Flux Braking

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philly

Senior Member
Can someone explain to me how flux braking works with a VFD? My brief understanding is that the drive increases the flux current to the motor which in turn increases the motor losses and allows quicker braking.

I understand how the increased flux current will produce greater I^2R losses in the stator however I dont see how this would help with braking the motor? Can anyone explain?
 

winnie

Senior Member
Location
Springfield, MA, USA
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Electric motor research
The kinetic energy of the system being braked has to go somewhere.

The 'proper' place to dump this kinetic energy is a braking resistor or back on the supply line using a regen VFD. But for infrequent braking a reasonable place to dump this kinetic energy is into heating up the motor.

Flux braking is simply a method of more rapidly dumping the kinetic energy into the motor heat so that you get better braking performance. The less efficient the motor is, the more rapidly it converts electrical power into heat. The more rapidly electrical power is being converted into heat, the more rapidly you can let the motor generate electricity without raising the DC bus voltage.

-Jon
 

philly

Senior Member
Flux braking is simply a method of more rapidly dumping the kinetic energy into the motor heat so that you get better braking performance. The less efficient the motor is, the more rapidly it converts electrical power into heat. The more rapidly electrical power is being converted into heat, the more rapidly you can let the motor generate electricity without raising the DC bus voltage.
-Jon

If I understand this correctly you are saying that by increasing flux current and motor losses this creates more heat in the motor which allows the kenetic engergy to be converted into heat energy? How does the presence of additional heat allow for the kenetic energy to be converted to heat energy? I'm not following this.
 

winnie

Senior Member
Location
Springfield, MA, USA
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Electric motor research
The heat itself does not help anything.

The breaking is caused by the regeneration of electricity. This regenerated electricity is going back to the DC bus.

Inefficiency in a motor means that electrical energy is being converted into heat. This electrical loss is being supplied by the DC bus.

Do this right, and the net result is that no electrical power is being stored on the DC bus, but instead all of the regenerated electricity is going right back to heating up the motor.

The heat is a side effect: you have to dump the regenerated energy _somewhere_, and a convenient place is to dump it by heating up the motor. Flux braking is a method of improving the rate at which the excess energy can be dumped into the motor as heating, essentially by increasing the inefficiency of the motor.

-Jon
 

philly

Senior Member
The heat is a side effect: you have to dump the regenerated energy _somewhere_, and a convenient place is to dump it by heating up the motor. Flux braking is a method of improving the rate at which the excess energy can be dumped into the motor as heating, essentially by increasing the inefficiency of the motor.

-Jon

I guess the part I'm getting hung up on is how supplying something additional from the drive (extra flux current) can aid in dissipating additional energy (kenetic) supplied by the load.

I understand how typically with no braking the kenetic energy is converted into electrical energy which is fed back to the DC bus and can trip it on overvoltage. I cant see however how increasing losses in the motor will help excess energy to be converted in the motor as heat instead of being fed back to the DC bus?
 

LarryFine

Master Electrician Electric Contractor Richmond VA
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Henrico County, VA
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If I understand this correctly you are saying that by increasing flux current and motor losses this creates more heat in the motor which allows the kenetic engergy to be converted into heat energy? How does the presence of additional heat allow for the kenetic energy to be converted to heat energy? I'm not following this.
Picture a ceiling fan that is spinning. Flip the reversing switch, then turn it off just as it's about to start spinning the other way.
 

winnie

Senior Member
Location
Springfield, MA, USA
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Electric motor research
I understand how typically with no braking the kenetic energy is converted into electrical energy which is fed back to the DC bus and can trip it on overvoltage. I cant see however how increasing losses in the motor will help excess energy to be converted in the motor as heat instead of being fed back to the DC bus?

Losses in the motor must be supplied from somewhere. Losses simply mean energy that the motor converts into some undesirable output (usually heat) rather than the desired torque.

Since we have a finite amount of kinetic energy in the load, any energy that goes to heating up the motor is energy that is not available for charging the DC bus.

-Jon
 
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ohmhead

Senior Member
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Well let me learn somthin today a motor has a field and rotor basic simple 3 phase motor with vfd what your saying is increase the field magnatizing force beyond its needed torque normally and then dump this energy back to the dc buss does the dc buss need to lower its voltage or freq during this flux braking during a braking control time and place ? please explain to this oldman with just a limited electrical background with flux braking & control . How do we perform this and dont use a paddle fan in reverse i can flux brake one by holding the blades with my hand . Are we on the same page as a motor generator effect kinda ?
 

philly

Senior Member
Since we have a finite amount of kinetic energy in the load, any energy that goes to heating up the motor is energy that is not available for charging the DC bus.
-Jon

Ok I think I might understand what you are saying now. Basically you are saying that the load is going to have a finite amount of kinetic energy that will be converted into a certain amount of electrical energy no matter what. This converted electrical energy will be placed on the DC bus which will add to the energy at the bus. This conversion will be constant and all the energy will be placed back on the bus.

