Engineering · Learn · Drives
Braking & Regenerative Control in VFDs
When a motor slows a heavy load, where does all that energy go?
A spinning load stores energy. To slow it, the motor has to take that energy out, and for a moment it works as a generator. The drive must absorb it, burn it in a resistor, or send it back to the supply, or it will trip on DC bus overvoltage. This page shows how the energy flows, the five common ways to deal with it, a simulator of the DC bus during a stop, a braking resistor calculator, and a selector that picks a method from your duty. Figures are typical; always confirm against your drive manual.
While motoring, energy flows from the supply, through the drive, into the motor and the load. While braking, it flows the other way: the load drives the motor, the motor generates, and the energy arrives at the drive’s DC bus. A standard drive has a diode rectifier, which cannot send power back to the supply. With nowhere to go, the energy raises the DC bus voltage until the drive trips, so you must give it somewhere to go.
Braking happens in the quadrants where torque and speed point opposite ways. Slowing down (decelerating) is a short burst. An overhauling load, such as a crane lowering a load or a downhill conveyor, keeps the motor in a braking quadrant continuously, which is a much heavier duty.
The braking chopper is a switch across the DC bus. When the bus voltage rises above a set level it turns on and connects the external resistor, which burns the extra energy as heat. When the voltage falls it turns off. Many small and medium drives have the chopper built in and only need the resistor. Large drives often take an external chopper unit. A DC bus voltage of about 1.35 to 1.41 times the mains voltage is normal (around 540 to 565 V on a 400 V supply); the chopper switches in much higher, as shown in the next sections.
A simplified model: the motor decelerates in a straight ramp, 85% of the released energy reaches the DC bus, and the bus capacitor, a small constant drive loss and the braking chopper decide what happens. The chopper switches the resistor on at the “chopper on” level and off about 15 V lower. If the bus reaches the overvoltage trip level the drive trips and the run is marked as tripped. Real drives vary in capacitor size, thresholds and control, so use the real values from the manual.
Stored energy E = ½ × J × ω², with ω = 2π × rpm ÷ 60. For a straight ramp the braking power falls from its peak to zero, so peak power ≈ 2 × E ÷ stop time, and average power over the whole cycle = E ÷ cycle time. The resistor value is R = V² ÷ Ppeak, using the voltage at which the chopper switches on, not the normal bus voltage. R must also stay at or above the minimum resistance the drive allows; a lower value overloads the chopper. The result is a first estimate: friction and load torque take some energy, and drive and motor losses reduce what reaches the resistor, but use the drive manufacturer’s sizing tables for the final choice.
| Parameter | Ramp-down | DC injection | Braking resistor | Regenerative (AFE) |
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| Application | Braking duty | Usual choice |
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Voltage levels, thresholds and savings are typical values for learning and first estimates. Braking resistors get very hot and braking circuits carry high DC voltage: follow the drive manual, protect and ventilate the resistor, and let qualified people do the wiring.