Horace Hub · the bottom of the chain

Why a flip makes heat: following the electrons

Your note of Mon 26 Oct asked why a transistor flip releases heat, and to see the electrons move. In short: every flip charges or drains a tiny capacitor through a transistor, and the carriers crossing that transistor hand their energy to the silicon lattice. Each flip leaves ½CV² of heat, whatever the transistor.

Built (week 44) on your earlier Horace work: the Hamming ladder (E60), the one-gate runs (E61), the flow-M flip (Q124), ASAP7 in SPICE (Q126) and the transistor ledger (Q128). Every number names its file under Sources. Voltages are the minion rail's 0.517 V unless marked; SPICE numbers come from ASAP7, an open 7 nm kit, not this chip's own devices.

1The supply pays CV²; half is heat at once, half on the way down

Charging a node: the supply pays the whole square, the node keeps only the lower triangle

Your open netlist's average toggling net, C = 2.233 fF, charged to 0.517 V.

all charge leaves the supply at 0.517 V heat in the pull-up ½CV² = 0.298 fJ stored on the node ½CV² = 0.298 fJ (heat on the fall) 0.517 V 0 V 0 Q = CV = 7,206 electrons charge pushed onto the node → voltage whole square: QV = CV² = 0.597 fJ drawn node voltage v = q / C one slice of charge
Follow the marked slice: that bit of charge left the supply at 0.517 V but landed on the node while the node sat at only 0.06 V, so almost all of its energy, the part above the diagonal (V − v), was lost on the way as heat. Charge that arrives late lands near 0.517 V and is mostly kept. The two triangles come out equal. On the fall nothing is drawn from the supply; the stored triangle drains to ground as heat. Textbook: Rabaey, ch. 5; Weste and Harris, ch. 5. Numbers: Q128 §2.

A gate's output is a node with a small capacitance: the next gates' inputs plus the wire. Raising it from 0 to V takes charge Q = CV, and every bit of that charge leaves the supply at the full V, so the supply pays QV = CV². The node keeps only ½CV², because the first charge went on while it was still near 0 V. The other half is lost in the pull-up transistor on the way. When the node falls, its stored ½CV² drains to ground through the pull-down and becomes heat too. So every flip, up or down, leaves ½CV². The split doesn't depend on the transistor: a weaker one charges the node more slowly and loses exactly the same.

Where it turns into heat: the field across the conducting channel speeds the carriers up, and they give that energy to the silicon lattice by scattering off it, setting the lattice vibrating (phonons). That's Joule heating. In a 7 nm transistor the channel is so short that many carriers cross it with few collisions and shed most of their energy at the drain end, mostly as optical phonons, which barely move and decay into the acoustic phonons that carry heat away (Pop, Sinha and Goodson, 2006). The wire's resistance takes its share too, in proportion to its share of the resistance on the path.

2The inverter, electron by electron

New to electronics? Section 7, the inverter in depth, starts from zero: the inverter's history, the five ideas of electricity you need, the transistor as a switch, and twelve step-by-step demos. Then come back here.

Each edge moves 7,206 electrons and leaves 0.298 fJ of heat; only the rise draws from the supply

A CMOS inverter at 0.517 V driving your open netlist's average net. Each dot is a packet of about 300 electrons.

The circuit

VDD rail · 0.517 V ground · 0 V PMOS pull-up S D NMOS pull-down D S input in = 1 output node out = 0.000 V load C 2.233 fF gates, wire supply lifts e⁻ 0.517 eV +

The same edge as electron energy

electron energy (higher is up) ground, 0 V VDD rail, 0.517 V 0.517 eV apart NMOS PMOS supply node

For an electron, ground is uphill of the rail. The node's level slides down as it charges. Early electrons fall far, late ones barely: on average half the 0.517 eV, so ½CV² per edge.

