Hi Fred

The temperature of the 7905 will be proportional to how much power it is
dissipating. Power here is the voltage drop across it from input to output
multiplied by the current it is supplying to the load.  It's own quiescent
current figures in too, about 10mA.

So in your "It gets very hot, can't keep your finger on it" circuit, the
poor little regulator sees (24in - 5out)* 28mA = 19V * .28A = 5.32 SMOKIN'
WATTS!

With 12 volts in, it sees 7V * .28A = 1.96 WATTS.  With Iq (10mA) call it
2W.  Yes, this will get hot.

Solutions?
1. Lower the input voltage.
2. If you can't do that, put something in to drop it. A resistor between
the 24V and the regulator can take the heat off the regulator. Of course,
then the resistor will get real hot.
3. Use a switcher. Look up buck converters.
4. If your load is fixed, you can bypass the regulator with a resistor to
supply some of the current that way and the regulator supplies the rest
keeping things at 5volts. Don't do this one unless you *really* understand
how it works.
5. Use two or more regulators in parallel with their outputs separated with
"ballast" resistors to make them share the load.

Good luck! Hope the blisters heal real soon.
Tom M.

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