Summary: I am using a LM317 to drive a liquid pump that draws 1A. Unfortunately my regulator is dissipating way too much power and getting too hot. Currently my input voltage to the regulator is 12V and I'm trying to create 7V so that's 5Watts! So I bought a heat sink BUT I would also like to drop the input voltage of the regulator a little lower, no need for it to be 12V.
Goal: To drop the 12V input voltage to around 9V for the regulator by using a zener diode, that way the power dissipation is lowered to about 2W.
Question:
1) Could I get a sanity check on my circuit I created?
2) Do I need to worry about the current going through the zener diode because its 1A?
3) Could I just use a 3.3ohm resistor instead of a zener diode at all that way 3.3ohmd*1A is 3.3V drop which should do the same thing.
Best Answer
In principle this will work- the zener would have to dissipate more than 3W, so it will have to be something like a 5W zener- not cheap. The heat has to go somewhere in a linear regulator, adding the zener will move part of it to the diode.
The resistor will work if the output current is a constant 1A current, however motors, especially those used in applications such as yours, tend to draw much higher current during starting so it may end up stalling your pump and burning up the resistor.
At this level, I strongly suggest you use a switching regulator (buck regulator), such as those based on the LM2596. It will run cool. Inexpensive modules are available from China which may or may not use genuine LM2596s, but seem to work well enough.