Electronic – Will 5W power on TIP41C/TIP42C require a heat sink

Yes, thermal resistance/conductance is a two-way street, for passive conductive cooling devices, with equal speed limits posted for both directions.

OK, now for the caveats: If your heat sink is rated for forced air flow, and you have no forced air flow, then the heat flow through the heat sink will be different. Your heat sink is rated for certain conditions at both ends. If you meet those conditions (at both ends), except for reversal of temperature conditions, then you can expect equal but opposite heat flow.

Say that your 10C/W heat sink is rated 10C/W for conduction of heat from a 1W source to still air, with the contact area with air being fins. Now, you put those fins INSIDE your enclosure, in contact with still air, and you place the outside end of your heat sink in contact with a device (say, a cold plate) that will keep that end of the heat sink 10C cooler than the inside air. In that case you will get 1W of energy flow from the fins of the heat sink to the cooling device (cold plate).

You would want to pay attention to such things as: Warm air rises and cooler air falls. Air fins are most effective when hot air can rise from them and allow cooler air to come in contact with the fins. Cooling fins, on the other hand, would be more efficient when placed such that cooled air can fall away from them.

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.

 E&OE

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 :-).