OK guys, I've been trying to do my homework on this. After this conversation I got worried about the regulator glowing red

. Sanity check my work here:
Parameters refresh
24v source, need 5v output
Load current requirements:
uP: < 1 mA
IR rcvr: = 1.5 mA
Call the total 2mA to be safe
No significant active cooling (i.e fan), but I have added a way to get some ventilation (passive vents in the devices small enclosure) so at least the PCB isn't in a airtight box that might self heat.
Since with 24v to 5v I certaintly don't need a low dropout unit, I looked at the National website and the old 7805 sends one to the LM340 as a replacement:
Data sheet:
It comes as:
LM340LAH-5.0 (Metal TO-39 can)
LM340LAZ-5.0 (TO-92 plastic)
Calcs:
Dropping 19v (24v-5v) at the LM340 at 2mA so it will disip. 38mW (19*2).
The LM340 has a max input of 35V - OK there. 11V to spare.
I think a conservative estimate of ambient can be 60degC (too hot to touch IIRC).
It's data sheet graphs say it's maximum power disp. with no heat sink, at 60 C ambient is ~600+ mW. At 45 C, 700mW (metal can version).
The plastic T092 version says it's max with .125" lead length from PCB is ~550mW.
So, it would appear to my laymen experience that with my 38mW requirement, I would be ~8% of the total power capability of a TO92, or ~5% of the metal can, with no heat sink. Nice and safe, No?
I do have a "dummy" 10mA LED to add, but I can source it's current from the 24v with an appropriate limiting resistor; turn it on and off with the PIC's current sourcing ability. But even if I powered the LED via 5v from the LM340 (via PIC), my load would then be: [(10mA+2mA)*19V]=228mW; still only ~50% of what the datasheet says the LM340 is rated for with no heatsink.
This look right?
Presuming I didn't make an error, the ambient temp of the device won't be close to 60 C, good bit less if it's running warm (I think), which would solidify the 550-700mW capability without sink. Plus the LM340 ground will be tied to a ground plane so that might add some heatsink capability. This, with 2 solid tantulum caps appear to make a good, small solution.
If I add a resistor to the LM340 input, I can derate it even further at it's highest currents, but that does not appear to be necessary.
Please let me know if I have this all wrong or made some noob error <big smile>
Cheers,
Scott