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I have seen a 500w motor which run on 1200 rpm but a 300w motor with over 10000rpm . I know using a gearbox can give required rpm or torque and torque is inversely proportion to speed but how can I increase speed and decrease torque keeping watts same. Please explain for dc and bldc motor.

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    \$\begingroup\$ Is this a schoolwork assignment? \$\endgroup\$ – Huisman Mar 27 '20 at 10:39
  • \$\begingroup\$ The watts drawn by a motor depends on the mechanical load you put on it. So it may be impossible to alter one parameter without changing others. \$\endgroup\$ – Simon B Mar 27 '20 at 10:46
  • \$\begingroup\$ Essentially, gearbox (or pulleys or CVT) is the answer. \$\endgroup\$ – user_1818839 Mar 27 '20 at 13:56
  • \$\begingroup\$ More turns and poles makes for lower RPM per volt, and more torque per current \$\endgroup\$ – DKNguyen Mar 27 '20 at 16:41
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how can I increase speed and decrease torque keeping watts same. Please explain for dc and bldc motor.

Brushed or brushless DC makes no difference, the principle is the same.

When the rotor/armature is spinning the motor acts as a generator, producing a voltage proportional to rpm (Faraday's Law). This voltage opposes the supply voltage. The difference is impressed across the resistance of the windings, causing a current flow determined by Ohm's Law. The result is that as motor speed increases the current (and torque, which is proportional to current) reduces until the torque produced matches the load torque. Maximum rpm occurs at no load, when the motor is generating (almost) as much voltage as the supply. The speed at this point is determined by magnet strength and physical construction of the motor (number of magnet poles, number of turns per winding etc.).

Output power = torque x speed. To make the motor spin faster while keeping watts the same you have to increase the supply voltage and reduce torque. One way to reduce torque is to use a gearbox or equivalent speed reducing mechanism. Other methods involve changing the load in some way, eg. smaller wheels on a vehicle, fewer/smaller blades on a propeller.

Note that motor output power is always less than input power due to losses in the motor (including 'copper' loss caused by power dissipated in the winding resistance, 'iron' loss caused by hysteresis and eddy currents in the magnetic core, friction from bearings and brushes etc.).

DC motors are often rated by input power, but losses vary with current and rpm so the efficiency is not the same at different operating points. The same motor run at higher rpm with a gearbox usually has higher efficiency (and therefore higher output power) due to the lower current draw. This is partially offset by losses in the gearbox.

A higher gear ratio comes at the cost of reduced efficiency, and higher motor rpm increases iron losses, so there is a limit to how much you can increase the voltage. The motor is also limited by maximum temperature and rpm (too hot and the magnets get demagnetized or winding insulation burns, too fast and the brushes bounce, armature throws windings, rotor flies apart).

In your example of a 500 watt motor at 1200 rpm vs a 300 watt motor at over 10000 rpm, with a suitable gearbox on the second they might be equivalent. However boosting the voltage on the first motor to get the same speed as the second will probably be less efficient and may not be safe. In general it is best to match the load to the motor's intended operating range.

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The speed of a d.c. motor is given by

$$ N = \frac { V - I_a.R_a}{kφ} $$

where 'V' is the applied voltage, 'Ia' the armature current, 'Ra' the armature resistance, 'k' a constant based on the number of poles, the number of conductors and number of parallel paths in the armature and 'φ' the flux per pole.

That speed could be either increased or decreased with the use of a gearbox.

The varying characteristics of motors are the result of design considerations made by the manufacturer over which you, as the user, have no control.

Should you have access to the armature and field terminals, the motor speed may be increased beyond rated speed by reducing the field current and consequently the torque. Likewise, the speed may be lowered by reducing the armature voltage keeping the field voltage constant.

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  • \$\begingroup\$ Does your last sentence mean to say, "You can only reduce from the rated speed by reducing the applied voltage from the rated voltage"? \$\endgroup\$ – Transistor Mar 27 '20 at 21:36
  • \$\begingroup\$ In general the speed of a DC shunt motor may be increased by field weakening. It is assumed that in the referred small motors there would be no access to the field winding. \$\endgroup\$ – vu2nan Mar 28 '20 at 2:08
  • \$\begingroup\$ I am sorry I misunderstood your question. Yes, I did mean that and have accordingly reworded my last sentence . Many thanks. \$\endgroup\$ – vu2nan Mar 28 '20 at 2:23

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