Drawn from the supply
0.000 fJ
Stored on the capacitor
0.000 fJ
Released as heat (PMOS · NMOS)
0.000 fJ
0 of 7,206 electrons moved this edge0 electrons drawn from the supply this cycleheat: PMOS 0.000 fJ · NMOS 0.000 fJ

The output is low. Press “Input 1→0” to raise it, or Play to watch both edges.

electrons (a packet of ~300)holes in the PMOS channela carrier heating the latticedrawnstoredheat, PMOSheat, NMOS
Meters are per cycle, reset at each rise; bars run to CV² = 0.597 fJ. Load: Q128 §2's average toggling signal net, 2.233 fF (0.485 fF transistor gates, 0.892 fF wire inside the cells, 0.856 fF wire between them). The node charges with an RC time constant, slowed tens of billions of times (the real edge takes about 20 ps); the flashes cluster at each channel's drain end. The inverter's own nodes add 0.045 fJ a cycle, 7.4% of the total (section 3).

When the input falls, the p-type transistor (PMOS) conducts and the output rises. Electrons leave the node's top plate, cross to the VDD rail and enter the supply, which lifts each one by 0.517 eV and pushes one out at ground, into the capacitor's bottom plate. When the input rises, the n-type transistor (NMOS) conducts and the output falls: electrons come up from ground through the NMOS and refill the top plate. Over a full cycle electrons go ground → node → rail, the opposite way to the current arrows, as your 26 Oct fix says.

Inside the PMOS the current is carried by holes, the gaps where electrons are missing, so there the hollow dots move down: the same current, as valence electrons hop up into the gaps. All the dots move together because a series loop carries one current. Real electrons drift slowly; what travels fast is the push, the way a full pipe delivers water at the far end the moment you push at the near end.

3What else heats it: three smaller sources ride on top

At 0.517 V the load's CV² is 92.6% of an inverter's switch; the inverter's own share is 7.4%

One INVx1 driving a 2.105 fF net, ASAP7 in SPICE at 0.517 V, per cycle (rise plus fall): 0.6077 fJ.

the net's CV²: 0.5626 fJ (92.6%)the inverter's own nodes and overlap current: 0.0451 fJ (7.4%)

Short circuit

input edge rail → ground time →

While the input ramps, both transistors are briefly half on and current runs straight from rail to ground. At the edges this design makes (23 ps rising, 17 ps falling), short circuit and gate-to-drain (Miller) coupling together add 1.3–3.0% to a switch. A slow 2 ns edge adds 7.5% at 0.517 V and 59% at 0.7 V.

Leakage

leakage, always switching time →

An off transistor still passes a trickle, switching or not, and it grows with temperature. One SPICE inverter leaks 0.035–0.043 nA at 0.517 V and 25 °C. For the card, your idle law is 12.6 W fixed plus a leakage term of 23.3 W at 80 °C that e-folds every 36 °C; E66 is measuring the two terms apart on aifoundry2 this morning.

Glitches

a b a⊕b glitch

When a gate's inputs arrive at different times, its output can flip and flip back before it settles, and each extra edge costs ½CV² like any other. With real gate delays your multiplier tree makes 1,871 physical toggles per random multiply-add against 1,244 at zero delay, about 1.4× the energy.

Sources: Q128 §1 (the stacked bar, the edge sweep, the inverter's leakage); the 21 Sep model.json idle law, as quoted in E66's registration; Q124's chip-diagram facts (glitches). No single file splits the card's watts among these, so there is no share bar for the card.

Short circuit and leakage are current that never charges a node; a glitch is a real flip that nobody needed. At 0.517 V the first is small because the band of input voltage where both transistors conduct is narrower than at 0.7 V, which is why it can't explain why one flip costs more than another on this chip.

4Inside the multiply-add: one input bit sets off hundreds of flips

One flipped bit of b toggles 381 wires in the multiplier, and each wire pays ½CV²

Your Q124 flip, followed through one lane: bit 11 of b, with a = π, in a gate-level simulation of a netlist synthesized from the chip's RTL.

1

One bit of b flips. In this multiplier b is the multiplicand; a is the one that gets recoded.

