Overhauling Load with Drive

[W@ttson]

Hello,

If I have a motor that is connected to a drive and it is for a crane application, I can have an idea of what the currents on the output of the drive would be as well as input of the drive.

If I require 100% torque for the load to pick it up and go rated speed, I would see roughly the rated current on the output of the drive and slightly less current on the input of the drive due to PF difference.

Now if I need 120% torque to get the load going to rated speed, I would again expect to see something of a 1.20x relationship on the output of the drive and similarly on the input of the drive from the rated values (ignoring that its not exact since the torque producing current would be the one being multiplied by 1.2 and through vectoral arithmetic the resultant current would be slightly less).

If I decide now to run the motor at 50% speed but at rated load, I can expect the same current on the output but roughly 50% current on the input. Due to the voltage remaining the same on the input but different voltage on the output and the application being a constant torque application.

Now, what happens on the lowering and the load becomes overhauling?

Say I am lowering the load at rated speed but the load becomes overhauling and the load begins to move faster than what the motor is telling it to go? (like car engine going down hill). The drive will have its DC bus voltage begin to raise. The chopper circuit will turn on an begin to dump energy into the dynamic braking resistor.

What is the input current to the drive? What is the output current to the drive?

How does that change when at 150% overhauling torque?
How does that change when at 50% speed?

 

[winnie]

I personally find it easier to think in terms of output power to the load, and then calculate the input current to deliver that output power. So to run the motor at full torque but 50% speed, the motor is getting roughly 50% power and that reflects as 50% current at the drive input (just as you said).

If the drive is dealing with an overhauling load, then the DC bus voltage rises, and the input rectifier simply stops conducting. So when the motor is operating at full speed but 150% overhauling torque, then you are operating at 150% rated power, all getting dumped to the braking resistor, with no input current. If you are operating at 50% speed and 150% overhauling torque, then you are operating at 75% rated power, all getting dumped into the braking resistor, with no input current.

Some drives are 'line regenerative'. If the DC bus voltage rises, then instead of a chopper circuit the input rectifier acts as an inverter and starts sourcing current back into the line. In this case you would calculate the regenerated power, reflect that as current at drive input voltage, and calculate the current.

-Jonathan

 

[Jraef]

During overhauling, the only current going into the drive is just what it takes to provide stator excitation, the current required to create the magnetic fields in the stator (and thereby the rotor). It varies by motor design, but is generally considered to be 20-35% of the rated current and doesn’t really change with load or speed.

 

[winnie]

During overhauling, the only current going into the drive is just what it takes to provide stator excitation, the current required to create the magnetic fields in the stator (and thereby the rotor). It varies my motor design, but is generally consistent to be 20-35% of the rated current and doesn’t really change with load or speed.

You have a bunch more experience in the field with normal drives; my experience is only with one off experimental units or the ones used to run our test dyno. But my experience has been that when a VFD driven motor goes into regeneration, the drive input current drops to zero (or goes negative with a regen capable drive), and the energy for stator excitation comes from the DC bus.

Many of the drives I've worked with power the drive control loads from the DC bus via a DCC converter, so if you have sufficient regen you don't even need the supply bus to be present.

 

[W@ttson]

I personally find it easier to think in terms of output power to the load, and then calculate the input current to deliver that output power. So to run the motor at full torque but 50% speed, the motor is getting roughly 50% power and that reflects as 50% current at the drive input (just as you said).

If the drive is dealing with an overhauling load, then the DC bus voltage rises, and the input rectifier simply stops conducting. So when the motor is operating at full speed but 150% overhauling torque, then you are operating at 150% rated power, all getting dumped to the braking resistor, with no input current. If you are operating at 50% speed and 150% overhauling torque, then you are operating at 75% rated power, all getting dumped into the braking resistor, with no input current.

Some drives are 'line regenerative'. If the DC bus voltage rises, then instead of a chopper circuit the input rectifier acts as an inverter and starts sourcing current back into the line. In this case you would calculate the regenerated power, reflect that as current at drive input voltage, and calculate the current.

-Jonathan

you hit on the crux of my question. Would the input drop to nothing or not. Thank you

 

[W@ttson]

During overhauling, the only current going into the drive is just what it takes to provide stator excitation, the current required to create the magnetic fields in the stator (and thereby the rotor). It varies by motor design, but is generally considered to be 20-35% of the rated current and doesn’t really change with load or speed.

Got it. So basically the No Load current rating. That typically is set as the drives q axis current. The utility will not be providing any torque producing current.

 

[W@ttson]

You have a bunch more experience in the field with normal drives; my experience is only with one off experimental units or the ones used to run our test dyno. But my experience has been that when a VFD driven motor goes into regeneration, the drive input current drops to zero (or goes negative with a regen capable drive), and the energy for stator excitation comes from the DC bus.

Many of the drives I've worked with power the drive control loads from the DC bus via a DCC converter, so if you have sufficient regen you don't even need the supply bus to be present.

Well in the future, I may be able to test it, so I will try to report back if I do get a chance.