VFD Braking Resistor

[Jonesy]

This is really an update on the Chekko Lathe ressurection.

Now the drive is operational... the motor and countershaft, and the flat belts are functionally tested, it's become apparent that I need to assist the vfd with respect to decellerarion. My little Randa slows to a stop in about 1.5 seconds so I asked the Chekko to do the same, but the countershaft and spindle pulleys, with a combined weight of around 50kg are having none of it. I can stop from around 1000 rpm (motor rpm) but any higher and I get errors. The 10" chuck (not yet fitted) will easily add another 30kg to this.

Quite obvious a braking resistor is required. But... how can they justify the rediculous prices they are asking for what is simply a big resistor?

The inverter spec stated a minimum 75ohm. That's no problem... but the wattage needs to be high. Very high. But how high?

In my scrap bins I found a heating element from an old towel rail. It's rated 600W @ 240V. But how does the 600W equate during braking when the voltages can be considerably higher? Anyway, it measures 93ohm so I figured what the hell?

As a test I connected it, with no heatsink, and got perfect braking up to the max setting 3000 rpm. It got warm, but not hot. With a chuck it'll have to work harder but early results are promising. Add to this the pulley ratio is now wrong as I had to remake the motor pulley at twice the original diameter so the max HZ will need to be reduced to approx half it's current value. So hopefully I'll be decellerating from closer to 1500rpm instead of 3000.

I've made a nice ally heatsink for the element and hope to test tomorrow with the chuck mounted.

Fingers crossed.

[J Tiers]

Not that high, likely. The energy is high, but the amount of time it is "on" is what determines the needed power capability.

Usually there is a duty cycle setting in the VFD. That goes from 1% or so, up to 5% or maybe even 15%. The calculated power is multiplied by the selected duty cycle to get the actual resistor power dissipation.

Normally it does not take that much actual power to keep the bus voltage in control. The circuit will not even be active unless it is over-voltaged, and then ot will operate at the duty cycle.

So, 75 ohms and an overvoltage of 400V suggests a 2200 watt resistor minimum. But if the duty cycle is 10%, that's 220 watts average. Quite a difference.

So you probably only need a 250W resistor, and can set the duty cycle to be under that.

What you DO need, is a resistor that can take a short term overload, which not all can. And you need to mount it on a heatsink (if a place is not provided in the VFD) which can dissipate the power and prevent the resistor from being overloaded if you stop several times.

The entire stopping deal is also controlled by how fast you want to stop. Often, a coast down is adequate, and that needs no help. Lengthening the stop time also helps avoid the need for a resistor, if you can tolerate that. Deceleration time is a parameter you can set.

[oxford]

I have never tried it but I have heard of people using a stove top coil from an electric range as a poor mans brake resistor.

[Jonesy]

Not that high, likely. The energy is high, but the amount of time it is "on" is what determines the needed power capability.

Usually there is a duty cycle setting in the VFD. That goes from 1% or so, up to 5% or maybe even 15%. The calculated power is multiplied by the selected duty cycle to get the actual resistor power dissipation.

Hi Mr T.

This is the closed I can find that may be what you're referring to.. but I don't think it is.

P555 Chopper P limitation
(Chopper power limitation) S
5 ... 100 %
{ 100 }
With this parameter it is possible to program a manual (peak) power limit for the brake resistor.
The switch-on delay (modulation level) for the chopper can only rise to a certain maximum
specified limit. Once this value has been reached, irrespective of the level of the link voltage, the
inverter switches off the current to the resistor.
The result would be an overvoltage switch-off of the FI.
The correct percentage value is calculated as follows: *100% [%] * 2
max
max
U
k = R P BW
R = Resistance of the brake resistor
PmaxBW = Momentary peak power of the brake resistor
Umax = FI chopper switching threshold
1~ 115/230 V ⇒ 440 V=
3~ 230 V ⇒ 500 V=
3~ 400 V ⇒ 1000 V=

I have never tried it but I have heard of people using a stove top coil from an electric range as a poor mans brake resistor.

I investigated that option. I have 3 different stove elements (spares for the range cooker in the kitchen) but all are higher wattage (1600+) which equates to a lower resistance (around 30-40ohm) so not suitable.

