Electronic – Locking up a DC motor with constant supply a bad thing

Power supplies do not "push" current; components (like motors) "draw" current from the voltage source. For ohmic (resistive) loads, the current drawn equals voltage divided by resistance.

Voltage sources (power supplies, batteries, etc) do not have the ability to provide an infinite current. Internal resistance in the source will make the voltage "sag" as the current goes up, because voltage loss over any resistance (including internal resistance, wiring, etc) equals current times resistance.

For example, a small 9V battery has a very high internal resistance. (Another related term is voltage source impedance.) Draw 100 mA, and it may lose 2V. This means we can calculate the internal impedance to about 20 Ohms. A LiPo battery intended for high current discharge has a much lower internal resistance. Let's say you get a 1V drop when you draw 25 Amps. This means we can calculate the source impedance to 40 milli-Ohms. A bench power supply may be capable of increasing its internal source voltage to compensate for sag, until you reach the maximum current. Thus, a 12V supply may supply an even 12V all the way up to a rated max current -- let's call that 5A. After that point is reached, the supply will lower its voltage until the current draw is no more than 5A. (This is a "constant current" supply.) This typically works because the current draw of most loads is proportional to the voltage. However, for a "true" short, the supply needs to lower the voltage all the way to 0, because even at very low voltages (0.1V, say) more than 5A may be drawn by the "load" that's the short.

So, if you want the motor to be able to provide all the torque it is rated for, to the point where it stalls, at the voltage that the motor is rated for, then you have to use a power supply that is capable of supplying the rated voltage, at the rated stall current. However, you don't want to do that for any real time, as the motor will quickly overheat and damage itself.

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:

• How can I implement regenerative braking of a DC motor?
• Why should I worry about a motor causing my supply voltage to shoot up when the back-EMF can't exceed the supply voltage?
• How can I measure back-EMF to infer the speed of a DC motor?