VFD Brake Resistor Operation

[philly]

I am trying to understand completely how a brake resistor on a VFD works. I know that when the DC bus level gets to a certain level the chopper circuit will turn on and dump the excess energy across the resistor.

Must every drive have this chopper circuit to dump this excess energy across the resistor? In other words, is there ever a situation where the drive is continuously connected to the resistor dumping energy across it? I was always under the impression that the resistor only came on when the Bus DC voltage reached a certain threshold.

Also when the resistor is turned on, does the excess energy actually help slow down the motor and load, or does it simply just provide a means for dumping the excess energy? I've heard the term brake used as if it was actually slowing the motor.

So if a drive has a 3min ramp down time and the DC bus voltage never exceeds its threshold, then will the brake resistor ever have any effect on slowing the motor or dumping the excess voltage?

 

[winnie]

The brake resistor doesn't slow the load.

Instead what happens is that the VFD causes the motor itself to operate as a generator. The motor becomes a mechanical load, turning kinetic energy stored in the inertia of the system into electrical energy.

With no place for this electrical energy to go, the DC rail capacitors start charging up, storing energy. If the DC rail voltage gets too high, then something breaks.

The brake resistor is simply set up to provide a place to dump this energy before you burn the capacitor bank.

I guess what I am trying to say is that the motor slows the load down, the VFD controls the motor to make this happen, and the brake resistor is a place that can use the electricity generated by the slowing operation.

If the slowing is gradual enough, then the generated electricity is less than the losses in the VFD and motor; so the resistor never sees any of the energy.

If you have another useful place for the electrical energy to go, then that can also be used for braking. I've heard of elevator systems where energy generated by one drive is used to power other drives.

-Jon

 

[StephenSDH]

It does break the motor because without it the drive will fault out on overvoltage, resulting in the motor coasting. If you want the drive to break the motor, you have to install a DB Resistor or have some sort of common dc buss system to offload power to another drive.

 

[LarryFine]

Diesel-electric train engines have dynamic braking resistor grids in them. They definitely are used to slow the load.

 

[Jraef]

Winnie's answer is great, I'll only add specifics to your questions.

When the resistor is in the circuit, the energy from the DC bus will always flow to it. So without the chopper, you will have a big DC resistance heater running all the time and no energy for the motor. Some smaller drives (i.e. under 3HP typically), use what are called IPMs (Integrated Power Modules) from the various transistor manufacturers instead of discrete components. So what you end up with is a single credit card sized device (thicker of course) that has the 6 diodes, 6 transistors AND the chopper transistor all in one integrated piece. So with those smaller drives, you don't need a separate chopper module only because it is already built-in.

As winnie said, you are actually braking the load using the motor as a generator, the resistors just accommodate the process. What the VFD does when you enable Braking is to monitor the motor speed and alter the output frequency on the fly to have the motor ALWAYS in a state of spinning faster than the connected power frequency. As the motor slows, the VFD output frequency lowers along with it, i.e. the motor is always overhauling and regenerating. This is referred to as "Dynamic Braking" because the more kinetic energy you have in the spinning load, the more available braking energy you have available. But as the load slows down, so then does the braking energy available and by the law of diminishing returns, a Dynamic Brake can technically never completely stop the load because at some point your system losses eat up what ever minuscule amount of kinetic energy is remaining. That is why VFDs have a DC Injection Brake that turns on when the motor speed gets close to zero; it has to finish the job started by the Dynamic Brake.

A "Line Regenerative VFD" is one where a second set of transistors replaces the diode bridge rectifier on the front-end, so that instead of dumping the excess motor energy into resistors, it fires the transistors to pump the energy back into the line. These are becoming more and more available and desirable because unlike the resistor method, there are almost no restrictions as to how often you can brake and for how long. Some people even make a big deal about energy recovery with them, especially on large spinning masses such as centrifuges. The type that only dump the energy into a DC bus to be used by7 other connected inverters is called a Common DC Bus Drive System, it has a standard front end rectifier.

 

[winnie]

I apologize for not being clear in my last post.

I was not trying to say that the resistor has nothing to with braking; after all it is called the 'dynamic braking' resistor.

I was instead trying to describe the function served by this resistor.

I said that on a VFD fed induction motor, the dynamic braking resistor does not _cause_ braking. If you break into the control circuitry and trigger the brake chopper, the motor would not change speed significantly. The resistor will get hot, and some energy from the mains will be wasted.

The braking resistor is there to dissipate excessive energy stored in the DC rail capacitors.

Since this energy is commonly the result of braking, then the reason that this resistor is installed is to permit braking via regeneration. But there are other (less effective) ways that a VFD can brake a motor, _without regeneration_ and thus _without a brake resistor_. The purpose of this resistor is not to brake the motor, but instead to deal with a side effect of braking the motor.

philly asked if the resistor could be continuously connected. A brake resistor is generally sized such that it would overload and quickly overheat if directly connected to the DC rail. Any resistor connected to the DC rail would continuously dissipate power, and thus be a cause of inefficiency. In general, however, there are small resistors connected across the DC rail to discharge and balance the capacitor banks. These resistors represent a small but continuous loss. These resistors may be enough if there isn't much in the way of regenerated energy on the DC bus.

