Conduction-Cooled Electronics Calculator

In vacuum there is no convection. A board that runs cool on the bench cooks in orbit, because the whole heat path is conduction through the copper into a frame or wedge-lock rail, plus a little radiation. This computes the resistance of each segment, the rise at each, the junction temperature, and the share of the heat that actually leaves by radiation.

Last updated 24 Sep 2026

Junction temperature through copper, a wedge lock and a little radiation

Board, rail and radiation → junction margin

Dissipation of the component you care about (W)The part whose junction temperature you want. Its heat goes through its own mounting into the board copper and then along with everything else.

Other heat crossing the same conduction path (W)Everything else on this side of the board whose heat has to travel the same route to the same rail. It does not raise the component’s own mounting rise, but it raises the board temperature that rise sits on top of — which is why a hot part next to a quiet one is a different problem from a hot part next to five others.

Board thickness (mm)Total finished thickness. Used with the copper below to split the board into a copper slab and a dielectric slab conducting in parallel.

Total copper thickness, all layers added (µm)Add up every layer’s finished copper and multiply by the fraction that is actually continuous in the direction of heat flow — a signal layer that is 40% copper contributes 40% of its thickness. 210 µm is six layers of 1 oz all continuous, which is what a conduction-cooled board is built to do. This is the single most effective input on the page.

Thermal conductivity of the copper (W/m·K)385 for electrodeposited copper foil at room temperature; 400 for the pure metal. Use a different figure here if the spreader is aluminium (about 170 for 6061) or an aluminium or copper core.

In-plane thermal conductivity of the dielectric (W/m·K)For FR-4 the two directions are quite different: roughly 0.29 to 0.34 W/m·K through the board and 0.81 to 1.06 W/m·K in the plane of the glass cloth. Heat travelling to the rail is in-plane, so the larger figure applies — and it still contributes almost nothing beside the copper. Polyimide is similar; a thermally conductive prepreg is higher.

Conduction path length from the component to the rail (mm)The distance the heat has to travel in the board. This is the input that board layout controls, and putting the dissipating parts near the rails is worth more than anything else on this page.

Width of that path (mm)How much of the board’s width is actually carrying the heat. For a part in the middle of a card with rails on two edges, heat leaves both ways: enter the width available to one path and halve the heat, or model the worse of the two. The spreading is treated as a straight bar here, which is optimistic for a small hot part far from the rail.

Component case to board copper (°C/W)The mounting. A surface-mount power package on a plain pad is poor; a dense array of filled and capped thermal vias under the pad is a great deal better; a solid copper coin pressed into the board is better again and is normal on conduction-cooled hardware. Take it from the package’s data sheet or from your own thermal model — this page will not guess it from a via count. Use the figures from your part’s datasheet; typical values vary widely between manufacturers.

θJC of the component, junction to case (°C/W)From the data sheet. JEDEC JESD51-12 warns that the junction-to-ambient figure on a data sheet’s front page belongs to a standard test board in still air and is no guide to anything here. Use the figures from your part’s datasheet; typical values vary widely between manufacturers.

Wedge-lock or rail interface figure (°C·inch/W)Card retainers are specified per unit engagement length, because the contact is a line rather than an area. Calmark state that the thermal resistance across the interface of a card clamped to a cold plate usually falls between 2 and 4 °C·inch/watt, with shorter three-piece retainers near 2 and retainers over six inches nearer 4, and five-piece models better because the clamping force is more evenly distributed. Use your retainer manufacturer’s own figure at your own torque.

Rail engagement length (mm)The length of card edge actually clamped. The interface resistance is the figure above divided by this, so a retainer that runs the full edge is worth a great deal more than a short one.

Is this in vacuum?Two separate things happen in vacuum and this input is only the second of them. The first is that convection stops entirely, which is why this page exists. The second is that the interface itself gets worse: Calmark state that at high altitude or near vacuum their measured figures increase by 10 to 40 per cent, because the air trapped in the microscopic gaps between the two surfaces was carrying part of the heat.

Rail or frame temperature (°C)Where the heat is going. On a spacecraft this is the chassis or the cold plate temperature at the card guide, which is itself an output of the box-level thermal model rather than a constant.

Emissivity of the board surfaceAround 0.85 to 0.9 for solder resist, conformal coating and most paints; 0.03 to 0.1 for bare polished aluminium; anodised aluminium is high. NASA’s Passive Thermal Control Engineering Guidebook notes aluminium emissivity ranging from 0.02 to 0.8 depending on the treatment, which is why this is an input.

Radiating area (mm²)The board area that can actually see something colder. Both faces of a 100 × 100 mm card is 20,000 mm², and that is a generous figure for a card in a cage: the faces are mostly looking at the neighbouring cards. Set this to zero to see the conduction-only answer.

