Another way - I think better - to drop the voltage prior to the 
regulator to 'spread the heat around' is to use a zener, which are 
easily obtained at 1W/3W/5W ratings.  So you could get a say 12V zener @ 
3W rating, and 100mA thru that is 0.1 * 12V = 1.2W.  Give it some copper 
area for it to dissipate heat, esp. if you use a SMT device.

The power resistor technique has the downside of... adding resistance to 
the power supply, yielding dips during current pulses, etc.  May or may 
not be an issue.  Also at low currents the regulator will see nearly 
24V, not a problem for heat-wise, but does mean the regulator has to be 
rated for 24V.  With a zener you'll always get the drop across it, so 
you can use a regulator with a lower voltage rating.

One can also use multiple diodes in series, each dropping the voltage 
0.5V or whatever.

Anyone have comments about this trick?  Anything I'm not considering 
with using a zener?

Thanks all!
J



Jinx wrote:
>> What else should I do to protect my PIC from this high power stuff?
> 
> What are you doing with /mclr ? If enabled, I'd suggest some RC, for
> example 10k to Vcc, 100n to Vss, rather than tie it directly to Vcc.
> RC will help to filter out any noise that gets onto Vcc
> 
>> Especially when I replace the 9V battery with a connection to +24V?
> 
> It would be a good idea to add some series resistance before the
> 78L05, especially if you start drawing any amounts of 5V current
> 
> If you're dropping 19V across the regulator (ie 24-5), then at 100mA
> output current that's 1.9W of heat to dissipate, which I'm sure exceeds
> the rating for the TO92 package. Reducing the 24V down to say 9V
> with a resistor will decrease the wattage in the regulator
> 
> So assuming 50mA current required, that's 55mA into the regulator
> (including its quiescent)
> 
> V = I * R
> (24-9) = 0.055 * R
> R = 272 ohms
> 
> P = I * I * R
> P = 0.055 * 0.055 * 272
> P = 0.8W
> 
> => 270R @ 1W (minimum) resistor
> 
> and the wattage in the regulator is reduced to < 0.25W
> 
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