As the title suggests I am looking for a way to drop 12V to 5V**. Which circuit should I use? And if I am going to use one, Why can't I use the others in the list ? I am looking for:
1. If the current going in is the same as the current going out.
2. Least heat dissipation.
My options :
Voltage Divider ; DCDC Converter ; Zener Diode ; Voltage Regulator IC.
** I am trying to run 4-5 servos (5V) using a microcontroller (PIC 18F452) and also 4 DC motors(12V).
Best Answer
First off, let's discuss your requirements. "Low drop of current" doesn't make much sense, typically you would specify something like "low dropout voltage", but that's not necessary in this case (12V is much more than 5V). Or perhaps you mean you want as much current out as goes in, which is easily met by a DC-DC converter. Regardless...
A voltage divider is not suitable for powering a load. As the current drawn by the load increases, the voltage decreases; and it will dissipate quite a lot of power.
Again, not appropriate except for very small loads, which motors are not. Large power dissipation for high load currents, and not very stable regulation. Zener diodes are typically used for references rather than regulators.
In the context of your question, these are the same thing. The distinction you are looking for is switching DC-DC converter vs. linear regulator (both of these are DC-DC regulators, and both are typically ICs). For your purposes, a switching converter is most appropriate as it will dissipate the least power, and will actually have more current out than in (a step down converter, 12V -> 5V, will be able to supply about twice as much current as the input current).