Herewith a short and simple example of how to select an inverter drive for a motor application.

For such an exercise – there are two requirements for selecting a correct drive:

  1. The application details of the motor in the process. (What the motor will be used for. What machine the motor will drive.)
  2. The motor nameplate details (for the voltage, current (FLA)

It is important to note that an inverter CANNOT be selected by the kW rate of the motor. This kW value is the mechanical power output of the motor. It is not the electrical input power required. For obvious reasons (such as losses) – these two are not the same. 

The Application Details

The application of motor refers to the process the motor will be applied to. It refers to the machine this motor will drive. All applications can be categorized into two types of applications:

  1. Variable Torgue applications – such as control systems and energy savings requirements (e.g. pumps, fans)
  2. Constant Torque applications – such as linear and high torque output requirements. (e.g cranes, conveyors, and crushers)

Other factors include what advanced option adaptors are to be added and communication requirements. 

The Motor Nameplate details

After the application details have been established – the next step would be to look at the motor nameplate details. It is not possible to select the correct inverter without these details.

(In cases where the motor nameplate details are not clear – then an oversized inverter will be selected to ensure all the FLA requirements are met.)

The two most important values read from the motor nameplate are:

  1. Voltage (from the mains supply)
  2. FLA (Full load current) – Amps (which relates to the torque requirements. 

(note kW rating is not a prerequisite requirement)

The selection process

The Motor above will be used for a constant torque application (thus, a drive without variable torque capability, will suffice).

  • No additional option adaptors will be required.
  • No ethernet-based communication protocols will be required.
  • The application is for an auger – thus not very high torque. (Super Light Duty)
  • Limited budget available – thus entry-level drive required.

This gives us an FR-CS80 drive selected.

As for the electrical requirements:

  • mains supply voltage: 220Vac 1ph
  • FLA: 3.61A

This gives us an FR-CS82S drive to be selected.
The exact model to be selected – the following steps are to be followed:

  • Step 1: Consult FR-CS80 Instruction Manual (Detailed). Goto Specifications – the tables
  • Step 2: Look at the 200Vac drives. Refer to the Output Specifications, specifically the Rated current (A). Look for the current that match
  • Step 3: After the required current is identified in the table – find the model number.

We see that FR-CS82S-042-60 is to be selected for this application. 

Reference and Information Resources

The following is an applicable set of information Resources: