I have heard people say that in motor control circuits, one must take precautions to keep the motor from feeding back into the power supply, causing the supply voltage to rise, consequently breaking things. But how can this be? Unless some external force is accelerating the motor, the back-EMF can never get higher than the supply voltage. How then could it ever drive the supply voltage higher?
Electronic – worry about a motor causing the supply voltage to shoot up when the back-EMF can’t exceed the supply voltage
back-emfboostmotor
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You probably already know the answer but the layout is confusing you.
simulate this circuit – Schematic created using CircuitLab
Figure 1 and 2. Motor arranged in H-bridge configuration but without limit switches.
If you had been asked to create a reversible motor with electronic control you would have created the H-bridge as shown in Figure 1. We can rearrange this as shown in figure two. Switch SW1A and SW1B are two poles of the same switch.
Now we just have to add in the limit switches.
Figure 3. H-bridge with limit switches.
Figure 3 shows the status when the system is at the reverse limit and SW1 has just been thrown to the FWD position.
The snubber diodes are always in circuit and always pointing the right direction.
The back EMF is generated in the wire that makes up the coils of the motor. When a wire is swept sideways thru a magnetic field, a voltage is generated along the length of the wire. Spin the motor with just a voltmeter connected, and you will see it make a voltage.
So yes, the resistance and the back EMF are actually distributed along the wire in the coil. There are lots (infinity, actually) of little resistances in series that each get a little voltage in series with them when the motor turns.
Viewed electrically from the outside, this can't be distinguished from a lumped resistance in series with a lumped voltage source. Since this is simpler to draw, think about, and analyze, that's how motors are usually shown.
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Best Answer
A motor driven by an H-bridge is also a boost converter. Here's an H-bridge:
Replace the motor with an inductor, resistance, and voltage source (back-EMF):
Let's just consider that we are driving the motor in one direction, and S3 is always open, and S4 is always closed:
Rotate V1, S1, and D1 (same circuit):
flip the whole thing left-for-right (still the same circuit):
We don't need active rectification, so we can delete S1. D2 also serves no purpose. We can also delete R1, since it's just a small resistance and doesn't change the function of the circuit other than to make it less efficient:
Looking pretty close, right? Of course, a real boost converter will have a capacitor on the output to make DC, and the load isn't a battery, but a resistor, and probably V1 isn't a motor's back-EMF but rather a battery. This step isn't necessary to demonstrate how the back-EMF can feed back into your power supply, but is provided just in case you don't recognize the boost converter:
QED.
It can also be shown that when the motor is being accelerated, an H-bridge is a buck converter. Consequently, it's easier to think about the interaction between the battery and the motor's kinetic energy in the frame of the law of conservation of energy. Neglecting non-ideal losses in the winding resistance, switching transistors, friction, etc, an H-bridge and a motor make an efficient energy converter. To increase the motor's kinetic energy, the battery must supply energy. To decrease the motor's kinetic energy, the battery must absorb energy.
If the battery, friction, or some other load can't convert the kinetic energy into heat or chemical energy, it will go somewhere else. Most likely, into your power supply decoupling capacitors, causing the power rail voltage to rise, because the energy stored in a capacitor is:
\$ E = \frac{1}{2}CV^2 \$
or equivalently,
\$ V = \sqrt{\dfrac{2E}{C}} \$
Where \$E\$ is energy in joules or watt-seconds, \$C\$ is the capacitance in farads, and \$V\$ is the electromotive force, in volts. To store more energy, the voltage must go up. It's not a mistake that this looks exactly like the formula for kinetic energy:
\$ E = \frac{1}{2}mv^2 \$
Where \$E\$ is energy in joules, \$m\$ is mass in kilograms, and \$v\$ is velocity in meters per second, or for rotating kinetic energy, \$m\$ is the moment of inertia in \$kg \cdot m^2\$ and \$v\$ is the angular velocity, in radians per second.
The point here is that you get regenerative braking even if you didn't want it. See How can I implement regenerative braking of a DC motor?