Static vs dynamic power dissipation: Worked examples

Prerequisites we lean on: CMOS Inverter operation, Threshold voltage and subthreshold conduction, Voltage scaling and Dennard scaling, Power gating and clock gating, Capacitance in interconnects. Parent: Static vs dynamic power (topic).

The scenario matrix

Before solving anything, let's map the whole territory. Each cell is a class of problem this topic can generate. Every worked example below is tagged with the cell it fills.

Cell	What makes it distinct	Filled by
A. Plain plug-in	All four dynamic quantities given, just substitute	Ex 1
B. Scaling knob	One variable changes ( $V_{D D}$ , $f$ , $α$ ) — predict the ratio	Ex 2
C. Degenerate / zero input	$α = 0$ (idle), or $f = 0$ — what survives?	Ex 3
D. Static leakage aggregation	Nanoamps × millions of gates → watts	Ex 4
E. Crossover point	Where does static equal dynamic? Solve for the boundary	Ex 5
F. Limiting behaviour	Exponential leakage as $V_{T} \to$ small; what dominates	Ex 6
G. Real-world word problem	Battery life of a phone block	Ex 7
H. Exam twist	Trap that mixes energy-vs-power, or resistance	Ex 8

Cell A — Plain plug-in

Worked example Example 1 — Dynamic power of a logic block

An adder toggles with $α = 0.15$ , drives $C_{L} = 2 pF$ , runs at $V_{D D} = 1.2 V$ , clock $f = 1 GHz$ . Find $P_{switching}$ .

Forecast: picofarads and gigahertz look tiny and huge — guess whether the answer lands in microwatts or milliwatts before reading on.

Write the formula. $P = α C_{L} V_{D D}^{2} f$ . Why this step? Every dynamic-switching problem starts here; we're just filling four slots.
Substitute numbers with their powers of ten. $P = 0.15 \times (2 \times 1 0^{- 12}) \times (1.2)^{2} \times (1 \times 1 0^{9})$ Why this step? Keeping powers of ten separate stops arithmetic slips.
Group the constants: $0.15 \times 2 \times 1.44 = 0.432$ , and the powers $1 0^{- 12} \times 1 0^{9} = 1 0^{- 3}$ . Why this step? Do the "nice" numbers and the "scale" numbers separately.
Combine: $P = 0.432 \times 1 0^{- 3} W = 0.432 mW$ .

Verify: Units — $F \cdot V^{2} \cdot Hz = (C/V) \cdot V^{2} \cdot (1/ s) = C \cdot V / s = J/s = W$ . ✓ Landed in milliwatts, matching intuition (bigger cap and voltage than the parent's Example 1).

Cell B — Scaling knob (predict the ratio)

Worked example Example 2 — Three knobs turned at once

Starting from Example 1, an engineer doubles $α$ , halves $V_{D D}$ , and triples $f$ . By what factor does $P_{switching}$ change?

Forecast: guess the net factor. Many people say " $2 \times \frac{1}{2} \times 3 = 3 \times$ " — is that right?

Identify how each knob enters. $α$ is linear, $V_{D D}$ is squared, $f$ is linear. Why this step? The exponent decides how a change propagates — the whole trap lives here.
Turn each change into a multiplier.
- $α$ : $\times 2$
- $V_{D D}$ : halved, but squared $\Rightarrow (\frac{1}{2})^{2} = \times \frac{1}{4}$
- $f$ : $\times 3$ Why this step? Halving the voltage is not halving the power; the square bites here.
Multiply the factors: $2 \times \frac{1}{4} \times 3 = \frac{6}{4} = 1.5$ . Why this step? Power is a product of these terms, so ratios of terms just multiply.
State the result: new power is $1.5 \times$ the old, i.e. $0.432 mW \times 1.5 = 0.648 mW$ .

Verify: Recompute from scratch: $α = 0.3, V_{D D} = 0.6, f = 3 \times 1 0^{9}$ . $P = 0.3 \times 2 \times 1 0^{- 12} \times 0.36 \times 3 \times 1 0^{9} = 0.648 mW$ . ✓ The naive " $3 \times$ " answer was wrong because it forgot to square the voltage change.

Figure — Static vs dynamic power dissipation

Look at the chalkboard bars: the pale-yellow $V_{D D}^{2}$ term shrinks four-fold while the linear knobs push the other way — the net is a modest $1.5 \times$ , not the $3 \times$ you'd guess by treating voltage linearly.

Cell C — Degenerate / zero input

Worked example Example 3 — The idle chip (what survives when nothing switches?)

A block sits idle: activity $α = 0$ (no node toggles), but the power is still on, $V_{D D} = 1.0 V$ , and each of its $5 \times 1 0^{6}$ gates leaks $I_{leak} = 20 nA$ . What total power does it draw?

