High Bandwidth Memory (HBM - HBM2 - HBM3): Worked examples

6.5.4 · D3 · Hardware › Advanced & Emerging Architectures › High Bandwidth Memory (HBM - HBM2 - HBM3)

Yahan sab kuch sirf teen ideas pe tika hai jo parent ne pehle hi build kar diye hain:

$N$ = data wires ki sankhya (bus ki "width").
$f$ = effective per-pin rate bits/second per wire mein (jo wire actually deliver karta hai).
Bit total ko 8 se divide karo kyunki ek byte mein 8 bits hote hain. Dekho the parent topic aur Memory Wall — kyun hum bandwidth ke peeche bhagte hain.

Scenario matrix

Har HBM numeric question inhi cells mein se kisi ek mein aata hai. Last column mein woh example number hai jo us cell ko cover karta hai.

Cell	Kya vary karta hai / twist kya hai	Covered by
A. Forward, single stack	$N$ , $f$ diye hain → BW nikalo	Ex 1
B. Scale to many stacks	Stack count se multiply karo	Ex 2
C. DDR clock trap	Ek clock diya hai, effective rate nahi	Ex 3
D. Inverse solve	BW diya hai → $f$ ya $N$ nikalo	Ex 4
E. Degenerate — ek die dies	Kam working channels ( $N$ shrink hota hai)	Ex 5
F. Zero / limiting case	Kya hota hai jab $f \to 0$ ya sirf ek wire ho	Ex 6
G. Comparison (HBM vs GDDR)	Same BW, alag $N, f$ → energy angle	Ex 7
H. Real-world word problem	"Kya memory compute ko feed kar sakti hai?"	Ex 8
I. Exam twist	Channels doubled but bus unchanged (HBM3)	Ex 9

Neeche ke figures woh do mental pictures anchor karte hain jo poori page ko chahiye.

Figure — High Bandwidth Memory (HBM - HBM2 - HBM3)

Example 1 — Cell A: forward, single stack

Worked example Base case (HBM2)

Statement: Ek HBM2 stack ka $1024$ -bit data bus hai aur har pin effective $2.0$ Gbps pe chalta hai. Is ek stack ki bandwidth nikalo.

Forecast: Pehle order of magnitude guess karo. Ek hazaar wires, har ek ~2 billion bits/s → lagbhag $2$ trillion bits/s. Bytes 8× chhote hote hain count mein, toh kuch hundred GB/s. Yeh guess pakad lo.

Step 1 — Width ko rate se multiply karo. Total bits/second $= N \times f = 1024 \times 2.0 \times 1 0^{9} = 2.048 \times 1 0^{12}$ bits/s. Yeh step kyun? $N$ wires mein se har ek independently $f$ bits/s deliver karta hai; woh parallel mein run karte hain, isliye deliveries simply add up ho jaati hain — yahi hai yahan multiplication ka matlab.

Step 2 — Bits ko bytes mein convert karo. $\frac{2.048 \times 1 0 ^{12}}{8} = 2.56 \times 1 0^{11}$ B/s $= 256$ GB/s. Yeh step kyun? Bandwidth bytes/s mein quote ki jaati hai, aur $8$ bits $= 1$ byte, isliye bit-count ko 8 se divide karte hain.

Verify: Units check — $wires \times \frac{bits}{wire \cdot s} \div \frac{bits}{byte} = \frac{bytes}{s}$ . ✓ Aur $256$ GB/s hamare "few hundred GB/s" forecast se match karta hai. ✓ Parent ke headline number se match karta hai.

Example 2 — Cell B: scale to many stacks

Worked example Char HBM3 stacks parallel mein

Statement: Ek accelerator $4$ HBM3 stacks use karta hai. HBM3 $1024$ -bit hai, $6.4$ Gbps per pin. Total bandwidth?

Forecast: Ek HBM2 stack 256 GB/s tha 2 Gbps pe. HBM3 per pin ~3.2× faster hai aur 4 stacks hain, toh roughly $256 \times 3.2 \times 4 \approx 3300$ GB/s. Guess ~3 TB/s.