2

Booth rows. Radix-4 Booth recoding reads a's significand three bits at a time and makes 16 rows of 33 bits, each 0, ±b or ±2b. The flipped bit of b changes a bit in every row that isn't zero. With a = π about half the rows are nonzero (9 of 16 by the recoding rule); with a = 1.0 only one or two are, one reason your quiet partner is cheaper.

3

Wallace tree. Four levels of 4:2 compressors, two full adders each, squeeze the 16 rows to two. A full adder's sum is a ⊕ b ⊕ c and its carry is the majority of the three, so a changed input always flips the sum and sometimes the carry, which moves one column left. On silicon: 248 full-adder cells, mostly the textbook 28-transistor mirror adder, plus XOR, majority and compound gates.

4

Product adder. Adds the last two rows; a change can ripple along the carry chain.

5

The rest of the seven-stage unit. Alignment, normalization, rounding and the accumulator see the changed product too.

381physical wires toggle in the multiplier (129 in stage f1, 252 in f2)
1,946named RTL signal bits change across the unit (482 distinct nets)
745 fJwhat the card pays for this flip, ± 56 (E65)

Adding it up (estimates, arithmetic shown). 381 wires × 0.2985 fJ (the average net's ½CV²) ≈ 0.11 pJ. Priced from the actual cells: 0.11–0.13 pJ at zero delay, and glitches add about 40%. Against the card's 745 fJ, the multiplier's own switching is 0.11–0.13 ÷ 0.745 = 15–17% of this flip, 21–24% with glitches.

Your ladder's average flip (a = 1.0) costs 116.7 fJ at the board, 87.5 fJ on the minion rail (E60). The open netlist prices the same flips at 46.3 fJ: 32.9 fJ of net switching, 32.9 ÷ 0.2985 ≈ 110 average-net toggles, plus 13.4 fJ inside the cells. That's 53% of the rail.

Against one XOR: a register-file bit into one XOR costs 20.8 fJ (E60), and the XOR itself is a few percent of it (E61): most is the bit's trip from the register file. The multiply-add input bit costs 116.7 ÷ 20.8 = 5.6× more because one bit fans out into hundreds of nodes.

Over a whole random multiply-add, charge onto transistor gates is 0.31 of the card's 5.78 pJ

transistor gates: 0.306 pJ (5%) wire inside the cells: 0.620 pJ (11%) wire between the cells: 0.590 pJ (10%) the cells' own nodes: 0.431 pJ (7%) not in the open count: 3.84 pJ (66%)
Per random fp32 multiply-add the placed open netlist makes about 4,100 net toggles: 3,623 on signal nets and 477 on the clock tree. Bottom-up it names 1.946 pJ of the card's 5.783 pJ over idle. Not yet counted: half the design's capacitance (clock-tree and repair nets added after simulation), coupling between wires, and everything outside the lane the board pays for, regulators included. Sources: Q124 facts in tensorfma32.json; Q126 open netlist; Q128 §2–3; E60; E61; E65.

I started with the inverter because it's the simplest whole CMOS gate: one pull-up, one pull-down, one output node. The multiply-add's reduction tree is built from full adders instead, but every node in them is the same kind of node, charged through a pull-up network and drained through a pull-down, so each flip costs ½CV² exactly as in section 2. The multiply-add simply has hundreds of such nodes per flipped bit, and four fifths of each node's capacitance is wire.

5Scale: from Landauer's floor to the card's watts

A flipped multiply-add bit sits 37 million times above Landauer's floor; the card's second, 22 powers of ten

Each bar's length is the number of factors of ten above kT ln 2 at 58 °C (3.17 zJ), the die temperature your E60 and E65 runs launched at.

Single-edge rungs are heat per edge (½CV²); "drawn per rise" figures are twice that. The board rungs include the regulators: the minion rail carries 0.75 of E60's board energy. Your Q128 ledger prices its rungs per rise against a 66 °C floor (3.25 zJ), which is why it prints 2,755× for the same transistor that sits at 1,410× here.

