7
\$\begingroup\$

I've done some research on brushless DC motor and have a few questions what would happen when a constant current is sent to a DC motor.

From my understanding, the current is directly related to the torque that the motor supplies, while the voltage would relate to the RPM. My question is, what would happen when you try to keep the current supplied constant, regardless of the load/rpm.

For example, let's say you have a weight attached to the shaft of the motor, like in this picture: https://www.clear.rice.edu/elec201/Book/images/img126.gif

You set current to be sent to the motor so that the torque applied is 10 in-lbs without any load. In this scenario, this current doesn't change. You attach a weight to the end of the pulley. What happens when the torque from the weight/load is 5, 10, and 15 in-lbs.

I assume, in the 5 in-lb scenario, the motor pull the weight up, at 10 in-lbs the motor would stall, and at 15 in-lbs, the weight would drop and it would actually be acting as a generator.

Now, I'm obviously ignoring the voltage and back-emf in this scenario. How would they play a role? Without any voltage, there won't be any current running through it, so there's no torque applied. But beyond that, would the voltage only affect the acceleration?

Sorry if this question has already been answered, but most of what I found related to a constant NET torque, not a constant applied torque from the generator.

\$\endgroup\$
  • \$\begingroup\$ A few points worth looking up and then clarifying in the question: (1) Brushless DC motors aren't driven by constant current drives, whereas brushed DC motors could be. So which is it? (2) The effect of voltage and of current aren't quite as perceived: The relevant parameter is power, i.e. voltage x current, and power relates to the generated torque. (3) This torque needs to overcome stiction, inertia (during speed transition anyway), bearing losses, vibration losses, and the actual load torque. At steady state, the generated torque must match the sum of the last few factors. \$\endgroup\$ – Anindo Ghosh Jan 8 '14 at 17:45
  • \$\begingroup\$ Useful reading on BLDC driving: BLDC Motor Control Algorithms, specially the first two lines: "Brushless motors are not self-commutating, and hence are more complicated to control. BLDC motor control requires knowledge of the rotor position and mechanism to commutate the motor." \$\endgroup\$ – Anindo Ghosh Jan 8 '14 at 17:53
  • \$\begingroup\$ The motor doesn't have to be brushless. I'll take a look at that article, thanks. It certainly makes sense that the voltage would play a role in the torque produced, but nothing clickedfor me as to how. Would a constant torque (ie. Overcoming all the resistance mentioned in number 3 EXCEPT the load torque) be possible? It just would require a more complicated control than simply a constant current ? \$\endgroup\$ – user35266 Jan 8 '14 at 19:25
  • \$\begingroup\$ Yes, a constant residual torque (after eliminating the various other forces opposing it) is certainly possible. Also, BLDC control algorithms are actually not very complex, as that same link above would have indicated. Constant torque would, simplistically, be easier to achieve than constant rotation speed, in some ways. The key part of the control for a BLDC is the commutation i.e. switching which coils get power in what sequence - one can't just be pushing a constant current through a single set of coils, else the motor would reach a static equilibrium and stop rotating. \$\endgroup\$ – Anindo Ghosh Jan 8 '14 at 19:31
  • \$\begingroup\$ Also, constant torque, constant rotation speed, constant acceleration to target set-point, and several other such control algorithms are available out of the box from a number of different BLDC controller ICs - it's far simpler doing what you need using one of those, compared to rolling your own. You might want to look up Texas Instruments InstaSpin-BLDC. It's just the family I am most familiar with, several other manufacturers have easy to use, sophisticated motor controllers as well. Also, similar integrated controllers exist for brushed DC too. \$\endgroup\$ – Anindo Ghosh Jan 8 '14 at 19:35
3
\$\begingroup\$

A constant current means, for an ideal motor, a constant torque. This is approximately true for real motors. It doesn't matter what you attach to the motor, or how fast it's turning.

What you seem to be missing is Newton's second law of motion. It states that force is the product of mass and acceleration:

$$ F = ma $$

The constant current you supply to the motor is one force. The weight opposes that force. The difference is the net force, \$F\$ in this equation, and \$m\$ is the mass of the weight, plus the mass of the rotor and the string and everything else the motor must move.

You set current to be sent to the motor so that the torque applied is 10 in-lbs without any load.

