Re: Gecko drives

> George
> I can't believe this.
> All the info is out there in plain sight and no one has pointed you
> to it.
> First for the Gecko drives go to
> http://www.geckodrives.com and the home page go to support, then
> Technical Libary and all the files arethere for power supplies and
> every thing associated with these drives.
>
> For a wiring diagram go to
> http://www.ktmarketing.com/CNC.html
> On the list is a wiring diagram for Gecko 320 drives using CNC Pro
> These are servo drives but they are not that far removed from the
> stepper circuit you will need.
>
> There is enough information on those two sites to answer all your
> question.
>
> John StevensonHi John,

Thanks for the link.

Gecko says the supply voltage should be between 3 to 25 times the
motor's rated voltage, so if the motor is rated 5.1V 1A, 3 -25x means
the voltage should be 15.3 - 127.5 volt. But another section says the
voltage must be between 24 VDC and 80 VDC. The two combined than says
that the voltage shoud be between 24VDC to 80VDC? I can then use any
voltage between 24V to 80V? What will be the effect? Is a lower
voltage or a higher voltage better? Does the voltage affect the speed
of the stepper, or is the speed of the stepper affected by the
software? Is the recommended supply for both stepper and servo drive?

When the motor is rated 5.1 volt 1A, is this rating for full winding
or half-winding?

The example states that if a 3.8A 1.2V half-winding connected motor
is connected for full-winding then it should be thought of as a 1.9A
2.4v motor instead. What is the difference between half-winding and
full winding. I believe it should be the resistance, full winding
being 2x the resistance of half winding.

What should be constant. I read somewhere the limiting factor for a
motor is the capability to disipate heat and heat = I2R which means
when the resistance is doubled, the current should be reduced by sqrt
(o.5) = 0.71. So if this is correct, then the above example should
be, if full winding the current should be reduced to 0.71x3.8 = 2.7A
(instead of 0.5x3.8 =1.9A as in the example) tO keep the heating rate
the same then the voltage should be 1.69v and not 2.4v?

So many things to consider. My head spins.

George