Read the bottom of the ladder as two factors, as your Q128 ledger does: one electron crossing the 0.517 V rail already leaves 26 times the floor, and the smallest transistor's gate takes 108 electrons per rise. Everything above is more of the same: more fins per net, wire that is four fifths of each net, hundreds of nets per flipped bit, about 4,100 net toggles per multiply-add, and 4.6 trillion multiply-adds a second at 9.18 TFLOPS.

6Could it be less?

Your read-port price follows V², not the clock

E60's ½CV² test: the read-port price on the minion rail on two 800 MHz cards, against aifoundry3 at 600 MHz and 0.522 V.

1.0 1.5 2.0 1.0 1.5 2.0 voltage squared, against aifoundry3 price, against aifoundry3 price ∝ V² (½CV²) if it also grew with the clock aifoundry3, 0.522 V card 0: 1.38× aifoundry2: 1.43×
Measured on the minion rail (E60, reduce.txt): aifoundry1’s card 0 at 0.604 V reads 1.384 against a V² of 1.339, and aifoundry2 at 0.617 V reads 1.426 against 1.400. A price that also grew with the clock (800 against 600 MHz) would sit near 1.79 and 1.87, the hollow circles.

7Inverter in depth: a tutorial from zero

The long way through section 2, for someone who knows programs and Boolean logic but no electronics. Twelve steps, each with a picture or a demo you can drive. The demos use this page's own numbers (0.517 V, the 2.233 fF average net, the 23.4 ps edge, all from Q128) unless a figure says it is illustrative.

1NOT is a function; in a chip, a bit is a voltage

In code, out = !in. In hardware there are no booleans, only wires, and a wire holds a voltage. A chip agrees on two levels: 0 V (ground) means 0, and the supply voltage, called VDD, means 1. On the ET-SoC-1's minion rail VDD is 0.517 V. An inverter is a circuit whose output sits at the opposite level from its input: it connects the output to VDD when the input is low, and to ground when the input is high.

The inverter's whole job, as a symbol and a table

Tap the button. The triangle means "drive"; the small circle on its tip means "invert".

in = 0 (0 V) out = 1 (0.517 V)
inout
0 (0 V)1 (0.517 V)
1 (0.517 V)0 (0 V)
Every other gate (AND, OR, XOR, the full adder of section 4) is built from the same two kinds of switch the inverter uses, so the inverter is the place to learn them.

2A short history: each new switch fixed the last one's problem

The inverter has been rebuilt from a better switch five times. Watch two problems get solved along the way: speed (moving parts, then hot glass) and power wasted while nothing happens.

From a clicking relay to 24 billion transistors

  1. 1830sThe relay. An electromagnet pulls a metal contact. A "normally closed" contact opens when the coil is powered: that is a NOT. Telephone exchanges ran on relays, and in his 1937 master's thesis Claude Shannon showed that relay circuits compute Boolean algebra. Slow (milliseconds), and the contacts wear out.
  2. 1904–06The vacuum tube. John Ambrose Fleming's valve (1904), then Lee de Forest's Audion (1906), in which a voltage on a wire grid controls a current through a vacuum. No moving parts, so thousands of times faster than a relay; one tube plus a resistor to the supply makes an inverter. ENIAC (1945) used about 17,500 tubes, which ran hot and burned out.
  3. 1925The field-effect idea. Julius Lilienfeld patents a switch in which a voltage on a plate controls the current in a solid. Nobody could make materials pure enough to build it for three decades.
  4. 1947The transistor. John Bardeen and Walter Brattain's point-contact transistor at Bell Labs, then William Shockley's junction transistor (1948). Small, cool and reliable.
  5. 1958–59The integrated circuit. Jack Kilby (Texas Instruments) and Robert Noyce (Fairchild) put several transistors and resistors on one piece of silicon. The Apollo Guidance Computer was later built from chips that each held two 3-input NOR gates.
  6. 1959The MOSFET. Mohamed Atalla and Dawon Kahng at Bell Labs build Lilienfeld's idea: a gate on a thin layer of oxide switching a channel in silicon. It is the switch of step 6, and of every modern chip.
  7. 1963CMOS. Frank Wanlass and Chih-Tang Sah at Fairchild pair an n-type and a p-type MOSFET so that one is always off, and the gate draws almost no current while it holds a value (step 7). RCA sold the first CMOS logic chips, the CD4000 series, in 1968; watches and calculators wanted the low power.
  8. 1970sNMOS microprocessors. Chips like the Intel 8080 and the MOS 6502 used only n-type transistors, with a load in place of the p-type one. Denser and faster at the time, but every gate with a low output draws current all the time.
  9. 1974Dennard scaling. Robert Dennard's rule: shrink every dimension and the voltage together, and the power per square millimetre stays flat. It held until the mid-2000s, when voltage stopped falling. Since then power, not transistor count, sets the limit, which is why your ladder counts flips.
  10. 1980sCMOS wins. As chips reached hundreds of thousands of gates, NMOS's idle current became the limit, and CMOS took over nearly all digital logic.
  11. 2011FinFETs. At 22 nm the channel stands up as a thin fin and the gate wraps it on three sides, so it can still shut the channel off at tiny sizes.
  12. TodayYour chip. The ET-SoC-1 is a 7 nm FinFET design of about 24 billion transistors, most of them in pairs doing what this section describes.