Not possible. There is nothing for the motor to "torque against". This is the mechanical equivalent of trying to develop 10 volts across a dead short. The motor will rapidly spin at its maximum speed, and the back-EMF will rise to the driving voltage such that your driving electronics are unable to supply enough voltage above the back-EMF to make enough current to have that much torque.

Let's just say you determine how much current is required for 10 in-lbs of torque, and you drive your motor with a constant-current supply set to that.

What happens when the torque from the weight/load is 5 in-lbs?

Assuming that the rotor and the string are massless and frictionless, the weight will be accelerated upwards by the net 5 in-lbs of torque (motor's 10 in-lbs, less 5 in-lbs from the weight). The rate of the acceleration is determined by the mass of the weight and Newton's law above.

As the speed of the motor changes (the weight is accelerating), the back-EMF also changes. Your constant-current supply to the motor will have to apply an increasing voltage to maintain the same current. Electrical power thus goes up, as does mechanical power.

What happens when the torque from the weight/load is 10 in-lbs?

Motor torque balances weight torque. However fast the weight is moving (if at all), it keeps doing that. Newton's first law applies.

What happens when the torque from the weight/load is 15 in-lbs?

The weight will accelerate downward, overpowering the motor. However, it won't be a free-fall. The motor cancels some of the force of the weight, resulting in a slower acceleration downwards.

If the weight overpowers the motor, then eventually it can get the motor to run backwards, relative to the way it would run if there were no load. When this happens, the back-EMF now adds (instead of subtracts) from the voltage you apply to the motor. At some point, your controller, which is attempting to maintain a constant current, must apply a negative voltage to maintain that current. In other words, the back-EMF is sufficient to create the necessary torque on its own: your controller must oppose it.

This is perfectly symmetrical with the first case, where the motor was overpowering the weight. In that case, electrical and mechanical power went up (without bound, if you let them). In this case, electrical and mechanical power go down (negative, if you let them). Energy is conserved because you are changing the gravitational potential of the weight.

The need to resist the back-EMF usually means storing electrical energy in a capacitor or battery, or using it to heat a resistor. If you can't do this fast enough, then the motor will create more torque than your desired 10 in-lbs, and you have hit the limits of your "constant current" driver.

Further reading:

\$\endgroup\$
  • \$\begingroup\$ Thanks for the detailed response and clearing up many misconceptions I had. It seems like I wasn't too far off. Great explanation on the third scenario as well. I will read those sources. Thanks again! \$\endgroup\$ – user35266 Jan 9 '14 at 1:11
1
\$\begingroup\$

Assuming the commutation electronics didn't have problems with the voltages involved, a brushless motor fed with a constant current would behave much as would a brush motor, and supply constant torque. If the combination of torque due to weight and friction were equal and opposite the torque produced by the motor, the motor would turn at constant speed. There would be a certain voltage drop due to DC resistance (equal to resistance times current), plus an additional voltage drop proportional to speed. If the motor's torque exceeded the torque due to weight and friction, the motor would continuously accelerate until that ceased to be the case; the motor's voltage drop would increase as this occurred. Speed would likely be limited by the supply's (or electronics') inability to supply current at higher voltages, by limited travel of the mechanics, or by mechanical breakdown.

In the scenario where the motor torque plus friction are insufficient to prevent the weight from falling, the weight would accelerate downward (or reduce its upward speed, if any). If the weight started by traveling upward (because of some external force, or because current had initially been higher) the motor would add energy to the weight as long as it continued moving upward. Between the time the weight started downward and the time the voltage produced by its speed was equal to the voltage loss due to resistance, the motor would be "plugging"--a condition in which the motor absorbs all mechanical and electrical power that goes into it (as the motor speeds up, the amount of mechanical power would increase and electrical power would decrease). Once the speed reaches the point that the voltage at the motor terminals would be zero, the motor will start acting as a generator, trying to push current in the direction it's already flowing [note that a the polarity would be the opposite of what typical constant-current supply would be equipped to handle, so such a supply would likely burn off all the generated power as heat].

\$\endgroup\$
  • \$\begingroup\$ Thanks for the response. Seems to be roughly what I expected. I am interestedabout the third case. Let's say the load is oscillating. Plugging would occur when the motor changes direction from positive to negative; would the same thing occur when the load decreases enough for it start spinning positively again? And it would it possible to set up a circuit that collects the power generated from the load reversing the direction of the shaft, correct? \$\endgroup\$ – user35266 Jan 8 '14 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.