Electrical and Control Systems: Supply, Drives, Safety Chain and Fire Service
This domain covers the power, drive and logic side of an elevating device: machine-room supply and disconnecting means, AC and DC drives, the safety chain, door operators, and the fire and emergency modes a mechanic must be able to verify.
Supply, disconnection and protection
- A dedicated feeder serves the machine room or machinery space, with a lockable mainline disconnect for the drive, separate control and lighting circuits, and pit and machine-room receptacles as required by the elevating-device requirements of the Canadian Electrical Code.
- Bonding of equipment, correct conductor and overcurrent sizing, and a traveling cable for the car are part of the installation, not accessories.
- Before work: open, lock, tag, and prove dead at the terminals. Drive DC-bus capacitors hold a charge after the disconnect is opened.
Drives and feedback
- Older DC equipment uses motor-generator sets or SCR drives; modern equipment uses a VVVF inverter with a DC bus, IGBT switching and a braking resistor to dissipate energy when the load overhauls the motor.
- An encoder supplies closed-loop speed and position feedback to the controller. Loss of feedback is a fault condition, not a reason to run slower.
Safety chain and operating modes
- The safety chain is a series string including the pit stop switch, car emergency stop, governor overspeed switch, final limit switch, and door interlock contacts, among others. Opening any contact drops the drive and sets the brake.
- Firefighters' service: Phase I - a lobby or machine-room smoke detector recalls cars to the designated landing and opens the doors. Phase II - a key switch in the car gives the firefighter hold-to-run control with call cancellation.
- Hydraulic units typically use battery lowering to bring the car to a landing on power failure.
Worked example
A car with rated load 1000 kg has car mass 1500 kg and a counterweight of 1950 kg (car plus 45% of rated load). Rated speed is 1.6 m/s and drive efficiency 0.85. Fully loaded, travelling up: imbalance = 1500 + 1000 − 1950 = 550 kg. Force = 550 × 9.81 = 5,396 N. Mechanical power = 5,396 × 1.6 = 8.6 kW. At 0.85 efficiency the motor input is about 8.6 ÷ 0.85 ≈ 10.2 kW. Travelling down with the same load, the load overhauls the machine and the drive regenerates or the braking resistor dissipates the energy.
Traps that cost marks
- Ignoring efficiency in power calculations, or forgetting the overhauling (regenerating) direction.
- Assuming the brake performs every normal stop; it is a stopping and holding means, not the primary deceleration control.
- Mixing up Phase I (automatic recall) and Phase II (in-car firefighter control).
- Jumping out safety-chain contacts to "test" a door or a switch.