3Electricity in five ideas

You need five quantities. Each has a programmer's analogy and one number on this chip.

Charge, Q

An amount of electrons, in coulombs. One electron carries 1.602 × 10⁻¹⁹ C, so a coulomb is 6.24 × 10¹⁸ electrons. Think: the number of items in a queue.

Voltage, V

Energy per unit of charge: 1 volt is 1 joule per coulomb. It works like height: charge "falling" through 0.517 V gives up 0.517 × 1.602 × 10⁻¹⁹ = 8.28 × 10⁻²⁰ J per electron. A voltage is always between two points; "the node is at 0.3 V" means 0.3 V above ground.

Current, I

Charge per second, in amperes; 1 A is 6.24 × 10¹⁸ electrons a second. Think: throughput. By convention current flows from + to −; electrons, being negative, move the other way (section 1 draws both).

Resistance, R

How hard a path is to push charge through. Ohm's law: V = I × R. Charge forced through a resistance turns energy into heat, at the rate I²R. A switched-on transistor behaves roughly like a resistor.

Capacitance, C

How much charge a node holds per volt: Q = C × V, in farads. Every wire and every transistor gate is a tiny capacitor; your average net is 2.233 fF (femtofarads, 10⁻¹⁵ F). Think: a bucket whose water level is the voltage.

The one sentence to keep: a logic gate does not send bits down a wire; it fills or empties the small capacitor that the wire is, through a switch that behaves like a resistor. Everything below follows from it.

4The water picture of an inverter

Two valves and a bucket: one valve fills it from the tank, the other empties it to the drain

Tap to flip the input. Exactly one valve is open at a time; the water level is the output voltage.

Tank: the supply, 0.517 V PMOS valve: open Bucket = output 0.517 V (a 1) NMOS valve: shut Drain: ground, 0 V in = 0
A vertical bar is an open valve, a horizontal one a shut valve. The input does not supply the water: it only turns the valves, which is why one inverter can drive several others. Where the analogy fails: a charge's energy is its voltage, not a height, and the "friction" in the valves (their resistance) is where the heat of section 1 is made.

5The capacitor: what a 1 costs

Filling the node takes Q = CV of charge; the supply pays CV², the node keeps ½CV²

Drag the sliders. The defaults are your average net on the minion rail.

7,206electrons moved per edge
0.597 fJdrawn from the supply on a rise (CV²)
0.298 fJheat per edge (½CV²)
Double the voltage and the heat goes up four times; double the capacitance and it doubles. That is why chips run at low voltage, and why section 6's first lever is V². Arithmetic: electrons = CV ÷ 1.602 × 10⁻¹⁹ C.