So to make room on the bus for this additional energy from the load/motor the drive tries to get rid of some of its stored energy and transfer it somewhere. To do this it increases its flux current which increases the losses in the motor. These additional losses in the motor will take this additional amount of energy from the DC bus and dump it as heat in the motor. So since the drive is getting rid of this extra energy and dumping it into the motor, it is making additional room for the converted kenetic energy being converted from the load/motor. Since the DC bus gets rid of energy in the motor it now has additional capacity for load energy. Is this correct?

I would expect then when looking at the drive during this braking you will see the flux current increase to a greater value. If flux braking in not enabled do you typically see an increased flux current value when the load/motor is regenerating energy back onto the drive bus?
 

Jraef

Moderator, OTD
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Ok I think I might understand what you are saying now. ...To do this it increases its flux current which increases the losses in the motor. These additional losses in the motor will take this additional amount of energy from the DC bus and dump it as heat in the motor. So since the drive is getting rid of this extra energy and dumping it into the motor, it is making additional room for the converted kenetic energy being converted from the load/motor. Since the DC bus gets rid of energy in the motor it now has additional capacity for load energy. Is this correct?
You are very close, but you are still not quite there. The area I highlighted is where you veered off just a little. It is not doing it to increase the losses, it is doing it to have a place to dump the energy by increasing the flux component of the output. The increased losses are a side effect. Think of a Dynamic Braking module where the drive triggers a chopper to "dump" excess DC bus energy into a resistor bank, which just dissipates the energy as waste heat. This is essentially the same thing, you are just substituting the motor flux as the dump site and expecting the motor to be the energy dissipation device instead of resistors.
 

ohmhead

Senior Member
Location
ORLANDO FLA
You are very close, but you are still not quite there. The area I highlighted is where you veered off just a little. It is not doing it to increase the losses, it is doing it to have a place to dump the energy by increasing the flux component of the output. The increased losses are a side effect. Think of a Dynamic Braking module where the drive triggers a chopper to "dump" excess DC bus energy into a resistor bank, which just dissipates the energy as waste heat. This is essentially the same thing, you are just substituting the motor flux as the dump site and expecting the motor to be the energy dissipation device instead of resistors.


Well how do we increase the flux component in more detail?
What happens to the drive during this process as far as the output wave shape meaning the pwm is the freq or the shape of the wave pulses changed in any way from drive output ?

What would a scope show on wave shape during this braking ?

How is dc buss effected when this happens or is the wave shape on output more of a lenghten dc pw on each cycle and the motor sees a constant dc pwm so it losses its torque kinda increasing field flux to rotor kinda and then please explain in more detail what happens to the actual wave shape of output of drive to the back feed of energy motor to buss dumping and explain the stored energy in the drive better ?
We understand the heat to resistance in motor windings thats ok with me is this the key iam missing higher current in field due to losses in torque like rotor lock ?

What or how we have stored energy in drive is it the dc power supply is that the store energy ?

Iam interested in knowing this as i see and can google lots of pages on this but they dont really get into the nuts and bolts of why they just kinda generally explain thinking we understand .
What happens electrically in the theory part of the wave freq or voltage or wave shape on only a flux braking condition ?

I enjoy learning this is a nice post which is rare so thats why iam asking help me to understand this better then a google .
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Well let me learn somthin today a motor has a field and rotor basic simple 3 phase motor with vfd what your saying is increase the field magnatizing force beyond its needed torque normally and then dump this energy back to the dc buss does the dc buss need to lower its voltage or freq during this flux braking during a braking control time and place ? please explain to this oldman with just a limited electrical background with flux braking & control . How do we perform this and dont use a paddle fan in reverse i can flux brake one by holding the blades with my hand . Are we on the same page as a motor generator effect kinda ?

To answer Ohmhead's questions:

Yes, we are talking about operating the motor as a generator in order to stop the mechanical load. A generator requires mechanical power input to generate the electrical output. In the case of braking, that mechanical power input comes from the inertia of the load, and slows it down.

