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I have some confusion on how to power a 1 hp DC winch motor that is intended to run off a 12 V battery. I need to precisely control the velocity of the motor and will need a PID controller to do this. Since speed of the motor is proportional to the supply voltage, is 12 V the maximum voltage and corresponding speed that the motor can reach? Can I apply a voltage greater than 12 V to it? I need to find a motor driver/controller that can handle the current draw. Does anyone know of a controller that can run this motor. Ideally it would have PID built in but it is not necessary. I believe my questions are from a lack of understanding of how voltage and current are related. Also the winch will not be spinning very fast(no more than a line speed of .5 feet per second). I am currently building a tachometer for the feedback control. Any help/suggestion is greatly appreciated.

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  • \$\begingroup\$ possible duplicate of Choosing a PID controller \$\endgroup\$ May 27 '11 at 18:25
  • \$\begingroup\$ not a duplicate but I had more specific questions to ask about the voltage and current. \$\endgroup\$
    – pdfj
    May 27 '11 at 18:50
  • \$\begingroup\$ This is like a better phrasing of your earlier question. (as identified by @Leon Heller). It would be good if you edited this detail into that question. \$\endgroup\$
    – JustJeff
    May 27 '11 at 19:55
  • \$\begingroup\$ If the motor is rated is 12Volt you should not go more than that- But there is always a margin. i suspect going up to 13.3 volts wont damage it - but it will cause undesired effects such as heat.. and possibly burn out the coils. \$\endgroup\$
    – Piotr Kula
    Jun 27 '11 at 17:21
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An unloaded DC engine runs with a speed proportional to the voltage, ideally without drawing any current (In practice, internal friction will cause the engine to draw some current). As you start loading the engine, the current will increase, and some power will be lost due to resistant losses in the windings. The applied voltage will now split into two parts. One part, the electric losses, is current times winding resistance, this part only generates heat in the engine. The other part is the voltage turning the engine. With load, current will increase, and your motor will slow down, unless you compensate by increasing the total voltage.

There are basically two things that can cause your engine to break down. If the engine is stalled, or you put too much energy into it, the resulting heat will burn of the windings and/or the brushes. A good way to avoid this is to have some current limit or fuse to limit the current if the engine is overloaded.

If you let the engine go too fast. Centrifugal forces will cause the rotor to collapse. Typically the windings gets pulled out, until they touch the stator and gets ripped apart. This means you can apply more than the rated voltage on a loaded engine as long as you A) Don't let the engine over-rev. B) Don't let the engine get to hot.

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  • \$\begingroup\$ A very good and comprehensive answer. People sometimes get confused about brushed DC motor voltage rating. While the hi-pot works until, say, 500 V, the working voltage is limited by unloaded motor speed. Similarly, too high voltage can produce too much current in case of a stall. \$\endgroup\$
    – SunnyBoyNY
    Nov 6 '14 at 3:44
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Motor velocity control of motors is usually done through Pulse Width Modulation where the voltage remains constant but is switched off/on many times per second. The more time it spends "off" the slower it goes.

A low-power PWM generator can be fed through something like one or more MOSFETs to switch the main load. The "duty cycle", or percentage of time the PWM generator spends in the "on" position, can be from any source you like - an analogue input (potentiometer), a PIC microcontroller or such (which often have a PWM controller built in), or whatever you like.

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