View factor to the cold surfaceThe fraction of the radiating surface’s hemisphere that is the cold surface. A card sandwiched between two neighbours is a small number; a board facing an open radiator or deep space approaches 1. This input and the one below are what decide whether radiation matters at all, and both are estimates.

Effective temperature of what it radiates to (°C)Not deep space unless it can actually see deep space. Inside a card cage the neighbouring cards are as hot as this one — put their temperature here and watch the radiation term collapse, which is the honest answer for most boxes. A board facing a cold radiator or an open aperture is a different case entirely.

Junction temperature you allow (°C)Your own limit rather than the part’s absolute maximum. NASA GSFC’s Preferred Reliability Practice PD-ED-1201 Table 1 — a US Government work — caps semiconductor junctions at 110 °C and microcircuit junctions at 100 °C and says those must not be exceeded in any ground, test or flight exposure.

A thermal network rather than an electrical one: every resistor is in degrees per watt, the dots are watts rather than amperes, and the ground symbols are the rail and whatever the board radiates to — the two reference temperatures. The heat leaves the junction through the package and the mounting into the board copper, then splits: most of it runs along the copper and through the rail interface, and a little radiates. In vacuum there is no third branch. The board run turns amber and red as the junction approaches and passes the temperature you allow.

19.1°CExample

a 5 W part 40 mm from the rail on a 1.6 mm card with 210 µm of continuous copper over a 120 mm path, 3 W more crossing the same route, a 3.0 °C·inch/W retainer engaged over 150 mm in vacuum, and a 40 °C rail

One nonlinear equation, solved by bisection

k_eff·t = k_Cu·t_Cu + k_diel·t_diel θ_board = L ÷ (k_eff·t·W)
θ_rail = (°C·inch/W) × vacuum factor ÷ (engagement in inches)
P = (T_board − T_rail) ÷ θ_path + εσAF(T_board⁴ − T_env⁴)
T_j = T_board + P_component·(θ_mount + θ_JC)

k_eff·t: copper and dielectric conduct in parallel for in-plane heat, so their conductance-thickness products add. The copper is essentially all of it
°C·inch/W: a wedge-lock retainer is specified per unit engagement length because the contact is a line, not an area. Divide by the engaged length in inches to get °C/W
F: view factor. The fraction of the radiating surface’s hemisphere occupied by the surface it is exchanging heat with. Inside a card cage it is small, and the surface it sees is another hot card
solved by bisection: the balance is quartic in the board temperature, and both terms increase with it, so the root is unique. Twenty-six halvings of a bracket that provably contains it

Worked example

a 5 W part 40 mm from the rail on a 1.6 mm card with 210 µm of continuous copper over a 120 mm path, 3 W more crossing the same route, a 3.0 °C·inch/W retainer engaged over 150 mm in vacuum, and a 40 °C rail
The copper and the dielectric conduct in parallel: 385 × 210 µm gives 0.0808 W/K per metre of width and the 1.39 mm of FR-4 gives 0.0011. The copper is 98.64% of the heat path, and the board's effective in-plane conductivity is 51.2 W/m·K
So θ along the board is 40 mm ÷ (0.081962 × 120 mm) = 4.067 °C/W. The retainer adds 3.0 × 1.25 ÷ (150 mm in inches) = 0.635 °C/W, giving 4.702 °C/W board to rail
Balancing conduction against radiation puts the board at 75.9 °C: 7.635 W leaves by conduction and 365.1 mW by radiation, so radiation is carrying 4.6% of the load. That is small here because the view factor is 0.15 and the surroundings are at 60 °C — not because radiation is weak at these temperatures. Give the same board a clear view of something cold and it becomes a major path
The rises: 31.1 °C along the copper and only 4.8 °C across the retainer. On a conduction-cooled card the board is almost always the problem and the interface almost never is, which is why moving the part towards the rail beats every other change
Then 5 W through 1.5 °C/W of mounting and 1.5 °C/W of θJC adds 15.0 °C, putting the junction at 90.9 °C — 19.1 °C below the 110 °C limit entered. Holding the other 3 W fixed and ignoring the radiation term, this part could dissipate 7.257 W before the junction reached that limit

