4.1.3 · Hardware › Memory Technologies
Ek DRAM bit bas ek tiny bucket of charge hai. Ek capacitor paani (bit) hold karta hai, aur ek transistor wo tap hai jo paani andar ya bahar jane deta hai. Yahi poora idea hai: 1 Transistor + 1 Capacitor = 1T1C . Baki sab kuch (leakage, refresh, sense amps) isliye exist karta hai kyunki bucket choti aur leaky hai.
Itne kam components kyun? Kyunki memory density sabse zaroori cheez hai. Agar har bit mein 6 transistors lagte hain (jaise SRAM mein), toh ek chip mein bahut kam bits fit honge. DRAM speed aur simplicity ki trade-off karke chhote se chhota cell banata hai — ideally 6 F 2 ya 4 F 2 bhi (F = feature size). Kam parts → zyada bits → per gigabyte sasta. Isliye aapki RAM DRAM hai aur CPU cache SRAM.
Ek 1T1C DRAM cell ek bit ko storage capacitor C s par charge ke roop mein store karta hai, jise ek single access transistor (ek NMOS pass transistor) ke through access kiya jaata hai.
Wordline (WL) → transistor ka gate. WL on karna = tap kholna.
Bitline (BL) → source/drain side; read/write ke dauran charge andar/bahar le jaati hai.
Storage node → capacitor ki woh plate jo actually bit hold karti hai.
Logic 1 = capacitor charged (near V D D ); logic 0 = capacitor discharged (near 0 ya V D D /2 ).
Capacitor charge store karta hai:
Q = C s ⋅ V ce l l
Yeh kyun? Capacitor ka defining relation hai Q = C V . Bit hi hai voltage V ce l l on C s . High V = "1", low V = "0".
BL ko desired voltage par drive karo (V D D for 1, 0 for 0).
WL raise karo → transistor conduct karta hai → C s charges/discharges ho ke BL se match karta hai.
WL lower karo → transistor off → charge isolated storage node par trap ho jaata hai.
WL last mein kyun lower karte hain? Jab tap band ho jaata hai, storage node float karta hai, toh charge ke paas jaane ki asaan jagah nahi hoti — ab woh "stored" hai.
Read karna wahan hai jahan physics interesting ho jaati hai. Bitline ki apni badi capacitance C B L hoti hai (yeh ek lamba wire hai jo kai cells ko touch karta hai), aur C B L ≫ C s .
Read karne se pehle, hum bitline ko V D D /2 par precharge karte hain. Phir WL raise karte hain aur C s ko C B L ke saath charge share karne dete hain.
Aise derive kyun karte hain? Transistor bas do capacitors ko connect karta hai; koi charge create ya destroy nahi hota, isliye charge conservation sab kuch de deta hai. Factor C s + C B L C s batata hai ki tiny cell sirf huge bitline ko kuch tens of millivolts se nudge karta hai — isliye humein ek sense amplifier chahiye jo Δ V ko full logic level par amplify kare.
"Destructive" kyun? Charge sharing ke baad, V ce l l apni original value par nahi raha (woh V D D /2 ki taraf dilute ho gaya). Read se bit kharab ho jaati hai, isliye sense amp ko har read ke baad bit write back karni padti hai.
Capacitor junction leakage aur "off" transistor mein subthreshold current ke through charge leak karta hai. Ek "1" dheere dheere "0" ki taraf decay karta hai. Isliye poori chip ko periodically read-and-rewrite karna padta hai — typically har 64 ms (ya high temperature par 32 ms). Woh periodic top-up hi refresh hai, aur isliye isse D ynamic RAM kaha jaata hai.
Leakage discharge ek RC decay follow karta hai:
V ce l l ( t ) = V D D e − t / ( R l e ak C s )
Exponential kyun? Ek R l e ak resistance se discharge hota charged capacitor C d t d V = − V / R l e ak satisfy karta hai, jiska solution upar wala exponential hai. Refresh tab hona chahiye jab V ce l l sense amp ke detectable margin se neeche girne se pehle.
Example 1 — Signal swing. Maano C s = 20 fF , C B L = 200 fF , V D D = 1.2 V , cell mein "1" stored hai (V ce l l = 1.2 ).
Δ V = 20 + 200 20 ( 1.2 − 0.6 ) = 220 20 × 0.6 ≈ 0.0545 V = 54.5 mV
Yeh step kyun? Hum derived swing formula mein values plug in karte hain. V D D /2 se ~55 mV upar ka nudge "1" signal karta hai — chota hai, isliye sense amp chahiye.
Example 2 — Reading a "0". Same caps, V ce l l = 0 .
Δ V = 220 20 ( 0 − 0.6 ) = − 0.0545 V
Negative kyun? Cell precharged bitline ko neeche kheenchta hai, isliye sense amp BL ko V D D /2 ke neeche dekhta hai → "0" detect karta hai. V D D /2 ke around symmetry isliye precharge wahan set hota hai.
