Torque at zero speed, braking resistor or regeneration

Two drive requirements follow from the active load of a lift: it must hold the car before the brake opens, and it must handle the energy the car gives back while it generates. This lesson is course section 3.2.1 and Tutorial 3.2.

Full torque at zero speed

With plain V/Hz control the flux takes a moment to build up after the drive starts. If the brake opened during that moment, the imbalance would pull the car the wrong way: rollback, or "load dip", which passengers feel and which can be dangerous. Lift drives therefore use closed-loop flux vector control (field-oriented control) and prove the torque first:

1. Magnetise: inject the magnetising current $I_d$ before the brake opens.
2. Build torque: compute the torque current $I_q$ from the load-weighing sensor, or from a pre-torque estimate.
3. Release the brake only once the motor torque is at least the load torque.

The same capability holds a car level while a forklift drives in (lesson 3).

Where regenerated energy goes

A lift generates for roughly half its duty cycle, in quadrants II and IV. The motor's energy comes back through the inverter's freewheeling diodes and charges the DC-link capacitor:

$$V_\text{DC}(t) = \frac{1}{C}\int i_\text{regen}\,dt + V_\text{initial}$$

The diode rectifier on the supply side cannot pass energy backwards. If nothing else takes it, the bus voltage keeps rising until it reaches the drive's limit, for example 800 V on a 400 V drive. The drive then trips on over-voltage: the motor coasts and the brake slams on. Two technologies prevent this.

• Dynamic braking. A chopper switches a braking resistor across the bus whenever the voltage crosses a threshold (around 750 to 760 V), and dissipates $P = V_\text{DC}^2 / R_\text{brake}$ as heat. Simple and reliable, but the energy is wasted and large resistor banks need ventilation.
• Regenerative drive (active front end, AFE). The diode rectifier is replaced by an IGBT bridge synchronised with the grid, which sends the energy back as clean AC power, $P = \sqrt{3}\,V_\text{line} I_\text{regen} \cos\varphi$. Standard in modern high-rise and green buildings.

The course's example is a 40-storey hotel. Without regeneration, every descending car must burn about 20 kW in resistors, and the machine room needs its own air conditioning. With an AFE, the same 20 kW feeds the lobby lighting and the coffee machines.

NothingChopper + resistorActive front end

Braking resistor

35 Ω

The 60 A chopper sets a floor of 12.7 Ω.

• DC bus voltage
• Over-voltage trip (800 V)
• Regenerated power
• Power in the resistor

The chopper holds the bus below the trip level: the energy becomes heat in the resistor.

Peak regenerated power

16.4 kW

Peak bus voltage

766 V

Energy burned

87.7 kJ

Chopper current

21.7 A

P = V² / R : 760² / 35 = 16.5 kW

Predict first

In the widget, the chopper and a 35 Ω resistor hold the bus. You swap in a 60 Ω resistor 'to be safe'. What happens?

Safer: less current, less heatThe bus voltage rises until the drive tripsNothing, since the chopper still switches

Tutorial 3.2: sizing the resistor

Rated load 1250 kg, car 1600 kg, 2.5 m/s, $B = 0.5$, total efficiency 0.85, bus threshold 760 V, 20 % safety margin, chopper rated 60 A.

1. Counterweight: $M_{cw} = 1600 + 0.5 \times 1250 = 2225$ kg.
2. Worst regenerating case, full load down (quadrant IV): car side $1600 + 1250 = 2850$ kg, so $M_\text{unb} = 2850 - 2225 = 625$ kg.
3. Mechanical power from gravity: $P_\text{mech} = 625 \times 9.81 \times 2.5 = 15.33$ kW.
4. Electrical power reaching the bus: the losses take their share first, so $P_\text{peak} = 15.33 \times 0.85 = 13.03$ kW.
5. Largest resistance that still absorbs it: $R_\text{max} = 760^2 / 13\,030 = 44.3$ Ω. With the margin: $R_\text{design} = 44.3 \times 0.8 = 35.5$ Ω.
6. Chopper limit: $R_\text{min} = 760 / 60 = 12.7$ Ω. At 35.5 Ω the current is 21.4 A, well below 60 A: a safe choice.

Lab III question 4 asks how sensitive the resistor is to efficiency. Try it in lesson 2's widget: at $\eta = 0.6$ (a worm gear) far less power reaches the bus, and the resistor can be larger.

FoundationStart here if this is new to you

A car going downhill needs brakes; the brakes turn the car's energy into heat. A lift drive is the same, with electricity in between: the motor turns the falling car's energy into electricity, and the resistor turns that electricity into heat. A regenerative drive is like an electric car that charges its battery while braking, except that the "battery" is the building's own supply.

ExplorerGo deeper: derivations and open questions

Resistor power rating. The resistor absorbs 13 kW only during the braking part of each trip. Using the trip in lesson 2's widget and a trip every 40 s, estimate the average power. Why is a resistor rated for its average power, with a separate check that its temperature rise during one long descent is acceptable?

Capacitor energy. How long does a 4.7 mF DC link take to go from 565 V to 800 V with 16 kW coming in and no chopper? What does that tell you about the chopper's switching speed?