AVR-Chat

Jay -

Some basic principles are in order here.

The (approximate) amount of power dissipated by the
regulator is the load current times the voltage difference
across the regulator. 

Temperature rise is directly proportional to power.

Thus, you have two choices with a given regulator and heat
sink: (1) reduce the load current and (2) reduce the
input-output voltage difference.

Moving the relay to the input side would help, but it would
be hard to adapt with such a large input voltage range
using a simple resistor. 

You might consider a second (6V?) regulator just for the
relay. 

The minimum input voltage of a 7805 is about 7.8V. Thus,
you can affort to "waste" 4V of your input. This is
governed by the minimum input of the regulator and the
minimum available source voltage. Adding 5 1N400x diodes in
series would probably help.

Can you use a relay with lower coil current? That would
help.

Your relay problems are PROBABLY the result of trying to
run a 6V relay at 5V. That is very typical of the behavior.
Can you replace it with a real 5V relay? Or, why not a 12V
relay. That would many are probably rated for voltages up
to 18V.

Jim
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