Heat dissipation and cooling solutions: Question bank

Before the traps, a few words you must have straight, because every item below leans on them:

Definition The anchors

Temperature difference $Δ T$ = the gap between the hot side and the cold side of a path, $Δ T = T_{hot} - T_{cold}$ , measured in $^{\circ} C$ (or K — the size of a degree is the same in both). Heat only flows when this gap is non-zero, from hot to cold. In Figure 1 (left) it is the height of the drop from top to bottom.
Heat flow $P$ = the rate at which heat energy moves through a path, in watts (W). It is the "current" in the plumbing analogy. (Some textbooks write this same quantity as $Q$ ; on this page we use $P$ throughout to match the parent note.)
Thermal resistance $R_{t h}$ = how hard heat finds it to flow, in $^{\circ} C/W$ . It is the " $Δ T$ per watt" a path costs you: push $P$ watts through it and it opens up a gap $Δ T = P \cdot R_{t h}$ . Low is good.
Thermal conductivity $k$ = a material property (copper vs air), in $W/m \cdot K$ . It is fixed by what the thing is made of.
Heat transfer coefficient $h$ = how well a moving fluid strips heat off a surface, in $W/m^{2} \cdot K$ . It depends on the flow, not just the fluid.
Surface (or cross-section) area $A$ = the area of the face heat flows through or is stripped off, in $m^{2}$ . Bigger $A$ = more parallel paths for heat, so a lower resistance. This is the $A$ inside $R_{co n v} = 1/ (h \cdot A)$ and $Δ T = P d / (k \cdot A)$ .

Trap fuel: people swap $k$ (a material fact) and $h$ (a flow fact) as if they were the same knob. They are not.

Three figures anchor every answer on this page. The captions tell you which item points back to which picture.

Figure 1. Left: the water-tower analogy — heat flow $P$ is current, $Δ T$ is the height of the drop, $R_{t h}$ is the pipe's narrowness. Right: why series stages add — each stage eats part of the total drop, and the drops stack top to bottom, exactly like resistors in a circuit.

Figure 2. The convection boundary layer. A thin blanket of nearly-still air clings to the hot surface and insulates it — that blanket is why $h$ for still air is small (~8). A fan blows the blanket thin, so heat crosses it faster and $h$ climbs to ~60. The material below never changed; only the flow did.

Figure 3. Why fins and why thin thermal paste. Left: one solid block touches little air; fins fan the same metal out into far more surface area $A$ , and $R_{co n v} = 1/ (h \cdot A)$ drops as $A$ grows. Right: Fourier's law $Δ T = P d / (k \cdot A)$ means the temperature drop across a layer grows with its thickness $d$ — so a thick paste layer builds a bigger, unwanted drop than a thin one.

True or false — justify

Copper always cools better than aluminium because its conductivity is higher

False — copper's

k

is ~2× higher, but conductivity only helps inside the metal; the final bottleneck is usually convection off the fins (

1/ (h \cdot A)

), and copper's extra weight can force smaller/fewer fins, sometimes losing the race.

A bigger heatsink base always lowers junction temperature

False — adding base thickness

d

raises the through-slab drop

Δ T = P d / (k \cdot A)

; a bigger base can help only if the extra material spreads heat sideways to more fins (heat-spreading), and past the useful spreading distance the extra mass just adds delay, not steady-state cooling.

Radiation can be ignored for a normal CPU cooler

True in typical conditions — for a metal cooler near room emissivity, ordinary areas and surface temperatures under ~100 °C, radiation is usually a few percent of the total; it is not a universal constant, but under normal PC conditions conduction plus convection carry essentially all the heat, and radiation only dominates in vacuum (space) where convection is impossible.

Liquid cooling wins because water conducts heat better than metal

False — water's

k

is worse than metal; liquid cooling wins on transport, i.e. its high heat capacity carries heat away from the chip to a large radiator, not on surface heat transfer.

