Regenerative Braking Calculator
Regenerative Braking Calculator
What happens when a motor becomes a generator: the braking torque and power, the energy that arrives back on the DC bus once both efficiencies are counted, the bus voltage rise that energy causes in the link capacitance — which is what trips an over-voltage fault — and the brake resistor you need when the bus cannot absorb it. The fraction actually recovered is always less than people expect.
Regenerative braking: bus rise and brake resistor
a servo axis with 0.0020 kg·m² at the shaft, braked from 3,000 rpm to rest in 0.3 s, with a 92% motor and a 97% inverter, on a 1,000 µF link at 320 V with the chopper at 380 V and the trip at 400 V
Energy in, voltage up, resistor out
Ebus = Ek·ηmotor·ηinverter
V₂ = √(V₁² + 2·Ebus ÷ C)
Rbrake = Vchop² ÷ Pbus,peak Pavg = Eresistor ÷ trepeat
reflected inertia of a mass: J = m·(r ÷ G)²
- η
- both efficiencies count in the generating direction, and they count again on the way back out. A 92% motor and a 97% inverter return 89% one way and 80% round trip, before the storage medium takes its share
- C
- DC link capacitance. The energy it can hold between the running voltage and the trip is ½C(Vtrip² − V₁²), which is usually a small fraction of the mechanical energy
- Rbrake
- sized so that at the chopper threshold it draws exactly the peak power arriving. Too large and the bus keeps rising; too small and the chopper transistor’s current rating decides instead
Worked example
a servo axis with 0.0020 kg·m² at the shaft, braked from 3,000 rpm to rest in 0.3 s, with a 92% motor and a 97% inverter, on a 1,000 µF link at 320 V with the chopper at 380 V and the trip at 400 V
ω₁ = 3,000 × π ÷ 30 = 314.16 rad/s, so the kinetic energy is ½ × 0.0020 × 314.16² = 98.7 J
Braking torque is J·Δω ÷ t = 2.094 N·m, and the peak braking power is that times the initial speed: 658 W at the shaft, falling linearly to zero as the machine stops. The average is half of it, 329 W
Both efficiencies take their cut on the way back, so only 89.24% of it reaches the bus: 88.08 J
Put that into 1,000 µF sitting at 320 V and the voltage goes to √(320² + 2 × 88.08 ÷ 1 mF) = 527.8 V — far past the 400 V trip. The capacitance can only take 28.8 J before tripping, which is 32.7% of what is coming
It gets there fast: the trip level arrives 53.89 ms into a 0.3 s brake, and the chopper threshold 38.2 ms in
The resistor therefore has to absorb 67.08 J per event. Sized to take the peak bus power at the chopper threshold, R = 380² ÷ 587.2 W = 245.9 Ω — so 587.2 W peak, but only 33.54 W averaged over a 2.0 s cycle, which is what the resistor's continuous rating has to cover
And the honest bottom line: with a brake resistor, none of the 98.7 J is recovered. Fit a battery instead and 89.2% of it reaches the bus, but only 79.6% of it comes back out again through the same two efficiencies when you re-accelerate — before the battery's own charge acceptance is counted
Why the DC link cannot absorb it
| Store | What a joule costs | Practical energy at a 540 V bus |
|---|---|---|
| DC link electrolytic capacitor | a few joules per kilogram, usable only over the voltage window between running and tripping | a 1,000 µF bank taken from 540 V to 750 V holds about 135 J |
| A moving machine | ½·J·ω² or ½·m·v², and both grow as the square of speed | a 1,200 kg vehicle at 50 km/h holds 116 kJ — three orders of magnitude more |
| Brake resistor | nothing to store, everything to dissipate: its continuous rating is the energy per event divided by the repetition period | the usual answer, and the reason drives have brake terminals |
| Battery or regenerative front end | genuinely stores it, at the cost of two conversion efficiencies each way | about 80% round trip through a good motor and inverter, before the battery’s own losses |
The energy has to go somewhere
Decelerate a motor faster than friction would and the machine becomes a generator. The torque reverses, current flows the other way through the inverter, and mechanical energy arrives on the DC link. Everything difficult about regenerative braking follows from one comparison: a DC link capacitor stores almost nothing compared with a moving machine. A 1,000 µF bank taken from its running voltage up to the over-voltage trip holds a few tens or hundreds of joules. A modest servo axis at 3,000 rpm holds about a hundred joules; a vehicle holds hundreds of kilojoules. So the voltage rise the page computes is usually absurd, and that absurdity is the answer.
Braking more slowly does not help. The energy to be removed is fixed by the speeds and the inertia: halving the deceleration halves the power but doubles the time, and the energy is identical. What a longer ramp buys is a lower peak power, which makes the chopper transistor and the resistor’s peak rating easier, and a longer time before the trip arrives. It does not reduce the joules. The only things that do are less inertia, a lower speed, or somewhere to put the energy.
