Motor calculations

[jevs]

What calculations could I use to figure out the voltage and current specs on these motors? I am trying to figure out what voltage to make the power supply and which drivers to get.

Fujitsu Fanuc LTD
Fanuc DC Motor 0 (z axis)
Type: 0-2000M
NO: D-771719
Date: 77 10
Permanent Magnet DC Servo Motor
Part Number A06B-0613-B031
Output Power: 0.4 KW (0.5 HP)
Rated Torque: 28 Kg-Cm
Max. Torque: 240 Kg-Cm
Max. Speed: 2000 Rpm
Rotor Inertia: 0.029 Kg-Cm-S 3
Back EMF Constant: 25 V/K RPM
Torque Constant: 2.44 Kg-Cm/Amp
Mechanical Time Constant: 25 mSec
Thermal Time Constant: 50 Min
Weight: 12 Kg

Fujitsu Fanuc LTD
Fanuc DC Motor 5 (x,y axis)
Type: 0-2000M
NO: D-771719
Date: 77 10
Permanent Magnet DC Servo Motor
Part Number A06B-0614-B031
Output Power: 0.8 KW (1 HP)
Rated Torque: 55 Kg-Cm
Max. Torque: 480 Kg-Cm
Max. Speed: 2000 Rpm
Rotor Inertia: 0.05 Kg-Cm-S 3
Back EMF Constant: 50 V/K RPM
Torque Constant: 4.87 Kg-Cm/Amp
Mechanical Time Constant: 15 mSec
Thermal Time Constant: 55 Min
Weight: 16 Kg

This is pretty much the only specs I will ever be able to get on these motors. I can't find any calculations on the net to derive a voltage and current rating for the motors so I can build a powersupply and order drivers. It took me a month of pestering Fanuc to even get them to finally fax me these specs. This project is turning into a nightmare just because of no specs on these motors.

[Al_The_Man]

I have the formula at the shop, I can get it tommorow, You have most of the information there, it is just a case of extracting it, DC motors when run from PWM drives are are pretty flexible when it comes to voltage rating as the current is controlled by the drive. The back EMF constant is just below what the terminal voltage is, were the motor run from a steady DC supply, in other words they are 2000 rpm motors so to run the 5 at 2000 rpm would be a tad over 100v (50V/Krpm).
Al

[ViperTX]

That is basically what I would do.....take the back emf voltage of 50 V / Krpm and multiply it by the max. number of K rpm = 100 volts. Take that voltage and divide it into the wattage of the motor and that will give you the rated current.....well close....one is 843 watts / 100 = 8.43 amps.

The other things I would do....is not buy motors without specs....*smile* or learn to do SWAGs.

[jevs]

Anything you can give me tommorow would be great. I would love to be able to finally order the drives and power supply parts.

Thanks

[Al_The_Man]

OK, to obtain the operating voltage, look at the BEMF constant, this is the voltage generated by the motor if the shaft was rotated at 1000rpm, in the case of the 5 motor it is 50v/1000 rpm, so at 2000 rpm the voltage generated would be 100vdc. This is because a motor is a generator also and will output a voltage whether it is by turning the shaft or be a voltage fed into it, This generated voltage opposes the applied voltage, so at a no-load condition on the shaft of the motor with 100vdc applied the opposing voltage will be almost 100v which cannot be exactly equal or no current will flow, only a small current will be flowing because of friction etc, and because torque is directly related to current the torque will be at minimum.
So we have a voltage of 100vdc required, but we also have losses in amplifiers and power supply losses so the required voltage in this case should be 100 * 12.5% to allow for this. so the DC supply should be around 112.5vdc.min.
Now to the required current, the Kt or Nm/amp shows that the motor will produce 4.87 Kg-Cm of torque per 1 amp of current, the rated torque is 55 Kg-Cm, so the current needed to produce this torque will be 55/4.87 = 11.29 amps.
If a PWM amplifier is used, these amplifiers typically use a 20Khz frequency, so the voltage will be a square wave at either 0v or 112v and the width of the pulse will govern the amount of current flowing.
Due to the characteristics of the armature, the current wave form will not resemble the voltage waveform at all, it will in fact be of an almost smooth/sawtooth nature the amplitude of which will move up or down depending on the width of the voltage pulse. The higher the current the higher the torque.
A PWM amplifier should not have to conduct at 100% pulse width, otherwise the maximum control may not be achieved, so in order to keep the operating range within the desired control, it is important to have a suffiently high DC power supply, most amplifiers have a current limit which can be set so that the rated torque is not exceeded.
So from the above, the DC supply should be a minimum of 112.5vdc and rated at a min of 12x2= 24amps to allow for momentary overload.
The amplifier should be rated for 120vdc and 12 amps max. continuos and 24amps peak.
You will also notice that the max torque is 480Kg-Cm, for this torque to occur the current would have to be 480/4.87 = 98 amps! This is the maximum current the motor would take before damage or demagnetization, and would be out of the range of the typical amplifier or power supply.
From this you should be able to calculate for the 0 motor.
Al