Series thermal resistances add, just like series electrical resistances

True — heat flows through junction→paste→heatsink→air one stage after another (Figure 1, right panel), so the drops stack:

R_{t o t a l} = R_{1} + R_{2} + R_{3}

, exactly the electrical analogue with

Δ T \leftrightarrow V

and

P \leftrightarrow I

More thermal paste means a better thermal connection

False — paste only exists to fill microscopic air gaps; a thick layer raises

R_{t h} = d / (k \cdot A)

because paste (

k \approx 3 - 12

) is far worse than the metal it separates. Thin wins.

A fan raises the heat transfer coefficient

h

but leaves the fin material's

k

untouched

True — the fan thins the slow boundary layer of air (Figure 2), boosting

h

from ~8 to ~60; the copper/aluminium's own conductivity is a material property the airflow cannot change.

Doubling fan RPM adds only a few decibels of noise

False — for a fan, radiated sound power rises very steeply with speed, modelled roughly as

\propto (RPM)^{5}

. In decibels that is

10 lo g_{10} (2^{5}) \approx 15 dB

added when RPM doubles — not a couple of dB, and a 15 dB jump sounds far more than "twice as loud".

Spot the error

R_{t h} = 1/ (h \cdot A)

, so I'll cut the heatsink area in half and add a stronger fan to compensate."

The trap: halving

A

doubles

R_{co n v}

, and you'd need

h

to double just to break even — but

h

from a fan saturates, so you usually lose. Area is the cheap, reliable lever; don't trade it away.

"Ambient is 25 °C and

Δ T = 90

°C, so the junction sits at 90 °C."

Error:

Δ T

is a rise above ambient, not an absolute temperature. Junction

= T_{ambi e n t} + Δ T = 25 + 90 = 115^{\circ} C

; forgetting to add ambient is the classic slip.

"Fourier's law

Δ T = P d / (k \cdot A)

says a thicker slab moves more heat."

Error: for a fixed heat flow

P

, thickness

d

is in the numerator of the temperature drop, so a thicker slab builds a bigger drop (worse cooling) — equivalently, for a fixed

Δ T

it moves less heat. This is exactly why thermal paste must be thin, not generous.

"My CPU idles cool, so the cooler is over-specified and I can remove a fan."

Error: idle power is a fraction of load power; cooling must handle peak dissipation (the Thermal Design Power (TDP) figure and above), or the chip will hit Thermal Throttling under load.

"Higher CFM fan → always lower temperature."

Error: CFM only helps until the heatsink surface is saturated — once the fins can't hand off heat any faster, extra airflow just adds noise. Beyond that point you need more area, not more air.

"Liquid metal TIM has

k \approx 70

, so I'll smear plenty on to be safe."

Error: liquid metal is electrically conductive — excess that spills onto pins or SMD components short-circuits them. Its whole advantage is achieved with a whisper-thin film.

"Blower (centrifugal) fans are worse than axial because they move less air."

Error: blowers trade raw flow for static pressure, letting them push through dense fins and cramped laptop channels where a high-CFM axial fan would just stall against the resistance.

Why questions

Why does convection depend on airflow but conduction does not?

Conduction happens inside a fixed solid whose

k

is set by material; convection depends on a boundary layer of stagnant fluid at the surface (Figure 2) — moving air thins that layer, so

h

(and thus cooling) is a flow property.

Why is the natural-convection

h

(~8 W/m²·K) so much smaller than the forced value (~60)?

In still air the only motion is the gentle rise of air warmed by the surface, so the insulating boundary layer stays thick and heat crawls across it; a fan drives air across the surface at metres per second, shearing that layer thin, and thinner blanket = faster heat crossing, so

h

climbs about 7–8× — these are measured, order-of-magnitude engineering values, not exact constants.

Why do heatsinks have fins instead of one solid block?

