Hello everyone,
Can I run a 5v motor from a 12v power source if I use a PWM (500Hz) duty cycle that does not exceed 50%?
TIA
Can I run a 5v motor from a 12v power source if I use a PWM (500Hz) duty cycle that does not exceed 50%?
PWM turns on and off the supply voltage (12V) frequently. It doesn't change the voltage. The motor will still see 12V when it sees any power.
Can I run a 5v motor from a 12v power source if I use a PWM (500Hz) duty cycle that does not exceed 50%?
I'd say yes, you can. Motor reacts on RMS value, as any inertial system (including heaters).
Hmm, well I was hoping for a better consensus... 
PWM turns on and off the supply voltage (12V) frequently. It doesn't change the voltage. The motor will still see 12V when it sees any power.
That was my fear, however, a mulitmeter does read the pwm output as half voltage.
Motor reacts on RMS value, as any inertial system (including heaters).
And that was my hope, that the peak would be quick enough that it wouldn't have time to cause damage.
Anyone else?
A PWM signal passed through a low-pass filter (which is what a motor effectively is) is what is known as a switching regulator 
That's basically how a switching regulator works. A square wave is generated and passed through a low pass filter to create a voltage. Feedback from the output voltage is used to set the duty cycle of the square wave to create a regulated output voltage.
Your PWM signal is basically an unregulated switched mode power supply.
That was my fear, however, a mulitmeter does read the pwm output as half voltage.
Because a multimeter typically can not react fast enough to the changing on/off state of the PWM pin.
Why not use the proper supply voltage? What (else) is the Arduino doing? How much amperage does the motor pull? What motor shield are you using?
Pokey:
Can I run a 5v motor from a 12v power source if I use a PWM (500Hz) duty cycle that does not exceed 50%?
Yes you can, although as PaulS says it is not an ideal solution.
Power dissipated in the motor is roughly proportional to voltage squared so to get a similar power to the '5V' case using 12V you will probably end up with a maximum duty cycle of around 17%. Your peak current requirements might go up substantially too, so make sure you check whether the supply and driver circuit can handle that.
Power dissipated in the motor is roughly proportional to voltage squared so to get a similar power to the '5V' case using 12V you will probably end up with a maximum duty cycle of around 17%.
This is false statement. I'd refer to microchip application note AN905.
http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=1824&appnote=en012149
When using digital control, a
pulse-width modulated (PWM) signal is used to gener-
ate an average voltage. The motor winding acts as a
low pass filter so a PWM waveform of sufficient
frequency will generate a stable current in the motor
winding. The relation between average voltage, the
supply voltage, and duty cycle is given by:
EQUATION 1:
VAVERAGE = D x VSUPPLY
Speed and duty cycle are proportional to one another.
The motor winding acts as a low pass filter so a PWM waveform of sufficient frequency will generate a stable current in the motor winding.
That's relative.
Is that "sufficient frequency" necessarily == 500 Hz?
-o-o-o-o-o-o-
"Generally, it is a good idea to experiment with the PWM frequency for a given motor to find a satisfactory frequency."
Still, I'm not getting that their argument is a blanket sanction for "over-volting"
Magician:
This is false statement. I'd refer to microchip application note AN905.
I disagree. Both the current rise against induction, and peak current constrained by resistance, are proportional to current. Dissipated power is proportional to the integral of voltage and current over time. If you use PWM the current will be less than if you use a constant voltage, but with or without PWM if you double the voltage you will get roughly double the current and roughly four times the dissipated power.
PeterH:
Magician:
This is false statement. I'd refer to microchip application note AN905.I disagree. Both the current rise against induction, and peak current constrained by resistance, are proportional to current. Dissipated power is proportional to the integral of voltage and current over time. If you use PWM the current will be less than if you use a constant voltage, but with or without PWM if you double the voltage you will get roughly double the current and roughly four times the dissipated power.
It depends on the PWM frequency and the inductance of the motor. In all cases, the power dissipation in the motor will be the integral of I^2 * R over time, where I is the motor current and R is the motor resistance. You can't calculate the power from the motor voltage so easily because the motor has inductance and back emf as well as resistance
With low frequency/low inductance, the current through the inductor will settle near V/R over most of the active part of the PWM cycle, where V is the applied voltage (12V less the voltage drop of the switching device) and R is the resistance of the motor. The flyback diode(s) will conduct for a very small part of the cycle. Torque will be proportional to the average current, and at 50% PWM power dissipation will be roughly twice as much as if the motor were powered continuously at half the voltage (i.e. 4x the power dissipation for half of the time).
With high frequency/high inductance, the motor inductance together with the flyback diode(s) will smooth out the motor current so that it is roughly constant throughout the cycle. In this case, there will be no increase in power dissipation in the motor compared to using a lower voltage and no PWM.
That's why the application note referred to "PWM of sufficient frequency".
dc42:
It depends on the PWM frequency and the inductance of the motor.
Sorry Magician, you were right.