I believe DC motors accelerates from 0 to maximum speed when power is applied. Let's say, it takes 5 seconds so that the DC motor could reach the maximum speed given certain power. Is it possible to rotate (or give torque) to the DC motors so it can reach the maximum speed directly (say at half a second)?
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2\$\begingroup\$ I I lad. :) Depends on the motor and the current. \$\endgroup\$– kennySep 9, 2011 at 11:21
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\$\begingroup\$ With an adequate electrical supply, the limiting factor with motor spin-up is rotor inertia. Load inertia may play into it somewhat, but usually loads on electric motors are geared down significantly, which causes the motor's own (ungeared) rotor inertia to dominate. \$\endgroup\$– JustJeffSep 10, 2011 at 2:15
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\$\begingroup\$ @JustJeff - Winding resistance may also pay a factor. If your power supply can supply a kiloamp, but the motor's internal resistance limits it to 10 amps, a bigger supply will not help. \$\endgroup\$– Connor WolfSep 10, 2011 at 6:08
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1\$\begingroup\$ @Fake Name - guess I was lumping all electrical effects together there. Certainly more current means more torque means faster spin up, but eventually, whatever you do electrically, you're up against this inertia. \$\endgroup\$– JustJeffSep 10, 2011 at 13:38
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3 Answers
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In an ideal DC motor, rev is proportional to voltage and torque proportional to the current. So if you connect it to a constant-voltage supply, it would ideally reach the maximum speed immediately. Of course, this does not happen in reality: the mechanical parts have a finite moment of inertia, so you get an increase of angular momentum -> torque -> current, which would be infinite for an immediate-response. Such a current is prevented by two factors:
- a non-arbitrarily powerful supply,
- the internal resistance of the copper wire inside the motor.
Assuming you have an over-sufficiently powerful supply, you can also overcome the latter factor to some degree: the voltage that drops off at the copper resistance simply follows Ohm's law. You can determine the resistance \$R\$ by applying a small voltage to the motor while this is blocked mechanically. Then, rather than simply using the constant voltage that corresponds to the desired rev in torque-free mode, you always add the voltage \$R\cdot I\$. You need to be careful not to produce resonances with such a feedback circuitry. Also, not all motors might cope very well with the large currents that may arise with this method.
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In order to speed up the start you need to apply a great power. There are two problems: heating by overcurrent and isolation by overvoltage. Most of motors can handle about 10% over the specification for small amount of time, 10 - 30 seconds. But this will reduce the life of the motor by almost 50%. It is like this becouse they are made to overcome the inertial load force on starts.
If you're not starting with full load applied, increasing the voltage/current may be a solution, but in this case try to use a good protection system. A speed controller can also be useful.
Another option, that is commonly used on big motors, is to use another smaller and faster motor in the same axis that will help the main motor to get out of inertia. This also reduce the main motor current consumption on start.
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If you apply a constant voltage, you get approximately an exponentially decaying current, from the stall current down to the running current. As the current decays, so too will the torque, and the consequent angular acceleration.
If instead you were to force a constant current, equal to the stall current, the torque wouldn't drop off, so neither would the acceleration, and the speed would come up more quickly. However, as the speed comes up, the back emf generated in the motor will also rise, which means that the voltage appearing across the motor will rise quite significantly, so you'd probably want to make sure that the motor's voltage rating isn't exceeded, lest the insulation break down.
Now, to keep the current constant, at any instant the voltage would have to be equal to the nominal rated voltage plus the emf due to the motor's speed at that instant. When the motor reaches rated speed at the rated voltage, its emf is almost equal to the rated voltage; so by the time your constant current drive gets the motor up to rated speed, the motor will have twice the rated voltage drop across it. Some motors may take this, some may not.