6The transistor: a switch worked by a voltage

A MOSFET has three terminals that matter: source, drain and gate. The gate sits on a layer of insulating oxide a few atoms thick, so no current flows into it; it acts on the silicon below only through its electric field, like one plate of a capacitor. In an n-type MOSFET (NMOS), a positive gate voltage pulls electrons up under the oxide until they form a thin conducting channel joining source and drain. Below a threshold voltage (a few tenths of a volt in modern chips) there is no channel and the switch is off; above it the channel conducts, like a resistor that gets smaller as the gate rises.

Raise the gate and a channel of electrons appears between source and drain

An NMOS transistor in cross-section. The 0.2 V threshold here is illustrative.

p-type silicon (the body, at ground) source (n) drain (n) oxide gate to ground to the output from the input

Off: below the threshold there is no channel, only a tiny leakage current.

In 3D: a FinFET, the gate straddling the fin on three sides

The shape your card actually uses. Drag the model to turn it; the far side of the gate is drawn and the near side is cut away so the fin and the channel stay visible. The slider above drives both views.

gatesource and drain, n⁺fin body, siliconchannel, once the gate is above threshold
The p-type MOSFET (PMOS) is the mirror image: p-type regions in n-type silicon, the carriers are holes (missing electrons, which move like positive charges), and it turns on when its gate is pulled below its source, that is, when the input is low. In a FinFET, the kind in your card, the channel is a thin vertical fin and the gate wraps it on three sides, which shuts it off more firmly; the electrical story is the same.

NMOS: on when the gate is high

drainsourcegate

PMOS: on when the gate is low

source (supply)draingate

7Two ways to build NOT, and why the second won

Design A, a switch and a resistor (resistor-transistor logic, and in spirit 1970s NMOS): a resistor ties the output up to the supply, and one NMOS can pull it down to ground. Input low: the switch is open and the resistor holds the output high. Input high: the switch closes and pulls the output low, but now current runs from the supply through the resistor and the switch to ground for as long as the input stays high, doing nothing useful. Design B, CMOS, replaces the resistor with a PMOS whose gate is tied to the same input. The two switches always disagree, one on and the other off, so there is never a standing path from the supply to ground; current flows only while the output changes.

Design A burns current whenever its output is low; CMOS only while it switches

Flip the input and see which design carries current while the gate sits still. Dark = switch on.

A: resistor + NMOS supply R 1 NMOS ground in=0 B: CMOS (PMOS + NMOS) supply PMOS 1 NMOS ground in=0
A: current while idle
none
B: current while idle
none
In design A the idle current is set by the resistor: a chip of a million such gates, half of them low at any moment, wastes power doing nothing. CMOS's idle current is only leakage (section 3), which is why it took over once chips grew.

8The CMOS inverter, slowly: sweep the input from 0 to 0.517 V

Real inputs don't jump, they ramp. Drag the input slowly and watch three things: how strongly each transistor conducts, where the output sits, and the current that runs straight through both while both are partly on. That current is the short-circuit current of section 3; in a sharp edge it lasts only a few picoseconds.

The output stays high, then drops steeply in the middle; current runs straight through only near the middle

Curve shapes and the 0.2 V thresholds are illustrative, not measured on this chip.

supply, 0.517 V PMOS on 0.517out NMOS off ground in
Output vs input high0 0input → 0.517 V dashed: through-current

Input low: the PMOS is on and the NMOS off, so the output is tied to the supply.

The steep middle is the inverter's gain: a small change of input makes a big change of output, and step 9 uses it to clean up noise. The dashed bump is the short-circuit current; the faster the input edge, the less time is spent in the middle, which is why section 3 found it at 1.3–3.0% of a switch at real edges.

9Why digital works: each inverter cleans the signal

Analog signals degrade as they pass through stages; digital ones get cleaner. The reason is the curve of step 8: near either end its slope is shallow, so an input that is a bit off still gives an output very close to the rail. Pass a sloppy level through a few inverters and it snaps to a clean 0 or 1. Only an input right in the middle stays undecided for long, which is why designers keep signals out of that region.

A noisy input snaps to a clean rail within a few stages

Set a messy input. Each bar is the voltage after one more inverter (the illustrative curve of step 8).