The element I found for the towel rail seems to fit the bill. And it was free. I like free. I am a very poor man

[CalM]

This is really an update on the Chekko Lathe ressurection.

Now the drive is operational... the motor and countershaft, and the flat belts are functionally tested, it's become apparent that I need to assist the vfd with respect to decellerarion. My little Randa slows to a stop in about 1.5 seconds so I asked the Chekko to do the same, but the countershaft and spindle pulleys, with a combined weight of around 50kg are having none of it. I can stop from around 1000 rpm (motor rpm) but any higher and I get errors. The 10" chuck (not yet fitted) will easily add another 30kg to this.

Quite obvious a braking resistor is required. But... how can they justify the rediculous prices they are asking for what is simply a big resistor?

The inverter spec stated a minimum 75ohm. That's no problem... but the wattage needs to be high. Very high. But how high?

In my scrap bins I found a heating element from an old towel rail. It's rated 600W @ 240V. But how does the 600W equate during braking when the voltages can be considerably higher? Anyway, it measures 93ohm so I figured what the hell?

As a test I connected it, with no heatsink, and got perfect braking up to the max setting 3000 rpm. It got warm, but not hot. With a chuck it'll have to work harder but early results are promising. Add to this the pulley ratio is now wrong as I had to remake the motor pulley at twice the original diameter so the max HZ will need to be reduced to approx half it's current value. So hopefully I'll be decellerating from closer to 1500rpm instead of 3000.

I've made a nice ally heatsink for the element and hope to test tomorrow with the chuck mounted.

Fingers crossed.

Does your VFD offer DC injection?
It's a way to send some of that extra energy back into the electric motor instead of the VFD and resistor bank.

[CalM]

Hi Mr T.

This is the closed I can find that may be what you're referring to..

P555 Chopper P limitation
(Chopper power limitation) S
5 ... 100 %
{ 100 }
With this parameter it is possible to program a manual (peak) power limit for the brake resistor.
The switch-on delay (modulation level) for the chopper can only rise to a certain maximum
specified limit. Once this value has been reached, irrespective of the level of the link voltage, the
inverter switches off the current to the resistor.
The result would be an overvoltage switch-off of the FI.
The correct percentage value is calculated as follows: *100% [%] * 2
max
max
U
k = R P BW
R = Resistance of the brake resistor
PmaxBW = Momentary peak power of the brake resistor
Umax = FI chopper switching threshold
1~ 115/230 V ⇒ 440 V=
3~ 230 V ⇒ 500 V=
3~ 400 V ⇒ 1000 V=



I investigated that option. I have 3 different elements (spares for the range cooker in the kitchen) but all are higher wattage (1600+) which equates to a lower resistance (around 30-40ohm) so not suitable.

The element I found for the towel rail seems to fit the bill. And it was free. I like free.

put two or three of those stove elements in series. Free is FREE ;-)

Add a "chicken lamp" aka brood heater IR bulb and then you can "SEE IT WORKING" ;-)

[Jonesy]

put two or three of those stove elements in series. Free is FREE ;-)

Add a "chicken lamp" aka brood heater IR bulb and then you can "SEE IT WORKING" ;-)

Hey Cal.

The stove elements weren't free. I paid for them! And it hurt.

Does your VFD offer DC injection?
It's a way to send some of that extra energy back into the electric motor instead of the VFD and resistor bank.

I thought DC injection was when it holds the motor still after it's stopped for a given time. I read that using this to slow the motor can cause excessive heat in the coils and damage the motor... which was also free, as was the VFD!!

[alanganes]

The pace I work has a Haas CNC bed mill that has a plain old stovetop heating element in a perforated metal box on top of the cabinet as a braking resistor. Came like that from the factory.

[mikey553]

Take a look at this link: https://www.ebay.com/itm/27414217746...sAAOSwy2pbrdZv
Usually you can find a right combination of resistance and power. I have two similar power resistors (300W, 70 ohm) and the quality appears to be good.

[Jonesy]

Hey mikey... was looking a similar resistor on ebay uk and ebay usa... 100ohm 500W. This is what got my old mind confused....

Ohms law gives a relation between amps, volts, current and power. Is that 500W at 240V, 120V or some other voltage?

My element is rated 600W at 240V. That equates to 96ohm. The actual reading is 93ohm. So that adds up.

