Selecting drives for electric vehicles
Industrial, electric-vehicle drives and controls vary according to vehicle design and duty cycle. Here is a primer on drive-system design of fork-lift and automatically guided vehicles.
Base-plate cooled AGV controller rated at 60-A continuous and 200 A peak.
Battery powered electric vehicles, such as electrically powered forklift trucks, automatically guided vehicles (AGVs), sweeping machines, personnel-transport vehicles, and delivery vans, offer clean, reliable, and economical transportation. Emissions from electric vehicles are low to nonexistent, their range is adequate, and their cost of operation is low.
These vehicles generally have similar components: a battery, a motor (or motors), a motor controller (or controllers), a speed or torque reference (accelerator pedal), and safety equipment such as fuses and contactors. Their differing drive requirements result from their duty cycles, which range from the heavily loaded fork-lift trucks to the lightly loaded AGVs.
Fork-lift trucks
Electric fork-lift trucks are available in many configurations, such as the familiar rubber-tired, four-wheel truck with rear drive wheels under a heavy frame that carries the battery and a counterweight to balance the load on the front forks. They are rated by how much load the forks can safely lift, usually in thousands of pounds with 3,000 and 5,000-lb ratings common.
Selecting and applying the appropriate drive system can be an engineering challenge. The masses of the load, battery, frame, and counterweight make acceleration and braking difficult. High torque is needed to get a 10,000 lb, or heavier, vehicle moving so the usual practice is to power the truck with a serieswound dc traction motor.
Motor characteristics. Series-wound traction motors have fields, connected in series with the armature, capable of carrying the motor armature current. These motors offer high torque at low speed or stall, and increasing speed as the torque load diminishes. The armature voltage in the circuit, Figure 1, is equal to the battery voltage less the voltage drop in the series field. The speed of the motor is equal to the armature voltage less the resistance voltage drop divided by a constant times the field flux:
where:
Ffld = Field flux
Ia = Armature current, A
kn = Speed-field flux constant
N = Motor speed, rpm
R = Resistance, ohm
Va = Armature voltage, V
Armature current flowing through the field coils produces field flux. Reducing armature current reduces the field flux, which increases the speed, assuming the motor produces enough torque to move the load.
The torque produced by this motor is proportional to the product of armature current and field flux:
T = ktIaFfld (2)
where:
kt = Torque constant
T = Motor torque
If the armature current and the field flux are low, the developed motor torque will also be low. Under these conditions, the motor is running at high speed and light load. Applying a torque load to the motor causes the armature to slow down and draw more current from the battery; the additional current increases the field flux and, along with increased armature current, produces more torque. The multiplying action of the torque caused by the product of field flux and armature current creates a large value of torque at stall. This is just what a heavy truck needs to accelerate from a standstill or climb a ramp.
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