Step 1 — Per-stack BW. $\frac{1024 \times 6.4 \times 1 0 ^{9}}{8} = 819.2 \times 1 0^{9}$ B/s $= 819.2$ GB/s. Yeh step kyun? Same $N f /8$ jaise Ex 1; sirf $f$ badla.

Step 2 — Stack count se multiply karo. $4 \times 819.2 = 3276.8$ GB/s $\approx 3.28$ TB/s. Yeh step kyun? Stacks independent channels pe baithe hain (GPU Architecture unhe ek saath access karta hai), isliye unki bandwidths add hoti hain — unhe bottleneck karne ke liye koi shared wire nahi hai.

Verify: $3276.8$ GB/s $= 3.2768$ TB/s, bilkul ~3 TB/s forecast mein. ✓ Sanity: 8 stacks karne pe ~6.5 TB/s milenge, jo top-end 2020s accelerators se match karta hai. ✓

Example 3 — Cell C: DDR clock trap

Worked example "Double" mat bhoolo

Statement: Ek datasheet kehta hai HBM2 memory clock $1.0$ GHz (DDR) hai. Effective per-pin rate kya hai, aur per-stack bandwidth kya hai?

Forecast: Jo log "DDR" bhool jaate hain unhe milta hai $1024 \times 1.0 G /8 = 128$ GB/s — real answer ka aadha. Predict karo kya yahi trap hai.

Step 1 — Clock ko effective rate mein badlo. DDR = Double Data Rate: rising clock edge pe ek bit, falling pe ek bit. Toh $f = 2 \times f_{clk} = 2 \times 1.0 GHz = 2.0$ Gbps. Yeh step kyun? Formula $N f /8$ ko effective bits-per-second per wire chahiye, clock frequency nahi. Ek DDR wire har clock cycle mein do bits deliver karta hai.

Step 2 — $N f /8$ apply karo. $\frac{1024 \times 2.0 \times 1 0 ^{9}}{8} = 256$ GB/s. Yeh step kyun? Ab $f$ honest delivered rate hai, isliye standard formula apply hota hai.

Verify: Yeh Ex 1 ke exactly barabar hai, kyunki "1 GHz DDR" aur "2.0 Gbps effective" ek hi cheez ke do naam hain. ✓ Trap answer $128$ GB/s exactly aadha hai — yahi tell-tale sign hai ki aap ×2 bhool gaye. ✓

Example 4 — Cell D: inverse solve

Example 5 — Cell E: degenerate case, ek channel fail hota hai

Worked example Kya hoga agar ek channel mar jaye?

Statement: Ek HBM2 stack mein normally $8$ channels $\times 128$ bits $= 1024$ bits hote hain. Maan lo ek channel disable ho gaya (ek defective die block, binning mein common). $2.0$ Gbps pe effective width aur bandwidth kya hogi?

Forecast: 8 mein se 1 channel lose karna matlab $1/8$ width lose karna, toh ~87.5% of 256 GB/s ≈ 224 GB/s. Yahi guess karo.

Step 1 — Naya width. $7$ working channels $\times 128 = 896$ bits. Toh $N = 896$ . Yeh step kyun? Ek dead channel zero wires contribute karta hai; bandwidth per channel build hoti hai, isliye hum bas uske 128 bits drop kar dete hain.

Step 2 — $N f /8$ apply karo. $\frac{896 \times 2.0 \times 1 0 ^{9}}{8} = 224 \times 1 0^{9} = 224$ GB/s. Yeh step kyun? Surviving wires ki rate unchanged hai; sirf $N$ shrink hua.

Verify: $\frac{224}{256} = \frac{7}{8} = 0.875$ — exactly channels ka bacha hua fraction. ✓ Forecast se match. ✓ Yahi wajah hai ki real products kabhi kabhi odd capacities/bandwidths pe ship hote hain: partial-good dies ek channel fuse off karke bech diye jaate hain.

Example 6 — Cell F: zero aur single-wire limits

Worked example Extremes tak push karna

Statement: Formula pe do limiting checks. (a) Agar ek wire ki rate $f = 0$ tak gir jaaye, toh bandwidth kya hogi? (b) Ek single wire ( $N = 1$ ) $2.0$ Gbps pe kya contribute karta hai?

