I have been rearranging my security alarm project a bit, trying to simplify it a little, without sacrifying any features. I got to the power supply, and wondered why I need so many 7805 regulators, so I found THIS PAGE, and regarding the circuit at the bottom for the high current adjustable regulator, if I swap the LM317 for a LM7805, will it cause the 2n3055's to output 5V regulated?
this would save me several 7805's!
~Travis
Nope, it will be about 4.3 V because of the BE diode drop. You could put a diode in the ground connection for the 7805 to compensate for it, though.
I hope that you have a hefty heat sink. Dropping 7 volts (12 volts - 5 volts) at 15 amps is 7x15=105 watts, a lot of heat.
Why not simply swap it for some DC-DC regulators?
I will absolutely consider a DC to DC converter for in the future.
Now would be a really good time, before you finalize your current design.
Or a 5V switchmode power supply if you're not "stuck" with a 12V battery or if there's another reason your stuck with 12V.
LM2596 module will go higher that 1A, but eventually you will need to change to something like an open-frame
supply: https://www.amazon.co.uk/Switching-Power-Supply-Driver-Adapter/dp/B007Q7EGSC/ref=sr_1_5?ie=UTF8&qid=1468006834&sr=8-5&keywords=100W+5V+power+supply
initially I won't be drawing the full 15A,
Yikes, thats a lot of current for an alarm system, have to ask what you are using that demands such a hefty current; is it some kind of electrified fence etc ?
Hi,
These days DCtoDC converter is the way to go.
The LM78XX series spec/data sheets do have a circuit to increase the output current of LM78XX type regulators, but not using the LM317 type addon you had.
The circuit that you showed, removed most of the regulation properties from the regulator.
But you have the problem of heat dissipation/efficiency.
Tom.... 
Deleted post - it duplicated a previous post. Sorry.