[Evodyne]

Let's start with torque. A motors shaft torque varies from a maximum when the shaft isn't allowed to turn to zero when running with no load. A small amount of "extra" torque is used to overcome friction within the motor, regardless of speed. The total torque is then the delivered shaft torque plus the torque used to overcome internal losses.

Current draw is proportional to torque. It will be at a maximum when the rotor is stalled, and at a minimum when the motor is running with no load. When at no load, the current draw will be just enough to generate the internal torque needed to overcome friction and keep the motor spinning. Load the shaft down with finger pressure and it slows because your are applying a counter torque in excess of its shaft torque. But as it slows both current draw and shaft torque climb. At some point the shaft torque balances the finger-induced torque and the motor runs at a new rpm until the equilibrium is again broken.

EMF, or Electro-Motive Force, is measured in Volts. Back EMF is the voltage that the motor itself generates and is opposite in polarity to the supply voltage and is proportional to rpm. It is at a minimum when the motor is stalled, and at a maximum when running unloaded. The difference between the supply voltage and back EMF is the effective voltage that is left to generate the current that drives the motor.

So both the current and back-EMF vary with rpm as it goes from the stalled condition to the no load condition. Electrical power is the product of current and voltage. Thus power is zero when stalled with zero back-EMF, and nearly nothing when running unloaded as the current is almost zero. The maximum output actually comes at the rpm half way between these two extremes. Power is also the product of rotational velocity times the torque. For power in Watts, torque in N-m, and speed in radians per second power is just speed times torque. For different units we multiply by these adjustment factors:

Torque, T, in oz-in, Speed in rpm: P = rpm x T x 0.00074
Torque, T, in lb-in, Speed in rpm: P = rpm x T x 0.0118
Torque, T in N-m, Speed in rpm: P = rpm x T x 0.1047

Also 1 HP = 746 W, so your 400 W motor is actually 0.54 HP

Torque conversions:1 kg-cm = 0.098 N-m = 0.868 lb-in = 13.9 oz-in, so...
Rated Torque = 28 Kg-Cm = 2.7 N-m = 24.3 lb-in = 389 oz-in.
Max. Torque = 240 Kg-Cm = 23.5 N-m = 208 lb-in = 3,336 oz-in
Torque Constant = 2.44 Kg-Cm/Amp = 0.23912 N-m/Amp = 2.118 lb-in/Amp = 33.92 oz-in/Amp
Max. Speed: 2000 Rpm
Back EMF Constant: 25 V/K RPM

We know that you don't want to exceed 2000 rpm, and we know that this speed will be obtained under a no load condition where back EMF is almost equal to the supply voltage. This is where the back EMF constant factors in: 2000 rpm x 25 V/1000 rpm = 2 x 25 = 50 volts. That's your supply voltage.

We also know that rated torque is developed at a speed one half of the maximum rpm and when the motor is consuming its rated power. So 389 oz-in of torque at 1000 rpm for 400 W of input power. Output power = 0.00074 x rpm x oz-in = 0.00074 x 1000 rpm x 389 oz-in = 287.85 Watts.

Efficiency then is power out divided by power in = 288 Watts / 400 Watts = 72%. This is a reasonable value for this type of motor.

Current at 389 oz-in is torque divided by the torque constant, or I = 389 oz-in x 1/33.92 Amp/oz-in = 11.468 Amps. Back EMF will be the back EMF constant times the actual speed. This yields 25 V/1000 rpm x 1000 rpm = 25 Volts. The efffective voltage is the supply minus the back EMF or 50 V - 25 V = 25 V. Power is effective voltage time current, thus P = 25 V x 11.5 Amps = 288 Watts, which agrees with the value determined above from torque and speed.

Extrapolating the current curve (which is a straight line) to zero rpm we find the supply current needed. Current is near zero at 2000 rpm, 11.5 amps at 1000 rpm, so 23 amps at 0 rpm. Safe design practices dictate that the supply has some extra capacity or headroom. Let's aim for 120% of our maximum non-stall current-our design current will be 23 Amps x 1.2 = 27.6 Amps. So around 30 Amps. Torque at 23 amps will be 33.92 oz-in/Amp x 23 Amp = 780 oz-in.