Forecast: (a) Ek wire jo zero baar toggle kare woh kuch move nahi karta → 0. (b) Ek wire puri bus ka $1/1024$ hai → lagbhag quarter GB/s.

Step 1 — Zero-rate case. $BW = N \times 0/8 = 0$ B/s, kisi bhi $N$ ke liye. Yeh step kyun? Width useless hai agar kuch toggle nahi karta. Yeh dikhata hai ki dono knobs non-zero hone chahiye — parent ka two-knob picture words mein.

Step 2 — Single-wire case. $\frac{1 \times 2.0 \times 1 0 ^{9}}{8} = 0.25 \times 1 0^{9} = 0.25$ GB/s $= 250$ MB/s. Yeh step kyun? Ek wire isolate karna HBM bandwidth ka "atom" dikhata hai. $1024$ se multiply karo aur tum poora 256 GB/s rebuild kar lo: $0.25 \times 1024 = 256$ . ✓

Verify: $0.25 \times 1024 = 256$ GB/s exactly Ex 1 recover karta hai. ✓ Aur $0$ hi sensible answer hai jab kuch move nahi karta. ✓ Yeh "degenerate input" guardrail hai — formula edges pe bhi theek behave karta hai.

Example 7 — Cell G: HBM vs GDDR, same bandwidth, alag energy

Worked example Width speed se power pe kyun jeet ti hai

Statement: $\approx 256$ GB/s reach karne ke do tarike compare karo: (i) HBM2-style, $N = 1024$ wires at $2.0$ Gbps; (ii) GDDR-style, $N = 128$ wires at $16$ Gbps. Confirm karo dono same bandwidth dete hain, phir argue karo kaun kam power peeta hai.

Forecast: Dono ~256 GB/s ke liye engineer kiye gaye hain, toh bandwidth tie honi chahiye. Power HBM ko favour karni chahiye kyunki har wire slower aur shorter hai.

Step 1 — HBM bandwidth. $1024 \times 2.0 G /8 = 256$ GB/s. Step 2 — GDDR bandwidth. $128 \times 16 G /8 = 256$ GB/s. Yeh steps kyun? Same formula, alag $(N, f)$ split — proof karta hai ki bandwidth product $N f$ se set hoti hai, sirf $f$ se nahi.

Step 3 — Power argument. Dynamic wire power $\propto C V^{2} f$ . HBM ke on-interposer wires short hain (chhhota $C$ ), low-swing (chhhota $V$ ), aur low- $f$ . GDDR ke long PCB traces $16$ Gbps pe bade $C$ , bade $V$ , aur $8 \times$ zyada $f$ per wire rakhte hain. Yeh step kyun? Yeh exam ke "which is more efficient?" ka jawab deta hai. Dekho Energy per Bit / pJ per bit aur DDR vs GDDR vs HBM measured figures ke liye.

Verify: $256 = 256$ — bandwidths exactly match karti hain, isliye single differentiator jo bachta hai woh energy hai. ✓ $N f$ products equal hain: $1024 \times 2 = 2048$ aur $128 \times 16 = 2048$ (Gbps·wires). ✓ Parent ke "wide + slow wins on power" se consistent. ✓

Example 8 — Cell H: real-world word problem

Worked example Kya memory GPU ko fed rakh sakti hai?

Statement: Ek GPU $200$ TFLOP/s useful work karta hai, aur uska workload (arithmetic intensity) lagbhag $2$ bytes of memory traffic per FLOP maangta hai. Uski memory $4$ HBM3 stacks hai (har ek $819.2$ GB/s). Kya memory bottleneck hai?

Forecast: $200$ TFLOP/s × 2 B/FLOP = 400 TB/s demanded. Char stacks ~3.3 TB/s dete hain. Yeh ~100× short hai — memory wall hai.

