Define:

- Tmax = hottest desired case (or heatsink) temperature.

Imax = max current for this design.

Tamb = ambient air temperature

Vin = Voltage from power supply

Vout = Voltage out of regulator.

Tj = junction temperature

Rjc - thermal resistance junction to case.

Rca = Heatsink thermal resistance.

Preg = Regulator power dissipation.

Required minimum heatsink = (Tmax-Tamb)/(Vin-Vout)/ Imax C/W

Junction temperature = (Vin-Vout)x Imax x (Rjc + Rca) + Tamb

Preg = (Vin - Vout) x Imax.

Add a series resistor to reduce regulator dissipation:

- Vinreg = Regulator input voltage.

R = Resistor resistance.
Pr = Resistor power dissipation.
Vdo = regulator dropout volatge

R <= (Vin - Vo_max_with_resistor - Vdo) x Imax.

Pr = Imax^2 x R
Vinreg = Vin - (Imax x R)
Pvreg = (Vin - Vinreg)x Imax.

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E&OE
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More anon if needed.

Your calculations are essentially correct (except as Mark points out, your 42W figures - this appears to be a mental typo - multiply by 0.5, not divide by 0.5).

Don't forget that there is an internal 5 C/W Rjc to allow for.

For limiting case assume junction max allowable is 125C and that internal thermal limiting will occur at that point.

To reduce power dissipation in IC for low Vout use a series resistor.

R <= (Vin - Vo_max_with_resistor - 2) x Imax.

eg For Vout max with a given resistor of say 8V and with 26V in and with I out max with this resistor of 600 mA -

- R <= (26-8-2)/0.6 <= 26.666 ohms. Say 27 ohms

At 0.6A it will drop 0.6 x 27 =+ 16 V.

Vin_reg = 26-16 = 10V.

This gives the regulator 2V headroom.

LM317 datasheet says headroom at 600 mA, warm ~= 1.8V (fig 3) so that's just marginal.

Resistor will drop V^2/R = (26-10)^2/27 = 9.5 Watt.

Regulator will drop (10-5) x .6 = 3 Watt.

It's time you got a switching power supply :-).

For interest, under these conditions the internal 5 C/W Rjc will drop 3 x 5 = 15C.

For junction JUST at 125C Tc = 125 - 15 = 110C.

Sizzles with wet finger.
Tca = (110-25) = 85C

Heatink needed = 85/3 ~= 25 C/W.

ie a modest heatsink will suffice if you don't Mind boiling water temperatures on the case and heat sink.
The resistor will be hot :-).

Not exactly 'hot', but if you fed (say) 30V into that circuit and set the output to 1.25V, the regulator would be dissipating about 150mW with the output disconnected. That's because the 240\$\Omega\$ resistor draws about 5.2mA from the output, and the regulator must dissipate (Vin - Vout) * 5.2mA even with the output disconnected.

A TO-220 has a thermal resistance of about 80°C/W so it would get noticeably warm.

It's not optional, BTW, to draw that much current. The LM317 needs some current internally to work, and if you don't draw a minimum current of around 5mA the output could go out of regulation (rise **above** the desired voltage).

P.S. If you use a 24.0 Ohm resistor (don't laugh, I've seen it happen, the markings or the color code is 2 4 0) the regulator **would** get hot.

## Best Answer

Dissipation is (9-5)V*170mA, or 680 mW. SOT223 thermal resistance is 62.5°C/W maximum. So the junction temperature is 62.5 * 0.680, or 42.5 degrees above ambient. And the thermal resistance to the case is about 15 °C/W, so the case will be 42.5-(15*0.680) or 32.3 degrees above ambient. If ambient is 25°C, the case is at 57°C. Does this sound right?

(You should look up and substitute your own thermal values, I just grabbed some from the Zetex datasheet. But I expect all SOT223 will have similar thermal characteristics.)

Notice that the 500 mA rating has nothing at all to do with it. Temperature is determined by the power dissipated, which is set by the input-output difference and the output current, and the thermal characteristics of the package and surroundings.