The braking resistor is normally triggered on DC rail voltage. This is because DC rail voltage is a perfect indication of excess energy being dumped into the system (say by dynamic braking). You could certainly trigger the braking resistor using some other method of detecting excess energy, but I don't know of any reason to use anything other than voltage. I suppose that if the DC rail is somehow directly connected to batteries, you might use the battery state of charge as your trigger. But really anything indicates excessive stored energy could trigger the braking resistor.

-Jon

 

[Besoeker]

It does break the motor because without it the drive will fault out on overvoltage, resulting in the motor coasting. If you want the drive to break the motor, you have to install a DB Resistor or have some sort of common dc buss system to offload power to another drive.

Hopefully, with a correctly dimensioned brake circuit, nothing will break....:grin:

 

[Besoeker]

Since this energy is commonly the result of braking, then the reason that this resistor is installed is to permit braking via regeneration.

Slight correction. Regenerative braking is when the energy goes back into the supply. With a dynamic braking resistor, the energy gets dissipated as heat.

We have installed systems with both. These are typically on systems with a number of drives fed from a common DC bus. Some drives have an overhauling load under normal running conditions and it would be impractical to continuously dissipate the excess energy in resistors. They use the regen mode to feed back into the DC system and that's fine as long as the net load on the DC is net positive.
But, for an emergency stop on all drives, there is no power being drawn from the DC so you can't use regen. Dynamic brake resistors are used for this operation.

 

[gar]

100213-1125 EST

For the moment erase from your thinking the operation of a simple braking resistor connected to a permanent magnetic DC motor.

Consider how some DC brushless, VFDs, and Vector drives work. There is a DC source or sink from which energy flows to the motor or is received from the motor. This would be classified as a four quadrant control. In one quadrant energy flows from the DC source thru the controller to the motor in the forward direction and the mechanical load on the motor is receiving energy. In a second quadrant of operation the motor is still moving forward but the control is trying to drive the motor in reverse and therefore is braking the motor. The other two quadrants of operation are with the motor running in reverse.

The easiest way to visually see what is happening is to consider a DC motor. With a battery applying a voltage of forward polarity and the motor has a mechanical load that is being driven and rotating in the forward direction there is a counter EMF generated in the motor that opposes the applied forward voltage. This counter EMF is slightly less than the applied source voltage by an amount that is just right to allow the necessary forward current to flow thru the motor to maintain the motor speed at the mechanical load (torque).

If we simply reverse the applied voltage, then the source voltage is adding to the counter EMF and the current to the motor reverses direction and now energy flows from the motor to the electric energy source and the motor is braked. All this without a braking resistor.

If the electric energy source is a capacitor, then since energy is flowing to it during this braking operation the capacitor's voltage will rise. Why does it rise? Because there is a one way device, diodes, that allows energy to flow from the AC primary lines to the capacitor, but does not allow energy to flow back to the AC lines. If the capacitors can not hold all of the energy flowing back into them without developing an excessive voltage, then some means is required to dump some of this energy. One method to achieve this dumping is to use a relay to connect a resistor across the capacitor bank at a predetermined threshold voltage, and then at a somewhat lower threshold voltage remove the resistor. This is a bang-bang servo to control maximum capacitor voltage.

Obviously the relay turn-on threshold has to be above the normal maximum voltage on the capacitor bank from a high AC line voltage.

There are other more complex regenerative means that transfer the excess capacitor energy back into the AC distribution system.

Of course there could be a big battery in parallel with the capacitor bank to receive the energy.

Most of what I have said here was already described. Just saying the same thing from a somewhat different perspective may help some better visualize what is occurring.

.

 

[philly]

Thanks for all of the great responses guys!. It makes things much clearer.

I have a VFD that has an extremely long ramp time of about 3min. The motor is a 125hp motor and the drive is 200hp. During this ramp down time I never see the DC bus voltage go above 700VDC and therefore suspect that neither the chopper nor brake ever get turned on. Is this a correct assumption? In this case does the long ramp down time help with the amount of kenetic energy that is converted to electrical energy thus not overdriving the bus?

What about when a VFD is given an E-Stop and therefore the drive is no longer performing a ramp down, and the motor is free spinning down? Will the motor only be able to regen for the duration of time that the magnetic field is still present in the motor which may only be a couple of seconds? After the field has decayed there can be no regenerative voltage driven from the motor thus the DC bus voltage should not increase during the remainder of the coast down, and the resistor will not be in the circuit?

What about when the braking resistor is not connected to the drive? Does the drive still brake the motor in the same way as described above only it has no where to dump additional energy across the resistor? In other words does it still monitor output speed in comparison with motor speed to always keep the motor in regen during decellarition with the only difference being no where to dump the excess voltage?

