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Tool · Electronics

Heatsink calculator

Pick a part or type its thermal resistances, give the voltages and current, and the calculator works out the power, the junction temperature bare and on a heat sink, and the largest heat sink thermal resistance that keeps it under the limit.

V
V
A
mA
°C
°C
°C/W
°C/W
°C/W
°C/W
Interface
Heat sink
Junction temperature115 °C
Heat to get rid of
Heat sink needed
Without a heat sink
Working

Datasheet figures: onsemi MC7800 (θJC 5, θJA 65 °C/W, 3.2 mA quiescent), onsemi LM317 (θJC 5, θJA 65 °C/W), onsemi MC78L00A (TO-92 θJA 200 °C/W), Infineon IRFZ44N (θJC 1.5, θCS 0.5, θJA 62 °C/W, 17.5 mΩ). Interface values from onsemi AN1040 Table 1 for TO-220; heat sink values are natural convection figures from the Aavid catalogue (577202B, 6021PB, 7021B, 7023B). Results are steady-state estimates in still air.

Power is what heats it

A linear regulator burns the voltage it drops times the current through it, plus its own quiescent current at the input voltage.

P = (Vin − Vout) I + VinIq

Thermal resistances add in series

Heat flows from the junction to the case, through the interface into the heat sink, and from the fins into the air. Each step adds a temperature rise.

TJ = TA + P(θJC + θCS + θSA)

Solve for the heat sink

Put the junction at its limit and solve for θSA. Any heat sink with a lower thermal resistance than that will do.

θSA = (TJ,max − TA)/P − θJC − θCS

How do you work out the heat sink you need?

The junction temperature is TJ = TA + P × (θJC + θCS + θSA), so the heat sink you need is θSA = (TJ,max − TA) ÷ P − θJC − θCS. A 7805 dropping 12 V to 5 V at 1 A turns 7.04 W into heat; with a 125 °C limit in 25 °C air it needs a heat sink of 8.2 °C/W or better.

For the full 7805 story, read does my 7805 need a heatsink?. To set an LM317's output, see the LM317 formula explained.

What the datasheets give

  • 7805 and LM317 in TO-220: θJC 5 °C/W and θJA 65 °C/W (onsemi MC7800 and LM317 datasheets). The MC7805C draws 3.2 mA quiescent current, typical.
  • Bare TO-220 limit: (125 − 25) ÷ 65 = 1.54 W in 25 °C air, if you keep to the 125 °C range the electrical tables are specified over.
  • 78L05 in TO-92: θJA 200 °C/W, and no θJC: onsemi says the plastic package sheds its heat through the leads.
  • IRFZ44N MOSFET: θJC 1.5 °C/W max, θCS 0.50 °C/W greased, θJA 62 °C/W, RDS(on) 17.5 mΩ max, 175 °C maximum junction.
  • Interface, TO-220 (onsemi AN1040): bare metal, dry 1.2, bare metal, greased 1, mica washer, greased 1.6, mica washer, dry 3.4 °C/W.

Nine heat problems

Each one loads the part, the heat sink (if any) and the thermometer into the calculator.

7805, 12 V to 5 V at 1 A

115 °C

(12 − 5) × 1 + 12 × 0.0032 = 7.04 W. Bare, that would be 482 °C on paper, so a heat sink is not optional. It needs θSA ≤ 8.2 °C/W.

A 6.8 °C/W finned sink with greased mounting gives 115 °C. A 12.5 °C/W channel sink would leave it at 155 °C, over the limit.

Power 7.04 W θSA needed 8.2 °C/W

7805, 9 V in, 0.5 A

157 °C bare

Only 2.03 W, but a bare TO-220 at 65 °C/W reaches 157 °C. Even a 43 °C/W heat sink is enough, so the smallest clip-on fixes it.

Power 2.03 W

7805, 12 V in, 100 mA

73 °C

0.74 W gives 73 °C at the junction: no heat sink needed.

Junction 73 °C

The same 7805 in a 50 °C box, on mica

144 °C

Warmer air and a greased mica washer (1.6 °C/W) push the junction to 144 °C on the 6.8 °C/W sink. It now needs θSA ≤ 4.1 °C/W.