Which page applies when

The dividing line is whether air is doing any work. If it is, the heat is leaving by convection and the two convective pages own the case. If it is not — in vacuum, or in a sealed potted assembly — then every watt leaves by solid conduction or by radiation, and there is nothing else.
Situation	What carries the heat	Page
A part on a heatsink in air, still or moving	conduction to the sink, then convection from the fins	the heatsink thermal resistance calculator
A vented or fan-cooled enclosure	air carrying heat out of the box: Q = P ÷ (ρ·c_p·ΔT)	the enclosure fan airflow calculator
A sealed box in air, no fan	natural convection and radiation from the box surface	neither, exactly — the fan page’s derivation is the closest starting point
A card in a rack in vacuum	conduction through the copper into the rail, plus a little radiation	this page
A box bolted to a spacecraft panel	conduction through the mounting feet into the panel, plus radiation to and from the surrounding structure	this page, with the board run replaced by the mounting-foot resistance

What the retainer figure means

Use your own retainer manufacturer’s figure at your own installed torque, and note that torque matters: the retainer is a mechanical clamp and its thermal performance is a function of the force it applies. These figures are quoted because the manufacturer publishes them openly.
Quantity	Typical published value	Source
Card-to-cold-plate interface, three-piece retainer	2 to 4 °C·inch/W, nearer 2 for short retainers and nearer 4 above six inches	Calmark Card-Lok thermal resistance data
Five-piece retainers (Series 260, 265)	lower, because the clamping force is greater and more evenly distributed	the same
At high altitude or near vacuum	add 10 to 40 per cent	the same — the air in the microscopic gaps was carrying part of the heat
Converting to °C/W	divide the °C·inch/W figure by the engaged length in inches	the specification is per unit length because the contact is a line

No air, so the copper is the heat sink

Every thermal calculation an electronics engineer learns has convection in it somewhere. A heatsink’s θSA is a convection number. A fan curve is a convection number. Even a bare board in still air loses most of its heat by free convection. In vacuum none of that exists. A heatsink in orbit is a lump of aluminium with nowhere to put its heat; the fins are pure mass. The whole path is solid conduction into whatever the box is bolted to, plus radiation — and radiation is only useful if there is somewhere genuinely cold to radiate to and a clear view of it.

The board is a copper heat spreader. For heat travelling in the plane of the board, the copper layers and the dielectric are slabs in parallel, so their conductance-thickness products add. Six layers of 1 oz copper is 210 µm at about 385 W/m·K, which is 0.081 W/K per metre of width. The 1.4 mm of FR-4 left over is 0.8 W/m·K in-plane — the glass cloth is anisotropic, and the through-board figure is smaller still at around 0.3 — which is 0.001 W/K. The dielectric carries about 1.4% of the heat. That single number explains most of what conduction-cooled design consists of: heavy continuous copper planes, thermal vias or a pressed copper coin under the dissipators, and the dissipators near the rails. It also explains why a two-layer board is hopeless in this application whatever its thickness.

The rail interface is specified per inch, and that is not a typo. A wedge-lock retainer clamps the card edge into a groove, and the contact is a line rather than an area, so the resistance falls as the engagement gets longer and the figure is quoted in °C·inch/W. Calmark publish 2 to 4 °C·inch/W for a card clamped to a cold plate — nearer 2 for short three-piece retainers, nearer 4 for ones over six inches, and better for five-piece designs because the clamping force is more evenly spread. Divide by the engaged length in inches and a 150 mm retainer at 3 °C·inch/W is 0.51 °C/W, which is usually small beside the board. The vacuum correction is the interesting part: Calmark state that the same measurements increase by 10 to 40 per cent at high altitude or near vacuum, because the air trapped in the microscopic gaps between the two metal surfaces was carrying some of the heat. Even a bolted joint is partly an air-cooled joint at sea level.

Radiation, without the usual overstatement. The energy balance is P = (T_board − T_rail) ÷ θ + εσAF(T_board⁴ − T_env⁴), which is quartic and is solved here by bisection. It is often said that radiation does nothing at electronics temperatures, and that is not quite right: a surface 50 °C above its surroundings radiates several hundred watts per square metre, and 0.02 m² of board with a clear view of something cold really will move a couple of watts. What makes radiation negligible inside a card cage is geometry, not temperature. The view factor from a card to anything cold is small because the neighbouring cards are in the way, and those neighbours are as hot as this card, so the net exchange is near zero. Put a realistic view factor and a realistic surround temperature into this page and the radiated share collapses; point the same board at an open radiator and it does not. The page takes both as inputs so that the answer comes from your geometry rather than from a slogan. One honest limitation: the board is treated as isothermal for the radiation term, which over-states the radiating temperature, so a large radiated share should be read as a signal that this calculation has reached its limit.

What this page leaves out. It is a one-dimensional steady-state model. A real board spreads heat in two dimensions and the constriction resistance around a small hot part is a significant term this bar model misses. It assumes the rail temperature is given, when in a real box the rail is warmed by everything on it. It says nothing about transients, and a duty-cycled load with a large thermal mass behaves very differently. And it does not model the board-to-component interface beyond the single resistance you type in. For the convective cases it deliberately does not touch, use the heatsink thermal resistance calculator and the enclosure fan airflow calculator; for the power the part is actually dissipating, the MOSFET loss calculator and the diode power loss calculator.