So what voltage is being referred to with regard to that resistor to get to a wattage of 500?

Are your heater elements rated at 230 / 240V over there?

If your running a lower voltages than the uk, are you using twice the current?

[PStechPaul]

223 volts = sqrt( 500 * 100 )

[Tundra Twin Track]

Is it possible to add a mechanical Brake,Disc or Drum,my 18x60 has a Drum Brake on Motor Pulley and will stop the 12” Bison Chuck in a split second.

[mikey553]

Hey mikey... was looking a similar resistor on ebay uk and ebay usa... 100ohm 500W. This is what got my old mind confused....

Ohms law gives a relation between amps, volts, current and power. Is that 500W at 240V, 120V or some other voltage?

My element is rated 600W at 240V. That equates to 96ohm. The actual reading is 93ohm. So that adds up.

So what voltage is being referred to with regard to that resistor to get to a wattage of 500?

Are your heater elements rated at 230 / 240V over there?

If your running a lower voltages than the uk, are you using twice the current?

Resistors are rated by their resistance and power they can dissipate. Voltage is irrelevant. Ohm's law is not applicable here. Think about it this way. You have rotating mass in your machine, which stores a certain amount of kinetic energy. When you stop the machine, your motor acts like a generator converting this kinetic energy into electrical one. VFD then passes the generated electrical current through power resistor. Resistor gets heated absorbing the electrical energy.

By the way my VFD (and many others) is powered by 240 V, 60 Hz single phase supply.

[J Tiers]

Resistors are rated by their resistance and power they can dissipate. Voltage is irrelevant. Ohm's law is not applicable here. Think about it this way. You have rotating mass in your machine, which stores a certain amount of kinetic energy. When you stop the machine, your motor acts like a generator converting this kinetic energy into electrical one. VFD then passes the generated electrical current through power resistor. Resistor gets heated absorbing the electrical energy.

By the way my VFD (and many others) is powered by 240 V, 60 Hz single phase supply.

Kind of.....

The switching is changed in phase so that the back EMF from the motor is used to charge up the bus capacitors. That takes energy out of the motor and drive , slowing them down.

The bus voltage rises, however. So the function of the resistor is to drain off the excess charge out of the bus capacitor. The power is the voltage squared divided by the resistance. That is for voltage applied 100% of the time (100% duty cycle).

If the voltage is applied only half the time, then the power in the resistor is reduced. That duty cycle is the percentage of the time that the resistor is connected to the bus voltage.

The more of the time it is connected, the faster it drains off charge, and the more energy it dissipates. That energy is what was taken out of the motor to slow it, and which was put into voltage on the capacitors in the VFD. You have to draw current from the capacitors to take that energy out, and lower the voltage. *

The resistor really does not need to be "on" all of the time during a stop. It really takes out energy, and for reasonable stops only a low duty cycle is needed. That one goes to 100%, but I'd not expect that to be needed.

yes, stove elements can work, if they are the right resistance, or are put in series to make up the right resistance. And they handle a lot of power, although they are a but clumsy to mount. A couple of them can likely handle 100% duty.

Most machines are good with a 250 watt or so resistor, operated at a lower duty cycle.

You can calculate all of this, if you know the energy of your spinning workpiece and chuck , drive system, etc. Usually, a lot of that is not easily knowable for a lathe, mill, etc. so some "cut and try" is needed.

But, if you can stop from 1000 rpm OK now, with no resistor, then you probably need very little more energy dissipation to stop at the highest speed, (depending on how fast that is.). If your top speed was 1400 rpm, you would need to dissipate only double the energy, which should be fairly easy.

If you can give the top speed of the machine, and some idea of the sort of chuck mass, and work type and mass that you want to use, that would help.

* If the voltage goes up to 400V, and you draw an ampere from the capacitors (which would be a 400 ohm resistor), that is drawing 400 watts, or approximately 1/2 horsepower. A 75 ohm resistor would then draw 400 / 75 = 5.3 amperes. Power drawn is then 400^2 / 75 = 2133 watts. That is just short of 3 HP.

[Sparky_NY]

Ebay is loaded with inexpensive braking resistors. I have had one of this type on my bridgeport for years and no issues.

https://www.ebay.com/itm/274142177462