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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?

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A motor driven by an H-bridge is also a boost converter. Here's an H-bridge:

schematic 1

Replace the motor with an inductor, resistance, and voltage source (back-EMF):

schematic 2

Let's just consider that we are driving the motor in one direction, and S3 is always open, and S4 is always closed:

schematic 3

Rotate V1, S1, and D1 (same circuit):

schematic 4

flip the whole thing left-for-right (still the same circuit):

schematic 5

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:

schematic 6

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:

schematic 7

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?

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  • 1
    \$\begingroup\$ +1. However, for the boost converter to work, S2 (last picture) must be turned on and off. Two cases apply. (1) You still apply a PWM and do something like "active braking". This will eventually lead to a boost converter. (2) No PWM to any transistor - only the diodes will act as rectifiers for the EMF, and the voltage will not rise to a dangerous level unless you externally turn the motor faster than it ran before having been switched off. \$\endgroup\$ – zebonaut Jan 26 '13 at 10:14
  • \$\begingroup\$ Circuit lab would be nice, you could probably use a 555 timer and a voltage source a cap that you are charging with some diodes that show in real time how it would work, but I do love circuit lab. \$\endgroup\$ – Kortuk Jan 26 '13 at 17:17
  • \$\begingroup\$ @zebonaut true, that if you stop switching the bridge, you can't raise the supply voltage. If you leave it switched low, the motor terminals are shorted, the motor current will be very high, and the kinetic energy is converted entirely into heat by the winding resistance and transistor losses. If you stop switching the bridge entirely, the motor freewheels and only friction is absorbing kinetic energy. Usually though, a PWM motor controller is between these two extremes, and every time the duty cycle decreases, you get regenerative braking, without doing anything fancy. \$\endgroup\$ – Phil Frost Jan 26 '13 at 19:47
  • \$\begingroup\$ @PhilFrost Just to clarify, this does not mean it's possible to pulse drive a motor in such a way to increase the voltage so that you can drive a motor rated for a higher voltage from a low-voltage source, right? You would really need a boost converter prior, correct? \$\endgroup\$ – horta Jun 6 '14 at 20:30
  • \$\begingroup\$ @horta Yeah, pretty much. The "boost" only happens when the motor's EMF exceeds the battery voltage, so the battery is the load. Since the EMF is also proportional to speed, this implies the motor is spinning faster than it would be in equilibrium, and so will be slowing down. \$\endgroup\$ – Phil Frost Jun 7 '14 at 3:24
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  1. What Phil said

2. This is not the EMF you are looking for. One problem is in your equating the voltage with the back EMF. This is not back EMF - this is energy stored in the system "demanding to be given a new home. I say demanding" because the energy WILL be transferred elsewhere and it will be delivered at a rate that the system wishes it to happen at. Get a bit behind in accepting the transfer and it will get more and more and more insistent. As required.

A rotating motor contains mechanical energy which is converted to electrical energy as the flux in the windings changes. When you brake it hard all the enegy is stored in the magnetic field and the magnetic field wants to share its bounty.
The field WILL collaps and the energy WILL get delivered to somewhere else.
So ...

One side of the motor is usually grounded (directly or via diodes) and in this case the other side is connected to supply. When the magnetic field delivers its energy if the supply is able to accept the energy at constant voltage (eg ideal battery or capacitor) then the magnetic field will not mind. It will stand and deliver.

However, if the supply will not accept energy at the rate that the field wishes to deliver it then the field will become a little more insistent - it will raise the voltage. If this does not work it will keep on raising the voltage until the energy is flowing out at the rate that its "wishes" it to.
It will go to infinity if it must.
In the real world there is always some capacitance (intended or not) and this will usually stop the voltage rise by storing the energy in the capacitor. Very small capacitor = very high voltage.


Added:

This is essentially a comment on Luc's answer, but is useful in its own right.

As above, the motor energy must "go somewhere.
If the motor is terminated in a load then the load will absorb the energy.
A snubber is one such load, but the power supply that Phil refers to is another.
IF the supply is "stiff" the supply voltage will not rise appreciably.
Stiffness may come from having other devices operating from the supply that can take the energy and/or enough capacitance to absorb the energy with modest voltage rise.

If the supply is not "stiff enough" its voltage will rise as the motor energy is transferred into it. In extreme cases the voltage rise may be enough to destroy the supply due to over voltage conditions.

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  • \$\begingroup\$ @PhilFrost - Yes. But also, as I noted "Stiffness may come from having other devices operating from the supply that can take the energy ... to absorb the energy with modest voltage rise." Some supplies are specifically design ed to dissipate energy if voltage rises too high or transfer it back into the supply (energy recovery). The "cleverest" of these takes DC from it's "load" and returns mains voltage and frequency AC back into the mains. \$\endgroup\$ – Russell McMahon Jan 26 '13 at 13:54
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I believe you are referring to the voltage spike that occurs when the current flowing into an inductive load (such as a motor, incandescent lamp, solenoid, etc) is interrupted suddenly. Due to the current-voltage relationship of an inductor, given by $$V_L(t) = L\frac{di_L(t)}{dt}$$ when the current is switched almost instantaneously, the voltage spikes to a very high level, which can cause damage to components attached to the inductive load (in this case a motor). The destruction of components is typically avoided by using what is called a snubber diode in order to provide a temporary path for the current generated by the coil.

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