Step 1 — Compute ki demanded bandwidth. $200 \times 1 0^{12} \frac{FLOP}{s} \times 2 \frac{B}{FLOP} = 400 \times 1 0^{12}$ B/s $= 400$ TB/s. Yeh step kyun? FLOP-rate ko bytes-per-FLOP se multiply karna "kitna math" ko "us math ko kitna data chahiye" mein convert karta hai — units FLOP cancel kar dete hain.

Step 2 — Supplied bandwidth. $4 \times 819.2 = 3276.8$ GB/s $= 3.2768$ TB/s (Ex 2 se). Yeh step kyun? Yeh physical ceiling hai jo HBM deliver kar sakta hai.

Step 3 — Compare karo. $\frac{400}{3.2768} \approx 122$ . Demand ~122× supply hai → memory-bound. Yeh step kyun? (Demand, supply) mein se jo bada ho aur system meet na kar sake woh bottleneck hai; yahan supply badly haart hai, jo exactly Memory Wall hai aur isliye hum data on-chip cache/reuse karte hain.

Verify: $3276.8 \times 122 \approx 3.998 \times 1 0^{5}$ GB/s $\approx 400$ TB/s — ratio check karta hai. ✓ Interpretation: heavy on-chip data reuse ke bina, yeh GPU HBM ka wait karte hue idle rehega. ✓

Example 9 — Cell I: exam twist (channels doubled, bus unchanged)

Worked example "HBM3 ne channels double kiye, toh kya bus twice as wide hai?"

Statement: HBM2 = $8$ channels $\times 128$ bits. HBM3 = $16$ channels $\times 64$ bits. Total data width dikhao, aur batao real HBM3 gain kahan se aata hai.

Forecast: "Twist" aapko $16 \times 128 = 2048$ bolne ki temptation deta hai. Predict karo kya yeh galat hai.

Step 1 — HBM2 width. $8 \times 128 = 1024$ bits. Step 2 — HBM3 width. $16 \times 64 = 1024$ bits. Yeh steps kyun? Width = channels × bits-per-channel. HBM3 ne bits-per-channel half kar diye aur channel count double kar diya, isliye product unchanged raha 1024 pe.

Step 3 — Gain kahan se aata hai. Dono $1024$ -bit hain. HBM3 ki zyada bandwidth ek bade per-pin $f$ se aati hai (e.g. $6.4$ Gbps vs $2.0$ ), wider bus se nahi. Zyada channels = finer-grained parallelism (zyada independent requests in flight), zyada wires nahi. Yeh step kyun? Yeh common misread ko khatam karta hai aur true lever ( $f$ ) ki taraf point karta hai, parent ke consistent.

Verify: $8 \times 128 = 1024 = 16 \times 64$ . ✓ Aur HBM3 BW $= 1024 \times 6.4 G /8 = 819.2$ GB/s vs HBM2 ka $256$ — $3.2 \times$ jump exactly ratio $6.4/2.0$ hai, confirm karta hai ki $f$ (na ki $N$ ) ne kaam kiya. ✓

Active recall

Recall Yeh kaun sa cell hai? (answers chhupao)

"1024-bit aur 2 Gbps diye hain, BW nikalo." ::: Cell A → 256 GB/s.
"Clock 1 GHz DDR hai, effective rate nikalo." ::: Cell C → 2.0 Gbps (×2).
"Stack 460.8 GB/s deliver karta hai, f nikalo." ::: Cell D → 3.6 Gbps.
"8 mein se ek channel disabled, BW nikalo." ::: Cell E → 224 GB/s.
"16×64 vs 8×128 — same width?" ::: Cell I → haan, dono 1024 bits.
"200 TFLOP/s ×2 B/FLOP vs 3.3 TB/s — bottleneck?" ::: Cell H → memory-bound (~122×).

Connections

Parent topic — woh theory jo yeh examples exercise karti hai.
Through-Silicon Vias (TSV) · Silicon Interposer · 2.5D and 3D Integration — isliye hi per stack 1024 wires routable hain.
GDDR Memory · DDR vs GDDR vs HBM · Energy per Bit / pJ per bit — Ex 7 mein comparison.
Memory Wall · GPU Architecture — Ex 8 mein word problem.