In order to achieve this 'dynamic braking', the VFD has to adjust its frequency, so that the rotor is spinning faster than the rotating magnetic field. This causes the motor to operate as a generator. The motor produces negative torque and slows the load down. The motor is also generating electricity.

The regenerated electricity has to go somewhere. With a normal VFD and normal regenerative braking, this electricity goes to charging up the DC bus. Very quickly, the voltage gets high enough that the drive trips on overvoltage (or the DC capacitors blow up).

The 'standard' method for dealing with this is to add a 'brake chopper' and 'braking resistor'; if the DC bus voltage gets high enough, then the brake chopper simply switches the resistor on to the DC bus. The resistor starts to discharge the bus, consuming some of this regenerated electricity. Another common method is to have a 'line regenerative' VFD, one capable of supplying power back to the AC mains. If the DC bus voltage gets high enough, then the VFD essentially becomes a grid tied inverter, and the regenerated electricity is used to supply other loads on the system.

What we have been talking about here are methods in which the regenerated electricity is used to heat the motor up, essentially by intentionally rendering the motor much less efficient. As you are aware, if a motor is operated at the wrong voltage, it becomes less efficient, meaning that it has greater losses (produces more heat) for the same output. The same is true when the motor is operated as a generator. If the machine is operated less efficiently, then for the same mechanical input, it produces less electrical output and heats up more.

In the case of 'flux braking' we are operating the motor as a generator, and additionally intentionally operating it as an _inefficient_ generator, so that the bulk of the mechanical input is converted to heat, with very little electrical output,

-Jon
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
On your second question about the 'nuts and bolts', the easiest way to see what the drive does is to think about what happens when you change the V/Hz ratio.

If you increase the V/Hz ratio of the voltage applied to the motor, the following things will happen: the magnetizing current will go up, the magnetic flux density in the motor will go up, the magnetizing losses will go up, and the torque produced at any given slip will go up (which for a fixed shaft torque slip will go down), and the 'torque producing current' needed for any particular shaft torque will go down.

This is true on the regenerative side of things as well as the motoring side of things; increase the voltage at any given frequency, and you get greater magnetizing current and greater magnetizing losses.

The nuts and bolts of how field oriented control actually changes V/Hz ratio is for another thread :)

Jaref, a slight verbage quibble: if you change the generator operating state so that it is less efficient, it seems equally fair to say that it has more internal heating and produces less electrical output as to say that it is regenerating to the DC bus and then the excess is being dumped from the DC bus back to heating the generator.

-Jon
 

philly

Senior Member
What we have been talking about here are methods in which the regenerated electricity is used to heat the motor up, essentially by intentionally rendering the motor much less efficient. As you are aware, if a motor is operated at the wrong voltage, it becomes less efficient, meaning that it has greater losses (produces more heat) for the same output. The same is true when the motor is operated as a generator. If the machine is operated less efficiently, then for the same mechanical input, it produces less electrical output and heats up more.

In the case of 'flux braking' we are operating the motor as a generator, and additionally intentionally operating it as an _inefficient_ generator, so that the bulk of the mechanical input is converted to heat, with very little electrical output,
-Jon

O.K. now that we are looking at it from the motor still being a generator when braking what is it about the increased flux that causes the generator action of the motor to be less efficient and wast energy as heat in the motor instead of putting it back on the VFD.

So we are still saying that the rotor is spinning faster than the the electric field and thus generating electricity back onto the cables and drives. This would continue to due this with normal flux level in motor. However now we increase the flux in the motor and the generator action by the motor becomes less efficient. The presence of the increased flux causing motor to be less efficient is maybe the part I am missing?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Philly,

Yup, we are still saying that the rotating magnetic field is moving slower than the rotor, so the machine is operating as a generator.

As to how increased flux makes the machine less efficient: let us consider a machine spinning at no load. The rotor is spinning at synchronous speed, and no torque is being produced. The bulk of the power flow is reactive VARs, where for part of the AC cycle each phase consumes power building up its magnetic field, and for part of the AC cycle each phase is generating power as the magnetic field dies down. There is a little bit of real power being consumed by the motor: the resistance losses in the wire and the core losses in the steel caused by the changing magnetic field.

In a perfect motor, at the no-load speed, losses are always zero and real power consumption is always zero. In a real motor you will have some power consumption even at no load. If you were to adjust the V/Hz of the motor in the no-load state, you would find that the losses change with voltage. As you increase the voltage, the no-load flux goes up, which increases core losses. Additionally, more magnetizing current flows, which increases resistance losses in the wire. Eventually you saturate the core, and losses start to climb even faster.

