Raw repair data壓縮前修復資料
16 Mb
16,000,000 bits
2,000,000 B · 1,953.125 KiB
APPLICATIONS / SRAM REPAIR · CAPACITY PLANNING應用 / SRAM 修復 · 先進節點容量規劃
Size your SRAM repair data把修復需求，算清楚
Explore compression and allocated OTP / eFuse capacity for large SRAM repair datasets.從大型 SRAM 到修復資料，掌握壓縮效益與 OTP / eFuse 配置容量。
Capacity overview容量總覽
Updated已更新Compressed payload壓縮後有效資料
160 kb
160,000 bits
20,000 B · 19.531 KiB
Capacity reduction資料容量減少
99 %
15,840,000 bits saved節省 15,840,000 bits
Compression 100 : 1壓縮倍率 100 : 1
How compression changes capacity壓縮倍率如何改變需求
Compare allocated capacity across compression ratios比較不同倍率下，兩種技術所需的配置容量
OTPeFuse● Current ratio● 目前倍率
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16 Gb SRAM at 1/1000 raw repair data and 100:1 compression requires a 160 kb payload.16 Gb SRAM 在 1/1000 修復資料比例、100:1 壓縮假設下，需 160 kb 有效修復資料。
ALLOCATION COMPARISON配置容量比較
One payload, two allocations相同資料，兩種配置方式
Adjust overhead, reserve and allocation blocks independently to calculate the required bits.分別調整額外開銷、預留空間與配置粒度，查看實際需要配置多少 bits。
| Capacity breakdown容量明細 | OTP | eFuse |
|---|
Both technologies currently have the same allocated capacity.目前兩種技術的配置容量相同。
Method and assumptions計算方法與假設
Data requirement資料需求
SRAM bits = ceil(input capacity × bits per unit)SRAM bits = ⌈輸入容量 × 單位位元數⌉
Raw repair bits = ceil(SRAM bits × repair fraction)原始修復 bits = ⌈SRAM bits × 原始修復資料比例⌉
Payload bits = ceil(raw repair bits / compression ratio)有效資料 bits = ⌈原始修復 bits / 壓縮倍率⌉
Capacity reduction = 1 − payload bits / raw repair bits資料容量減少比例 = 1 − 有效資料 bits / 原始修復 bits
Allocated storage實際配置
Overhead bits = ceil(payload bits × overhead %)開銷 bits = ⌈有效資料 bits × 開銷 %⌉
Reserve bits = ceil((payload bits + overhead bits) × reserve %)預留 bits = ⌈（有效資料 bits + 開銷 bits）× 預留 %⌉
Required bits = payload bits + overhead bits + reserve bits需求 bits = 有效資料 bits + 開銷 bits + 預留 bits
Allocated bits = ceil(required bits / block bits) × block bits配置 bits = ⌈需求 bits / 粒度 bits⌉ × 粒度 bits
Reading this estimate如何閱讀這份試算
Each stage rounds up to whole bits. Bytes round up as ceil(bits / 8). kb / Mb / Gb are decimal; Kib / Mib / Gib are binary, with 1 Gib = 2³⁰ bits. Lowercase b means bits; uppercase B means bytes.⌈ ⌉ 表示向上取整。各階段取整至完整位元；B 顯示容納位元所需的完整位元組（⌈bits / 8⌉）。kb / Mb / Gb 採十進位；Kib / Mib / Gib 採二進位，1 Gib = 2³⁰ bits。小寫 b 是位元、大寫 B 是位元組。
The defaults of 16 Gb, 1/1000 and 100:1 are adjustable planning assumptions, not IP specifications. Both technologies start with 0% overhead, 0% reserve and 1-bit blocks to show payload demand. Overhead can approximate ECC, indexes and control data; use actual encoding and macro capacities for implementation.預設 16 Gb、1/1000 與 100:1 為可調的規劃情境；OTP / eFuse 的開銷、預留均為 0%，粒度為 1 bit，用來呈現純資料需求，並非特定 IP 規格。開銷可概估 ECC、索引與控制資訊；正式配置請依實際編碼及巨集容量設定。
This model compares storage capacity. It excludes bitcell area, peripheral circuits, reliability, power and cost, so results do not establish area advantage, repair success rate or technology superiority. The 16 Gb example is not a validated area crossover.本模型比較儲存容量，未納入每位元面積、周邊電路、可靠度、功耗或成本，因此容量結果不代表面積優勢、修復成功率或特定技術的優劣。16 Gb 是範例容量，並非已驗證的面積交叉點。
Calculations run in your browser. Refreshing restores the defaults; export CSV to save your inputs and results.參數僅於目前瀏覽器頁面中運算；重新整理會回到預設值。匯出 CSV 可保留本次輸入與結果。