Using an air cooled series R before regulator to drop Vin to headroom limit at max I reduces regulator dissipation to Vheadroom x Imax.
2W resistor would nominally be OK (under 50% dissipation) but depending on available room a 5W ceramic blob would be very happy.
How hard can it be to 'design' a resistor? Let's find out.
What the resistor would be doing is emanating some of the heat rather
than the regulator, but total heat is the same - right? –
Total max waste heat is (Vin-Vout) x Imax .
BUT air cooled resistors rated at far more than you are dissipating are cheap and common and allow regulator dissipation to be reduced to a minimum - so heatsinking and/or package used can be less significant.
Resistors dissipating significant power should be run at a maximum of well below their rated power for reliability. Here you have 700 mW max. The regulator needs SOME headroom so say 500 mW in resistor max. So a 2W resistor would probably be OK BUT If room etc not an issue I'd use a 5W ceramic block resistor - cheap and common.
fyi 5w ceramic block resistors about $0.20/ea, and metal film 2w 47
ohm resistor about $0.03/ea. Am I right in aiming at 47 ohm here? for
ceramics, seem to be limited to 22, 47 and 75 ohm.
The aim is to distribute the heat most benignly.
First you MUST take maximum current draw EVER and maximum Vin ever.
If Imax is 100 mA usually but sometimes 100 or 120 mA more than very momentarily then use the worst case unless it is certain to be only for short enough period to not matter thermally.
Then look at 12V supply - is it 1% regulated or 5% or 10% or ???.
A brief surge at startup does not matter thermally (probably) but it if can run 10% high worst case then that's what must be used.
Say worst case is 13V and 120 mA (probably High).
If vout = 5V then you must deal with (13V-5V) x 120 mA
= 8 * 120mA = 960 mW heat.
If it was 12V and 100 mA then heat dissipation = (12-5) x 100 mA = 700 mW
so the 13V/120 mA would be about 35% more heat if it applied.
Then look at the onboard regulator. What Vmin does it need worst case at max possible current? Some regulators safely allow 1V or less "headroom" while some may need 2 to 3V. Say it was a modernish LDO and allow 2V. So Lowest possible max regulator heat is 2V x 120 mA = 240 mW. Then resistor COULD drop regulator input voltage to within 2V of 5V = 7V. Now you need MINIMUM 12v rail to calculate resistor so headroom is maintained at 12V_min_level. Say it's always at least 12V.
So resistor can be R = V/I = (12-7)/120 mA = 41.666 Ohms.
You can use smaller but not larger to maintain 2V headroom.
If Imaxpossible = 100 mA then Rmax = (12-7)/100 mA = 50 Ohms so 47R is OK.
Now look at regulator - is the above 240 mW worst case OK with heatsink etc used.
If it's easily able to handle that then all is well.
But, if it will get too hot then you need more heatsinking or an external regulator.
That 240 mW was based on 120 mA worst case and 2V dropout allowance - and it MAY be much lower. SO finally calculate Rexternal based on all above
3 cents for a 2W metal film sounds nicely low cost in US. Easy enough in China. Here R dissipation is 960 mW max based on above and probably lower. Work out actual worst case figures. There is no hard rule to how close a resistor can be run to its wattage ratings and notionally 100% is OK BUT a look at the data sheet may show that that rating is only for impossible air conditions etc. So look CLOSELY at data sheet and housing etc. I'd try to not run a resistor above 50% of it's allowable ratings in a given application and lower is nicer.
Running a 5W resistor at under 1W gives me a comfortable feeling.
Running a 2W after rating for conditions RESISTOR (eg maybe 3W nominal) at approaching 1 watt sustained worst case would leave me far less comfortable.
Wise? Do they pay you for callouts? :-).
Please provide links to resistors concerned if available.