voltage, current, torque and speed in DC motors

Question

First off, please don't accuse me of not doing proper research before asking--I have done much research, but I could not find any answers that made sense to me. I have been playing with some DC motors, but I am confused as to the relation between voltage, current, torque and speed. I have noticed that sometimes, the motor seems to have little to no torque, yet when I limit the current to the motor, it seems to have a different behavior. Unlike this question, I want an answer that explains (preferably with formulas) what determines voltage, current, torque and speed, in simple terms. thanks!

Answer 1

I have noticed that sometimes, the motor seems to have little to no torque, yet when I limit the current to the motor, it seems to have a different behavior.

In a standard (brushed, permanent magnet) DC motor, torque is directly proportional to current. So if you find that the motor sometimes has less torque it's simply because it is drawing less current.

I want an answer that explains (preferably with formulas) what determines voltage, current, torque and speed, in simple terms.

The other question has an answer with formulas, but perhaps the explanation wasn't simple enough for you?

In simplest terms:-

Voltage is set by the supply. When spinning the motor generates an internal voltage called back-emf, which is proportional to rpm. The difference between the back-emf voltage and the supply voltage is dropped across the motor's internal resistance (Rm), resulting in a current draw according to Ohm's Law (I = V/R). The current produces torque which causes the armature to spin faster, increasing back-emf and reducing current until the torque produced is balanced by internal and external loading.

If there were no internal losses an unloaded motor would spin fast enough to produce a back-emf equal to the supply voltage. However some current is required to overcome internal mechanical losses (brush and bearing friction, wind resistance) and magnetic losses (eddy currents, hysteresis) which are roughly proportional to speed. The motor's no-load current accounts for all of these internal losses. No-load current does not contribute to output torque, and slightly reduces no-load speed due to voltage drop across Rm.

When an external load is applied current increases to produce the torque required to match it, and speed reduces due to the increased voltage drop across Rm. As load is further increased the motor slows down more, eventually reaching stall (zero rpm) where there is no back-emf and all the supply voltage is dropped across Rm, resulting in maximum current and torque.

As loading increases torque increases linearly but rpm decreases linearly, so maximum output power occurs at 50% rpm and torque. Motor efficiency is the ratio of output power (torque x rpm) to input power (Volts x Amps).

Plotting all these interactions on a graph gives you this:-

Answer 2

The 'simple' answer is that for any given motor, it has a mostly fixed operating voltage and speed (I am not going into Variable Frequency Drives or other complex servo-loop drives). For a given normal load it pulls so many amps of current to maintain the torque (physical energy as angular momentum).

Things stay the same until you increase the load on the motor, like putting a lot more clothes in your clothes washer. It attempts to maintain the same speed but must pull more current to increase or maintain the torque it needs to work. Remember that work is equal to joules, which is equal to 1 watt over 1 second.

Watts RMS for AC motors is voltage times the amps consumed. 746 watts is equal to one horsepower. (I will leave out the details of phase lag here). If you limit the current a motor can draw then you limit the maximum torque it can apply to do extra work.

Normally all motors have a fuse or circuit breaker to limit the current to safe levels or the motor would burn up trying to move an extra heavy load.

That was the short answer. See @replete's comment and link above for the fine details.