Fins multiply the surface area

A

exposed to air (Figure 3, left), and

R_{co n v} = 1/ (h \cdot A)

falls as

A

grows; a solid block of equal mass touches far less air, so it dumps heat slower despite holding the same metal.

Why is the tiny junction-to-case resistance often the hardest to reduce?

It lives inside the chip package, fixed by the silicon and lid geometry — you cannot open it up; it sets a floor you can't cross no matter how good your external cooler is.

Why does the parent compare

Δ T = P \cdot R_{t h}

to Ohm's law

V = I \cdot R

Because the maths is identical: temperature difference is the "voltage", heat flow

P

(watts) is the "current", and thermal resistance is the "resistance" — so all the series/parallel intuition from circuits carries straight over.

Why does forced convection give roughly a 7–8× improvement over still air?

Because

R_{t h} = 1/ (h \cdot A)

and the fan lifts

h

from ~8 to ~60 W/m²·K (about 7.5×) while

A

is unchanged — the resistance drops by exactly that ratio.

Why is black anodizing on a heatsink mostly cosmetic for cooling?

It improves radiation, but under normal PC conditions radiation is only a few percent of heat removal below 100 °C, so the real benefit is oxidation resistance and looks — not a meaningful temperature drop.

Edge cases

What cools an electronic device in the vacuum of space, where there is no air?

With no fluid there is zero convection, so the chain is: conduction carries heat through the board and mounting straps to external radiator panels, which then dump it as radiation (the

T^{4}

term) into cold space. Spacecraft add heat pipes, loop heat pipes and cold plates to conduct heat from parts to those panels — nothing removes heat without either a conducting path to a radiator or the radiator's own emission.

At steady state with a perfectly matched cooler, what happens to the temperature over time?

It stops rising and holds constant: heat in equals heat out, so

Δ T

settles at

P \cdot R_{t h}

— the "bucket" fills only until the hole drains water as fast as it pours in.

If the fan fails (h drops to the natural-convection value) on a chip designed for forced cooling, what happens?

R_{t h}

jumps ~7–8×, so

Δ T

jumps the same factor; the chip races toward its thermal limit and either throttles hard (see Thermal Throttling) or shuts down to protect itself.

What is the thermal effect of a zero-thickness (ideal) thermal paste layer?

In the limit

d \to 0

, the paste's own resistance

d / (k \cdot A) \to 0

— but you still need some paste to fill air gaps, so the real optimum is "as thin as possible while still filling", not literally zero.

What happens to

R_{t h} = 1/ (h \cdot A)

as heatsink area grows without bound?

It approaches zero, so in theory cooling becomes perfect — but in practice fins crowd, airflow stalls between them, and the effective

h

falls, so infinite area gives diminishing, not infinite, returns.

Why does a thick heatsink base still help sometimes, despite the thickness penalty?

Because a base is not a pure 1-D slab: heat entering under the small chip footprint spreads sideways through the base to reach fins far from the chip. A thicker base lowers this spreading resistance and feeds outer fins better — the benefit lasts only until spreading gain is outweighed by the through-thickness drop

P d / (k \cdot A)

If two chips share one heatsink, how does the total heat load change the base temperature?

The powers add before the shared resistance:

Δ T = (P_{1} + P_{2}) \cdot R_{t h}

, so the base runs hotter than either chip alone would drive it — a real trap when combining a CPU and VRM under one cooler.

Recall Quick self-test

The single knob a fan changes ::: The heat transfer coefficient $h$ (never the material's $k$ ). $Δ T = 90$ °C rise, ambient 25 °C → junction ::: $115^{\circ} C$ (rise plus ambient). Why thin thermal paste beats thick ::: $R_{t h} = d / (k \cdot A)$ grows with thickness $d$ ; paste is a poor conductor, so less of it is better once gaps are filled. Doubling fan RPM adds how much noise (sound power $\propto RPM^{5}$ ) ::: About $10 lo g_{10} (2^{5}) \approx 15$ dB.