Sizing the brake resistor is two separate calculations. Its resistance is set by the peak: at the chopper threshold it must draw the peak power arriving at the bus, so R = Vchop² ÷ Ppeak. Any larger and the bus keeps climbing past the threshold; much smaller and the chopper transistor’s current rating becomes the limit instead. Its wattage is set by the average: the energy per braking event divided by the time between events. Those two numbers are often wildly different — a resistor may need to take 600 W for a third of a second every two seconds, which is 100 W continuous but needs the thermal mass to survive the pulse. A wirewound part’s short-term overload rating, not its continuous rating, is what to check. For the inverter-side thermal question see the heatsink calculator and for the chopper device’s own loss the MOSFET loss calculator.
How much is actually recovered is the number people get wrong. With a brake resistor the answer is zero: the point of electrical braking there is controllable deceleration and no brake wear, not energy. With a battery or a regenerative front end the energy genuinely returns — but the motor’s and the inverter’s efficiencies both apply on the way in, and both apply again on the way back out when you re-accelerate. A 92% motor and a 97% inverter return 89% one way and 80% round trip, and the battery’s own charge and discharge losses come off that. For a vehicle there is a further reduction that this page does not model: the friction brakes still take a share of any hard stop, because a motor sized for cruise cannot absorb an emergency deceleration, and below a few kilometres an hour regeneration stops working at all. For the vehicle-level energy picture see the e-bike range calculator, for the machine’s own torque and back-EMF the DC motor calculator, and for the envelope the drive has to brake within the BLDC torque-speed envelope calculator.
Reading the chart. The capacitance-only line is drawn only to just past the trip level, because past there the drive has already faulted and the rest of the curve is fiction — the point is how quickly it gets there, not how absurdly high it would go. Where that line crosses the trip level is the moment the drive faults; where the chopper line flattens is the moment the chopper starts conducting.
What this model assumes. Constant deceleration, so the braking torque is constant and the power falls linearly to zero — which is what a drive with a speed ramp does, and not what a vehicle’s brake pedal does. Both efficiencies are taken as constants, when in reality copper loss goes with the square of torque so a hard brake is less efficient than a gentle one. Friction and windage are ignored, which makes the energy figure slightly pessimistic — they help you brake. And the bus is assumed to have nothing else drawing from it; in a multi-axis machine one axis braking while another accelerates is the cheapest energy recovery there is, and it needs no chopper at all.
Frequently asked questions
Why does my drive trip on over-voltage when I decelerate?
Because the energy coming back has nowhere to go but the link capacitor, and the capacitor’s energy window between the running voltage and the trip is tiny compared with the machine’s kinetic energy. The page computes both, and the ratio is usually ten or a hundred to one. The fixes are a brake chopper and resistor, a longer ramp (which delays the trip but does not prevent it), or a front end that can return power to the supply.
Does a longer braking ramp avoid the need for a brake resistor?
Not usually. The energy is fixed by the speeds and the inertia, so a longer ramp only lowers the power. It helps if the bus has something else drawing from it — another axis accelerating, a big load — because then the returned power can be consumed as it arrives rather than stored. Against a capacitor alone it just takes longer to trip.
How do I size the brake resistor?
Two numbers. The resistance comes from the peak: V_chop² ÷ peak power arriving at the bus, so that at the chopper threshold the resistor draws exactly what is coming. The wattage comes from the average: energy per event divided by the time between events. Then check the part’s short-term overload curve, because the pulse is usually far above the continuous rating.
How much energy does regenerative braking actually recover?
With a brake resistor, none — it all becomes heat. With a battery or a regenerative front end, the motor and inverter efficiencies apply on the way in and again on the way out, so a 92% motor and a 97% inverter give about 89% one way and 80% round trip, before the battery’s own losses. Vehicle figures quoted as 60–70% of braking energy recovered are counting only the part of the stop the motor actually handles.
Can I just fit a bigger DC link capacitor?
Arithmetically yes, practically almost never. The page reports the capacitance you would need and the multiple of what you have; it is commonly ten to a hundred times. Electrolytic capacitors store only a few joules per kilogram, and only over the voltage window you are allowed to use. Supercapacitors are a genuine answer at the cost of a DC-DC converter to match their voltage.
Does the motor brake at all if the drive cannot take the energy?
It tries, which is the dangerous part. In a permanent-magnet machine the back-EMF at speed can exceed the bus voltage, and if the drive stops controlling current the machine rectifies through the inverter’s body diodes straight into the link — uncontrolled, and limited only by the winding resistance. That is a genuine failure mode for a field-weakened drive and it is why an over-voltage crowbar or a resistor is a safety item, not an efficiency one.
Related calculators
References
- Mohan N, Undeland T M, Robbins W P. Power Electronics: Converters, Applications and Design, 3rd ed. Wiley, 2003. Four-quadrant operation of DC and AC drives, the dynamic-braking chopper across the DC link, and the energy stored in the link capacitance. Edition and publisher verified by search; chapter numbers are not quoted because they were not.
- Bose B K. Modern Power Electronics and AC Drives. Prentice Hall, 2002. Regenerative and dynamic braking of AC drives, the braking chopper, and the distinction between a regenerative front end and a dissipative one.
- IEC 61800-2, Adjustable speed electrical power drive systems — Part 2: General requirements — Rating specifications for low voltage adjustable frequency AC power drive systems. The DC-link and braking ratings a drive datasheet declares against. Copyrighted; cited, not reproduced.