Forecast: with $α = 0$ , is the power exactly zero?

Kill the dynamic term. $P_{switching} = α C_{L} V_{D D}^{2} f$ with $α = 0$ gives $P_{switching} = 0$ . Why this step? Dynamic power is paid only when the output changes — no change, no charge sloshing, no cost.
Short-circuit term also vanishes. $P_{sc} \approx α f \overset{ˉ}{I}_{sc} V_{D D} t_{sc}$ also has $α = 0 \Rightarrow 0$ . Why this step? Crowbar current only flows during transitions; idle means no transitions.
Only leakage remains. Total leakage current $I = 5 \times 1 0^{6} \times 20 \times 1 0^{- 9} = 0.1 A$ . Why this step? Leakage flows always, independent of switching — this is the whole point of "static".
Compute static power: $P_{static} = I V_{D D} = 0.1 \times 1.0 = 0.1 W = 100 mW$ .

Verify: Sanity — an idle block drawing $100 mW$ is exactly why designers add power gating to cut $V_{D D}$ off entirely. If we had also set $V_{D D} = 0$ (true power-off), even leakage would vanish since $P_{static} = I_{leak} V_{D D} \to 0$ . ✓ "Idle" is not "zero power".

Cell D — Static leakage aggregation

Cell E — Crossover point (solve for the boundary)

Worked example Example 5 — Where does static equal dynamic?

A gate has $C_{L} = 1.5 pF$ , $V_{D D} = 1.0 V$ , $α = 0.1$ , and leaks $I_{leak} = 8 nA$ . Below what clock frequency does static power exceed dynamic?

Forecast: below or above ~1 GHz? Guess first.

Set the two powers equal at the boundary. $α C_{L} V_{D D}^{2} f = I_{leak} V_{D D}$ Why this step? The crossover is by definition where dynamic = static; solving the equation finds that frequency.
Cancel one $V_{D D}$ and solve for $f$ . $f = \frac{I _{leak}}{α C _{L} V _{D D}}$ Why this step? Divide both sides by $α C_{L} V_{D D}^{2}$ ; one $V_{D D}$ survives in the denominator.
Substitute: $f = \frac{8 \times 1 0 ^{- 9}}{0.1 \times 1.5 \times 1 0 ^{- 12} \times 1.0}$ . Why this step? Every symbol now has a number; just plug in.
Evaluate: denominator $= 0.15 \times 1 0^{- 12} = 1.5 \times 1 0^{- 13}$ ; so $f = \frac{8 \times 1 0 ^{- 9}}{1.5 \times 1 0 ^{- 13}} \approx 5.33 \times 1 0^{4} Hz \approx 53.3 kHz$ .

Verify: Below $\approx 53 kHz$ , static wins; above it, dynamic wins. Sanity check by plugging $f = 53.3 kHz$ back: dynamic $= 0.1 \times 1.5 \times 1 0^{- 12} \times 1 \times 5.33 \times 1 0^{4} = 8 \times 1 0^{- 9} W$ ; static $= 8 \times 1 0^{- 9} \times 1.0 = 8 \times 1 0^{- 9} W$ . Equal ✓. A high-performance chip at GHz is deep in dynamic-dominated territory; a sleepy sensor at kHz is leakage-dominated.

The crossing point on the chalkboard plot is where the rising blue dynamic line meets the flat pink static line — everything left of it is a leakage problem, everything right is a switching problem.

Cell F — Limiting behaviour (exponential leakage)

Worked example Example 6 — Dropping the threshold voltage

Subthreshold leakage follows $I_{s u b} \propto e^{- V_{T} / (n V_{t h})}$ with $n = 1.5$ and thermal voltage $V_{t h} = 26 mV$ at 300 K. A design lowers $V_{T}$ from $0.35 V$ to $0.25 V$ for speed. By what factor does leakage grow?

Forecast: dropping $V_{T}$ by $0.1 V$ — does leakage roughly double, or explode by an order of magnitude?

Take the ratio of the two leakages. The proportionality constant cancels: $\frac{I _{s u b}^{new}}{I _{s u b}^{old}} = \frac{e ^{- V_{T}^{new} / (n V_{t h})}}{e ^{- V_{T}^{old} / (n V_{t h})}} = e^{(V_{T}^{old} - V_{T}^{new}) / (n V_{t h})}$ Why this step? We don't know the constant, but ratios erase it — leaving only the change in $V_{T}$ .
Insert the numbers into the exponent. $\frac{Δ V _{T}}{n V _{t h}} = \frac{0.35 - 0.25}{1.5 \times 0.026} = \frac{0.10}{0.039} \approx 2.564$ Why this step? The exponential's argument is the only thing that matters; compute it carefully.
Exponentiate: ratio $= e^{2.564} \approx 12.99$ . Why this step? $e^{2.564}$ turns the linear voltage drop into a multiplicative blow-up.

