From what I read here and on the internet I understood that shorting or adding a big load to the contacts of a DC motor will make it very hard to turn as a generator.
I have a 500W brushed DC motor that is considerably hard to turn by hand as is. Upon shorting its contacts I imagined it would be impossible to turn, however in reality I observe that it is just as easy (or as hard) as before. Why is that and how can I prevent the motor from turning (a.k.a. parking brake)?
Best Answer
You are not getting much breaking torque because your motor does not generate enough back EMF at low velocities, and because it has a bad power factor when connected to a small real load.
simulate this circuit – Schematic created using CircuitLab
This is a simplified equivalent circuit showing a DC motor on the left side and a load on the right side connected through a switch, this circuit does not include any parasitic effects.
In an ideal DC motor V_BackEMF is proportional to the angular velocity of the motor, the proportionality factor depends on the specific motor.
In an ideal DC motor the current through L_coils and R_coils is proportional to the torque imparted on the shaft of the motor, again the proportionality factor depends on the specific motor.
When you turn the shaft manually without the coils being energized you create a voltage V_BackEMF which gives rise to a current through L_coils and R_coils and which in turn gives rise to a torque on the shaft of the motor.
The voltage V_BackEMF can be easily measured by turning the shaft at a fixed angular velocity and measureing between the poles of the motor with an RMS voltmeter or an oscilloscope.
The resistance R_coils can be easily measured with a ohm-meter.
The current is a bit more complicated to understand, because due to the inductance L_coils the current is not in phase with the voltage, if you look up the equation for an inductor in the time domain you will find that the voltage across the inductor is the product of the derivative of the current through the inductor and the inductor value. If we swap the equation around we get that the current is the product of the integral of the voltage and the reciprocal of the inductor value. Now if we let the voltage be a cosine function then we know that the integral of a cosine function is a sine function, which corresponds to a phase shift of pi/2. In other words the current comes pi/2 (90 deg.) after the voltage.
Due to the voltage and current being out of phase the power factor of the motor is not very high when connected to a very small real load (a short), consisting of R_coils and R_load in series, and hence the current is not going to be as high as possible. This can be compensated by ether passive or active power factor correction, but I am not going to go into that now, as it is a very large topic all on its own.
So with this background let me give my educated guess as to why your motor is not generating much breaking torque;
1) Your motor probably has a low voltage to angular velocity factor (as most brushed DC motors)
2) Your motor probably has a bad power factor when connected to a small real load (as most brushed DC motors)
The solution is definitely NOT as Reroute suggests to put a larger real ohmic load on (as 10 ohm) this is only going to reduce the current and hence the breaking torque. A better solution is to put an appropriately sized capacitor across the poles instead of a short, but even if you do that I don't believe that you are going to get much breaking torque at low speed.
The reason some motor drivers can provide "parking breake" is because they can put in energy to increase the breaking current in the motor, and they can also actively correct the power factor.