3.2.5 · Hardware › CMOS Circuit Design
Ek CMOS inverter basically ek voltage-controlled switch pair hai. Jab aap input voltage ko 0 se V D D tak sweep karte ho, toh pull-up (PMOS) aur pull-down (NMOS) transistors apne kaam badal lete hain . VTC bas V o u t vs V in ki woh picture hai jo is hand-off ko capture karti hai. Digital logic ka poora magic us hand-off ki sharpness mein rehta hai — ek sharp transition matlab clean 0s aur 1s jo noise ke baad bhi theek rahein.
Voltage Transfer Characteristic (VTC) woh plot hai jo ek gate (canonically CMOS inverter) ke liye steady-state output voltage V o u t ko input voltage V in ke against dikhata hai, jahan dono transistors ki currents equal rakhni padti hain (load ki taraf koi current nahi jaati, DC condition).
Jo defining physical constraint hai woh hai Kirchhoff's Current Law output node par:
I D N = I D P
NMOS drain current (neeche pull karna) ko PMOS drain current (upar pull karna) ke barabar hona chahiye, kyunki steady state mein koi net current output node ko charge nahi karti.
Currents equal KYU hoti hain? Output capacitor charging khatam kar chuka hai, toh C d V / d t = 0 . Current ka ek hi raasta hai — dono transistors ke through series mein V D D se GND tak. Series → same current. Dono ko equal karke aur V o u t solve karke hame curve milta hai.
Define karo, NMOS ke liye: V GS n = V in , V D S n = V o u t , threshold V T n > 0 .
PMOS ke liye: V S Gp = V D D − V in , V S D p = V D D − V o u t , threshold ∣ V T p ∣ .
Har transistor cutoff / saturation / triode mein ho sakta hai:
Region
V in range
NMOS
PMOS
V o u t
A
0 ≤ V in < V T n
cutoff
triode
≈ V D D
B
badhta hua
saturation
triode
high, girta hua
C
V M ke paas
saturation
saturation
steep drop
D
badhta hua
triode
saturation
low, girta hua
E
$V_{in}>V_{DD}-
V_{Tp}
$
triode
Region C woh near-vertical hissa hai. Dono devices saturation mein current sources ki tarah kaam karte hain, toh V in mein thodi si change ke liye V o u t mein bahut badi change chahiye taaki currents balanced rahein. Woh steep slope (gain ≫ 1 ) exactly wahi cheez hai jo weak logic levels ko regenerate karti hai.
Har step kyun?
V M par dono saturated hain → kyunki V in = V o u t ka matlab hai V D S n = V o u t > V in − V T n = V GS n − V T n ; saturation inequality hold karta hai. → square-law equations use karo.
Currents equal set karo → output par KCL (defining constraint).
Square roots lo → dono sides perfect squares hain; lena V M mein linearize kar deta hai.
β n ( V M − V T n ) = β p ( V D D − V M − ∣ V T p ∣ )
β n se divide karo, V M collect karo → boxed formula milta hai.
Symmetric inverter. Maano V T n = ∣ V T p ∣ aur hum chahte hain V M = V D D /2 (equal noise margins). r solve karo:
V M = V D D /2 set karo jab V T n = ∣ V T p ∣ = V T :
2 V D D = 1 + r V T + r ( V D D − V T ) ⇒ r = 1
Yeh step kyun? V M = V D D /2 plug karo aur simplify karo; dono sides symmetrically balance hone ka ek hi tarika hai woh hai r = 1 , matlab β n = β p . Kyunki k n ′ ≈ 2 – 3 k p ′ (electrons holes se faster hote hain), hame chahiye W p ≈ 2 – 3 W n taaki PMOS ko NMOS jitna "strong" banaya ja sake.
Numerical V M . V D D = 1.8 V, V T n = 0.4 V, ∣ V T p ∣ = 0.4 V, r = β p / β n = 1.5 .
V M = 1 + 1.5 0.4 + 1.5 ( 1.8 − 0.4 ) = 2.5 0.4 + 2.1 = 2.5 2.5 = 1.0 V
Yeh step kyun? Bada r (stronger PMOS) V M ko upar V D D ki taraf kheenchta hai, kyunki ek strong pull-up tab tak tug-of-war jeetta hai jab tak V in high ho jaaye. Yahan V M = 1.0 > 0.9 V shift confirm karta hai.
Unity-gain KYU? Jab ∣ slope ∣ > 1 hota hai toh gate noise ko attenuate karta hai (output swing, input wiggle se chhoti → signal restore ho jaata hai). − 1 slope ke points us boundary ko mark karte hain jo "clean up karna" aur "amplify karna" ke beech hai. Jo bhi us range ke andar hai woh safely transition mein hai aur kisi rail ki taraf push ho jaata hai. Bade noise margins = robust logic.
"V M hamesha V D D /2 hota hai."
