DC bus overvoltage on a decelerating centrifuge drive
A 75 kW, 400 V, 4-pole, 1485 rpm motor drives a centrifuge. The total moment of inertia referred to the motor shaft is 22 kg-m2. The drive is a 6-pulse diode-front-end VFD with a 565 V nominal DC bus, 3 x 3300 microfarad capacitors and a DC bus overvoltage trip at 800 V. The process requires a controlled stop from 1480 rpm to 300 rpm in 12 s. The drive is set to ramp down in 12 s but trips on 'DC bus overvoltage' every time, so the operator has to coast down, which takes 6 minutes because of the load's windage. The site wants a quick fix. A braking chopper with a resistor has been quoted, but the project engineer proposes 'just adding more DC bus capacitance to ride through the deceleration', arguing that capacitors are cheaper and maintenance-free.
1. Quantify the energy recovered during the stop and show what DC bus voltage it would produce without a chopper. Is the extra-capacitance proposal workable?
Read the worked answer
Angular speeds: 1480 rpm = 154.99 rad/s, 300 rpm = 31.42 rad/s. The kinetic energy to be removed is E = 0.5 x J x (w1^2 - w2^2) = 0.5 x 22 x (24,022 - 987) = 11 x 23,035 = 253 kJ. Spread over 12 s that is an average of 21 kW.
Not all of it reaches the DC bus: the motor's copper, iron and windage losses and the drive's losses absorb perhaps 10-15%, so roughly 18 kW, or about 215 kJ, is pushed back into the capacitors during the ramp.
Now apply E = 0.5 x C x (V2^2 - V1^2) with C = 3 x 3300 microfarad = 9.9 mF: V2^2 = 565^2 + 2 x 253,000/0.0099 = 319,225 + 51.1e6, giving V2 of about 7,160 V. Even using only the 215 kJ that actually reaches the bus, the result is thousands of volts - nine times the 800 V trip level.
To hold the bus below 800 V the capacitance would have to be around 1 F. The proposal is not workable: capacitors store energy as 0.5CV^2, so at 565 V a capacitor bank is an extremely poor energy sink compared with a resistor that dissipates the same energy. A chopper or a regenerative front end is required.
2. Size the braking chopper resistor, including the duty rating, and state the values you would specify.
Read the worked answer
Set the chopper to switch on at about 700 V, comfortably above the 565 V nominal and below the 800 V trip, leaving margin for the DC bus ripple and the minimum trip hysteresis.
Average braking power during the ramp, after losses, is about 215 kJ / 12 s = 18 kW. The resistor value for that power at 700 V is R = V^2/P = 700^2/18,000 = 27 ohm. Choose a standard 27 ohm resistor: peak current = 700/27 = 26 A, peak power 18 kW during the ramp.
Duty: if the machine cycle is 60 s (12 s deceleration, 48 s running), the average dissipation is 215 kJ / 60 s = 3.6 kW. Specify a 27 ohm, 5 kW continuously rated wirewound resistor to give margin, but the critical check is the manufacturer's pulse-energy rating in kJ, not the kW figure. Most wirewound resistors have a thermal time constant of minutes, so a 5 kW unit can absorb far more than 5 kW for 12 s, but this must be confirmed from the datasheet at the actual 12 s pulse.
Also specify: a chopper module rated for the peak current with hysteresis, a freewheel diode, a resistor mounted clear of combustible material with an over-temperature trip, and a drive parameter that stops the ramp if the chopper fails so the drive does not trip on overvoltage.
3. Check the braking torque requirement electrically and mechanically, and compare the chopper with the realistic alternatives.
Read the worked answer
Braking torque = J x (w1 - w2)/t = 22 x (154.99 - 31.42)/12 = 22 x 10.30 = 227 N-m. Motor rated torque = 75,000/(1485 x 2 x pi/60) = 75,000/155.5 = 482 N-m. The required braking torque is 47% of rated, well within the motor's capability and within a 150% drive current limit, so the 12 s stop is not current-limited - the only problem is where the energy goes. At low speed the motor's self-ventilation is poor, but 12 s is short enough that thermal effects are negligible, and the drive's I2t protection covers it.
Alternatives to compare:
- Extend the deceleration. At 30 s the average power drops to about 8.4 kW, allowing a much smaller 47 ohm resistor, but the process cycle time is lost - only valid if the 12 s is a wish rather than a requirement.
- Regenerative front end. It recovers the 215 kJ instead of burning it, worth about 0.06 kWh per stop, which is trivial unless the cycle is very frequent (at 60 s cycles, 60 stops/h, it is 3.6 kW of recovered power and worth considering on energy and cabinet-heat grounds).
- DC bus common coupling with another drive that is motoring, but this is unreliable if the other drive is not always running.
- Mechanical brake, which introduces wear and maintenance.
For a single 12 s stop per minute the chopper is the correct, lowest-risk choice; the regenerative front end becomes attractive only if the stop frequency or the decelerating power rises further.