Example 3 — Stored charge & refresh margin. C s = 20 fF , V D D = 1.2 V :
Q = C s V = 20 × 1 0 − 15 × 1.2 = 2.4 × 1 0 − 14 C = 24 fC
Care kyun karein? Yeh roughly 1.5 × 1 0 5 electrons hai — leakage se thodi si fraction bhi kho jaaye toh bit flip ho jaata hai, isliye refresh timing isi baat se set hoti hai ki yeh tiny charge kitni tezi se bleed hota hai.
"DRAM read non-destructive hota hai, jaise ek variable read karna."
Kyun sahi lagta hai: Software aur SRAM mein reading kabhi data change nahi karti. Fix: DRAM read charge-sharing hai, jo cell voltage ko dilute karta hai — bit destroy ho jaata hai aur sense amp ko har access ke baad rewrite karna padta hai.
"Bada C B L better hai kyunki bitlines zyada carry karti hain."
Kyun sahi lagta hai: Zyada capacitance matlab zyada signal lagta hai. Fix: Δ V = C s + C B L C s ( … ) — bada C B L signal swing ko shrink karta hai. Designers bitlines ko chota rakhte hain (segmented arrays) taaki C B L , C s ke relative zyada bada na ho.
"Refresh isliye chahiye kyunki transistor kharab hai."
Kyun sahi lagta hai: Hum tap ko leaking ke liye blame karte hain. Fix: Ek near-ideal transistor mein bhi junction & subthreshold leakage hoti hai, aur C s density ke liye deliberately tiny hai. Chota charge + koi bhi leakage = data decay. Refresh ek fundamental consequence hai, koi defect nahi.
"0 V par precharge karo, phir read karo."
Kyun sahi lagta hai: Empty se start karo, reason karna easy hai. Fix: V D D /2 par precharge karo taaki stored "1" upar swing kare aur "0" neeche symmetrically — yeh worst-case swing ko aadha kar deta hai aur differential sense amp ko ek reference ke against compare karne deta hai.
Recall Feynman: explain to a 12-year-old
Socho ki har bit ek choti si paani ki cup hai jisme ek tap laga hai. Bhari cup = 1, khaali cup = 0. Cup read karne ke liye, use ek giant tank se connect karte ho; paani ka level thoda hi badalta hai, isliye tumhe ek super-sensitive gauge chahiye (sense amp) jo bataye "kya yeh bhari thi ya khaali thi?". Lekin tank se connect karne par cup spill ho jaati hai, isliye turant refill karna padta hai. Saath hi, cups dheere dheere leak karti hain, isliye har 64 milliseconds mein ek helper saari bhari cups ko top up karne jaata hai. Woh topping-up hi refresh kehlaata hai, aur isliye isse Dynamic RAM kehte hain.
"One Tap, One Cup — Dilute, Sense, Refill, Refresh."
1 T ransistor (Tap) + 1 C apacitor (Cup)
D ilute (charge share) → S ense (amplify) → R efill (write-back) → R efresh (leaky cup).
1T1C DRAM cell ke do components kya hain? Ek access (NMOS) transistor aur ek storage capacitor C s .
Transistor gate se kaunsi line connect hoti hai? Wordline (WL).
Read/write ke dauran charge andar/bahar kaunsi line le jaati hai? Bitline (BL).
DRAM read ko "destructive" kyun kaha jaata hai? C s aur C B L ke beech charge sharing cell voltage ko dilute kar deta hai, isliye sense amp ko bit rewrite karni padti hai.
Bitline par signal-swing formula likhiye. Δ V = C s + C B L C s ( V ce l l − 2 V D D ) .
Bitline ko V D D /2 par precharge kyun karte hain? Taaki "1" upar swing kare aur "0" neeche symmetrically, differential sense amp ke liye.
Swing formula small Δ V kyun produce karta hai? Kyunki C B L ≫ C s hai, isliye tiny cell sirf big bitline ko tens of mV se nudge karta hai.
DRAM mein "D" kya stand karta hai aur kyun? Dynamic — capacitor leak karta hai, isliye data ko periodically refresh karna padta hai.
Typical refresh interval? ~64 ms (≈32 ms at high temperature).
SRAM ke 6T ki jagah 1T1C kyun? Bahut zyada density (per bit kam transistors) → per bit sasta.
"1" hold karne wale cell par stored charge? Q = C s V D D .
Choti bitline signal ko full logic level par kya amplify karta hai? Sense amplifier.
SRAM 6T cell — fast, non-destructive, low-density counterpart.
Sense Amplifiers — zaroorat hai kyunki Δ V tiny hota hai.
DRAM Refresh — capacitor leakage ka direct consequence.
Charge Sharing — destructive reads ke peeche ki physics.
Bitline and Wordline Architecture — cells arrays mein kaise wired hote hain.
Capacitor Q=CV — woh fundamental relation jo bit store karta hai.
because CBL greater than Cs
Destructive read needs rewrite