Good day, everyone.
I am currently working on a project that uses an LM317 adjustable voltage regulator with a 12-15 volt battery pack (not exactly sure which, yet). I know that if you want to be able to adjust the output voltage by hand, you use a pot connected to the 317. My question, though, is this: As the battery voltage drops, how much will the position of the potentiometer have to change to keep it at a specific voltage? In other words, let's say I have a 15 volt supply, and I have my potentiometer at the point where I get 9 volts from the regulator. When the supply voltage drops to, say, 12 volts, would the pot have to be in a different position to still get 9 volts from the regulator? I don't imagine this would be the case--It would make more sense if the variable resistor would stay in the same spot, but I just want to make sure.
Thanks in advance
Der Strom
You are correct. If the supply voltage drops below about 11 volts in your case the output may be below 9V. This drop out voltage depends upon the temperature of the LM317 and the load current. For your supply I would use a 120Ω resistor between the output terminal and the adjust terminal and a 1KΩ pot between the adjust terminal and ground.
The 317 will regulate to a chosen value so long as there is about 2-3volts across the regulator to give it voltage to work with, and the 317 hasn't gone into thermal limiting. Any input voltage from ~11-12v to less than 37v will provide that 9v output. However, as the voltage across the 317 increases, its power dissipation increases too. More dissipation = more heat.