Verify: A mere $100 mV$ cut in $V_{T}$ multiplied leakage by about $13 \times$ . This exponential sensitivity (not linear!) is exactly why aggressive threshold scaling made static power a monster. Limit check: as $V_{T} \to 0$ , the exponent $\to$ (positive, large) and $I_{s u b} \to$ huge — the transistor stops being a good "off" switch. ✓

Cell G — Real-world word problem

Worked example Example 7 — Phone screen driver, battery budget

A display block draws $P_{active} = 180 mW$ while animating (10% of the day) and $P_{idle} = 20 mW$ leakage the rest of the time (90%). The battery holds $10 Wh$ . If this block were the only load, how many hours until empty?

Forecast: closer to 50 h or 250 h? Guess.

Average the power over a day (weighted by time fraction). $P_{avg} = 0.10 \times 180 + 0.90 \times 20 [mW]$ Why this step? Energy = power × time; over a mixed day the time-weighted average power sets the drain.
Compute: $0.10 \times 180 = 18$ , $0.90 \times 20 = 18$ , sum $= 36 mW = 0.036 W$ . Why this step? Notice leakage (idle) contributes as much as the active bursts — the takeaway.
Battery life $= \frac{energy}{power} = \frac{10 Wh}{0.036 W}$ . Why this step? Watt-hours ÷ watts = hours; the units cancel cleanly.
Evaluate: $10/0.036 \approx 277.8 h$ .

Verify: $\approx 278$ hours. Units: $Wh / W = h$ ✓. Note the punchline — half the drain is idle leakage, so power-gating the idle state to near-zero would roughly halve $P_{avg}$ and nearly double battery life.

Cell H — Exam twist (energy vs power, and resistance)

Worked example Example 8 — The resistance trap

A cap $C_{L} = 0.5 pF$ charges to $V_{D D} = 1.0 V$ then discharges, once. (a) How much energy is dissipated as heat over the full up-then-down cycle? (b) If the PMOS on-resistance were doubled, how does that heat change?

Forecast: most students think doubling $R$ doubles the heat. True or false?

Energy dissipated charging $= \frac{1}{2} C_{L} V_{D D}^{2}$ (half the supply energy; the other half is stored on the cap). Why this step? From the parent derivation — the "½ each way" magic.
Energy dissipated discharging $= \frac{1}{2} C_{L} V_{D D}^{2}$ (the stored half is now dumped through the NMOS). Why this step? Discharge burns exactly what charge stored.
Full-cycle heat $= \frac{1}{2} C_{L} V_{D D}^{2} + \frac{1}{2} C_{L} V_{D D}^{2} = C_{L} V_{D D}^{2}$ . $= 0.5 \times 1 0^{- 12} \times (1.0)^{2} = 5 \times 1 0^{- 13} J = 0.5 pJ$ Why this step? The two halves add to the full supply energy per cycle.
(b) Effect of doubling $R$ : none. Total heat $= C_{L} V_{D D}^{2}$ contains no $R$ . Why this step? Bigger $R$ means smaller current for longer time; the product $\int i^{2} R d t$ stays $\frac{1}{2} C_{L} V_{D D}^{2}$ . $R$ sets speed (the time constant $τ = R C$ ), not energy.

Verify: Full-cycle heat $= 0.5 pJ$ , independent of $R$ ✓. The trap ("double $R$ → double heat") confuses instantaneous power $I^{2} R$ with total energy; over the transient, current falls exactly enough to cancel the larger $R$ . This matches [!mistake] #1 in the parent note.

Recall — did every cell stick?

Recall Quick self-test across the matrix

With $α = 0$ , what power remains? ::: Only static (leakage) $P_{static} = I_{leak} V_{D D}$ ; both dynamic terms vanish. Halving $V_{D D}$ changes switching power by what factor? ::: $\times \frac{1}{4}$ (quadratic in $V_{D D}$ ). $5 nA$ leakage across $2 \times 1 0^{8}$ gates at $0.9 V$ gives? ::: $0.9 W$ . Crossover frequency formula (static = dynamic)? ::: $f = I_{leak} / (α C_{L} V_{D D})$ . Dropping $V_{T}$ by $0.1 V$ ( $n = 1.5$ , $V_{t h} = 26 mV$ ) multiplies leakage by about? ::: $\approx 13 \times$ (exponential). Does doubling transistor $R$ change per-cycle switching energy? ::: No — energy is $C_{L} V_{D D}^{2}$ , $R$ only sets speed.