θSA needed 4.1 °C/W

LM317, 12 V to 3.3 V at 1.5 A

13 W

8.7 V × 1.5 A = 13.05 W. It would need θSA ≤ 1.66 °C/W, and even a 4.4 °C/W sink leaves the junction at 161 °C. This is a job for a switching regulator.

θSA needed 1.66 °C/W

IRFZ44N switching 10 A

2.8 W

10² × 17.5 mΩ × 1.6 = 2.8 W, using the datasheet's normalized curve (about 1.6 times at 100 °C). Bare it reaches 199 °C, past 175 °C; a 24.4 °C/W clip brings it to 99 °C.

Bare 199 °C

IRFZ44N at 20 A

11.2 W

Twice the current is four times the heat: 11.2 W. θSA must be 11.4 °C/W or less; a 6.8 °C/W sink gives 124 °C.

θSA needed 11.4 °C/W

78L05 in TO-92

72 °C

9 V to 5 V at 50 mA is 0.23 W, about 72 °C. At 12 V and 100 mA it would be 0.75 W and 174 °C: the TO-92 tops out near 0.5 W in 25 °C air.

Bare limit 0.5 W

The Aavid catalogue example

10.7 °C/W

6 W, θJC 3 °C/W, 0.5 °C/W interface, 150 °C limit and 65 °C air: (150 − 65) ÷ 6 − 3 − 0.5 = 10.7 °C/W, the same answer the catalogue gives.

θSA 10.7 °C/W

7805 power and heat sink needed (25 °C air, 125 °C limit, greased)

Input0.1 A0.5 A1 A
7 V0.22 W, none1.02 W, none2.02 W, ≤ 43.4
9 V0.43 W, none2.03 W, ≤ 43.34.03 W, ≤ 18.8
12 V0.74 W, none3.54 W, ≤ 22.37.04 W, ≤ 8.2
15 V1.05 W, none5.05 W, ≤ 13.810.05 W, ≤ 4
24 V1.98 W, ≤ 44.69.58 W, ≤ 4.419.1 W, too much

Power in watts, then the heat sink thermal resistance needed in °C/W; "none" means the bare TO-220 stays under 125 °C.

How the calculator works

Power: (Vin − Vout) × I + Vin × Iq for a linear regulator, I² × RDS(on) × a hot factor for a MOSFET, or the watts you type.

Bare: TJ = TA + P × θJA. On a heat sink: TJ = TA + P × (θJC + θCS + θSA), with the case and heat sink temperatures in between.

The heat sink needed is the θSA that puts TJ exactly at your limit; the calculator suggests the first catalogue sink at or below it.

What the model leaves out

  • Airflow and orientation. Catalogue θSA values are for natural convection; a fan lowers them, an enclosure raises them.
  • Board copper. θJA depends on the board; surface-mount parts lean on copper area, which this does not model.
  • Transients. Short pulses heat the part less than the steady-state figure suggests; this is the steady state.

Common misconceptions

"A hot case means a failing part." In the 1 A example the tab sits at 80 °C while the junction, at 115 °C, is still inside its 125 °C limit.

"A small load needs no heat sink." It is the voltage dropped times the current that counts: a 7805 from 9 V at 0.5 A already passes 125 °C bare.

"A bigger heat sink always fixes it." θJC and the interface set a floor; at high power no heat sink is enough.

Frequently asked questions

How do I calculate the heat sink I need?

θSA = (TJ,max − TA) ÷ P − θJC − θCS. Pick a heat sink with a thermal resistance at or below that number.

How much power can a TO-220 handle without a heat sink?

With θJA 65 °C/W, about 1.5 W in 25 °C air for a 125 °C junction.

Does a 7805 need a heat sink?

Above about 1.5 W, yes. That is roughly (Vin − 5) × I; 12 V in at 1 A is 7 W and needs about 8 °C/W.

What is θCS?

The thermal resistance between the part's tab and the heat sink. onsemi measures about 1.0 °C/W greased and 1.6 °C/W with a greased mica washer for TO-220.

Why does a MOSFET get hotter than its RDS(on) suggests?

RDS(on) rises with temperature. The IRFZ44N datasheet curve shows about 1.6 times the 25 °C value at 100 °C.

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