Autotransformer question

Ignoramus28517 · 2010-06-28T14:28:38+00:00

I am still thinking about the best way to provide power for servodrives. As of now I went back to using 30A8 drives, which require 20to 80VDC. One alternative that I have is an autotransformer that was in theBridgeport control cabinet. This is a 3 kVa autotransformer that hasmultiple taps. Its job was to take 220v and convert it to 380 voltsfor the Bosch servo drives. Being 3 kVa, it is a heavy item, weighing perhaps 50 lbs or evenmore. As I said, it has many taps, and among them specifically it has a 0vtap, 460v tap and 180v tap.If I connect 124 VAC that I get from the wall, to 0-460v taps, I wouldthen get 48 VAC on the 0-180v taps. When rectified and filtered, itwould be 68 volts DC. That's quite an acceptable voltage, providing abit of margin of safety.However, it would not be isolated. I will call Advanced Motion to confirm that 30A8 drives can be usedwith non-isolated DC voltage. I have two questions. 1) Is that somehow harmful to use non-isolated DC voltage as input forthose drives2) What exactly does 3 kVa rating mean when applied to thisautotransformer, and how many amps can I get from the 0-180v tap inthe above described configuration.Thanksi

Jun 28, 2010 47 Replies

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Ignoramus28517 16 years ago

I may be mistaken in my calculation. Let me redo. Stall torque of the motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

Now, I contend that 300 lbs is a lot of force when it comes to milling. It is far more than I am able to exert on my manual Bridgeport.

I may have an error in my calculation. I just did an experiment. I ran my X table at 30 volts. It moved at appx. 1 inch per second at 1.45 amps.

I pushed against the table fairly hard (being somewhat space constrained). Maybe I exerted 70 lbs force. The table did not slow down at all (of course) and the amps rose from 1.45 to something like

1.67. i

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Jim Wilkins 16 years ago

At the maximum handle force I would apply to a 1/2" end mill, a spring scale between the spindle and table reads 60 lbs.

jsw

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Ignoramus8895 16 years ago

Makes perfect sense to me.

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Jim Wilkins 16 years ago

This is the spring scale, much capacity for little money:

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You can multiply the range with rope and pulleys at the cost of less accuracy from pulley friction.

jsw

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Jon Elson 16 years ago

These calculations based on slope of the screw thread make my head hurt. A leadscrew can be made equvalent to a drum with a rope around it. For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This assumes the rope is infinitely thin.) Then you can convert torque to linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =

40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the leadscrew. 80/0.0318 = 2513 linear pounds force on the table. I **think** I have done this calculation correctly, I have been using this scheme for a decade.

Jon

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Ignoramus8895 16 years ago

Jon, your calculation makes some intuitive sense. Yesterday at night, I realized that I forgot 2*pi in my calculation, which would then yield a number similar to yours.

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