What is a very small ramp time (couple of seconds) on a high inertial load is setup? Should this load be able to be drawn to a complete stop by the drive? Would it operate the same way by monitoring motor speed and quickly changing output speed to slow motor?

It appears that you guys have confirmed my beliefs above in stating that the resistor is not what is performing the actual braking, however it is a means of dissipating excess energy caused by the braking action of the drive when the motor is forced to covert its kenetic energy to electrical energy.

 

[StephenSDH]

What about when a VFD is given an E-Stop and therefore the drive is no longer performing a ramp down, and the motor is free spinning down?

Category 0 Stop - The drive is immediately disabled on estop in hardwire and the motor/load will just coast.
Category 1 Stop - The machine will ramp down under power when the estop is pressed, once everything is stopped power is removed by a delayed relay. (this is often used where there is alot of inertia so it is more safe to stop the machine under power then it is to let everything coast.)

You likely have a Cat 0 stop so the drive will do no breaking.

What about when the braking resistor is not connected to the drive? Does the drive still brake the motor in the same way as described above only it has no where to dump additional energy across the resistor? In other words does it still monitor output speed in comparison with motor speed to always keep the motor in regen during decellarition with the only difference being no where to dump the excess voltage?

Yes. You will see the bus voltage rise up when you are applying deceleration torque to the load. Rotational Energy becomes stored electrical energy. The drive will fault when it exceeds voltage limit and let the motor coast.

What is a very small ramp time (couple of seconds) on a high inertial load is setup? Should this load be able to be drawn to a complete stop by the drive? Would it operate the same way by monitoring motor speed and quickly changing output speed to slow motor?.

It really depends on your system. If you don't need to start and stop quickly it is more gentle on the shafts/couplings to have a long ramp. Web lines will ramp up slowly to prevent shock to the system. Prevent shocking PID tension/temperature loops or ripping the the web.

 

[winnie]

During this ramp down time I never see the DC bus voltage go above 700VDC and therefore suspect that neither the chopper nor brake ever get turned on. Is this a correct assumption? In this case does the long ramp down time help with the amount of kenetic energy that is converted to electrical energy thus not overdriving the bus?

It _may_ be that the voltage is being clamped at 700V because of the brake chopper, but it is also possible that the brake chopper never turns on. You can look for voltage across the brake chopper resistor to confirm one way or the other.

The long ramp down time does not change the total amount of energy that needs to be dissipated; the kinetic energy of the rotating load is exactly the same. The long ramp time does change the rate of this energy dissipation; if the rate is low enough than various other loss terms in the system are sufficient 'consume' the regenerated energy. The capacitor bleed resistors, losses in the motors, even the inverter control circuitry (if powered from the DC bus) can all consume the regenerated energy, if the regeneration power is low enough.

Finally, braking by running the motor as a generator is not the only possible braking technique. A drive can also run DC current through the motor windings, or can actively drive the motor in the reverse direction. Both of these approaches prevent regeneration, but end up dumping all of the kinetic energy as motor heating.

-Jon

 

[philly]

It _may_ be that the voltage is being clamped at 700V because of the brake chopper, but it is also possible that the brake chopper never turns on. You can look for voltage across the brake chopper resistor to confirm one way or the other.

Can you measure this DC voltage by putting your meter across the leads going to the resistor? I'm assuming these leads would be after the chopper so you wouldn't be measuring jut the bus voltage. Would current measured simply be the voltage divided by the resistor value?

The long ramp down time does not change the total amount of energy that needs to be dissipated; the kinetic energy of the rotating load is exactly the same. The long ramp time does change the rate of this energy dissipation; if the rate is low enough than various other loss terms in the system are sufficient 'consume' the regenerated energy. The capacitor bleed resistors, losses in the motors, even the inverter control circuitry (if powered from the DC bus) can all consume the regenerated energy, if the regeneration power is low enough.

So a very short ramp time will stop the motor quickly but the rate of the energy dissipation will be very quick and and therefore a great amount of regenerated voltage will be dumped across the DC bus?

Will this short ramp time necessarily stop the motor in the designated ramp time or is it possible that the motor may still turn after the given ramp time?

Are the other losses you are talking about such things as cable losses represented by V^2/R or I^2R? So for small rate of dissipation these losses are enough to prevent overvoltage?

Why do some drive applications not require braking resistors? Such as conveyors? Isnt there still a braking effect converting kenetic energy to electrical energy?

-Jon[/QUOTE]

 

[StephenSDH]

Why do some drive applications not require braking resistors? Such as conveyors?

Conveyors have a large amount of friction and a very low stored kinetic energy. They coast stop very quickly, and generally do not pose a safety hazard when coast stopping. If you have a large mass on the conveyor on an incline you might want to add a breaking resistor.

Don't think of decelerating as always braking the load. If the load naturally stops faster then the set decel rate the drive will actually do work to keep the load moving. Decelerating doesn't always mean the drive is regenerating/braking.