These are losses associated with creating the magnetic field. These losses happen even if no torque is produced. At any given drive frequency, the drive voltage is a dial which controls the magnitude of the magnetizing losses. The lower the drive voltage, the weaker the magnetic field and the lower the losses associated with creating the magnetic field.

When we run the motor to produce torque, we have a trade-off. The stronger the magnetic field, the less 'torque producing current' is needed to create a given torque. So that means that we can get the same torque with a strong magnetic field and a low 'torque producing current' or a weak magnetic field and a high 'torque producing current'. Well you have resistance losses associated with your torque producing current. You you can trade the losses associated with your magnetizing current and the losses associated with your torque producing current.

For any giving mechanical output (rotor torque and rotor speed) there will be an ideal drive frequency and voltage which will minimize the total sum of losses. You can get the same mechanical output with a different combination of voltage and frequency, but the machine will be less efficient. For example, you could raise the drive voltage, which would raise the magnetizing current and magnetic field strength, and increase the losses associated with creating the magnetic field; this would reduce the required torque producing current, and reduce the losses associated with the torque producing current. But we care about _total_ losses; any loss is simply heat in the motor. Raise the drive voltage above optimum, and for the same mechanical output the motor heats up more.

When the motor is operated as a generator, it is possible to make it so inefficient that the net output power is zero. Just raise the drive voltage enough, and the magnetizing current will get very high with lots of resistance and core losses. The motor consumes mechanical power (torque and speed), generates electrical power, and all of this electrical power is wasted creating the excessive magnetic field.

-Jon
 

ohmhead

Senior Member
Location
ORLANDO FLA
Winnie thanks for the response i will read this over but it helps me better than the google sites ive looked at ! You always explain this better when i ask a question i dont ask many but you always come back with a good clear picture for me thanks we didnt have this in electricians school i have a basic understanding now this is a lot different than the old electric brake method let me read more and i know ill have one more question for you . take care :)

Good post Philly!
 
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Jraef

Moderator, OTD
Staff member
Location
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Electrical Engineer
...
Jaref, a slight verbage quibble: if you change the generator operating state so that it is less efficient, it seems equally fair to say that it has more internal heating and produces less electrical output as to say that it is regenerating to the DC bus and then the excess is being dumped from the DC bus back to heating the generator.

-Jon
That was ohmhead's verbage. But both of you have done a much better job of the "nuts and bolts".
 

philly

Senior Member
When the motor is operated as a generator, it is possible to make it so inefficient that the net output power is zero. Just raise the drive voltage enough, and the magnetizing current will get very high with lots of resistance and core losses. The motor consumes mechanical power (torque and speed), generates electrical power, and all of this electrical power is wasted creating the excessive magnetic field.

-Jon

Thanks for the great explanation I'm finally starting to get it.

I understand by raising the voltage in the drive you will increase the magnetising current and terefore the flux and by doing that you will increase the losses in the motor. Now while this is happening and the motor is regenerating how does the generated electrical power feed into the excessive magnetic field.

In other words by increasing the voltage at the drive end we are creating this excessive magnatic field in the motor. Now when the mechanical energy from the motor is converted into electrical energy how is it wasted by feeding into this excessive magnatic field created in the motor? Does it have to do with sauturation?

Thanks again for the help.
 

ohmhead

Senior Member
Location
ORLANDO FLA
Thanks for the great explanation I'm finally starting to get it.

I understand by raising the voltage in the drive you will increase the magnetising current and terefore the flux and by doing that you will increase the losses in the motor. Now while this is happening and the motor is regenerating how does the generated electrical power feed into the excessive magnetic field.

In other words by increasing the voltage at the drive end we are creating this excessive magnatic field in the motor. Now when the mechanical energy from the motor is converted into electrical energy how is it wasted by feeding into this excessive magnatic field created in the motor? Does it have to do with sauturation?

Thanks again for the help.

Well from what i see here and please correct me but when the motor turns faster than the drive feeding it its now a generator the voltage is now fed back in the dc buss when the capacitors charge up fully the drive sees a over voltage and shuts down the output of the drive then the motor coast to a stop with the load helping its kinda instant .

So from that if you up the voltage above normal voltage at same freq it gets a higher current in field more magnetic force but the motor can not use this to turn because the field has expanded outward past the point of doing any good over saturation in my uneducated thinking . Winnie please help me out here !!!! Did i make sense or am i off base
 
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