bananasiong

Hi,
I am always wondering why does the DC motor need more current when the load added and draw the max current when it stalls.
These are my own opinion:
The DC motor that is usually used is 3-pole type. When the power is turned on, one of the coil is conducted and electromagnet created. Before the current rises to the max, the second coil is conducted and the first coil is no longer connected to the brushed due to the pushing of the magnetic field causing the changes of the brushes connection. Then the same thing happen to the 2nd and 3rd pole as the motor turns around. So this constant current is the current required by the DC motor without load.
When the load added, the speed becomes slower and hence, the conduct time of the coil is increased and the current has enough time to rise higher.
When the motor stalls, the brushed are connected to only one coil, and this is the max current required by the motor.
So basically the max current is defined as:
Imax = Vsupply/Rcoil
Vsupply = supplied voltage
Rcoil = coil resistance

Are the explanation correct?

Thanks

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Exam T_T.T

Leftyretro

More to it then that, there is a counter-EMF force that comes into play.

Here is good explanation of simple DC motor

http://en.wikipedia.org/wiki/Brushed_DC_Electric_Motor

Lefty

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Measurement changes behavior

dknguyen

The movement of the magnets passing the windings in the motor produces a back EMF which is like a voltage that opposes the voltage applied to the motor. The back EMF is greatest when the motor is spinning the fastest. So when the motor is spinning fast (ie. low load) the two voltage oppose each other the most and the net voltage across the windings is the least. As the motor slows down due to increased load, the back EMF voltage is reduced and therefore opposes the applied voltage less which makes the net voltage across the motor windings larger thereby increasing the current. At stall, there is zero back EMF generated so the net voltage is the applied voltage and it's entirety appears across the windings resistance which I think is the stall current: I = V/R.

I'm not sure what part your explanation (about how the windings switch before the current can completely rise due to motor inductance) plays in this, but for a brushless motor where you "manually choose which windings to energize " rather than rely on brushes to automatically do the commutation, if you apply simply apply current across one winding it shorts out. It is the exact same effect as stalling out a brushed motor except in a brushless motor there are no brushes to try commutate and change the windings that are being energized when you do this. So, like a brushed motor you only ever momentarily energize the windings and then commutate to another winding at a rate where the motor inductance is able to prevent the current from ever rising to short-circuit (aka stall levels).

So the effect of increasing current as the motor slows down is due both to the current rise-time via the motor inductance AND due to the back EMF. Both are dependent on speed.

THe one that governs how fast a motor spins though, is the back EMF effect. The inductance-induced-current-rise-time does not explain why a DC motor has no-load speed limits. The motor spins at a rate where the induced back EMF generated cancels the applied voltage.

survivingbrain

Thinking it in another perspective:

The free running motor (be it any kind of electric motor!!) doesnt have any load, that means no work is done by the motor.

Let the electromagnetic flux required for a free running motor coil be 'x'.
Since the motor has no load, less flux is required to 'lift' the armature and to rotate the shaft.

When load acts, more flux has to be linked with the coil!! (to overcome the load)
more flux requires more current!!