Frequently asked questions

Why can I not just use a heatsink calculator with a very low airflow?

Because a heatsink’s θSA is a convection figure, and in vacuum convection is not small — it is zero. There is no airflow low enough to represent it. A finned heatsink in orbit does nothing except add mass and a little radiating area; the heat has to leave by solid conduction into the structure or by radiation, and this page is about those two.

How much does the FR-4 contribute to the heat path?

Almost nothing in the plane of the board. In-plane FR-4 is around 0.8 to 1.1 W/m·K against copper’s 385, so 210 µm of copper outperforms 1.4 mm of dielectric by a factor of about seventy. The page shows the split. The practical consequence is that the copper you can make continuous along the path — not the copper you have, the copper that is unbroken in the direction of heat flow — is the design variable.

Why is the wedge-lock resistance given in °C·inch per watt?

Because the retainer clamps a line of card edge rather than an area, so the resistance is inversely proportional to the engaged length. Divide the published figure by the engagement in inches to get °C/W. Calmark’s published range is 2 to 4 °C·inch/W for a card clamped to a cold plate, with shorter retainers at the better end.

Does the interface really get worse in vacuum?

Yes, and by a measurable amount: Calmark put it at 10 to 40 per cent. Two clamped metal surfaces touch only at asperities, and at sea level the gaps between those asperities are full of air that conducts. Pump the air out and that parallel path goes away. It is the same reason a thermal interface material matters more in vacuum than on the bench.

How much heat actually leaves by radiation?

It depends entirely on the view factor and on how cold the thing it can see is, which is why both are inputs. Inside a card cage, with neighbours as hot as the card, the answer is usually a few per cent. Facing an open radiator it can be most of the load. The common claim that radiation is negligible at electronics temperatures is a statement about geometry dressed up as a statement about physics.

My part is in the middle of the card with rails on both edges.

Then the heat splits between two paths. Model the worse of the two: enter the distance and width of one path and the share of the heat that goes that way. If the two are symmetric, half the heat travels half as far as the card width, and the answer improves by a factor of about four against a single-sided path — which is why two rails are normal on conduction-cooled cards.

Related calculators

References

Calmark (Birtcher) Card-Lok retainers, published thermal resistance data. States that testing has shown the thermal resistance across the interface of a card clamped to a cold plate usually falls within the range of 2 to 4 °C·inch/watt, with shorter three-piece Card-Loks close to 2 and ones over six inches closer to 4; that five-piece models (Series 260 and 265) give significantly reduced gradients because the clamping forces are greater and more evenly distributed; and that at high altitude or near vacuum the results can increase by 10 to 40 per cent. The recommended installed torque for the Series 265 is 68 N·cm (6 in-lb), plus 11 to 22 N·cm with the locking-element option.
NASA. Passive Thermal Control Engineering Guidebook, Revision 4.0, 25 September 2023. A US Government work. Section 4.2.2 confirms that convection does not occur in the space vacuum environment; section 4.2.5 covers board and component thermal analysis, chip-to-board conductance and board material stack-up; sections 4.2.1 and 4.4.4 cover contact conductance for bolted interfaces; section 4.1.9 notes aluminium emissivity ranging from 0.02 to 0.8 and solar absorptivity from 0.1 to 0.9 depending on treatment, which is why emissivity is an input here.
Gilmore DG (ed.). Spacecraft Thermal Control Handbook, Volume I: Fundamental Technologies, 2nd ed. The Aerospace Press / AIAA, 2002. The standard reference for the radiative and conductive network modelling this page reduces to a one-dimensional path, and for the interface conductance data behind bolted and clamped joints.
FR-4 laminate thermal properties: through-plane thermal conductivity around 0.29 to 0.34 W/m·K and in-plane around 0.81 to 1.06 W/m·K for typical laminates. The anisotropy comes from the glass cloth and is the reason this page asks for the in-plane figure — heat travelling to a card rail is travelling in the plane. Use your own laminate supplier’s data sheet; the spread between products is substantial.
JEDEC. JESD51-12: Guidelines for Reporting and Using Electronic Package Thermal Information. The reason a θJA from a data sheet’s front page cannot be used here: it is measured on a standard test board in still air, which is the one environment this page is explicitly not about. Copyrighted; cited, not reproduced.
The Stefan-Boltzmann constant σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴), exact by the 2019 SI definition of the kelvin, the kilogram and the second.