Power Supply

So I want to make a power supply and came across a schematic that has both a current and voltage regulating device using a pair of LM317s. I was more than curious how can one set both current and voltage at a power source or does one dominate the other?

http://www.google.com/imgres?imgurl=http://www2.michaelgregg.com/wp-content/uploads/2011/11/lm317-lab-power-supply.png&imgrefurl=http://www2.michaelgregg.com/?page_id%3D40&h=647&w=1082&sz=74&tbnid=azvSJxAe2LVwtM:&tbnh=73&tbnw=122&zoom=1&usg=__RwXJJSgf14v3Ii-LPcRbvd4BrLw=&docid=GOizfKRXdPl6OM&hl=en&sa=X&ei=_mmZUYPjIYu89gSY1YCwDQ&ved=0CDAQ9QEwAA&dur=2648

I looked at the schematic. The right LM317 provides a constant voltage but the left LM317 provides a current limiting safety function, so it is not a constant current source. The total circuit of two LM 317s is gives a constant voltage output with a current limit that can be set low enough for safety in the load.

Interesting ckt. Normally, you make an LM317 constant-current source by essentially
having the 1K pot wiper set at the top. Therefore, the output current of the 1st
stage would be 1.25V/Rsc [forgetting about the 1K pot, D1, D2, Q1, which you
don't have in a normal constant-current ckt].

It's a little unclear exactly why this ckt functions as a current-limiter rather than
constant-current source, but my first guess is that excess current [above that drawn
ultimately by the load] actually gets shunted off through D1+D2, and the thing would
croak without them in there.

For my part, I wouldn't even try using this thing without doing a lot of bench-testing
"first".

Also, no way you could use this thing over the range as specified. Eg, if Vin=40V and
Vo is set to 5V and Iout=1.5V, then each LM317 will be dissipating about

Pd = 1/2 * (40-5-1.25)V * 1.5A = 25 watts --> toast

Not hardly.

Thanks for the info! I was more confused about the concept of a constant current and constant voltage source at the same time. I came across this resource and may start here: http://www.scribd.com/doc/43459427/Analog-Power-Supply-Design Something like the diagram on page 62, but I would use some sort of modern transformer to give me DC power: no reason to shock oneself and they are probably cheaper to buy than make considering the cell phone industry...