WHY it exists: every wire + gate input has capacitance. To make the output go HIGH you must dump charge onto CL; to make it LOW you drain that charge to ground. Both cost energy.
Derivation — energy per rising transition.
When the PMOS turns on, the supply VDD charges CL from 0 to VDD.
Energy drawn from the supply:
Esupply=∫0∞VDDi(t)dt=VDD∫0∞CLdtdVoutdt=VDDCL∫0VDDdVout=CLVDD2
Why this step? Current into the cap is i=CLdVout/dt; substituting turns a time integral into a voltage integral, and the voltage limits are 0→VDD.
Energy actually stored on the capacitor at the end:
Estored=∫0VDDVoutidt=CL∫0VDDVoutdVout=21CLVDD2
So half the supply energy is stored, the other half 21CLVDD2 is burned as heat in the PMOS resistance. On the falling edge the stored 21CLVDD2 is dumped through the NMOS as heat.
From energy to power. If the gate makes f complete cycles per second:
Pswitching=CLVDD2f
Real circuits don't toggle every clock. Introduce the activity factorα = probability a node makes a rising transition per clock:
WHY: the input ramps through the region VTn<Vin<VDD−∣VTp∣ where neither transistor is fully off. A pulse of current Isc flows straight through.
Psc≈αfIˉscVDDtsc
where tsc is proportional to input rise/fall time. HOW to reduce: keep edges sharp (fast transitions) — slow inputs widen the crowbar window. Usually 5–15% of dynamic power.
Subthreshold leakage — a transistor "off" (VGS<VT) still passes a small current that grows exponentially as VT shrinks:
Isub∝e(VGS−VT)/(nVth),Vth=qkT≈26 mV at 300KWHY it matters: as chips scaled, engineers lowered VT for speed — which blew up Isub exponentially.
Gate-oxide tunneling — electrons quantum-tunnel through ultra-thin gate oxide.
High-VT transistors, power gating, body biasing, high-k/thicker oxide.
Why did static power become dominant in nanometer nodes?
Scaling lowered VT to keep speed, and subthreshold leakage rises exponentially with lower VT.
Recall Feynman: explain to a 12-year-old
Imagine a water bucket (the wire) you fill and empty over and over. Dynamic power is the effort of filling and dumping it — you only work when you actually pour. The higher you lift the water (voltage), the much harder it is — twice the height is four times the work. Static power is a tiny hole in the bucket that always drips, even when you're resting. In old buckets the hole was tiny; in new super-thin buckets the hole leaks a lot, so even doing nothing wastes water.
Dekho, CMOS gate mein power do tarah se waste hoti hai. Pehla hai dynamic power — ye tabhi lagti hai jab output badalta hai. Jab output HIGH hota hai to load capacitor CL ko VDD tak charge karna padta hai, aur jab LOW hota hai to discharge. Iska formula hai P=αCLVDD2f. Yaha α activity factor hai (kitni baar switch hota hai), f frequency hai. Sabse important baat: VDDsquare mein aata hai, isliye voltage aadha karo to dynamic power ek-chauthai ho jaati hai — ye hi low-power design ka number one trick hai.
Ek mazedaar physics point: capacitor ko charge karne mein supply se CLVDD2 energy nikalti hai, lekin usme se aadhi (21CLVDD2) capacitor pe store hoti hai aur aadhi transistor mein heat ban jaati hai. Discharge pe wo stored aadhi bhi heat ban jaati hai. To poore cycle mein CLVDD2 heat — aur ye transistor ki resistance pe depend nahi karti! Resistance sirf speed decide karti hai, total energy nahi.
Dusra hai static power ya leakage — ye tab bhi behti hai jab kuch switch nahi ho raha, chip idle hai. Transistor "off" hone par bhi thoda subthreshold current leak karta hai, aur ye current VT kam karne par exponentially badhta hai. Purane bade chips mein leakage ignorable thi, par aaj ke nanometer chips mein VT chhota hota hai (speed ke liye), isliye leakage bahut badh gayi — kabhi kabhi total power ka aadha! Isliye phones mein "power gating" use hoti hai — idle block ko poori tarah supply se kaat do.
Yaad rakho: dynamic = "kaam karte waqt paseena" (switching pe lagti hai, frequency ke saath badhti hai), static = "idle bijli ka bill" (hamesha behti hai). Exam aur real design dono mein ye distinction bahut poocha jaata hai.