There is no magic, the current your PIC will consume at 5V will come
from the 12V Battery, so the 78?05 will drain *the PIC current* from the
battery, as 12V, and will deliver it as 5V to the PIC. It means that the
78?05 will be wasting 140% of what the PIC will consume, and a little
bit more for the 78?05 internal functions. This solution efficience is
5/12 = around 41%.

To reduce this problem, you have few possibilities;
a) Use of a DC-DC conversion based on capacitors (charge pump
converters).
b) Use of a DC-DC conversion based on coiled step down.

Both systems can be built using a single chip, DIP 8 or SOIC 8, and will
cost you around $5 including the coil or the capacitors.  The coiled
(step-down) is more efficient than the capacitive version (charge pump),
it can goes up to 90% of efficience, it means that if the PIC power
consume is 5Vdc @ 5mA (25mW), the whole conversion system will consume
less than 28mW from the battery, or around 2.5mA!!!

The charge pump solution (capacitors) works pretty well for low
currents, but have not a linear efficience curve along with output
voltage or current.

I use both solutions in my circuits. Each time I take few factors in
considerations to choose one or another.  Normaly I do bench test for
both circuits. Theory here is merely a guidance...

Wagner

Mike M wrote:
>
> Well its a simple alarm circuit, once the car is started or anything in the ca
r activated other then door locks opening and closing it is ASSUMED that the loc
k has been disabled.  So i would think that im pretty safe as far as a clean sig
nal screwing up the function of the system.  Unless it will damage the chip or r
egulator?? u think??
>
> As far as the 7805 for some reason i cant get over the idea of currrent.  If a
nyones going to reply to this to the group i think i would suggest changing the
SUBJECT line...Tell me if this is correct...It doesnt matter how much current yo
ur batter or what not can supply...it matters how much you are drawing out of it
 (?)
>
> MikE