Need Help!- Can we overdrive servos

[EMMKAY]

Can we over drive servos the way we over drive steppers i.e. higher than rated voltage supply with current control. Does that have any adverse affect on the servo motors.

e.g. I have a 500W servo motor rated at 24V and max 20amps on loading. can I drive the servo at voltages exceeding the stated voltage such as around 70Volts or more.

cheers all!
EmmKay

[TOTALLYRC]

Originally Posted by EMMKAY Can we over drive servos the way we over drive steppers i.e. higher than rated voltage supply with current control. Does that have any adverse affect on the servo motors. e.g. I have a 500W servo motor rated at 24V and max 20amps on loading. can I drive the servo at voltages exceeding the stated voltage such as around 70Volts or more. cheers all! EmmKay

You can but a few problems will arise.
You are looking a triple the voltage which will burn the comutator in short order and might even cause the motor to toss it windings.

I have gone above rated voltage in the past, but only by a little bit, not the extreme amount that you are looking to go.

Mike

[Al_The_Man]

Two things can occur, higher voltage equals higher rpm, if you exceed the maximum rpm of the motor centrifugal forces may destroy it.
If the motor is allowed to draw excessive current due to the higher voltage, you may possibly exceed the motor peak current which will also destroy the motor due to demagnetization etc.
A PWM servo supply can be from 10% to 50% higher than the motor rated voltage, this is only to ensure than the drive can operate the motor up to it maximum rpm and rated torque current. Otherwise there is no distinct advantage.
The reason a stepper supply is higher is to enable the motor operating current to be maintained at high rpm's to overcome the current reducing effects of increased impedance as rpm rises.
Al.

[EMMKAY]

Thank you both for your replys..
Perhaps 70V is much over the top.. realistically I am looking at enough of a voltage increase which would give me the rated max torque within a 5ms time period. I have not measured the motor response time in ms at the rated voltage for the max torque the motor can supply!...

emmkay

[Al_The_Man]

You do not necessarily require max rated voltage to achieve max torque, As a servo motor typically has maximum torque at zero rpm with fairly flat torque up to maximum rated speed, bit there are many other factors involved to achieve a the results you are looking for, a couple of which are the motor to load inertia ratio at the required accel/decal time.
You may want to look at plugging the numbers into one of the free sizing programs to get the idea of the results.
http://www.electromate.com/technical...morgansoftware
Al.

[MrWild]

In looking at the torque curves on a Pitman 14207, I'm trying to figure out what would be good for pulley ratios. No load speed is 3,200rpm, The current and torque constants are 7.06A/50 oz.in. The graph actually shows 60ozin at 7A, and at that torque the motor is spinning about 2,700rpm. The servos will put out 410ozin at 40A stall. The drives red line at 30A and 15A continuous(adjustable for less). At 15A it says 150inoz/2000rpm and 30A/300inoz./800rpm. Rapids will take about 180 ozin, and machining about 300ozin, and Z will take 500.

What happens if I run a 7.08A continuous servo at 15A continuous? As the servo will be doing a range of speeds and torque, not always at the limit and loafing at times, is the continuous written in stone, or can the drive be set for higher amps continuous than what is on the servo data sheet? I'd love to run the servos at just 2-1 ratios on a 5/8-5 ball screw just to do the Tim Tailor if fast is good, super fast is better. The way I read it though, if I want 300ozin I'll have to run a 5 or 6-1 pulley ratio to keep it in the 7A area. At 6-1 if it hit 30A 'd have 1,800ozin which would really screw things up.

The confusing part is that machining may need 300ozin, but even doing super light cuts and 60ipm the ball screw is only running 300 rpm. At 2-1 that's just 600rpm servo speed. So how do you figure out pulleys? Does 300inoz require less A if it is at 600rpm versus say 1700rpm?

[Al_The_Man]

The peak current value should never be reached or run at that current, this is usually the point that demagnetization can occur.
Look at the maximum gearing reduction that will still give you the required max. rapid/feed rate based on the motor rpm.
The torque is increased by the factor of the reduction and the motor/load inertia ratio is decreased by the square of the reduction.
Al.

[MrWild]

Originally Posted by Al_The_Man The torque is increased by the factor of the reduction and the motor/load inertia ratio is decreased by the square of the reduction. Al.

With the peak amps of the drive at 30A (and for just 2 seconds before shut down) the servo peak stall amps will never be met. I follow the torque multiplication of 2-1 giving a 300oz in output from a servo 150ozin input, but this other you mention has me confused. What is inertia ratio, and what does it mean? Can you give an example?

Also can a servo be over driven with more continuous amps than the data sheet says? Does it really matter? If continuous amps is supposed to be 7A but the drive needs to put out 15 for accel/deccel (or more), how is this addressed?

Thanks!

[Al_The_Man]

Originally Posted by MrWild If continuous amps is supposed to be 7A but the drive needs to put out 15 for accel/deccel (or more), how is this addressed? Thanks!

This is where the motor/load inertia ratio comes into play, for a given motor and load, the inertia ratio is affected not only by the motor and machine mechanics, but the rate of required accel/decel required.
The industry standard has usually been to keep this number down to less then 10:1.
You may have seen post here where someone has mentioned that the torque required to move a gantry by turning their ballscrew by hand is very low, often motor size is based on this, this is breakaway torque only.
Once you try and accelerate this load at some high required value then the fact of inertia changes it drastically.
If you can get a Graphical motor sizing program, you can plug in all the mechanical constraints and it will give the optimum motor size based on your required accel/decel, and you can see the results just by changing this value.
http://www.electromate.com/technical...morgansoftware
An OK result will typically show the motor load inertia ratio at under 10:1.
As I mentioned earlier, the advantage of reduction is not only higher torque but the ratio is decreased by the square of the reduction, so if you have a 4:1 reduction this decreases the inertia ratio by a factor 16.
If your motor cannot keep up with the demands of a particular accel decel and has to run at higher than rated current then either the motor is undersized or the reduction is too low.
The effect on steppers is the loss of steps.
Someone mentioned that the s/w above will not seem to run on win7?
Al.

[MrWild]

I have W7 on the new Dell, the laptop has XP. 3-1 feels about right, but that's just a Pink Fuzzy method of engineering. NC-cams would be all over me for that notion. WAGS drove him crazy. I'll download the program and see what I need to measure for the inputs. As I have the parts already, selection of the proper gearing is important to make it all work.

AMC 30A8T
Pitman 14207
Mesa 5i20, 7i33, 7i37
33v 20A supply
EMC2