Kyun sahi lagta hai: textbooks symmetric inverter draw karte hain, aur 0/1 "should be" symmetric.
Sach: V M = V D D /2 tabhi hota hai jab β n = β p (aur V T n = ∣ V T p ∣ ). Real PMOS weak hota hai, toh equal W ke saath inverter ka r < 1 hota hai aur V M < V D D /2 . Usse center karne ke liye PMOS ko upsize karna padega.
"Region C mein output undefined/floating hai."
Kyun sahi lagta hai: curve almost vertical hai, unstable lagti hai.
Fix: yeh bilkul defined hai — dono transistors saturated current sources hain aur current balance V o u t ko pin karta hai. Yeh bas bahut sensitive hai (high gain), floating nahi. Is region mein ek short-circuit current spike bhi aati hai (dono devices ON hain) — jo ek real power cost hai.
"Slope ki sign noise margin ke liye matter nahi karti."
Kyun sahi lagta hai: log "gain > 1" likhte hain.
Fix: ek inverter ka gain negative hota hai; noise-margin condition hai d V in d V o u t = − 1 , matlab magnitude 1. + 1 use karne par curve par kuch nahi milta.
Recall Test yourself (hidden)
VTC ka har point kaunsi constraint define karta hai? ⇒ I D N = I D P (KCL).
Kis region mein DONO transistors saturated hote hain? ⇒ transition region (C), V M par/ke paas.
V M ko V D D ki taraf upar kya shift karta hai? ⇒ stronger PMOS (bada r = β p / β n ).
V I L , V I H define karo. ⇒ woh points jahan slope = − 1 ho.
Recall Feynman: explain to a 12-year-old
Socho ek see-saw hai jisme ek taraf "0-kid" aur doosri taraf "1-kid" hai, dono output ko push kar rahe hain. Tum ek lever (input) ko left se right slide karte ho. Pehle 1-kid bilkul jeet jaata hai aur output HIGH pe stuck rehta hai. Jaise-jaise tum slide karte ho, 0-kid stronger hota jaata hai, aur ek special jagah par dono exactly tie ho jaate hain — ek tiny sa dhakka output ko HIGH se LOW super fast tip kar deta hai. Woh fast tipping hi reason hai ki computers ek clean "1" ko ek clean "0" se tab bhi pehchaan lete hain jab wire thoda noisy ho. Dono kids ko equally strong banana tipping point ko exactly beech mein rakhta hai — sabse fair aur safe jagah.
"CS-SS-ST-SS-CS reads the same backwards" — jab V in badhta hai toh (NMOS state, PMOS state) ki region sequence symmetric hoti hai: NMOS ke liye C utoff→S at→S at(dono)→T riode→cut, PMOS ko mirror karta hai. Middle = S aturation S aturation = steep switching.
Aur: "Weaker P → V_M lower; boost W_p to center."
CMOS inverter VTC ke har point par kaunsi DC constraint hold karti hai? I D N = I D P (KCL: series transistors, output cap fully charged).
V M par simultaneously kaun si do device states hoti hain?NMOS aur PMOS dono saturation mein.
r = β p / β n ke saath switching threshold V M ka formula?V M = 1 + r V T n + r ( V D D − ∣ V T p ∣ ) .
Symmetric inverter ke liye r par kya condition hoti hai (V M = V D D /2 , equal V T )? r = 1 , matlab β n = β p .
Centered V M ke liye W p > W n kyun hona chahiye? Hole mobility < electron mobility (k p ′ < k n ′ ), toh PMOS width ko strength equal karne ke liye badhana padta hai.
VTC par V I L aur V I H kaise define hote hain? Woh points jahan d V o u t / d V in = − 1 ho (unity-gain points).
VTC levels ke terms mein noise margins? N M H = V O H − V I H , N M L = V I L − V O L .
Agar PMOS ko stronger banaya jaaye (r badhta hai) toh V M ka kya hoga? V M badh jaata hai (V D D ki taraf move karta hai).
Transition region itna steep kyun hota hai? Dono devices saturated current sources ki tarah kaam karte hain; chhota Δ V in currents equal rakhne ke liye bada Δ V o u t maangta hai → high gain.
Transition region mein operate karne ki extra cost kya hai? Short-circuit (crowbar) current, kyunki dono transistors simultaneously conduct karte hain.
CMOS Inverter — woh device jiska VTC yeh hai.
MOSFET IV Characteristics — triode/saturation square-law equations jo derivation mein use hui hain.
Noise Margins — V I L , V I H , V O H , V O L se compute hote hain.
Static CMOS Logic Gates — VTC effective W/L ke zariye NAND/NOR tak generalize hoti hai.
Short-circuit Power Dissipation — both-ON transition region se aati hai.
Propagation Delay — slope/gain switching speed se relate karta hai.
Ratio r = sqrt of beta_p/beta_n