1 — budget: $\Delta T_{\text{max}}=100-25=75\text{°C}$ (unchanged). Step 2 — new total allowed: $R_{\theta ,\text{total}}=\frac{75}{190}\approx 0.3947\text{°C/W}$ Step 3 — fixed stages subtract karo: $R_{\theta ,\text{SA}}\le 0.3947-0.2-0.05=0.1447\text{°C/W}$ Verdict: tumhe ab $\le 0.145\text{°C/W}$ chahiye, lekin old cooler 0.35 °C/W hai — bahut weak. Zyada power resistance budget ko dramatically shrink kar deta hai. Overclock ke liye bahut strong cooler chahiye (high-end AIO ya custom loop).

TDP sirf CPU heat measure karta hai. Wall reading poore system ki hai:

• CPU: 15 W
• Discrete GPU: ~10 W
• Display backlight: ~8 W
• SSD / RAM / board: ~5 W
• PSU conversion loss (~10 %): ~5 W

Sum: $15+10+8+5+5=43\text{W}\approx 45\text{W}$. Koi contradiction nahi — TDP ek component ka thermal load hai, system energy draw nahi.

CPU 1 (cool air, 22 °C): $R_{\theta ,\text{total}}=\frac{90-22}{200}=\frac{68}{200}=0.34\text{°C/W}$ $R_{\theta ,\text{SA}}\le 0.34-0.25=0.09\text{°C/W}$ CPU 2 (pre-warmed air, 35 °C): $R_{\theta ,\text{total}}=\frac{90-35}{200}=\frac{55}{200}=0.275\text{°C/W}$ $R_{\theta ,\text{SA}}\le 0.275-0.25=0.025\text{°C/W}$ Verdict: CPU 2 ko $\le 0.025\text{°C/W}$ chahiye — nearly 4× stricter. Downstream chip ko cool karna bahut zyada mushkil hai, yahi reason hai ki rack layout aur hot-aisle/cold-aisle design itna matter karta hai.

Step 1 — full chain resistance: $R_{\theta ,\text{total}}=0.20+0.15=0.35\text{°C/W}$. Step 2 — budget: $\Delta T_{\text{max}}=100-25=75\text{°C}$. Step 3 — $\Delta T=P{R}_{\theta }$ ko power ke liye invert karo: $P_{\text{max}}=\frac{\Delta T_{\text{max}}}{R_{\theta ,\text{total}}}=\frac{75}{0.35}\approx 214.3\text{W}$ Answer: $\approx 214\text{W}$ sustained. Is tak TDP wala koi bhi chip survive karta hai; ek 250 W part throttle kar jaata.

Step 1 — power ratio (frequency fixed hai, isliye sirf $V^2$ change hota hai): $\frac{P_{\text{new}}}{P_{\text{old}}}=\frac{V_{\text{new}}^2}{V_{\text{old}}^2}=\frac{(1.15{)}^2}{(1.30{)}^2}=\frac{1.3225}{1.69}\approx 0.7825$ Step 2 — apply karo: $P_{\text{new}}\approx 0.7825\times 180\approx 140.9\text{W}$. Verdict: haan — undervolting power ko $\approx 141\text{W}$ tak drop kar deta hai, comfortably 150 W cooler limit ke neeche, bina kisi clock cost ke. Yahi reason hai ki undervolting sabse sasta thermal fix hai: power $V^2$ ke saath scale karta hai, isliye chhote voltage cuts quadratically pay off karte hain.

TDP, $T_{\text{j,max}}$, $T_{\text{ambient}}$, aur fixed chain resistances diye gaye hon, kaun si single inequality tumhara heatsink decide karti hai? Answer ::: $R_{\theta ,\text{SA}}\le \frac{T_{\text{j,max}}-T_{\text{ambient}}}{\text{TDP}}-R_{\theta ,\text{JC}}-R_{\theta ,\text{CS}}$