Lowering voltage: voltage divider, voltage regulator or switching?

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1

Hi!
A question by an electronic newbie.
I need to power a vibra motor which needs 3.3 volts and 60 mA.
My power supply is a 9 volt battery.
I know I can drop voltage in 3 different ways:

  1. a voltage divider (using 95 and 55 ohm resistors);
  2. a voltage regulator (for example the LM1117);
  3. a switching voltage regulator (for example the Pololu 3.3 D24V3F3).

My question is very general: can you explain me what is exactly the differences in these methods?
I guess it is about saving current: method 1) wastes current; method 2) wastes current but less than first method; method 3) is the most efficient.

Please, can you confirm my statements? I'm not sure about them.
Thanks a lot.
Valerio

2

valerio_sperati:
My question is very general: can you explain me what is exactly the differences in these methods?
I guess it is about saving current: method 1) wastes current; method 2) wastes current but less than first method; method 3) is the most efficient.

Please, can you confirm my statements? I'm not sure about them.
Thanks a lot.
Valerio

Substitute the word "power" or "energy" for current, and you are correct. But there is also the difference that (1) has poor voltage regulation, compared with (2) and (3).

3

Note that all those diagrams of different methods of voltage reduction are rubbish because there are no capacitors in any of the circuits. You need capacitors when doing this sort of thing to keep the circuits stable.

4

  1. A motor with a constant load (current) could be supplied through a current limiting resistor. One resistor.
    The one to ground is just using power.
    You only need to drop the excess voltage. 9 - 3.3 = 5.7volt. Resistor needed is 5.7volt / 0.06Amp = 95ohm.
    Your calculation for that resistor did not include the 55ohm resistor.
    That one would have used 3.3volt / 55ohm = 60mA. Nothing left for the motor.
    A resistor is not very good when the battery goes down.

  2. A lineair regulator controls current to try to keep the voltage across the motor constant.
    The excess voltage (5.7volt), times the current (motor and ~4-10mA for the chip) is, like the single resistor, converted into heat.

  3. A switching regulator is more a power converter. Like a transformer is for AC.
    It lowers the voltage, and increases the current.
    The motor uses 3.3volt * 0.06A = ~0.2watt.
    Battery draw is 0.2watt / the converter's efficiency.
    Battery current draw, unlike 1) and 2), is lower.
    Could be 0.2watt / 0.8 = 0.25watt = 0.25watt / 9volt = ~28mA.
    Use micro power converters for small loads.
    Bigger converters could have bad efficiency without, or with light loads.
    Leo..

5

Resistor divider is for measuring only, never for power.

switch-mode is the most efficient, but generates noise - not suitable for sensitive analog
circuitry (RF receivers, microphone amplifiers).

linear voltage regulator typically gives low-noise stable output, always used for quality analog
circuitry, but anything less critical the power wastage means switch-mode is better (but can
be more expensive)