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Let's say that a VFD is powering a PMSM and the motor is rotating at rated speed. The motor is connected to a high inertial load, so it cannot decelerate quickly. Now if the VFD commands a quick deceleration, how does the regen current from motor behave?

Equivalently, I have simplified the circuit to a DC permanent magnet motor with the below ciricuit diagram:

enter image description here

Initially, the VFD's main capacitor (C1) still has the rectified (through D2) supply voltage (V1) in it and SW1 is fully on so the motor is running at rated speed (with voltage V1 across it).

Now, to simplify the deceleration scenario, if I switch off SW1. I believe that the flyback diode (D1) and the motor inductance together forms a boost converter, so the main capacitor is charged to a higher voltage than V1.

But I am unable to realize what the current (or braking torque) waveform through the flyback diode (D1) looks like? Is it constant or intermittent spikes or a gradual drop over the deceleration period? What would the back current equation be?

A follow up question is, what does speed graph of motor during this regeneration period look like? Does it ramp down with constant deceleration?

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    \$\begingroup\$ Is the load rotational inertia only? If so, then turning off SW1 will simply cause the motor to coast down. It is not a regen situation. The diode is needed to prevent inductive kickback, but once the inductor energy has been dumped into the capacitor, you will have very low current. In order to stop the rotor quickly, you need to apply a lower voltage to the motor than the back EMF. \$\endgroup\$
    – user57037
    Jan 22, 2022 at 0:19

3 Answers 3

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Now if the VFD commands a quick deceleration, how does the regen current from motor behave?

The VFD commanding deceleration results first in a reduction of frequency. Along with that, the motor voltage would be reduced by reducing the PWM pulse widths. The DC bus capacitor across the inverter input would normally maintain a constant voltage supported by a rectifier connected to the AC input. Reducing the motor frequency causes in to transition to generator operation. The phase relationship between the motor voltage and current would shift as the motor starts supplying power to the VFD. As a result the DC current at the inverter input reverses. If there is no system to dissipate the power returned to the DC bus, the capacitors will charge to a higher voltage and fail rather quickly.

There are a number of options for dealing with the situation. At least one would be built into any VFD that is a viable product. Some options are:

  1. Detect that capacitor voltage has reached the safe limit, de-energize the motor and let it coast.
  2. Override the speed command and only command a speed reduction for which existing motor and VFD losses can dissipate the energy.
  3. Perform 2, but connect across the DC bus a resistor of modest power dissipation capability that will increase the braking power dissipation.
  4. Perform 3, but connect a resistor bank with sufficient power dissipation capability to stop the load quickly.
  5. Provide a regenerative input power section rather than a rectifier so that the motor can act continuously as a generator at the motor's rated torque.

There are a lot of alternatives in the details of point 2. That could be implemented as motor torque control or DC current control or in other ways.

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  • \$\begingroup\$ Is it correct to believe that the regenerated current from the motor flows back to the capacitor when the switch is ON and not through the flyback diode? Or is it through both the diode and the IGBT switch? I understand the mechanisms available to protect the capacitor. But I am not sure how the regen current graph looks like when it is charging the capacitor? \$\endgroup\$
    – gmancity
    Jan 22, 2022 at 12:17
  • \$\begingroup\$ There is a diode connected inverse parallel with the transistor in each IGBT. The regenerated current flows through those diodes. \$\endgroup\$
    – user80875
    Jan 22, 2022 at 14:57
  • \$\begingroup\$ Ultimately, by understanding how the generated reverse current waveform looks like during deceleration, I am trying to figure out the braking torque on the PMSM motor. Would the braking torque be constant over the deceleration period or would it be decreasing linearly (or as a higher order function) ? Once I find this out, I can then calculate the error between commanded deceleration ramp time by VFD and the actual motor stopping time. \$\endgroup\$
    – gmancity
    Jan 22, 2022 at 22:35
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    \$\begingroup\$ Understanding the waveform is not likely to help much. The motor should be capable of providing constant braking torque at rated torque or higher from rated speed down to 10% of rated speed or lower. The limit on that will be the VFD's capability for absorbing the energy. \$\endgroup\$
    – user80875
    Jan 22, 2022 at 23:37
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You've simplified your circuit too far. The VFD has (or at least I expect it to have) a half-bridge on each of its three terminals. The equivalent circuit with a brushed DC motor for acceleration and deceleration in one direction is this:

schematic

simulate this circuit – Schematic created using CircuitLab

When the motor is accelerating, SW1 and the motor's armature inductance make a buck converter, with power going from V1 to the armature's back-EMF, Va.

When the motor is decelerating, SW1 and the motor's armature inductance makes a boost converter, with power going from the armature's back-EMF to C1, with D1 blocking V1 from having power forced backwards into it.

What's missing from this is that for real regenerative braking of the motor you need somewhere for the braking energy to go, or you can't do regenerative breaking. So there's either a circuit that turns SW1 off entirely if the voltage on C1 gets too high, or there's a braking resistor connected to the node common to D1 and C1 that gets switched in when C1 reaches some voltage, or if you're getting really fancy (in, for example, a car), there's a battery charger or something that does something else with the regenerative energy than just heating up a resistor.

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  • \$\begingroup\$ Ultimately, I am trying to figure out the braking torque on the PMSM motor by understanding the regenerated current by this motor during deceleration. Would the braking torque be constant over the deceleration period or would it be decreasing linearly (or as a higher order function) ? Once I find this out, I can then calculate the error between commanded deceleration ramp time by VFD and the actual motor stopping time. \$\endgroup\$
    – gmancity
    Jan 22, 2022 at 12:37
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The cap. won’t store much energy , so either it must be pumped back into a battery or switched to a brake resistor . This is missing in your bridge.

The stored inertial energy must be converted to heat energy either in motor or switched IGBT or a series R, in proportion to the values.

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  • \$\begingroup\$ I have clarity on the capacitor protection methods available, but I am unable to see how the regen current waveform would look like when it is charging the capacitor. Does this charging with reverse current generated by the motor happen when the IGBTs are ON (with lower frequency) or through the flyback diodes only? Any reference on the theory or calculating the regenerated current waveform would clear this up for me. \$\endgroup\$
    – gmancity
    Jan 22, 2022 at 12:25
  • \$\begingroup\$ The Caps are only good for high frequency, but motor current is like DC so a low R brake heat dissipator and switch must loop the current from the generator , where current controls the rate of deacceleration, just the opposite for the motor. So the supply must be able to absorb this (battery) or dumped into a heater coil resistor.. Estimate the battery as a capacitor with ten thousand Farads so it can handle DC for an hour or so, Otherwise if you short the motor, it gets too hot unless added cooling mechanism or pulsed by IGBT switches \$\endgroup\$ Jan 22, 2022 at 13:55

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