At 23 Amps and 50 Volts we are pumping 23 x 50 or 1150 Watts into the motor. This condition occurs when it is loaded to a very low, near zero rpm, but not stalled. However at very low rpm, virtual no output power is generated. Thus 1150 Watts is dissipated within the motor and shows up as heat-think of a 1,150 Watt hair dryer!

What happens at stall? We also know the torque at stall (which is the maximum torque) is 3,336 oz-in. If we divide by the torque constant by inverting and multiplying we get 3,336 oz-in x 1/33.92 Amp/oz-in = 98.3 Amps! Your supply would either need to be able to supply this current or bow out gracefully (current limiting or fuse popping). Why is the stall so big? Any time the motor is turning there are additional loss factor collectively known as viscous effects. These are in addition to winding, brush, and brush to winding resistance. But the instant the rotor stalls these additional viscous losses go away, so there is much less impeding the flow of current, and the value skyrockets.

As far as drives the popular Geckodrives go into current limiting at 20 Amps. With a 50 Volt supply this means you would see 679 oz-in of torque vs. 780 oz-in. Rutex has drives that can handle the motor at it's full 30 Amp rating and a lot of guys like these drives. Both are step and direction type servo drives.

Hopefully this has helped. You get to do the other motor! I'm sorry if this was too long.

Lance

[Evodyne]

Hey Al-we each did a motor! Looks like he's got no homework. Nice job.

[jevs]

Thanks for the info. I am going to go with the Rutex drives that are 200V/40 Amps. Since I have 2 of the 5 motors and one of the 0 motors, how would you guys suggest I set up the power supply? It would be nice to buy only one transformer with two secondaries?One for the 0 motor and one for the 5 motors? Or do I need a transformer with 3 secondaries. One for each motor drive? Or should I just get 3 transformers? I will be using 220V single phase to run this machine. Any suggestions on where to get or what part number to buy for this transformer? You say Minimum voltage for the 5 motor is needing to be 112.5V. Should I bump it up a little bit more just to be safe? If so what is your suggestion? Using the same equation for the 0 motor I guess I need 56.25 Volts for the 0 motor supply? Should I bump that one up a little also? If so to what? Just going for a final clarification on supply voltages since I have picked the drives now. I am going to call Tom and order them now.

[Al_The_Man]

Generally I have found that with PWM drives, the voltage on the motor is not critical,within reason.
The enemy of Servo motors is excess current (Heat) and excess speed, I might even be tempted to go with one supply for both as long as you can set the current limit to the required level and still have control, Although the voltage amplitude will be high, the mean current will automatically be reduced by the smaller pulse width.
I have no experience with the Rutex drives, normally I use an amplifier that uses the current mode of operation +- 10v analogue, the max voltage in (10v) can be set to represent the max. current allowed for the motor. The control signal is directly proportional to the current out.
Al

[Al_The_Man]

Originally Posted by Al_The_Man , I might even be tempted to go with one supply for both as long as you can set the current limit to the required level and still have control, Although the voltage amplitude will be high, the mean current will automatically be reduced by the smaller pulse width.

On second thoughts that might be pushing it, I have used about 40% voltage higher but this would be 100% so maybe stick with the original plan. I would put together a linear unregulated supply using a couple of Toroidal transformers, they usually come up on ebay and if you need to tailor the secondary voltage, usually adding or subtracting turns is easily done, there are previous links here that show how.
Al

[jevs]

OK, so just to clarify. Build two power supplies using single primary, single secondary Toroidal transformers. Make sure one can handle the current draw of both 5 motors, and make the other one for the 0 motor. Make the one for the 5 motors output 112.5VDC after the rectifier and capacitor(s) do their job. Make the other supply output 56.25 Volts after it is rectified and smoothed. If this is correct I think I got a plan. Would it hurt to bump it up to about 120 Volts and 60 Volts just for even numbers to work with and just a tad extra in case I need it? The rutex drives do have a current limiting trim pot that I can adjust. The rutex drives do want unregeulated supplies, so this is nice also. I was told not to use regulated supplies.

[Al_The_Man]

2 supplies for this rated at 24amps 2.5% ripple 60% duty cycle
120vdc = 86.2vac, 24,000µfd 2.5Kva Txfr
60vdc = 44vac, 47,000µfd 1.3Kva Txfr.
Thats for 1 motor each.
Al