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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
If I have a DC motor rated at 1hp at 1750 rpm and I attach a heavy load such that it draws 10A at 300rpm, what will happen if I use a 2:1 gear reduction so that I am spinning the same load at 600rpm. That is, the load is moving at 300rpm when the motor is now going at 600rpm? Will the current be the same or is there less current draw when the motor gets to its rated speed?
Registered Member #56
Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445
That depends on the efficiency of the motor, most motors have considerably more efficiency at at closer to their rated RPMs, so you can expect a considerable drop in current draw from the motor.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
"Most motors" doesn't help here, this is a DC motor, which I assume means permanent magnet (PM).
For a perfect PM motor(and 1hp is getting to be big enough for that to be a good first approximation), the terminal voltage is proportional to the speed, and the current consumption is proprortional to the output torque. It's that simple.
The first correction for losses means more current is needed to supply friction, eddy and viscous torques, and more voltage is needed to overcome IR drops in the windings. Both easily figured from "real efficiency < perfect efficiency".
If it's not a PM but a wound field motor, then a shunt winding behaves just like a PM, except it draws additional power. If it's a series wound motor (car starter motor for instance), then the speed/torque curve becomes very lazy (as opposed to stiff for a PM/shunt motor), delivering very high speed off load and high torque at modest current at the stall.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
When you used a gearbox to double the motor RPM, the motor torque would be cut in half, according to conservation of energy in the gearbox.
The equations for an ideal PM motor are simple:
Torque = Ki * current RPM = Kv * voltage
where the torque constant Ki and back EMF constant Kv come from your motor manufacturer, or you work them out yourself from locked-rotor and free running tests.
So the motor would need twice the voltage in order to reach twice the RPM that it did before, and when it got there, would draw half the current to provide half the torque that it did before. Twice the voltage and half the current works out to the same power, so again Mr. Conservation Of Energy is kept happy.
In real life, small motors and gearboxes are quite inefficient, so the behaviour is not as clear-cut as this. But you could still expect the motor to need more voltage and draw less current.
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DC motors and current consuption