Below 30% of the supply a stage reads its input as 0, above 70% as 1, and between them it is undecided (the shaded band). The level alternates from stage to stage, because each one inverts; a bar in the accent colour reads as a 1.

10Speed: the output fills through a resistor

An on transistor is roughly a resistor R and the output node is a capacitor C, so the output follows a capacitor charging through a resistor: v(t) = V(1 − e−t/RC). The product RC, the time constant, sets the speed: the output reaches half the supply after 0.69 RC (the propagation delay) and goes from 20% to 80% in 1.39 RC. From Q128's 23.4 ps rise on the 2.233 fF average net, RC ≈ 16.9 ps, so the pull-up acts like about 7.6 kΩ.

More load is slower, a wider transistor is faster, and the heat per edge doesn't depend on R

Load multiplies C (more gates and wire to fill); drive strength divides R (a wider transistor). Derived from Q128's 23.4 ps edge.

0.517half0 V 0200 ps time after the input switches (grey: average net, × 1 drive)
11.7 psdelay to half the supply (0.69 RC)
23.4 ps20–80% edge (1.39 RC)
0.298 fJheat per edge: grows with the load, the same at any drive
A stronger driver finishes sooner but loses the same ½CV² (section 1). It costs energy elsewhere, though: its own bigger gate is a bigger capacitor for the stage before it. Real transistors aren't constant resistors, so this is the first-order picture; the 23.4 ps is Q128's SPICE result.

11Power: flips per second times energy per flip

Each edge leaves ½CV² of heat. A node that switches τ times per clock cycle at clock frequency f therefore dissipates P = τ × f × ½CV². Summed over every node in the chip, this is the "dynamic power" of textbooks (usually written αCV²f, with α the fraction of cycles in which the node rises). It is why a program's data shows up in the power bill: the data decides how many nodes flip, which is what your Hamming ladder controls.

One net's power, and what a crowd of them costs

Defaults: your average net, the minion rail, 600 MHz (E60's clock on aifoundry3).

89.5 nWone average net
89.5 mWa million such nets
A clock wire flips twice a cycle (τ = 2); a data wire carrying random bits, about once every other cycle (τ ≈ 0.5). Power rises with the square of the voltage, and a faster clock usually needs a higher voltage too, which is why power climbs so steeply with frequency.

12From one inverter to a computer

Swap the single PMOS and NMOS for small networks and you get every other gate. NAND: two NMOS in series to ground (the output falls only if both inputs are 1) and two PMOS in parallel to the supply. NOR is the reverse. NAND alone can build any logic, so a chip is, to first order, many copies of the inverter's idea: a pull-up network and a pull-down network that never conduct at the same time.

An odd ring of inverters can never settle, so it oscillates

Each step is one gate delay. Highlighted nodes have an input that disagrees with what they show, so they flip next.

A ring of 3 repeats every 6 gate delays.

A ring of N inverters (N odd) repeats every 2N gate delays; chips use one to measure their own speed. With an even number the ring is stable instead, and two inverters feeding each other is how a bit is stored: the 6-transistor SRAM cell in every cache is a cross-coupled inverter pair plus two access switches.

Two inverters in a loop hold a bit

10
Each holds the other in place: that is SRAM.

Glossary

VDD
the supply voltage; a 1
node
a wire and everything on it; a capacitor
threshold
the gate voltage at which a channel forms
fan-out
how many gate inputs one output drives
delay
time for the output to cross half the supply
leakage
the small current through an off switch

← Back to section 2, the inverter animated electron by electron. History: Shannon's 1937 thesis; Atalla and Kahng (1959); Wanlass and Sah, ISSCC 1963; Dennard et al., IEEE JSSC 1974. Device physics: Rabaey, ch. 3 and 5; Weste and Harris, ch. 2.

Sources

Conventions: energies at 0.517 V (the open flow's minion rail) unless marked; an edge or toggle leaves ½CV² of heat; a rise draws CV² from the supply and a fall draws nothing; electrons are charge ÷ e. SPICE: ASAP7's TT corner at 25 °C.