Resistor vs. voltage regulator question

[robert gagnon]

I have this 12 volt/5000 rpm motor that I want to run from 6 volt at 450
rpm.
A 6 ohm resistor connected to the battery does just that and the motor draws
..750 amp. from the battery. However I made a simple adjustable voltage
regulator with a LM317 that draws the same amperage at 450 rpm. My question
is: will the battery lasts as long with either method or is one method
better than the other? Also are there other more efficient ways to do that?
I'm a newcomer in electronics so any somewhat simple answers would be
appreciated. Thanks , BobG.

 

[Eric Immel]

robert gagnon wrote:

I have this 12 volt/5000 rpm motor that I want to run from 6 volt at 450
rpm.
A 6 ohm resistor connected to the battery does just that and the motor draws
.750 amp. from the battery. However I made a simple adjustable voltage
regulator with a LM317 that draws the same amperage at 450 rpm. My question
is: will the battery lasts as long with either method or is one method
better than the other? Also are there other more efficient ways to do that?
I'm a newcomer in electronics so any somewhat simple answers would be
appreciated. Thanks , BobG.

A much more efficient way to do this is to use a pulse width modulator.

Your current setup simply burns off a bunch of power by dissipating it
in either the regulator chip or the resistor. You can figure out how
much power is being used by the formula P=IE. That is, multiply the
current through the circuit .75A, by the voltage dropped across the
limiting device 12V(source)-6V(motor)=6V(resistor or regulator). This
means that the wasted power is 6V*.75A=4.5W this is true with either
method. Dissipating this much power not only drains your battery, it
creates a large thermal load on the limiting device, you'd need a 10W
resistor, or a good sized heat sink on your regulator. You can calculate
the electrical efficiency of this circuit by dividing the useful power
by the total power. Total power = 12V*.75A = 9W. Electrical efficiency =
4.5W/9W = 50%. How to improve this?

A pulse width modulator (PWM) limits the average power by delivering
full power for some time, then delivering no power for some time in a
repating periodic fashion. In other words, the limiting device connects
the full 12V to the motor for a while, then disconnects it completely
for a while, then connects, then disconnects, etc. This is repeated
quickly enough that, to the device being powered, the applpied power
looks continuous. The ratio of time on to time off determines the
average power delivered to the load. To see why this is more efficient
repeat the calculations above for each of the two states of the limiting
device. Power disipated by the device when connected (assume 1A motor
current) = 12V(source) - 12V(motor) = 0V(limiting device). Wasted power
= 0V*1A)=0W. Power disipated by the device when disconnected (no motor
current) = 12V(source) - 0V(motor) = 12V(limiting device). Wasted power
= 12V*0A=0W. If the limiting device connects the battery to the motor
half the time, the average power to the motor will be the same as
applying 6V. If you calculate the efficiency for this PWM, you come up
with 100%! Yay! Free energy! Not so fast. This is a description of an
ideal device. Such things exist only in our dreams. Here in the real
world, the limiting device will drop some voltage when connected say
about .2 to 1 volt depending on the type of device. Also the circuitry
that generates the timing will use some energy. PWM drivers can provide
very high efficiencies, but there's no such thing as a free lunch.

For more detail, google PWM motor control. Additional keywords will
probably be necessary.

EI