WHY 50%? Because the "logic threshold" (the voltage the next gate decides is a 1 vs a 0) sits
roughly halfway. Measuring at 50% is a consistent, technology-independent reference point.
Note the difference from transition (rise/fall) time: that is measured between the 10% and
90% points of one signal and describes how sharp the edge is — a different quantity.
A CMOS gate output drives a load capacitanceCL (wire capacitance + input capacitances of the
gates it feeds). When the output switches, a transistor must move charge onto or off that capacitor
through its channel resistanceR.
Model the switching transistor as a resistor R charging capacitor CL from supply VDD.
By KCL, the current charging the cap equals the current through the resistor:
CLdtdV=RVDD−V
Why this step? The same current flows through R and into CL (series path), and a capacitor's
current is CdV/dt by definition.
Separate variables and integrate from V=0:
VDD−VdV=RCLdt⇒−ln(VDD−V)=RCLt+k
Applying V(0)=0 gives k=−lnVDD, so:
V(t)=VDD(1−e−t/RCL)
Why this step? This is the standard exponential charging curve — the output does not jump; it
eases up to VDD with time constant τ=RCL.
Now find tpLH: the time to reach the 50% point V=VDD/2:
2VDD=VDD(1−e−t/RCL)⇒e−t/RCL=21⇒t=RCLln2
Consequences you can now predict (Forecast-then-Verify):
Drive more gates → larger CL → more delay. (Fan-out costs time.)
Use a bigger/stronger transistor → smaller R → less delay (but bigger input cap for the previous gate!).
What is the propagation delay in terms of R and C_L?
t_pd ≈ 0.69·R·C_L (since ln2 ≈ 0.693).
Why does delay increase with fan-out?
More gates driven = larger load capacitance C_L, and delay ∝ C_L.
How do delays combine through a chain of gates?
They ADD along a path; the longest (critical) path sets the max clock frequency f_max = 1/t_critical.
Difference between propagation delay and rise time?
Propagation delay is 50%(in)→50%(out) between two signals; rise/fall time is 10%→90% of ONE signal (edge sharpness).
What is a static hazard (glitch)?
A brief false output pulse caused by unequal path delays making inputs to a gate arrive at different times.
Does making the driving transistor bigger always speed the circuit?
No — smaller R helps, but its larger input capacitance slows the previous stage; it's a trade-off.
Recall Feynman: explain to a 12-year-old
Imagine filling a cup with water through a thin straw. Even if you decide instantly to fill it,
the water still takes a moment to rise to the halfway line. A logic gate is like that: the "water"
is electric charge, the "cup" is a tiny capacitor, and the "straw" is the transistor's resistance.
The gate knows the answer immediately, but it takes a little time to actually push the voltage up
to where the next gate can read it. Chain many cups and straws together, and the waiting times add
up — that's why fast computers work hard to keep those straws fat and the cups small.
Dekho, logic gate ka truth table batata hai output kya hoga, lekin ye kabhi nahi batata ki output
kab milega. Real hardware me har gate transistors se bana hota hai, aur transistor ke through ek
chhoti capacitor charge karni padti hai. Capacitor ko charge hone me time lagta hai — bilkul jaise
patli straw se glass bharne me time lagta hai. Isi waiting time ko hum propagation delay (tpd)
kehte hain, aur ye input ke 50% se output ke 50% tak measure hota hai.
Formula derive karna simple hai: current R aur CL dono me se same behta hai, to
CLdV/dt=(VDD−V)/R. Isko solve karo to output exponential tarah upar chadhta hai:
V(t)=VDD(1−e−t/RCL). Jab output 50% pe pahunchta hai tab t=RCLln2, yaani
tpd≈0.69RCL. Matlab jitni zyada load capacitance (zyada gates drive karo, lambe wires),
utni zyada delay. Ye baat yaad rakho — fan-out free nahi hota, time lagta hai.
Aur ek important cheez: agar gates chain me lage hain, to delays add hoti hain, average nahi.
Sabse lamba path (critical path) hi decide karta hai ki chip kitni fast clock chala sakti hai:
fmax=1/tcritical. Jab do inputs alag-alag time pe pahunchte hain (unequal delay), to ek
chhota galat pulse aa sakta hai jise glitch ya static hazard kehte hain. Isliye engineers
transistor sizing aur wiring bahut dhyan se karte hain — logic sahi ho ye kaafi nahi, timing bhi
sahi honi chahiye.