Why does the rotational speed of a DC Motor drop when the torque increases?

subatomic particle

Joined May 8, 2018
76
Hello everyone,
i have a basic understandig of electric maschines and i am trying to understand some concepts better.

I understand the torque for rotational movement the same way i understand the Force for translational movement.
This means that if i applied a higher torque (more force), then i would make the rotor rotate faster, or?

Can someone explain this to me please?

ronsimpson

Joined Oct 7, 2019
2,797
Your data: This motor will turn at 4000rpms with no load. (0Nm)
If you load down the motor and pull out of the motor 0.4Nm the motor will spin at 2000rpm.

Maybe you are reading the graph backwards. The motor is turning the shaft. The load is trying to stop the turning. If there is no load the torque=0.

Irving

Joined Jan 30, 2016
3,548
As per Newton, in the linear world, Force = Mass x Acceleration, in the rotational world Torque = Inertia x Angular acceleration. Given a fixed inertia, more torque accelerates the rotation.

However your chart illustrates another relationship.... Power = torque x rotational velocity, so for fixed power, speed is ∝ 1/torque.

As @ronsimpson says, low torque loading =high speed. Obviously 0 torque = ∞ speed! so the nominal no load torque actually is not zero but represents the minimal inertia of the motor's own rotating parts. At the other end, 0 rpm is the stall torque.

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MrAl

Joined Jun 17, 2014
10,909
Hello everyone,
i have a basic understandig of electric maschines and i am trying to understand some concepts better.
View attachment 284182

I understand the torque for rotational movement the same way i understand the Force for translational movement.
This means that if i applied a higher torque (more force), then i would make the rotor rotate faster, or?

Can someone explain this to me please?
You have part of that right. Torque is for rotational movement and Force for translational movement. They work the same way except one is rotating and the other is just moving ahead.

The torque is developed through the applied voltage and resistance of the motor, and is reduced by the friction.
With the presence of a load, some torque is subtracted from the torque of the motor, so the more load the less torque. The final torque amount had to overcome that friction and with less and less torque with more and more load the speed reduces more and more.

It's interesting that when you apply more mass alone to the motor as part of the total load, it does not change the speed of the motor unless there is more friction with it. The extra mass will slow the motor down, but only temporarily until it starts spinning up to the normal speed. Same as translational speed with a larger mass. This of course means that larger masses take longer to reach full speed than smaller masses in either system. Friction, on the other hand, in either system could eventually stop the driver (motor).

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Tonyr1084

Joined Sep 24, 2015
7,556
This means that if i applied a higher torque (more force), then i would make the rotor rotate faster, or?
I've seen power supplies that can increase voltage in a variable voltage power source.
I've seen power supplies that can increase current in a variable current power source.
But I've yet to see a power supply that can apply higher torque.

As @ronsimpson says, low torque loading = high speed. Obviously 0 torque = ∞ speed! so the nominal no load torque actually is not zero but represents the minimal inertia of the motor's own rotating parts. At the other end, 0 rpm is the stall torque.
The armature (rotating mass) is driven by magnetic forces. As the commutator makes contact with a coil the magnetic force pulls the armature toward the magnet. As it gets closer the magnetic draw grows stronger, but the angle of the magnetism of the magnet (or stator coil) approaches its nearest the amount of developed torque drops. But the motor can still accelerate. Before the fields have finished attracting each other the commutator switches that coil off and turns the next on and the process continues.

When you add a load to the shaft you hold back the attraction and it continues to pull. The more you resist the further back the magnetic fields need to pull at each other. As you approach a stall, the magnetic fields get so strong that in increase in torque goes up greatly while the motor slows further and further. Since there is no knob to turn up the torque you would have to redesign the motor with possibly a moveable magnet or commutator brushes. There's a sweet spot where the motor is designed to operate at its optimum for the purpose it's designed. A motor with a slow acceleration is well suited for high speeds. A motor with rapid acceleration is not suited for high speeds but rather suited for pulling heavier loads. All that is taken into consideration when a motor is engineered. The angular declination of the magnetic forces as well as the intensity of those forces determines whether a motor is going to accelerate something from a standing start to max speed quickly or whether it's going to develop the speed over a longer period of time but resulting in a higher speed.

The more you bog down a motor the more current the motor will draw. Stalling a motor will result in the motor drawing max current, and possibly burning up.

Joined Jul 18, 2013
28,052
The armature (rotating mass) is driven by magnetic forces. As the commutator makes contact with a coil the magnetic force pulls the armature toward the magnet. As it gets closer the magnetic draw grows stronger, but the angle of the magnetism of the magnet (or stator coil) approaches its nearest the amount of developed torque drops. But the motor can still accelerate. Before the fields have finished attracting each other the commutator switches that coil off and turns the next on and the process continues.
No mention of the generated BEMF opposing the applied EMF?

Tonyr1084

Joined Sep 24, 2015
7,556
No mention of the generated BEMF opposing the applied EMF?
No. I tend to think pretty basic and simplistic. It's a major reason why I haven't gone far in my electronics hobby. But yes - there is BEMF. I just forgot to mention it.

MrAl

Joined Jun 17, 2014
10,909
I think it is also interesting the way torque is interpreted.
Of course the physical torque is the torque measured at the load, but in some models the torque is converted into speed (and speed into position), so if you have no speed you have no net torque because the torque produced by the motor is equaled by the torque of the load. This is like a force applied to a thick brick wall, where the force of the wall pushes back on the driving force equally and opposite in direction.

PaulEE

Joined Dec 23, 2011
474
Given a fixed voltage, you have fixed current. Given fixed current, you have a some maximum torque ability. The magnetic field opposing the stationary magnets in a typical DC motor is proportional to the current through the windings.

A motor rated for an amp will start on far less, but you can lightly pinch the shaft with your finger with current limited to 0.1A and stop it. Turn the current up and you'll burn your finger trying to stop it.

I used to work in the automatic door industry; the company I worked for used 90VDC motors. This is precisely how we made sure we didn't seriously hurt grandma; the H-bridge driver had a current shunt on the bottom, and we monitored it and were able to decrease the PWM duty-cycle to limit the average current so grandma didn't get plowed down.

I invite anyone reading this to hook up a DC motor and tune your power supply to (whatever it's rated) volts and limit the current to 1/10th of what it is rated and try it. It's really interesting to experience.

As for the original question - when you load a motor down, you demand that maximum torque, and often will max out the power supply's ability to supply any more current. This often causes the voltage to start to droop if you drag the power supply out of compliance, which slows it down...

Paul
KI5VNH

GetDeviceInfo

Joined Jun 7, 2009
2,180
As a motor is a power converter, from electrical to mechanical, you can calculate its power rating , at that voltage, from your data. It then goes, with an increase in torque, a corresponding decrease in speed is required at rated power.

Ian0

Joined Aug 7, 2020
8,947
Is it a series-wound or shunt-wound motor?

Sensacell

Joined Jun 19, 2012
3,330
Consider a theoretical perfect (impossible) brushed DC motor, driven by a perfect voltage source.

The motor would always spin at a constant RPM where the BEMF equals the supply voltage.

Loading the motor would cause current to increase, but the speed would remain constant.
The current could go infinitely high to hold the speed.

But now consider a real, impefect motor with coil resistance; loading the motor does increase the current, but the coil resistance now prevents the current from increasing to the level required to maintain the original RPM - the RPM drops to a lower equilibrium level.